Rest in Peace to Aaron Swartz beloved friend of the internet, defender for freedom of speech and information, May your struggle not be in vain, the earth is 4.5 billion years, mankind has been around 160,000 yrs and the average span of an individual is just 70, but the idea of free universal knowledge can never die.
Index of Applications Engineering and Physical Sciences Acceleration, 124, 128, 156, 158, 176, 253, 906 Air pressure, 431 Air traffic control, 154, 745, 650, 850 Aircraft glide path, 193 Angle of elevation, 151, 155, 156 Angular rate of change, 374 Architecture, 694 Area, 116, 126, 153, 256, 603, 674 Asteroid Apollo, 738 Atmospheric pressure and altitude, 327, 353, 951 Automobile aerodynamics, 30 Average speed, 40, 89 Average temperature, 984, 1034 Average velocity, 112 Beam deflection, 693 Beam strength, 35, 222 Billiard balls and normal lines, 927 Boiling temperature, 35 Boyle’s Law, 485, 504 Braking load, 774 Breaking strength of a steel cable, 364 Bridge design, 694 Building design, 445, 556, 1008, 1035, 1064 Buoyant force, 501 Cable tension, 757, 765 Capillary action, 1008 Car performance, 35 Carbon dating, 413 Center of mass, of glass, 496 Center of pressure on a sail, 1001 Centripetal acceleration, 850 Centripetal force, 850 Centroid, 494, 495, 502, 519 Chemical mixture problem, 427, 429 Chemical reaction, 391, 422, 550, 962 Circular motion, 840, 848 Comet Hale-Bopp, 741 Construction, 154, 765 Cycloidal motion, 839, 849 Depth of gasoline in a tank, 503 of water in a swimming pool, 153 of water in a vase, 29 Distance, 241 Einstein’s Special Theory of Relativity and Newton’s First Law of Motion, 204 Electric circuit, 406, 426, 429 Electric force, 485 Electric force fields, 1041 Electric potential, 878 Electrical charge, 1105 Electrical circuits, 1147
Electrical resistance, 185 Electricity, 155, 303 Electromagnetic theory, 577 Emptying a tank of oil, 481 Error in volume of a ball bearing, 233 in volume and surface area of a cube, 236 Explorer 18, 694, 741 Explorer 55, 694 Falling object, 34, 315, 426, 429 Ferris wheel, 866 Flow rate, 286, 355, 1105 Fluid force, 541 on a circular plate, 502 of gasoline, 501, 502 on a stern of a boat, 502 in a swimming pool, 504, 506 on a tank wall, 501, 502 of water, 501 Force, 289, 501, 771 Force field, 1130 Free-falling object, 69, 82, 91 Frictional force, 858, 862 Gauss’s Law, 1103 Gravitational fields, 1041 Gravitational force, 577 Halley’s comet, 694, 737 Harmonic motion, 36, 38, 138, 353 Heat flux, 1123 Heat transfer, 336 Heat-seeking particle, 921 Heat-seeking path, 926 Height of a baseball, 29 of a basketball, 32 Highway design, 169, 193, 866 Honeycomb, 169 Horizontal motion, 355 Hyperbolic detection system, 691 Hyperbolic mirror, 695 Ideal Gas Law, 879, 898, 914 Illumination, 222, 241 Inflating balloon, 150 Kepler’s Laws, 737, 738, 862 Kinetic and potential energy, 1071, 1074 Law of Conservation of Energy, 1071 Lawn sprinkler, 169 Length, 603 of a catenary, 473, 503 of pursuit, 476 of a stream, 475 Linear and angular velocity, 158 Linear vs. angular speed, 156 Load supported by a beam, 1155 Load supports, 765 Lunar gravity, 253 Magnetic field of Earth, 1050 Map of the ocean floor, 926
Mass, 1055, 1061 on the surface of Earth, 486 Maximum area, 219, 220, 221, 222, 224, 240, 242, 949 Maximum cross-sectional area of an irrigation canal, 223 Maximum volume, 221, 222, 223 of a box, 215, 216, 220, 222, 944, 949, 958 of a can buoy, 959 of a package, 222 Minimum length, 218, 221, 222, 240 Minimum surface area, 222 Minimum time, 222, 230 Motion of a liquid, 1118, 1119 of a particle, 712 pendulum, 1155 spring, 1154 Moving ladder, 154 Moving shadow, 156, 158, 160 Muzzle velocity, 756, 757 Navigation, 695, 757 Newton’s Law of Gravitation, 1041 Orbit of Earth, 708 Orbital speed, 850 Parabolic reflector, 684 Parachute jump, 1148 Particle motion, 128, 287, 290, 823, 831, 833, 839, 840, 849, 850, 861 Path of a ball, 838 of a baseball, 837, 838, 860 of a bomb, 839, 865 of a football, 839 of a projectile, 182, 712, 838, 839, 964 of a shot-put throw, 839 Pendulum, 138, 237, 906, 1155 Planetary motion, 741 Planetary orbits, 687 Planimeter, 1122 Power, 169, 906 Projectile motion, 237, 675, 705, 757, 836, 838, 839, 847, 849, 850, 860, 865, 913 Radioactive decay, 356, 409, 413, 421, 431 Refraction of light, 959 Refrigeration, 158 Resultant force, 754, 756 Ripples in a pond, 149 Rolling a ball bearing, 185 Satellite antenna, 742 Satellite orbit, 694, 866 Satellites, 127 Sending a space module into orbit, 480, 571 Solar collector, 693 Sound intensity, 40, 327, 414
(continued on back inside cover)
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Tear out Formula Cards for Homework Success.
DERIVATIVES AND INTEGRALS Basic Differentiation Rules 1. 4. 7. 10. 13. 16. 19. 22. 25. 28. 31. 34.
d 关cu兴 cu dx d u vu uv dx v v2 d 关x兴 1 dx d u 关e 兴 eu u dx d 关sin u兴 共cos u兲u dx d 关cot u兴 共csc2 u兲u dx d u 关arcsin u兴 dx 冪1 u2 d u 关arccot u兴 dx 1 u2 d 关sinh u兴 共cosh u兲u dx d 关coth u兴 共csch2 u兲u dx d u 关sinh1 u兴 dx 冪u2 1 d u 关coth1 u兴 dx 1 u2
2.
冤冥
5. 8. 11. 14. 17. 20. 23. 26. 29. 32. 35.
d 关u ± v兴 u ± v dx d 关c兴 0 dx d u 关u兴 共u 兲, u 0 dx u d u 关loga u兴 dx 共ln a兲u d 关cos u兴 共sin u兲u dx d 关sec u兴 共sec u tan u兲u dx d u 关arccos u兴 dx 冪1 u2 d u 关arcsec u兴 dx u 冪u2 1 d 关cosh u兴 共sinh u兲u dx d 关sech u兴 共sech u tanh u兲u dx d u 关cosh1 u兴 dx 冪u2 1 d u 关sech1 u兴 dx u冪1 u2
ⱍⱍ
3. 5. 7.
© Brooks/Cole, Cengage Learning
9. 11. 13. 15. 17. 19.
冕 冕 冕 冕 冕 冕 冕 冕 冕 冕
冕
ⱍⱍ
kf 共u兲 du k f 共u兲 du
2.
du u C
4.
du ln u C u 1 au du au C ln a
ⱍⱍ
6.
冢 冣
8.
cos u du sin u C
ⱍ
10.
ⱍ
cot u du ln sin u C
ⱍ
12.
ⱍ
csc u du ln csc u cot u C
14.
csc2 u du cot u C
16.
csc u cot u du csc u C
18.
du 1 u arctan C 2 u a a
20.
a2
6. 9.
ⱍⱍ
Basic Integration Formulas 1.
3.
冕 冕 冕 冕 冕 冕 冕 冕 冕 冕
12. 15. 18. 21. 24. 27. 30. 33. 36.
d 关uv兴 uv vu dx d n 关u 兴 nu n1u dx d u 关ln u兴 dx u d u 关a 兴 共ln a兲au u dx d 关tan u兴 共sec2 u兲u dx d 关csc u兴 共csc u cot u兲u dx d u 关arctan u兴 dx 1 u2 d u 关arccsc u兴 dx u 冪u2 1 d 关tanh u兴 共sech2 u兲u dx d 关csch u兴 共csch u coth u兲u dx d u 关tanh1 u兴 dx 1 u2 d u 关csch1 u兴 dx u 冪1 u2
关 f 共u兲 ± g共u兲兴 du u n du
ⱍⱍ
ⱍⱍ
冕
f 共u兲 du ±
冕
g共u兲 du
u n1 C, n 1 n1
eu du eu C
sin u du cos u C
ⱍ
ⱍ
tan u du ln cos u C
ⱍ
ⱍ
sec u du ln sec u tan u C sec2 u du tan u C sec u tan u du sec u C du
u C a du 1 u arcsec C 2 2 a a u冪u a 冪a2 u2
arcsin
ⱍⱍ
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TRIGONOMETRY Definition of the Six Trigonometric Functions Opposite
Right triangle definitions, where 0 < < 兾2. opp hyp e sin csc nus hyp opp e t po Hy adj hyp cos sec θ hyp adj Adjacent opp adj tan cot adj opp Circular function definitions, where is any angle. y y r sin csc r = x2 + y2 r y (x, y) x r r cos sec θ y r x x x y x tan cot x y
Reciprocal Identities 1 cos x 1 cos x sec x sec x
sin x cos x
(− 1, 0) π 180° 210°
1 cot x 1 cot x tan x tan x
cos x sin x
Pythagorean Identities sin2 x cos2 x 1 1 tan2 x sec2 x
330°
(− 23 , − 12) 76π 5π 225°240° 300°315°7π 116π ( 23 , − 21) (− 22 , − 22 ) 4 43π 270° 32π 53π 4 ( 22 , − 22 ) 1 3 (0, − 1) ( 2 , − 2 ) (− 12 , − 23 ) sin 2u 2 sin u cos u cos 2u cos2 u sin2 u 2 cos2 u 1 1 2 sin2 u 2 tan u tan 2u 1 tan2 u
1 cot2 x csc2 x
1 cos 2u 2 1 cos 2u 2 cos u 2 1 cos 2u tan2 u 1 cos 2u sin2 u
Cofunction Identities
Sum-to-Product Formulas
冢2 x冣 cos x csc冢 x冣 sec x 2 sec冢 x冣 csc x 2
sin u sin v 2 sin
冢2 x冣 sin x tan冢 x冣 cot x 2 cot冢 x冣 tan x 2
cos
Even/Odd Identities sin共x兲 sin x csc共x兲 csc x sec共x兲 sec x
x
Power-Reducing Formulas
cot x
sin
0° 0 360° 2π (1, 0)
Double-Angle Formulas
Quotient Identities tan x
(− 12 , 23 ) π (0, 1) ( 12 , 23 ) 90° (− 22 , 22 ) 3π 23π 2 π3 π ( 22 , 22 ) 120° 60° 4 π 3 1 45° , ) (− 23 , 12) 56π 4150°135° ( 6 2 2 30°
cos共x兲 cos x tan共x兲 tan x cot共x兲 cot x
Sum and Difference Formulas sin共u ± v兲 sin u cos v ± cos u sin v cos共u ± v兲 cos u cos v sin u sin v tan u ± tan v tan共u ± v兲 1 tan u tan v
冢u 2 v冣 cos冢u 2 v冣 uv uv sin u sin v 2 cos冢 sin冢 冣 2 2 冣 uv uv cos u cos v 2 cos冢 cos冢 2 冣 2 冣 uv uv cos u cos v 2 sin冢 sin冢 冣 2 2 冣 Product-to-Sum Formulas 1 sin u sin v 关cos共u v兲 cos共u v兲兴 2 1 cos u cos v 关cos共u v兲 cos共u v兲兴 2 1 sin u cos v 关sin共u v兲 sin共u v兲兴 2 1 cos u sin v 关sin共u v兲 sin共u v兲兴 2
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© Brooks/Cole, Cengage Learning
1 csc x 1 csc x sin x sin x
y
Multivariable Calculus 10e
Ron Larson The Pennsylvania State University The Behrend College
Bruce Edwards University of Florida
Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States
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Multivariable Calculus Tenth Edition Larson/Edwards Publisher: Liz Covello Senior Development Editor: Carolyn Lewis Assistant Editor: Liza Neustaetter Editorial Assistant: Stephanie Kreuz
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Library of Congress Control Number: 2012948318 ISBN-13: 978-1-285-06029-3 ISBN-10: 1-285-06029-6 Brooks/Cole 20 Channel Center Street Boston, MA 02210 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local office at: international.cengage.com/region Cengage Learning products are represented in Canada by Nelson Education, Ltd. For your course and learning solutions, visit www.cengage.com. Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com. Instructors: Please visit login.cengage.com and log in to access instructor-specific resources.
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Contents 11
Vectors and the Geometry of Space 11.1 11.2 11.3 11.4 11.5 11.6 11.7
12
Vector-Valued Functions 12.1 12.2 12.3 12.4 12.5
13
747
Vectors in the Plane 748 Space Coordinates and Vectors in Space 758 The Dot Product of Two Vectors 766 The Cross Product of Two Vectors in Space 775 Lines and Planes in Space 783 Section Project: Distances in Space 793 Surfaces in Space 794 Cylindrical and Spherical Coordinates 804 Review Exercises 811 P.S. Problem Solving 813
815
Vector-Valued Functions 816 Section Project: Witch of Agnesi 823 Differentiation and Integration of Vector-Valued Functions 824 Velocity and Acceleration 832 Tangent Vectors and Normal Vectors 841 Arc Length and Curvature 851 Review Exercises 863 P.S. Problem Solving 865
Functions of Several Variables
867
13.1 13.2 13.3
Introduction to Functions of Several Variables 868 Limits and Continuity 880 Partial Derivatives 890 Section Project: Moiré Fringes 899 13.4 Differentials 900 13.5 Chain Rules for Functions of Several Variables 907 13.6 Directional Derivatives and Gradients 915 13.7 Tangent Planes and Normal Lines 927 Section Project: Wildflowers 935 13.8 Extrema of Functions of Two Variables 936 13.9 Applications of Extrema 944 Section Project: Building a Pipeline 951 13.10 Lagrange Multipliers 952 Review Exercises 960 P.S. Problem Solving 963
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iv
Contents
14
Multiple Integration 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8
15
15.4
15.5 15.6 15.7 15.8
16
Iterated Integrals and Area in the Plane 966 Double Integrals and Volume 974 Change of Variables: Polar Coordinates 986 Center of Mass and Moments of Inertia 994 Section Project: Center of Pressure on a Sail 1001 Surface Area 1002 Section Project: Capillary Action 1008 Triple Integrals and Applications 1009 Triple Integrals in Other Coordinates 1020 Section Project: Wrinkled and Bumpy Spheres 1026 Change of Variables: Jacobians 1027 Review Exercises 1034 P.S. Problem Solving 1037
Vector Analysis 15.1 15.2 15.3
16.4
1039
Vector Fields 1040 Line Integrals 1051 Conservative Vector Fields and Independence of Path 1065 Green’s Theorem 1075 Section Project: Hyperbolic and Trigonometric Functions 1083 Parametric Surfaces 1084 Surface Integrals 1094 Section Project: Hyperboloid of One Sheet 1105 Divergence Theorem 1106 Stokes’s Theorem 1114 Review Exercises 1120 Section Project: The Planimeter 1122 P.S. Problem Solving 1123
Additional Topics in Differential Equations 16.1 16.2 16.3
965
1125
Exact First-Order Equations 1126 Second-Order Homogeneous Linear Equations 1133 Second-Order Nonhomogeneous Linear Equations 1141 Section Project: Parachute Jump 1148 Series Solutions of Differential Equations 1149 Review Exercises 1153 P.S. Problem Solving 1155
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Contents
v
Appendices Appendix A: Proofs of Selected Theorems A2 Appendix B: Integration Tables A3 Appendix C: Precalculus Review (Web)* C.1 Real Numbers and the Real Number Line C.2 The Cartesian Plane C.3 Review of Trigonometric Functions Appendix D: Rotation and the General Second-Degree Equation Appendix E: Complex Numbers (Web)* Appendix F: Business and Economic Applications (Web)* Answers to All Odd-Numbered Exercises and Tests Index A119
A85
*Available at the text-specific website www.cengagebrain.com
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(Web)*
Preface Welcome to Calculus, Tenth Edition. We are proud to present this new edition to you. As with all editions, we have been able to incorporate many useful comments from you, our user. For this edition, we have introduced some new features and revised others. You will still find what you expect – a pedagogically sound, mathematically precise, and comprehensive textbook. We are pleased and excited to offer you something brand new with this edition – a companion website at LarsonCalculus.com. This site offers many resources that will help you as you study calculus. All of these resources are just a click away. Our goal for every edition of this textbook is to provide you with the tools you need to master calculus. We hope that you find the changes in this edition, together with LarsonCalculus.com, will accomplish just that. In each exercise set, be sure to notice the reference to CalcChat.com. At this free site, you can download a step-by-step solution to any odd-numbered exercise. Also, you can talk to a tutor, free of charge, during the hours posted at the site. Over the years, thousands of students have visited the site for help. We use all of this information to help guide each revision of the exercises and solutions.
New To This Edition NEW LarsonCalculus.com This companion website offers multiple tools and resources to supplement your learning. Access to these features is free. Watch videos explaining concepts or proofs from the book, explore examples, view three-dimensional graphs, download articles from math journals and much more.
NEW Chapter Opener Each Chapter Opener highlights real-life applications used in the examples and exercises.
NEW Interactive Examples Examples throughout the book are accompanied by Interactive Examples at LarsonCalculus.com. These interactive examples use Wolfram’s free CDF Player and allow you to explore calculus by manipulating functions or graphs, and observing the results.
NEW Proof Videos Watch videos of co-author Bruce Edwards as he explains the proofs of theorems in Calculus, Tenth Edition at LarsonCalculus.com.
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Preface
vii
NEW How Do You See It? The How Do You See It? feature in each section presents a real-life problem that you will solve by visual inspection using the concepts learned in the lesson. This exercise is excellent for classroom discussion or test preparation.
REVISED Remark These hints and tips reinforce or expand upon concepts, help you learn how to study mathematics, caution you about common errors, address special cases, or show alternative or additional steps to a solution of an example.
118.
HOW DO YOU SEE IT? The figure shows the graphs of the position, velocity, and acceleration functions of a particle. y 16 12 8 4 −1
1
4 5 6 7
t
REVISED Exercise Sets The exercise sets have been carefully and extensively examined to ensure they are rigorous and relevant and include all topics our users have suggested. The exercises have been reorganized and titled so you can better see the connections between examples and exercises. Multi-step, real-life exercises reinforce problem-solving skills and mastery of concepts by giving students the opportunity to apply the concepts in real-life situations.
(a) Copy the graphs of the functions shown. Identify each graph. Explain your reasoning. To print an enlarged copy of the graph, go to MathGraphs.com. (b) On your sketch, identify when the particle speeds up and when it slows down. Explain your reasoning.
Table of Content Changes Appendix A (Proofs of Selected Theorems) now appears in video format at LarsonCalculus.com. The proofs also appear in text form at CengageBrain.com.
Trusted Features Applications Carefully chosen applied exercises and examples are included throughout to address the question, “When will I use this?” These applications are pulled from diverse sources, such as current events, world data, industry trends, and more, and relate to a wide range of interests. Understanding where calculus is (or can be) used promotes fuller understanding of the material.
Writing about Concepts Writing exercises at the end of each section are designed to test your understanding of basic concepts in each section, encouraging you to verbalize and write answers and promote technical communication skills that will be invaluable in your future careers.
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viii
Preface
Theorems Theorems provide the conceptual framework for calculus. Theorems are clearly stated and separated from the rest of the text by boxes for quick visual reference. Key proofs often follow the theorem and can be found at LarsonCalculus.com.
Definitions Definition of Definite Integral If f is defined on the closed interval 关a, b兴 and the limit of Riemann sums over partitions ⌬ n
lim
兺 f 共c 兲 ⌬ x
储⌬储→0 i⫽1
i
i
exists (as described above), then f is said to be integrable on 关a, b兴 and the limit is denoted by
兺 f 共c 兲 ⌬ x ⫽ 冕
b
n
lim
储⌬储→0 i⫽1
i
As with theorems, definitions are clearly stated using precise, formal wording and are separated from the text by boxes for quick visual reference.
i
f 共x兲 dx.
a
The limit is called the definite integral of f from a to b. The number a is the lower limit of integration, and the number b is the upper limit of integration.
Explorations Explorations provide unique challenges to study concepts that have not yet been formally covered in the text. They allow you to learn by discovery and introduce topics related to ones presently being studied. Exploring topics in this way encourages you to think outside the box.
Historical Notes and Biographies Historical Notes provide you with background information on the foundations of calculus. The Biographies introduce you to the people who created and contributed to calculus.
Technology Throughout the book, technology boxes show you how to use technology to solve problems and explore concepts of calculus. These tips also point out some pitfalls of using technology.
Section Projects Projects appear in selected sections and encourage you to explore applications related to the topics you are studying. They provide an interesting and engaging way for you and other students to work and investigate ideas collaboratively.
Putnam Exam Challenges Putnam Exam questions appear in selected sections. These actual Putnam Exam questions will challenge you and push the limits of your understanding of calculus.
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Additional Resources Student Resources • Student Solutions Manual for Multivariable Calculus (Chapters 11–16 of Calculus): ISBN 1-285-08575-2 These manuals contain worked-out solutions for all odd-numbered exercises. www.webassign.net Printed Access Card: ISBN 0-538-73807-3 Online Access Code: ISBN 1-285-18421-1 Enhanced WebAssign is designed for you to do your homework online. This proven and reliable system uses pedagogy and content found in this text, and then enhances it to help you learn calculus more effectively. Automatically graded homework allows you to focus on your learning and get interactive study assistance outside of class. Enhanced WebAssign for Calculus, 10e contains the Cengage YouBook, an interactive eBook that contains video clips, highlighting and note-taking features, and more!
CourseMate is a perfect study tool for bringing concepts to life with interactive learning, study, and exam preparation tools that support the printed textbook. CourseMate includes: an interactive eBook, videos, quizzes, flashcards, and more! • CengageBrain.com—To access additional materials including CourseMate, visit www.cengagebrain.com. At the CengageBrain.com home page, search for the ISBN of your title (from the back cover of your book) using the search box at the top of the page. This will take you to the product page where these resources can be found.
Instructor Resources www.webassign.net Exclusively from Cengage Learning, Enhanced WebAssign offers an extensive online program for Calculus, 10e to encourage the practice that is so critical for concept mastery. The meticulously crafted pedagogy and exercises in our proven texts become even more effective in Enhanced WebAssign, supplemented by multimedia tutorial support and immediate feedback as students complete their assignments. Key features include: • Thousands of homework problems that match your textbook’s end-of-section exercises • Opportunities for students to review prerequisite skills and content both at the start of the course and at the beginning of each section • Read It eBook pages, Watch It Videos, Master It tutorials, and Chat About It links • A customizable Cengage YouBook with highlighting, note-taking, and search features, as well as links to multimedia resources • Personal Study Plans (based on diagnostic quizzing) that identify chapter topics that students will need to master • A WebAssign Answer Evaluator that recognizes and accepts equivalent mathematical responses in the same way you grade assignments • A Show My Work feature that gives you the option of seeing students’ detailed solutions • Lecture videos, and more!
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x
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Acknowledgements We would like to thank the many people who have helped us at various stages of Calculus over the last 39 years. Their encouragement, criticisms, and suggestions have been invaluable.
Reviewers of the Tenth Edition Denis Bell, University of Northern Florida; Abraham Biggs, Broward Community College; Jesse Blosser, Eastern Mennonite School; Mark Brittenham, University of Nebraska; Mingxiang Chen, North Carolina A & T State University; Marcia Kleinz, Atlantic Cape Community College; Maxine Lifshitz, Friends Academy; Bill Meisel, Florida State College at Jacksonville; Martha Nega, Georgia Perimeter College; Laura Ritter, Southern Polytechnic State University; Chia-Lin Wu, Richard Stockton College of New Jersey
Reviewers of Previous Editions Stan Adamski, Owens Community College; Alexander Arhangelskii, Ohio University; Seth G. Armstrong, Southern Utah University; Jim Ball, Indiana State University; Marcelle Bessman, Jacksonville University; Linda A. Bolte, Eastern Washington University; James Braselton, Georgia Southern University; Harvey Braverman, Middlesex County College; Tim Chappell, Penn Valley Community College; Oiyin Pauline Chow, Harrisburg Area Community College; Julie M. Clark, Hollins University; P.S. Crooke, Vanderbilt University; Jim Dotzler, Nassau Community College; Murray Eisenberg, University of Massachusetts at Amherst; Donna Flint, South Dakota State University; Michael Frantz, University of La Verne; Sudhir Goel, Valdosta State University; Arek Goetz, San Francisco State University; Donna J. Gorton, Butler County Community College; John Gosselin, University of Georgia; Shahryar Heydari, Piedmont College; Guy Hogan, Norfolk State University; Ashok Kumar, Valdosta State University; Kevin J. Leith, Albuquerque Community College; Douglas B. Meade, University of South Carolina; Teri Murphy, University of Oklahoma; Darren Narayan, Rochester Institute of Technology; Susan A. Natale, The Ursuline School, NY; Terence H. Perciante, Wheaton College; James Pommersheim, Reed College; Leland E. Rogers, Pepperdine University; Paul Seeburger, Monroe Community College; Edith A. Silver, Mercer County Community College; Howard Speier, Chandler-Gilbert Community College; Desmond Stephens, Florida A&M University; Jianzhong Su, University of Texas at Arlington; Patrick Ward, Illinois Central College; Diane Zych, Erie Community College Many thanks to Robert Hostetler, The Behrend College, The Pennsylvania State University, and David Heyd, The Behrend College, The Pennsylvania State University, for their significant contributions to previous editions of this text. We would also like to thank the staff at Larson Texts, Inc., who assisted in preparing the manuscript, rendering the art package, typesetting, and proofreading the pages and supplements. On a personal level, we are grateful to our wives, Deanna Gilbert Larson and Consuelo Edwards, for their love, patience, and support. Also, a special note of thanks goes out to R. Scott O’Neil. If you have suggestions for improving this text, please feel free to write to us. Over the years we have received many useful comments from both instructors and students, and we value these very much. Ron Larson Bruce Edwards
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Your Course. Your Way. Calculus Textbook Options The traditional calculus course is available in a variety of textbook configurations to address the different ways instructors teach—and students take—their classes.
TOPICS COVERED 3-semester
The book can be customized to meet your individual needs and is available through CengageBrain.com.
APPROACH Late Transcendental Functions
Early Transcendental Functions
Calculus 10e
Calculus Early Transcendental Functions 5e
Accelerated coverage
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Essential Calculus
C A L C U L U S EARLY TRANSCENDENTAL FUNCTIONS
LARSON
Single Variable Only
Calculus 10e Single Variable
EDWARDS
F I F T H
E D I T I O N
Calculus: Early Transcendental Functions 5e Single Variable
Calculus I with Precalculus 3e
CALCULUS OF A SINGLE VARIABLE EARLY TRANSCENDENTAL FUNCTIONS
LARSON
Multivariable
Custom All of these textbook choices can be customized to fit the individual needs of your course.
EDWARDS
F I F T H
E D I T I O N
Calculus 10e Multivariable
Calculus 10e Multivariable
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Calculus: Early Transcendental Functions 5e
Essential Calculus
Calculus I with Precalculus 3e
C A L C U L U S EARLY TRANSCENDENTAL FUNCTIONS
LARSON
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E D I T I O N
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Multivariable Calculus 10e
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11 11.1 11.2 11.3 11.4 11.5 11.6 11.7
Vectors and the Geometry of Space Vectors in the Plane Space Coordinates and Vectors in Space The Dot Product of Two Vectors The Cross Product of Two Vectors in Space Lines and Planes in Space Surfaces in Space Cylindrical and Spherical Coordinates
Geography (Exercise 45, p. 803)
Torque (Exercise 29, p. 781)
Work (Exercise 64, p. 774)
Auditorium Lights (Exercise 101, p. 765) Navigation (Exercise 84, p. 757) 747 Clockwise from top left, Denis Tabler/Shutterstock.com; Elena Elisseeva/Shutterstock.com; Losevsky Photo and Video/Shutterstock.com; Mikael Damkier/Shutterstock.com; Ziva_K/iStockphoto.com
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748
Chapter 11
Vectors and the Geometry of Space
11.1 Vectors in the Plane Write the component form of a vector. Perform vector operations and interpret the results geometrically. Write a vector as a linear combination of standard unit vectors.
Component Form of a Vector Q Terminal point P
PQ
Initial point
A directed line segment Figure 11.1
Many quantities in geometry and physics, such as area, volume, temperature, mass, and time, can be characterized by a single real number that is scaled to appropriate units of measure. These are called scalar quantities, and the real number associated with each is called a scalar. Other quantities, such as force, velocity, and acceleration, involve both magnitude and direction and cannot be characterized completely by a single real number. A directed line segment is used to represent such a quantity, as shown in Figure 11.1. The directed line segment PQ has initial point P and terminal point Q, and its length (or magnitude) is denoted by PQ . Directed line segments that have the same length and direction are equivalent, as shown in Figure 11.2. The set of all directed line segments that are equivalent to a given directed line segment PQ is a vector in the plane and is denoted by \
\
\
\
v PQ .
Equivalent directed line segments Figure 11.2
In typeset material, vectors are usually denoted by lowercase, boldface letters such as u, v, and w. When written by hand, however, vectors are often denoted by letters with u,→ v , and → w. arrows above them, such as → Be sure you understand that a vector represents a set of directed line segments (each having the same length and direction). In practice, however, it is common not to distinguish between a vector and one of its representatives.
Vector Representation: Directed Line Segments Let v be represented by the directed line segment from 0, 0 to 3, 2, and let u be represented by the directed line segment from 1, 2 to 4, 4. Show that v and u are equivalent. Solution Let P0, 0 and Q3, 2 be the initial and terminal points of v, and let R1, 2 and S4, 4 be the initial and terminal points of u, as shown in Figure 11.3. You can use the Distance Formula to show that PQ and RS have the same length. \
\
\
PQ 3 0 2 2 0 2 13 RS 4 1 2 4 2 2 13 \
Both line segments have the same direction, because they both are directed toward the upper right on lines having the same slope. 20 2 30 3
\
Slope of PQ
y
\
Slope of RS \
42 2 41 3
S
u
3
2
and
(4, 4)
4
(3, 2)
(1, 2) R
1
Q
v
\
Because PQ and RS have the same length and direction, you can conclude that the two vectors are equivalent. That is, v and u are equivalent.
P (0, 0) 1
x
2
3
4
The vectors u and v are equivalent. Figure 11.3
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11.1 y
Vectors in the Plane
749
The directed line segment whose initial point is the origin is often the most convenient representative of a set of equivalent directed line segments such as those shown in Figure 11.3. This representation of v is said to be in standard position. A directed line segment whose initial point is the origin can be uniquely represented by the coordinates of its terminal point Qv1, v2 , as shown in Figure 11.4.
4
3
(v1, v2) 2
Q
Definition of Component Form of a Vector in the Plane If v is a vector in the plane whose initial point is the origin and whose terminal point is v1, v2 , then the component form of v is
v
1
v = 〈v1, v2 〉
(0, 0) P
x
1
2
3
v v1, v2 .
4
The coordinates v1 and v2 are called the components of v. If both the initial point and the terminal point lie at the origin, then v is called the zero vector and is denoted by 0 0, 0.
A vector in standard position Figure 11.4
This definition implies that two vectors u u1, u 2 and v v1, v2 are equal if and only if u1 v1 and u 2 v2. The procedures listed below can be used to convert directed line segments to component form or vice versa. 1. If P p1, p2 and Q q1, q2 are the initial and terminal points of a directed line segment, then the component form of the vector v represented by PQ is \
v1, v2 q1 p1, q2 p2 . Moreover, from the Distance Formula, you can see that the length (or magnitude) of v is v q1 p12 q2 p22 v21 v22.
Length of a vector
2. If v v1, v2 , then v can be represented by the directed line segment, in standard position, from P0, 0 to Q v1, v2 . The length of v is also called the norm of v. If v 1, then v is a unit vector. Moreover, v 0 if and only if v is the zero vector 0.
Component Form and Length of a Vector
y
Find the component form and length of the vector v that has initial point 3, 7 and terminal point 2, 5.
Q (− 2, 5) 6 4
Solution Let P3, 7 p1, p2 and Q2, 5 q1, q2 . Then the components of v v1, v2 are x
−6
−4
−2
2 −2
4
and
v
−4 −6 −8
v1 q1 p1 2 3 5
6
v2 q2 p2 5 7 12. P (3, −7)
Component form of v: v 5, 12 Figure 11.5
So, as shown in Figure 11.5, v 5, 12, and the length of v is v 5 2 122 169 13.
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750
Chapter 11
Vectors and the Geometry of Space
Vector Operations Definitions of Vector Addition and Scalar Multiplication Let u u1, u2 and v v1, v2 be vectors and let c be a scalar. 1. The vector sum of u and v is the vector u v u1 v1, u2 v2. 2. The scalar multiple of c and u is the vector cu cu1, cu 2 . 3. The negative of v is the vector v
1 v 2
2v
3 − v −v 2
v 1v v1, v2 . 4. The difference of u and v is u v u v u1 v1, u2 v2 .
Geometrically, the scalar multiple of a vector v and a scalar c is the vector that is
The scalar multiplication of v Figure 11.6
c times as long as v, as shown in Figure 11.6. If c is positive, then cv has the same
direction as v. If c is negative, then cv has the opposite direction. The sum of two vectors can be represented geometrically by positioning the vectors (without changing their magnitudes or directions) so that the initial point of one coincides with the terminal point of the other, as shown in Figure 11.7. The vector u v, called the resultant vector, is the diagonal of a parallelogram having u and v as its adjacent sides.
v u+v u
u
u+v
u
v
WILLIAM ROWAN HAMILTON (1805–1865)
Some of the earliest work with vectors was done by the Irish mathematician William Rowan Hamilton. Hamilton spent many years developing a system of vector-like quantities called quaternions. It wasn’t until the latter half of the nineteenth century that the Scottish physicist James Maxwell (1831–1879) restructured Hamilton’s quaternions in a form useful for representing physical quantities such as force, velocity, and acceleration. See LarsonCalculus.com to read more of this biography.
v
To find u v,
(1) move the initial point of v (2) move the initial point of u to the terminal point of u, or to the terminal point of v.
Figure 11.7
Figure 11.8 shows the equivalence of the geometric and algebraic definitions of vector addition and scalar multiplication, and presents (at far right) a geometric interpretation of u v. (ku1, ku2) (u1, u2)
(u1 + v1, u2 + v2) ku
u+v u2
(u1, u2) u
(v1, v2) v2
v v1
−v
ku2
u
u1
Vector addition Figure 11.8
u + (−v)
u1
u−v
u
u2
v
ku1
Scalar multiplication
Vector subtraction
The Granger Collection, New York
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11.1
Vectors in the Plane
751
Vector Operations For v 2, 5 and w 3, 4, find each of the vectors. 1 a. 2 v
b. w v
c. v 2w
Solution 1 1 1 5 a. 2v 22, 25 1, 2 b. w v w1 v1, w2 v2 3 2, 4 5 5, 1 c. Using 2w 6, 8, you have
v 2w 2, 5 6, 8 2 6, 5 8 4, 13. Vector addition and scalar multiplication share many properties of ordinary arithmetic, as shown in the next theorem. THEOREM 11.1 Properties of Vector Operations Let u, v, and w be vectors in the plane, and let c and d be scalars.
EMMY NOETHER (1882–1935)
One person who contributed to our knowledge of axiomatic systems was the German mathematician Emmy Noether. Noether is generally recognized as the leading woman mathematician in recent history.
FOR FURTHER INFORMATION
For more information on Emmy Noether, see the article “Emmy Noether, Greatest Woman Mathematician” by Clark Kimberling in Mathematics Teacher. To view this article, go to MathArticles.com.
1. 2. 3. 4. 5. 6. 7. 8.
uvvu u v w u v w u0u u u 0 cdu cd u c d u cu du cu v cu cv 1u u, 0u 0
Commutative Property Associative Property Additive Identity Property Additive Inverse Property
Distributive Property Distributive Property
Proof The proof of the Associative Property of vector addition uses the Associative Property of addition of real numbers.
u v w u1, u2 v1, v2 w1, w2 u1 v1, u2 v2 w1, w2 u1 v1 w1, u2 v2 w2 u1 v1 w1, u2 v2 w2 u1, u2 v1 w1, v2 w2 u v w The other properties can be proved in a similar manner. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Any set of vectors (with an accompanying set of scalars) that satisfies the eight properties listed in Theorem 11.1 is a vector space.* The eight properties are the vector space axioms. So, this theorem states that the set of vectors in the plane (with the set of real numbers) forms a vector space.
* For more information about vector spaces, see Elementary Linear Algebra, Seventh Edition, by Ron Larson (Boston, Massachusetts: Brooks/Cole, Cengage Learning, 2013). The Granger Collection, NYC
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752
Chapter 11
Vectors and the Geometry of Space
THEOREM 11.2 Length of a Scalar Multiple Let v be a vector and let c be a scalar. Then
c is the absolute value of c.
cv c v.
Proof
Because cv cv1, cv2 , it follows that
cv cv1, cv2 cv12 cv22 c 2 v12 c 2 v22 c 2v12 v22
c v .
c v12 v22
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
In many applications of vectors, it is useful to find a unit vector that has the same direction as a given vector. The next theorem gives a procedure for doing this. THEOREM 11.3 Unit Vector in the Direction of v If v is a nonzero vector in the plane, then the vector v 1 v v v
u
has length 1 and the same direction as v.
Proof Because 1 v is positive and u 1 v v, you can conclude that u has the same direction as v. To see that u 1, note that u
1v v 1v v v1 v 1.
So, u has length 1 and the same direction as v.
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
In Theorem 11.3, u is called a unit vector in the direction of v. The process of multiplying v by 1 v to get a unit vector is called normalization of v.
Finding a Unit Vector Find a unit vector in the direction of v 2, 5 and verify that it has length 1. Solution
From Theorem 11.3, the unit vector in the direction of v is
v 2 2, 5 1 5 2, 5 , . v 22 52 29 29 29 This vector has length 1, because
229 529 294 2925 2929 1. 2
2
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11.1 y
Vectors in the Plane
753
Generally, the length of the sum of two vectors is not equal to the sum of their lengths. To see this, consider the vectors u and v as shown in Figure 11.9. With u and v as two sides of a triangle, the length of the third side is u v , and
v
u v u v . u
u+v
Equality occurs only when the vectors u and v have the same direction. This result is called the triangle inequality for vectors. (You are asked to prove this in Exercise 77, Section 11.3.)
Standard Unit Vectors
x
The unit vectors 1, 0 and 0, 1 are called the standard unit vectors in the plane and are denoted by
Triangle inequality Figure 11.9
i 1, 0 y
and
j 0, 1
Standard unit vectors
as shown in Figure 11.10. These vectors can be used to represent any vector uniquely, as follows. v v1, v2 v1, 0 0, v2 v11, 0 v2 0, 1 v1i v2 j
2
The vector v v1 i v2 j is called a linear combination of i and j. The scalars v1 and v2 are called the horizontal and vertical components of v.
j = 〈0, 1〉
1
Writing a Linear Combination of Unit Vectors
i = 〈1, 0〉
x
1
Let u be the vector with initial point 2, 5 and terminal point 1, 3, and let v 2i j. Write each vector as a linear combination of i and j.
2
Standard unit vectors i and j Figure 11.10
a. u
b. w 2u 3v
Solution a. u q1 p1, q2 p2 1 2, 3 5 3, 8 3i 8j b. w 2u 3v 23i 8j 32i j 6i 16j 6i 3j 12i 19j
If u is a unit vector and is the angle (measured counterclockwise) from the positive x-axis to u, then the terminal point of u lies on the unit circle, and you have u cos , sin cos i sin j
y
u
cos θ
v v cos , sin v cos i v sin j.
sin θ
θ −1
as shown in Figure 11.11. Moreover, it follows that any other nonzero vector v making an angle with the positive x-axis has the same direction as u, and you can write
(cos θ , sin θ )
1
x
Writing a Vector of Given Magnitude and Direction
1
−1
The angle from the positive x-axis to the vector u Figure 11.11
Unit vector
The vector v has a magnitude of 3 and makes an angle of 30 6 with the positive x-axis. Write v as a linear combination of the unit vectors i and j. Solution write
Because the angle between v and the positive x-axis is 6, you can
v v cos i v sin j 3 cos
3 33 i 3 sin j i j. 6 6 2 2
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754
Chapter 11
Vectors and the Geometry of Space
Vectors have many applications in physics and engineering. One example is force. A vector can be used to represent force, because force has both magnitude and direction. If two or more forces are acting on an object, then the resultant force on the object is the vector sum of the vector forces.
Finding the Resultant Force Two tugboats are pushing an ocean liner, as shown in Figure 11.12. Each boat is exerting a force of 400 pounds. What is the resultant force on the ocean liner? y
Solution Using Figure 11.12, you can represent the forces exerted by the first and second tugboats as
400 cos(− 20°) F2
− 20° 400
400 sin(−20°) x
F1 400 20°
F1 400cos 20, sin 20 400 cos20 i 400 sin20 j F2 400cos20, sin20 400 cos20 i 400 sin20 j. The resultant force on the ocean liner is
400 sin(20°)
F F1 F2 400 cos20 i 400 sin20 j 400 cos20 i 400 sin20 j
800 cos20 i 752i.
400 cos(20°)
The resultant force on the ocean liner that is exerted by the two tugboats Figure 11.12
So, the resultant force on the ocean liner is approximately 752 pounds in the direction of the positive x-axis. In surveying and navigation, a bearing is a direction that measures the acute angle that a path or line of sight makes with a fixed north-south line. In air navigation, bearings are measured in degrees clockwise from north. y
Finding a Velocity
N W
See LarsonCalculus.com for an interactive version of this type of example.
E
An airplane is traveling at a fixed altitude with a negligible wind factor. The airplane is traveling at a speed of 500 miles per hour with a bearing of 330, as shown in Figure 11.13(a). As the airplane reaches a certain point, it encounters wind with a velocity of 70 miles per hour in the direction N 45 E (45 east of north), as shown in Figure 11.13(b). What are the resultant speed and direction of the airplane?
S v1 120° x
Solution
Using Figure 11.13(a), represent the velocity of the airplane (alone) as
v1 500 cos120 i 500 sin120 j. (a) Direction without wind
The velocity of the wind is represented by the vector v2 70 cos45 i 70 sin45 j.
y
v2
The resultant velocity of the airplane (in the wind) is
N W
v v1 v2 500 cos120 i 500 sin120 j 70 cos45 i 70 sin45 j 200.5i 482.5j.
E S
v Wind
v1
To find the resultant speed and direction, write v v cos i sin j. Because v 200.52 482.52 522.5, you can write
θ x
(b) Direction with wind
Figure 11.13
v 522.5
482.5 i j 522.5 cos112.6 i sin112.6 j . 200.5 522.5 522.5
The new speed of the airplane, as altered by the wind, is approximately 522.5 miles per hour in a path that makes an angle of 112.6 with the positive x-axis.
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11.1
11.1 Exercises
y
4 3 2 1
4 3
v
2
(1, 2)
1 −1
(3, 4)
22. 2u
v
23. v x
x
1
2
3
4
−1 −2
5
y
3.
1 2
4
4
2
(2, −3)
vector v and its initial point are given. Find the terminal point. 27. v 1, 3; Initial point: 4, 2
v (2, 1)
28. v 4, 9; Initial point: 5, 3
1 x
−6
x
Finding a Terminal Point In Exercises 27 and 28, the
v (−4, − 3)
v
u
26. u 2v
x
2
1 24. 2 v
y
(−1, 3)
−4 −2
y
25. u v
4 5 6
(3, −2)
4.
2
Sketching a Vector In Exercises 21–26, use the figure to sketch a graph of the vector. To print an enlarged copy of the graph, go to MathGraphs.com. 21. u
y
2.
(5, 4)
755
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Sketching a Vector In Exercises 1–4, (a) find the component form of the vector v and (b) sketch the vector with its initial point at the origin. 1.
Vectors in the Plane
−2 −1
1
2
Finding a Magnitude of a Vector In Exercises 29–34, find the magnitude of v.
Equivalent Vectors In Exercises 5–8, find the vectors u and v whose initial and terminal points are given. Show that u and v are equivalent. 5. u: 3, 2, 5, 6
6. u: 4, 0, 1, 8
v: 1, 4, 3, 8
v: 2, 1, 7, 7
7. u: 0, 3, 6, 2
8. u: 4, 1, 11, 4
v: 3, 10, 9, 5
v: 10, 13, 25, 10
Writing a Vector in Different Forms In Exercises 9–16, the initial and terminal points of a vector v are given. (a) Sketch the given directed line segment, (b) write the vector in component form, (c) write the vector as the linear combination of the standard unit vectors i and j, and (d) sketch the vector with its initial point at the origin. Initial Point 9. 2, 0
Terminal Point
Initial Point
Terminal Point
5, 5
10. 4, 6
3, 6
11. 8, 3
6, 1
12. 0, 4
5, 1
13. 6, 2
6, 6
14. 7, 1
3, 1
16. 0.12, 0.60
0.84, 1.25
15.
32, 43
12, 3
Sketching Scalar Multiples In Exercises 17 and 18, sketch each scalar multiple of v.
(b) 3v
7 (c) 2 v
2 (d) 3 v
18. v 2, 3 (a) 4v
1 (b) 2v
30. v 3i
31. v 4, 3
32. v 12, 5
33. v 6i 5j
34. v 10i 3j
Finding a Unit Vector In Exercises 35–38, find the unit vector in the direction of v and verify that it has length 1. 35. v 3, 12 37. v
36. v 5, 15
3 5 2, 2
38. v 6.2, 3.4
Finding Magnitudes In Exercises 39–42, find the following. (a) u (d)
uu
(b) v (e)
vv
(c) u v (f)
uu vv
39. u 1, 1, v 1, 2 40. u 0, 1, v 3, 3 1 41. u 1, 2, v 2, 3
42. u 2, 4, v 5, 5
Using the Triangle Inequality In Exercises 43 and 44, sketch a graph of u, v, and u ⴙ v. Then demonstrate the triangle inequality using the vectors u and v. 43. u 2, 1, v 5, 4
44. u 3, 2, v 1, 2
Finding a Vector In Exercises 45–48, find the vector v with the given magnitude and the same direction as u.
17. v 3, 5 (a) 2v
29. v 7i
(c) 0v
(d) 6v
Using Vector Operations In Exercises 19 and 20, find (a) 23u, (b) 3v, (c) v ⴚ u, and (d) 2u ⴙ 5v. 19. u 4, 9, v 2, 5
Magnitude
Direction
45. v 6
u 0, 3
46. v 4
u 1, 1
47. v 5
u 1, 2
48. v 2
u 3, 3
20. u 3, 8, v 8, 25
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
756
Chapter 11
Vectors and the Geometry of Space
Finding a Vector In Exercises 49–52, find the component form of v given its magnitude and the angle it makes with the positive x-axis.
Finding Values In Exercises 61–66, find a and b such that
49. v 3, 0
50. v 5, 120
61. v 2, 1
62. v 0, 3
51. v 2, 150
52. v 4, 3.5
63. v 3, 0
64. v 3, 3
65. v 1, 1
66. v 1, 7
Finding a Vector In Exercises 53–56, find the component form of u ⴙ v given the lengths of u and v and the angles that u and v make with the positive x-axis. 53. u 1, u 0
54. u 4, u 0
v 3, v 45
v 2, v 60
55. u 2, u 4
56. u 5, u 0.5
v 1, v 2
v 5, v 0.5
v ⴝ au ⴙ bw, where u ⴝ 1, 2 and w ⴝ 1, ⴚ1.
Finding Unit Vectors In Exercises 67–72, find a unit vector (a) parallel to and (b) perpendicular to the graph of f at the given point. Then sketch the graph of f and sketch the vectors at the given point. 67. f x x2,
3, 9
68. f x x2 5, 1, 4
69. f x x3,
1, 1
70. f x x3,
71. f x 25 x 2, 72. f x tan x,
WRITING ABOUT CONCEPTS
4 , 1
57. Scalar and Vector In your own words, state the difference between a scalar and a vector. Give examples of each.
Finding a Vector
58. Scalar or Vector Identify the quantity as a scalar or as a vector. Explain your reasoning.
73. u 1, 45
In Exercises 73 and 74, find the component form of v given the magnitudes of u and u ⴙ v and the angles that u and u ⴙ v make with the positive x-axis.
(b) The price of a company’s stock (d) The weight of a car
y
59. Using a Parallelogram Three vertices of a parallelogram are 1, 2, 3, 1, and 8, 4. Find the three possible fourth vertices (see figure).
500 lb 180 N
30°
y
x
− 45°
6 5 4 3 2 1
(8, 4)
θ
275 N
x
200 lb
(1, 2) (3, 1) x
60.
u v 6 , 120
75. Resultant Force Forces with magnitudes of 500 pounds and 200 pounds act on a machine part at angles of 30 and 45, respectively, with the x-axis (see figure). Find the direction and magnitude of the resultant force.
(c) The air temperature in a room
− 4 − 3− 2 − 1
74. u 4, 30
u v 2 , 90
(a) The muzzle velocity of a gun
2, 8
3, 4
1 2 3 4 5 6 7 8 9 10
HOW DO YOU SEE IT? Use the figure to determine whether each statement is true or false. Justify your answer. b
Figure for 75
Figure for 76
76. Numerical and Graphical Analysis Forces with magnitudes of 180 newtons and 275 newtons act on a hook (see figure). The angle between the two forces is degrees. (a) When 30, find the direction and magnitude of the resultant force. (b) Write the magnitude M and direction of the resultant force as functions of , where 0 180.
t
(c) Use a graphing utility to complete the table. a
w
d c u
s v
0
30
60
90
120
150
180
M
(a) a d
(b) c s
(c) a u c
(d) v w s
(d) Use a graphing utility to graph the two functions M and .
(e) a d 0
(f ) u v 2b t
(e) Explain why one of the functions decreases for increasing values of , whereas the other does not.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.1 77. Resultant Force Three forces with magnitudes of 75 pounds, 100 pounds, and 125 pounds act on an object at angles of 30, 45, and 120, respectively, with the positive x-axis. Find the direction and magnitude of the resultant force. 78. Resultant Force Three forces with magnitudes of 400 newtons, 280 newtons, and 350 newtons act on an object at angles of 30, 45, and 135, respectively, with the positive x-axis. Find the direction and magnitude of the resultant force. 79. Think About It acting on a point.
Consider two forces of equal magnitude
(a) When the magnitude of the resultant is the sum of the magnitudes of the two forces, make a conjecture about the angle between the forces. (b) When the resultant of the forces is 0, make a conjecture about the angle between the forces. (c) Can the magnitude of the resultant be greater than the sum of the magnitudes of the two forces? Explain. 80. Cable Tension Determine the tension in each cable supporting the given load for each figure. (a) A
(b) 30°
50°
10 in.
B
Vectors in the Plane
757
83. Navigation A plane is flying with a bearing of 302. Its speed with respect to the air is 900 kilometers per hour. The wind at the plane’s altitude is from the southwest at 100 kilometers per hour (see figure). What is the true direction of the plane, and what is its speed with respect to the ground? 84. Navigation A plane flies at a constant groundspeed of 400 miles per hour due east and encounters a 50-mile-per-hour wind from the northwest. Find the airspeed and compass direction that will allow the plane to maintain its groundspeed and eastward direction.
True or False? In Exercises 85–90, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 85. If u and v have the same magnitude and direction, then u and v are equivalent.
20 in.
A
B
C
86. If u is a unit vector in the direction of v, then v v u. 87. If u ai bj is a unit vector, then a 2 b 2 1.
24 in.
88. If v ai bj 0, then a b.
3000 lb
89. If a b, then a i bj 2 a.
C
90. If u and v have the same magnitude but opposite directions, then u v 0.
5000 lb
81. Projectile Motion A gun with a muzzle velocity of 1200 feet per second is fired at an angle of 6 above the horizontal. Find the vertical and horizontal components of the velocity.
91. Proof
Prove that
u cos i sin j and
v sin i cos j
are unit vectors for any angle .
82. Shared Load To carry a 100-pound cylindrical weight, two workers lift on the ends of short ropes tied to an eyelet on the top center of the cylinder. One rope makes a 20 angle away from the vertical and the other makes a 30 angle (see figure).
92. Geometry Using vectors, prove that the line segment joining the midpoints of two sides of a triangle is parallel to, and one-half the length of, the third side.
(a) Find each rope’s tension when the resultant force is vertical.
94. Proof Prove that the vector w u v v u bisects the angle between u and v.
(b) Find the vertical component of each worker’s force. 20°
95. Using a Vector Consider the vector u x, y. Describe the set of all points x, y such that u 5.
N 30°
100 lb
W
E S
PUTNAM EXAM CHALLENGE
100 km/hr
96. A coast artillery gun can fire at any angle of elevation between 0 and 90 in a fixed vertical plane. If air resistance is neglected and the muzzle velocity is constant v0, determine the set H of points in the plane and above the horizontal which can be hit.
900 km/hr 32°
93. Geometry Using vectors, prove that the diagonals of a parallelogram bisect each other.
45°
This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
Figure for 82
Figure for 83
Mikael Damkier/Shutterstock.com
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758
Chapter 11
Vectors and the Geometry of Space
11.2 Space Coordinates and Vectors in Space Understand the three-dimensional rectangular coordinate system. Analyze vectors in space.
Coordinates in Space z
xz-plane
yz-plane
y x
xy-plane
The three-dimensional coordinate system Figure 11.14
Up to this point in the text, you have been primarily concerned with the two-dimensional coordinate system. Much of the remaining part of your study of calculus will involve the three-dimensional coordinate system. Before extending the concept of a vector to three dimensions, you must be able to identify points in the three-dimensional coordinate system. You can construct this system by passing a z-axis perpendicular to both the x- and y-axes at the origin, as shown in Figure 11.14. Taken as pairs, the axes determine three coordinate planes: the xy-plane, the xz-plane, and the yz-plane. These three coordinate planes separate three-space into eight octants. The first octant is the one for which all three coordinates are positive. In this three-dimensional system, a point P in space is determined by an ordered triple 共x, y, z兲, where x, y, and z are as follows. x directed distance from yz-plane to P y directed distance from xz-plane to P z directed distance from xy-plane to P Several points are shown in Figure 11.15. z 6 5 4
(2, − 5, 3)
3
(− 2, 5, 4)
−6 −5 −4 −3
2 −8
1
−4
−2
REMARK The threedimensional rotatable graphs that are available at LarsonCalculus.com can help you visualize points or objects in a three-dimensional coordinate system.
3
(1, 6, 0)
4
y
8
5 6
(3, 3, − 2)
x
Points in the three-dimensional coordinate system are represented by ordered triples. Figure 11.15
A three-dimensional coordinate system can have either a right-handed or a lefthanded orientation. To determine the orientation of a system, imagine that you are standing at the origin, with your arms pointing in the direction of the positive x- and y-axes, and with the z-axis pointing up, as shown in Figure 11.16. The system is right-handed or left-handed depending on which hand points along the x-axis. In this text, you will work exclusively with the right-handed system.
z
z
y
x
x
y
Right-handed system Figure 11.16
Left-handed system
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11.2 z
(x2, y2, z2)
Q
d
⏐z2 − z1⏐
P x
759
Many of the formulas established for the two-dimensional coordinate system can be extended to three dimensions. For example, to find the distance between two points in space, you can use the Pythagorean Theorem twice, as shown in Figure 11.17. By doing this, you will obtain the formula for the distance between the points 共x1, y1, z1兲 and 共x2, y2, z 2 兲.
y
(x1, y1, z1)
Space Coordinates and Vectors in Space
d 冪共x2 x1兲2 共y2 y1兲2 共z2 z1兲2
(x2, y2, z1)
Distance Formula
(x2 − x1)2 + (y2 − y1)2
Finding the Distance Between Two Points in Space
The distance between two points in space Figure 11.17
Find the distance between the points 共2, 1, 3兲 and 共1, 0, 2兲. Solution d 冪共1 2兲2 共0 1兲2 共2 3兲2 冪1 1 25 冪27 3冪3
A sphere with center at 共x0 , y0 , z0兲 and radius r is defined to be the set of all points 共x, y, z兲 such that the distance between 共x, y, z兲 and 共x0 , y0 , z0兲 is r. You can use the Distance Formula to find the standard equation of a sphere of radius r, centered at 共x0 , y0 , z0兲. If 共x, y, z兲 is an arbitrary point on the sphere, then the equation of the sphere is
z
(x, y, z) r (x0, y0, z 0 )
共x x0兲2 共y y0兲2 共z z0兲2 r2 y
x
Distance Formula
Equation of sphere
as shown in Figure 11.18. Moreover, the midpoint of the line segment joining the points 共x1, y1, z1兲 and 共x2, y2, z2兲 has coordinates
Figure 11.18
冢x
1
x2 y1 y2 z1 z2 , , . 2 2 2
冣
Midpoint Formula
Finding the Equation of a Sphere Find the standard equation of the sphere that has the points
共5, 2, 3兲 and 共0, 4, 3兲 as endpoints of a diameter. Solution
Using the Midpoint Formula, the center of the sphere is
冢5 2 0, 22 4, 3 2 3冣 冢52, 1, 0冣.
Midpoint Formula
By the Distance Formula, the radius is r
冪冢0 25冣
2
共4 1兲2 共3 0兲2
冪974
冪97
2
.
Therefore, the standard equation of the sphere is
冢x 25冣
2
共 y 1兲2 z2
97 . 4
Equation of sphere
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760
Chapter 11
Vectors and the Geometry of Space
z
Vectors in Space
〈v1, v2, v3 〉
In space, vectors are denoted by ordered triples v 具v1, v2, v3 典. The zero vector is denoted by 0 具0, 0, 0典. Using the unit vectors
v 〈0, 0, 1〉
i 具1, 0, 0典, j 具0, 1, 0典, and k 具0, 0, 1典
k j 〈0, 1, 0〉 i
the standard unit vector notation for v is y
〈1, 0, 0〉
v v1i v2 j v3k as shown in Figure 11.19. If v is represented by the directed line segment from P共 p1, p2, p3兲 to Q共q1, q2, q3兲, as shown in Figure 11.20, then the component form of v is written by subtracting the coordinates of the initial point from the coordinates of the terminal point, as follows.
x
The standard unit vectors in space Figure 11.19
v 具v1, v2, v3典 具q1 p1, q2 p2, q3 p3 典
z
Vectors in Space Let u 具u1, u2, u3 典 and v 具v1, v2, v3 典 be vectors in space and let c be a scalar.
Q(q1, q2, q3) P(p1, p2, p3)
v
y
1. Equality of Vectors: u v if and only if u1 v1, u2 v2, and u3 v3. 2. Component Form: If v is represented by the directed line segment from P共 p1, p2, p3兲 to Q共q1, q2, q3兲, then v 具v1, v2, v3 典 具q1 p1, q2 p2, q3 p3 典. 3. Length: 储v储 冪v12 v22 v32
x
v = 〈q1 − p1, q2 − p2, q3 − p3 〉
Figure 11.20
冢 冣
v 1 具v1, v2, v3典, 储 v储 储v储 5. Vector Addition: v u 具v1 u1, v2 u2, v3 u3 典 6. Scalar Multiplication: cv 具cv1, cv2, cv3 典 4. Unit Vector in the Direction of v:
v0
Note that the properties of vector operations listed in Theorem 11.1 (see Section 11.1) are also valid for vectors in space.
Finding the Component Form of a Vector in Space See LarsonCalculus.com for an interactive version of this type of example.
Find the component form and magnitude of the vector v having initial point 共2, 3, 1兲 and terminal point 共0, 4, 4兲. Then find a unit vector in the direction of v. Solution
The component form of v is
v 具q1 p1, q2 p2, q3 p3 典 具0 共2兲, 4 3, 4 1典 具2, 7, 3典 which implies that its magnitude is 储v储 冪22 共7兲2 32 冪62. The unit vector in the direction of v is v 储 v储 1 具2, 7, 3典 冪62
u
冬冪262, 冪762, 冪362冭.
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11.2 y
761
Recall from the definition of scalar multiplication that positive scalar multiples of a nonzero vector v have the same direction as v, whereas negative multiples have the direction opposite of v. In general, two nonzero vectors u and v are parallel when there is some scalar c such that u cv. For example, in Figure11.21, the vectors u, v, and w are parallel because
u = 2v w = −v
u
Space Coordinates and Vectors in Space
u 2v and w v.
v x
w
Definition of Parallel Vectors Two nonzero vectors u and v are parallel when there is some scalar c such that u cv.
Parallel vectors Figure 11.21
Parallel Vectors Vector w has initial point 共2, 1, 3兲 and terminal point 共4, 7, 5兲. Which of the following vectors is parallel to w? a. u 具3, 4, 1典 b. v 具12, 16, 4典 Solution
Begin by writing w in component form.
w 具4 2, 7 共1兲, 5 3典 具6, 8, 2典 a. Because u 具3, 4, 1典 12 具6, 8, 2典 12 w, you can conclude that u is parallel to w. b. In this case, you want to find a scalar c such that 具12, 16, 4典 c具6, 8, 2典. To find c, equate the corresponding components and solve as shown. 12 6c 16 8c 4
c 2 c 2 c 2
2c
Note that c 2 for the first two components and c 2 for the third component. This means that the equation 具12, 16, 4典 c具6, 8, 2典 has no solution, and the vectors are not parallel.
Using Vectors to Determine Collinear Points
z
(1, − 2, 3)
Determine whether the points
P 4
P共1, 2, 3兲,
2
(2, 1, 0) Q
and R共4, 7, 6兲
are collinear.
2 4
\
Solution
6
6 x
Q共2, 1, 0兲,
8
8
y
\
The component forms of PQ and PR are
\
PQ 具2 1, 1 共2兲, 0 3典 具1, 3, 3典 and
(4, 7, −6) R
The points P, Q, and R lie on the same line. Figure 11.22
\
PR 具4 1, 7 共2兲, 6 3典 具3, 9, 9典. These two vectors have a common initial point. So, P, Q, and R lie on the same line if and only if PQ and PR are parallel—which they are because PR 3 PQ , as shown in Figure 11.22. \
\
\
\
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
762
Chapter 11
Vectors and the Geometry of Space
Standard Unit Vector Notation a. Write the vector v 4i 5k in component form. b. Find the terminal point of the vector v 7i j 3k, given that the initial point is P共2, 3, 5兲. c. Find the magnitude of the vector v 6i 2j 3k. Then find a unit vector in the direction of v. Solution a. Because j is missing, its component is 0 and v 4i 5k 具4, 0, 5典. b. You need to find Q共q1, q2, q3兲 such that \
v PQ 7i j 3k. This implies that q1 共2兲 7, q2 3 1, and q3 5 3. The solution of these three equations is q1 5, q2 2, and q3 8. Therefore, Q is 共5, 2, 8兲. c. Note that v1 6, v2 2, and v3 3. So, the magnitude of v is 储v储 冪共6兲2 22 共3兲2 冪49 7. The unit vector in the direction of v is 1 7 共6i
2j 3k兲 67i 27j 37k.
Measuring Force A television camera weighing 120 pounds is supported by a tripod, as shown in Figure 11.23. Represent the force exerted on each leg of the tripod as a vector.
z
P (0, 0, 4) Q3
)
)
3 1 − , ,0 2 2
Solution Let the vectors F1, F2, and F3 represent the forces exerted on the three legs. From Figure 11.23, you can determine the directions of F1, F2, and F3 to be as follows. \
PQ 1 具0 0, 1 0, 0 4典 具0, 1, 4典 冪3 冪3 1 1 0, 0, 0 4 , , 4 PQ 2 2 2 2 2 冪3 冪3 1 1 PQ 3 0, 0, 0 4 , , 4 2 2 2 2
Q1 (0, − 1, 0)
\
y
Q2
)
)
3 1 , ,0 2 2
\
x
Figure 11.23
冬 冬
冭 冬 冭 冬
冭
冭
Because each leg has the same length, and the total force is distributed equally among the three legs, you know that 储F1 储 储 F2 储 储F3 储. So, there exists a constant c such that
冬冪23, 12, 4冭,
F1 c具0, 1, 4典, F2 c
and
冬
F3 c
冪3 1
2
冭
, , 4 . 2
Let the total force exerted by the object be given by F 具0, 0, 120典. Then, using the fact that F F1 F2 F3 you can conclude that F1, F2, and F3 all have a vertical component of 40. This implies that c共4兲 40 and c 10. Therefore, the forces exerted on the legs can be represented by F1 具0, 10, 40典, F2 具 5冪3, 5, 40典, and F3 具 5冪3, 5, 40典.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.2
11.2 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Plotting Points In Exercises 1–4, plot the points in the same three-dimensional coordinate system. 1. (a) 共2, 1, 3兲
(b) 共1, 2, 1兲
2. (a) 共3, 2, 5兲
(b)
3. (a) 共5, 2, 2兲
(b) 共5, 2, 2兲
4. (a) 共0, 4, 5兲
(b) 共4, 0, 5兲
763
Space Coordinates and Vectors in Space
共32, 4, 2兲
Finding Coordinates of a Point In Exercises 5–8, find the coordinates of the point. 5. The point is located three units behind the yz-plane, four units to the right of the xz-plane, and five units above the xy-plane. 6. The point is located seven units in front of the yz-plane, two units to the left of the xz-plane, and one unit below the xy-plane. 7. The point is located on the x-axis, 12 units in front of the yz-plane.
31. Think About It The triangle in Exercise 27 is translated five units upward along the z-axis. Determine the coordinates of the translated triangle. 32. Think About It The triangle in Exercise 28 is translated three units to the right along the y-axis. Determine the coordinates of the translated triangle.
Finding the Midpoint In Exercises 33 –36, find the coordinates of the midpoint of the line segment joining the points. 33. 共3, 4, 6兲, 共1, 8, 0兲
34. 共7, 2, 2兲, 共5, 2, 3兲
35. 共5, 9, 7兲, 共2, 3, 3兲
36. 共4, 0, 6兲, 共8, 8, 20兲
Finding the Equation of a Sphere In Exercises 37–40, find the standard equation of the sphere. 38. Center: 共4, 1, 1兲
37. Center: 共0, 2, 5兲 Radius: 2
Radius: 5
8. The point is located in the yz-plane, three units to the right of the xz-plane, and two units above the xy-plane.
39. Endpoints of a diameter: 共2, 0, 0兲, 共0, 6, 0兲
9. Think About It the xy-plane?
What is the z-coordinate of any point in
Finding the Equation of a Sphere In Exercises 41–44,
10. Think About It the yz-plane?
What is the x-coordinate of any point in
11. z 6
12. y 2
13. x 3
14. z 52
15. y < 0
16. x > 0
ⱍⱍ
19. xy > 0,
z 3
21. xyz < 0
complete the square to write the equation of the sphere in standard form. Find the center and radius. 41. x 2 y 2 z 2 2x 6y 8z 1 0
Using the Three-Dimensional Coordinate System In Exercises 11–22, determine the location of a point 冇x, y, z冈 that satisfies the condition(s).
17. y 3
40. Center: 共3, 2, 4兲, tangent to the yz-plane
42. x2 y2 z2 9x 2y 10z 19 0 43. 9x 2 9y 2 9z 2 6x 18y 1 0 44. 4x 2 4y 2 4z 2 24x 4y 8z 23 0
18. x > 4
Finding the Component Form of a Vector in Space In Exercises 45–48, (a) find the component form of the vector v, (b) write the vector using standard unit vector notation, and (c) sketch the vector with its initial point at the origin.
20. xy < 0, z 4
45.
ⱍⱍ
z
22. xyz > 0
Finding the Distance Between Two Points in Space In Exercises 23–26, find the distance between the points.
2
24. 共2, 3, 2兲, 共2, 5, 2兲
25. 共1, 2, 4兲, 共6, 2, 2兲
26. 共2, 2, 3兲, 共4, 5, 6兲
6
6 4
23. 共0, 0, 0兲, 共4, 2, 7兲
4
v
(4, 2, 1)
(4, 0, 3) 2
(2, 4, 3)
of the sides of the triangle with the indicated vertices, and determine whether the triangle is a right triangle, an isosceles triangle, or neither.
2 4
z
47.
30. 共4, 1, 1兲, 共2, 0, 4兲, 共3, 5, 1兲
z
48. 6
6 4 2 2 4 6 x
y
6
x
27. 共0, 0, 4兲, 共2, 6, 7兲, 共6, 4, 8兲 29. 共1, 0, 2兲, 共1, 5, 2兲, 共3, 1, 1兲
4
6
6
28. 共3, 4, 1兲, 共0, 6, 2兲, 共3, 5, 6兲
v (0, 5, 1)
y
6
x
Classifying a Triangle In Exercises 27–30, find the lengths
z
46.
(0, 3, 3)
4
(2, 3, 4)
2
v 4
(3, 3, 0)
6
y
2 4
v (2, 3, 0)
4
6
6 x
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
y
764
Chapter 11
Vectors and the Geometry of Space
Finding the Component Form of a Vector in Space In Exercises 49 and 50, find the component form and magnitude of the vector v with the given initial and terminal points. Then find a unit vector in the direction of v. 49. Initial point: 共3, 2, 0兲
50. Initial point: 共1, 2, 4兲
Terminal point: 共4, 1, 6兲
Terminal point: 共2, 4, 2兲
68. 共0, 0, 0兲, 共1, 3, 2兲, 共2, 6, 4兲
Verifying a Parallelogram In Exercises 69 and 70, use vectors to show that the points form the vertices of a parallelogram. 69. 共2, 9, 1兲, 共3, 11, 4兲, 共0, 10, 2兲, 共1, 12, 5兲 70. 共1, 1, 3兲, 共9, 1, 2兲, 共11, 2, 9兲, 共3, 4, 4兲
Writing a Vector in Different Forms In Exercises 51 and 52, the initial and terminal points of a vector v are given. (a) Sketch the directed line segment, (b) find the component form of the vector, (c) write the vector using standard unit vector notation, and (d) sketch the vector with its initial point at the origin. 51. Initial point: 共1, 2, 3兲
52. Initial point: 共2, 1, 2兲
Terminal point: 共3, 3, 4兲
Terminal point: 共4, 3, 7兲
Finding a Terminal Point In Exercises 53 and 54, the vector v and its initial point are given. Find the terminal point. 53. v 具3, 5, 6典
54. v 具 1,
Initial point: 共0, 6, 2兲
典 共0, 2, 52 兲
23, 12
Initial point:
Finding Scalar Multiples In Exercises 55 and 56, find each scalar multiple of v and sketch its graph. 55. v 具1, 2, 2典 (c)
3 2v
71. v 具0, 0, 0典
72. v 具1, 0, 3典
73. v 3j 5k
74. v 2i 5j k
75. v i 2j 3k
76. v 4i 3j 7k
Finding Unit Vectors In Exercises 77–80, find a unit vector (a) in the direction of v and (b) in the direction opposite of v. 77. v 具2, 1, 2典
78. v 具6, 0, 8典
79. v 4i 5j 3k
80. v 5i 3j k
81. Using Vectors Consider the two nonzero vectors u and v, and let s and t be real numbers. Describe the geometric figure generated by the terminal points of the three vectors tv, u tv, and su tv.
56. v 具2, 2, 1典
(b) v
(a) 2v
Finding the Magnitude In Exercises 71– 76, find the magnitude of v.
(a) v
(d) 0v
(c)
1 2v
(b) 2v
82.
5 (d) 2v
Finding a Vector In Exercises 57–60, find the vector z, given that u ⴝ 具1, 2, 3典, v ⴝ 具2, 2, ⴚ1典, and w ⴝ 具4, 0, ⴚ4典. 57. z u v 2w
58. z 5u 3v 12w
59. 2z 3u w
60. 2u v w 3z 0
HOW DO YOU SEE IT? Determine 共x, y, z兲 for each figure. Then find the component form of the vector from the point on the x-axis to the point 共x, y, z兲.
(a)
(b) z
z
(0, 3, 3)
(4, 0, 8) (x, y, z)
(x, y, z)
Parallel Vectors In Exercises 61–64, determine which of the vectors is (are) parallel to z. Use a graphing utility to confirm your results. 61. z 具3, 2, 5典
62. z 12i 23j 34k
(a) 具6, 4, 10典 (b) 具 2,
4 3,
10 3
(a) 6i 4j 9k
典
(b) i
4 3j
y
y
(0, 4, 0)
(0, 3, 0) (4, 0, 0)
(3, 0, 0) x
x
3 2k
(c) 具6, 4, 10典
(c) 12i 9k
Finding a Vector In Exercises 83–86, find the vector v with
(d) 具1, 4, 2典
3 9 (d) 4i j 8k
the given magnitude and the same direction as u.
63. z has initial point 共1, 1, 3兲 and terminal point 共2, 3, 5兲. (a) 6i 8j 4k
(b) 4j 2k
64. z has initial point 共5, 4, 1兲 and terminal point 共2, 4, 4兲. (a) 具7, 6, 2典
(b) 具14, 16, 6典
Using Vectors to Determine Collinear Points In Exercises 65–68, use vectors to determine whether the points are collinear. 65. 共0, 2, 5兲, 共3, 4, 4兲, 共2, 2, 1兲 66. 共4, 2, 7兲, 共2, 0, 3兲, 共7, 3, 9兲
Magnitude
Direction
83. 储v储 10
u 具0, 3, 3典
84. 储v储 3
u 具1, 1, 1典
3 85. 储v储 2
u 具2, 2, 1典
86. 储v储 7
u 具4, 6, 2典
Sketching a Vector In Exercises 87 and 88, sketch the vector v and write its component form. 87. v lies in the yz-plane, has magnitude 2, and makes an angle of 30 with the positive y-axis.
67. 共1, 2, 4兲, 共2, 5, 0兲, 共0, 1, 5兲
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.2
Space Coordinates and Vectors in Space
765
101. Auditorium Lights
88. v lies in the xz-plane, has magnitude 5, and makes an angle of 45 with the positive z-axis.
The lights in an auditorium are 24-pound discs of radius 18 inches. Each disc is supported by three equally spaced cables that are L inches long (see figure).
Finding a Point Using Vectors In Exercises 89 and 90, use vectors to find the point that lies two-thirds of the way from P to Q. 89. P共4, 3, 0兲, Q共1, 3, 3兲 90. P共1, 2, 5兲, Q共6, 8, 2兲
L
91. Using Vectors Let u i j, v j k, and w au bv.
18 in.
(a) Sketch u and v. (b) If w 0, show that a and b must both be zero. (c) Find a and b such that w i 2j k.
(a) Write the tension T in each cable as a function of L. Determine the domain of the function.
(d) Show that no choice of a and b yields w i 2j 3k.
(b) Use a graphing utility and the function in part (a) to complete the table.
92. Writing The initial and terminal points of the vector v are 共x1, y1, z1兲 and 共x, y, z兲. Describe the set of all points 共x, y, z兲 such that 储v储 4.
L
WRITING ABOUT CONCEPTS
20
25
30
35
40
45
50
T
93. Describing Coordinates A point in the threedimensional coordinate system has coordinates 共x0, y0, z0兲. Describe what each coordinate measures.
(c) Use a graphing utility to graph the function in part (a). Determine the asymptotes of the graph.
94. Distance Formula Give the formula for the distance between the points 共x1, y1, z1兲 and 共x2, y2, z2兲.
(d) Confirm the asymptotes of the graph in part (c) analytically.
95. Standard Equation of a Sphere Give the standard equation of a sphere of radius r, centered at 共x0, y0, z0兲.
(e) Determine the minimum length of each cable when a cable is designed to carry a maximum load of 10 pounds.
96. Parallel Vectors State the definition of parallel vectors. 97. Using a Triangle and Vectors Let A, B, and C be vertices of a triangle. Find AB BC CA . \
\
\
98. Using Vectors Let r 具x, y, z典 and r0 具1, 1, 1典. Describe the set of all points 共x, y, z兲 such that 储r r0 储 2. 99. Diagonal of a Cube Find the component form of the unit vector v in the direction of the diagonal of the cube shown in the figure.
102. Think About It Suppose the length of each cable in Exercise 101 has a fixed length L a, and the radius of each disc is r0 inches. Make a conjecture about the limit lim T r0 →a and give a reason for your answer. 103. Load Supports Find the tension in each of the supporting cables in the figure when the weight of the crate is 500 newtons. z
45 cm
z
D
z
x
Figure for 99
x
A
B
8 ft 10 ft
y
75
Figure for 100
100. Tower Guy Wire The guy wire supporting a 100-foot tower has a tension of 550 pounds. Using the distances shown in the figure, write the component form of the vector F representing the tension in the wire. Losevsky Photo and Video/Shutterstock.com
D 6 ft
− 50
⏐⏐v⏐⏐ = 1
18 ft
115 cm A
y
C
60 cm y
x
v
70 cm B
65 cm
100
C
Figure for 103
Figure for 104
104. Construction A precast concrete wall is temporarily kept in its vertical position by ropes (see figure). Find the total force exerted on the pin at position A. The tensions in AB and AC are 420 pounds and 650 pounds. 105. Geometry Write an equation whose graph consists of the set of points P共x, y, z兲 that are twice as far from A共0, 1, 1兲 as from B共1, 2, 0兲. Describe the geometric figure represented by the equation.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
766
Chapter 11
Vectors and the Geometry of Space
11.3 The Dot Product of Two Vectors Use properties of the dot product of two vectors. Find the angle between two vectors using the dot product. Find the direction cosines of a vector in space. Find the projection of a vector onto another vector. Use vectors to find the work done by a constant force.
The Dot Product So far, you have studied two operations with vectors—vector addition and multiplication by a scalar—each of which yields another vector. In this section, you will study a third vector operation, the dot product. This product yields a scalar, rather than a vector. Definition of Dot Product The dot product of u u 1, u 2 and v v1, v2 is
REMARK Because the dot product of two vectors yields a scalar, it is also called the scalar product (or inner product) of the two vectors.
u
The dot product of u u 1, u 2, u 3 and v v1, v2, v3 is u v u 1v1 u 2v2 u 3v3.
THEOREM 11.4 Properties of the Dot Product Let u, v, and w be vectors in the plane or in space and let c be a scalar.
Exploration Interpreting a Dot Product Several vectors are shown below on the unit circle. Find the dot products of several pairs of vectors. Then find the angle between each pair that you used. Make a conjecture about the relationship between the dot product of two vectors and the angle between the vectors. 90°
120°
v u 1v1 u 2v2.
1. u v v u 2. u v w u v u w 3. cu v cu v u cv 4. 0 v 0 5. v v v 2 Proof
Commutative Property Distributive Property
To prove the first property, let u u 1, u 2, u 3 and v v1, v2, v3. Then
u v u 1v1 u 2v2 u 3v3 v1u 1 v2u 2 v3u 3 v u. For the fifth property, let v v1, v2, v3. Then v
60°
v v12 v22 v32 v12 v22 v32 2 v2.
Proofs of the other properties are left to you. 30°
150° 180°
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
0°
210°
330° 240°
270°
300°
Finding Dot Products Let u 2, 2, v 5, 8, and w 4, 3. a. b. c. d.
u v 2, 2 5, 8 25 28 6 u vw 64, 3 24, 18 u 2v 2u v 26 12 w 2 w w 4, 3 4, 3 44 33 25
Notice that the result of part (b) is a vector quantity, whereas the results of the other three parts are scalar quantities.
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11.3
767
The Dot Product of Two Vectors
Angle Between Two Vectors The angle between two nonzero vectors is the angle , 0 , between their respective standard position vectors, as shown in Figure 11.24. The next theorem shows how to find this angle using the dot product. (Note that the angle between the zero vector and another vector is not defined here.) v−u
u
v
θ
Origin
The angle between two vectors Figure 11.24
THEOREM 11.5 Angle Between Two Vectors If is the angle between two nonzero vectors u and v, where 0 , then uv cos . u v
Proof Consider the triangle determined by vectors u, v, and v u, as shown in Figure 11.24. By the Law of Cosines, you can write v u2 u2 v2 2u v cos . Using the properties of the dot product, the left side can be rewritten as v u2 v u v u v u v v u u vvuvvuuu v2 2u v u 2 and substitution back into the Law of Cosines yields v2 2u v u2 u2 v2 2u v cos 2u v 2u v cos uv . cos u v See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Note in Theorem 11.5 that because u and v are always positive, u v and cos will always have the same sign. Figure 11.25 shows the possible orientations of two vectors. Opposite direction θ
u
u v0 u θ
θ
v
2 cos 0
Same direction
u v
0 < < 2 0 < cos < 1
v
0 cos 1
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768
Chapter 11
Vectors and the Geometry of Space
From Theorem 11.5, you can see that two nonzero vectors meet at a right angle if and only if their dot product is zero. Two such vectors are said to be orthogonal. Definition of Orthogonal Vectors The vectors u and v are orthogonal when u
v 0.
REMARK The terms “perpendicular,” “orthogonal,” and “normal” all mean essentially the same thing––meeting at right angles. It is common, however, to say that two vectors are orthogonal, two lines or planes are perpendicular, and a vector is normal to a line or plane. From this definition, it follows that the zero vector is orthogonal to every vector u, because 0 u 0. Moreover, for 0 , you know that cos 0 if and only if 2. So, you can use Theorem 11.5 to conclude that two nonzero vectors are orthogonal if and only if the angle between them is 2.
Finding the Angle Between Two Vectors See LarsonCalculus.com for an interactive version of this type of example.
For u 3, 1, 2, v 4, 0, 2, w 1, 1, 2, and z 2, 0, 1, find the angle between each pair of vectors. a. u and v
b. u and w
c. v and z
Solution uv 12 0 4 8 4 u v 1420 2145 70 4 Because u v < 0, arccos 2.069 radians. 70 uw 314 0 b. cos 0 u w 146 84 Because u w 0, u and w are orthogonal. So, 2. vz 10 8 0 2 c. cos 1 v z 205 100 Consequently, . Note that v and z are parallel, with v 2z. a. cos
REMARK The angle between u and v in Example 3(a) can also be written as approximately 118.561 .
When the angle between two vectors is known, rewriting Theorem 11.5 in the form u v u v cos
Alternative form of dot product
produces an alternative way to calculate the dot product.
Alternative Form of the Dot Product Given that u 10, v 7, and the angle between u and v is 4, find u v. Solution
Use the alternative form of the dot product as shown.
u v u v cos 107 cos
352 4
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.3
The Dot Product of Two Vectors
769
Direction Cosines
REMARK Recall that , , and are the Greek letters alpha, beta, and gamma, respectively.
z
For a vector in the plane, you have seen that it is convenient to measure direction in terms of the angle, measured counterclockwise, from the positive x-axis to the vector. In space, it is more convenient to measure direction in terms of the angles between the nonzero vector v and the three unit vectors i, j, and k, as shown in Figure 11.26. The angles , , and are the direction angles of v, and cos , cos , and cos are the direction cosines of v. Because v
i v i cos v cos
and v i v1, v2, v3
1, 0, 0 v1
k γ
v β
α
j y
i
x
Direction angles Figure 11.26
it follows that cos v1v. By similar reasoning with the unit vectors j and k, you have v1 v v cos 2 v v cos 3 . v cos
is the angle between v and i. is the angle between v and j. is the angle between v and k.
Consequently, any nonzero vector v in space has the normalized form v v v v 1 i 2 j 3 k cos i cos j cos k v v v v and because vv is a unit vector, it follows that cos 2 cos 2 cos 2 1.
Finding Direction Angles Find the direction cosines and angles for the vector v 2i 3j 4k, and show that cos 2 cos 2 cos 2 1. α = angle between v and i β = angle between v and j γ = angle between v and k
Solution
4 3
γ v = 2i + 3j + 4k
1
1 2 3 4
β
α
v1 2 v 29 v 3 cos 2 v 29 v 4 cos 3 v 29 cos
z
2
Because v 22 32 42 29, you can write the following.
2
The direction angles of v Figure 11.27
Angle between v and i
56.1
Angle between v and j
42.0
Angle between v and k
Furthermore, the sum of the squares of the direction cosines is
1
x
68.2
4 9 16 29 29 29 29 29 1.
cos 2 cos 2 cos 2
3 4
y
See Figure 11.27.
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770
Chapter 11
Vectors and the Geometry of Space
Projections and Vector Components You have already seen applications in which two vectors are added to produce a resultant vector. Many applications in physics and engineering pose the reverse problem—decomposing a vector into the sum of two vector components. The following physical example enables you to see the usefulness of this procedure. Consider a boat on an inclined ramp, as shown in Figure 11.28. The force F due to gravity pulls the boat down the ramp and against the ramp. These two forces, w1 and w2, are orthogonal—they are called the vector components of F.
w1
F w1 w2
The forces w1 and w2 help you analyze the effect of gravity on the boat. For example, w1 indicates the force necessary to keep the boat from rolling down the ramp, whereas w2 indicates the force that the tires must withstand.
w2
F
Vector components of F
The force due to gravity pulls the boat against the ramp and down the ramp. Figure 11.28
Definitions of Projection and Vector Components Let u and v be nonzero vectors. Moreover, let u w1 w2 where w1 is parallel to v and w2 is orthogonal to v, as shown in Figure 11.29. 1. w1 is called the projection of u onto v or the vector component of u along v, and is denoted by w1 projvu. 2. w2 u w1 is called the vector component of u orthogonal to v.
θ is acute.
u
w2
θ is obtuse.
u
w2 θ
θ
v
v w1
w1
w1 projvu projection of u onto v vector component of u along v w2 vector component of u orthogonal to v Figure 11.29
Finding a Vector Component of u Orthogonal to v
y
Find the vector component of u 5, 10 that is orthogonal to v 4, 3, given that
(5, 10)
10
w1 projvu 8, 6
8
(8, 6) u
(−3, 4) 4
u 5, 10 w1 w2.
(4, 3)
w2 2
v
−2
2
Solution Because u w1 w2, where w1 is parallel to v, it follows that w2 is the vector component of u orthogonal to v. So, you have
w1 x
−4
−2
u w1 w2 Figure 11.30
4
and
6
8
w2 u w1 5, 10 8, 6 3, 4. Check to see that w2 is orthogonal to v, as shown in Figure 11.30.
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11.3
The Dot Product of Two Vectors
771
From Example 5, you can see that it is easy to find the vector component w2 once you have found the projection, w1, of u onto v. To find this projection, use the dot product in the next theorem, which you will prove in Exercise 78. THEOREM 11.6 Projection Using the Dot Product If u and v are nonzero vectors, then the projection of u onto v is
REMARK Note the distinction between the terms “component” and “vector component.” For example, using the standard unit vectors with u u1i u2 j, u1 is the component of u in the direction of i and u1i is the vector component in the direction of i.
projv u
u v v v. 2
The projection of u onto v can be written as a scalar multiple of a unit vector in the direction of v. That is,
uv v v uv v vv k vv. 2
The scalar k is called the component of u in the direction of v. So, k
uv u cos . v
z
Decomposing a Vector into Vector Components 4
w2
Find the projection of u onto v and the vector component of u orthogonal to v for
u
2
u 3i 5j 2k and v 7i j 2k. w1
u = 3i − 5j + 2k v = 7i + j − 2k
−2
6
−4
8 x
v
Solution The projection of u onto v is 2
y
w1 projv u
14 2 4 7i j 2k i j k.
u v v v 12 54 9 9 9 2
The vector component of u orthogonal to v is the vector w2 u w1 3i 5j 2k
u w1 w2 Figure 11.31
149 i 92 j 94 k 139 i 479 j 229 k.
See Figure 11.31.
Finding a Force A 600-pound boat sits on a ramp inclined at 30 , as shown in Figure 11.32. What force is required to keep the boat from rolling down the ramp? Solution Because the force due to gravity is vertical and downward, you can represent the gravitational force by the vector F 600j. To find the force required to keep the boat from rolling down the ramp, project F onto a unit vector v in the direction of the ramp, as follows. v cos 30 i sin 30 j v
w1
1 i j 2 2
Unit vector along ramp
Therefore, the projection of F onto v is
30° F w1 = projv(F)
Figure 11.32
3
w1 projv F
Fv v v F vv 600 12 v 300 23 i 21 j .
2
The magnitude of this force is 300, and therefore a force of 300 pounds is required to keep the boat from rolling down the ramp.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Work The work W done by the constant force F acting along the line of motion of an object is given by \
W magnitude of forcedistance F PQ as shown in Figure 11.33(a). When the constant force F is not directed along the line of motion, you can see from Figure 11.33(b) that the work W done by the force is W projPQ F PQ cos F PQ F PQ . \
\
\
\
F
F
θ
projPQ F P
Q P Work = ⎜⎜F ⎜⎜⎜⎜PQ ⎜⎜
Q
Work = ⎜⎜proj PQ F⎜⎜⎜⎜PQ⎜⎜ (b) Force acts at angle with the line of motion.
(a) Force acts along the line of motion.
Figure 11.33
This notion of work is summarized in the next definition. Definition of Work The work W done by a constant force F as its point of application moves along the vector PQ is one of the following. \
\
1. W projPQ F PQ
Projection form
2. W F PQ
Dot product form
\
\
Finding Work To close a sliding door, a person pulls on a rope with a constant force of 50 pounds at a constant angle of 60 , as shown in Figure 11.34. Find the work done in moving the door 12 feet to its closed position.
12 ft P
projPQ F
Q
60° F
12 ft
Figure 11.34
Solution
Using a projection, you can calculate the work as follows.
1 W projPQ F PQ cos60 F PQ 5012 300 foot-pounds 2 \
\
\
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.3
11.3 Exercises
The Dot Product of Two Vectors
773
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding Dot Products In Exercises 1–8, find (a) u v, u, (c) u 2, (d) u v v, and (e) u 2v .
(b) u
1. u 3, 4, v 1, 5
2. u 4, 10, v 2, 3
3. u 6, 4, v 3, 2
4. u 4, 8, v 7, 5
31. u 3i 2j 2k
32. u 4i 3j 5k
33. u 0, 6, 4
34. u 1, 5, 2
5. u 2, 3, 4, v 0, 6, 5 6. u i, v i
Finding the Projection of u onto v In Exercises 35–42, (a) find the projection of u onto v, and (b) find the vector component of u orthogonal to v.
7. u 2i j k
35. u 6, 7, v 1, 4
8. u 2i j 2k
vik
v i 3j 2k
Finding the Angle Between Two Vectors In Exercises 9–16, find the angle between the vectors (a) in radians and (b) in degrees.
36. u 9, 7, v 1, 3
37. u 2i 3j, v 5i j 38. u 2i 3j, v 3i 2j 39. u 0, 3, 3, v 1, 1, 1 40. u 8, 2, 0, v 2, 1, 1
9. u 1, 1, v 2, 2
10. u 3, 1, v 2, 1
11. u 3i j, v 2i 4j
41. u 2i j 2k, v 3j 4k 42. u i 4k, v 3i 2k
3 3 i sin j, v cos i sin j 12. u cos 6 6 4 4
WRITING ABOUT CONCEPTS
13. u 1, 1, 1
14. u 3i 2j k
43. Dot Product Define the dot product of vectors u and v.
v 2i 3j
44. Orthogonal Vectors State the definition of orthogonal vectors. When vectors are neither parallel nor orthogonal, how do you find the angle between them? Explain.
v 2, 1, 1 15. u 3i 4j
16. u 2i 3j k
v 2j 3k
v i 2j k
Alternative Form of Dot Product In Exercises 17 and 18,
use the alternative form of the dot product to find u v.
45. Using Vectors Determine which of the following are defined for nonzero vectors u, v, and w. Explain your reasoning.
17. u 8, v 5, and the angle between u and v is 3. 18. u 40, v 25, and the angle between u and v is 56.
Comparing Vectors In Exercises 19 – 24, determine whether u and v are orthogonal, parallel, or neither. 19. u 4, 3, v
1 2,
23
21. u j 6k
20. u
2j, v 2i 4j
22. u 2i 3j k
v i 2j k 23. u 2, 3, 1
13i
v 2i j k 24. u cos , sin , 1 v sin , cos , 0
v 1, 1, 1
(a) u
v w uvw
(b) u vw
(c)
(d) u
v w
46. Direction Cosines Describe direction cosines and direction angles of a vector v. 47. Projection Give a geometric description of the projection of u onto v. 48. Projection What can be said about the vectors u and v when (a) the projection of u onto v equals u and (b) the projection of u onto v equals 0? 49. Projection When the projection of u onto v has the same magnitude as the projection of v onto u, can you conclude that u v ? Explain.
Classifying a Triangle In Exercises 25–28, the vertices of a triangle are given. Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle. Explain your reasoning.
HOW DO YOU SEE IT? What is known about , the angle between two nonzero vectors u and v, when
50.
25. 1, 2, 0, 0, 0, 0, 2, 1, 0 26. 3, 0, 0, 0, 0, 0, 1, 2, 3 27. 2, 0, 1, 0, 1, 2, 0.5, 1.5, 0
(a) u
v 0?
(b) u
28. 2, 7, 3, 1, 5, 8, 4, 6, 1
Finding Direction Angles In Exercises 29–34, find the direction cosines and angles of u , and demonstrate that the sum of the squares of the direction cosines is 1. 29. u i 2j 2k
u
v
θ
(c) u v < 0?
> 0?
v
Origin
30. u 5i 3j k
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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51. Revenue The vector u 3240, 1450, 2235 gives the numbers of hamburgers, chicken sandwiches, and cheeseburgers, respectively, sold at a fast-food restaurant in one week. The vector v 2.25, 2.95, 2.65 gives the prices (in dollars) per unit for the three food items. Find the dot product u v, and explain what information it gives. 52. Revenue Repeat Exercise 51 after increasing prices by 4%. Identify the vector operation used to increase prices by 4%.
Orthogonal Vectors In Exercises 53–56, find two vectors in opposite directions that are orthogonal to the vector u. (The answers are not unique.) 53. u 14 i 32 j
54. u 9i 4j
55. u 3, 1, 2
56. u 4, 3, 6
57. Finding an Angle Find the angle between a cube’s diagonal and one of its edges. 58. Finding an Angle Find the angle between the diagonal of a cube and the diagonal of one of its sides. 59. Braking Load A 48,000-pound truck is parked on a 10 slope (see figure). Assume the only force to overcome is that due to gravity. Find (a) the force required to keep the truck from rolling down the hill and (b) the force perpendicular to the hill.
63. Work A car is towed using a force of 1600 newtons. The chain used to pull the car makes a 25 angle with the horizontal. Find the work done in towing the car 2 kilometers. 64. Work A sled is pulled by exerting a force of 100 newtons on a rope that makes a 25 angle with the horizontal. Find the work done in pulling the sled 40 meters.
True or False? In Exercises 65 and 66, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 65. If u v u w and u 0, then v w. 66. If u and v are orthogonal to w, then u v is orthogonal to w.
Using Points of Intersection In Exercises 67–70, (a) find all points of intersection of the graphs of the two equations, (b) find the unit tangent vectors to each curve at their points of intersection, and (c) find the angles 0 90 between the curves at their points of intersection. 67. y x2,
y x13
69. y 1
x2,
Weight = 48,000 lb
60. Braking Load A 5400-pound sport utility vehicle is parked on an 18 slope. Assume the only force to overcome is that due to gravity. Find (a) the force required to keep the vehicle from rolling down the hill and (b) the force perpendicular to the hill. 61. Work An object is pulled 10 feet across a floor, using a force of 85 pounds. The direction of the force is 60 above the horizontal (see figure). Find the work done.
85 lb 20°
60° 10 ft
70. y 1 x, y x3 1
72. Proof Use vectors to prove that a parallelogram is a rectangle if and only if its diagonals are equal in length. 73. Bond Angle Consider a regular tetrahedron with vertices 0, 0, 0, k, k, 0, k, 0, k, and 0, k, k, where k is a positive real number. (a) Sketch the graph of the tetrahedron. (b) Find the length of each edge. (c) Find the angle between any two edges. (d) Find the angle between the line segments from the centroid k2, k2, k2 to two vertices. This is the bond angle for a molecule such as CH 4 or PbCl 4, where the structure of the molecule is a tetrahedron. 74. Proof Consider the vectors u cos , sin , 0 and v cos , sin , 0, where > . Find the dot product of the vectors and use the result to prove the identity cos cos cos sin sin .
Not drawn to scale
Figure for 62
62. Work A toy wagon is pulled by exerting a force of 25 pounds on a handle that makes a 20 angle with the horizontal (see figure). Find the work done in pulling the wagon 50 feet. Ziva_K/iStockphoto.com
1
2
71. Proof Use vectors to prove that the diagonals of a rhombus are perpendicular.
10°
Figure for 61
y
y x13
68. y x3, x2
75. Proof
Prove that u v 2 u 2 v 2 2u v.
76. Proof
Prove the Cauchy-Schwarz Inequality,
u v u v . 77. Proof
Prove the triangle inequality u v u v .
78. Proof
Prove Theorem 11.6.
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11.4
The Cross Product of Two Vectors in Space
775
11.4 The Cross Product of Two Vectors in Space Find the cross product of two vectors in space. Use the triple scalar product of three vectors in space.
The Cross Product Exploration Geometric Property of the Cross Product Three pairs of vectors are shown below. Use the definition to find the cross product of each pair. Sketch all three vectors in a three-dimensional system. Describe any relationships among the three vectors. Use your description to write a conjecture about u, v, and
Many applications in physics, engineering, and geometry involve finding a vector in space that is orthogonal to two given vectors. In this section, you will study a product that will yield such a vector. It is called the cross product, and it is most conveniently defined and calculated using the standard unit vector form. Because the cross product yields a vector, it is also called the vector product. Definition of Cross Product of Two Vectors in Space Let u ⫽ u 1i ⫹ u 2 j ⫹ u3 k
and
v ⫽ v 1i ⫹ v 2 j ⫹ v 3k
be vectors in space. The cross product of u and v is the vector u ⫻ v ⫽ u 2v3 ⫺ u 3v2 i ⫺ u 1v3 ⫺ u 3v1 j ⫹ u 1v2 ⫺ u 2v1 k.
u ⫻ v.
a. u ⫽ 3, 0, 3, v ⫽ 3, 0, ⫺3 It is important to note that this definition applies only to three-dimensional vectors. The cross product is not defined for two-dimensional vectors. A convenient way to calculate u ⫻ v is to use the determinant form with cofactor expansion shown below. (This 3 ⫻ 3 determinant form is used simply to help remember the formula for the cross product—it is technically not a determinant because not all the entries of the corresponding matrix are real numbers.)
z 3 2
−2
u 1
−3
1
1
2
3 x
−3
y
3
−3
v
k u3 v3
i ⫽ u1 v1
j u2 v2
k u3 i ⫺ v3
z 3 2 −3
−2
1
1
2 3 x
−2
−3
u 2
⫽ y
3
−2
c. u ⫽ 3, 3, 0, v ⫽ 3, ⫺3, 0
−3
v
−2
−2
1
u x
−3
−3
2
u3 u i⫺ 1 v3 v1
i u1 v1
j u2 v2
u3 j⫹ v3
k u3 j ⫹ v3 u1 v1
i u1 v1
u2 k v2
j u2 v2
k u3 k v3
a c
b ⫽ ad ⫺ bc. d
Here are a couple of examples.
1
2
u2 v2
Put “v” in Row 3.
Note the minus sign in front of the j-component. Each of the three 2 ⫻ 2 determinants can be evaluated by using the diagonal pattern
z
2
Put “u” in Row 2.
⫽ u 2v 3 ⫺ u 3v 2 i ⫺ u 1v 3 ⫺ u 3v 1 j ⫹ u1v2 ⫺ u 2v 1 k
−3
3
j u2 v2
b. u ⫽ 0, 3, 3, v ⫽ 0, ⫺3, 3 v
i u ⫻ v ⫽ u1 v1
3
2 3
y
and
4 ⫺6
4 ⫽ 2⫺1 ⫺ 43 ⫽ ⫺2 ⫺ 12 ⫽ ⫺14 ⫺1
0 ⫽ 43 ⫺ 0⫺6 ⫽ 12 3
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NOTATION FOR DOT AND CROSS PRODUCTS
The notation for the dot product and cross product of vectors was first introduced by the American physicist Josiah Willard Gibbs (1839–1903). In the early 1880s, Gibbs built a system to represent physical quantities called “vector analysis.” The system was a departure from Hamilton’s theory of quaternions.
Finding the Cross Product For u ⫽ i ⫺ 2j ⫹ k and v ⫽ 3i ⫹ j ⫺ 2k, find each of the following. a. u ⫻ v
b. v ⫻ u
Solution
c. v
⫻
v
i a. u ⫻ v ⫽ 1 3
j ⫺2 1
k 1 ⫺2
⫺2 1 1 1 1 ⫺2 i⫺ j⫹ k 1 ⫺2 3 ⫺2 3 1 ⫽ 4 ⫺ 1 i ⫺ ⫺2 ⫺ 3 j ⫹ 1 ⫹ 6 k ⫽ 3i ⫹ 5j ⫹ 7k
⫽
i b. v ⫻ u ⫽ 3 1
j 1 ⫺2
k ⫺2 1
1 ⫺2 3 ⫺2 3 1 i⫺ j⫹ k ⫺2 1 1 1 1 ⫺2 ⫽ 1 ⫺ 4i ⫺ 3 ⫹ 2j ⫹ ⫺6 ⫺ 1k ⫽ ⫺3i ⫺ 5j ⫺ 7k
⫽
REMARK Note that this result is the negative of that in part (a).
c. v
⫻
i v⫽ 3 3
j 1 1
k ⫺2 ⫽ 0 ⫺2
The results obtained in Example 1 suggest some interesting algebraic properties of the cross product. For instance, u ⫻ v ⫽ ⫺ v ⫻ u, and v ⫻ v ⫽ 0. These properties, and several others, are summarized in the next theorem. THEOREM 11.7 Algebraic Properties of the Cross Product Let u, v, and w be vectors in space, and let c be a scalar. 1. 2. 3. 4. 5. 6.
u ⫻ v ⫽ ⫺ v ⫻ u u ⫻ v ⫹ w ⫽ u ⫻ v ⫹ u ⫻ w cu ⫻ v ⫽ cu ⫻ v ⫽ u ⫻ cv u⫻0⫽0⫻u⫽0 u⫻u⫽0 u ⭈ v ⫻ w ⫽ u ⫻ v ⭈ w To prove Property 1, let u ⫽ u 1i ⫹ u 2 j ⫹ u 3k and v ⫽ v 1i ⫹ v 2 j ⫹ v 3k.
Proof Then, u
⫻
v ⫽ u 2v 3 ⫺ u 3v 2i ⫺ u 1v 3 ⫺ u 3v 1j ⫹ u 1v 2 ⫺ u 2v 1k
v
⫻
u ⫽ v 2u 3 ⫺ v3u 2i ⫺ v 1u 3 ⫺ v3u 1j ⫹ v1u 2 ⫺ v2u 1k
and which implies that u ⫻ v ⫽ ⫺ v exercises (see Exercises 51–54).
⫻
u. Proofs of Properties 2, 3, 5, and 6 are left as
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
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11.4
The Cross Product of Two Vectors in Space
777
Note that Property 1 of Theorem 11.7 indicates that the cross product is not commutative. In particular, this property indicates that the vectors u ⫻ v and v ⫻ u have equal lengths but opposite directions. The next theorem lists some other geometric properties of the cross product of two vectors. THEOREM 11.8 Geometric Properties of the Cross Product Let u and v be nonzero vectors in space, and let be the angle between u and v.
REMARK It follows from Properties 1 and 2 in Theorem 11.8 that if n is a unit vector orthogonal to both u and v, then u ⫻ v ⫽ ± u v sin n.
1. 2. 3. 4.
Proof
u ⫻ v is orthogonal to both u and v. u ⫻ v ⫽ u v sin u ⫻ v ⫽ 0 if and only if u and v are scalar multiples of each other. u ⫻ v ⫽ area of parallelogram having u and v as adjacent sides. To prove Property 2, note because cos ⫽ u ⭈ v u v , it follows that
u v sin ⫽ u v1 ⫺ cos 2
⫽ u v
1⫺
u ⭈ v 2 u 2 v 2
⫽ u 2 v 2 ⫺ u ⭈ v 2 ⫽ u12 ⫹ u22 ⫹ u32v12 ⫹ v22 ⫹ v32 ⫺ u 1v1 ⫹ u 2v2 ⫹ u 3v3 2 ⫽ u 2v3 ⫺ u 3v2) 2 ⫹ u 1v3 ⫺ u 3v1 2 ⫹ u 1v2 ⫺ u 2v12 ⫽ u ⫻ v. v
⎜⎜v ⎜⎜ sin θ θ
u
The vectors u and v form adjacent sides of a parallelogram. Figure 11.35
To prove Property 4, refer to Figure 11.35, which is a parallelogram having v and u as adjacent sides. Because the height of the parallelogram is v sin , the area is Area ⫽ baseheight ⫽ u v sin ⫽ u ⫻ v. Proofs of Properties 1 and 3 are left as exercises (see Exercises 55 and 56). See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Both u ⫻ v and v ⫻ u are perpendicular to the plane determined by u and v. One way to remember the orientations of the vectors u, v, and u ⫻ v is to compare them with the unit vectors i, j, and k ⫽ i ⫻ j, as shown in Figure 11.36. The three vectors u, v, and u ⫻ v form a right-handed system, whereas the three vectors u, v, and v ⫻ u form a left-handed system. k=i×j
u×v
j i
xy-plane
v u Plane determined by u and v
Right-handed systems Figure 11.36
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Using the Cross Product See LarsonCalculus.com for an interactive version of this type of example. z
Find a unit vector that is orthogonal to both
(− 3, 2, 11)
u ⫽ i ⫺ 4j ⫹ k
12
and 10
v ⫽ 2i ⫹ 3j. Solution u and v.
8
6
(1, −4, 1)
⫻
v, as shown in Figure 11.37, is orthogonal to both
i j k u ⫻ v ⫽ 1 ⫺4 1 2 3 0 ⫽ ⫺3i ⫹ 2j ⫹ 11k
u×v 4
The cross product u
−4
2
u
Cross product
Because 2
4
v
2
u
y
(2, 3, 0)
4
⫻
v ⫽ ⫺3 2 ⫹ 2 2 ⫹ 11 2 ⫽ 134
a unit vector orthogonal to both u and v is
x
The vector u ⫻ v is orthogonal to both u and v. Figure 11.37
u⫻v 3 2 11 ⫽⫺ i⫹ j⫹ k. u ⫻ v 134 134 134 In Example 2, note that you could have used the cross product v ⫻ u to form a unit vector that is orthogonal to both u and v. With that choice, you would have obtained the negative of the unit vector found in the example.
Geometric Application of the Cross Product The vertices of a quadrilateral are listed below. Show that the quadrilateral is a parallelogram, and find its area. A ⫽ 5, 2, 0 C ⫽ 2, 4, 7
B ⫽ 2, 6, 1 D ⫽ 5, 0, 6
Solution From Figure 11.38, you can see that the sides of the quadrilateral correspond to the following four vectors.
z
8
\
\
AB ⫽ ⫺3i ⫹ 4j ⫹ k AD ⫽ 0i ⫺ 2j ⫹ 6k \
6
C = (2, 4, 7)
\
\
\
\
\
\
So, AB is parallel to CD and AD is parallel to CB , and you can conclude that the quadrilateral is a parallelogram with AB and AD as adjacent sides. Moreover, because
D = (5, 0, 6) 2
\
2
AB 4
6
⫻
y
A = (5, 2, 0)
x
\
j 4 ⫺2
k 1 6
\
\
Cross product
⫽ 26i ⫹ 18j ⫹ 6k the area of the parallelogram is \
The area of the parallelogram is approximately 32.19. Figure 11.38
i AD ⫽ ⫺3 0
B = (2, 6, 1) 6
\
CD ⫽ 3i ⫺ 4j ⫺ k ⫽ ⫺AB CB ⫽ 0i ⫹ 2j ⫺ 6k ⫽ ⫺AD
AB
⫻
\
AD ⫽ 1036 32.19.
Is the parallelogram a rectangle? You can determine whether it is by finding the angle between the vectors AB and AD . \
\
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.4
The Cross Product of Two Vectors in Space
779
In physics, the cross product can be used to measure torque—the moment M of a force F about a point P, as shown in Figure 11.39. If the point of application of the force is Q, then the moment of F about P is \
M ⫽ PQ
M
⫻
F.
Moment of F about P \
P
The magnitude of the moment M measures the tendency of the vector PQ to rotate counterclockwise (using the right-hand rule) about an axis directed along the vector M.
PQ Q
An Application of the Cross Product
F
A vertical force of 50 pounds is applied to the end of a one-foot lever that is attached to an axle at point P, as shown in Figure 11.40. Find the moment of this force about the point P when ⫽ 60⬚.
The moment of F about P Figure 11.39
Solution
z
Represent the 50-pound force as
F ⫽ ⫺50k
Q
and the lever as 3 1 PQ ⫽ cos60⬚ j ⫹ sin60⬚k ⫽ j ⫹ k. 2 2
F
\
60° y x
A vertical force of 50 pounds is applied at point Q. Figure 11.40
The moment of F about P is
P
\
M ⫽ PQ
⫻
i
j
k
F⫽ 0
1 2 0
3
0
2 ⫺50
⫽ ⫺25i.
Moment of F about P
The magnitude of this moment is 25 foot-pounds. In Example 4, note that the moment (the tendency of the lever to rotate about its axle) is dependent on the angle . When ⫽ 2, the moment is 0. The moment is greatest when ⫽ 0.
The Triple Scalar Product For vectors u, v, and w in space, the dot product of u and v u ⭈ v
⫻
⫻
w
w
is called the triple scalar product, as defined in Theorem 11.9. The proof of this theorem is left as an exercise (see Exercise 59). THEOREM 11.9 The Triple Scalar Product For u ⫽ u1i ⫹ u2 j ⫹ u3 k, v ⫽ v1 i ⫹ v2 j ⫹ v3 k, and w ⫽ w1i ⫹ w2 j ⫹ w3k, the triple scalar product is
u1 u ⭈ v ⫻ w ⫽ v1 w1
u2 v2 w2
u3 v3 . w3
Note that the value of a determinant is multiplied by ⫺1 when two rows are interchanged. After two such interchanges, the value of the determinant will be unchanged. So, the following triple scalar products are equivalent. u
⭈ v ⫻ w ⫽ v ⭈ w ⫻ u ⫽ w ⭈ u ⫻ v
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Vectors and the Geometry of Space
v×w
If the vectors u, v, and w do not lie in the same plane, then the triple scalar product u ⭈ v ⫻ w) can be used to determine the volume of the parallelepiped (a polyhedron, all of whose faces are parallelograms) with u, v, and w as adjacent edges, as shown in Figure 11.41. This is established in the next theorem. u
THEOREM 11.10 Geometric Property of the Triple Scalar Product The volume V of a parallelepiped with vectors u, v, and w as adjacent edges is w
V ⫽ u ⭈ v
v
⎜⎜projv × wu ⎜⎜
Area of base ⫽ v ⫻ w Volume of parallelepiped ⫽ u Figure 11.41
⭈ v ⫻ w
w .
⫻
Proof In Figure 11.41, note that the area of the base is v ⫻ w and the height of the parallelpiped is projv ⫻ wu. Therefore, the volume is V ⫽ heightarea of base ⫽ projv ⫻ wu v ⫻ w ⫽
u ⭈ v ⫻ w v ⫻ w v ⫻ w
⫽ u ⭈ v
⫻
w .
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Volume by the Triple Scalar Product Find the volume of the parallelepiped shown in Figure 11.42 having
z
(3, − 5, 1)
(3, 1, 1) 2
u
u ⫽ 3i ⫺ 5j ⫹ k v ⫽ 2j ⫺ 2k
1
w v 3
and
x
(0, 2, − 2)
w ⫽ 3i ⫹ j ⫹ k
The parallelepiped has a volume of 36. Figure 11.42
as adjacent edges. Solution
y
6
By Theorem 11.10, you have
V ⫽ u ⭈ v
w
⫻
⫺5 2 1
3 ⫽ 0 3
Triple scalar product
1 ⫺2 1
⫽3
2 1
⫺2 0 ⫺ ⫺5 1 3
⫺2 0 ⫹ 1 1 3
2 1
⫽ 34 ⫹ 56 ⫹ 1⫺6 ⫽ 36.
A natural consequence of Theorem 11.10 is that the volume of the parallelepiped is 0 if and only if the three vectors are coplanar. That is, when the vectors u ⫽ u1, u2, u3 , v ⫽ v1, v2, v3 , and w ⫽ w1, w2, w3 have the same initial point, they lie in the same plane if and only if u ⭈ v
⫻
u1 w ⫽ v1 w1
u2 v2 w2
u3 v3 ⫽ 0. w3
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.4
11.4 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Cross Product of Unit Vectors In Exercises 1–6, find the
29. Torque
cross product of the unit vectors and sketch your result.
A child applies the brakes on a bicycle by applying a downward force of 20 pounds on the pedal when the crank makes a 40⬚ angle with the horizontal (see figure). The crank is 6 inches in length. Find the torque at P.
1. j ⫻ i
2. i ⫻ j
3. j ⫻ k
4. k ⫻ j
5. i ⫻ k
6. k ⫻ i
Finding Cross Products In Exercises 7–10, find (a) u ⴛ v, (b) v ⴛ u, and (c) v ⴛ v. 7. u ⫽ ⫺2i ⫹ 4j
8. u ⫽ 3i ⫹ 5k
v ⫽ 3i ⫹ 2j ⫹ 5k
v ⫽ 2i ⫹ 3j ⫺ 2k
9. u ⫽ 7, 3, 2
781
The Cross Product of Two Vectors in Space
10. u ⫽ 3, ⫺2, ⫺2
v ⫽ 1, ⫺1, 5
v ⫽ 1, 5, 1 6 in.
Finding a Cross Product In Exercises 11–16, find u ⴛ v
F = 20 lb
and show that it is orthogonal to both u and v. 11. u ⫽ 12, ⫺3, 0
40° P
12. u ⫽ ⫺1, 1, 2
v ⫽ ⫺2, 5, 0
v ⫽ 0, 1, 0
13. u ⫽ 2, ⫺3, 1
14. u ⫽ ⫺10, 0, 6
v ⫽ 1, ⫺2, 1
v ⫽ 5, ⫺3, 0
15. u ⫽ i ⫹ j ⫹ k
16. u ⫽ i ⫹ 6j
v ⫽ 2i ⫹ j ⫺ k
v ⫽ ⫺2i ⫹ j ⫹ k
30. Torque Both the magnitude and the direction of the force on a crankshaft change as the crankshaft rotates. Find the torque on the crankshaft using the position and data shown in the figure.
Finding a Unit Vector In Exercises 17–20, find a unit vector that is orthogonal to both u and v. 18. u ⫽ ⫺8, ⫺6, 4
19. u ⫽ ⫺3i ⫹ 2j ⫺ 5k
20. u ⫽ 2k
v ⫽ i ⫺ j ⫹ 4k
ft
v ⫽ 10, ⫺12, ⫺2
180 lb
16
v ⫽ 2, 5, 3
B
0.
17. u ⫽ 4, ⫺3, 1
F θ
60°
12 in.
2000 lb
v ⫽ 4i ⫹ 6k
that has the given vectors as adjacent sides. Use a computer algebra system or a graphing utility to verify your result. 21. u ⫽ j
22. u ⫽ i ⫹ j ⫹ k
v⫽j⫹k
v⫽j⫹k
23. u ⫽ 3, 2, ⫺1
24. u ⫽ 2, ⫺1, 0
v ⫽ 1, 2, 3
v ⫽ ⫺1, 2, 0
Area In Exercises 25 and 26, verify that the points are the vertices of a parallelogram, and find its area. 25. A0, 3, 2, B1, 5, 5, C6, 9, 5, D5, 7, 2 26. A2, ⫺3, 1, B6, 5, ⫺1, C7, 2, 2, D3, ⫺6, 4
A
15 in.
Area In Exercises 21–24, find the area of the parallelogram Figure for 30
Figure for 31
31. Optimization A force of 180 pounds acts on the bracket shown in the figure. \
(a) Determine the vector AB and the vector F representing the force. (F will be in terms of .) (b) Find the magnitude of the moment about A by evaluating AB ⫻ F . \
(c) Use the result of part (b) to determine the magnitude of the moment when ⫽ 30⬚. (d) Use the result of part (b) to determine the angle when the magnitude of the moment is maximum. At that angle, what is the relationship between the vectors F and AB ? Is it what you expected? Why or why not? \
Area In Exercises 27 and 28, find the area of the triangle with
the given vertices. Hint: u ⴛ v is the area of the triangle having u and v as adjacent sides. 27. A0, 0, 0, B1, 0, 3, C⫺3, 2, 0 1 2
28. A2, ⫺3, 4, B0, 1, 2, C⫺1, 2, 0 Elena Elisseeva/Shutterstock.com
(e) Use a graphing utility to graph the function for the magnitude of the moment about A for 0⬚ ⱕ ⱕ 180⬚. Find the zero of the function in the given domain. Interpret the meaning of the zero in the context of the problem.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Chapter 11
Vectors and the Geometry of Space 42. Using Dot and Cross Products When u ⫻ v ⫽ 0 and u ⭈ v ⫽ 0, what can you conclude about u and v?
32. Optimization A force of 56 pounds acts on the pipe wrench shown in the figure.
A 18 in.
(a) Find the magnitude of the moment about O by evaluating OA ⫻ F . Use a graphing utility to graph the resulting function of .
WRITING ABOUT CONCEPTS θ
\
F
30° O
(b) Use the result of part (a) to determine the magnitude of the moment when ⫽ 45⬚. (c) Use the result of part (a) to determine the angle when the magnitude of the moment is maximum. Is the answer what you expected? Why or why not?
Finding a Triple Scalar Product In Exercises 33–36, find u ⭈ v ⴛ w . 33. u ⫽ i
34. u ⫽ 1, 1, 1
v⫽j
v ⫽ 2, 1, 0
w⫽k
w ⫽ 0, 0, 1
43. Cross Product and v.
Define the cross product of vectors u
44. Cross Product cross product.
State the geometric properties of the
45. Magnitude When the magnitudes of two vectors are doubled, how will the magnitude of the cross product of the vectors change? Explain.
46.
HOW DO YOU SEE IT? The vertices of a triangle in space are x1, y1, z1, x2, y2, z2, and x3, y3, z3. Explain how to find a vector perpendicular to the triangle. z
35. u ⫽ 2, 0, 1
36. u ⫽ 2, 0, 0
5
v ⫽ 0, 3, 0
v ⫽ 1, 1, 1
4
w ⫽ 0, 0, 1
w ⫽ 0, 2, 2
3
(x3, y3, z3)
2 1
Volume In Exercises 37 and 38, use the triple scalar product
−2
to find the volume of the parallelepiped having adjacent edges u, v, and w. 38. u ⫽ 1, 3, 1
v⫽j⫹k
v ⫽ 0, 6, 6
w⫽i⫹k
w ⫽ ⫺4, 0, ⫺4
z
statement is true or false. If it is false, explain why or give an example that shows it is false.
v
4
1
2
v u
2
2
y
4 x
(x2, y2 z2)
(x1, y1, z1)
True or False? In Exercises 47–50, determine whether the
6
w
y
x
z
2
5
3 4
37. u ⫽ i ⫹ j
1
2
47. It is possible to find the cross product of two vectors in a two-dimensional coordinate system.
u 6
8
y
w
48. If u and v are vectors in space that are nonzero and nonparallel, then u ⫻ v ⫽ v ⫻ u. 49. If u ⫽ 0 and u ⫻ v ⫽ u ⫻ w, then v ⫽ w.
x
Volume In Exercises 39 and 40, find the volume of the parallelepiped with the given vertices. 39. 0, 0, 0, 3, 0, 0, 0, 5, 1, 2, 0, 5
3, 5, 1, 5, 0, 5, 2, 5, 6, 5, 5, 6 40. 0, 0, 0, 0, 4, 0, ⫺3, 0, 0, ⫺1, 1, 5
⫺3, 4, 0, ⫺1, 5, 5, ⫺4, 1, 5, ⫺4, 5, 5 41. Comparing Dot Products Identify the dot products that are equal. Explain your reasoning. (Assume u, v, and w are nonzero vectors.) (a) u
⭈ v ⫻ w
(b) v ⫻ w ⭈ u
(c) u ⫻ v ⭈ w
(d) u ⫻ ⫺w ⭈ v
(e) u ⭈ w ⫻ v
(f) w ⭈ v ⫻ u
(g) ⫺u ⫻ v ⭈ w
(h) w ⫻ u ⭈ v
50. If u ⫽ 0, u ⭈ v ⫽ u ⭈ w, and u ⫻ v ⫽ u ⫻ w, then v ⫽ w.
Proof In Exercises 51–56, prove the property of the cross product. 51. u ⫻ v ⫹ w ⫽ u ⫻ v ⫹ u ⫻ w 52. cu ⫻ v ⫽ cu ⫻ v ⫽ u ⫻ cv 53. u ⫻ u ⫽ 0
54. u
⭈ v ⫻ w ⫽ u ⫻ v ⭈ w
55. u ⫻ v is orthogonal to both u and v. 56. u ⫻ v ⫽ 0 if and only if u and v are scalar multiples of each other. 57. Proof Prove that u ⫻ v ⫽ u v if u and v are orthogonal. 58. Proof
Prove that u ⫻ v ⫻ w ⫽ u
59. Proof
Prove Theorem 11.9.
⭈ w v ⫺ u ⭈ v w.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.5
783
Lines and Planes in Space
11.5 Lines and Planes in Space Write a set of parametric equations for a line in space. Write a linear equation to represent a plane in space. Sketch the plane given by a linear equation. Find the distances between points, planes, and lines in space.
Lines in Space z
In the plane, slope is used to determine the equation of a line. In space, it is more convenient to use vectors to determine the equation of a line. In Figure 11.43, consider the line L through the point P共x1, y1, z1兲 and parallel to the vector v ⫽ 具a, b, c典. The vector v is a direction vector for the line L, and a, b, and c are direction numbers. One way of describing the line L is to say that it consists of all points Q共x, y, z兲 for which the vector PQ is parallel to v. This means that PQ is a scalar multiple of v and you can write PQ ⫽ t v, where t is a scalar (a real number).
Q(x, y, z) L
P(x1, y1, z1)
\
v = 〈a, b, c〉
\
\
\
PQ ⫽ 具x ⫺ x1, y ⫺ y1, z ⫺ z1 典 ⫽ 具at, bt, ct典 ⫽ t v
y
PQ = tv
By equating corresponding components, you can obtain parametric equations of a line in space.
x
THEOREM 11.11 Parametric Equations of a Line in Space A line L parallel to the vector v ⫽ 具a, b, c典 and passing through the point P共x1, y1, z1兲 is represented by the parametric equations
Line L and its direction vector v Figure 11.43
x ⫽ x1 ⫹ at,
y ⫽ y1 ⫹ bt, and
z ⫽ z1 ⫹ ct.
If the direction numbers a, b, and c are all nonzero, then you can eliminate the parameter t to obtain symmetric equations of the line. x ⫺ x1 y ⫺ y1 z ⫺ z1 ⫽ ⫽ a b c
Symmetric equations
z
Finding Parametric and Symmetric Equations
(1, −2, 4) 4
−4
2
Find parametric and symmetric equations of the line L that passes through the point 共1, ⫺2, 4兲 and is parallel to v ⫽ 具2, 4, ⫺4典, as shown in Figure 11.44.
−4 −2
Solution To find a set of parametric equations of the line, use the coordinates x1 ⫽ 1, y1 ⫽ ⫺2, and z1 ⫽ 4 and direction numbers a ⫽ 2, b ⫽ 4, and c ⫽ ⫺4. x ⫽ 1 ⫹ 2t, y ⫽ ⫺2 ⫹ 4t, z ⫽ 4 ⫺ 4t
2 2
4 x
4
L
y
Because a, b, and c are all nonzero, a set of symmetric equations is x⫺1 y⫹2 z⫺4 ⫽ ⫽ . 2 4 ⫺4
v = 〈2, 4, −4〉
The vector v is parallel to the line L. Figure 11.44
Parametric equations
Symmetric equations
Neither parametric equations nor symmetric equations of a given line are unique. For instance, in Example 1, by letting t ⫽ 1 in the parametric equations, you would obtain the point 共3, 2, 0兲. Using this point with the direction numbers a ⫽ 2, b ⫽ 4, and c ⫽ ⫺4 would produce a different set of parametric equations x ⫽ 3 ⫹ 2t, y ⫽ 2 ⫹ 4t, and
z ⫽ ⫺4t.
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Vectors and the Geometry of Space
Parametric Equations of a Line Through Two Points See LarsonCalculus.com for an interactive version of this type of example.
Find a set of parametric equations of the line that passes through the points
共⫺2, 1, 0兲 and 共1, 3, 5兲. Solution Begin by using the points P共⫺2, 1, 0兲 and Q共1, 3, 5兲 to find a direction vector for the line passing through P and Q. \
v ⫽ PQ ⫽ 具1 ⫺ 共⫺2兲, 3 ⫺ 1, 5 ⫺ 0典 ⫽ 具3, 2, 5典 ⫽ 具a, b, c典 Using the direction numbers a ⫽ 3, b ⫽ 2, and c ⫽ 5 with the point P共⫺2, 1, 0兲, you can obtain the parametric equations x ⫽ ⫺2 ⫹ 3t, y ⫽ 1 ⫹ 2t, and
z ⫽ 5t.
REMARK As t varies over all real numbers, the parametric equations in Example 2 determine the points 共x, y, z兲 on the line. In particular, note that t ⫽ 0 and t ⫽ 1 give the original points 共⫺2, 1, 0兲 and 共1, 3, 5兲.
Planes in Space You have seen how an equation of a line in space can be obtained from a point on the line and a vector parallel to it. You will now see that an equation of a plane in space can be obtained from a point in the plane and a vector normal (perpendicular) to the plane. Consider the plane containing the point P共x1, y1, z1兲 having a nonzero normal vector
z
n
P Q
n ⫽ 具a, b, c典
y
n · PQ = 0
as shown in Figure 11.45. This plane consists of all points Q共x, y, z兲 for which vector PQ is orthogonal to n. Using the dot product, you can write the following.
x
\
The normal vector n is orthogonal to each vector PQ in the plane. Figure 11.45 \
n ⭈ PQ ⫽ 0 具a, b, c典 ⭈ 具x ⫺ x1, y ⫺ y1, z ⫺ z1 典 ⫽ 0 a共x ⫺ x1兲 ⫹ b共 y ⫺ y1兲 ⫹ c共z ⫺ z1兲 ⫽ 0 \
The third equation of the plane is said to be in standard form. THEOREM 11.12 Standard Equation of a Plane in Space The plane containing the point 共x1, y1, z1兲 and having normal vector n ⫽ 具a, b, c典 can be represented by the standard form of the equation of a plane a共x ⫺ x1兲 ⫹ b共 y ⫺ y1兲 ⫹ c共z ⫺ z1兲 ⫽ 0.
By regrouping terms, you obtain the general form of the equation of a plane in space. ax ⫹ by ⫹ cz ⫹ d ⫽ 0
General form of equation of plane
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.5
Lines and Planes in Space
785
Given the general form of the equation of a plane, it is easy to find a normal vector to the plane. Simply use the coefficients of x, y, and z and write n ⫽ 具a, b, c典.
Finding an Equation of a Plane in Three-Space z
Find the general equation of the plane containing the points (− 2, 1, 4)
共2, 1, 1兲, 共0, 4, 1兲, and 共⫺2, 1, 4兲.
5 4
Solution To apply Theorem 11.12, you need a point in the plane and a vector that is normal to the plane. There are three choices for the point, but no normal vector is given. To obtain a normal vector, use the cross product of vectors u and v extending from the point 共2, 1, 1兲 to the points 共0, 4, 1兲 and 共⫺2, 1, 4兲, as shown in Figure 11.46. The component forms of u and v are
v
3 −3
2
−2
1
(2, 1, 1) 2
3
x
u ⫽ 具0 ⫺ 2, 4 ⫺ 1, 1 ⫺ 1典 ⫽ 具⫺2, 3, 0典 v ⫽ 具⫺2 ⫺ 2, 1 ⫺ 1, 4 ⫺ 1典 ⫽ 具⫺4, 0, 3典
(0, 4, 1)
u
2
4
5
A plane determined by u and v Figure 11.46
y
and it follows that
ⱍ
ⱍ
n⫽u⫻v i j k ⫽ ⫺2 3 0 ⫺4 0 3 ⫽ 9i ⫹ 6j ⫹ 12k ⫽ 具a, b, c典
is normal to the given plane. Using the direction numbers for n and the point 共x1, y1, z1兲 ⫽ 共2, 1, 1兲, you can determine an equation of the plane to be a共x ⫺ x1兲 ⫹ b共 y ⫺ y1兲 ⫹ c共z ⫺ z1兲 ⫽ 0 9共x ⫺ 2兲 ⫹ 6共 y ⫺ 1兲 ⫹ 12共z ⫺ 1兲 ⫽ 0 9x ⫹ 6y ⫹ 12z ⫺ 36 ⫽ 0 3x ⫹ 2y ⫹ 4z ⫺ 12 ⫽ 0.
Standard form General form Simplified general form
REMARK In Example 3, check to see that each of the three original points satisfies the equation 3x ⫹ 2y ⫹ 4z ⫺ 12 ⫽ 0.
n1
θ
Two distinct planes in three-space either are parallel or intersect in a line. For two planes that intersect, you can determine the angle 共0 ⱕ ⱕ 兾2兲 between them from the angle between their normal vectors, as shown in Figure 11.47. Specifically, if vectors n1 and n2 are normal to two intersecting planes, then the angle between the normal vectors is equal to the angle between the two planes and is
n2
cos ⫽
ⱍn1 ⭈ n2ⱍ. 储n1储 储n2储
Angle between two planes
θ
Consequently, two planes with normal vectors n1 and n2 are The angle between two planes Figure 11.47
1. perpendicular when n1 ⭈ n2 ⫽ 0. 2. parallel when n1 is a scalar multiple of n2.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Vectors and the Geometry of Space
Finding the Line of Intersection of Two Planes Find the angle between the two planes x ⫺ 2y ⫹ z ⫽ 0 and
2x ⫹ 3y ⫺ 2z ⫽ 0.
Then find parametric equations of their line of intersection (see Figure 11.48). z Line of intersection
REMARK The threedimensional rotatable graphs that are available at LarsonCalculus.com can help you visualize surfaces such as those shown in Figure 11.48. If you have access to these graphs, you should use them to help your spatial intuition when studying this section and other sections in the text that deal with vectors, curves, or surfaces in space.
Plane 1
Plane 2
θ
y
x
Figure 11.48
Solution Normal vectors for the planes are n1 ⫽ 具1, ⫺2, 1典 and n2 ⫽ 具2, 3, ⫺2典. Consequently, the angle between the two planes is determined as follows. cos ⫽
ⱍn1 ⭈ n2ⱍ
储 n1 储 储n2 储
⫽
ⱍ⫺6ⱍ
冪6 冪17
⫽
6 冪102
⬇ 0.59409
This implies that the angle between the two planes is ⬇ 53.55⬚. You can find the line of intersection of the two planes by simultaneously solving the two linear equations representing the planes. One way to do this is to multiply the first equation by ⫺2 and add the result to the second equation. x ⫺ 2y ⫹ z ⫽ 0 2x ⫹ 3y ⫺ 2z ⫽ 0
⫺2x ⫹ 4y ⫺ 2z ⫽ 0 2x ⫹ 3y ⫺ 2z ⫽ 0 7y ⫺ 4z ⫽ 0
y⫽
4z 7
Substituting y ⫽ 4z兾7 back into one of the original equations, you can determine that x ⫽ z兾7. Finally, by letting t ⫽ z兾7, you obtain the parametric equations x ⫽ t, y ⫽ 4t, and z ⫽ 7t
Line of intersection
which indicate that 1, 4, and 7 are direction numbers for the line of intersection. Note that the direction numbers in Example 4 can be obtained from the cross product of the two normal vectors as follows. n1
⫻
ⱍ ⱍ
i n2 ⫽ 1 2 ⫽
ⱍ
⫺2 3
j ⫺2 3
k 1 ⫺2
ⱍ ⱍ ⱍ ⱍ ⱍ
1 1 i⫺ ⫺2 2
1 1 j⫹ ⫺2 2
⫺2 k 3
⫽ i ⫹ 4j ⫹ 7k
This means that the line of intersection of the two planes is parallel to the cross product of their normal vectors.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.5
787
Lines and Planes in Space
Sketching Planes in Space If a plane in space intersects one of the coordinate planes, then the line of intersection is called the trace of the given plane in the coordinate plane. To sketch a plane in space, it is helpful to find its points of intersection with the coordinate axes and its traces in the coordinate planes. For example, consider the plane 3x ⫹ 2y ⫹ 4z ⫽ 12.
Equation of plane
You can find the xy-trace by letting z ⫽ 0 and sketching the line 3x ⫹ 2y ⫽ 12
xy-trace
in the xy-plane. This line intersects the x-axis at 共4, 0, 0兲 and the y-axis at 共0, 6, 0兲. In Figure 11.49, this process is continued by finding the yz-trace and the xz-trace, and then shading the triangular region lying in the first octant. z
z
z
(0, 0, 3)
(0, 0, 3) (0, 6, 0)
(0, 6, 0)
(0, 6, 0)
y
y
y
(4, 0, 0)
(4, 0, 0)
(4, 0, 0) x
x
x
xy-trace 共z ⫽ 0兲: yz-trace 共x ⫽ 0兲: 3x ⫹ 2y ⫽ 12 2y ⫹ 4z ⫽ 12 Traces of the plane 3x ⫹ 2y ⫹ 4z ⫽ 12 Figure 11.49
xz-trace 共 y ⫽ 0兲: 3x ⫹ 4z ⫽ 12
If an equation of a plane has a missing variable, such as
z
Plane: 2x + z = 1 (0, 0, 1)
2x ⫹ z ⫽ 1 then the plane must be parallel to the axis represented by the missing variable, as shown in Figure 11.50. If two variables are missing from an equation of a plane, such as ax ⫹ d ⫽ 0 then it is parallel to the coordinate plane represented by the missing variables, as shown in Figure 11.51. z
( 12 , 0, 0) y x
Plane 2x ⫹ z ⫽ 1 is parallel to the y-axis. Figure 11.50
z
z
)0, 0, − dc ) )0, − db , 0) y x
)− da, 0, 0)
Plane ax ⫹ d ⫽ 0 is parallel to the yz-plane. Figure 11.51
y x
Plane by ⫹ d ⫽ 0 is parallel to the xz-plane.
y x
Plane cz ⫹ d ⫽ 0 is parallel to the xy-plane.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
788
Chapter 11
Vectors and the Geometry of Space
Distances Between Points, Planes, and Lines Q
n
D
projn PQ
Consider two types of problems involving distance in space: (1) finding the distance between a point and a plane, and (2) finding the distance between a point and a line. The solutions of these problems illustrate the versatility and usefulness of vectors in coordinate geometry: the first problem uses the dot product of two vectors, and the second problem uses the cross product. The distance D between a point Q and a plane is the length of the shortest line segment connecting Q to the plane, as shown in Figure 11.52. For any point P in the plane, you can find this distance by projecting the vector PQ onto the normal vector n. The length of this projection is the desired distance. \
P
D = ⎜⎜projn PQ ⎜⎜
THEOREM 11.13 Distance Between a Point and a Plane The distance between a plane and a point Q (not in the plane) is
The distance between a point and a plane Figure 11.52
ⱍPQ ⭈ nⱍ \
\
D ⫽ 储 projnPQ 储 ⫽
储 n储
where P is a point in the plane and n is normal to the plane.
To find a point in the plane ax ⫹ by ⫹ cz ⫹ d ⫽ 0, where a ⫽ 0, let y ⫽ 0 and z ⫽ 0. Then, from the equation ax ⫹ d ⫽ 0, you can conclude that the point
冢⫺ ad, 0, 0冣 lies in the plane.
Finding the Distance Between a Point and a Plane Find the distance between the point Q共1, 5, ⫺4兲 and the plane 3x ⫺ y ⫹ 2z ⫽ 6.
REMARK In the solution to Example 5, note that the choice of the point P is arbitrary. Try choosing a different point in the plane to verify that you obtain the same distance.
Solution You know that n ⫽ 具3, ⫺1, 2典 is normal to the plane. To find a point in the plane, let y ⫽ 0 and z ⫽ 0, and obtain the point P共2, 0, 0兲. The vector from P to Q is \
PQ ⫽ 具1 ⫺ 2, 5 ⫺ 0, ⫺4 ⫺ 0典 ⫽ 具⫺1, 5, ⫺4典. Using the Distance Formula given in Theorem 11.13 produces
ⱍPQ ⭈ nⱍ ⫽ ⱍ具⫺1, 5, ⫺4典 ⭈ 具3, ⫺1, 2典ⱍ ⫽ ⱍ⫺3 ⫺ 5 ⫺ 8ⱍ ⫽ \
D⫽
储n 储
冪9 ⫹ 1 ⫹ 4
冪14
16 冪14
⬇ 4.28.
From Theorem 11.13, you can determine that the distance between the point Q共x0, y0, z0兲 and the plane ax ⫹ by ⫹ cz ⫹ d ⫽ 0 is D⫽
ⱍa共x0 ⫺ x1兲 ⫹ b共 y0 ⫺ y1兲 ⫹ c共z0 ⫺ z1兲ⱍ 冪a2 ⫹ b2 ⫹ c2
or
D⫽
ⱍax0 ⫹ by0 ⫹ cz0 ⫹ dⱍ 冪a2 ⫹ b2 ⫹ c2
Distance between a point and a plane
where P共x1, y1, z1兲 is a point in the plane and d ⫽ ⫺ 共ax1 ⫹ by1 ⫹ cz1兲.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.5
Lines and Planes in Space
789
z
3x − y + 2z − 6 = 0
Finding the Distance Between Two Parallel Planes 3
Two parallel planes, 3x ⫺ y ⫹ 2z ⫺ 6 ⫽ 0 and 6x ⫺ 2y ⫹ 4z ⫹ 4 ⫽ 0, are shown in Figure 11.53. To find the distance between the planes, choose a point in the first plane, such as 共x0, y0, z0兲 ⫽ 共2, 0, 0兲. Then, from the second plane, you can determine that a ⫽ 6, b ⫽ ⫺2, c ⫽ 4, and d ⫽ 4, and conclude that the distance is
−6
(2, 0, 0) y
2
x
D⫽
D
⫽
6x − 2y + 4z + 4 = 0
The distance between the parallel planes is approximately 2.14. Figure 11.53
⫽
ⱍax0 ⫹ by0 ⫹ cz0 ⫹ dⱍ 冪a2 ⫹ b2 ⫹ c2
ⱍ6共2兲 ⫹ 共⫺2兲共0兲 ⫹ 共4兲共0兲 ⫹ 4ⱍ 冪62 ⫹ 共⫺2兲2 ⫹ 42
16 冪56
⫽
8 冪14
⬇ 2.14.
The formula for the distance between a point and a line in space resembles that for the distance between a point and a plane—except that you replace the dot product with the length of the cross product and the normal vector n with a direction vector for the line. THEOREM 11.14 Distance Between a Point and a Line in Space The distance between a point Q and a line in space is \
储 PQ ⫻ u储 D⫽ 储 u储 where u is a direction vector for the line and P is a point on the line.
Point
Proof In Figure 11.54, let D be the distance between the point Q and the line. Then D ⫽ 储PQ 储 sin , where is the angle between u and PQ . By Property 2 of Theorem 11.8, you have 储u储 储PQ 储 sin ⫽ 储 u ⫻ PQ 储 ⫽ 储PQ ⫻ u储. Consequently,
Q
\
\
\
D = ⎜⎜PQ ⎜⎜ sin θ
\
\
\
储 PQ ⫻ u储 D ⫽ 储PQ 储 sin ⫽ . 储 u储 \
θ u
P
Line
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
The distance between a point and a line Figure 11.54
Finding the Distance Between a Point and a Line z
Find the distance between the point Q共3, ⫺1, 4兲 and the line
6
D
x ⫽ ⫺2 ⫹ 3t, y ⫽ ⫺2t, and
5
Q = (3, − 1, 4)
Solution Using the direction numbers 3, ⫺2, and 4, a direction vector for the line is u ⫽ 具3, ⫺2, 4典. To find a point on the line, let t ⫽ 0 and obtain P ⫽ 共⫺2, 0, 1兲. So,
3
\
PQ ⫽ 具3 ⫺ 共⫺2兲, ⫺1 ⫺ 0, 4 ⫺ 1典 ⫽ 具5, ⫺1, 3典
2 −2
−2
4 x
3
2
z ⫽ 1 ⫹ 4t.
ⱍ ⱍ
and you can form the cross product
1 −1
1 2
\
PQ
3
⫻
4 5
y
The distance between the point Q and the line is 冪6 ⬇ 2.45. Figure 11.55
i u⫽ 5 3
j ⫺1 ⫺2
k 3 ⫽ 2i ⫺ 11j ⫺ 7k ⫽ 具2, ⫺11, ⫺7典. 4
Finally, using Theorem 11.14, you can find the distance to be \
D⫽
储 PQ ⫻ u储 冪174 ⫽ ⫽ 冪6 ⬇ 2.45. 储u 储 冪29
See Figure 11.55.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
790
Chapter 11
Vectors and the Geometry of Space
11.5 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Checking Points on a Line In Exercises 1 and 2, determine whether each point lies on the line. 1. x ⫽ ⫺2 ⫹ t, y ⫽ 3t, z ⫽ 4 ⫹ t (a) 共0, 6, 6兲 2.
(b) 共2, 3, 5兲
x⫺3 y⫺7 ⫽ ⫽z⫹2 2 8 (a) 共7, 23, 0兲
(b) 共1, ⫺1, ⫺3兲
Finding Parametric and Symmetric Equations In Exercises 3–8, find sets of (a) parametric equations and (b) symmetric equations of the line through the point parallel to the given vector or line (if possible). (For each line, write the direction numbers as integers.) Point
Using Parametric and Symmetric Equations In Exercises 21–24, find the coordinates of a point P on the line and a vector v parallel to the line. 21. x ⫽ 3 ⫺ t, y ⫽ ⫺1 ⫹ 2t, z ⫽ ⫺2 22. x ⫽ 4t, y ⫽ 5 ⫺ t, z ⫽ 4 ⫹ 3t 23.
v ⫽ 具3, 1, 5典
4. 共0, 0, 0兲
v ⫽ 具 ⫺2, 52, 1典
5. 共⫺2, 0, 3兲
v ⫽ 2i ⫹ 4j ⫺ 2k
6. 共⫺3, 0, 2兲
v ⫽ 6j ⫹ 3k
7. 共1, 0, 1兲
x ⫽ 3 ⫹ 3t, y ⫽ 5 ⫺ 2t, z ⫽ ⫺7 ⫹ t
8. 共⫺3, 5, 4兲
x⫺1 y⫹1 ⫽ ⫽z⫺3 3 ⫺2
24.
25. L1: x ⫽ 6 ⫺ 3t, y ⫽ ⫺2 ⫹ 2t, z ⫽ 5 ⫹ 4t L2: x ⫽ 6t, y ⫽ 2 ⫺ 4t, z ⫽ 13 ⫺ 8t L3: x ⫽ 10 ⫺ 6t, y ⫽ 3 ⫹ 4t, z ⫽ 7 ⫹ 8t L4: x ⫽ ⫺4 ⫹ 6t, y ⫽ 3 ⫹ 4t, z ⫽ 5 ⫺ 6t 26. L1: x ⫽ 3 ⫹ 2t, y ⫽ ⫺6t, z ⫽ 1 ⫺ 2t L2: x ⫽ 1 ⫹ 2t, y ⫽ ⫺1 ⫺ t, z ⫽ 3t L3: x ⫽ ⫺1 ⫹ 2t, y ⫽ 3 ⫺ 10t, z ⫽ 1 ⫺ 4t L4: x ⫽ 5 ⫹ 2t, y ⫽ 1 ⫺ t, z ⫽ 8 ⫹ 3t 27. L1:
x⫺8 y⫹5 z⫹9 ⫽ ⫽ 4 ⫺2 3
L2:
x⫹7 y⫺4 z⫹6 ⫽ ⫽ 2 1 5
L3:
x ⫹ 4 y ⫺ 1 z ⫹ 18 ⫽ ⫽ ⫺8 4 ⫺6
L4:
x⫺2 y⫹3 z⫺4 ⫽ ⫽ ⫺2 1 1.5
28. L1:
x⫺3 y⫺2 z⫹2 ⫽ ⫽ 2 1 2
Finding Parametric Equations In Exercises 13–20, find a set of parametric equations of the line.
L2:
x⫺1 y⫺1 z⫹3 ⫽ ⫽ 4 2 4
13. The line passes through the point 共2, 3, 4兲 and is parallel to the xz-plane and the yz-plane.
L3:
x⫹2 y⫺1 z⫺3 ⫽ ⫽ 1 0.5 1
14. The line passes through the point 共⫺4, 5, 2兲 and is parallel to the xy-plane and the yz-plane.
L4:
x⫺3 y⫹1 z⫺2 ⫽ ⫽ 2 4 ⫺1
Finding Parametric and Symmetric Equations In Exercises 9–12, find sets of (a) parametric equations and (b) symmetric equations of the line through the two points (if possible). (For each line, write the direction numbers as integers.) 9. 共5, ⫺3, ⫺2兲, 共⫺ 23, 23, 1兲 11. 共7, ⫺2, 6兲, 共⫺3, 0, 6兲
10. 共0, 4, 3兲, 共⫺1, 2, 5兲 12. 共0, 0, 25兲, 共10, 10, 0兲
15. The line passes through the point 共2, 3, 4兲 and is perpendicular to the plane given by 3x ⫹ 2y ⫺ z ⫽ 6. 16. The line passes through the point 共⫺4, 5, 2兲 and is perpendicular to the plane given by ⫺x ⫹ 2y ⫹ z ⫽ 5. 17. The line passes through the point 共5, ⫺3, ⫺4兲 and is parallel to v ⫽ 具2, ⫺1, 3典. 18. The line passes through the point 共⫺1, 4, ⫺3兲 and is parallel to v ⫽ 5i ⫺ j.
x⫹3 y z⫺3 ⫽ ⫽ 5 8 6
Determining Parallel Lines In Exercises 25 – 28, determine whether any of the lines are parallel or identical.
Parallel to
3. 共0, 0, 0兲
x⫺7 y⫹6 ⫽ ⫽z⫹2 4 2
Finding a Point of Intersection In Exercises 29–32, determine whether the lines intersect, and if so, find the point of intersection and the cosine of the angle of intersection. 29. x ⫽ 4t ⫹ 2, y ⫽ 3,
z ⫽ ⫺t ⫹ 1
x ⫽ 2s ⫹ 2, y ⫽ 2s ⫹ 3, z ⫽ s ⫹ 1 30. x ⫽ ⫺3t ⫹ 1, y ⫽ 4t ⫹ 1, z ⫽ 2t ⫹ 4 x ⫽ 3s ⫹ 1, y ⫽ 2s ⫹ 4, z ⫽ ⫺s ⫹ 1
19. The line passes through the point 共2, 1, 2兲 and is parallel to the line x ⫽ ⫺t, y ⫽ 1 ⫹ t, z ⫽ ⫺2 ⫹ t.
31.
20. The line passes through the point 共⫺6, 0, 8兲 and is parallel to the line x ⫽ 5 ⫺ 2t, y ⫽ ⫺4 ⫹ 2t, z ⫽ 0.
x y⫺2 ⫽ ⫽ z ⫹ 1, 3 ⫺1
32.
x⫺2 y⫺2 ⫽ ⫽ z ⫺ 3, ⫺3 6
x⫺1 z⫹3 ⫽y⫹2⫽ 4 ⫺3 x⫺3 z⫹2 ⫽y⫹5⫽ 2 4
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
11.5
Checking Points on a Plane In Exercises 33 and 34, determine whether the plane passes through each point. 33. x ⫹ 2y ⫺ 4z ⫺ 1 ⫽ 0 (a) 共⫺7, 2, ⫺1兲
Finding an Equation of a Plane In Exercises 53–56, find an equation of the plane that contains all the points that are equidistant from the given points. 53. 共2, 2, 0兲, 共0, 2, 2兲
54. 共1, 0, 2兲,
55. 共⫺3, 1, 2兲,
56. 共⫺5, 1, ⫺3兲,
(b) 共⫺1, 5, ⫺1兲
Comparing Planes In Exercises 57–62, determine whether the planes are parallel, orthogonal, or neither. If they are neither parallel nor orthogonal, find the angle of intersection.
Finding an Equation of a Plane In Exercises 35–40, find an equation of the plane passing through the point perpendicular to the given vector or line. Point
791
(b) 共5, 2, 2兲
34. 2x ⫹ y ⫹ 3z ⫺ 6 ⫽ 0 (a) 共3, 6, ⫺2兲
Lines and Planes in Space
Perpendicular to
共6, ⫺2, 4兲
57. 5x ⫺ 3y ⫹ z ⫽ 4
共2, 0, 1兲 共2, ⫺1, 6兲
58. 3x ⫹ y ⫺ 4z ⫽ 3
x ⫹ 4y ⫹ 7z ⫽ 1
⫺9x ⫺ 3y ⫹ 12z ⫽ 4
59. x ⫺ 3y ⫹ 6z ⫽ 4
60. 3x ⫹ 2y ⫺ z ⫽ 7
35. 共1, 3, ⫺7兲
n⫽j
36. 共0, ⫺1, 4兲
n⫽k
37. 共3, 2, 2兲
n ⫽ 2i ⫹ 3j ⫺ k
38. 共0, 0, 0兲
n ⫽ ⫺3i ⫹ 2k
39. 共⫺1, 4, 0兲
x ⫽ ⫺1 ⫹ 2t, y ⫽ 5 ⫺ t, z ⫽ 3 ⫺ 2t
Sketching a Graph of a Plane In Exercises 63–70, sketch
40. 共3, 2, 2兲
x⫺1 z⫹3 ⫽y⫹2⫽ 4 ⫺3
a graph of the plane and label any intercepts.
5x ⫹ y ⫺ z ⫽ 4
x ⫺ 4y ⫹ 2z ⫽ 0
61. x ⫺ 5y ⫺ z ⫽ 1
62. 2x ⫺ z ⫽ 1
5x ⫺ 25y ⫺ 5z ⫽ ⫺3
4x ⫹ y ⫹ 8z ⫽ 10
63. 4x ⫹ 2y ⫹ 6z ⫽ 12
64. 3x ⫹ 6y ⫹ 2z ⫽ 6
Finding an Equation of a Plane In Exercises 41–52, find
65. 2x ⫺ y ⫹ 3z ⫽ 4
66. 2x ⫺ y ⫹ z ⫽ 4
an equation of the plane.
67. x ⫹ z ⫽ 6
68. 2x ⫹ y ⫽ 8
41. The plane passes through 共0, 0, 0兲, 共2, 0, 3兲, and 共⫺3, ⫺1, 5兲.
69. x ⫽ 5
70. z ⫽ 8
42. The plane passes through 共3, ⫺1, 2兲, 共2, 1, 5兲, and 共1, ⫺2, ⫺2兲. 43. The plane passes through 共1, 2, 3兲, 共3, 2, 1兲, and 共⫺1, ⫺2, 2兲. 44. The plane passes through the point 共1, 2, 3兲 and is parallel to the yz-plane. 45. The plane passes through the point 共1, 2, 3兲 and is parallel to the xy-plane. 46. The plane contains the y-axis and makes an angle of 兾6 with the positive x-axis. 47. The plane contains the lines given by x⫺1 ⫽y⫺4⫽z ⫺2 and x⫺2 y⫺1 z⫺2 ⫽ ⫽ . ⫺3 4 ⫺1 48. The plane passes through the point 共2, 2, 1兲 and contains the line given by x y⫺4 ⫽ ⫽ z. 2 ⫺1 49. The plane passes through the points 共2, 2, 1兲 and 共⫺1, 1, ⫺1兲 and is perpendicular to the plane 2x ⫺ 3y ⫹ z ⫽ 3. 50. The plane passes through the points 共3, 2, 1兲 and 共3, 1, ⫺5兲 and is perpendicular to the plane 6x ⫹ 7y ⫹ 2z ⫽ 10. 51. The plane passes through the points 共1, ⫺2, ⫺1兲 and 共2, 5, 6兲 and is parallel to the x-axis. 52. The plane passes through the points 共4, 2, 1兲 and 共⫺3, 5, 7兲 and is parallel to the z-axis.
Parallel Planes In Exercises 71–74, determine whether any of the planes are parallel or identical. 72. P1: 2x ⫺ y ⫹ 3z ⫽ 8
71. P1: ⫺5x ⫹ 2y ⫺ 8z ⫽ 6 P2: 15x ⫺ 6y ⫹ 24z ⫽ 17
P2: 3x ⫺ 5y ⫺ 2z ⫽ 6
P3: 6x ⫺ 4y ⫹ 4z ⫽ 9
P3: 8x ⫺ 4y ⫹ 12z ⫽ 5
P4: 3x ⫺ 2y ⫺ 2z ⫽ 4
P4: ⫺4x ⫺ 2y ⫹ 6z ⫽ 11
73. P1: 3x ⫺ 2y ⫹ 5z ⫽ 10 P2: ⫺6x ⫹ 4y ⫺ 10z ⫽ 5 P3: ⫺3x ⫹ 2y ⫹ 5z ⫽ 8 P4: 75x ⫺ 50y ⫹ 125z ⫽ 250 74. P1: ⫺60x ⫹ 90y ⫹ 30z ⫽ 27 P2: 6x ⫺ 9y ⫺ 3z ⫽ 2 P3: ⫺20x ⫹ 30y ⫹ 10z ⫽ 9 P4: 12x ⫺ 18y ⫹ 6z ⫽ 5
Intersection of Planes In Exercises 75 and 76, (a) find the angle between the two planes, and (b) find a set of parametric equations for the line of intersection of the planes. 75. 3x ⫹ 2y ⫺ z ⫽ 7
76. 6x ⫺ 3y ⫹ z ⫽ 5
x ⫺ 4y ⫹ 2z ⫽ 0
⫺x ⫹ y ⫹ 5z ⫽ 5
Intersection of a Plane and a Line In Exercises 77–80, find the point(s) of intersection (if any) of the plane and the line. Also, determine whether the line lies in the plane. 77. 2x ⫺ 2y ⫹ z ⫽ 12, 78. 2x ⫹ 3y ⫽ ⫺5,
x⫺
1 y ⫹ 共3兾2兲 z ⫹ 1 ⫽ ⫽ 2 ⫺1 2
x⫺1 y z⫺3 ⫽ ⫽ 4 2 6
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
792
Chapter 11
Vectors and the Geometry of Space
79. 2x ⫹ 3y ⫽ 10,
x⫺1 y⫹1 ⫽ ⫽z⫺3 3 ⫺2
80. 5x ⫹ 3y ⫽ 17,
x⫺4 y⫹1 z⫹2 ⫽ ⫽ 2 ⫺3 5
Finding the Distance Between a Point and a Plane In Exercises 81–84, find the distance between the point and the plane. 81. 共0, 0, 0兲
82. 共0, 0, 0兲
2x ⫹ 3y ⫹ z ⫽ 12
WRITING ABOUT CONCEPTS
99. Normal Vector Let L1 and L2 be nonparallel lines that do not intersect. Is it possible to find a nonzero vector v such that v is normal to both L1 and L2? Explain your reasoning.
HOW DO YOU SEE IT? Match the general equation with its graph. Then state what axis or plane the equation is parallel to.
100.
5x ⫹ y ⫺ z ⫽ 9
83. 共2, 8, 4兲
84. 共1, 3, ⫺1兲
2x ⫹ y ⫹ z ⫽ 5
3x ⫺ 4y ⫹ 5z ⫽ 6
Finding the Distance Between Two Parallel Planes In Exercises 85–88, verify that the two planes are parallel, and find the distance between the planes. 85. x ⫺ 3y ⫹ 4z ⫽ 10
( continued )
(a) ax ⫹ by ⫹ d ⫽ 0
(b) ax ⫹ d ⫽ 0
(c) cz ⫹ d ⫽ 0
(d) ax ⫹ cz ⫹ d ⫽ 0
z
(i)
z
(ii)
86. 4x ⫺ 4y ⫹ 9z ⫽ 7
x ⫺ 3y ⫹ 4z ⫽ 6
4x ⫺ 4y ⫹ 9z ⫽ 18
87. ⫺3x ⫹ 6y ⫹ 7z ⫽ 1
88. 2x ⫺ 4z ⫽ 4
6x ⫺ 12y ⫺ 14z ⫽ 25
2x ⫺ 4z ⫽ 10
Finding the Distance Between a Point and a Line In
y
y
x
x
z
(iii)
z
(iv)
Exercises 89–92, find the distance between the point and the line given by the set of parametric equations. 89. 共1, 5, ⫺2兲;
x ⫽ 4t ⫺ 2,
90. 共1, ⫺2, 4兲;
x ⫽ 2t,
91. 共⫺2, 1, 3兲;
x ⫽ 1 ⫺ t, y ⫽ 2 ⫹ t, z ⫽ ⫺2t
92. 共4, ⫺1, 5兲;
x ⫽ 3, y ⫽ 1 ⫹ 3t, z ⫽ 1 ⫹ t
y ⫽ 3,
y ⫽ t ⫺ 3,
z ⫽ ⫺t ⫹ 1 z ⫽ 2t ⫹ 2
Finding the Distance Between Two Parallel Lines In Exercises 93 and 94, verify that the lines are parallel, and find the distance between them. 93. L1: x ⫽ 2 ⫺ t, y ⫽ 3 ⫹ 2t, z ⫽ 4 ⫹ t L2: x ⫽ 3t, y ⫽ 1 ⫺ 6t, z ⫽ 4 ⫺ 3t 94. L1: x ⫽ 3 ⫹ 6t, y ⫽ ⫺2 ⫹ 9t, z ⫽ 1 ⫺ 12t L2: x ⫽ ⫺1 ⫹ 4t, y ⫽ 3 ⫹ 6t,
z ⫽ ⫺8t
WRITING ABOUT CONCEPTS 95. Parametric and Symmetric Equations Give the parametric equations and the symmetric equations of a line in space. Describe what is required to find these equations. 96. Standard Equation of a Plane in Space Give the standard equation of a plane in space. Describe what is required to find this equation. 97. Intersection of Two Planes Describe a method of finding the line of intersection of two planes. 98. Parallel and Perpendicular Planes Describe a method for determining when two planes, a1x ⫹ b1y ⫹ c1z ⫹ d1 ⫽ 0 and a2 x ⫹ b2 y ⫹ c2z ⫹ d2 ⫽ 0, are (a) parallel and (b) perpendicular. Explain your reasoning.
y x
y x
101. Modeling Data Personal consumption expenditures (in billions of dollars) for several types of recreation from 2005 through 2010 are shown in the table, where x is the expenditures on amusement parks and campgrounds, y is the expenditures on live entertainment (excluding sports), and z is the expenditures on spectator sports. (Source: U.S. Bureau of Economic Analysis) Year
2005
2006
2007
2008
2009
2010
x
36.4
39.0
42.4
44.7
43.0
45.2
y
15.3
16.6
17.4
17.5
17.0
17.3
z
16.4
18.1
20.0
20.5
20.1
21.4
A model for the data is given by 0.46x ⫹ 0.30y ⫺ z ⫽ 4.94. (a) Complete a fourth row in the table using the model to approximate z for the given values of x and y. Compare the approximations with the actual values of z. (b) According to this model, increases in expenditures on recreation types x and y would correspond to what kind of change in expenditures on recreation type z?
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11.5 102. Mechanical Design The figure shows a chute at the top of a grain elevator of a combine that funnels the grain into a bin. Find the angle between two adjacent sides.
793
105. Finding a Point of Intersection Find the point of intersection of the plane 3x ⫺ y ⫹ 4z ⫽ 7 and the line through 共5, 4, ⫺3兲 that is perpendicular to this plane. 106. Finding the Distance Between a Plane and a Line Show that the plane 2x ⫺ y ⫺ 3z ⫽ 4 is parallel to the line x ⫽ ⫺2 ⫹ 2t, y ⫽ ⫺1 ⫹ 4t, z ⫽ 4, and find the distance between them.
8 in.
8 in.
Lines and Planes in Space
8 in.
6 in. 6 in.
103. Distance Two insects are crawling along different lines in three-space. At time t (in minutes), the first insect is at the point 共x, y, z兲 on the line x ⫽ 6 ⫹ t, y ⫽ 8 ⫺ t, z ⫽ 3 ⫹ t. Also, at time t, the second insect is at the point 共x, y, z兲 on the line x ⫽ 1 ⫹ t, y ⫽ 2 ⫹ t, z ⫽ 2t. Assume that distances are given in inches. (a) Find the distance between the two insects at time t ⫽ 0. (b) Use a graphing utility to graph the distance between the insects from t ⫽ 0 to t ⫽ 10. (c) Using the graph from part (b), what can you conclude about the distance between the insects? (d) How close to each other do the insects get? 104. Finding an Equation of a Sphere Find the standard equation of the sphere with center 共⫺3, 2, 4兲 that is tangent to the plane given by 2x ⫹ 4y ⫺ 3z ⫽ 8.
107. Finding a Point of Intersection Find the point of intersection of the line through 共1, ⫺3, 1兲 and 共3, ⫺4, 2兲 and the plane given by x ⫺ y ⫹ z ⫽ 2. 108. Finding Parametric Equations Find a set of parametric equations for the line passing through the point 共1, 0, 2兲 that is parallel to the plane given by x ⫹ y ⫹ z ⫽ 5 and perpendicular to the line x ⫽ t, y ⫽ 1 ⫹ t, z ⫽ 1 ⫹ t.
True or False? In Exercises 109–114, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 109. If v ⫽ a1i ⫹ b1j ⫹ c1k is any vector in the plane given by a2x ⫹ b2y ⫹ c2z ⫹ d2 ⫽ 0, then a1a2 ⫹ b1b2 ⫹ c1c2 ⫽ 0. 110. Every two lines in space are either intersecting or parallel. 111. Two planes in space are either intersecting or parallel. 112. If two lines L1 and L2 are parallel to a plane P, then L1 and L2 are parallel. 113. Two planes perpendicular to a third plane in space are parallel. 114. A plane and a line in space are either intersecting or parallel.
Distances in Space You have learned two distance formulas in this section—one for the distance between a point and a plane, and one for the distance between a point and a line. In this project, you will study a third distance problem—the distance between two skew lines. Two lines in space are skew if they are neither parallel nor intersecting (see figure). (a) Consider the following two lines in space. L1: x ⫽ 4 ⫹ 5t, y ⫽ 5 ⫹ 5t, z ⫽ 1 ⫺ 4t L2: x ⫽ 4 ⫹ s, y ⫽ ⫺6 ⫹ 8s, z ⫽ 7 ⫺ 3s (i) Show that these lines are not parallel. (ii) Show that these lines do not intersect, and therefore are skew lines.
(c) Use the procedure in part (a) to find the distance between the lines. L1: x ⫽ 3t, y ⫽ 2 ⫺ t, z ⫽ ⫺1 ⫹ t L2: x ⫽ 1 ⫹ 4s, y ⫽ ⫺2 ⫹ s, z ⫽ ⫺3 ⫺ 3s (d) Develop a formula for finding the distance between the skew lines. L1: x ⫽ x1 ⫹ a1t, y ⫽ y1 ⫹ b1t, z ⫽ z1 ⫹ c1t L2: x ⫽ x2 ⫹ a2s, y ⫽ y2 ⫹ b2s, z ⫽ z 2 ⫹ c2s L1
(iii) Show that the two lines lie in parallel planes. (iv) Find the distance between the parallel planes from part (iii). This is the distance between the original skew lines. (b) Use the procedure in part (a) to find the distance between the lines.
L2
L1: x ⫽ 2t, y ⫽ 4t, z ⫽ 6t L2: x ⫽ 1 ⫺ s, y ⫽ 4 ⫹ s, z ⫽ ⫺1 ⫹ s
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794
Chapter 11
Vectors and the Geometry of Space
11.6 Surfaces in Space Recognize and write equations of cylindrical surfaces. Recognize and write equations of quadric surfaces. Recognize and write equations of surfaces of revolution.
Cylindrical Surfaces The first five sections of this chapter contained the vector portion of the preliminary work necessary to study vector calculus and the calculus of space. In this and the next section, you will study surfaces in space and alternative coordinate systems for space. You have already studied two special types of surfaces. 1. Spheres: 共x ⫺ x0兲2 ⫹ 共 y ⫺ y0兲2 ⫹ 共z ⫺ z0兲2 ⫽ r 2 2. Planes: ax ⫹ by ⫹ cz ⫹ d ⫽ 0 z
Section 11.2 Section 11.5
A third type of surface in space is a cylindrical surface, or simply a cylinder. To define a cylinder, consider the familiar right circular cylinder shown in Figure 11.56. The cylinder was generated by a vertical line moving around the circle x2 ⫹ y 2 ⫽ a2 in the xy-plane. This circle is a generating curve for the cylinder, as indicated in the next definition. y
x
Right circular cylinder: x 2 + y 2 = a2
Rulings are parallel to z-axis Figure 11.56
Definition of a Cylinder Let C be a curve in a plane and let L be a line not in a parallel plane. The set of all lines parallel to L and intersecting C is a cylinder. The curve C is the generating curve (or directrix) of the cylinder, and the parallel lines are rulings.
Without loss of generality, you can assume that C lies in one of the three coordinate planes. Moreover, this text restricts the discussion to right cylinders–– cylinders whose rulings are perpendicular to the coordinate plane containing C, as shown in Figure 11.57. Note that the rulings intersect C and are parallel to the line L. For the right circular cylinder shown in Figure 11.56, the equation of the generating curve in the xy-plane is x2 ⫹ y2 ⫽ a2.
Rulings intersect C and are parallel to L.
x
z
Generating curve C
L intersects C. y
Right cylinder: A cylinder whose rulings are perpendicular to the coordinate plane containing C Figure 11.57
To find an equation of the cylinder, note that you can generate any one of the rulings by fixing the values of x and y and then allowing z to take on all real values. In this sense, the value of z is arbitrary and is, therefore, not included in the equation. In other words, the equation of this cylinder is simply the equation of its generating curve. x2 ⫹ y 2 ⫽ a2
Equation of cylinder in space
Equations of Cylinders The equation of a cylinder whose rulings are parallel to one of the coordinate axes contains only the variables corresponding to the other two axes.
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11.6
Surfaces in Space
795
Sketching a Cylinder Sketch the surface represented by each equation. a. z ⫽ y 2
b. z ⫽ sin x, 0 ⱕ x ⱕ 2
Solution a. The graph is a cylinder whose generating curve, z ⫽ y 2, is a parabola in the yz-plane. The rulings of the cylinder are parallel to the x-axis, as shown in Figure 11.58(a). b. The graph is a cylinder generated by the sine curve in the xz-plane. The rulings are parallel to the y-axis, as shown in Figure 11.58(b). Generating curve C lies in xz-plane
Generating curve C z lies in yz-plane
z 1 y
π
y
x
x
Cylinder: z = y 2 (a) Rulings are parallel to x-axis.
Cylinder: z = sin x (b) Rulings are parallel to y-axis.
Figure 11.58
Quadric Surfaces The fourth basic type of surface in space is a quadric surface. Quadric surfaces are the three-dimensional analogs of conic sections. Quadric Surface The equation of a quadric surface in space is a second-degree equation in three variables. The general form of the equation is Ax2 ⫹ By2 ⫹ Cz2 ⫹ Dxy ⫹ Exz ⫹ Fyz ⫹ Gx ⫹ Hy ⫹ Iz ⫹ J ⫽ 0. There are six basic types of quadric surfaces: ellipsoid, hyperboloid of one sheet, hyperboloid of two sheets, elliptic cone, elliptic paraboloid, and hyperbolic paraboloid.
The intersection of a surface with a plane is called the trace of the surface in the plane. To visualize a surface in space, it is helpful to determine its traces in some well-chosen planes. The traces of quadric surfaces are conics. These traces, together with the standard form of the equation of each quadric surface, are shown in the table on the next two pages. In the table on the next two pages, only one of several orientations of each quadric surface is shown. When the surface is oriented along a different axis, its standard equation will change accordingly, as illustrated in Examples 2 and 3. The fact that the two types of paraboloids have one variable raised to the first power can be helpful in classifying quadric surfaces. The other four types of basic quadric surfaces have equations that are of second degree in all three variables.
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796
Chapter 11
Vectors and the Geometry of Space
Ellipsoid
z
x2 2
a
y x
⫹
y2 2
b
Trace Ellipse Ellipse Ellipse
⫹
z2 2
c
z
yz-trace
⫽1
Plane Parallel to xy-plane Parallel to xz-plane Parallel to yz-plane
xz-trace
y x
xy-trace
The surface is a sphere when a ⫽ b ⫽ c ⫽ 0. Hyperboloid of One Sheet
z
x2 a2 Trace Ellipse Hyperbola Hyperbola y
⫹
y2 b2
⫺
z2 c2
z
⫽1
Plane Parallel to xy-plane Parallel to xz-plane Parallel to yz-plane xy-trace
The axis of the hyperboloid corresponds to the variable whose coefficient is negative.
x
y
x
yz-trace
xz-trace
Hyperboloid of Two Sheets
z
yz-trace
z
xz-trace
z2 x2 y2 ⫺ 2⫺ 2⫽1 2 c a b
x
Trace Ellipse Hyperbola Hyperbola
Plane Parallel to xy-plane Parallel to xz-plane Parallel to yz-plane
y
parallel to xy-plane x
The axis of the hyperboloid corresponds to the variable whose coefficient is positive. There is no trace in the coordinate plane perpendicular to this axis.
no xy-trace y
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11.6
Elliptic Cone
z
x2
2
797
Surfaces in Space
z
xz-trace
2
y z ⫹ ⫺ ⫽0 a2 b2 c2 Trace Ellipse Hyperbola Hyperbola y x
Plane Parallel to xy-plane Parallel to xz-plane Parallel to yz-plane
The axis of the cone corresponds to the variable whose coefficient is negative. The traces in the coordinate planes parallel to this axis are intersecting lines.
xy-trace (one point) y x
parallel to xy-plane
yz-trace
Elliptic Paraboloid
z
z
Trace Ellipse Parabola Parabola
Plane Parallel to xy-plane Parallel to xz-plane Parallel to yz-plane
The axis of the paraboloid corresponds to the variable raised to the first power.
xy-trace (one point)
x
Hyperbolic Paraboloid
z
z⫽
y x
parallel to xy-plane
y
x
xz-trace
yz-trace
x2 y2 z⫽ 2⫹ 2 a b
Trace Hyperbola Parabola Parabola
y2 b2
⫺
y
z
yz-trace
x2 a2
Plane Parallel to xy-plane Parallel to xz-plane Parallel to yz-plane
y x
The axis of the paraboloid corresponds to the variable raised to the first power.
parallel to xy-plane xz-trace
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798
Chapter 11
Vectors and the Geometry of Space
To classify a quadric surface, begin by writing the equation of the surface in standard form. Then, determine several traces taken in the coordinate planes or taken in planes that are parallel to the coordinate planes.
Sketching a Quadric Surface Classify and sketch the surface 4x 2 ⫺ 3y 2 ⫹ 12z2 ⫹ 12 ⫽ 0. Solution
y2 z2 − =1 4 1
4x 2 ⫺ 3y 2 ⫹ 12z 2 ⫹ 12 ⫽ 0 x2 y2 ⫹ ⫺ z2 ⫺ 1 ⫽ 0 ⫺3 4 y 2 x 2 z2 ⫺ ⫺ ⫽1 4 3 1
z
y2 3
4
−
x2 3
=1
2 1
4 x
2
3
Begin by writing the equation in standard form.
1
2
y
Hyperboloid of two sheets: y2 x2 − − z2 = 1 4 3
Write original equation. Divide by ⫺12. Standard form
From the table on pages 796 and 797, you can conclude that the surface is a hyperboloid of two sheets with the y-axis as its axis. To sketch the graph of this surface, it helps to find the traces in the coordinate planes. y2 x2 ⫺ ⫽1 4 3 2 x z2 xz-trace 共 y ⫽ 0兲: ⫹ ⫽ ⫺1 3 1 2 y z2 yz-trace 共x ⫽ 0兲: ⫺ ⫽1 4 1 xy-trace 共z ⫽ 0兲:
Hyperbola
No trace
Hyperbola
The graph is shown in Figure 11.59.
Figure 11.59
Sketching a Quadric Surface Classify and sketch the surface
Elliptic paraboloid: x = y 2 + 4z 2
x ⫺ y 2 ⫺ 4z 2 ⫽ 0.
z
Solution Because x is raised only to the first power, the surface is a paraboloid. The axis of the paraboloid is the x-axis. In standard form, the equation is
2
x = y2
−4 2
4
y
x ⫽ y2 ⫹ 4z2.
Standard form
Some convenient traces are listed below. y2 z2 + =1 4 1
10 x
x = 4z 2
Figure 11.60
xy-trace 共 z ⫽ 0兲: xz-trace 共 y ⫽ 0兲:
x ⫽ y2 x ⫽ 4z2
parallel to yz-plane 共x ⫽ 4兲:
2
Parabola Parabola
2
y z ⫹ ⫽1 4 1
Ellipse
The surface is an elliptic paraboloid, as shown in Figure 11.60. Some second-degree equations in x, y, and z do not represent any of the basic types of quadric surfaces. For example, the graph of x2 ⫹ y2 ⫹ z2 ⫽ 0
Single point
is a single point, and the graph of x2 ⫹ y2 ⫽ 1
Right circular cylinder
is a right circular cylinder.
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11.6
799
Surfaces in Space
For a quadric surface not centered at the origin, you can form the standard equation by completing the square, as demonstrated in Example 4.
A Quadric Surface Not Centered at the Origin See LarsonCalculus.com for an interactive version of this type of example.
(x − 2)2 (y + 1)2 (z − 1)2 + + =1 2 4 4
Classify and sketch the surface
z
x 2 ⫹ 2y 2 ⫹ z2 ⫺ 4x ⫹ 4y ⫺ 2z ⫹ 3 ⫽ 0. 3
Solution x2
Begin by grouping terms and factoring where possible.
⫺ 4x ⫹ 2共y2 ⫹ 2y兲 ⫹ z2 ⫺ 2z ⫽ ⫺3
Next, complete the square for each variable and write the equation in standard form.
(2, − 1, 1)
1 −1 5 x
An ellipsoid centered at 共2, ⫺1, 1兲 Figure 11.61
y
共x2 ⫺ 4x ⫹ 兲 ⫹ 2共 y 2 ⫹ 2y ⫹ 兲 ⫹ 共z2 ⫺ 2z ⫹ 兲 ⫽ ⫺3 共x2 ⫺ 4x ⫹ 4兲 ⫹ 2共 y 2 ⫹ 2y ⫹ 1兲 ⫹ 共z2 ⫺ 2z ⫹ 1兲 ⫽ ⫺3 ⫹ 4 ⫹ 2 ⫹ 1 共x ⫺ 2兲2 ⫹ 2共 y ⫹ 1兲2 ⫹ 共z ⫺ 1兲2 ⫽ 4 共x ⫺ 2兲2 共 y ⫹ 1兲2 共z ⫺ 1兲2 ⫹ ⫹ ⫽1 4 2 4 From this equation, you can see that the quadric surface is an ellipsoid that is centered at 共2, ⫺1, 1兲. Its graph is shown in Figure 11.61.
TECHNOLOGY A 3-D graphing utility can help you visualize a surface in space.* Such a graphing utility may create a three-dimensional graph by sketching several traces of the surface and then applying a “hidden-line” routine that blocks out portions of the surface that lie behind other portions of the surface. Two examples of figures that were generated by Mathematica are shown below. z
z
y y x
x Generated by Mathematica
Elliptic paraboloid y 2 z2 x⫽ ⫹ 2 2
Generated by Mathematica
Hyperbolic paraboloid y2 x2 z⫽ ⫺ 16 16
Using a graphing utility to graph a surface in space requires practice. For one thing, you must know enough about the surface to be able to specify a viewing window that gives a representative view of the surface. Also, you can often improve the view of a surface by rotating the axes. For instance, note that the elliptic paraboloid in the figure is seen from a line of sight that is “higher” than the line of sight used to view the hyperbolic paraboloid. * Some 3-D graphing utilities require surfaces to be entered with parametric equations. For a discussion of this technique, see Section 15.5.
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800
Chapter 11
Vectors and the Geometry of Space
Surfaces of Revolution Circular cross section
z
The fifth special type of surface you will study is a surface of revolution. In Section 7.4, you studied a method for finding the area of such a surface. You will now look at a procedure for finding its equation. Consider the graph of the radius function
Generating curve y = r (z) (0, 0, z) (0, r (z), z)
(x, y, z)
y ⫽ r 共z兲
Generating curve
in the yz-plane. When this graph is revolved about the z-axis, it forms a surface of revolution, as shown in Figure 11.62. The trace of the surface in the plane z ⫽ z 0 is a circle whose radius is r 共z0兲 and whose equation is
r (z)
y
x 2 ⫹ y 2 ⫽ 关r 共z 0兲兴 2.
Circular trace in plane: z ⫽ z 0
Replacing z 0 with z produces an equation that is valid for all values of z. In a similar manner, you can obtain equations for surfaces of revolution for the other two axes, and the results are summarized as follows.
x
Figure 11.62
Surface of Revolution If the graph of a radius function r is revolved about one of the coordinate axes, then the equation of the resulting surface of revolution has one of the forms listed below. 1. Revolved about the x-axis: y 2 ⫹ z2 ⫽ 关r共x兲兴 2 2. Revolved about the y-axis: x 2 ⫹ z2 ⫽ 关r共 y兲兴 2 3. Revolved about the z-axis: x 2 ⫹ y 2 ⫽ 关r 共z兲兴 2
Finding an Equation for a Surface of Revolution Find an equation for the surface of revolution formed by revolving (a) the graph of y ⫽ 1兾z about the z-axis and (b) the graph of 9x2 ⫽ y3 about the y-axis. Solution a. An equation for the surface of revolution formed by revolving the graph of y⫽
1 z
Radius function
about the z-axis is x2 ⫹ y2 ⫽ 关r 共z兲兴 2 1 2 x2 ⫹ y2 ⫽ . z
Revolved about the z-axis
冢冣
z
Surface: x 2 + z 2 = 19 y 3
Substitute 1兾z for r 共z兲.
b. To find an equation for the surface formed by revolving the graph of 9x2 ⫽ y3 about the y-axis, solve for x in terms of y to obtain 1 x ⫽ y 3兾2 ⫽ r 共 y兲. 3 y
x
So, the equation for this surface is x2 ⫹ z2 ⫽ 关r 共 y兲兴 2 1 3兾2 x2 ⫹ z2 ⫽ y 3
冢
Generating curve 9x 2 = y 3
Figure 11.63
Radius function
1 x2 ⫹ z2 ⫽ y 3. 9
Revolved about the y-axis
冣
2
Substitute 3 y 3兾2 for r 共y兲. 1
Equation of surface
The graph is shown in Figure 11.63.
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11.6
Surfaces in Space
801
The generating curve for a surface of revolution is not unique. For instance, the surface x2 ⫹ z2 ⫽ e⫺2y can be formed by revolving either the graph of x ⫽ e⫺y about the y-axis or the graph of z ⫽ e⫺y about the y-axis, as shown in Figure 11.64. z z
Surface: x 2 + z 2 = e − 2y
Generating curve in yz-plane z = e−y
y y x x
Generating curve in xy-plane x = e−y
Figure 11.64
Finding a Generating Curve Find a generating curve and the axis of revolution for the surface x2 ⫹ 3y2 ⫹ z2 ⫽ 9. Solution
The equation has one of the forms listed below.
x ⫹ y ⫽ 关r 共z兲兴 2 y2 ⫹ z2 ⫽ 关r 共x兲兴 2 x2 ⫹ z2 ⫽ 关r 共 y兲兴 2 2
2
Revolved about z-axis Revolved about x-axis Revolved about y-axis
Surface: x 2 + 3y 2 + z 2 = 9
2
Because the coefficients of x and z2 are equal, you should choose the third form and write
z
Generating curve in yz-plane z = 9 − 3y 2
x2 ⫹ z2 ⫽ 9 ⫺ 3y 2. The y-axis is the axis of revolution. You can choose a generating curve from either of the traces x2 ⫽ 9 ⫺ 3y2
Trace in xy-plane y
or z2 ⫽ 9 ⫺ 3y 2.
Trace in yz-plane
x
For instance, using the first trace, the generating curve is the semiellipse x ⫽ 冪9 ⫺ 3y2.
Generating curve in xy-plane x = 9 − 3y 2
Generating curve
The graph of this surface is shown in Figure 11.65.
Figure 11.65
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802
Chapter 11
Vectors and the Geometry of Space
11.6 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Matching In Exercises 1–6, match the equation with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] z
(a)
25. Cylinder State the definition of a cylinder.
z
(b)
6
26. Trace of a Surface What is meant by the trace of a surface? How do you find a trace?
3 2
4
27. Quadric Surfaces Identify the six quadric surfaces and give the standard form of each.
2
3 x
56
4
y
x
z
(c)
y
4
−3
28. Classifying an Equation What does the equation z ⫽ x 2 represent in the xz-plane? What does it represent in three-space? 29. Classifying an Equation What does the equation 4x2 ⫹ 6y2 ⫺ 3z2 ⫽ 12 represent in the xy-plane? What does it represent in three-space?
z
(d)
WRITING ABOUT CONCEPTS
4 4 2
−5
2
4
5
x
y
z
(e)
6
x
3
1.
HOW DO YOU SEE IT? The four figures are graphs of the quadric surface z ⫽ x 2 ⫹ y 2. Match each of the four graphs with the point in space from which the paraboloid is viewed. The four points are 共0, 0, 20兲, 共0, 20, 0兲, 共20, 0, 0兲, and 共10, 10, 20兲.
30.
z
(f) 3 2
3 2 1
4 x
y
(a)
2 4
y 5 x
−3
x2 y2 z2 ⫹ ⫹ ⫽1 9 16 9
4
4
z
z
(b)
y
2. 15x 2 ⫺ 4y 2 ⫹ 15z 2 ⫽ ⫺4 y
3. 4x 2 ⫺ y 2 ⫹ 4z 2 ⫽ 4
4. y 2 ⫽ 4x 2 ⫹ 9z 2
5. 4x 2 ⫺ 4y ⫹ z 2 ⫽ 0
6. 4x 2 ⫺ y 2 ⫹ 4z ⫽ 0
Sketching a Surface in Space
y
x
(c)
z
(d)
In Exercises 7–12,
describe and sketch the surface. 7. y ⫽ 5 9. y 2 ⫹ z 2 ⫽ 9 11.
4x 2
⫹
y
8. z ⫽ 2 y2
⫽4
10. y 2 ⫹ z ⫽ 6 12. y 2 ⫺ z 2 ⫽ 16
x x
Sketching a Quadric Surface In Exercises 13–24, classify and sketch the quadric surface. Use a computer algebra system or a graphing utility to confirm your sketch. 13. x 2 ⫹
y2 ⫹ z2 ⫽ 1 4
14.
x2 y2 z2 ⫹ ⫹ ⫽1 16 25 25
15. 16x 2 ⫺ y 2 ⫹ 16z 2 ⫽ 4
16. ⫺8x2 ⫹ 18y2 ⫹ 18z2 ⫽ 2
17. 4x2 ⫺ y2 ⫺ z2 ⫽ 1
18. z 2 ⫺ x 2 ⫺
19. x 2 ⫺ y ⫹ z 2 ⫽ 0
20. z ⫽ x 2 ⫹ 4y 2
21. x 2 ⫺ y 2 ⫹ z ⫽ 0
22. 3z ⫽ ⫺y 2 ⫹ x 2
23. z 2 ⫽ x 2 ⫹
y2 9
y2 ⫽1 4
Finding an Equation of a Surface of Revolution In Exercises 31–36, find an equation for the surface of revolution formed by revolving the curve in the indicated coordinate plane about the given axis. Equation of Curve
Coordinate Plane
Axis of Revolution
31. z 2 ⫽ 4y
yz-plane
y-axis
32. z ⫽ 3y
yz-plane
y-axis
33. z ⫽ 2y
yz-plane
z-axis
24. x 2 ⫽ 2y 2 ⫹ 2z 2
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11.6 Equation of Curve
Surfaces in Space
803
46. Machine Design The top of a rubber bushing designed to absorb vibrations in an automobile is the surface of revolution generated by revolving the curve
Coordinate Plane
Axis of Revolution
34. 2z ⫽ 冪4 ⫺ x 2
xz-plane
x-axis
35. xy ⫽ 2
xy-plane
x-axis
z ⫽ 12 y2 ⫹ 1
36. z ⫽ ln y
yz-plane
z-axis
for 0 ⱕ y ⱕ 2 in the yz-plane about the z-axis.
Finding a Generating Curve In Exercises 37 and 38, find
(a) Find an equation for the surface of revolution.
an equation of a generating curve given the equation of its surface of revolution.
(b) All measurements are in centimeters and the bushing is set on the xy-plane. Use the shell method to find its volume.
37. x 2 ⫹ y 2 ⫺ 2z ⫽ 0
(c) The bushing has a hole of diameter 1 centimeter through its center and parallel to the axis of revolution. Find the volume of the rubber bushing.
38. x 2 ⫹ z 2 ⫽ cos2 y
Finding the Volume of a Solid In Exercises 39 and 40, use the shell method to find the volume of the solid below the surface of revolution and above the xy-plane. 39. The curve z ⫽ 4x ⫺ x 2 in the xz-plane is revolved about the z-axis. 40. The curve z ⫽ sin y 共0 ⱕ y ⱕ 兲 in the yz-plane is revolved about the z-axis.
Analyzing a Trace In Exercises 41 and 42, analyze the trace when the surface z ⴝ 12 x 2 ⴙ 14 y 2
41. Find the lengths of the major and minor axes and the coordinates of the foci of the ellipse generated when the surface is intersected by the planes given by and (b) z ⫽ 8.
42. Find the coordinates of the focus of the parabola formed when the surface is intersected by the planes given by (a) y ⫽ 4
z⫽
Determine the
y2 x2 ⫺ b2 a2
with the plane bx ⫹ ay ⫺ z ⫽ 0. (Assume a, b > 0.兲 48. Intersection of Surfaces intersection of the surfaces
Explain why the curve of
x 2 ⫹ 3y 2 ⫺ 2z 2 ⫹ 2y ⫽ 4 and
is intersected by the indicated planes.
(a) z ⫽ 2
47. Using a Hyperbolic Paraboloid intersection of the hyperbolic paraboloid
and (b) x ⫽ 2.
2x 2 ⫹ 6y 2 ⫺ 4z 2 ⫺ 3x ⫽ 2 lies in a plane.
True or False? In Exercises 49–52, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 49. A sphere is an ellipsoid. 50. The generating curve for a surface of revolution is unique.
Finding an Equation of a Surface In Exercises 43 and
51. All traces of an ellipsoid are ellipses.
44, find an equation of the surface satisfying the conditions, and identify the surface.
52. All traces of a hyperboloid of one sheet are hyperboloids.
43. The set of all points equidistant from the point 共0, 2, 0兲 and the plane y ⫽ ⫺2
53. Think About It Three types of classic “topological” surfaces are shown below. The sphere and torus have both an “inside” and an “outside.” Does the Klein bottle have both an inside and an outside? Explain.
44. The set of all points equidistant from the point 共0, 0, 4兲 and the xy-plane 45. Geography Because of the forces caused by its rotation, Earth is an oblate ellipsoid rather than a sphere. The equatorial radius is 3963 miles and the polar radius is 3950 miles. Find an equation of the ellipsoid. (Assume that the center of Earth is at the origin and that the trace formed by the plane z ⫽ 0 corresponds to the equator.)
Sphere
Torus
Klein bottle
Klein bottle
Denis Tabler/Shutterstock.com
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804
Chapter 11
Vectors and the Geometry of Space
11.7 Cylindrical and Spherical Coordinates Use cylindrical coordinates to represent surfaces in space. Use spherical coordinates to represent surfaces in space.
Cylindrical Coordinates
Cylindrical coordinates: r2 = x2 + y2 y tan θ = x z=z
z
Rectangular coordinates: x = r cos θ y = r sin θ z=z
(x, y, z) P (r, θ , z)
x
y
θ
x
r
y
You have already seen that some two-dimensional graphs are easier to represent in polar coordinates than in rectangular coordinates. A similar situation exists for surfaces in space. In this section, you will study two alternative space-coordinate systems. The first, the cylindrical coordinate system, is an extension of polar coordinates in the plane to three-dimensional space. The Cylindrical Coordinate System In a cylindrical coordinate system, a point P in space is represented by an ordered triple 共r, , z兲. 1. 共r, 兲 is a polar representation of the projection of P in the xy-plane. 2. z is the directed distance from 共r, 兲 to P.
To convert from rectangular to cylindrical coordinates (or vice versa), use the conversion guidelines for polar coordinates listed below and illustrated in Figure 11.66.
Figure 11.66
Cylindrical to rectangular: x r cos ,
y r sin ,
zz
y tan , x
zz
Rectangular to cylindrical: r2 x2 y2,
(x, y, z) = (− 2 3, 2, 3)
The point 共0, 0, 0兲 is called the pole. Moreover, because the representation of a point in the polar coordinate system is not unique, it follows that the representation in the cylindrical coordinate system is also not unique.
P
z
Cylindrical-to-Rectangular Conversion
z
4
−4
3
Convert the point 共r, , z兲 共4, 5兾6, 3兲 to rectangular coordinates.
−3 2
Solution
−2
r
1
−1 1 x
−1
θ
(
(r, θ , z) = 4,
1
2
5π ,3 6
Figure 11.67
(
Using the cylindrical-to-rectangular conversion equations produces
3
4
y
冢
冣
冪3 5 2冪3 4 2 6 5 1 2 y 4 sin 4 6 2 z 3.
x 4 cos
冢冣
So, in rectangular coordinates, the point is 共x, y, z兲 共2冪3, 2, 3兲, as shown in Figure 11.67.
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11.7
Cylindrical and Spherical Coordinates
805
Rectangular- to - Cylindrical Conversion Convert the point
共x, y, z兲 共1, 冪3, 2兲 to cylindrical coordinates. z
Solution (x, y, z) = (1,
3, 2)
r ± 冪1 3 ± 2
3
r=2
2
π θ= 3
2
y
3
You have two choices for r and infinitely many choices for . As shown in Figure 11.68, two convenient representations of the point are
3 x
n 3
z2 z=2
2
arctan 共冪3 兲 n
tan 冪3
1
1
Use the rectangular-to-cylindrical conversion equations.
π 4π (r, θ , z) = 2, , 2 or −2, , 2 3 3
(
( (
(
冢2, 3 , 2冣
r > 0 and in Quadrant I
冢2, 43 , 2冣.
r < 0 and in Quadrant III
and
Figure 11.68
Cylindrical coordinates are especially convenient for representing cylindrical surfaces and surfaces of revolution with the z-axis as the axis of symmetry, as shown in Figure 11.69. x 2 + y 2 = 4z r=2 z
x2 + y2 = 9 r=3 z
x2 + y2 = z2 r=z
x2 + y2 − z2 = 1 r2 = z2 + 1
z
z
z
y
y
x
y
x
Cylinder Figure 11.69
Paraboloid
y
x
x
Cone
Hyperboloid
Vertical planes containing the z-axis and horizontal planes also have simple cylindrical coordinate equations, as shown in Figure 11.70. z
z
Vertical plane: θ =c
θ =c
Horizontal plane: z=c
y y
x x
Figure 11.70
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806
Chapter 11
Vectors and the Geometry of Space
Rectangular-to-Cylindrical Conversion Find an equation in cylindrical coordinates for the surface represented by each rectangular equation.
Rectangular: x 2 + y 2 = 4z 2
a. x 2 y 2 4z 2 b. y 2 x
Cylindrical: r 2 = 4z 2
z
Solution
3
a. From Section 11.6, you know that the graph of x 2 y 2 4z 2
x
4
6
4
6
y
is an elliptic cone with its axis along the z-axis, as shown in Figure 11.71. When you replace x 2 y 2 with r 2, the equation in cylindrical coordinates is x 2 y 2 4z 2 r 2 4z 2.
Figure 11.71
Rectangular equation Cylindrical equation
b. The graph of the surface y2 x Rectangular: y2 = x
is a parabolic cylinder with rulings parallel to the z-axis, as shown in Figure 11.72. To obtain the equation in cylindrical coordinates, replace y 2 with r 2 sin2 and x with r cos , as shown.
Cylindrical: r = csc θ cot θ z
y2 r 2 sin2 r共r sin2 cos 兲 r sin2 cos
x r cos 0 0 cos r 2 sin r csc cot
2 1
x
4
2
y
Figure 11.72
Rectangular equation Substitute r sin for y and r cos for x. Collect terms and factor. Divide each side by r. Solve for r. Cylindrical equation
Note that this equation includes a point for which r 0, so nothing was lost by dividing each side by the factor r. Converting from cylindrical coordinates to rectangular coordinates is less straightforward than converting from rectangular coordinates to cylindrical coordinates, as demonstrated in Example 4. Cylindrical: r 2 cos 2θ + z 2 + 1 = 0
Cylindrical-to-Rectangular Conversion
z
Find an equation in rectangular coordinates for the surface represented by the cylindrical equation
3
r 2 cos 2 z 2 1 0. Solution 3
r 2 cos 2 z 2 1 0 共 sin2 兲 z 2 1 0 r 2 cos 2 r 2 sin2 z 2 1 x 2 y 2 z 2 1 y2 x2 z2 1
2
−1
x
−2 −3
Rectangular: y2 − x2 − z2 = 1
Figure 11.73
2
3
y
r2
cos 2
Cylindrical equation Trigonometric identity
Replace r cos with x and r sin with y. Rectangular equation
This is a hyperboloid of two sheets whose axis lies along the y-axis, as shown in Figure 11.73.
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11.7
Cylindrical and Spherical Coordinates
807
Spherical Coordinates z
In the spherical coordinate system, each point is represented by an ordered triple: the first coordinate is a distance, and the second and third coordinates are angles. This system is similar to the latitude-longitude system used to identify points on the surface of Earth. For example, the point on the surface of Earth whose latitude is 40
North (of the equator) and whose longitude is 80 West (of the prime meridian) is shown in Figure 11.74. Assuming that Earth is spherical and has a radius of 4000 miles, you would label this point as
Prime meridian y
80° W 40° N
共4000, 80 , 50 兲. x
Radius
80 clockwise from prime meridian
50 down from North Pole
Equator
The Spherical Coordinate System In a spherical coordinate system, a point P in space is represented by an ordered triple 共, , 兲, where is the lowercase Greek letter rho and is the lowercase Greek letter phi.
Figure 11.74
1. is the distance between P and the origin, 0. 2. is the same angle used in cylindrical coordinates for r 0. 3. is the angle between the positive z-axis and the line segment OP , 0 . \
Note that the first and third coordinates, and , are nonnegative.
z
r = ρ sin φ =
The relationship between rectangular and spherical coordinates is illustrated in Figure 11.75. To convert from one system to the other, use the conversion guidelines listed below.
x2 + y2
Spherical to rectangular:
z P
φ
O
ρ
(ρ, θ , φ ) (x, y, z)
x sin cos ,
y sin sin ,
z cos
y
θ
r x
x
Rectangular to spherical:
y
Spherical coordinates Figure 11.75
冢
z y 2 x2 y2 z2, tan , arccos x 冪x2 y2 z2
冣
To change coordinates between the cylindrical and spherical systems, use the conversion guidelines listed below. Spherical to cylindrical 冇r 0冈: r2 2 sin2 ,
, z cos
Cylindrical to spherical 冇r 0冈:
冪r2 z2, , arccos
冢冪r z z 冣 2
2
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808
Chapter 11
Vectors and the Geometry of Space
The spherical coordinate system is useful primarily for surfaces in space that have a point or center of symmetry. For example, Figure 11.76 shows three surfaces with simple spherical equations. z
z
z
φ=c
c
y
x x
θ=c
y
y x
Half-cone: π 0 0. (c) Show that the volume of the solid in part (b) is equal to one-half the product of the area of the base times the altitude, as shown in the figure.
0
z
Base
(a) Use a graphing utility to graph the function on the interval 2 x 2. (b) Find a unit vector parallel to the graph of f at the point 0, 0.
Altitude
(c) Find a unit vector perpendicular to the graph of f at the point 0, 0. (d) Find the parametric equations of the tangent line to the graph of f at the point 0, 0. 3. Proof Using vectors, prove that the line segments joining the midpoints of the sides of a parallelogram form a parallelogram (see figure).
y x
8. Volume (a) Use the disk method to find the volume of the sphere x 2 y 2 z 2 r 2. (b) Find the volume of the ellipsoid 9. Proof
x2 y2 z2 2 2 1. 2 a b c
Prove the following property of the cross product.
u v w z u v zw u v wz 10. Using Parametric Equations by the parametric equations 4. Proof Using vectors, prove that the diagonals of a rhombus are perpendicular (see figure).
x t 3,
Consider the line given
y 12t 1, z 2t 1
and the point 4, 3, s for any real number s. (a) Write the distance between the point and the line as a function of s. (b) Use a graphing utility to graph the function in part (a). Use the graph to find the value of s such that the distance between the point and the line is minimum.
5. Distance (a) Find the shortest distance between the point Q2, 0, 0 and the line determined by the points P10, 0, 1 and P20, 1, 2. (b) Find the shortest distance between the point Q2, 0, 0 and the line segment joining the points P10, 0, 1 and P20, 1, 2.
(c) Use the zoom feature of a graphing utility to zoom out several times on the graph in part (b). Does it appear that the graph has slant asymptotes? Explain. If it appears to have slant asymptotes, find them. 11. Sketching Graphs Sketch the graph of each equation given in spherical coordinates. (a) 2 sin
(b) 2 cos
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814
Chapter 11
Vectors and the Geometry of Space
12. Sketching Graphs Sketch the graph of each equation given in cylindrical coordinates. (a) r 2 cos
(b) z r 2 cos 2
13. Tetherball A tetherball weighing 1 pound is pulled outward from the pole by a horizontal force u until the rope makes an angle of degrees with the pole (see figure).
16. Latitude-Longitude System Los Angeles is located at 34.05 North latitude and 118.24 West longitude, and Rio de Janeiro, Brazil, is located at 22.90 South latitude and 43.23 West longitude (see figure). Assume that Earth is spherical and has a radius of 4000 miles. Prime meridian
z
(a) Determine the resulting tension in the rope and the magnitude of u when 30.
y
Los Angeles
(b) Write the tension T in the rope and the magnitude of u as functions of . Determine the domains of the functions. (c) Use a graphing utility to complete the table.
0
10
20
30
40
50
60
x
T
Equator
u Rio de Janeiro
(d) Use a graphing utility to graph the two functions for 0 60. (e) Compare T and u as increases. (f) Find (if possible)
lim T
→ 2
and
lim u . Are the
→ 2
results what you expected? Explain.
(b) Find the rectangular coordinates for the location of each city. (c) Find the angle (in radians) between the vectors from the center of Earth to the two cities. (d) Find the great-circle distance s between the cities. Hint: s r
θ
(e) Repeat parts (a)–(d) for the cities of Boston, located at 42.36 North latitude and 71.06 West longitude, and Honolulu, located at 21.31 North latitude and 157.86 West longitude.
θ u
θ
1 lb
Figure for 13
(a) Find the spherical coordinates for the location of each city.
17. Distance Between a Point and a Plane Consider the plane that passes through the points P, R, and S. Show that the distance from a point Q to this plane is
Figure for 14
Distance
14. Towing A loaded barge is being towed by two tugboats, and the magnitude of the resultant is 6000 pounds directed along the axis of the barge (see figure). Each towline makes an angle of degrees with the axis of the barge. (a) Find the tension in the towlines when 20.
ax by cz d1 0
Distance
30
40
50
60
T (d) Use a graphing utility to graph the tension function. (e) Explain why the tension increases as increases. 15. Proof Consider the vectors u cos , sin , 0 and v cos , sin , 0, where > . Find the cross product of the vectors and use the result to prove the identity
\
18. Distance Between Parallel Planes distance between the parallel planes
(c) Use a graphing utility to complete the table. 20
\
where u PR , v PS , and w PQ .
is
10
u v
\
(b) Write the tension T of each line as a function of . Determine the domain of the function.
u v w
d1 d2
Show that the
and ax by cz d2 0
a2 b2 c 2
.
19. Intersection of Planes Show that the curve of intersection of the plane z 2y and the cylinder x 2 y 2 1 is an ellipse. 20. Vector Algebra Read the article “Tooth Tables: Solution of a Dental Problem by Vector Algebra” by Gary Hosler Meisters in Mathematics Magazine. (To view this article, go to MathArticles.com.) Then write a paragraph explaining how vectors and vector algebra can be used in the construction of dental inlays.
sin sin cos cos sin .
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12 12.1 12.2 12.3 12.4 12.5
Vector-Valued Functions Vector-Valued Functions Differentiation and Integration of Vector-Valued Functions Velocity and Acceleration Tangent Vectors and Normal Vectors Arc Length and Curvature
Speed (Exercise 68, p. 861)
Air Traffic Control (Exercise 65, p. 850)
Football (Exercise 32, p. 839)
Shot-Put Throw (Exercise 42, p. 839) Playground Slide (Exercise 81, p. 823) 815 Clockwise from top left, 06photo/Shutterstock.com; Elena Aliaga/Shutterstock.com; Jamie Roach/Shutterstock.com; Jack.Q/Shutterstock.com; Nicholas Moore/Shutterstock.com
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816
Chapter 12
Vector-Valued Functions
12.1 Vector-Valued Functions Analyze and sketch a space curve given by a vector-valued function. Extend the concepts of limits and continuity to vector-valued functions.
Space Curves and Vector-Valued Functions In Section 10.2, a plane curve was defined as the set of ordered pairs f t, g t together with their defining parametric equations x f t and
y gt
where f and g are continuous functions of t on an interval I. This definition can be extended naturally to three-dimensional space. A space curve C is the set of all ordered triples f t, gt, ht together with their defining parametric equations x f t,
y gt,
and z ht
where f, g, and h are continuous functions of t on an interval I. Before looking at examples of space curves, a new type of function, called a vector-valued function, is introduced. This type of function maps real numbers to vectors. Definition of Vector-Valued Function A function of the form rt f t i gt j
Plane
rt f t i gt j ht k
Space
or
y
is a vector-valued function, where the component functions f, g, and h are real-valued functions of the parameter t. Vector-valued functions are sometimes denoted as
r(t2) C
r(t1) r(t0)
rt f t, gt
Plane
rt f t, gt, ht.
Space
or x
Curve in a plane
Technically, a curve in a plane or in space consists of a collection of points and the defining parametric equations. Two different curves can have the same graph. For instance, each of the curves rt sin t i cos t j
z
Curve in space r(t2) r(t1) r(t0)
C y
x
Curve C is traced out by the terminal point of position vector rt. Figure 12.1
and rt sin t 2 i cos t 2 j
has the unit circle as its graph, but these equations do not represent the same curve— because the circle is traced out in different ways on the graphs. Be sure you see the distinction between the vector-valued function r and the real-valued functions f, g, and h. All are functions of the real variable t, but rt is a vector, whereas f t, gt, and ht are real numbers for each specific value of t. Vector-valued functions serve dual roles in the representation of curves. By letting the parameter t represent time, you can use a vector-valued function to represent motion along a curve. Or, in the more general case, you can use a vector-valued function to trace the graph of a curve. In either case, the terminal point of the position vector rt coincides with the point x, y or x, y, z on the curve given by the parametric equations, as shown in Figure 12.1. The arrowhead on the curve indicates the curve’s orientation by pointing in the direction of increasing values of t.
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12.1
Vector-Valued Functions
817
Unless stated otherwise, the domain of a vector-valued function r is considered to be the intersection of the domains of the component functions f, g, and h. For instance, the domain of r t ln t i 1 t j t k is the interval 0, 1. y
Sketching a Plane Curve 2
Sketch the plane curve represented by the vector-valued function
1
rt 2 cos t i 3 sin t j, 0 t 2. x
−3
−1
1
3
Solution
Vector-valued function
From the position vector rt, you can write the parametric equations
x 2 cos t and
y 3 sin t.
Solving for cos t and sin t and using the identity cos 2 t sin2 t 1 produces the rectangular equation r(t) = 2 cos ti − 3 sin tj
The ellipse is traced clockwise as t increases from 0 to 2. Figure 12.2
x2 y 2 2 1. 22 3
Rectangular equation
The graph of this rectangular equation is the ellipse shown in Figure 12.2. The curve has a clockwise orientation. That is, as t increases from 0 to 2, the position vector rt moves clockwise, and its terminal point traces the ellipse.
Sketching a Space Curve See LarsonCalculus.com for an interactive version of this type of example.
Sketch the space curve represented by the vector-valued function rt 4 cos t i 4 sin t j tk, 0 t 4. Solution equations
Vector-valued function
From the first two parametric
x 4 cos t and
z
(4, 0, 4 π )
y 4 sin t
Cylinder: x 2 + y 2 = 16
4π
you can obtain x 2 y 2 16.
Rectangular equation
This means that the curve lies on a right circular cylinder of radius 4, centered about the z-axis. To locate the curve on this cylinder, you can use the third parametric equation z t. In 1953, Francis Crick and James D. Watson discovered the double helix structure of DNA.
In Figure 12.3, note that as t increases from 0 to 4, the point x, y, z spirals up the cylinder to produce a helix. A real-life example of a helix is shown in the drawing at the left.
(4, 0, 0) x
4
y
r(t) = 4 cos ti + 4 sin tj + tk
As t increases from 0 to 4, two spirals on the helix are traced out. Figure 12.3
In Examples 1 and 2, you were given a vector-valued function and were asked to sketch the corresponding curve. The next two examples address the reverse problem— finding a vector-valued function to represent a given graph. Of course, when the graph is described parametrically, representation by a vector-valued function is straightforward. For instance, to represent the line in space given by x 2 t, y 3t, and z 4 t, you can simply use the vector-valued function r t 2 t i 3tj 4 t k. When a set of parametric equations for the graph is not given, the problem of representing the graph by a vector-valued function boils down to finding a set of parametric equations.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
818
Chapter 12
Vector-Valued Functions
Representing a Graph: Vector-Valued Function y
t = −2
Represent the parabola
t = −1
y x2 1
t=2
5 4
by a vector-valued function.
3
Solution Although there are many ways to choose the parameter t, a natural choice is to let x t. Then y t 2 1 and you have
t=1
2
t=0
y=
x2
r t ti t 2 1 j.
+1 x
−2
−1
1
2
There are many ways to parametrize this graph. One way is to let x t. Figure 12.4
Vector-valued function
Note in Figure 12.4 the orientation produced by this particular choice of parameter. Had you chosen x t as the parameter, the curve would have been oriented in the opposite direction.
Representing a Graph: Vector-Valued Function Sketch the space curve C represented by the intersection of the semiellipsoid x2 y2 z2 1, 12 24 4
z 0
and the parabolic cylinder y x 2. Then find a vector-valued function to represent the graph. Solution The intersection of the two surfaces is shown in Figure 12.5. As in Example 3, a natural choice of parameter is x t. For this choice, you can use the given equation y x2 to obtain y t 2. Then it follows that z2 x2 y2 t2 t4 24 2t 2 t 4 6 t24 t2 1 1 . 4 12 24 12 24 24 24 Because the curve lies above the xy-plane, you should choose the positive square root for z and obtain the parametric equations x t, y t 2,
and z
6 t 6 4 t . 2
2
The resulting vector-valued function is r t ti t 2 j
6 t 6 4 t k, 2
2
2 t 2.
Vector-valued function
Note that the k-component of rt implies 2 t 2. From the points 2, 4, 0 and 2, 4, 0 shown in Figure 12.5, you can see that the curve is traced as t increases from 2 to 2. z
Parabolic cylinder
C: x = t y = t2
(0, 0, 2) 2
z=
Ellipsoid
REMARK Curves in space can be specified in various ways. For instance, the curve in Example 4 is described as the intersection of two surfaces in space.
(6 + t 2 )(4 − t 2 ) 6
Curve in space (−2, 4, 0)
4 x
(2, 4, 0)
5
y
The curve C is the intersection of the semiellipsoid and the parabolic cylinder. Figure 12.5
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12.1
Vector-Valued Functions
819
Limits and Continuity Many techniques and definitions used in the calculus of real-valued functions can be applied to vector-valued functions. For instance, you can add and subtract vectorvalued functions, multiply a vector-valued function by a scalar, take the limit of a vector-valued function, differentiate a vector-valued function, and so on. The basic approach is to capitalize on the linearity of vector operations by extending the definitions on a component-by-component basis. For example, to add two vectorvalued functions (in the plane), you can write r1t r2t f1t i g1t j f2t i g2t j
Sum
f1t f2t i g1t g2t j. To subtract two vector-valued functions, you can write r1t r2t f1t i g1t j f2t i g2t j
Difference
f1t f2t i g1t g2t j. Similarly, to multiply a vector-valued function by a scalar, you can write cr t c f1t i g1t j
Scalar multiplication
cf1t i cg1t j. To divide a vector-valued function by a scalar, you can write r t f1t i g1tj , c0 c c f t g t 1 i 1 j. c c
Scalar division
This component-by-component extension of operations with real-valued functions to vector-valued functions is further illustrated in the definition of the limit of a vector-valued function. Definition of the Limit of a Vector-Valued Function 1. If r is a vector-valued function such that r t f t i gt j, then
−L
L
r (t)
lim r t lim f t i lim g t j
O
t→a
t→a
t→a
Plane
provided f and g have limits as t → a. 2. If r is a vector-valued function such that r t f t i gt j ht k, then
r(t)
lim r t lim f t i lim g t j lim ht k t→a
t→a
t→a
t→a
Space
provided f, g, and h have limits as t → a. L O r(t)
As t approaches a, rt approaches the limit L. For the limit L to exist, it is not necessary that ra be defined or that ra be equal to L. Figure 12.6
If rt approaches the vector L as t → a, then the length of the vector rt L approaches 0. That is, rt L → 0 as
t → a.
This is illustrated graphically in Figure 12.6. With this definition of the limit of a vector-valued function, you can develop vector versions of most of the limit theorems given in Chapter 1. For example, the limit of the sum of two vector-valued functions is the sum of their individual limits. Also, you can use the orientation of the curve rt to define one-sided limits of vector-valued functions. The next definition extends the notion of continuity to vector-valued functions.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
820
Chapter 12
Vector-Valued Functions
Definition of Continuity of a Vector-Valued Function A vector-valued function r is continuous at the point given by t a when the limit of rt exists as t → a and lim rt ra. t→a
A vector-valued function r is continuous on an interval I when it is continuous at every point in the interval.
From this definition, it follows that a vector-valued function is continuous at t a if and only if each of its component functions is continuous at t a. z
Continuity of a Vector-Valued Function 16
a = −4
Discuss the continuity of the vector-valued function a=4
14
r t ti aj a 2 t 2k
a is a constant.
at t 0.
12
Solution
As t approaches 0, the limit is
10
lim rt lim t i lim a j lim a 2 t 2 k t→0
8
t→0
t→0
0i aj a 2 k aj a 2k.
6 4
Because
2 −4
t→0
2
4
y
4 x
r0 0 i a j a 2k aj a 2 k you can conclude that r is continuous at t 0. By similar reasoning, you can conclude that the vector-valued function r is continuous at all real-number values of t.
a=0 a = −2
a=2
For each value of a, the curve represented by the vector-valued function rt t i aj a2 t2k is a parabola. Figure 12.7
For each value of a, the curve represented by the vector-valued function in Example 5 rt ti aj a 2 t 2k
a is a constant.
is a parabola. You can think of each parabola as the intersection of the vertical plane y a and the hyperbolic paraboloid y2 x2 z
TECHNOLOGY Almost any type of three-dimensional sketch is difficult to do by hand, but sketching curves in space is especially difficult. The problem is trying to create the illusion of three dimensions. Graphing utilities use a variety of techniques to add “three-dimensionality” to graphs of space curves: one way is to show the curve on a surface, as in Figure 12.7.
as shown in Figure 12.7.
Continuity of a Vector-Valued Function Determine the interval(s) on which the vector-valued function rt ti t 1 j t 2 1k is continuous. Solution The component functions are f t t, gt t 1, and ht t 2 1. Both f and h are continuous for all real-number values of t. The function g, however, is continuous only for t 1. So, r is continuous on the interval 1, .
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12.1
12.1 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding the Domain In Exercises 1–8, find the domain of the vector-valued function. 1. rt
1 t i j 3tk t1 2
Matching In Exercises 19–22, match the equation with its graph. [The graphs are labeled (a), (b), (c), and (d).] z
(a)
2. rt 4 t 2 i t 2 j 6t k 3. rt ln t i e t j t k
z
(b) 4
4
2
2 y
4. rt sin t i 4 cos t j t k
−2
5. rt Ft Gt, where
2
−2
4
Gt cos t i sin t j
6. rt Ft Gt, where Ft ln t i 5t j 3t 2 k,
z
(c)
Gt i 4t j 3t 2 k
1
2 1
8. rt Ft Gt, where
x
1 3 t i Ft t 3 i t j t k, Gt j t 2 k t1
Evaluating a Function In Exercises 9–12, evaluate (if possible) the vector-valued function at each given value of t.
y
20. rt cos t i sin t j t 2 k, 1 t 1 21. rt t i t 2 j e0.75t k, 2 t 2 2t k, 3
0.1 t 5
represented by the vector-valued function and give the orientation of the curve.
(b) r4
(c) r
(d) r6 t r6 1 11. rt ln t i j 3t k t (c) rt 4
(d) r1 t r1 12. rt t i t 32 j et4 k (b) r4
4
Sketching a Curve In Exercises 23–38, sketch the curve
10. rt cos t i 2 sin t j
(a) r0
x
19. rt t i 2t j t 2 k, 2 t 2
(c) rs 1
(d) r2 t r2
(b) r3
2
y
1
22. rt t i ln t j
9. rt 12 t 2 i t 1 j
(a) r2
y
2
4
Ft sin t i cos t j, Gt sin t j cos t k
(a) r0
2
z
(d)
7. rt Ft Gt, where
(b) r0
x
x
Ft cos t i sin t j t k,
(a) r1
821
Vector-Valued Functions
(c) rc 2
(d) r9 t r9
Writing a Vector-Valued Function In Exercises 13–16, represent the line segment from P to Q by a vector-valued function and by a set of parametric equations. 13. P0, 0, 0, Q3, 1, 2 14. P0, 2, 1, Q4, 7, 2 15. P2, 5, 3, Q1, 4, 9 16. P1, 6, 8, Q3, 2, 5
Think About It In Exercises 17 and 18, find r t u t . Is the result a vector-valued function? Explain. 17. rt 3t 1 i 14 t 3 j 4k, ut t 2 i 8j t 3 k 18. rt 3 cos t, 2 sin t, t 2, ut 4 sin t, 6 cos t, t 2
t 23. rt i t 1 j 4
24. rt 5 ti t j
25. rt t 3 i t 2j
26. rt t 2 ti t 2 tj
27. r cos i 3 sin j 28. rt 2 cos t i 2 sin t j 29. r 3 sec i 2 tan j 30. rt 2 cos3 t i 2 sin3 tj 31. rt t 1 i 4t 2 j 2t 3 k 32. rt t i 2t 5 j 3t k 33. rt 2 cos t i 2 sin t j t k 34. rt t i 3 cos t j 3 sin t k 35. rt 2 sin t i 2 cos t j et k 3 36. rt t 2 i 2tj 2 tk 2 37. rt t, t 2, 3 t 3
38. rt cos t t sin t, sin t t cos t, t
Identifying a Common Curve In Exercises 39–42, use a computer algebra system to graph the vector-valued function and identify the common curve. 3 2 1 39. rt t 2 i t j t k 2 2
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
822
Chapter 12
40. rt t i
3
2
41. rt sin t i
Vector-Valued Functions
1 t2j t2k 2
23 cos t 21tj 12 cos t 23 k
42. rt 2 sin t i 2 cos t j 2 sin t k
Think About It In Exercises 43 and 44, use a computer algebra system to graph the vector-valued function r t . For each u t , make a conjecture about the transformation (if any) of the graph of r t . Use a computer algebra system to verify your conjecture. 43. rt 2 cos t i 2 sin tj 12 tk 1 2t k
(b) ut 2 cos t i 2 sin t j 2t k 2 sin t j 2 cos t k
t→
t 2 1 j 1t k 1 cos t k 65. lim t i 3t j t ln t 1 j k 66. lim t i t 1 t1 sin t j e k 67. lim e i t 1 t k 68. lim e i j t t 1 64. lim 3ti t→2
2
2
t→0
2
t
t
t→
t
2
Continuity of a Vector-Valued Function In Exercises 69–74, determine the interval(s) on which the vector-valued function is continuous.
1 (e) ut 6 cos t i 6 sin t j 2t k 1 44. rt t i t 2j 2 t 3 k 1 (a) ut t i t 2 2 j 2 t 3 k
69. rt t i
1 (b) ut t 2 i t j 2 t 3 k
1 j t
1 (c) ut t i t 2j 2 t 3 4k
70. rt t i t 1 j
1 (d) ut t i t 2j 8t 3k
71. rt t i arcsin t j t 1 k
(e) ut ti t 2j
1 3 2 t k
72. rt 2et i et j lnt 1 k
Representing a Graph by a Vector-Valued Function In Exercises 45–52, represent the plane curve by a vectorvalued function. (There are many correct answers.) 45. y x 5
46. 2x 3y 5 0
47. y x 22
48. y 4 x 2
49. x 2 y 2 25
50. x 22 y 2 4
51.
63. lim t i cos t j sin t k
t→0
1 (c) ut 2 cost i 2 sint j 2t k
(d) ut
Finding a Limit In Exercises 63–68, find the limit (if it exists).
t→1
(a) ut 2cos t 1i 2 sin t j
1 2t i
62. Sketching a Curve Show that the vector-valued function rt et cos t i et sin tj et k lies on the cone z 2 x 2 y 2. Sketch the curve.
x2 y2 1 16 4
52.
x2 y2 1 9 16
73. rt et, t 2, tan t
3 t 74. rt 8, t,
WRITING ABOUT CONCEPTS Writing a Transformation In Exercises 75–78, consider the vector-valued function r t ⴝ t 2 i ⴙ t ⴚ 3 j ⴙ tk. Write a vector-valued function s t that is the specified transformation of r.
Representing a Graph by a Vector-Valued Function
75. A vertical translation three units upward
In Exercises 53–60, sketch the space curve represented by the intersection of the surfaces. Then represent the curve by a vector-valued function using the given parameter.
76. A vertical translation four units downward
Surfaces 53. z
x2
Parameter
y , xy0 2
xt
54. z x 2 y 2, z 4
x 2 cos t
55. x 2 y 2 4,
x 2 sin t
z x2
56. 4x 2 4y 2 z 2 16, 57.
x2
y z 4, 2
2
58. x 2 y 2 z 2 10,
x z2
xz2 xy4
zt x 1 sin t x 2 sin t
y2 z2 4
x t first octant
60. x 2 y 2 z 2 16, xy 4
x t first octant
59. x 2 z 2 4,
61. Sketching a Curve Show that the vector-valued function rt t i 2t cos tj 2t sin t k lies on the cone 4x 2 y 2 z 2. Sketch the curve.
77. A horizontal translation two units in the direction of the negative x-axis 78. A horizontal translation five units in the direction of the positive y-axis 79. Continuity of a Vector-Valued Function State the definition of continuity of a vector-valued function. Give an example of a vector-valued function that is defined but not continuous at t 2. 80. Comparing Functions Which of the following vectorvalued functions represent the same graph? (a) rt 3 cos t 1i 5 sin t 2j 4k (b) rt 4i 3 cos t 1j 5 sin t 2k (c) rt 3 cos t 1i 5 sin t 2j 4k (d) rt 3 cos 2t 1i 5 sin 2t 2j 4k
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12.1 81. Playground Slide
Vector-Valued Functions
823
Particle Motion In Exercises 87 and 88, two particles travel along the space curves r t and u t . A collision will occur at the point of intersection P when both particles are at P at the same time. Do the particles collide? Do their paths intersect?
The outer edge of a playground slide is in the shape of a helix of radius 1.5 meters. The slide has a height of 2 meters and makes one complete revolution from top to bottom. Find a vector-valued function for the helix. Use a computer algebra system to graph your function. (There are many correct answers.)
87. rt t2i 9t 20j t2k ut 3t 4i t2j 5t 4k 88. rt ti t2j t3k ut 2t 3i 8tj 12t 2k
Think About It In Exercises 89 and 90, two particles travel along the space curves r t and u t . 89. If rt and ut intersect, will the particles collide? 90. If the particles collide, do their paths rt and ut intersect?
True or False? In Exercises 91–94, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.
HOW DO YOU SEE IT? The four figures
82.
below are graphs of the vector-valued function rt 4 cos t i 4 sin t j t4k. Match each of the four graphs with the point in space from which the helix is viewed. The four points are 0, 0, 20, 20, 0, 0, 20, 0, 0, and 10, 20, 10. z
(a)
91. If f, g, and h are first-degree polynomial functions, then the curve given by x f t, y gt, and z ht is a line. 92. If the curve given by x f t, y gt, and z ht is a line, then f, g, and h are first-degree polynomial functions of t.
z
(b)
93. Two particles travel along the space curves rt and ut. The intersection of their paths depends only on the curves traced out by rt and ut, while collision depends on the parametrizations. 94. The vector-valued function rt t2 i t sin t j t cos t k lies on the paraboloid x y 2 z2.
x y
y Generated by Mathematica
Generated by Mathematica
(c)
z
(d)
Witch of Agnesi In Section 3.5, you studied a famous curve called the Witch of Agnesi. In this project, you will take a closer look at this function.
y
y
x Generated by Mathematica
Generated by Mathematica
83. Proof Let rt and ut be vector-valued functions whose limits exist as t → c. Prove that lim rt ut lim rt lim ut. t→c
t→c
t→c
84. Proof Let rt and ut be vector-valued functions whose limits exist as t → c. Prove that lim rt ut lim rt lim ut. t→c
t→c
t→c
85. Proof Prove that if r is a vector-valued function that is continuous at c, then r is continuous at c. 86. Verifying a Converse Verify that the converse of Exercise 85 is not true by finding a vector-valued function r such that r is continuous at c but r is not continuous at c.
Consider a circle of radius a centered on the y-axis at 0, a. Let A be a point on the horizontal line y 2a, let O be the origin, and let B be the point where the segment OA intersects the circle. A point P is on the Witch of Agnesi when P lies on the horizontal line through B and on the vertical line through A. (a) Show that the point A is traced out by the vector-valued function rA 2a cot i 2aj for 0 < < , where is the angle that OA makes with the positive x-axis. (b) Show that the point B is traced out by the vector-valued function rB a sin 2 i a1 cos 2 j for 0 < < . (c) Combine the results of parts (a) and (b) to find the vectorvalued function r for the Witch of Agnesi. Use a graphing utility to graph this curve for a 1. (d) Describe the limits lim r and lim r .
→0
→
(e) Eliminate the parameter and determine the rectangular equation of the Witch of Agnesi. Use a graphing utility to graph this function for a 1 and compare your graph with that obtained in part (c). Jack.Q/Shutterstock.com
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
824
Chapter 12
12.2
Vector-Valued Functions
Differentiation and Integration of Vector-Valued Functions Differentiate a vector-valued function. Integrate a vector-valued function.
Differentiation of Vector-Valued Functions In Sections 12.3–12.5, you will study several important applications involving the calculus of vector-valued functions. In preparation for that study, this section is devoted to the mechanics of differentiation and integration of vector-valued functions. The definition of the derivative of a vector-valued function parallels the definition for real-valued functions. Definition of the Derivative of a Vector-Valued Function The derivative of a vector-valued function r is r⬘ 共t兲 ⫽ lim
⌬t→0
REMARK In addition to r⬘ 共t兲, other notations for the derivative of a vector-valued function are d 关r共t兲兴, dt
dr , and dt
r共t ⫹ ⌬t兲 ⫺ r共t兲 ⌬t
for all t for which the limit exists. If r⬘ 共t兲 exists, then r is differentiable at t. If r⬘ 共t兲 exists for all t in an open interval I, then r is differentiable on the interval I. Differentiability of vector-valued functions can be extended to closed intervals by considering one-sided limits.
Dt 关r共t兲兴. Differentiation of vector-valued functions can be done on a component-bycomponent basis. To see why this is true, consider the function r 共t兲 ⫽ f 共t兲i ⫹ g共t兲 j. Applying the definition of the derivative produces the following. r 共t ⫹ ⌬t兲 ⫺ r 共t兲 ⌬t→0 ⌬t f 共t ⫹ ⌬t兲 i ⫹ g共t ⫹ ⌬t兲 j ⫺ f 共t兲i ⫺ g 共t兲 j ⫽ lim ⌬t→0 ⌬t f 共t ⫹ ⌬t兲 ⫺ f 共t兲 g共t ⫹ ⌬t兲 ⫺ g共t兲 ⫽ lim i⫹ j ⌬t→0 ⌬t ⌬t f 共t ⫹ ⌬t兲 ⫺ f 共t兲 g共t ⫹ ⌬t兲 ⫺ g共t兲 ⫽ lim i ⫹ lim ⌬t→0 ⌬t→0 ⌬t ⌬t
r⬘共t兲 ⫽ lim z
r(t + Δt) − r(t)
冦冤
r′(t) r(t + Δt)
冦 冤
r(t) y
x
Figure 12.8
⫽ f⬘ 共t兲 i ⫹ g⬘共t兲 j
冥 冤 冥冧 冦 冤
冥冧
冥冧 j
This important result is listed in the theorem shown below. Note that the derivative of the vector-valued function r is itself a vector-valued function. You can see from Figure 12.8 that r⬘ 共t兲 is a vector tangent to the curve given by r共t兲 and pointing in the direction of increasing t-values. THEOREM 12.1
Differentiation of Vector-Valued Functions
1. If r 共t兲 ⫽ f 共t兲 i ⫹ g共t兲 j, where f and g are differentiable functions of t, then r⬘ 共t兲 ⫽ f⬘共t兲 i ⫹ g⬘ 共t兲 j.
Plane
2. If r 共t兲 ⫽ f 共t兲 i ⫹ g共t兲 j ⫹ h 共t兲k, where f, g, and h are differentiable functions of t, then r⬘ 共t兲 ⫽ f⬘共t兲 i ⫹ g⬘ 共t兲 j ⫹ h⬘ 共t兲k.
Space
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12.2 y
r(t) = ti + (t 2 + 2)j
See LarsonCalculus.com for an interactive version of this type of example.
For the vector-valued function
5
r′(1)
4
r共t兲 ⫽ ti ⫹ 共t2 ⫹ 2兲j find r⬘共t兲. Then sketch the plane curve represented by r共t兲 and the graphs of r共1兲 and r⬘共1兲.
(1, 3)
3
r(1)
Solution 1 x −3
−2
−1
825
Differentiation of a Vector-Valued Function
6
Figure 12.9
Differentiation and Integration of Vector-Valued Functions
1
2
3
Differentiate on a component-by-component basis to obtain
r⬘共t兲 ⫽ i ⫹ 2tj.
Derivative
From the position vector r共t兲, you can write the parametric equations x ⫽ t and y ⫽ t2 ⫹ 2. The corresponding rectangular equation is y ⫽ x2 ⫹ 2. When t ⫽ 1, r共1兲 ⫽ i ⫹ 3j and r⬘共1兲 ⫽ i ⫹ 2j. In Figure 12.9, r共1兲 is drawn starting at the origin, and r⬘共1兲 is drawn starting at the terminal point of r共1兲. Higher-order derivatives of vector-valued functions are obtained by successive differentiation of each component function.
Higher-Order Differentiation For the vector-valued function r 共t兲 ⫽ cos t i ⫹ sin tj ⫹ 2tk find each of the following. a. b. c. d.
r⬘ 共t兲 r⬙ 共t兲 r⬘ 共t兲 ⭈ r⬙ 共t兲 r⬘ 共t兲 ⫻ r⬙ 共t兲
Solution a. r⬘ 共t兲 ⫽ ⫺sin ti ⫹ cos tj ⫹ 2 k b. r⬙ 共t兲 ⫽ ⫺cos ti ⫺ sin tj ⫹ 0 k ⫽ ⫺cos ti ⫺ sin tj c. r⬘共t兲 ⭈ r⬙ 共t兲 ⫽ sin t cos t ⫺ sin t cos t ⫽ 0
ⱍ
i j d. r⬘ 共t兲 ⫻ r⬙ 共t兲 ⫽ ⫺sin t cos t ⫺cos t ⫺sin t ⫽
ⱍ
cos t ⫺sin t
k 2 0
ⱍ ⱍ
ⱍ
2 ⫺sin t i⫺ 0 ⫺cos t
⫽ 2 sin ti ⫺ 2 cos tj ⫹ k
First derivative
Second derivative Dot product
Cross product
ⱍ ⱍ
ⱍ
2 ⫺sin t cos t j⫹ k 0 ⫺cos t ⫺sin t
In Example 2(c), note that the dot product is a real-valued function, not a vector-valued function.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
826
Chapter 12
Vector-Valued Functions
The parametrization of the curve represented by the vector-valued function r 共t兲 ⫽ f 共t兲 i ⫹ g共t兲j ⫹ h 共t兲k is smooth on an open interval I when f⬘, g⬘ , and h⬘ are continuous on I and r⬘ 共t兲 ⫽ 0 for any value of t in the interval I.
Finding Intervals on Which a Curve Is Smooth y
Find the intervals on which the epicycloid C given by r 共t兲 ⫽ 共5 cos t ⫺ cos 5t兲i ⫹ 共5 sin t ⫺ sin 5t兲j,
6 4
π t= 2
is smooth. Solution
2
t=0
t=π −6
−4
x
−2
2 −2 −4
0 ⱕ t ⱕ 2
4
t = 2π
6
π t=3 2
−6
r(t) = (5 cos t − cos 5t)i + (5 sin t − sin 5t)j
The epicycloid is not smooth at the points where it intersects the axes. Figure 12.10
The derivative of r is
r⬘ 共t兲 ⫽ 共⫺5 sin t ⫹ 5 sin 5t兲i ⫹ 共5 cos t ⫺ 5 cos 5t兲j. In the interval 关0, 2兴, the only values of t for which r⬘ 共t兲 ⫽ 0i ⫹ 0j are t ⫽ 0, 兾2, , 3兾2, and 2. Therefore, you can conclude that C is smooth on the intervals
冢0, 2 冣, 冢2 , 冣, 冢, 32冣,
and
冢32, 2冣
as shown in Figure 12.10. In Figure 12.10, note that the curve is not smooth at points at which the curve makes abrupt changes in direction. Such points are called cusps or nodes. Most of the differentiation rules in Chapter 2 have counterparts for vector-valued functions, and several of these are listed in the next theorem. Note that the theorem contains three versions of “product rules.” Property 3 gives the derivative of the product of a real-valued function w and a vector-valued function r, Property 4 gives the derivative of the dot product of two vector-valued functions, and Property 5 gives the derivative of the cross product of two vector-valued functions (in space). THEOREM 12.2 Properties of the Derivative Let r and u be differentiable vector-valued functions of t, let w be a differentiable real-valued function of t, and let c be a scalar. 1. 2.
REMARK Note that Property 5 applies only to three-dimensional vector-valued functions because the cross product is not defined for two-dimensional vectors.
3. 4. 5. 6. 7.
d 关cr 共t兲兴 ⫽ cr⬘ 共t兲 dt d 关r 共t兲 ± u共t兲兴 ⫽ r⬘ 共t兲 ± u⬘ 共t兲 dt d 关w 共t兲r 共t兲兴 ⫽ w 共t兲r⬘ 共t兲 ⫹ w⬘ 共t兲r 共t兲 dt d 关r 共t兲 ⭈ u共t兲兴 ⫽ r 共t兲 ⭈ u⬘ 共t兲 ⫹ r⬘ 共t兲 ⭈ u共t兲 dt d 关r 共t兲 ⫻ u 共t兲兴 ⫽ r 共t兲 ⫻ u⬘ 共t兲 ⫹ r⬘ 共t兲 ⫻ u 共t兲 dt d 关r共w 共t兲兲兴 ⫽ r⬘ 共w 共t兲兲w⬘共t兲 dt If r 共t兲 ⭈ r 共t兲 ⫽ c, then r 共t兲 ⭈ r⬘ 共t兲 ⫽ 0.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12.2
Proof
Differentiation and Integration of Vector-Valued Functions
827
To prove Property 4, let
r 共t兲 ⫽ f1共t兲i ⫹ g1共t兲j and
u共t兲 ⫽ f2共t兲i ⫹ g2共t兲j
where f1, f2, g1, and g2 are differentiable functions of t. Then r共t兲 ⭈ u共t兲 ⫽ f1共t兲 f2共t兲 ⫹ g1共t兲g2共t兲 and it follows that
Exploration Let r共t兲 ⫽ cos ti ⫹ sin tj. Sketch the graph of r共t兲. Explain why the graph is a circle of radius 1 centered at the origin. Calculate r共兾4兲 and r⬘共兾4兲. Position the vector r⬘共兾4兲 so that its initial point is at the terminal point of r共兾4兲. What do you observe? Show that r共t兲 ⭈ r共t兲 is constant and that r共t兲 ⭈ r⬘共t兲 ⫽ 0 for all t. How does this example relate to Property 7 of Theorem 12.2?
d 关r共t兲 ⭈ u共t兲兴 ⫽ f1共t兲 f2⬘ 共t兲 ⫹ f1⬘ 共t兲 f2共t兲 ⫹ g1共t兲 g2⬘ 共t兲 ⫹ g1⬘ 共t兲 g2共t兲 dt ⫽ 关 f1共t兲 f2⬘ 共t兲 ⫹ g1共t兲 g2⬘ 共t兲兴 ⫹ 关 f1⬘ 共t兲 f2共t兲 ⫹ g1⬘ 共t兲 g2共t兲兴 ⫽ r 共t兲 ⭈ u⬘ 共t兲 ⫹ r⬘ 共t兲 ⭈ u共t兲. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Proofs of the other properties are left as exercises (see Exercises 67–71 and Exercise 74).
Using Properties of the Derivative For r 共t兲 ⫽ a.
1 i ⫺ j ⫹ ln tk and u共t兲 ⫽ t 2 i ⫺ 2tj ⫹ k, find t
d 关r共t兲 ⭈ u共t兲兴 and dt
b.
d 关u共t兲 ⫻ u⬘ 共t兲兴. dt
Solution 1 1 a. Because r⬘ 共t兲 ⫽ ⫺ 2 i ⫹ k and u⬘ 共t兲 ⫽ 2ti ⫺ 2j, you have t t d 关r共t兲 ⭈ u共t兲兴 dt ⫽ r共t兲 ⭈ u⬘共t兲 ⫹ r⬘共t兲 ⭈ u共t兲 1 1 1 ⫽ i ⫺ j ⫹ ln tk ⭈ 共2ti ⫺ 2j兲 ⫹ ⫺ 2 i ⫹ k t t t 1 ⫽ 2 ⫹ 2 ⫹ 共⫺1兲 ⫹ t 1 ⫽3⫹ . t
冢
冣
冢
冣 ⭈ 共t i ⫺ 2tj ⫹ k兲 2
b. Because u⬘ 共t兲 ⫽ 2ti ⫺ 2 j and u⬙ 共t兲 ⫽ 2i, you have d 关u共t兲 ⫻ u⬘ 共t兲兴 ⫽ 关u共t兲 ⫻ u⬙ 共t兲兴 ⫹ 关u⬘ 共t兲 ⫻ u⬘ 共t兲兴 dt
ⱍⱍ ⱍ ⱍⱍ
i j ⫽ t 2 ⫺2t 2 0
k 1 ⫹0 0
⫺2t 1 t2 i⫺ 0 0 2 ⫽ 0i ⫺ 共⫺2兲j ⫹ 4tk ⫽ 2j ⫹ 4tk. ⫽
ⱍ ⱍ ⱍ
1 t 2 ⫺2t j⫹ k 0 2 0
Try reworking parts (a) and (b) in Example 4 by first forming the dot and cross products and then differentiating to see that you obtain the same results.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
828
Chapter 12
Vector-Valued Functions
Integration of Vector-Valued Functions The next definition is a consequence of the definition of the derivative of a vectorvalued function. Definition of Integration of Vector-Valued Functions 1. If r 共t兲 ⫽ f 共t兲i ⫹ g共t兲j, where f and g are continuous on 关a, b兴, then the indefinite integral (antiderivative) of r is
冕
r 共t兲 dt ⫽
冤冕f 共t兲 dt冥i ⫹ 冤冕g共t兲 dt冥 j
Plane
and its definite integral over the interval a ⱕ t ⱕ b is
冕
b
冤冕
b
r 共t兲 dt ⫽
a
冥
f 共t兲 dt i ⫹
a
冤冕
b
冥
g共t兲 dt j.
a
2. If r 共t兲 ⫽ f 共t兲i ⫹ g共t兲j ⫹ h共t兲k, where f, g, and h are continuous on 关a, b兴, then the indefinite integral (antiderivative) of r is
冕
r 共t兲 dt ⫽
冤冕f 共t兲 dt冥i ⫹ 冤冕g共t兲 dt冥 j ⫹ 冤冕h 共t兲 dt冥k
Space
and its definite integral over the interval a ⱕ t ⱕ b is
冕
b
冤冕 f 共t兲 dt冥 i ⫹ 冤冕 g共t兲 dt冥 j ⫹ 冤冕 h 共t兲 dt冥k. b
r 共t兲 dt ⫽
a
b
a
b
a
a
The antiderivative of a vector-valued function is a family of vector-valued functions all differing by a constant vector C. For instance, if r 共t兲 is a three-dimensional vectorvalued function, then for the indefinite integral 兰r共t兲 dt, you obtain three constants of integration
冕
f 共t兲 dt ⫽ F共t兲 ⫹ C1,
冕
g 共t兲 dt ⫽ G 共t兲 ⫹ C2,
冕
h 共t兲 dt ⫽ H 共t兲 ⫹ C3
where F⬘ 共t兲 ⫽ f 共t兲, G⬘ 共t兲 ⫽ g 共t兲, and H⬘ 共t兲 ⫽ h 共t兲. These three scalar constants produce one vector constant of integration
冕
r 共t兲 dt ⫽ 关F共t兲 ⫹ C1兴 i ⫹ 关G 共t兲 ⫹ C2兴 j ⫹ 关H 共t兲 ⫹ C3兴 k ⫽ 关F共t兲i ⫹ G 共t兲 j ⫹ H 共t兲k兴 ⫹ 关C1i ⫹ C2 j ⫹ C3k兴 ⫽ R共t兲 ⫹ C
where R⬘ 共t兲 ⫽ r 共t兲.
Integrating a Vector-Valued Function Find the indefinite integral
冕
共t i ⫹ 3j兲 dt.
Solution
冕
Integrating on a component-by-component basis produces
共t i ⫹ 3j兲 dt ⫽
t2 i ⫹ 3tj ⫹ C. 2
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12.2
Differentiation and Integration of Vector-Valued Functions
829
Example 6 shows how to evaluate the definite integral of a vector-valued function.
Definite Integral of a Vector-Valued Function Evaluate the integral
冕
1
0
3 ti ⫹ 冪
0
Solution
冕
冕冢 1
r 共t兲 dt ⫽
1
冢冕
冣 冢冕
1
r 共t兲 dt ⫽
0
1
t1兾3 dt i ⫹
0
⫽
冣
1 j ⫹ e⫺t k dt. t⫹1
0
冤 冢34冣 t 冥
1
4兾3
0
冣 冢冕
1 dt j ⫹ t⫹1
冤 ⱍ
冣
e⫺t dt k
0
ⱍ冥 0 j ⫹ 冤 ⫺e⫺t冥 0 k
i ⫹ ln t ⫹ 1
冢
1
1
1
冣
3 1 ⫽ i ⫹ 共ln 2兲 j ⫹ 1 ⫺ k 4 e As with real-valued functions, you can narrow the family of antiderivatives of a vector-valued function r⬘ down to a single antiderivative by imposing an initial condition on the vector-valued function r. This is demonstrated in the next example.
The Antiderivative of a Vector-Valued Function Find the antiderivative of r⬘ 共t兲 ⫽ cos 2ti ⫺ 2 sin tj ⫹
1 k 1 ⫹ t2
that satisfies the initial condition r共0兲 ⫽ 3i ⫺ 2j ⫹ k. Solution r 共t兲 ⫽ ⫽ ⫽
冕 冢冕
r⬘ 共t兲 dt
冣 冢冕 ⫺2 sin t dt冣 j ⫹ 冢冕 1 ⫹1 t dt冣k
cos 2t dt i ⫹
2
冢12 sin 2t ⫹ C 冣i ⫹ 共2 cos t ⫹ C 兲j ⫹ 共arctan t ⫹ C 兲k 1
2
3
Letting t ⫽ 0, you can write r 共0兲 ⫽ 共0 ⫹ C1兲i ⫹ 共2 ⫹ C2兲j ⫹ 共0 ⫹ C3兲k. Using the fact that r共0兲 ⫽ 3i ⫺ 2j ⫹ k, you have
共0 ⫹ C1兲i ⫹ 共2 ⫹ C2兲j ⫹ 共0 ⫹ C3兲k ⫽ 3i ⫺ 2j ⫹ k. Equating corresponding components produces C1 ⫽ 3,
2 ⫹ C2 ⫽ ⫺2, and C3 ⫽ 1.
So, the antiderivative that satisfies the initial condition is r 共t兲 ⫽
冢12 sin 2t ⫹ 3冣i ⫹ 共2 cos t ⫺ 4兲j ⫹ 共arctan t ⫹ 1兲k.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
830
Chapter 12
Vector-Valued Functions
12.2 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Differentiation of Vector-Valued Functions In Exercises 1–6, find r⬘冇t冈, r冇t0冈, and r⬘冇t0冈 for the given value of t0. Then sketch the plane curve represented by the vector-valued function, and sketch the vectors r冇t0冈 and r⬘ 冇t0冈. Position the vectors such that the initial point of r 冇t0冈 is at the origin and the initial point of r⬘ 冇t0冈 is at the terminal point of r 冇t0冈. What is the relationship between r⬘ 冇t0冈 and the curve? 2. r共t兲 ⫽ 共1 ⫹ t兲i ⫹ t 3j,
t0 ⫽ 1
3. r 共t兲 ⫽ cos ti ⫹ sin tj,
t0 ⫽
6. r共t兲 ⫽
t0 ⫽ 0
具e⫺t,
et典,
2
2t 2t 2 1 i⫹ j 35. r共t兲 ⫽ 共t ⫺ 1兲i ⫹ j ⫺ t 2k 3 8⫹t 8 ⫹ t3 t
36. r共t兲 ⫽ et i ⫺ e⫺t j ⫹ 3tk 37. r 共t兲 ⫽ ti ⫺ 3tj ⫹ tan tk 1 38. r 共t兲 ⫽ 冪t i ⫹ 共t 2 ⫺ 1兲 j ⫹ 4tk
Using Properties of the Derivative In Exercises 39 and
sketch the space curve represented by the vector-valued function, and sketch the vectors r冇t0冈 and r⬘ 冇t0冈. 3 7. r 共t兲 ⫽ 2 cos ti ⫹ 2 sin tj ⫹ tk, t0 ⫽ 2 t0 ⫽ 2
40, use the properties of the derivative to find the following. (a) r⬘ 冇t冈 (d)
9. r共t兲 ⫽ t3i ⫺ 3tj
12. r共t兲 ⫽ 具t cos t, ⫺2 sin t典
13. r共t兲 ⫽ 6ti ⫺ 7t 2j ⫹ t 3 k
1 t2 14. r 共t兲 ⫽ i ⫹ 16tj ⫹ k t 2
15. r 共t兲 ⫽ a cos 3 t i ⫹ a sin 3 tj ⫹ k 16. r 共t兲 ⫽ 4冪t i ⫹ t 2冪t j ⫹ ln t 2 k 17. r 共t兲 ⫽ e⫺t i ⫹ 4j ⫹ 5tet k
18. r 共t兲 ⫽ 具t3, cos 3t, sin 3t典
19. r 共t兲 ⫽ 具t sin t, t cos t, t典 20. r 共t兲 ⫽ 具arcsin t, arccos t, 0典
d [3r冇t冈 ⴚ u冇t冈] dt
(c)
(e)
d [r冇t冈 ⴛ u冇t冈] dt
(f)
21. r共t兲 ⫽
u共t兲 ⫽
1 i ⫹ 2 sin tj ⫹ 2 cos tk t
Using Two Methods
In Exercises 41 and 42, find d d (a) [r冇t冈 ⭈ u冇t冈] and (b) [r冇t冈 ⴛ u冇t冈] in two different ways. dt dt (i) Find the product first, then differentiate. (ii) Apply the properties of Theorem 12.2. 41. r共t兲 ⫽ ti ⫹ 2t 2 j ⫹ t 3k,
u共t兲 ⫽ t 4k
42. r共t兲 ⫽ cos t i ⫹ sin t j ⫹ t k, u共t兲 ⫽ j ⫹ tk
Finding an Indefinite Integral In Exercises 43– 50, find
⫹
43.
1 2 2t j
22. r共t兲 ⫽ 共 ⫹ t兲i ⫹ 共t 2 ⫺ t兲 j t2
45.
23. r共t兲 ⫽ 4 cos ti ⫹ 4 sin tj 24. r共t兲 ⫽ 8 cos t i ⫹ 3 sin tj
47.
Higher-Order Differentiation In Exercises 25–28, find (a) r⬘ 冇t冈, (b) r⬙ 冇t冈, (c) r⬘ 冇t冈 ⭈ r⬙ 冇t冈, and (d) r⬘冇t冈 ⴛ r⬙ 冇t冈.
48.
25. r共t兲 ⫽ 12 t 2 i ⫺ tj ⫹ 16t 3k 26. r共t兲 ⫽ t 3 i ⫹ 共2t 2 ⫹ 3兲 j ⫹ 共3t ⫺ 5兲k 27. r共t兲 ⫽ 具cos t ⫹ t sin t, sin t ⫺ t cos t, t典 28. r共t兲 ⫽ 具e⫺t, t 2, tan t典
d r冇2t冈 dt
u共t兲 ⫽ 4ti ⫹ t 2j ⫹ t 3k
the indefinite integral.
Higher-Order Differentiation In Exercises 21–24, find (a) r⬘ 冇t冈, (b) r⬙ 冇t冈, and (c) r⬘ 冇t冈 ⭈ r⬙ 冇t冈.
d 冇5t冈u冇t冈 dt
40. r 共t兲 ⫽ ti ⫹ 2 sin tj ⫹ 2 cos tk
10. r共t兲 ⫽ 冪t i ⫹ 共1 ⫺ t3兲j
11. r共t兲 ⫽ 具2 cos t, 5 sin t典
d [r冇t冈 ⭈ u冇t冈] dt
(b)
39. r 共t兲 ⫽ ti ⫹ 3tj ⫹ t 2k,
Finding a Derivative In Exercises 9–20, find r⬘ 冇t冈.
t 3i
1 i ⫹ 3tj t⫺1
31. r 共兲 ⫽ 2 cos 3 i ⫹ 3 sin 3 j
34. r 共t兲 ⫽
Differentiation of Vector-Valued Functions In Exercises 7 and 8, find r⬘冇t冈, r冇t0冈, and r⬘冇t0冈 for the given value of t0. Then
8. r 共t兲 ⫽ ti ⫹ t 2j ⫹ 32k,
30. r共t兲 ⫽
33. r 共兲 ⫽ 共 ⫺ 2 sin 兲 i ⫹ 共1 ⫺ 2 cos 兲j
2
4. r共t兲 ⫽ 3 sin ti ⫹ 4 cos tj, t0 ⫽ t0 ⫽ 0
29. r 共t兲 ⫽ t 2 i ⫹ t 3j
32. r 共兲 ⫽ 共 ⫹ sin 兲i ⫹ 共1 ⫺ cos 兲j
1. r 共t兲 ⫽ t 2 i ⫹ tj, t0 ⫽ 2
5. r共t兲 ⫽ 具et, e2t典,
Finding Intervals on Which a Curve Is Smooth In Exercises 29–38, find the open interval(s) on which the curve given by the vector-valued function is smooth.
49. 50.
冕 冕冢 冕冤 冕 冕冢 冕
共2ti ⫹ j ⫹ k兲 dt
44.
冣
1 i ⫹ j ⫺ t 3兾2 k dt t
46.
冕共 冕冢
4t 3 i ⫹ 6tj ⫺ 4冪t k兲 dt ln ti ⫹
冣
1 j ⫹ k dt t
冥
共2t ⫺ 1兲i ⫹ 4t 3j ⫹ 3冪t k dt
共et i ⫹ sin tj ⫹ cos tk兲 dt sec2 ti ⫹
冣
1 j dt 1 ⫹ t2
共e⫺t sin ti ⫹ e⫺t cos tj兲 dt
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12.2
Differentiation and Integration of Vector-Valued Functions
Evaluating a Definite Integral In Exercises 51– 56, evaluate the definite integral.
冕 冕 冕 冕
冕
1
51.
1
共8ti ⫹ tj ⫺ k兲 dt
52.
⫺1
0
兾2
53.
共ti ⫹ t 3j ⫹ 冪3 t k兲 dt
关共a cos t兲 i ⫹ 共a sin t兲 j ⫹ k兴 dt
0
兾4
54.
关共sec t tan t兲i ⫹ 共tan t兲j ⫹ 共2 sin t cos t兲k兴 dt
0
冕
2
55.
73.
56.
0
储t i ⫹ t 2 j储 dt
0
Finding an Antiderivative In Exercises 57– 62, find r冇t冈
d 再r 共t兲 ⭈ 关u共t兲 ⫻ v共t兲兴冎 ⫽ r⬘ 共t兲 ⭈ 关u共t兲 ⫻ v 共t兲兴 ⫹ dt r 共t兲 ⭈ 关u⬘ 共t兲 ⫻ v 共t兲兴 ⫹ r 共t兲 ⭈ 关u共t兲 ⫻ v⬘ 共t兲兴
74. If r 共t兲 ⭈ r 共t兲 is a constant, then r 共t兲 ⭈ r⬘ 共t兲 ⫽ 0. 75. Particle Motion A particle moves in the xy-plane along the curve represented by the vector-valued function r共t兲 ⫽ 共t ⫺ sin t兲i ⫹ 共1 ⫺ cos t兲j. (a) Use a graphing utility to graph r. Describe the curve. (b) Find the minimum and maximum values of 储r⬘ 储 and 储r⬙ 储.
3
共ti ⫹ et j ⫺ te t k兲 dt
831
76. Particle Motion A particle moves in the yz-plane along the curve represented by the vector-valued function r共t兲 ⫽ 共2 cos t兲j ⫹ 共3 sin t兲k.
that satisifies the initial condition(s).
(a) Describe the curve.
57. r⬘ 共t兲 ⫽ 4e2t i ⫹ 3et j, r共0兲 ⫽ 2i
(b) Find the minimum and maximum values of 储r⬘ 储 and 储r⬙ 储.
58. r⬘ 共t兲 ⫽ 3t 2 j ⫹ 6冪t k, 59. r⬙ 共t兲 ⫽ ⫺32j,
r 共0兲 ⫽ i ⫹ 2j
r⬘ 共0兲 ⫽ 600冪3i ⫹ 600j, r共0兲 ⫽ 0
60. r⬙ 共t兲 ⫽ ⫺4 cos tj ⫺ 3 sin tk,
r⬘ 共0兲 ⫽ 3k, r 共0兲 ⫽ 4 j
1 61. r⬘ 共t兲 ⫽ te⫺t i ⫺ e⫺t j ⫹ k, r 共0兲 ⫽ 2i ⫺ j ⫹ k
77. Perpendicular Vectors Consider the vector-valued function r共t兲 ⫽ 共et sin t兲i ⫹ 共et cos t兲j. Show that r共t兲 and r⬙共t兲 are always perpendicular to each other.
2
62. r⬘ 共t兲 ⫽
1 1 1 i ⫹ 2 j ⫹ k, r共1兲 ⫽ 2i 1 ⫹ t2 t t
78.
HOW DO YOU SEE IT? The graph shows a vector-valued function r共t兲 for 0 ⱕ t ⱕ 2 and its derivative r⬘共t兲 for several values of t. y
WRITING ABOUT CONCEPTS 63. Differentiation State the definition of the derivative of a vector-valued function. Describe how to find the derivative of a vector-valued function and give its geometric interpretation.
4
π t=5 6
−5
65. Using a Derivative The three components of the derivative of the vector-valued function u are positive at t ⫽ t0. Describe the behavior of u at t ⫽ t0.
−2 −1 −1
t= 1
2
π 4
x
3
−2
π t=5 4
66. Using a Derivative The z-component of the derivative of the vector-valued function u is 0 for t in the domain of the function. What does this imply about the graph of u?
−4
(a) For each derivative shown in the graph, determine whether each component is positive or negative. (b) Is the curve smooth on the interval 关0, 2兴? Explain.
Proof In Exercises 67–74, prove the property. In each case,
d 67. 关cr共t兲兴 ⫽ cr⬘ 共t兲 dt
2 1
64. Integration How do you find the integral of a vectorvalued function?
assume r, u, and v are differentiable vector-valued functions of t in space, w is a differentiable real-valued function of t, and c is a scalar.
3
True or False? In Exercises 79–82, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.
68.
d 关r共t兲 ± u共t兲兴 ⫽ r⬘ 共t兲 ± u⬘ 共t兲 dt
79. If a particle moves along a sphere centered at the origin, then its derivative vector is always tangent to the sphere.
69.
d 关w 共t兲r共t兲兴 ⫽ w 共t兲r⬘ 共t兲 ⫹ w⬘共t兲r共t兲 dt
80. The definite integral of a vector-valued function is a real number.
70.
d 关r 共t兲 ⫻ u共t兲兴 ⫽ r 共t兲 ⫻ u⬘ 共t兲 ⫹ r⬘ 共t兲 ⫻ u共t兲 dt
81.
71.
d 关r共w 共t兲兲兴 ⫽ r⬘ 共w 共t兲兲w⬘共t兲 dt
82. If r and u are differentiable vector-valued functions of t, then
72.
d 关r 共t兲 ⫻ r⬘ 共t兲兴 ⫽ r 共t兲 ⫻ r⬙ 共t兲 dt
d 关储r共t兲储兴 ⫽ 储r⬘共t兲储 dt
d 关r共t兲 ⭈ u共t兲兴 ⫽ r⬘共t兲 ⭈ u⬘共t兲. dt
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
832
Chapter 12
Vector-Valued Functions
12.3 Velocity and Acceleration Describe the velocity and acceleration associated with a vector-valued function. Use a vector-valued function to analyze projectile motion.
Velocity and Acceleration Exploration Exploring Velocity the circle given by
Consider
rt cos ti sin tj. (The symbol is the Greek letter omega.) Use a graphing utility in parametric mode to graph this circle for several values of . How does affect the velocity of the terminal point as it traces out the curve? For a given value of , does the speed appear constant? Does the acceleration appear constant? Explain your reasoning. 2
−3
3
−2
You are now ready to combine your study of parametric equations, curves, vectors, and vector-valued functions to form a model for motion along a curve. You will begin by looking at the motion of an object in the plane. (The motion of an object in space can be developed similarly.) As an object moves along a curve in the plane, the coordinates x and y of its center of mass are each functions of time t. Rather than using the letters f and g to represent these two functions, it is convenient to write x xt and y yt. So, the position vector rt takes the form rt xti ytj.
Position vector
The beauty of this vector model for representing motion is that you can use the first and second derivatives of the vector-valued function r to find the object’s velocity and acceleration. (Recall from the preceding chapter that velocity and acceleration are both vector quantities having magnitude and direction.) To find the velocity and acceleration vectors at a given time t, consider a point Qxt t, yt t that is approaching the point Pxt, yt along the curve C given by rt xti ytj, as shown in Figure 12.11. As t → 0, the direction of the vector PQ (denoted by r) approaches the direction of motion at time t. \
r rt t rt r rt t rt t t r rt t rt lim lim t→0 t t→0 t If this limit exists, it is defined as the velocity vector or tangent vector to the curve at point P. Note that this is the same limit used to define r t. So, the direction of r t gives the direction of motion at time t. Moreover, the magnitude of the vector r t r t xti ytj xt 2 yt 2 gives the speed of the object at time t. Similarly, you can use r t to find acceleration, as indicated in the definitions at the top of the next page. y
y
P C
Δr
Δt → 0
Velocity vector at time t
Velocity vector at time t Q
r(t) r(t + Δt) x
x
r approaches the velocity vector. t Figure 12.11 As t → 0,
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12.3
Velocity and Acceleration
833
Definitions of Velocity and Acceleration If x and y are twice-differentiable functions of t, and r is a vector-valued function given by rt xti ytj, then the velocity vector, acceleration vector, and speed at time t are as follows. Velocity vt rt xti ytj Acceleration at r t x ti y tj Speed vt rt xt 2 yt 2
For motion along a space curve, the definitions are similar. That is, for rt xti ytj ztk, you have Velocity vt r t xti ytj ztk Acceleration at r t x ti y tj z tk Speed vt r t xt 2 yt 2 zt 2.
Velocity and Acceleration Along a Plane Curve REMARK In Example 1, note
Find the velocity vector, speed, and acceleration vector of a particle that moves along that the velocity and acceleration the plane curve C described by vectors are orthogonal at any t t point in time. This is characteristic rt 2 sin i 2 cos j. Position vector 2 2 of motion at a constant speed. (See Exercise 53.) Solution The velocity vector is t t vt rt cos i sin j. 2 2
Velocity vector
The speed (at any time) is rt
cos 2t sin 2t 1. 2
2
Speed
The acceleration vector is
Circle: x 2 + y 2 = 4
t 1 t 1 at r t sin i cos j. 2 2 2 2
y
Acceleration vector
2
a(t)
The parametric equations for the curve in Example 1 are
v(t)
1
x 2 sin x −2
−1
1
2
−1
and
t y 2 cos . 2
By eliminating the parameter t, you obtain the rectangular equation x 2 y 2 4.
Rectangular equation
So, the curve is a circle of radius 2 centered at the origin, as shown in Figure 12.12. Because the velocity vector
−2
r(t) = 2 sin
t 2
t t i + 2 cos j 2 2
The particle moves around the circle at a constant speed. Figure 12.12
t t vt cos i sin j 2 2 has a constant magnitude but a changing direction as t increases, the particle moves around the circle at a constant speed.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
834
Chapter 12
Vector-Valued Functions
Velocity and Acceleration Vectors in the Plane
r(t) = (t 2 − 4)i + tj
Sketch the path of an object moving along the plane curve given by
y
rt t 2 4i t j
4
v(2)
3
and find the velocity and acceleration vectors when t 0 and t 2.
a(2)
v(0)
Solution Using the parametric equations x t 2 4 and y t, you can determine that the curve is a parabola given by
1
a(0)
x − 3 −2 − 1 −1
1
2
3
Position vector
x y2 4
4
Rectangular equation
as shown in Figure 12.13. The velocity vector (at any time) is −3
vt rt 2t i j
x = y2 − 4
−4
Velocity vector
and the acceleration vector (at any time) is
At each point on the curve, the acceleration vector points to the right. Figure 12.13
at r t 2i.
Acceleration vector
When t 0, the velocity and acceleration vectors are v0 20i j j and
y
a0 2i.
When t 2, the velocity and acceleration vectors are v2 22i j 4i j and Sun
a2 2i.
For the object moving along the path shown in Figure 12.13, note that the acceleration vector is constant (it has a magnitude of 2 and points to the right). This implies that the speed of the object is decreasing as the object moves toward the vertex of the parabola, and the speed is increasing as the object moves away from the vertex of the parabola. This type of motion is not characteristic of comets that travel on parabolic paths through our solar system. For such comets, the acceleration vector always points to the origin (the sun), which implies that the comet’s speed increases as it approaches the vertex of the path and decreases as it moves away from the vertex. (See Figure 12.14.)
x
a
At each point in the comet’s orbit, the acceleration vector points toward the sun. Figure 12.14
Velocity and Acceleration Vectors in Space See LarsonCalculus.com for an interactive version of this type of example.
Sketch the path of an object moving along the space curve C given by rt t i t 3j 3tk, t 0
Position vector
and find the velocity and acceleration vectors when t 1. Solution Using the parametric equations x t and y t 3, you can determine that the path of the object lies on the cubic cylinder given by
Curve: r(t) = ti + t 3 j + 3tk, t ≥ 0 C
z
y x3.
v(1)
6
(1, 1, 3) 4
y
a(1)
Moreover, because z 3t, the object starts at 0, 0, 0 and moves upward as t increases, as shown in Figure 12.15. Because rt t i t 3j 3tk, you have
10
2
Rectangular equation
vt rt i 3t 2j 3k
Velocity vector
at r t 6tj.
Acceleration vector
and 2
y= 4 x
Figure 12.15
x3
When t 1, the velocity and acceleration vectors are v1 r1 i 3j 3k and
a1 r 1 6j.
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12.3
Velocity and Acceleration
835
So far in this section, you have concentrated on finding the velocity and acceleration by differentiating the position vector. Many practical applications involve the reverse problem—finding the position vector for a given velocity or acceleration. This is demonstrated in the next example.
Finding a Position Vector by Integration An object starts from rest at the point 1, 2, 0 and moves with an acceleration of at j 2k
Acceleration vector
where at is measured in feet per second per second. Find the location of the object after t 2 seconds. Solution From the description of the object’s motion, you can deduce the following initial conditions. Because the object starts from rest, you have v0 0. Moreover, because the object starts at the point x, y, z 1, 2, 0, you have r0 x0i y0j z0k 1i 2j 0k i 2j. To find the position vector, you should integrate twice, each time using one of the initial conditions to solve for the constant of integration. The velocity vector is vt
at dt
j 2k dt
tj 2tk C where C C1i C2 j C3k. Letting t 0 and applying the initial condition v0 0, you obtain v0 C1i C2 j C3k 0
C1 C2 C3 0.
So, the velocity at any time t is vt t j 2tk.
Velocity vector
Integrating once more produces Curve:
r(t) = i +
rt
(t2 + 2( j + t k 2
2
z
6
4 2
2 4
r(2)
(1, 2, 0) t=0
vt dt
t j 2tk dt
t2 j t2k C 2
where C C4i C5 j C6k. Letting t 0 and applying the initial condition r0 i 2j, you have
(1, 4, 4) t=2
6
y
6 x
The object takes 2 seconds to move from point 1, 2, 0 to point 1, 4, 4 along the curve. Figure 12.16
r0 C4i C5 j C6k i 2j
C4 1, C5 2, C6 0.
So, the position vector is rt i
t2 2 j t 2 k. 2
Position vector
The location of the object after t 2 seconds is given by r2 i 4j 4k as shown in Figure 12.16.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
836
Chapter 12
Vector-Valued Functions
Projectile Motion y
You now have the machinery to derive the parametric equations for the path of a projectile. Assume that gravity is the only force acting on the projectile after it is launched. So, the motion occurs in a vertical plane, which can be represented by the xy-coordinate system with the origin as a point on Earth’s surface, as shown in Figure 12.17. For a projectile of mass m, the force due to gravity is
v0 = Initial velocity v(t1)
v0 = v(0)
a
a a
F mgj
v(t2)
Force due to gravity
where the acceleration due to gravity is g 32 feet per second per second, or 9.81 meters per second per second. By Newton’s Second Law of Motion, this same force produces an acceleration a at and satisfies the equation F ma. Consequently, the acceleration of the projectile is given by ma mg j, which implies that
Initial height x
Figure 12.17
a gj.
Acceleration of projectile
Derivation of the Position Vector for a Projectile A projectile of mass m is launched from an initial position r0 with an initial velocity v0. Find its position vector as a function of time. Solution vt rt
Begin with the acceleration at gj and integrate twice.
at dt vt dt
g j dt gt j C1
1 gtj C1 dt gt 2j C1t C2 2
You can use the facts that v0 v0 and r0 r0 to solve for the constant vectors C1 and C2. Doing this produces C1 v0
and C2 r0.
Therefore, the position vector is 1 rt gt 2j t v0 r0. 2
In many projectile problems, the constant vectors r0 and v0 are not given explicitly. Often you are given the initial height h, the initial speed v0, and the angle at which the projectile is launched, as shown in Figure 12.18. From the given height, you can deduce that r0 hj. Because the speed gives the magnitude of the initial velocity, it follows that v0 v0 and you can write
⎜⎜v0 ⎜⎜= v0 = initial speed ⎜⎜r0 ⎜⎜= h = initial height
v0 x i y j v0 cos i v0 sin j v0 cos i v0 sin j.
y
v0
So, the position vector can be written in the form
yj θ
1 rt gt2j tv0 r0 2
xi h
Position vector
Position vector
r0 x
x = ⎜⎜v0 ⎜⎜cos θ y = ⎜⎜v0 ⎜⎜sin θ
Figure 12.18
1 gt 2j tv0 cos i tv0 sin j hj 2
1 v0 cos t i h v0 sin t gt 2 j. 2
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12.3
Velocity and Acceleration
837
THEOREM 12.3 Position Vector for a Projectile Neglecting air resistance, the path of a projectile launched from an initial height h with initial speed v0 and angle of elevation is described by the vector function rt v0 cos ti h v0 sin t 12gt 2 j where g is the acceleration due to gravity.
Describing the Path of a Baseball A baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground, as shown in Figure 12.19. Find the maximum height reached by the baseball. Will it clear a 10-foot-high fence located 300 feet from home plate? Solution
10 ft 45° 300 ft 3 ft
You are given
h 3, v0 100, and
45 .
Figure 12.19
So, using Theorem 12.3 with g 32 feet per second per second produces
t i 3 100 sin t 16t 2 j 4 4 2 502ti 3 502t 16t j.
rt 100 cos
The velocity vector is vt rt 502i 502 32tj. The maximum height occurs when yt 502 32t is equal to 0, which implies that t
252
2.21 seconds. 16
So, the maximum height reached by the ball is y 3 502 649 8
81 feet.
252 252 16 16 16
2
Maximum height when t 2.21 seconds
The ball is 300 feet from where it was hit when 300 xt
300 502t.
Solving this equation for t produces t 32 4.24 seconds. At this time, the height of the ball is y 3 502 32 1632 303 288 15 feet. Height when t 4.24 seconds 2
Therefore, the ball clears the 10-foot fence for a home run.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
838
Chapter 12
Vector-Valued Functions
12.3 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding Velocity and Acceleration Along a Plane Curve
20. at 2i 3k, v0 4j,
r0 0
In Exercises 1–8, the position vector r describes the path of an object moving in the xy-plane.
21. at t j t k, v1 5j,
r1 0
(a) Find the velocity vector, speed, and acceleration vector of the object.
23. at cos t i sin t j, v0 j k, r0 i
(b) Evaluate the velocity vector and acceleration vector of the object at the given point.
Projectile Motion In Exercises 25–38, use the model for
(c) Sketch a graph of the path, and sketch the velocity and acceleration vectors at the given point. Position Vector
Point
1. rt 3t i t 1j
3, 0
2. rt t i
1, 3
4 j
t2
3. rt
t2i
4, 2
tj
1 4. rt 4 t3 1i tj
3, 2
5. rt 2 cos t i 2 sin t j
2, 2
6. rt 3 cos t i 2 sin t j
3, 0
7. rt t sin t, 1 cos t
, 2
8. rt et, et
1, 1
Finding Velocity and Acceleration Vectors In Exercises 9–18, the position vector r describes the path of an object moving in space.
22. at 32 k,
v0 3i 2j k, r0 5j 2k
24. at et i 8k, v0 2i 3j k, r0 0 projectile motion, assuming there is no air resistance. 25. A baseball is hit from a height of 2.5 feet above the ground with an initial velocity of 140 feet per second and at an angle of 22 above the horizontal. Find the maximum height reached by the baseball. Determine whether it will clear a 10-foot-high fence located 375 feet from home plate. 26. Determine the maximum height and range of a projectile fired at a height of 3 feet above the ground with an initial velocity of 900 feet per second and at an angle of 45 above the horizontal. 27. A baseball, hit 3 feet above the ground, leaves the bat at an angle of 45 and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the initial speed of the ball, and how high does it rise?
(a) Find the velocity vector, speed, and acceleration vector of the object.
28. A baseball player at second base throws a ball 90 feet to the player at first base. The ball is released at a point 5 feet above the ground with an initial velocity of 50 miles per hour and at an angle of 15 above the horizontal. At what height does the player at first base catch the ball?
(b) Evaluate the velocity vector and acceleration vector of the object at the given value of t.
29. Eliminate the parameter t from the position vector for the motion of a projectile to show that the rectangular equation is
Position Vector
Time
y
16 sec2 2 x tan x h. v02
9. rt t i 5tj 3t k
t1
10. rt 4t i 4t j 2t k
t3
1 11. rt t i t 2j 2t 2k
t4
y x 0.005x 2.
t2
Use the result of Exercise 29 to find the position vector. Then find the speed and direction of the ball at the point at which it has traveled 60 feet horizontally.
12. rt 3t i t j
1 2 4t k
13. rt t i t j 9
t2
k
t0
14. rt t 2 i t j 2t 32 k
t4
15. rt 4t, 3 cos t, 3 sin t
t
16. rt 2 cos t, 2 sin t, t2
t
17. rt et cos t, et sin t, et
t0
1 18. rt ln t, , t 4 t
4
t2
Finding a Position Vector by Integration In Exercises 19–24, use the given acceleration vector to find the velocity and position vectors. Then find the position at time t ⴝ 2. 19. at i j k, v0 0,
r0 0
30. The path of a ball is given by the rectangular equation
31. The Rogers Centre in Toronto, Ontario, has a center field fence that is 10 feet high and 400 feet from home plate. A ball is hit 3 feet above the ground and leaves the bat at a speed of 100 miles per hour. (a) The ball leaves the bat at an angle of 0 with the horizontal. Write the vector-valued function for the path of the ball. (b) Use a graphing utility to graph the vector-valued function for 0 10 , 0 15 , 0 20 , and 0 25 . Use the graphs to approximate the minimum angle required for the hit to be a home run. (c) Determine analytically the minimum angle required for the hit to be a home run.
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12.3 32. Football The quarterback of a football team releases a pass at a height of 7 feet above the playing field, and the football is caught by a receiver 30 yards directly downfield at a height of 4 feet. The pass is released at an angle of 35 with the horizontal. (a) Find the speed of the football when it is released. (b) Find the maximum height of the football. (c) Find the time the receiver has to reach the proper position after the quarterback releases the football.
Velocity and Acceleration
839
38. Find the angles at which an object must be thrown to obtain (a) the maximum range and (b) the maximum height.
Projectile Motion In Exercises 39 and 40, use the model for projectile motion, assuming there is no air resistance. [ g ⴝ ⴚ9.8 meters per second per second] 39. Determine the maximum height and range of a projectile fired at a height of 1.5 meters above the ground with an initial velocity of 100 meters per second and at an angle of 30 above the horizontal. 40. A projectile is fired from ground level at an angle of 8 with the horizontal. The projectile is to have a range of 50 meters. Find the minimum initial velocity necessary. 41. Shot-Put Throw The path of a shot thrown at an angle is
1 rt v0 cos t i h v0 sin t gt2 j 2 33. A bale ejector consists of two variable-speed belts at the end of a baler. Its purpose is to toss bales into a trailing wagon. In loading the back of a wagon, a bale must be thrown to a position 8 feet above and 16 feet behind the ejector.
where v0 is the initial speed, h is the initial height, t is the time in seconds, and g is the acceleration due to gravity. Verify that the shot will remain in the air for a total of v0 sin v02 sin2 2gh seconds g
(a) Find the minimum initial speed of the bale and the corresponding angle at which it must be ejected from the baler.
t
(b) The ejector has a fixed angle of 45 . Find the initial speed required.
and will travel a horizontal distance of
34. A bomber is flying at an altitude of 30,000 feet at a speed of 540 miles per hour (see figure). When should the bomb be released for it to hit the target? (Give your answer in terms of the angle of depression from the plane to the target.) What is the speed of the bomb at the time of impact? 540 mi/h
30,000 ft
35. A shot fired from a gun with a muzzle velocity of 1200 feet per second is to hit a target 3000 feet away. Determine the minimum angle of elevation of the gun. 36. A projectile is fired from ground level at an angle of 12 with the horizontal. The projectile is to have a range of 200 feet. Find the minimum initial velocity necessary. 37. Use a graphing utility to graph the paths of a projectile for the given values of and v0. For each case, use the graph to approximate the maximum height and range of the projectile. (Assume that the projectile is launched from ground level.) (a) 10 , v0 66 ftsec (b) 10 , v0 146 ftsec (c) 45 , v0 66 ftsec (d) 45 , v0 146 ftsec (e) 60 , v0 66 ftsec
v02 cos
sin g
feet. sin 2gh v 2
2 0
42. Shot-Put Throw A shot is thrown from a height of h 6 feet with an initial speed of v0 45 feet per second and at an angle of
42.5 with the horizontal. Use the result of Exercise 41 to find the total time of travel and the total horizontal distance traveled.
Cycloidal Motion In Exercises 43 and 44, consider the motion of a point (or particle) on the circumference of a rolling circle. As the circle rolls, it generates the cycloid rt ⴝ b t ⴚ sin ti ⴙ b1 ⴚ cos tj where is the constant angular velocity of the circle and b is the radius of the circle. 43. Find the velocity and acceleration vectors of the particle. Use the results to determine the times at which the speed of the particle will be (a) zero and (b) maximized. 44. Find the maximum speed of a point on the circumference of an automobile tire of radius 1 foot when the automobile is traveling at 60 miles per hour. Compare this speed with the speed of the automobile. Nicholas Moore/Shutterstock.com; Jamie Roach/Shutterstock.com
(f) 60 , v0 146 ftsec
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Circular Motion In Exercises 45–48, consider a particle moving on a circular path of radius b described by rt ⴝ b cos t i ⴙ b sin t j, where ⴝ du/dt is the constant angular velocity. 45. Find the velocity vector and show that it is orthogonal to rt.
55. Investigation A particle moves on an elliptical path given by the vector-valued function rt 6 cos t i 3 sin t j. (a) Find vt, vt , and at. (b) Use a graphing utility to complete the table.
46. (a) Show that the speed of the particle is b.
t
(b) Use a graphing utility in parametric mode to graph the circle for b 6. Try different values of . Does the graphing utility draw the circle faster for greater values of ?
0
4
2
2 3
Speed
47. Find the acceleration vector and show that its direction is always toward the center of the circle.
(c) Graph the elliptical path and the velocity and acceleration vectors at the values of t given in the table in part (b).
48. Show that the magnitude of the acceleration vector is b2.
(d) Use the results of parts (b) and (c) to describe the geometric relationship between the velocity and acceleration vectors when the speed of the particle is increasing, and when it is decreasing.
Circular Motion In Exercises 49 and 50, use the results of Exercises 45–48. 49. A stone weighing 1 pound is attached to a two-foot string and is whirled horizontally (see figure). The string will break under a force of 10 pounds. Find the maximum speed the stone can attain without breaking the string. Use F ma, where 1 m 32 .
1 lb 30 mi/h
2 ft
56. Particle Motion Consider a particle moving on an elliptical path described by rt a cos t i b sin t j, where d dt is the constant angular velocity. (a) Find the velocity vector. What is the speed of the particle? (b) Find the acceleration vector and show that its direction is always toward the center of the ellipse. 57. Path of an Object When t 0, an object is at the point 0, 1 and has a velocity vector v0 i. It moves with an acceleration of at sin t i cos t j. Show that the path of the object is a circle.
58. 300 ft Figure for 49
Figure for 50
50. A 3400-pound automobile is negotiating a circular interchange of radius 300 feet at 30 miles per hour (see figure). Assuming the roadway is level, find the force between the tires and the road such that the car stays on the circular path and does not skid. (Use F ma, where m 340032.) Find the angle at which the roadway should be banked so that no lateral frictional force is exerted on the tires of the automobile.
HOW DO YOU SEE IT? The graph shows the path of a projectile and the velocity and acceleration vectors at times t1 and t2. Classify the angle between the velocity vector and the acceleration vector at times t1 and t2. Is the speed increasing or decreasing at times t1 and t2? Explain your reasoning. y
v(t1) a(t1)
WRITING ABOUT CONCEPTS 51. Velocity and Speed In your own words, explain the difference between the velocity of an object and its speed. 52. Particle Motion Consider a particle that is moving on the path r1t xt i yt j zt k.
a(t2)
v(t2)
x
(a) Discuss any changes in the position, velocity, or acceleration of the particle when its position is given by the vector-valued function r2t r12t.
True or False? In Exercises 59–62, determine whether the
(b) Generalize the results for the vector-valued function r3t r1t.
statement is true or false. If it is false, explain why or give an example that shows it is false. 59. The acceleration of an object is the derivative of the speed.
53. Proof Prove that when an object is traveling at a constant speed, its velocity and acceleration vectors are orthogonal. 54. Proof Prove that an object moving in a straight line at a constant speed has an acceleration of 0.
60. The velocity of an object is the derivative of the position. 61. The velocity vector points in the direction of motion. 62. If a particle moves along a straight line, then the velocity and acceleration vectors are orthogonal.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12.4
Tangent Vectors and Normal Vectors
841
12.4 Tangent Vectors and Normal Vectors Find a unit tangent vector and a principal unit normal vector at a point on a space curve. Find the tangential and normal components of acceleration.
Tangent Vectors and Normal Vectors In the preceding section, you learned that the velocity vector points in the direction of motion. This observation leads to the next definition, which applies to any smooth curve—not just to those for which the parameter represents time. Definition of Unit Tangent Vector Let C be a smooth curve represented by r on an open interval I. The unit tangent vector Tt at t is defined as Tt
rt , rt
rt 0.
Recall that a curve is smooth on an interval when r is continuous and nonzero on the interval. So, “smoothness” is sufficient to guarantee that a curve has a unit tangent vector.
Finding the Unit Tangent Vector Find the unit tangent vector to the curve given by rt ti t 2j when t 1.
y
Solution 4
rt i 2tj.
3
Tt T(1)
1
x −1
Derivative of rt
So, the unit tangent vector is
2
−2
The derivative of rt is
1
2
r(t) = ti + t 2 j
The direction of the unit tangent vector depends on the orientation of the curve. Figure 12.20
rt rt 1 1 4t 2
Definition of Tt
i 2t j.
Substitute for r t.
When t 1, the unit tangent vector is T1
1 i 2j 5
as shown in Figure 12.20. In Example 1, note that the direction of the unit tangent vector depends on the orientation of the curve. For the parabola described by rt t 2i t 2 2j T1 would still represent the unit tangent vector at the point 1, 1, but it would point in the opposite direction. Try verifying this.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Vector-Valued Functions
The tangent line to a curve at a point is the line that passes through the point and is parallel to the unit tangent vector. In Example 2, the unit tangent vector is used to find the tangent line at a point on a helix.
Finding the Tangent Line at a Point on a Curve Find Tt and then find a set of parametric equations for the tangent line to the helix given by rt 2 cos t i 2 sin t j tk
. 4
at the point 2, 2,
The derivative of rt is
Solution
rt 2 sin t i 2 cos t j k which implies that rt 4 sin2 t 4 cos2 t 1 5. Therefore, the unit tangent vector is
Curve: r(t) = 2 cos ti + 2 sin tj + tk
rt rt 1 2 sin t i 2 cos t j k. 5
Tt
z 6
Unit tangent vector
At the point 2, 2, 4, t 4 and the unit tangent vector is
5
C
T Tangent line
4 15 2 22 i 2 22 j k
1 5
2 i 2 j k.
Using the direction numbers a 2, b 2, and c 1, and the point x1, y1, z1 2, 2, 4, you can obtain the parametric equations (given with parameter s) listed below.
−3
3 x
)
2,
π 2, 4
)
3
y
The tangent line to a curve at a point is determined by the unit tangent vector at the point. Figure 12.21
x x1 as 2 2s y y1 bs 2 2s z z1 cs s 4 This tangent line is shown in Figure 12.21. In Example 2, there are infinitely many vectors that are orthogonal to the tangent vector Tt. One of these is the vector Tt. This follows from Property 7 of Theorem 12.2. That is, Tt Tt Tt 2 1
Tt Tt 0.
By normalizing the vector Tt, you obtain a special vector called the principal unit normal vector, as indicated in the next definition. Definition of Principal Unit Normal Vector Let C be a smooth curve represented by r on an open interval I. If Tt 0, then the principal unit normal vector at t is defined as Nt
Tt . Tt
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12.4
Tangent Vectors and Normal Vectors
Finding the Principal Unit Normal Vector Find Nt and N1 for the curve represented by rt 3ti 2t 2j. Solution
By differentiating, you obtain
rt 3i 4tj which implies that rt 9 16t2. So, the unit tangent vector is rt rt 1 3i 4t j. 9 16t2
Tt
Unit tangent vector
Using Theorem 12.2, differentiate Tt with respect to t to obtain 1 16t 4j 3i 4t j 2 9 16t 232 9 16t 12 4t i 3j 9 16t 232
Tt
which implies that 16t 99 16t 2
Tt 12
2 3
12 . 9 16t 2
Therefore, the principal unit normal vector is Tt Tt 1 4ti 3j. 9 16t 2
Nt
Principal unit normal vector
When t 1, the principal unit normal vector is 1 N1 4i 3j 5 as shown in Figure 12.22. y
3
Curve: r(t) = 3ti + 2t 2 j
C
N(1) = 15 (− 4i + 3j)
2
1
T(1) = 15 (3i + 4j) x 1
2
3
The principal unit normal vector points toward the concave side of the curve. Figure 12.22
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
843
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Vector-Valued Functions
The principal unit normal vector can be difficult to evaluate algebraically. For plane curves, you can simplify the algebra by finding Tt xti ytj
Unit tangent vector
and observing that Nt must be either N1t yti xtj
N2t yti xtj.
or
Because xt 2 yt 2 1, it follows that both N1t and N 2t are unit normal vectors. The principal unit normal vector N is the one that points toward the concave side of the curve, as shown in Figure 12.22 (see Exercise 76). This also holds for curves in space. That is, for an object moving along a curve C in space, the vector Tt points in the direction the object is moving, whereas the vector Nt is orthogonal to Tt and points in the direction in which the object is turning, as shown in Figure 12.23. z
C
T
x
N
y
At any point on a curve, a unit normal vector is orthogonal to the unit tangent vector. The principal unit normal vector points in the direction in which the curve is turning. Figure 12.23
Helix: r(t) = 2 cos ti + 2 sin tj + tk z
Finding the Principal Unit Normal Vector
2π
Find the principal unit normal vector for the helix rt 2 cos t i 2 sin tj tk. 3π 2
Solution Tt
π
−1
Tt
x
Unit tangent vector
1 5
2 cos t i 2 sin t j.
Because Tt 25, it follows that the principal unit normal vector is
−2 −1
Tt Tt 1 2 cos t i 2 sin t j 2 cos t i sin t j.
Nt
1 2
1 2 sin t i 2 cos t j k. 5
So, Tt is given by
π 2 −2
From Example 2, you know that the unit tangent vector is
1 2
y
Nt is horizontal and points toward the z-axis. Figure 12.24
Principal unit normal vector
Note that this vector is horizontal and points toward the z-axis, as shown in Figure 12.24.
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12.4
Tangent Vectors and Normal Vectors
845
Tangential and Normal Components of Acceleration In the preceding section, you considered the problem of describing the motion of an object along a curve. You saw that for an object traveling at a constant speed, the velocity and acceleration vectors are perpendicular. This seems reasonable, because the speed would not be constant if any acceleration were acting in the direction of motion. You can verify this observation by noting that r t rt 0 when rt is a constant. (See Property 7 of Theorem 12.2.) For an object traveling at a variable speed, however, the velocity and acceleration vectors are not necessarily perpendicular. For instance, you saw that the acceleration vector for a projectile always points down, regardless of the direction of motion. In general, part of the acceleration (the tangential component) acts in the line of motion, and part of it (the normal component) acts perpendicular to the line of motion. In order to determine these two components, you can use the unit vectors Tt and Nt, which serve in much the same way as do i and j in representing vectors in the plane. The next theorem states that the acceleration vector lies in the plane determined by Tt and Nt. THEOREM 12.4 Acceleration Vector If rt is the position vector for a smooth curve C and Nt exists, then the acceleration vector at lies in the plane determined by Tt and Nt. Proof To simplify the notation, write T for Tt, T for Tt, and so on. Because T rr vv, it follows that v vT. By differentiating, you obtain a v
Product Rule
d v T vT dt d T v T vT dt T d v T v T N. dt
N T T
Because a is written as a linear combination of T and N, it follows that a lies in the plane determined by T and N. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
The coefficients of T and N in the proof of Theorem 12.4 are called the tangential and normal components of acceleration and are denoted by aT
d v
dt
and aN v T . So, you can write at aTTt aNNt. The next theorem lists some convenient formulas for aN and a T.
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Chapter 12
Vector-Valued Functions
THEOREM 12.5
Tangential and Normal Components of Acceleration If rt is the position vector for a smooth curve C [for which Nt exists], then the tangential and normal components of acceleration are as follows. d va v a T dt v v a aN v T a N a2 a T2 v
aT a
a•T>0
T
N a•N
Note that aN 0. The normal component of acceleration is also called the centripetal component of acceleration.
T a•N N a
a•T 0.
Solution
aN = b
vt rt b sin t i b cos t j ck vt b 2 sin2 t b 2 cos2 t c2 b 2 c 2 at r t b cos t i b sin t j
z
Velocity vector Speed Acceleration vector
By Theorem 12.5, the tangential component of acceleration is b
aT
v a b2 sin t cos t b 2 sin t cos t 0 0. v b 2 c 2
Tangential component of acceleration
Moreover, because a b2 cos2 t b2 sin2 t b y
you can use the alternative formula for the normal component of acceleration to obtain aN a 2 aT2 b2 02 b.
x
The normal component of acceleration is equal to the radius of the cylinder around which the helix is spiraling. Figure 12.26
Normal component of acceleration
Note that the normal component of acceleration is equal to the magnitude of the acceleration. In other words, because the speed is constant, the acceleration is perpendicular to the velocity. See Figure 12.26.
Projectile Motion r(t) = (50
2t)i + (50
The position vector for the projectile shown in Figure 12.27 is
2t − 16t 2)j
rt 502 ti 502t 16t2j.
y
Find the tangential components of acceleration when t 0, 1, and 25216.
100 75 50
t=1
t=
Solution
25 2 16
vt 502 i 502 32t j
25
t=0
Position vector
x 25
50
75 100 125 150
The path of a projectile Figure 12.27
Velocity vector
vt 250 2 16502t 16 2t 2 at 32j
Speed Acceleration vector
The tangential component of acceleration is vt at 32502 32t . vt 2502 16502t 162t 2 At the specified times, you have aTt
Tangential component of acceleration
32502 162 22.6 100 32502 32 a T 1 15.4 250 2 16502 16 2 a T 0
aT
2516 2 3250502 2 50
2
0.
You can see from Figure 12.27 that at the maximum height, when t 25216, the tangential component is 0. This is reasonable because the direction of motion is horizontal at the point and the tangential component of the acceleration is equal to the horizontal component of the acceleration.
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Chapter 12
Vector-Valued Functions
12.4 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding the Unit Tangent Vector In Exercises 1–6, find the unit tangent vector to the curve at the specified value of the parameter. 1. rt t 2 i 2tj,
2. rt t3 i 2t 2j,
t1
3. rt 4 cos ti 4 sin tj,
t
4
4. rt 6 cos ti 2 sin tj,
t
3
t1
Finding a Tangent Line In Exercises 7–12, find the unit tangent vector T t and find a set of parametric equations for the line tangent to the space curve at point P. 8. rt t 2 i t j 43 k,
P1, 1, 43
9. rt 3 cos t i 3 sin tj t k, P3, 0, 0 P1, 1, 3
12. rt 2 sin t, 2 cos t, 4
sin2 t,
32. When the angular velocity is halved, by what factor is a N changed?
Sketching a Graph and Vectors In Exercises 33–36,
Time
1 33. rt t i j t
P1, 3, 1
t0 2
34. rt t3 i tj
Finding the Principal Unit Normal Vector In Exercises 13–20, find the principal unit normal vector to the curve at the specified value of the parameter.
6 14. rt ti j, t
31. Determine the speed of the object at any time t and explain its value relative to the value of a T .
Vector-Valued Function
11. rt 2 cos t, 2 sin t, 4, P2, 2, 4
13. rt ti 12 t 2j,
29. Find Tt, Nt, a T , and a N.
sketch the graph of the plane curve given by the vector-valued function, and, at the point on the curve determined by r t0, sketch the vectors T and N. Note that N points toward the concave side of the curve.
P0, 0, 0
10. rt t, t, 4 t2 ,
r t ⴝ a cos t i ⴙ a sin t j.
30. Determine the directions of T and N relative to the position vector r.
5. rt 3t i ln t j, t e 6. rt et cos ti etj, t 0
7. rt t i t 2j tk,
Circular Motion In Exercises 29–32, consider an object moving according to the position vector
t2
t0 1
35. rt 2t 1i
t0 2
t2j
36. rt 2 cos t i 2 sin tj
t0
4
Finding Vectors In Exercises 37–42, find T t, N t, a T , and aN at the given time t for the space curve r t. [Hint: Find a t, T t, aT, and aN. Solve for N in the equation a t ⴝ aT T ⴙ aN N.]
t3
15. rt ln t i t 1 j, t 2
Vector-Valued Function
Time
16. rt cos t i sin tj, t 6
37. rt t i 2t j 3t k
t1
17. rt t i t2j ln t k, t 1
38. rt cos ti sin tj 2tk
t
18. rt 2 t i et j et k,
t0
39. rt t i t 2 j
3 19. rt 6 cos t i 6 sin tj k, t 4 20. rt cos 3t i 2 sin 3t j k,
t1
22. rt t2 i 2t j,
23. rt t t3i 2t2j, t 1 24. rt t3 4ti t2 1j, t 0 25. rt et i e2t j, 26. rt
et i
et j
t0
t
t1
41. rt
et
42. rt
et i
sin t i
et
2tj
cos t j
t2 et k
et k
t0 t0
WRITING ABOUT CONCEPTS 43. Definitions Define the unit tangent vector, the principal unit normal vector, and the tangential and normal components of acceleration. 44. Unit Tangent Vector How is the unit tangent vector related to the orientation of a curve? Explain. 45. Acceleration Describe the motion of a particle when the normal component of acceleration is 0.
t k, t 0
27. rt et cos t i et sin t j,
t1
40. rt 2t 1i t2j 4tk
t
Finding Tangential and Normal Components of Acceleration In Exercises 21–28, find T t, N t, a T , and a N at the given time t for the plane curve r t. 1 21. rt t i j, t
t2 k 2
3
2
46. Acceleration Describe the motion of a particle when the tangential component of acceleration is 0.
28. rt 4 cos 3t i 4 sin 3t j, t
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12.4 47. Finding Vectors
An object moves along the path given by
849
Tangent Vectors and Normal Vectors
Finding a Binormal Vector In Exercises 51–56, find the
rt 3ti 4tj.
vectors T and N, and the binormal vector B ⴝ T ⴛ N, for the vector-valued function r t at the given value of t.
Find vt, at, Tt, and Nt (if it exists). What is the form of the path? Is the speed of the object constant or changing?
51. rt 2 cos t i 2 sin t j 52. rt t i t 2 j
HOW DO YOU SEE IT? The figures show the
48.
paths of two particles. y
(i)
y
55. rt 4 sin t i 4 cos t j 2t k,
z
t s
s
x
t
Nⴝ
The figure also shows the vectors vtvt and at at at the indicated values of t. (a) Find aT and aN at t 12, t 1, and t 32. (b) Determine whether the speed of the particle is increasing or decreasing at each of the indicated values of t. Give reasons for your answers. y
t=1 x x
t=2
Figure for 49
3 4
alternative formula
rt t sin t, 1 cos t.
t = 32
t0
Alternative Formula for the Principal Unit Normal Vector In Exercises 57–60, use the vector-valued function r t to find the principal unit normal vector N t using the
49. Cycloidal Motion The figure shows the path of a particle modeled by the vector-valued function
t=1
t0 0
x
(b) Which vector, y or z, represents the principal unit normal vector? Explain.
t = 12
4
56. rt 3 cos 2t i 3 sin 2t j t k, t0
(a) Which vector, s or t, represents the unit tangent vector?
y
2
t0 1
54. rt 2et i et cos t j et sin t k, y
y
t0
53. rt i sin t j cos t k, t0 (ii)
z
t3 k, 3
t k, 2
Figure for 50
50. Motion Along an Involute of a Circle The figure shows a particle moving along a path modeled by
v v
va ⴚ v av . va v av
57. rt 3ti 2t2j 58. rt 3 cos 2ti 3 sin 2tj 59. rt 2ti 4tj t2k 60. rt 5 cos ti 5 sin tj 3tk 61. Projectile Motion Find the tangential and normal components of acceleration for a projectile fired at an angle with the horizontal at an initial speed of v0. What are the components when the projectile is at its maximum height? 62. Projectile Motion Use your results from Exercise 61 to find the tangential and normal components of acceleration for a projectile fired at an angle of 45 with the horizontal at an initial speed of 150 feet per second. What are the components when the projectile is at its maximum height? 63. Projectile Motion A projectile is launched with an initial velocity of 120 feet per second at a height of 5 feet and at an angle of 30 with the horizontal. (a) Determine the vector-valued function for the path of the projectile. (b) Use a graphing utility to graph the path and approximate the maximum height and range of the projectile. (c) Find vt, vt , and at. (d) Use a graphing utility to complete the table.
rt cos t t sin t, sin t t cos t.
t
The figure also shows the vectors vt and at for t 1 and t 2.
Speed
(a) Find a T and aN at t 1 and t 2. (b) Determine whether the speed of the particle is increasing or decreasing at each of the indicated values of t. Give reasons for your answers.
0.5
1.0
1.5
2.0
2.5
3.0
(e) Use a graphing utility to graph the scalar functions a T and aN. How is the speed of the projectile changing when a T and aN have opposite signs?
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Chapter 12
Vector-Valued Functions
64. Projectile Motion A projectile is launched with an initial velocity of 220 feet per second at a height of 4 feet and at an angle of 45 with the horizontal. (a) Determine the vector-valued function for the path of the projectile. (b) Use a graphing utility to graph the path and approximate the maximum height and range of the projectile. (c) Find vt, vt , and at.
0.5
1.0
1.5
2.0
2.5
In Exercises 69–72, use the result of Exercise 68 to find the speed necessary for the given circular orbit around Earth. Let GM ⴝ 9.56 ⴛ 104 cubic miles per second per second, and assume the radius of Earth is 4000 miles. 69. The orbit of the International Space Station 255 miles above the surface of Earth 70. The orbit of the Hubble telescope 360 miles above the surface of Earth 71. The orbit of a heat capacity mapping satellite 385 miles above the surface of Earth
(d) Use a graphing utility to complete the table. t
Orbital Speed
3.0
Speed 65. Air Traffic Control Because of a storm, ground controllers instruct the pilot of a plane flying at an altitude of 4 miles to make a 90 turn and climb to an altitude of 4.2 miles. The model for the path of the plane during this maneuver is rt 10 cos 10 t, 10 sin 10 t, 4 4t, 0 t
72. The orbit of a communications satellite r miles above the surface of Earth that is in geosynchronous orbit. [The satellite completes one orbit per sidereal day (approximately 23 hours, 56 minutes), and therefore appears to remain stationary above a point on Earth.]
True or False? In Exercises 73 and 74, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 73. If a car’s speedometer is constant, then the car cannot be accelerating. 74. If aN 0 for a moving object, then the object is moving in a straight line. 1 20
75. Motion of a Particle modeled by
A particle moves along a path
where t is the time in hours and r is the distance in miles.
rt coshbti sinhbtj
(a) Determine the speed of the plane.
where b is a positive constant.
(b) Calculate a T and a N. Why is one of these equal to 0?
(a) Show that the path of the particle is a hyperbola. (b) Show that at b2 rt.
66. Projectile Motion A plane flying at an altitude of 36,000 feet at a speed of 600 miles per hour releases a bomb. Find the tangential and normal components of acceleration acting on the bomb. 67. Centripetal Acceleration An object is spinning at a constant speed on the end of a string, according to the position vector given in Exercises 29–32. (a) When the angular velocity is doubled, how is the centripetal component of acceleration changed? (b) When the angular velocity is unchanged but the length of the string is halved, how is the centripetal component of acceleration changed? 68. Centripetal Force An object of mass m moves at a constant speed v in a circular path of radius r. The force required to produce the centripetal component of acceleration is called the centripetal force and is given by F mv 2r. Newton’s Law of Universal Gravitation is given by F GMmd 2, where d is the distance between the centers of the two bodies of masses M and m, and G is a gravitational constant. Use this law to show that the speed required for circular motion is v GMr.
76. Proof Prove that the principal unit normal vector N points toward the concave side of a plane curve. 77. Proof Prove that the vector Tt is 0 for an object moving in a straight line. v a . v
78. Proof
Prove that aN
79. Proof
Prove that aN a2 aT2.
PUTNAM EXAM CHALLENGE 80. A particle of unit mass moves on a straight line under the action of a force which is a function f v of the velocity v of the particle, but the form of this function is not known. A motion is observed, and the distance x covered in time t is found to be connected with t by the formula x at bt2 ct3, where a, b, and c have numerical values determined by observation of the motion. Find the function f v for the range of v covered by the experiment. This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
Elena Aliaga/Shutterstock.com
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12.5
Arc Length and Curvature
851
12.5 Arc Length and Curvature Find the arc length of a space curve. Use the arc length parameter to describe a plane curve or space curve. Find the curvature of a curve at a point on the curve. Use a vector-valued function to find frictional force.
Arc Length In Section 10.3, you saw that the arc length of a smooth plane curve C given by the parametric equations x xt and y yt, a t b, is
Exploration Arc Length Formula The formula for the arc length of a space curve is given in terms of the parametric equations used to represent the curve. Does this mean that the arc length of the curve depends on the parameter being used? Would you want this to be true? Explain your reasoning. Here is a different parametric representation of the curve in Example 1.
b
s
xt 2 yt 2 dt.
a
In vector form, where C is given by rt xti ytj, you can rewrite this equation for arc length as
b
s
rt dt.
a
The formula for the arc length of a plane curve has a natural extension to a smooth curve in space, as stated in the next theorem. THEOREM 12.6 Arc Length of a Space Curve If C is a smooth curve given by rt xti ytj ztk on an interval
a, b , then the arc length of C on the interval is
4 1 rt t 2 i t 3 j t 4 k 3 2
b
s
a
Find the arc length from t 0 to t 2 and compare the result with that found in Example 1.
b
xt 2 yt 2 zt 2 dt
rt dt.
a
Finding the Arc Length of a Curve in Space See LarsonCalculus.com for an interactive version of this type of example.
z
Find the arc length of the curve given by
r(t) = ti + 43 t 3/2 j + 12 t 2 k
4 1 rt t i t 3 2 j t 2 k 3 2
2
from t 0 to t 2, as shown in Figure 12.28.
1
t=0
C
4 1 Solution Using xt t, yt 3t 3 2, and zt 2t 2, you obtain xt 1, 1 2 yt 2t , and zt t. So, the arc length from t 0 to t 2 is given by
t=2
1 x
2
2
−1
3
4
y
s
xt 2 yt 2 zt 2 dt
Formula for arc length
0
As t increases from 0 to 2, the vector rt traces out a curve. Figure 12.28
2
1 4t t 2 dt
0 2
Integration tables (Appendix B), Formula 26
t 22 3 dt
0
t2 3 t 22 3 ln t 2 t 22 3 2 2 3 3 213 ln4 13 1 ln 3 2 2 4.816.
0 2
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852
Chapter 12
Curve: r(t) = b cos ti + b sin tj +
Vector-Valued Functions
Finding the Arc Length of a Helix
1 − b 2 tk
Find the length of one turn of the helix given by
z
t = 2π
rt b cos ti b sin t j 1 b 2 tk as shown in Figure 12.29. Solution
Begin by finding the derivative.
rt b sin ti b cos tj 1 b 2 k
C
Derivative
Now, using the formula for arc length, you can find the length of one turn of the helix by integrating rt from 0 to 2. s
2
rt dt
Formula for arc length
0
t=0
b
b
2
b 2sin2 t cos2 t 1 b 2 dt
0
y
x
2
dt
0
One turn of a helix Figure 12.29
2
t
0
2 So, the length is 2 units.
Arc Length Parameter You have seen that curves can be represented by vector-valued functions in different ways, depending on the choice of parameter. For motion along a curve, the convenient parameter is time t. For studying the geometric properties of a curve, however, the convenient parameter is often arc length s.
s(t) =
∫
t
[x′(u)]2 + [y′(u)]2 + [z′(u)]2 du
a
z
Definition of Arc Length Function Let C be a smooth curve given by rt defined on the closed interval a, b . For a t b, the arc length function is
t
t=b
C t
st
t
ru du
a
xu 2 yu 2 zu 2 du.
a
The arc length s is called the arc length parameter. (See Figure 12.30.)
t=a y
Note that the arc length function s is nonnegative. It measures the distance along C from the initial point xa, ya, za to the point xt, yt, zt. Using the definition of the arc length function and the Second Fundamental Theorem of Calculus, you can conclude that
x
Figure 12.30
ds rt . dt
Derivative of arc length function
In differential form, you can write ds rt dt.
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12.5
Arc Length and Curvature
853
Finding the Arc Length Function for a Line Find the arc length function st for the line segment given by
y
rt 3 3ti 4t j,
r(t) = (3 − 3t)i + 4tj 0≤t≤1
4
0 t 1
and write r as a function of the parameter s. (See Figure 12.31.) Solution
3
Because rt 3i 4j and
rt 32 42 5
2
you have 1
t
st x 1
2
3
The line segment from 3, 0 to 0, 4 can be parametrized using the arc length parameter s. Figure 12.31
ru du
0 t
5 du
0
5t. Using s 5t (or t s 5), you can rewrite r using the arc length parameter as follows.
3 4 rs 3 s i s j, 0 s 5 5 5 One of the advantages of writing a vector-valued function in terms of the arc length parameter is that rs 1. For instance, in Example 3, you have
rs
3 5
2
4 5
2
1.
So, for a smooth curve C represented by r(s, where s is the arc length parameter, the arc length between a and b is
b
Length of arc
rs ds
a b
ds
a
ba length of interval. Furthermore, if t is any parameter such that rt 1, then t must be the arc length parameter. These results are summarized in the next theorem, which is stated without proof. THEOREM 12.7 Arc Length Parameter If C is a smooth curve given by rs xsi ysj
Plane curve
rs xsi ysj zsk
Space curve
or
where s is the arc length parameter, then
rs 1. Moreover, if t is any parameter for the vector-valued function r such that
rt 1, then t must be the arc length parameter.
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854
Chapter 12
Vector-Valued Functions
y
Curvature
C
Q
P x
Curvature at P is greater than at Q. Figure 12.32
Definition of Curvature Let C be a smooth curve (in the plane or in space) given by rs, where s is the arc length parameter. The curvature K at s is
y
T2
C
Q
T3
An important use of the arc length parameter is to find curvature—the measure of how sharply a curve bends. For instance, in Figure 12.32, the curve bends more sharply at P than at Q, and you can say that the curvature is greater at P than at Q. You can calculate curvature by calculating the magnitude of the rate of change of the unit tangent vector T with respect to the arc length s, as shown in Figure 12.33.
ddsT Ts .
K
T1 P x
The magnitude of the rate of change of T with respect to the arc length is the curvature of a curve. Figure 12.33
A circle has the same curvature at any point. Moreover, the curvature and the radius of the circle are inversely related. That is, a circle with a large radius has a small curvature, and a circle with a small radius has a large curvature. This inverse relationship is made explicit in the next example.
Finding the Curvature of a Circle Show that the curvature of a circle of radius r is 1 K . r
y
K= T
1 r
Solution Without loss of generality, you can consider the circle to be centered at the origin. Let x, y be any point on the circle and let s be the length of the arc from r, 0 to x, y, as shown in Figure 12.34. By letting be the central angle of the circle, you can represent the circle by
(x, y) r
θ
s (r, 0)
x
r r cos i r sin j.
Using the formula for the length of a circular arc s r, you can rewrite r in terms of the arc length parameter as follows. s s rs r cos i r sin j r r
The curvature of a circle is constant. Figure 12.34
is the parameter.
Arc length s is the parameter.
s s So, rs sin i cos j, and it follows that rs 1, which implies that the r r unit tangent vector is Ts
rs s s sin i cos j
rs r r
and the curvature is K Ts
1r cos sr i 1r sin sr j 1r
at every point on the circle. Because a straight line doesn’t curve, you would expect its curvature to be 0. Try checking this by finding the curvature of the line given by
rs 3
3 4 s i sj. 5 5
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12.5 T(t)
855
In Example 4, the curvature was found by applying the definition directly. This requires that the curve be written in terms of the arc length parameter s. The next theorem gives two other formulas for finding the curvature of a curve written in terms of an arbitrary parameter t. The proof of this theorem is left as an exercise [see Exercise 84, parts (a) and (b)].
ΔT T(t + Δt)
T(t) Δs C
THEOREM 12.8 Formulas for Curvature If C is a smooth curve given by rt, then the curvature K of C at t is K
T(t) Δs
C
Arc Length and Curvature
T(t)
ΔT T(t + Δt)
Tt rt r t .
rt
rt 3
Because rt ds dt, the first formula implies that curvature is the ratio of the rate of change in the tangent vector T to the rate of change in arc length. To see that this is reasonable, let t be a “small number.” Then, Tt
Tt t Tt t Tt t Tt T . ds dt
st t st t st t st s In other words, for a given s, the greater the length of T, the more the curve bends at t, as shown in Figure 12.35.
Finding the Curvature of a Space Curve
Figure 12.35
Find the curvature of the curve given by 1 rt 2t i t 2j t 3k. 3 Solution It is not apparent whether this parameter represents arc length, so you should use the formula K Tt rt . rt 2i 2t j t 2k
rt 4 4t 2 t 4 t2 2 rt Tt
rt
Length of rt
2i 2t j t 2k t2 2
t 2 22j 2tk 2t2i 2t j t 2 k t 2 22 4t i 4 2t 2j 4t k t 2 22
Tt
Tt
16t 2 16 16t 2 4t 4 16t 2
t 2 22
2 2 t 2 22 2 2 t 2
t2
Length of Tt
Therefore, K
2
Tt 2 .
rt t 22
Curvature
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856
Chapter 12
Vector-Valued Functions
The next theorem presents a formula for calculating the curvature of a plane curve given by y f x. THEOREM 12.9 Curvature in Rectangular Coordinates If C is the graph of a twice-differentiable function given by y f x, then the curvature K at the point x, y is K
y .
1 y 2 3 2
Proof By representing the curve C by rx xi f xj 0k (where x is the parameter), you obtain rx i fxj,
rx 1 fx 2 and r x f xj. Because rx r x f xk, it follows that the curvature is K
rx r x
rx 3
f x 1 fx 2 3 2 y 2 3 2.
y
r = radius of curvature K=
P
1 y
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
1 r
r x
Center of curvature C
The circle of curvature Figure 12.36
Let C be a curve with curvature K at point P. The circle passing through point P with radius r 1 K is called the circle of curvature when the circle lies on the concave side of the curve and shares a common tangent line with the curve at point P. The radius is called the radius of curvature at P, and the center of the circle is called the center of curvature. The circle of curvature gives you a nice way to estimate the curvature K at a point P on a curve graphically. Using a compass, you can sketch a circle that lies against the concave side of the curve at point P, as shown in Figure 12.36. If the circle has a radius of r, then you can estimate the curvature to be K 1 r.
Finding Curvature in Rectangular Coordinates Find the curvature of the parabola given by y x 14x 2 at x 2. Sketch the circle of curvature at 2, 1.
y = x − 14 x 2 y
P(2, 1)
1
Solution
The curvature at x 2 is as follows.
Q(4, 0) x
−1
1 −1
2
(2, − 1)
3
y 1 y
−2
K
−3 −4
r=
1 =2 K
The circle of curvature Figure 12.37
1 2
x 2
y 0 y
y 2 3 2
1 y
K
1 2
1 2
Because the curvature at P2, 1 is 12, it follows that the radius of the circle of curvature at that point is 2. So, the center of curvature is 2, 1, as shown in Figure 12.37. [In the figure, note that the curve has the greatest curvature at P. Try showing that the curvature at Q4, 0 is 1 25 2 0.177.]
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12.5
The amount of thrust felt by passengers in a car that is turning depends on two things––the speed of the car and the sharpness of the turn. Figure 12.38
REMARK Note that Theorem 12.10 gives additional formulas for aT and aN.
Arc Length and Curvature
857
Arc length and curvature are closely related to the tangential and normal components of acceleration. The tangential component of acceleration is the rate of change of the speed, which in turn is the rate of change of the arc length. This component is negative as a moving object slows down and positive as it speeds up— regardless of whether the object is turning or traveling in a straight line. So, the tangential component is solely a function of the arc length and is independent of the curvature. On the other hand, the normal component of acceleration is a function of both speed and curvature. This component measures the acceleration acting perpendicular to the direction of motion. To see why the normal component is affected by both speed and curvature, imagine that you are driving a car around a turn, as shown in Figure 12.38. When your speed is high and the turn is sharp, you feel yourself thrown against the car door. By lowering your speed or taking a more gentle turn, you are able to lessen this sideways thrust. The next theorem explicitly states the relationships among speed, curvature, and the components of acceleration. THEOREM 12.10 Acceleration, Speed, and Curvature If rt is the position vector for a smooth curve C, then the acceleration vector is given by at
d 2s ds 2 TK N 2 dt dt
where K is the curvature of C and ds dt is the speed.
Proof
For the position vector rt, you have
at aTT aNN d
v T v T N dt d 2s ds 2 T v KN dt dt 2s ds 2 d 2TK N. dt dt
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Tangential and Normal Components of Acceleration Find aT and aN for the curve given by rt 2t i t 2j 13 t 3k. Solution
From Example 5, you know that
ds rt t 2 2 dt
and K
2 . t 2 22
Therefore, aT
d 2s 2t dt 2
Tangential component
and aN K
dsdt
2
2 t 2 22 2. t 2 22
Normal component
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858
Chapter 12
Vector-Valued Functions
Application There are many applications in physics and engineering dynamics that involve the relationships among speed, arc length, curvature, and acceleration. One such application concerns frictional force. A moving object with mass m is in contact with a stationary object. The total force required to produce an acceleration a along a given path is F ma m
ddt sT mKdsdt N 2
2
2
maTT maNN. The portion of this total force that is supplied by the stationary object is called the force of friction. For example, when a car moving with constant speed is rounding a turn, the roadway exerts a frictional force that keeps the car from sliding off the road. If the car is not sliding, the frictional force is perpendicular to the direction of motion and has magnitude equal to the normal component of acceleration, as shown in Figure 12.39. The potential frictional force of a road around a turn can be increased by banking the roadway.
Force of friction
The force of friction is perpendicular to the direction of motion. Figure 12.39
Frictional Force 60 km/h
A 360-kilogram go-cart is driven at a speed of 60 kilometers per hour around a circular racetrack of radius 12 meters, as shown in Figure 12.40. To keep the cart from skidding off course, what frictional force must the track surface exert on the tires? Solution The frictional force must equal the mass times the normal component of acceleration. For this circular path, you know that the curvature is
12 m
K
1 . 12
Curvature of circular racetrack
Therefore, the frictional force is maN mK
dsdt
2
360 kg Figure 12.40
m 121m60,000 3600 sec
2
8333 kgm sec2.
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12.5
Arc Length and Curvature
SUMMARY OF VELOCITY, ACCELERATION, AND CURVATURE Unless noted otherwise, let C be a curve (in the plane or in space) given by the position vector rt xti ytj
Curve in the plane
rt xti ytj ztk
Curve in space
or
where x, y, and z are twice-differentiable functions of t. Velocity vector, speed, and acceleration vector vt rt ds
vt rt dt at r t aTTt aNNt d 2s ds 2 2 Tt K Nt dt dt
Velocity vector Speed Acceleration vector
K is curvature and
ds is speed. dt
Unit tangent vector and principal unit normal vector rt
rt Tt Nt
Tt Tt
Unit tangent vector
Principal unit normal vector
Components of acceleration aT a T
v a d 2s 2
v dt
aN a N
v a
v a 2 aT2 K
ds dt
Normal component of acceleration
2
K is curvature and
Formulas for curvature in the plane y K
1 y 2 3 2 xy yx K
x 2 y 2 3 2
Tangential component of acceleration
ds is speed. dt
C given by y f x C given by x xt, y yt
Formulas for curvature in the plane or in space K Ts r s
Tt rt r t K
rt
rt 3 at Nt K
vt 2
s is arc length parameter. t is general parameter.
Cross product formulas apply only to curves in space.
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859
860
Chapter 12
Vector-Valued Functions
12.5 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding the Arc Length of a Plane Curve In Exercises 1–6, sketch the plane curve and find its length over the given interval. 1. rt 3ti tj, 0, 3 2. rt t i t 2j, 0, 4
5. rt a
ti a
t
0
4. rt t 1i t2 j, 0, 6 sin3 t j,
(a) Write the length of the arc s on the helix as a function of t by evaluating the integral s
3. rt t 3 i t2 j, 0, 1 cos3
17. Investigation Consider the helix represented by the vector-valued function rt 2 cos t, 2 sin t, t.
xu 2 yu 2 zu 2 du.
(b) Solve for t in the relationship derived in part (a), and substitute the result into the original set of parametric equations. This yields a parametrization of the curve in terms of the arc length parameter s.
0, 2
6. rt a cos t i a sin t j, 0, 2 7. Projectile Motion A baseball is hit 3 feet above the ground at 100 feet per second and at an angle of 45 with respect to the ground.
(c) Find the coordinates of the point on the helix for arc lengths s 5 and s 4.
(a) Find the vector-valued function for the path of the baseball.
18. Investigation Repeat Exercise 17 for the curve represented by the vector-valued function
(b) Find the maximum height.
(d) Verify that rs 1.
rt 4sin t t cos t, 4cos t t sin t, 32t2.
(c) Find the range. (d) Find the arc length of the trajectory. 8. Projectile Motion Repeat Exercise 7 for a baseball that is hit 4 feet above the ground at 80 feet per second and at an angle of 30 with respect to the ground.
Finding the Arc Length of a Curve in Space In Exercises 9–14, sketch the space curve and find its length over the given interval. Vector-Valued Function 9. rt t i 4t j 3t k
Interval
Finding Curvature In Exercises 19–22, find the curvature K of the curve, where s is the arc length parameter.
19. rs 1
2
2
2
s j
20. rs 3 si j 21. Helix in Exercise 17: rt 2 cos t, 2 sin t, t 22. Curve in Exercise 18:
10. rt i t 2 j t3 k
0, 2
11. rt 4t, cos t, sin t
12. rt 2 sin t, 5t, 2 cos t
0,
23. rt 4t i 2t j,
13. rt a cos t i a sin t j bt k
0, 2
25. rt t i
14. rt cos t t sin t, sin t t cos t, t 2
0, 2
27. rt t, sin t,
3 2
2
rt 4sin t t cos t, 4cos t t sin t, 32 t 2
0, 1
0,
s i 1
15. Investigation Consider the graph of the vector-valued function rt t i 4 t 2j t3 k on the interval 0, 2 . (a) Approximate the length of the curve by finding the length of the line segment connecting its endpoints. (b) Approximate the length of the curve by summing the lengths of the line segments connecting the terminal points of the vectors r0, r0.5, r1, r1.5, and r2.
Finding Curvature In Exercises 23–28, find the curvature K of the plane curve at the given value of the parameter. t1
1 j, t 1 t t
24. rt t 2i j, t 2 1 26. rt t i t3 j, t 2 9
2
28. rt 5 cos t, 4 sin t, t
Finding Curvature In Exercises 29–36, find the curvature K of the curve. 29. rt 4 cos 2 t i 4 sin 2 t j 30. rt 2 cos t i sin t j
(c) Describe how you could obtain a more accurate approximation by continuing the processes in parts (a) and (b).
31. rt a cos t i a sin t j
(d) Use the integration capabilities of a graphing utility to approximate the length of the curve. Compare this result with the answers in parts (a) and (b).
33. rt t i t 2 j
16. Investigation Repeat Exercise 15 for the vector-valued function rt 6 cos t 4 i 2 sin t 4 j t k.
3
32. rt a cos t i b sin t j t2 k 2
1 34. rt 2t 2 i tj t 2 k 2
35. rt 4t i 3 cos t j 3 sin t k 36. rt e2t i e2t cos t j e2t sin tk
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
12.5 Finding Curvature In Exercises 37–40, find the curvature K of the curve at the point P. 37. rt 3ti 2t2j,
(c) Describe the curvature of C as t changes from t 0 to t 2. 65. Investigation given by
P1, 0, 1
Finding Curvature in Rectangular Coordinates In Exercises 41–48, find the curvature and radius of curvature of the plane curve at the given value of x. 41. y 3x 2, x a
4 42. y 2x , x
43. y 2x 2 3, x 1
3 44. y 416 x 2,
45. y cos 2x, x 2
46. y
47. y x3,
48. y xn, x 1, n 2
x2
64. Motion of a Particle A particle moves along the plane curve C described by rt ti t2j. (b) Find the curvature K of the plane curve at t 0, t 1, and t 2.
t3 k, P2, 4, 2 4
40. rt et cos ti et sin tj et k,
e3x,
x1 x0
x0
y1 axb x
49. y x 1 3
50. y
x3
51. y x 2 3
52. y
1 x
53. y ln x
54. y e x
Find all a and b such that the two curves
and y2
x x2
intersect at only one point and have a common tangent line and equal curvature at that point. Sketch a graph for each set of values of a and b.
66.
Maximum Curvature In Exercises 49–54, (a) find the point on the curve at which the curvature K is a maximum, and (b) find the limit of K as x → ⴥ. 2
861
(a) Find the length of C on the interval 0 t 2.
P3, 2
38. rt et i 4tj, P1, 0 39. rt ti t2j
Arc Length and Curvature
HOW DO YOU SEE IT? Using the graph of the ellipse, at what point(s) is the curvature the least and the greatest? y
x 2 + 4y 2 = 4 2
x −1
1 −2
Curvature In Exercises 55–58, find all points on the graph of the function such that the curvature is zero. 55. y 1 x3
56. y x 13 3
57. y cos x
58. y sin x
WRITING ABOUT CONCEPTS 59. Arc Length Give the formula for the arc length of a smooth curve in space. 60. Curvature Give the formulas for curvature in the plane and in space. 61. Curvature Describe the graph of a vector-valued function for which the curvature is 0 for all values of t in its domain. 62. Curvature Given a twice-differentiable function y f x, determine its curvature at a relative extremum. Can the curvature ever be greater than it is at a relative extremum? Why or why not? 63. Investigation
Consider the function f x x 4 x 2.
(a) Use a computer algebra system to find the curvature K of the curve as a function of x. (b) Use the result of part (a) to find the circles of curvature to the graph of f when x 0 and x 1. Use a computer algebra system to graph the function and the two circles of curvature. (c) Graph the function Kx and compare it with the graph of f x. For example, do the extrema of f and K occur at the same critical numbers? Explain your reasoning.
67. Sphere and Paraboloid A sphere of radius 4 is dropped into the paraboloid given by z x 2 y 2. (a) How close will the sphere come to the vertex of the paraboloid? (b) What is the radius of the largest sphere that will touch the vertex? 68. Speed The smaller the curvature of a bend in a road, the faster a car can travel. Assume that the maximum speed around a turn is inversely proportional to the square root of the curvature. A car moving on the path y 13x3, where x and y are measured in miles, can safely go 30 miles per hour at 1, 13 . How fast can it go at 32, 98 ? 69. Center of Curvature Let C be a curve given by y f x. Let K be the curvature K 0 at the point Px0, y0 and let z
1 fx02 . f x0
Show that the coordinates , of the center of curvature at P are , x0 fx0z, y0 z.
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70. Center of Curvature Use the result of Exercise 69 to find the center of curvature for the curve at the given point. (a) y e x, 0, 1
(b) y
x2 1 , 1, 2 2
(c) y x2, 0, 0
71. Curvature A curve C is given by the polar equation r f . Show that the curvature K at the point r, is K
2r2 rr r 2.
r 2 r 2 3 2
Hint: Represent the curve by r r cos i r sin j. 72. Curvature Use the result of Exercise 71 to find the curvature of each polar curve. (a) r 1 sin
(b) r
(c) r a sin
(d) r e
83. Curvature Verify that the curvature at any point x, y on the graph of y cosh x is 1 y2. 84. Formulas for Curvature Use the definition of curvature in space, K Ts r s , to verify each formula. (a) K
Tt
rt
(b) K
rt r t
rt 3
(c) K
at Nt
vt 2
True or False? In Exercises 85–88, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.
73. Curvature Given the polar curve r ea, a > 0, find the curvature K and determine the limit of K as (a) → and (b) a → .
85. The arc length of a space curve depends on the parametrization.
74. Curvature at the Pole Show that the formula for the curvature of a polar curve r f given in Exercise 71 reduces to K 2 r for the curvature at the pole.
88. The normal component of acceleration is a function of both speed and curvature.
Curvature at the Pole In Exercises 75 and 76, use the result of Exercise 74 to find the curvature of the rose curve at the pole. 75. r 4 sin 2
76. r 6 cos 3
77. Proof For a smooth curve given by the parametric equations x f t and y gt, prove that the curvature is given by K
ftg t gtf t . ft 2 g t 23 2
78. Horizontal Asymptotes Use the result of Exercise 77 to find the curvature K of the curve represented by the parametric equations xt t3 and yt 12t 2. Use a graphing utility to graph K and determine any horizontal asymptotes. Interpret the asymptotes in the context of the problem. 79. Curvature of a Cycloid Use the result of Exercise 77 to find the curvature K of the cycloid represented by the parametric equations x a sin and
y a1 cos .
What are the minimum and maximum values of K? 80. Tangential and Normal Components of Acceleration Use Theorem 12.10 to find aT and aN for each curve given by the vector-valued function. (a) rt
3t 2 i
3t j t3
1 (b) rt t i t 2 j 2 t 2 k
81. Frictional Force A 5500-pound vehicle is driven at a speed of 30 miles per hour on a circular interchange of radius 100 feet. To keep the vehicle from skidding off course, what frictional force must the road surface exert on the tires? 82. Frictional Force A 6400-pound vehicle is driven at a speed of 35 miles per hour on a circular interchange of radius 250 feet. To keep the vehicle from skidding off course, what frictional force must the road surface exert on the tires?
86. The curvature of a circle is the same as its radius. 87. The curvature of a line is 0.
Kepler’s Laws In Exercises 89–96, you are asked to verify Kepler’s Laws of Planetary Motion. For these exercises, assume that each planet moves in an orbit given by the vectorvalued function r. Let r ⴝ r , let G represent the universal gravitational constant, let M represent the mass of the sun, and let m represent the mass of the planet. 89. Prove that r r r
dr . dt
90. Using Newton’s Second Law of Motion, F ma, and Newton’s Second Law of Gravitation F
GmM r r3
show that a and r are parallel, and that rt rt L is a constant vector. So, rt moves in a fixed plane, orthogonal to L.
91. Prove that
d r 1 3 r r r. dt r r
92. Show that
r r
L e is a constant vector. GM r
93. Prove Kepler’s First Law: Each planet moves in an elliptical orbit with the sun as a focus. 94. Assume that the elliptical orbit r
ed 1 e cos
is in the xy-plane, with L along the z-axis. Prove that
L r2
d . dt
95. Prove Kepler’s Second Law: Each ray from the sun to a planet sweeps out equal areas of the ellipse in equal times. 96. Prove Kepler’s Third Law: The square of the period of a planet’s orbit is proportional to the cube of the mean distance between the planet and the sun.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Review Exercises
Review Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Domain and Continuity In Exercises 1–4, (a) find the domain of r, and (b) determine the values (if any) of t for which the function is continuous. 1 2. r共t兲 冪t i jk t4
1. r共t兲 tan t i j t k 3. r共t兲 ln t i t j t k
Evaluating a Function In Exercises 5 and 6, evaluate (if possible) the vector-valued function at each given value of t. 5. r共t兲 共2t 1兲 i t 2 j 冪t 2 k (b) r共2兲
2
冢冣
21. r共t兲 2t3i 4tj t2k 22. r共t兲 共4t 3兲i t2j 共2t 2 4兲k
(a) r冇t冈 (d)
(c) r共s 兲
d [r冇t冈 u冇t冈] dt
(b)
d [u冇t冈 ⴚ 2r冇t冈] dt
(c)
d 冇3t冈r冇t冈 dt
(e)
d [r冇t冈 ⴛ u冇t冈] dt
(f)
d u冇2t冈 dt
23. r共t兲 3t i 共t 1兲 j,
(d) r共 t兲 r共兲
Writing a Vector-Valued Function In Exercises 7 and 8, represent the line segment from P to Q by a vector-valued function and by a set of parametric equations.
u共t兲 t i t 2 j 23t 3 k
24. r共t兲 sin t i cos t j t k, u共t兲 sin t i cos t j
1 k t
Finding an Indefinite Integral In Exercises 25–28, find the indefinite integral.
7. P共3, 0, 5兲, Q共2, 2, 3兲 8. P共2, 3, 8兲,
20. r共t兲 5 cos t i 2 sin t j
24, use the properties of the derivative to find the following.
(c) r共c 1兲
6. r共t兲 3 cos t i 共1 sin t兲 j t k (b) r
19. r共t兲 共t 2 4t兲 i 3t 2j
Using Properties of the Derivative In Exercises 23 and
(d) r共1 t兲 r共1兲
(a) r共0兲
Higher-Order Differentiation In Exercises 19 and 20, find (a) r冇t冈, (b) r 冇t冈, and (c) r冇t冈 r 冇t冈.
Higher-Order Differentiation In Exercises 21 and 22, find (a) r冇t冈, (b) r 冇t冈, (c) r冇t冈 r 冇t冈, and (d) r冇t冈 r 冇t冈.
4. r共t兲 共2t 1兲 i t 2 j t k
(a) r共0兲
863
Q共5, 1, 2兲
25.
Sketching a Curve In Exercises 9–12, sketch the curve represented by the vector-valued function and give the orientation of the curve. 9. r共t兲 具 cos t, sin t典
10. r共t兲 具t 2,
t2
1典
11. r共t兲 共t 1兲i 共3t 1兲j 2t k
26. 27. 28.
12. r共t兲 2 cos t i t j 2 sin t k
Representing a Graph by a Vector-Valued Function In Exercises 13 and 14, represent the plane curve by a vectorvalued function. (There are many correct answers.) 13. 3x 4y 12 0
共i 3j 4tk兲 dt 共t 2 i 5tj 8t3k兲 dt
冣
2 3冪t i j k dt t
共sin t i cos t j e2t k兲 dt
Evaluating a Definite Integral In Exercises 29 –32, evaluate the definite integral.
冕 冕 冕 冕
2
29.
2
14. y 9 x2
Representing a Graph by a Vector-Valued Function
冕 冕 冕冢 冕
共3t i 2t 2 j t 3 k兲 dt
1
30.
共t i 冪t j 4t k兲 dt
0
In Exercises 15 and 16, sketch the space curve represented by the intersection of the surfaces. Use the parameter x ⴝ t to find a vector-valued function for the space curve.
31.
15. z x 2 y 2, x y 0
32.
2
共e t兾2 i 3t 2 j k兲 dt
0
兾3
共2 cos t i sin t j 3k兲 dt
0
16. x 2 z 2 4, x y 0
Finding a Limit In Exercises 17 and 18, find the limit.
Finding an Antiderivative In Exercises 33 and 34, find r冇t冈 that satisfies the initial condition(s).
17. lim 共t i 冪4 t j k兲
33. r共t兲 2t i et j et k,
t→4
冢
sin 2t 18. lim i et j et k t→0 t
冣
r共0兲 i 3j 5k
34. r共t兲 sec t i tan t j t 2 k,
r共0兲 3k
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Vector-Valued Functions
Finding Velocity and Acceleration Vectors In Exercises 35–38, the position vector r describes the path of an object moving in space. (a) Find the velocity vector, speed, and acceleration vector of the object. (b) Evaluate the velocity vector and acceleration vector of the object at the given value of t. Position Vector 35. r共t兲 4ti
t3j
Time
36. r共t兲 冪t i 5tj 2t2k
t4
37. r共t兲 具cos3 t, sin3 t, 3t典
t
38. r共t兲 具t, tan t, et 典
t0
2 3
Finding Tangential and Normal Components of Acceleration In Exercises 51 and 52, find T冇t冈, N冇t冈, aT, and aN at the given time t for the plane curve r冇t冈. 3 51. r共t兲 i 6t j, t
t3
52. r共t兲 3 cos 2t i 3 sin 2t j, t
t1
tk
50. r共t兲 4 cos ti 4 sin tj k, t
6
Finding the Arc Length of a Plane Curve In Exercises 53–56, sketch the plane curve and find its length over the given interval.
Projectile Motion In Exercises 39–42, use the model for projectile motion, assuming there is no air resistance. [a冇t冈 ⴝ ⴚ32 feet per second per second or a冇t冈 ⴝ ⴚ9.8 meters per second per second] 39. A projectile is fired from ground level with an initial velocity of 84 feet per second at an angle of 30 with the horizontal. Find the range of the projectile. 40. A baseball is hit from a height of 3.5 feet above the ground with an initial velocity of 120 feet per second and at an angle of 30 above the horizontal. Find the maximum height reached by the baseball. Determine whether it will clear an 8-foot-high fence located 375 feet from home plate. 41. A projectile is fired from ground level at an angle of 20 with the horizontal. The projectile has a range of 95 meters. Find the minimum initial velocity. 42. Use a graphing utility to graph the paths of a projectile for v0 20 meters per second, h 0 and (a) 30 , (b) 45 , and (c) 60 . Use the graphs to approximate the maximum height and range of the projectile for each case.
Vector-Valued Function
Interval
53. r共t兲 2ti 3tj
关0, 5兴
54. r共t兲 t i 2tk
关0, 3兴
2
55. r共t兲 10 cos t i 10 sin t j
关0, 2兴
56. r共t兲 10 cos t i 10 sin t j
关0, 2兴
3
3
Finding the Arc Length of a Curve in Space In Exercises 57–60, sketch the space curve and find its length over the given interval. Vector-Valued Function
Interval
57. r共t兲 3t i 2tj 4tk
关0, 3兴
58. r共t兲 ti t 2 j 2tk
关0, 2兴
59. r共t兲 具8 cos t, 8 sin t, t典
冤0, 2 冥
60. r共t兲 具2共sin t t cos t兲, 2共cos t t sin t兲, t典
冤0, 2 冥
Finding Curvature In Exercises 61–64, find the curvature K of the curve.
Finding the Unit Tangent Vector In Exercises 43 and 44, find the unit tangent vector to the curve at the specified value of the parameter. 43. r共t兲 3t i
3t3 j,
63. r共t兲 2ti
1 2 2t j
62. r共t兲 2冪t i 3tj
t2k
64. r共t兲 2ti 5 cos tj 5 sin tk
t1
44. r共t兲 2 sin t i 4 cos t j, t
6
Finding Curvature In Exercises 65 and 66, find the curvature K of the curve at the point P.
Finding a Tangent Line In Exercises 45 and 46, find the unit tangent vector T冇t冈 and find a set of parametric equations for the line tangent to the space curve at point P.
冢
45. r共t兲 2 cos t i 2 sin t j t k, P 1, 冪3, 46. r共t兲 t i t 2 j 23t3 k,
61. r共t兲 3ti 2tj
P共2, 4, 16 3兲
3
65. r共t兲 12t2i tj 13t3k,
P共12, 1, 13 兲
66. r共t兲 4 cos t i 3 sin t j t k, P共4, 0, 兲
Finding Curvature in Rectangular Coordinates In Exercises 67–70, find the curvature and radius of curvature of the plane curve at the given value of x.
冣
67. y 12 x 2 2, x 4
68. y ex兾2, x 0
47–50, find the principal unit normal vector to the curve at the specified value of the parameter.
69. y ln x,
70. y tan x,
47. r共t兲 2t i 3t2 j,
71. Frictional Force A 7200-pound vehicle is driven at a speed of 25 miles per hour on a circular interchange of radius 150 feet. To keep the vehicle from skidding off course, what frictional force must the road surface exert on the tires?
Finding the Principal Unit Normal Vector In Exercises
t1
48. r共t兲 t i ln t j,
49. r共t兲 3 cos 2t i 3 sin 2t j 3 k, t
4
t2
x1
x
4
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
P.S. Problem Solving
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
P.S. Problem Solving 1. Cornu Spiral
冕
t
x共t兲
0
The cornu spiral is given by
冕
t
冢2u 冣 du 2
cos
y共t兲
and
冢2u 冣 du. 2
sin
0
865
4. Projectile Motion Repeat Exercise 3 for the case in which the bomber is facing away from the launch site, as shown in the figure. y
The spiral shown in the figure was plotted over the interval t .
4000
3200 Bomb 1600
Projectile θ
x
Cannon
5. Cycloid
5000
Consider one arch of the cycloid
r共 兲 共 sin 兲i 共1 cos 兲j, 0 2
Generated by Mathematica
(a) Find the arc length of this curve from t 0 to t a. (b) Find the curvature of the graph when t a. (c) The cornu spiral was discovered by James Bernoulli. He found that the spiral has an amazing relationship between curvature and arc length. What is this relationship?
as shown in the figure. Let s共 兲 be the arc length from the highest point on the arch to the point 共x共 兲, y共 兲兲, and let
共 兲 1兾K be the radius of curvature at the point 共x共 兲, y共 兲兲. Show that s and are related by the equation s 2 2 16. (This equation is called a natural equation for the curve.) y
2. Radius of Curvature Let T be the tangent line at the point P共x, y兲 to the graph of the curve x 2兾3 y2兾3 a 2兾3, a > 0, as shown in the figure. Show that the radius of curvature at P is three times the distance from the origin to the tangent line T.
(x(θ ), y(θ ))
y x
a
π
P(x, y)
6. Cardioid
x
−a
a
Consider the cardioid
r 1 cos ,
T −a
3. Projectile Motion A bomber is flying horizontally at an altitude of 3200 feet with a velocity of 400 feet per second when it releases a bomb. A projectile is launched 5 seconds later from a cannon at a site facing the bomber and 5000 feet from the point that was directly beneath the bomber when the bomb was released, as shown in the figure. The projectile is to intercept the bomb at an altitude of 1600 feet. Determine the required initial speed and angle of inclination of the projectile. (Ignore air resistance.)
2π
0 2
as shown in the figure. Let s共 兲 be the arc length from the point 共2, 兲 on the cardioid to the point 共r, 兲, and let p共 兲 1兾K be the radius of curvature at the point 共r, 兲. Show that s and are related by the equation s 2 9 2 16. (This equation is called a natural equation for the curve.) π 2
(r, θ ) (2, π )
0
1
y 4000
3200
7. Proof If r共t兲 is a nonzero differentiable function of t, prove that
Bomb 1600
1 d 共储r共t兲储兲 r 共t兲 r共t兲. dt 储r共t兲储
Projectile θ
Cannon
x 5000
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Chapter 12
Vector-Valued Functions
8. Satellite A communications satellite moves in a circular orbit around Earth at a distance of 42,000 kilometers from the center of Earth. The angular velocity d radian per hour dt 12
13. Arc Length and Curvature function r共t兲 具t cos t, t sin t典,
Consider the vector-valued
0 t 2.
(a) Use a graphing utility to graph the function. (b) Find the length of the arc in part (a).
is constant. (a) Use polar coordinates to show that the acceleration vector is given by d 2r d 2r d a 2 r dt dt 2 dt
d 2 dr d ur r 2 2 u dt dt dt
冢 冣冥
冤
2
冤
冥
where ur cos i sin j is the unit vector in the radial direction and u sin i cos j. (b) Find the radial and angular components of acceleration for the satellite.
Binormal Vector In Exercises 9 –11, use the binormal vector defined by the equation B ⴝ T ⴛ N. 9. Find the unit tangent, unit normal, and binormal vectors for the helix r共t兲 4 cos ti 4 sin tj 3tk
(c) Find the curvature K as a function of t. Find the curvatures for t-values of 0, 1, and 2. (d) Use a graphing utility to graph the function K. (e) Find (if possible) lim K. t→
(f) Using the result of part (e), make a conjecture about the graph of r as t → . 14. Ferris Wheel You want to toss an object to a friend who is riding a Ferris wheel (see figure). The following parametric equations give the path of the friend r1共t兲 and the path of the object r2共t兲. Distance is measured in meters and time is measured in seconds.
冢
r1共t兲 15 sin
t t i 16 15 cos j 10 10
冣 冢
冣
r2共t兲 关22 8.03共t t0兲兴 i 关1 11.47共t t0兲 4.9共t t0兲2兴 j
at t 兾2. Sketch the helix together with these three mutually orthogonal unit vectors. 10. Find the unit tangent, unit normal, and binormal vectors for the curve r共t兲 cos ti sin tj k at t 兾4. Sketch the curve together with these three mutually orthogonal unit vectors. 11. (a) Prove that there exists a scalar , called the torsion, such that dB兾ds N. (b) Prove that
dN K T B. ds
(The three equations dT兾ds K N, dN兾ds K T B, and dB兾ds N are called the Frenet-Serret formulas.) 12. Exit Ramp A highway has an exit ramp that begins at the origin of a coordinate system and follows the curve y
(a) Locate your friend’s position on the Ferris wheel at time t 0.
1 5兾2 x 32
to the point 共4, 1兲 (see figure). Then it follows a circular path whose curvature is that given by the curve at 共4, 1兲. What is the radius of the circular arc? Explain why the curve and the circular arc should have the same curvature at 共4, 1兲. Circular arc
y
4
2
(b) Determine the number of revolutions per minute of the Ferris wheel. (c) What are the speed and angle of inclination (in degrees) at which the object is thrown at time t t0? (d) Use a graphing utility to graph the vector-valued functions using a value of t0 that allows your friend to be within reach of the object. (Do this by trial and error.) Explain the significance of t0. (e) Find the approximate time your friend should be able to catch the object. Approximate the speeds of your friend and the object at that time.
1 5/2 y = 32 x
(4, 1) x
2
4
6
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 13.10
Functions of Several Variables Introduction to Functions of Several Variables Limits and Continuity Partial Derivatives Differentials Chain Rules for Functions of Several Variables Directional Derivatives and Gradients Tangent Planes and Normal Lines Extrema of Functions of Two Variables Applications of Extrema Lagrange Multipliers
Hardy-Weinberg Law (Exercise 15, p. 949)
Ocean Floor (Exercise 74, p. 926)
Wind Chill (Exercise 31, p. 906)
Marginal Costs (Exercise 110, p. 898) Forestry (Exercise 75, p. 878) Clockwise from top left, Sashkin/Shutterstock.com; Brandelet/Shutterstock.com; Amy Walters/Shutterstock.com; Val Thoermer/Shutterstock.com; Roca/Shutterstock.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
867
868
Chapter 13
Functions of Several Variables
13.1 Introduction to Functions of Several Variables Understand the notation for a function of several variables. Sketch the graph of a function of two variables. Sketch level curves for a function of two variables. Sketch level surfaces for a function of three variables. Use computer graphics to graph a function of two variables.
Functions of Several Variables Exploration Without using a graphing utility, describe the graph of each function of two variables. a. z ⫽ x2 ⫹ y2 b. z ⫽ x ⫹ y c. z ⫽ x2 ⫹ y d. z ⫽ 冪x2 ⫹ y2 e. z ⫽ 冪1 ⫺ x2 ⫹ y2
So far in this text, you have dealt only with functions of a single (independent) variable. Many familiar quantities, however, are functions of two or more variables. Here are three examples. 1. The work done by a force, W ⫽ FD, is a function of two variables. 2. The volume of a right circular cylinder, V ⫽ r 2h, is a function of two variables. 3. The volume of a rectangular solid, V ⫽ lwh, is a function of three variables. The notation for a function of two or more variables is similar to that for a function of a single variable. Here are two examples. z ⫽ f 共x, y兲 ⫽ x2 ⫹ xy
Function of two variables
2 variables
and w ⫽ f 共x, y, z兲 ⫽ x ⫹ 2y ⫺ 3z
Function of three variables
3 variables
Definition of a Function of Two Variables Let D be a set of ordered pairs of real numbers. If to each ordered pair 共x, y兲 in D there corresponds a unique real number f 共x, y兲, then f is a function of x and y. The set D is the domain of f, and the corresponding set of values for f 共x, y兲 is the range of f. For the function z ⫽ f 共x, y兲 x and y are called the independent variables and z is called the dependent variable.
MARY FAIRFAX SOMERVILLE (1780–1872)
Somerville was interested in the problem of creating geometric models for functions of several variables. Her most well-known book, The Mechanics of the Heavens, was published in 1831. See LarsonCalculus.com to read more of this biography.
Similar definitions can be given for functions of three, four, or n variables, where the domains consist of ordered triples 共x1, x2, x3兲, quadruples 共x1, x2, x3, x4兲, and n-tuples 共x1, x2, . . . , xn兲. In all cases, the range is a set of real numbers. In this chapter, you will study only functions of two or three variables. As with functions of one variable, the most common way to describe a function of several variables is with an equation, and unless it is otherwise restricted, you can assume that the domain is the set of all points for which the equation is defined. For instance, the domain of the function f 共x, y兲 ⫽ x 2 ⫹ y 2 is the entire xy-plane. Similarly, the domain of f 共x, y兲 ⫽ ln xy is the set of all points 共x, y兲 in the plane for which xy > 0. This consists of all points in the first and third quadrants. Mary Evans Picture Library/The Image Works
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.1
Introduction to Functions of Several Variables
869
Domains of Functions of Several Variables y
Find the domain of each function.
4
a. f 共x, y兲 ⫽
2
x
−2 −1 −1
1
2
冪9 ⫺
x2
x ⫺ y2 ⫺ z2
4
a. The function f is defined for all points 共x, y兲 such that x ⫽ 0 and x 2 ⫹ y 2 ⱖ 9. So, the domain is the set of all points lying on or outside the circle x 2 ⫹ y 2 ⫽ 9, except those points on the y-axis, as shown in Figure 13.1. b. The function g is defined for all points 共x, y, z兲 such that
−2
−4
x 2 ⫹ y 2 ⫹ z 2 < 9.
Domain of
f(x, y) =
x
b. g 共x, y, z兲 ⫽
Solution
1 −4
冪x 2 ⫹ y 2 ⫺ 9
x2
+ x
y2
−9
Consequently, the domain is the set of all points 共x, y, z兲 lying inside a sphere of radius 3 that is centered at the origin.
Figure 13.1
Functions of several variables can be combined in the same ways as functions of single variables. For instance, you can form the sum, difference, product, and quotient of two functions of two variables as follows.
共 f ± g兲共x, y兲 ⫽ f 共x, y兲 ± g共x, y兲 共 f g兲共x, y兲 ⫽ f 共x, y兲g共x, y兲 f f 共x, y兲 共x, y兲 ⫽ , g 共x, y兲 ⫽ 0 g g 共x, y兲
Sum or difference Product Quotient
You cannot form the composite of two functions of several variables. You can, however, form the composite function 共 g ⬚ h 兲共x, y兲, where g is a function of a single variable and h is a function of two variables.
共g ⬚ h兲共x, y兲 ⫽ g共h共x, y兲兲
Composition
The domain of this composite function consists of all 共x, y兲 in the domain of h such that h 共x, y兲 is in the domain of g. For example, the function f 共x, y兲 ⫽ 冪16 ⫺ 4x 2 ⫺ y 2 can be viewed as the composite of the function of two variables given by h 共x, y兲 ⫽16 ⫺ 4x 2 ⫺ y 2 and the function of a single variable given by g共u兲 ⫽ 冪u. The domain of this function is the set of all points lying on or inside the ellipse 4x 2 ⫹ y 2 ⫽ 16. A function that can be written as a sum of functions of the form cx m y n (where c is a real number and m and n are nonnegative integers) is called a polynomial function of two variables. For instance, the functions f 共x, y兲 ⫽ x 2 ⫹ y 2 ⫺ 2xy ⫹ x ⫹ 2 and
g 共x, y兲 ⫽ 3xy 2 ⫹ x ⫺ 2
are polynomial functions of two variables. A rational function is the quotient of two polynomial functions. Similar terminology is used for functions of more than two variables.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
870
Chapter 13
Functions of Several Variables
The Graph of a Function of Two Variables Surface: z = f(x, y)
z
(x, y, z) f(x, y) y
Domain: D
x
(x, y)
As with functions of a single variable, you can learn a lot about the behavior of a function of two variables by sketching its graph. The graph of a function f of two variables is the set of all points 共x, y, z兲 for which z ⫽ f 共x, y兲 and 共x, y兲 is in the domain of f. This graph can be interpreted geometrically as a surface in space, as discussed in Sections 11.5 and 11.6. In Figure 13.2, note that the graph of z ⫽ f 共x, y兲 is a surface whose projection onto the xy-plane is D, the domain of f. To each point 共x, y兲 in D there corresponds a point 共x, y, z兲 on the surface, and, conversely, to each point 共x, y, z兲 on the surface there corresponds a point 共x, y兲 in D.
Figure 13.2
Describing the Graph of a Function of Two Variables What is the range of f 共x, y兲 ⫽ 冪16 ⫺ 4x 2 ⫺ y 2 ? Describe the graph of f. Solution that
The domain D implied by the equation of f is the set of all points 共x, y兲 such
16 ⫺ 4x 2 ⫺ y 2 ⱖ 0. So, D is the set of all points lying on or inside the ellipse
16 − 4x 2 − y 2
Surface: z = z
x2 y2 ⫹ ⫽ 1. 4 16
Trace in plane z = 2
Ellipse in the xy-plane
The range of f is all values z ⫽ f 共x, y兲 such that 0 ⱕ z ⱕ 冪16, or
4
0 ⱕ z ⱕ 4.
Range of f
A point 共x, y, z兲 is on the graph of f if and only if Range 3 y
4
x
Domain
The graph of f 共x, y兲 ⫽ 冪16 ⫺ 4x2 ⫺ y2 is the upper half of an ellipsoid. Figure 13.3
z=
16 −
4x 2
z ⫽ 冪16 ⫺ 4x 2 ⫺ y 2 z 2 ⫽ 16 ⫺ 4x 2 ⫺ y 2 4x 2 ⫹ y 2 ⫹ z 2 ⫽ 16 x2 y2 z2 ⫹ ⫹ ⫽ 1, 0 ⱕ z ⱕ 4. 4 16 16 From Section 11.6, you know that the graph of f is the upper half of an ellipsoid, as shown in Figure 13.3.
−
To sketch a surface in space by hand, it helps to use traces in planes parallel to the coordinate planes, as shown in Figure 13.3. For example, to find the trace of the surface in the plane z ⫽ 2, substitute z ⫽ 2 in the equation z ⫽ 冪16 ⫺ 4x 2 ⫺ y 2 and obtain
y2
z
2 ⫽ 冪16 ⫺ 4x 2 ⫺ y 2
x2 y2 ⫹ ⫽ 1. 3 12
So, the trace is an ellipse centered at the point 共0, 0, 2兲 with major and minor axes of lengths 4冪3
y
x
Figure 13.4
and
2冪3.
Traces are also used with most three-dimensional graphing utilities. For instance, Figure 13.4 shows a computer-generated version of the surface given in Example 2. For this graph, the computer took 25 traces parallel to the xy-plane and 12 traces in vertical planes. If you have access to a three-dimensional graphing utility, use it to graph several surfaces.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.1
871
Introduction to Functions of Several Variables
Level Curves A second way to visualize a function of two variables is to use a scalar field in which the scalar z ⫽ f 共x, y兲
20°
°
20°
°
30
30 °
1008
40 °
°
30
20
100 8 100 4 100 0 100 4 100 8 10 101 16 2
20° 1004
40
101 2
1008
100 4
101 2
is assigned to the point 共x, y兲. A scalar field can be characterized by level curves (or contour lines) along which the value of f 共x, y兲 is constant. For instance, the weather map in Figure 13.5 shows level curves of equal pressure called isobars. In weather maps for which the level curves represent points of equal temperature, the level curves are called isotherms, as shown in Figure 13.6. Another common use of level curves is in representing electric potential fields. In this type of map, the level curves are called equipotential lines.
50
1008
°
°
04 10 60
°
1008
08 10
80
80°
°
00 10
°
0
70
0 10
90°
Level curves show the lines of equal pressure (isobars), measured in millibars. Figure 13.5
Level curves show the lines of equal temperature (isotherms), measured in degrees Fahrenheit. Figure 13.6
Contour maps are commonly used to show regions on Earth’s surface, with the level curves representing the height above sea level. This type of map is called a topographic map. For example, the mountain shown in Figure 13.7 is represented by the topographic map in Figure 13.8.
Figure 13.7
Figure 13.8
A contour map depicts the variation of z with respect to x and y by the spacing between level curves. Much space between level curves indicates that z is changing slowly, whereas little space indicates a rapid change in z. Furthermore, to produce a good three-dimensional illusion in a contour map, it is important to choose c-values that are evenly spaced. Alfred B. Thomas/Earth Scenes/Animals Animals; USGS
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872
Chapter 13
Functions of Several Variables
Sketching a Contour Map The hemisphere f 共x, y兲 ⫽ 冪64 ⫺ x 2 ⫺ y 2 is shown in Figure 13.9. Sketch a contour map of this surface using level curves corresponding to c ⫽ 0, 1, 2, . . . , 8. Solution For each value of c, the equation f 共x, y兲 ⫽ c is a circle (or point) in the xy-plane. For example, when c1 ⫽ 0, the level curve is x 2 ⫹ y 2 ⫽ 64
Circle of radius 8
which is a circle of radius 8. Figure 13.10 shows the nine level curves for the hemisphere. Surface: 64 −
f (x, y) =
x2
−
y
c1 = 0 c2 = 1 c3 = 2 c4 = 3
y2
z
c5 = 4 c6 = 5 c7 = 6 c8 = 7
8
4 8
c9 = 8 x
−8
8 x
−4
4
8
−4
y
8
−8
Hemisphere Figure 13.9
Contour map Figure 13.10
z
Sketching a Contour Map
12
See LarsonCalculus.com for an interactive version of this type of example.
10 8
The hyperbolic paraboloid
6
z ⫽ y2 ⫺ x2
4
is shown in Figure 13.11. Sketch a contour map of this surface.
2 x
4
4
y
Solution For each value of c, let f 共x, y兲 ⫽ c and sketch the resulting level curve in the xy-plane. For this function, each of the level curves 共c ⫽ 0兲 is a hyperbola whose asymptotes are the lines y ⫽ ± x. c=0 c = 12 When c < 0, the transverse axis is horizontal. c = −2 y c=2 c = −4 For instance, the level curve for c ⫽ ⫺4 is
c = −6 c = −8 c = −10 c = −12
4
Surface: z = y2 − x2
Hyperbolic paraboloid Figure 13.11
x2 y2 ⫺ ⫽ 1. 22 22 When c > 0, the transverse axis is vertical. For instance, the level curve for c ⫽ 4 is x2 y2 ⫺ 2 ⫽ 1. 2 2 2 When c ⫽ 0, the level curve is the degenerate conic representing the intersecting asymptotes, as shown in Figure 13.12.
x −4
4
−4
Hyperbolic level curves (at increments of 2) Figure 13.12
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.1
Introduction to Functions of Several Variables
873
One example of a function of two variables used in economics is the CobbDouglas production function. This function is used as a model to represent the numbers of units produced by varying amounts of labor and capital. If x measures the units of labor and y measures the units of capital, then the number of units produced is f 共x, y兲 ⫽ Cx a y 1⫺a where C and a are constants with 0 < a < 1.
The Cobb-Douglas Production Function A toy manufacturer estimates a production function to be
z = 100x 0.6 y 0.4 y
f 共x, y兲 ⫽ 100x 0.6 y 0.4
c = 80,000 c = 160,000
2000 1500
(2000, 1000)
where x is the number of units of labor and y is the number of units of capital. Compare the production level when x ⫽ 1000 and y ⫽ 500 with the production level when x ⫽ 2000 and y ⫽ 1000. Solution
1000
When x ⫽ 1000 and y ⫽ 500, the production level is
f 共1000, 500兲 ⫽ 100共1000 0.6兲共500 0.4兲 ⬇ 75,786.
500 x
500
1000 1500 2000
(1000, 500)
Level curves (at increments of 10,000) Figure 13.13
When x ⫽ 2000 and y ⫽ 1000, the production level is f 共2000, 1000兲 ⫽ 100共2000 0.6兲共1000 0.4兲 ⫽ 151,572. The level curves of z ⫽ f 共x, y兲 are shown in Figure 13.13. Note that by doubling both x and y, you double the production level (see Exercise 79).
Level Surfaces The concept of a level curve can be extended by one dimension to define a level surface. If f is a function of three variables and c is a constant, then the graph of the equation f 共x, y, z兲 ⫽ c is a level surface of the function f, as shown in Figure 13.14. f (x, y, z) = c3 f (x, y, z) = c2
z
f (x, y, z) = c1
y x
Level surfaces of f Figure 13.14
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874
Chapter 13
Functions of Several Variables
Level Surfaces Describe the level surfaces of f 共x, y, z兲 ⫽ 4x 2 ⫹ y 2 ⫹ z 2. Solution z
Level surfaces: 4x 2 + y 2 + z 2 = c
4x 2 ⫹ y 2 ⫹ z 2 ⫽ c.
Equation of level surface
So, the level surfaces are ellipsoids (whose cross sections parallel to the yz-plane are circles). As c increases, the radii of the circular cross sections increase according to the square root of c. For example, the level surfaces corresponding to the values c ⫽ 0, c ⫽ 4, and c ⫽ 16 are as follows.
c=4
c=0 y x
c = 16
Figure 13.15
Each level surface has an equation of the form
4x 2 ⫹ y 2 ⫹ z 2 ⫽ 0 x2 y2 z2 ⫹ ⫹ ⫽1 1 4 4 x2 y2 z2 ⫹ ⫹ ⫽1 4 16 16
Level surface for c ⫽ 0 (single point) Level surface for c ⫽ 4 (ellipsoid) Level surface for c ⫽ 16 (ellipsoid)
These level surfaces are shown in Figure 13.15. If the function in Example 6 represented the temperature at the point 共x, y, z兲, then the level surfaces shown in Figure 13.15 would be called isothermal surfaces.
Computer Graphics The problem of sketching the graph of a surface in space can be simplified by using a computer. Although there are several types of three-dimensional graphing utilities, most use some form of trace analysis to give the illusion of three dimensions. To use such a graphing utility, you usually need to enter the equation of the surface and the region in the xy-plane over which the surface is to be plotted. (You might also need to enter the number of traces to be taken.) For instance, to graph the surface f 共x, y兲 ⫽ 共x 2 ⫹ y 2兲e 1⫺x
2
⫺y 2
you might choose the following bounds for x, y, and z. ⫺3 ⱕ x ⱕ 3 ⫺3 ⱕ y ⱕ 3 0 ⱕ z ⱕ 3
Bounds for x Bounds for y Bounds for z
Figure 13.16 shows a computer-generated graph of this surface using 26 traces taken parallel to the yz-plane. To heighten the three-dimensional effect, the program uses a “hidden line” routine. That is, it begins by plotting the traces in the foreground (those corresponding to the largest x-values), and then, as each new trace is plotted, the program determines whether all or only part of the next trace should be shown. The graphs on the next page show a variety of surfaces that were plotted by computer. If you have access to a computer drawing program, use it to reproduce these surfaces. Remember also that the three-dimensional graphics in this text can be viewed and rotated. These rotatable graphs are available at LarsonCalculus.com.
f (x, y) = (x 2 + y 2)e 1 − x
2 − y2
z
x
y
Figure 13.16
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.1
z
875
Introduction to Functions of Several Variables
z
z
x
x y
x
y y
Three different views of the graph of f 共x, y兲 ⫽ 共2 ⫺ y2 ⫹ x2兲e1⫺x
2
z
⫺ 共 y2兾4兲
z
y
x
y
x
y
x
Single traces
Double traces
Level curves
Traces and level curves of the graph of f 共x, y兲 ⫽
⫺4x x2 ⫹ y2 ⫹ 1
z
z
z
y
x y y
x x
f (x, y) = sin x sin y
f(x, y) = −
1 x2 + y2
f (x, y) =
1 − x2 − y2 ⎪1 − x 2 − y 2⎪
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
876
Chapter 13
Functions of Several Variables
13.1 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Determining Whether a Graph Is a Function In Exercises 1 and 2, use the graph to determine whether z is a function of x and y. Explain. z
1.
13. f 共x, y兲 ⫽ x sin y (a) 共2, 兾4兲 (b) 共3, 1兲 (c) 共⫺3, 兾3兲 (d) 共4, 兾2兲 14. V共r, h兲 ⫽ r 2h (a) 共3, 10兲 (b) 共5, 2兲 (c) 共4, 8兲 (d) 共6, 4兲
冕
y
2
15. g 共x, y兲 ⫽
共2t ⫺ 3兲 dt
x
3 (a) 共4, 0兲 (b) 共4, 1兲 (c) 共4, 2 兲 (d)
冕
y
16. g 共x, y兲 ⫽
3
4
4 x
1 dt t
x
y
共32, 0兲
(a) 共4, 1兲 (b) 共6, 3兲 (c) 共2, 5兲 (d)
共12, 7兲
17. f 共x, y兲 ⫽ 2x ⫹ y2
2.
z
f 共x ⫹ ⌬x, y兲 ⫺ f 共x, y兲 ⌬x
(a)
3
(b)
f 共x, y ⫹ ⌬y兲 ⫺ f 共x, y兲 ⌬y
(b)
f 共x, y ⫹ ⌬y兲 ⫺ f 共x, y兲 ⌬y
18. f 共x, y兲 ⫽ 3x2 ⫺ 2y 5
f 共x ⫹ ⌬x, y兲 ⫺ f 共x, y兲 ⌬x
(a)
x
Finding the Domain and Range of a Function In
5 y
Exercises 19–30, find the domain and range of the function.
Determining Whether an Equation Is a Function In Exercises 3–6, determine whether z is a function of x and y.
19. f 共x, y兲 ⫽ x 2 ⫹ y 2
20. f 共x, y兲 ⫽ e xy y 22. g 共x, y兲 ⫽ 冪x xy 24. z ⫽ x⫺y
21. g 共x, y兲 ⫽ x冪y x⫹y xy
3. x 2z ⫹ 3y2 ⫺ x y ⫽ 10
4. xz 2 ⫹ 2x y ⫺ y 2 ⫽ 4
23. z ⫽
x2 y2 ⫹ ⫹ z2 ⫽ 1 4 9
6. z ⫹ x ln y ⫺ 8yz ⫽ 0
25. f 共x, y兲 ⫽ 冪4 ⫺ x 2 ⫺ y 2
26. f 共x, y兲 ⫽ 冪4 ⫺ x 2 ⫺ 4y 2
27. f 共x, y兲 ⫽ arccos共x ⫹ y兲
28. f 共x, y兲 ⫽ arcsin共 y兾x兲
29. f 共x, y兲 ⫽ ln共4 ⫺ x ⫺ y兲
30. f 共x, y兲 ⫽ ln共xy ⫺ 6兲
5.
Evaluating a Function In Exercises 7–18, find and simplify the function values. 7. f 共x, y兲 ⫽ xy (a) 共3, 2兲
(b) 共⫺1, 4兲
(d) 共5, y兲
(e) 共x, 2兲
(c) 共30, 5兲
(f) 共5, t兲
8. f 共x, y兲 ⫽ 4 ⫺ x 2 ⫺ 4y 2 (a) 共0, 0兲
(b) 共0, 1兲
(c) 共2, 3兲
(d) 共1, y兲
(e) 共x, 0兲
(f) 共t, 1兲
31. Think About It The graphs labeled (a), (b), (c), and (d) are graphs of the function f 共x, y兲 ⫽ ⫺4x兾共x 2 ⫹ y 2 ⫹ 1兲. Match the four graphs with the points in space from which the surface is viewed. The four points are 共20, 15, 25兲, 共⫺15, 10, 20兲, 共20, 20, 0兲, and 共20, 0, 0兲. z
(a)
z
(b) x
9. f 共x, y兲 ⫽ xey (a) 共5, 0兲
(b) 共3, 2兲
(c) 共2, ⫺1兲
(d) 共5, y兲
(e) 共x, 2兲
(f) 共t, t兲
ⱍ
ⱍ
y y
10. g 共x, y兲 ⫽ ln x ⫹ y (a) 共1, 0兲
(b) 共0, ⫺1兲
(c) 共0, e兲
(d) 共1, 1兲
(e) 共e, e兾2兲
(f) 共2, 5兲
11. h 共x, y, z兲 ⫽ (a) 共2, 3, 9兲
Generated by Maple
z
(c)
Generated by Maple
(d)
z
xy z (b) 共1, 0, 1兲
(c) 共⫺2, 3, 4兲 (d) 共5, 4, ⫺6兲
x
12. f 共x, y, z兲 ⫽ 冪x ⫹ y ⫹ z (a) 共0, 5, 4兲 (b) 共6, 8, ⫺3兲 (c) 共4, 6, 2兲 (d) 共10, ⫺4, ⫺3兲
y
y x Generated by Maple
Generated by Maple
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.1 32. Think About It
Use the function given in Exercise 31.
ⱍ
ⱍ
47. f 共x, y兲 ⫽ ln y ⫺ x 2
(a) Find the domain and range of the function.
877
Introduction to Functions of Several Variables 48. f 共x, y兲 ⫽ cos
2
⫹ 2y 2 4
冣
z
z
(b) Identify the points in the xy-plane at which the function value is 0.
冢x
5
4
(c) Does the surface pass through all the octants of the rectangular coordinate system? Give reasons for your answer.
−6 y 3 2 5 4
Sketching a Surface In Exercises 33 – 40, sketch the x
surface given by the function. 33. f 共x, y兲 ⫽ 4
34. f 共x, y兲 ⫽ 6 ⫺ 2x ⫺ 3y
35. f 共x, y兲 ⫽
1 36. g 共x, y兲 ⫽ 2 y
y2
39. f 共x, y兲 ⫽ e⫺x
冦
49. z ⫽ x ⫹ y, c ⫽ ⫺1, 0, 2, 4
x ⱖ 0, y ⱖ 0 x < 0 or y < 0
xy, 40. f 共x, y兲 ⫽ 0,
10 x
y
Sketching a Contour Map In Exercises 49–56, describe the level curves of the function. Sketch a contour map of the surface using level curves for the given c-values.
1 38. z ⫽ 2冪x 2 ⫹ y 2
37. z ⫽ ⫺x 2 ⫺ y 2
4 5 6
−2
50. z ⫽ 6 ⫺ 2x ⫺ 3y, c ⫽ 0, 2, 4, 6, 8, 10 51. z ⫽ x2 ⫹ 4y2,
Graphing a Function Using Technology In Exercises 41–44, use a computer algebra system to graph the function. 41. z ⫽ y 2 ⫺ x 2 ⫹ 1
1 冪144 ⫺ 16x 2 ⫺ 9y 2 42. z ⫽ 12
43. f 共x, y兲 ⫽ x 2e共⫺xy兾2兲
44. f 共x, y兲 ⫽ x sin y
c ⫽ 0, 1, 2, 3, 4
52. f 共x, y兲 ⫽ 冪9 ⫺ x2 ⫺ y2, 53. f 共x, y兲 ⫽ xy, 54. f 共x, y兲 ⫽
e xy兾2,
c ⫽ 0, 1, 2, 3
c ⫽ ± 1, ± 2, . . . , ± 6 c ⫽ 2, 3, 4, 12, 13, 14
55. f 共x, y兲 ⫽ x兾共x2 ⫹ y2兲,
c ⫽ ± 12, ± 1, ± 32, ± 2
Matching In Exercises 45–48, match the graph of the
56. f 共x, y兲 ⫽ ln共x ⫺ y兲,
surface with one of the contour maps. [The contour maps are labeled (a), (b), (c), and (d).]
Graphing Level Curves In Exercises 57–60, use a graphing
y
(a)
c ⫽ 0, ± 12, ± 1, ± 32, ± 2
utility to graph six level curves of the function.
y
(b)
x
x
ⱍ ⱍ
57. f 共x, y兲 ⫽ x 2 ⫺ y 2 ⫹ 2
58. f 共x, y兲 ⫽ xy
8 59. g共x, y兲 ⫽ 1 ⫹ x2 ⫹ y2
60. h共x, y兲 ⫽ 3 sin共 x ⫹ y 兲
ⱍⱍ ⱍⱍ
WRITING ABOUT CONCEPTS 61. Function of Two Variables What is a graph of a function of two variables? How is it interpreted geometrically? Describe level curves.
y
(c)
62. Using Level Curves All of the level curves of the surface given by z ⫽ f 共x, y兲 are concentric circles. Does this imply that the graph of f is a hemisphere? Illustrate your answer with an example.
y
(d)
x
x
63. Creating a Function Construct a function whose level curves are lines passing through the origin. 64. Conjecture and y ⱖ 0.
45. f 共x, y兲 ⫽ e1⫺x
2
46. f 共x, y兲 ⫽ e1⫺x
⫺y2
2
(b) Make a conjecture about the relationship between the graphs of f and g 共x, y兲 ⫽ f 共x, y兲 ⫺ 3. Explain your reasoning.
6
3
(c) Make a conjecture about the relationship between the graphs of f and g 共x, y兲 ⫽ ⫺f 共x, y兲. Explain your reasoning. 3
x
(a) Sketch the graph of the surface given by f. ⫹y2
z
z
Consider the function f 共x, y兲 ⫽ xy, for x ⱖ 0
y
3 4 x
3 4
y
(d) Make a conjecture about the relationship between the graphs of f and g 共x, y兲 ⫽ 12 f 共x, y兲. Explain your reasoning. (e) On the surface in part (a), sketch the graph of z ⫽ f 共x, x兲.
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878
Chapter 13
Functions of Several Variables
Writing In Exercises 65 and 66, use the graphs of the level curves (c-values evenly spaced) of the function f to write a description of a possible graph of f. Is the graph of f unique? Explain. 65.
y
y
66.
x x
67. Investment In 2012, an investment of $1000 was made in a bond earning 6% compounded annually. Assume that the buyer pays tax at rate R and the annual rate of inflation is I. In the year 2022, the value V of the investment in constant 2012 dollars is
冤
1 ⫹ 0.06共1 ⫺ R兲 V共I, R兲 ⫽ 1000 1⫹I
冥
74. f 共x, y, z兲 ⫽ sin x ⫺ z, c ⫽ 0 75. Forestry The Doyle Log Rule is one of several methods used to determine the lumber yield of a log (in board-feet) in terms of its diameter d (in inches) and its length L (in feet). The number of board-feet is N共d, L兲 ⫽
冢d ⫺4 4冣 L. 2
(a) Find the number of board-feet of lumber in a log 22 inches in diameter and 12 feet in length. (b) Find N共30, 12兲.
.
Inflation Rate 0
c⫽0
10
Use this function of two variables to complete the table.
Tax Rate
73. f 共x, y, z兲 ⫽ 4x 2 ⫹ 4y 2 ⫺ z 2,
0.03
76. Queuing Model The average length of time that a customer waits in line for service is W 共x, y兲 ⫽
0.05
1 , x⫺y
x > y
where y is the average arrival rate, written as the number of customers per unit of time, and x is the average service rate, written in the same units. Evaluate each of the following.
0 0.28
(a) W 共15, 9兲 (b) W 共15, 13兲
0.35
(c) W 共12, 7兲 (d) W 共5, 2兲
68. Investment A principal of $5000 is deposited in a savings account that earns interest at a rate of r (written as a decimal), compounded continuously. The amount A共r, t兲 after t years is
77. Temperature Distribution The temperature T (in degrees Celsius) at any point 共x, y兲 in a circular steel plate of radius 10 meters is
A共r, t兲 ⫽ 5000ert.
T ⫽ 600 ⫺ 0.75x 2 ⫺ 0.75y 2
Use this function of two variables to complete the table.
where x and y are measured in meters. Sketch some of the isothermal curves.
Number of Years Rate
5
10
15
20
78. Electric Potential 共x, y兲 is
The electric potential V at any point
5
0.02
V共x, y兲 ⫽
0.03
Sketch the equipotential curves for V ⫽ 12, V ⫽ 13, and V ⫽ 14.
.
79. Cobb-Douglas Production Function Use the CobbDouglas production function (see Example 5) to show that when the number of units of labor and the number of units of capital are doubled, the production level is also doubled.
0.04 0.05
Sketching a Level Surface In Exercises 69–74, sketch the graph of the level surface f 冇x, y, z冈 ⴝ c at the given value of c. 69. f 共x, y, z兲 ⫽ x ⫺ y ⫹ z,
冪25 ⫹ x2 ⫹ y2
c⫽1
70. f 共x, y, z兲 ⫽ 4x ⫹ y ⫹ 2z, c ⫽ 4 71. f 共x, y, z兲 ⫽ x 2 ⫹ y 2 ⫹ z 2,
c⫽9
1 72. f 共x, y, z兲 ⫽ x 2 ⫹ 4 y 2 ⫺ z, c ⫽ 1
80. Cobb-Douglas Production Function Show that the Cobb-Douglas production function z ⫽ Cx ay1⫺a can be rewritten as ln
x z ⫽ ln C ⫹ a ln . y y
Val Thoermer/Shutterstock.com
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13.1 81. Ideal Gas Law According to the Ideal Gas Law, PV ⫽ kT where P is pressure, V is volume, T is temperature (in kelvins), and k is a constant of proportionality. A tank contains 2000 cubic inches of nitrogen at a pressure of 26 pounds per square inch and a temperature of 300 K.
Introduction to Functions of Several Variables
85. Construction Cost A rectangular box with an open top has a length of x feet, a width of y feet, and a height of z feet. It costs $1.20 per square foot to build the base and $0.75 per square foot to build the sides. Write the cost C of constructing the box as a function of x, y, and z.
86.
(a) Determine k. (b) Write P as a function of V and T and describe the level curves. 82. Modeling Data The table shows the net sales x (in billions of dollars), the total assets y (in billions of dollars), and the shareholder’s equity z (in billions of dollars) for Apple for the years 2006 through 2011. (Source: Apple Inc.) Year
2006
2007
2008
2009
2010
2011
x
19.3
24.6
37.5
42.9
65.2
108.2
y
17.2
24.9
36.2
47.5
75.2
116.4
z
10.0
14.5
22.3
31.6
47.8
76.6
879
HOW DO YOU SEE IT? The contour map shown in the figure was computer generated using data collected by satellite instrumentation. Color is used to show the “ozone hole” in Earth’s atmosphere. The purple and blue areas represent the lowest levels of ozone, and the green areas represent the highest levels. (Source: National Aeronautics and Space Administration)
A model for these data is z ⫽ f 共x, y兲 ⫽ 0.035x ⫹ 0.640y ⫺ 1.77. (a) Use a graphing utility and the model to approximate z for the given values of x and y. (b) Which of the two variables in this model has the greater influence on shareholder’s equity? Explain. (c) Simplify the expression for f 共x, 150兲 and interpret its meaning in the context of the problem. 83. Meteorology Meteorologists measure the atmospheric pressure in millibars. From these observations, they create weather maps on which the curves of equal atmospheric pressure (isobars), are drawn (see figure). On the map, the closer the isobars, the higher the wind speed. Match points A, B, and C with (a) highest pressure, (b) lowest pressure, and (c) highest wind velocity.
(a) Do the level curves correspond to equally spaced ozone levels? Explain. (b) Describe how to obtain a more detailed contour map.
True or False? In Exercises 87–90, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 87. If f 共x0, y0兲 ⫽ f 共x1, y1兲, then x0 ⫽ x1 and y0 ⫽ y1. 88. If f is a function, then f 共ax, ay兲 ⫽ a 2f 共x, y兲.
B
89. A vertical line can intersect the graph of z ⫽ f 共x, y兲 at most once. C
90. Two different level curves of the graph of z ⫽ f 共x, y兲 can intersect.
A
PUTNAM EXAM CHALLENGE 91. Let f: ⺢2 → ⺢ be a function such that Figure for 83
Figure for 84
84. Acid Rain The acidity of rainwater is measured in units called pH. A pH of 7 is neutral, smaller values are increasingly acidic, and larger values are increasingly alkaline. The map shows curves of equal pH and gives evidence that downwind of heavily industrialized areas, the acidity has been increasing. Using the level curves on the map, determine the direction of the prevailing winds in the northeastern United States. NASA
f 共x, y兲 ⫹ f 共 y, z兲 ⫹ f 共z, x兲 ⫽ 0 for all real numbers x, y, and z. Prove that there exists a function g: ⺢ → ⺢ such that f 共x, y兲 ⫽ g共x兲 ⫺ g共 y兲 for all real numbers x and y. This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
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880
Chapter 13
Functions of Several Variables
13.2 Limits and Continuity Understand the definition of a neighborhood in the plane. Understand and use the definition of the limit of a function of two variables. Extend the concept of continuity to a function of two variables. Extend the concept of continuity to a function of three variables.
Neighborhoods in the Plane In this section, you will study limits and continuity involving functions of two or three variables. The section begins with functions of two variables. At the end of the section, the concepts are extended to functions of three variables. Your study of the limit of a function of two variables begins by defining a two-dimensional analog to an interval on the real number line. Using the formula for the distance between two points
共x, y兲 and 共x0, y0兲 in the plane, you can define the -neighborhood about 共x0, y0兲 to be the disk centered at 共x0, y0兲 with radius > 0
再共x, y兲: SONYA KOVALEVSKY (1850–1891)
Much of the terminology used to define limits and continuity of a function of two or three variables was introduced by the German mathematician Karl Weierstrass (1815–1897).Weierstrass’s rigorous approach to limits and other topics in calculus gained him the reputation as the “father of modern analysis.” Weierstrass was a gifted teacher. One of his best-known students was the Russian mathematician Sonya Kovalevsky, who applied many of Weierstrass’s techniques to problems in mathematical physics and became one of the first women to gain acceptance as a research mathematician.
冎
冪共x x0兲2 共y y0兲2 <
Open disk
as shown in Figure 13.17. When this formula contains the less than inequality sign, 0 there corresponds a > 0 such that
ⱍ f 共x, y兲 Lⱍ <
whenever 0 < 冪共x x0兲2 共 y y0兲2 < .
Graphically, the definition of the limit of a function of two variables implies that for any point 共x, y兲 共x0, y0兲 in the disk of radius , the value f 共x, y兲 lies between L and L , as shown in Figure 13.19. The definition of the limit of a function of two variables is similar to the definition of the limit of a function of a single variable, yet there is a critical difference. To determine whether a function of a single variable has a limit, you need only test the approach from two directions—from the right and from the left. When the function approaches the same limit from the right and from the left, you can conclude that the limit exists. For a function of two variables, however, the statement
z
L +ε L L−ε
y x
(x1, y1) (x0, y0 )
Disk of radius δ
For any 共x, y兲 in the disk of radius , the value f 共x, y兲 lies between L and L . Figure 13.19
共x, y兲 → 共x0, y0兲 means that the point 共x, y兲 is allowed to approach 共x0, y0兲 from any direction. If the value of lim
共x, y兲 → 共x0, y0 兲
f 共x, y兲
is not the same for all possible approaches, or paths, to 共x0, y0兲, then the limit does not exist.
Verifying a Limit by the Definition Show that
lim
共x, y兲 → 共a, b兲
x a.
Solution Let f 共x, y兲 x and L a. You need to show that for each > 0, there exists a -neighborhood about 共a, b兲 such that
ⱍ f 共x, y兲 Lⱍ ⱍx aⱍ <
whenever 共x, y兲 共a, b兲 lies in the neighborhood. You can first observe that from 0 < 冪共x a兲2 共 y b兲2 < it follows that
ⱍ f 共x, y兲 aⱍ ⱍx aⱍ
冪共x a兲2 冪共x a兲2 共 y b兲2 < .
So, you can choose , and the limit is verified.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
882
Chapter 13
Functions of Several Variables
Limits of functions of several variables have the same properties regarding sums, differences, products, and quotients as do limits of functions of single variables. (See Theorem 1.2 in Section 1.3.) Some of these properties are used in the next example.
Verifying a Limit Evaluate 5x 2y . y2
lim
共x, y兲 → 共1, 2兲 x 2
Solution
By using the properties of limits of products and sums, you obtain
lim
5x 2y 5共12兲共2兲 10
lim
共x 2 y 2兲 共12 22兲 5.
共x, y兲 → 共1, 2兲
and 共x, y兲 → 共1, 2兲
Because the limit of a quotient is equal to the quotient of the limits (and the denominator is not 0), you have 5x 2y 10 2. 共x, y兲 → 共1, 2兲 x 2 y 2 5 lim
Verifying a Limit Evaluate
5x 2y . 共x, y兲 → 共0, 0兲 x y 2 lim
2
Solution In this case, the limits of the numerator and of the denominator are both 0, and so you cannot determine the existence (or nonexistence) of a limit by taking the limits of the numerator and denominator separately and then dividing. From the graph of f in Figure 13.20, however, it seems reasonable that the limit might be 0. So, you can try applying the definition to L 0. First, note that
ⱍyⱍ 冪x 2 y 2 and x2 1. x y2 2
z
Then, in a -neighborhood about 共0, 0兲, you have
7
0 < 冪x 2 y 2 <
6
and it follows that, for 共x, y兲 共0, 0兲,
5
ⱍ f 共x, y兲 0ⱍ −5 − 4
2
3
4
5
5
y
x
<
Surface: 5x 2y f (x, y) = 2 x + y2
Figure 13.20
ⱍ ⱍ
5x 2y y2 x2 5y 2 x y2 5y 5冪x 2 y 2 5. x2
ⱍ ⱍ冢 ⱍⱍ
冣
So, you can choose 兾5 and conclude that 5x 2y 0. 共x, y兲 → 共0, 0兲 x y 2 lim
2
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.2 z
Limits and Continuity
883
For some functions, it is easy to recognize that a limit does not exist. For instance, it is clear that the limit lim
4
共x, y兲 → 共0, 0兲
1 x2 y2
does not exist because the values of f 共x, y兲 increase without bound as 共x, y兲 approaches 共0, 0兲 along any path (see Figure 13.21). For other functions, it is not so easy to recognize that a limit does not exist. For instance, the next example describes a limit that does not exist because the function approaches different values along different paths. 3 x
y
A Limit That Does Not Exist
3
1 does not exist. x2 y 2 Figure 13.21
See LarsonCalculus.com for an interactive version of this type of example.
lim
共x, y兲 → 共0, 0兲
Show that the limit does not exist. lim
共x, y兲 → 共0, 0兲
Solution
冢xx
2 2
y2 y2
冣
2
The domain of the function
f 共x, y兲
冢xx
2 2
y2 y2
冣
2
consists of all points in the xy-plane except for the point 共0, 0兲. To show that the limit as 共x, y兲 approaches 共0, 0兲 does not exist, consider approaching 共0, 0兲 along two different “paths,” as shown in Figure 13.22. Along the x-axis, every point is of the form
共x, 0兲 and the limit along this approach is lim
共x, 0兲 → 共0, 0兲
冢xx
2 2
02 02
冣
2
lim
12 1.
Limit along x-axis
共x, 0兲 → 共0, 0兲
However, when 共x, y兲 approaches 共0, 0兲 along the line y x, you obtain lim
共x, x兲 → 共0, 0兲
冢
x2 x2 x2 x2
冣
2
lim
共x, x兲 → 共0, 0兲
冢 冣 0 2x 2
2
0.
Limit along line y x
This means that in any open disk centered at 共0, 0兲, there are points 共x, y兲 at which f takes on the value 1, and other points at which f takes on the value 0. For instance,
z Along x-axis: (x, 0) → (0, 0)
Limit is 1. 2
f 共x, y兲 1 at 共1, 0兲, 共0.1, 0兲, 共0.01, 0兲, and 共0.001, 0兲, and
3 x
3
f 共x, y兲 0 at 共1, 1兲, 共0.1, 0.1兲, 共0.01, 0.01兲, and 共0.001, 0.001兲. So, f does not have a limit as 共x, y兲 approaches 共0, 0兲.
y
Along y = x: (x, x) → (0, 0) Limit is 0.
lim
共x, y兲 → 共0, 0兲
冢xx
2 2
y2 y2
冣
2
does not exist.
Figure 13.22
In Example 4, you could conclude that the limit does not exist because you found two approaches that produced different limits. Be sure you understand that when two approaches produce the same limit, you cannot conclude that the limit exists. To form such a conclusion, you must show that the limit is the same along all possible approaches.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
884
Chapter 13
Functions of Several Variables
Continuity of a Function of Two Variables Notice in Example 2 that the limit of f 共x, y兲 5x 2 y兾共x 2 y 2兲 as 共x, y兲 → 共1, 2兲 can be evaluated by direct substitution. That is, the limit is f 共1, 2兲 2. In such cases, the function f is said to be continuous at the point 共1, 2兲.
REMARK This definition of continuity can be extended to boundary points of the open region R by considering a special type of limit in which 共x, y兲 is allowed to approach 共x0, y0兲 along paths lying in the region R. This notion is similar to that of one-sided limits, as discussed in Chapter 1.
Definition of Continuity of a Function of Two Variables A function f of two variables is continuous at a point 冇x0, y0冈 in an open region R if f 共x0, y0兲 is equal to the limit of f 共x, y兲 as 共x, y兲 approaches 共x0, y0兲. That is, lim
共x, y兲 → 共x0 , y0 兲
f 共x, y兲 f 共x0, y0兲.
The function f is continuous in the open region R if it is continuous at every point in R.
In Example 3, it was shown that the function f 共x, y兲
5x 2y y2
x2
is not continuous at 共0, 0兲. Because the limit at this point exists, however, you can remove the discontinuity by defining f at 共0, 0兲 as being equal to its limit there. Such a discontinuity is called removable. In Example 4, the function f 共x, y兲
冢xx
2 2
y2 y2
冣
2
was also shown not to be continuous at 共0, 0兲, but this discontinuity is nonremovable. THEOREM 13.1 Continuous Functions of Two Variables If k is a real number and f and g are continuous at 共x0, y0兲, then the following functions are also continuous at 共x0, y0兲. 1. Scalar multiple: kf 3. Product: fg
2. Sum or difference: f ± g 4. Quotient: f兾g, g共x0, y0兲 0
Theorem 13.1 establishes the continuity of polynomial and rational functions at every point in their domains. Furthermore, the continuity of other types of functions can be extended naturally from one to two variables. For instance, the functions whose graphs are shown in Figures 13.23 and 13.24 are continuous at every point in the plane. z
z
Surface: f (x, y) = 12 sin(x 2 + y 2)
Surface: f (x, y) = cos(y 2)e −
x2 + y2
2
x
y
The function f is continuous at every point in the plane. Figure 13.23
x
2
2
y
The function f is continuous at every point in the plane. Figure 13.24
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.2
885
Limits and Continuity
The next theorem states conditions under which a composite function is continuous. THEOREM 13.2 Continuity of a Composite Function If h is continuous at 共x0, y0兲 and g is continuous at h共x0, y0兲, then the composite function given by 共g h兲共x, y兲 g共h共x, y兲兲 is continuous at 共x0, y0兲. That is, lim
共x, y兲 → 共x0, y0兲
g共h共x, y兲兲 g共h共x0, y0兲兲.
Note in Theorem 13.2 that h is a function of two variables and g is a function of one variable.
Testing for Continuity Discuss the continuity of each function. a. f 共x, y兲
x 2y x2 y 2
b. g共x, y兲
2 y x2
Solution a. Because a rational function is continuous at every point in its domain, you can conclude that f is continuous at each point in the xy-plane except at 共0, 0兲, as shown in Figure 13.25. b. The function g共x, y兲
2 y x2
is continuous except at the points at which the denominator is 0, which is given by the equation y x2 0. So, you can conclude that the function is continuous at all points except those lying on the parabola y x2. Inside this parabola, you have y > x 2, and the surface represented by the function lies above the xy-plane, as shown in Figure 13.26. Outside the parabola, y < x 2, and the surface lies below the xy-plane. z
z
g(x, y) = 5
5
2 y − x2
4 3 2
4 3
4
x
y
y 5 x
f (x, y) =
x − 2y x2 + y2
The function f is not continuous at 共0, 0兲. Figure 13.25
y = x2
The function g is not continuous on the parabola y x2. Figure 13.26
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886
Chapter 13
Functions of Several Variables
Continuity of a Function of Three Variables The preceding definitions of limits and continuity can be extended to functions of three variables by considering points 共x, y, z兲 within the open sphere
共x x0兲2 共y y0兲2 共z z0兲2 < 2.
Open sphere
The radius of this sphere is , and the sphere is centered at 共x0, y0, z0兲, as shown in Figure 13.27. z
(x0, y0, z0)
δ
y x
Open sphere in space Figure 13.27
A point 共x0, y0, z0兲 in a region R in space is an interior point of R if there exists a -sphere about 共x0, y0, z0兲 that lies entirely in R. If every point in R is an interior point, then R is called open. Definition of Continuity of a Function of Three Variables A function f of three variables is continuous at a point 冇x0, y0, z0冈 in an open region R if f 共x0, y0, z0兲 is defined and is equal to the limit of f 共x, y, z兲 as 共x, y, z兲 approaches 共x0, y0, z0兲. That is, lim
共x, y, z兲 → 共x0 , y0, z 0兲
f 共x, y, z兲 f 共x0, y0, z0兲.
The function f is continuous in the open region R if it is continuous at every point in R.
Testing Continuity of a Function of Three Variables Discuss the continuity of f 共x, y, z兲
x2
1 . y2 z
Solution The function f is continuous except at the points at which the denominator is 0, which are given by the equation x2 y2 z 0. So, f is continuous at each point in space except at the points on the paraboloid z x2 y2.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.2
13.2 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Verifying a Limit by the Definition In Exercises 1–4, use the definition of the limit of a function of two variables to verify the limit. 1. 3.
lim
共x, y兲 → 共1, 0兲
x1
2.
y 3
lim
共x, y兲 → 共1, 3兲
4.
lim
共x, y兲 → 共4, 1兲
lim
共x, y兲 → 共a, b兲
x4
yb
Using Properties of Limits In Exercises 5–8, find the
34.
lim
共x, y, z兲 → 共0, 0, 0兲
xy yz2 xz2 x2 y2 z2
Continuity In Exercises 35 and 36, discuss the continuity of the function and evaluate the limit of f 冇x, y冈 (if it exists) as 冇x, y冈 → 冇0, 0冈. 35. f 共x, y兲 exy z
indicated limit by using the limits lim
冇x, y冈 → 冇a, b冈
5. 6. 7. 8.
f 冇x, y冈 4 and
lim
关 f 共x, y兲 g共x, y兲兴
lim
冤 g共x, y兲 冥
lim
关 f 共x, y兲g共x, y兲兴
lim
冤 f 共x, yf兲共x,yg兲共x, y兲冥
共x, y兲 → 共a, b兲
共x, y兲 → 共a, b兲 共x, y兲 → 共a, b兲
共x, y兲 → 共a, b兲
lim
冇x, y冈 → 冇a, b冈
g冇x, y冈 3.
7
5 f 共x, y兲
Limit and Continuity In Exercises 9–22, find the limit and discuss the continuity of the function. 9. 11. 13. 15. 17. 19. 21.
lim
共2x2 y兲
10.
lim
e xy
12.
lim
x y
14.
共x, y兲 → 共2, 1兲 共x, y兲 → 共1, 2兲
共x, y兲 → 共0, 2兲
xy 共x, y兲 → 共1, 1兲 x 2 y 2 lim
lim
共x, y兲 → 共 兾4, 2兲
lim
共x, y兲 → 共0, 1兲
887
Limits and Continuity
y cos xy
arcsin xy 1 xy
lim
共x, y, z兲 → 共1, 3, 4兲
冪x y z
lim
共x 4y 1兲
lim
xy x2 1
lim
xy xy
共x, y兲 → 共0, 0兲 共x, y兲 → 共2, 4兲
共x, y兲 → 共1, 2兲
1
20.
36. f 共x, y兲 1
22.
cos共x 2 y 2兲 x2 y 2 z 2
lim
lim
共x, y兲 → 共0, 1兲
arccos共x兾y兲 1 xy
lim
共x, y, z兲 → 共2, 1, 0兲
y
3
x
x 共x, y兲 → 共1, 1兲 冪x y x lim sin 18. 共x, y兲 → 共2 , 4兲 y 16.
2
3
1
5
4
3
4
y
5
x
xeyz
Finding a Limit In Exercises 23–34, find the limit (if it exists). If the limit does not exist, explain why. 23. 25.
lim
xy 1 1 xy
24.
lim
1 xy
26.
共x, y兲 → 共1, 1兲
共x, y兲 → 共0, 0兲
xy 共x, y兲 → 共0, 0兲 冪x 冪y xy lim 29. 共x, y兲 → 共0, 0兲 x 2 y 27.
31. 32. 33.
lim
lim
共x, y兲 → 共0, 0兲
lim
共x, y兲 → 共0, 0兲
lim
共
x2
x2 1兲共 y2 1兲
ln共x2 y2兲
共x, y, z兲 → 共0, 0, 0兲
xy yz xz x2 y2 z2
lim
x 2y
共x, y兲 → 共1, 1兲
lim
共x, y兲 → 共0, 0兲
1 xy2
1 x2y2
xy1 共x, y兲 → 共2, 1兲 冪x y 1 x lim 30. 共x, y兲 → 共0, 0兲 x 2 y 2 28.
lim
Limit and Continuity In Exercises 37–40, use a graphing utility to make a table showing the values of f 冇x, y冈 at the given points for each path. Use the result to make a conjecture about the limit of f 冇x, y冈 as 冇x, y冈 → 冇0, 0冈. Determine analytically whether the limit exists and discuss the continuity of the function. 37. f 共x, y兲
xy x2 y2
Path: y 0 Points: 共1, 0兲, 共0.5, 0兲, 共0.1, 0兲, 共0.01, 0兲, 共0.001, 0兲 Path: y x Points: 共1, 1兲, 共0.5, 0.5兲, 共0.1, 0.1兲, 共0.01, 0.01兲, 共0.001, 0.001兲
z 2 2
2 x
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y
888
Chapter 13
38. f 共x, y兲
x2
Functions of Several Variables z
xy 2 y4
Finding a Limit Using Polar Coordinates In Exercises 43–48, use polar coordinates to find the limit. [Hint: Let x r cos and y r sin , and note that 冇x, y冈 → 冇0, 0冈 implies r → 0.]
2
Path: x y 2 Points: 共1, 1兲, 共0.25, 0.5兲, 共0.01, 0.1兲, 共0.0001, 0.01兲, 共0.000001, 0.001兲 Path: x
3
y
43.
4 x
45.
y 2
Points: 共1, 1兲, 共0.25, 0.5兲, 共0.01, 0.1兲, 共0.0001, 0.01兲, 共0.000001, 0.001兲 39. f 共x, y兲
x2
47.
3 2
52. x
Points: 共1, 1兲, 共0.5, 0.5兲, 共0.1, 0.1兲, 共0.01, 0.01兲, 共0.001, 0.001兲
46.
lim
cos共x2 y2兲
48.
共x, y兲 → 共0, 0兲
x2 y2
lim
共x, y兲 → 共0, 0兲
lim
共x, y兲 → 共0, 0兲
冪x2 y2
sin冪x2 y2
3
3
lim
lim
共x, y兲 → 共0, 0兲
共x2 y2兲ln 共x2 y2兲
Continuity In Exercises 53–58, discuss the continuity of the function. 53. f 共x, y, z兲
z
2x y 2 2x2 y
4
Points: 共1, 0兲, 共0.25, 0兲, 共0.01, 0兲, 共0.001, 0兲, 共0.000001, 0兲
−3
z x2 y2 4
55. f 共x, y, z兲
sin z ex ey
57. f 共x, y兲 2
Path: y x
3
y
x
−4
Comparing Continuity In Exercises 41 and 42, discuss the continuity of the functions f and g. Explain any differences.
共x, y兲 共0, 0兲 共x, y兲 共0, 0兲 共x, y兲 共0, 0兲 共x, y兲 共0, 0兲
x 2 2xy 2 y 2 , 共x, y兲 共0, 0兲 42. f 共x, y兲 x2 y2 0, 共x, y兲 共0, 0兲 x 2 2xy 2 y 2 , 共x, y兲 共0, 0兲 x2 y2 1, 共x, y兲 共0, 0兲
1 冪x 2 y 2 z 2
54. f 共x, y, z兲
−3 −2
Points: 共1, 1兲, 共0.25, 0.25兲, 共0.01, 0.01兲, 共0.001, 0.001兲, 共0.0001, 0.0001兲
sin共x 2 y 2兲 共x, y兲 → 共0, 0兲 x2 y 2 lim
y
Path: y 0
g共x, y兲
x2y2 x2 y2
sin冪x2 y2 50. 共x, y兲 → 共0, 0兲 冪x2 y2 1 cos共x2 y2兲 51. lim 共x, y兲 → 共0, 0兲 x2 y2 49.
4
Path: y x
x 4 y4 , g 共x, y兲 x 2 y 2 1,
lim
共x, y兲 → 共0, 0兲
x3 y3 共x, y兲 → 共0, 0兲 x 2 y 2 lim
z
Points: 共1, 0兲, 共0.5, 0兲, 共0.1, 0兲, 共0.01, 0兲, 共0.001, 0兲
冦 冦 冦 冦
44.
49–52, use polar coordinates and L’Hôpital’s Rule to find the limit.
y y2
x 4 y4 , 41. f 共x, y兲 x 2 y 2 0,
xy 2 y2
共x, y兲 → 共0, 0兲 x 2
Finding a Limit Using Polar Coordinates In Exercises
Path: y 0
40. f 共x, y兲
lim
冦 冦
56. f 共x, y, z兲 xy sin z
sin xy , xy 0 xy 1, xy 0
sin共x2 y2兲 , x2 y2 x2 y2 58. f 共x, y兲 1, x2 y2
Continuity of a Composite Function In Exercises 59–62, discuss the continuity of the composite function f g. 59. f 共t兲 t 2
60. f 共t兲
g共x, y兲 2x 3y 61. f 共t兲
1 t
1 t
g共x, y兲 x 2 y 2 62. f 共t兲
g共x, y兲 2x 3y
1 1t
g共x, y兲 x 2 y 2
Finding a Limit In Exercises 63–68, find each limit. (a) lim
f 冇x x, y冈 f 冇x, y冈 x
(b) lim
f 冇x, y y冈 f 冇x, y冈 y
x→0
y→0
63. f 共x, y兲 x 2 4y 65. f 共x, y兲
x y
67. f 共x, y兲 3x xy 2y
64. f 共x, y兲 x 2 y 2 66. f 共x, y兲
1 xy
68. f 共x, y兲 冪y 共 y 1兲
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13.2
True or False? In Exercises 69–72, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 69. If 70. If
lim
f 共x, y兲 0, then lim f 共x, 0兲 0.
lim
f 共0, y兲 0, then
共x, y兲 → 共0, 0兲 共x, y兲 → 共0, 0兲
76.
lim
共x, y, z兲 → 共0, 0, 0兲
x→0
lim
共x, y兲 → 共0, 0兲
f 共x, y兲 0.
lim
共x, y兲 → 共0, 1兲
72. If g and h are continuous functions of x and y, and f 共x, y兲 g共x兲 h共 y兲, then f is continuous.
1 y2 z2
2
冥
Find the following limit. x2 1 共 y 1兲2
冤x
tan1
冥
78. Continuity For the function
共x, y兲 → 共0, 0兲
Consider
2
77. Finding a Limit
71. If f is continuous for all nonzero x and y, and f 共0, 0兲 0, then lim f 共x, y兲 0.
73. Limit
冤x
tan1
889
Limits and Continuity
f 共x, y兲 xy
y2 y2
冢xx
2 2
冣
define f 共0, 0兲 such that f is continuous at the origin.
x2 y2 (see figure). 共x, y兲 → 共0, 0兲 xy lim
79. Proof
z
Prove that
lim
共x, y兲 → 共a, b兲
20
关 f 共x, y兲 g共x, y兲兴 L1 L2
where f 共x, y兲 approaches L1 and g共x, y兲 approaches L 2 as 共x, y兲 → 共a, b兲.
20
20 x
80. Proof Prove that if f is continuous and f 共a, b兲 < 0, then there exists a -neighborhood about 共a, b兲 such that f 共x, y兲 < 0 for every point 共x, y兲 in the neighborhood.
y
WRITING ABOUT CONCEPTS 81. Limit Define the limit of a function of two variables. Describe a method for showing that
(a) Determine (if possible) the limit along any line of the form y ax.
lim
共x, y兲 → 共x0, y0 兲
does not exist.
(b) Determine (if possible) the limit along the parabola y x 2.
82. Continuity State the definition of continuity of a function of two variables.
(c) Does the limit exist? Explain. 74. Limit
Consider
lim
共x, y兲 → 共0, 0兲
x 2y (see figure). x 4 y2
83. Limits and Function Values (a) If f 共2, 3兲 4, can you conclude anything about f 共x, y兲? Give reasons for your answer. lim
z
共x, y兲 → 共2, 3兲
y
1
f 共x, y兲
(b) If
lim
共x, y兲 → 共2, 3兲
f 共x, y兲 4, can you conclude anything
about f 共2, 3兲? Give reasons for your answer. −1
−1
1
x
84.
HOW DO YOU SEE IT? The figure shows the graph of f 共x, y兲 ln x2 y2. From the graph, does it appear that the limit at each point exists? z
(a) Determine (if possible) the limit along any line of the form y ax. (b) Determine (if possible) the limit along the parabola y x2. (c) Does the limit exist? Explain.
Finding a Limit Using Spherical Coordinates In
−8
Exercises 75 and 76, use spherical coordinates to find the limit. [Hint: Let x sin cos , y sin sin , and z cos , and note that 冇x, y, z冈 → 冇0, 0, 0冈 implies → 0 .] 75.
lim
共x, y, z兲 → 共0, 0, 0兲 x2
xyz y2 z2
−6 8
−4 6
4
2
4
6
8
x −5
(a) 共1, 1兲
(b) 共0, 3兲
(c) 共0, 0兲
(d) 共2, 0兲
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
y
890
Chapter 13
Functions of Several Variables
13.3 Partial Derivatives Find and use partial derivatives of a function of two variables. Find and use partial derivatives of a function of three or more variables. Find higher-order partial derivatives of a function of two or three variables.
Partial Derivatives of a Function of Two Variables In applications of functions of several variables, the question often arises, “How will the value of a function be affected by a change in one of its independent variables?” You can answer this by considering the independent variables one at a time. For example, to determine the effect of a catalyst in an experiment, a chemist could conduct the experiment several times using varying amounts of the catalyst, while keeping constant other variables such as temperature and pressure. You can use a similar procedure to determine the rate of change of a function f with respect to one of its several independent variables. This process is called partial differentiation, and the result is referred to as the partial derivative of f with respect to the chosen independent variable.
JEAN LE ROND D’ALEMBERT (1717–1783)
The introduction of partial derivatives followed Newton’s and Leibniz’s work in calculus by several years. Between 1730 and 1760, Leonhard Euler and Jean Le Rond d’Alembert separately published several papers on dynamics, in which they established much of the theory of partial derivatives.These papers used functions of two or more variables to study problems involving equilibrium, fluid motion, and vibrating strings. See LarsonCalculus.com to read more of this biography.
Definition of Partial Derivatives of a Function of Two Variables If z ⫽ f 共x, y兲, then the first partial derivatives of f with respect to x and y are the functions fx and fy defined by fx共x, y兲 ⫽ lim
f 共x ⫹ ⌬x, y兲 ⫺ f 共x, y兲 ⌬x
Partial derivative with respect to x
fy 共x, y兲 ⫽ lim
f 共x, y ⫹ ⌬y兲 ⫺ f 共x, y兲 ⌬y
Partial derivative with respect to y
⌬x→0
and ⌬y→0
provided the limits exist. This definition indicates that if z ⫽ f 共x, y兲, then to find fx, you consider y constant and differentiate with respect to x. Similarly, to find fy , you consider x constant and differentiate with respect to y.
Finding Partial Derivatives a. To find fx for f 共x, y兲 ⫽ 3x ⫺ x2y2 ⫹ 2x3y, consider y to be constant and differentiate with respect to x. fx共x, y兲 ⫽ 3 ⫺ 2xy2 ⫹ 6x2y
Partial derivative with respect to x
To find fy, consider x to be constant and differentiate with respect to y. fy共x, y兲 ⫽ ⫺2x2y ⫹ 2x3 b. To find fx for f 共x, y兲 ⫽ 共ln x兲共sin with respect to x.
Partial derivative with respect to y
兲, consider y to be constant and differentiate
x2y
fx 共x, y兲 ⫽ 共ln x兲共cos x2y兲共2xy兲 ⫹
sin x2y x
Partial derivative with respect to x
To find fy, consider x to be constant and differentiate with respect to y. fy共x, y兲 ⫽ 共ln x兲共cos x2y兲共x2兲
Partial derivative with respect to y
Gianni Dagli Orti/The Art Archive/Alamy
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.3
REMARK The notation ⭸z兾⭸x is read as “the partial derivative of z with respect to x,” and ⭸z兾⭸y is read as “the partial derivative of z with respect to y.”
Partial Derivatives
891
Notation for First Partial Derivatives For z ⫽ f 共x, y兲, the partial derivatives fx and fy are denoted by ⭸ ⭸z f 共x, y兲 ⫽ fx共x, y兲 ⫽ z x ⫽ ⭸x ⭸x
Partial derivative with respect to x
⭸ ⭸z f 共x, y兲 ⫽ fy共x, y兲 ⫽ z y ⫽ . ⭸y ⭸y
Partial derivative with respect to y
and
The first partials evaluated at the point 共a, b兲 are denoted by ⭸z ⭸x and ⭸z ⭸y
ⱍ ⱍ
共a, b兲
共a, b兲
⫽ fx共a, b兲
⫽ fy共a, b兲.
Finding and Evaluating Partial Derivatives For f 共x, y兲 ⫽ xe x y, find fx and fy, and evaluate each at the point 共1, ln 2兲. 2
Solution z
(x 0 , y0 , z 0 )
Because
fx共x, y兲 ⫽ xe x y共2xy兲 ⫹ e x 2
2y
Partial derivative with respect to x
the partial derivative of f with respect to x at 共1, ln 2兲 is fx共1, ln 2兲 ⫽ e ln 2共2 ln 2兲 ⫹ e ln 2 ⫽ 4 ln 2 ⫹ 2. Because fy共x, y兲 ⫽ xe x y共x 2兲 2 ⫽ x3ex y 2
y
x
the partial derivative of f with respect to y at 共1, ln 2兲 is
Plane: y = y0
fy共1, ln 2兲 ⫽ e ln 2 ⫽ 2.
⭸f ⫽ slope in x-direction ⭸x Figure 13.28
z
Partial derivative with respect to y
The partial derivatives of a function of two variables, z ⫽ f 共x, y兲, have a useful geometric interpretation. If y ⫽ y0, then z ⫽ f 共x, y0兲 represents the curve formed by intersecting the surface z ⫽ f 共x, y兲 with the plane y ⫽ y0, as shown in Figure 13.28. Therefore,
(x 0 , y0 , z 0 )
fx共x0, y0兲 ⫽ lim
⌬x→0
f 共x0 ⫹ ⌬x, y0兲 ⫺ f 共x0, y0兲 ⌬x
represents the slope of this curve at the point 共x0, y0, f 共x0, y0 兲兲. Note that both the curve and the tangent line lie in the plane y ⫽ y0. Similarly, y
x
Plane: x = x 0
⭸f ⫽ slope in y-direction ⭸y Figure 13.29
fy共x0, y0兲 ⫽ lim
⌬y→0
f 共x0, y0 ⫹ ⌬y兲 ⫺ f 共x0, y0兲 ⌬y
represents the slope of the curve given by the intersection of z ⫽ f 共x, y兲 and the plane x ⫽ x0 at 共x0, y0, f 共x0, y0兲兲, as shown in Figure 13.29. Informally, the values of ⭸f兾⭸x and ⭸f兾⭸y at the point 共x0, y0, z0兲 denote the slopes of the surface in the x- and y-directions, respectively.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
892
Chapter 13
Functions of Several Variables
Finding the Slopes of a Surface See LarsonCalculus.com for an interactive version of this type of example.
Find the slopes in the x-direction and in the y-direction of the surface f 共x, y兲 ⫽ ⫺
x2 25 ⫺ y2 ⫹ 2 8
at the point 共 2, 1, 2兲. 1
Solution
The partial derivatives of f with respect to x and y are
fx共x, y兲 ⫽ ⫺x and
fy共x, y兲 ⫽ ⫺2y.
Partial derivatives
So, in the x-direction, the slope is fx
冢12, 1冣 ⫽ ⫺ 21
Figure 13.30
and in the y-direction, the slope is fy
冢12, 1冣 ⫽ ⫺2.
Figure 13.31
z
z
Surface: 4
f (x, y) = −
(
25 x2 − y2 + 8 2
( 12 , 1, 2(
(
1 , 1, 2 2
2 3
y
2
Slope in x-direction: 1 1 fx , 1 = − 2 2
x
4
( (
Figure 13.30
3
y
Slope in y-direction: 1 fy , 1 = −2 2
x
( (
Figure 13.31
Finding the Slopes of a Surface Find the slopes of the surface f 共x, y兲 ⫽ 1 ⫺ 共x ⫺ 1兲2 ⫺ 共 y ⫺ 2兲 2
Surface:
at the point 共1, 2, 1兲 in the x-direction and in the y-direction.
f (x, y) = 1 − (x − 1)2 − (y − 2)2 z 1
(1, 2, 1)
Solution
fx(x, y)
fx共x, y兲 ⫽ ⫺2共x ⫺ 1兲 and
fy(x, y)
fy共x, y兲 ⫽ ⫺2共 y ⫺ 2兲.
Partial derivatives
So, at the point 共1, 2, 1兲, the slope in the x-direction is
1 2 3 4
x
The partial derivatives of f with respect to x and y are
y
fx共1, 2兲 ⫽ ⫺2共1 ⫺ 1兲 ⫽ 0 and the slope in the y-direction is fy共1, 2兲 ⫽ ⫺2共2 ⫺ 2兲 ⫽ 0 as shown in Figure 13.32.
Figure 13.32
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13.3
Partial Derivatives
893
No matter how many variables are involved, partial derivatives can be interpreted as rates of change.
Using Partial Derivatives to Find Rates of Change
a
A = ab sin θ
a sin θ
The area of a parallelogram with adjacent sides a and b and included angle is given by A ⫽ ab sin , as shown in Figure 13.33.
. 6 b. Find the rate of change of A with respect to for a ⫽ 10, b ⫽ 20, and ⫽ . 6 a. Find the rate of change of A with respect to a for a ⫽ 10, b ⫽ 20, and ⫽
θ b
The area of the parallelogram is ab sin . Figure 13.33
Solution a. To find the rate of change of the area with respect to a, hold b and constant and differentiate with respect to a to obtain ⭸A ⫽ b sin . ⭸a
Find partial derivative with respect to a.
For a ⫽ 10, b ⫽ 20, and ⫽ 兾6, the rate of change of the area with respect to a is ⭸A ⫽ 20 sin ⫽ 10. ⭸a 6
Substitute for b and .
b. To find the rate of change of the area with respect to , hold a and b constant and differentiate with respect to to obtain ⭸A ⫽ ab cos . ⭸
Find partial derivative with respect to .
For a ⫽ 10, b ⫽ 20, and ⫽ 兾6, the rate of change of the area with respect to is ⭸A ⫽ 200 cos ⫽ 100冪3. ⭸ 6
Substitute for a, b, and .
Partial Derivatives of a Function of Three or More Variables The concept of a partial derivative can be extended naturally to functions of three or more variables. For instance, if w ⫽ f 共x, y, z兲, then there are three partial derivatives, each of which is formed by holding two of the variables constant. That is, to define the partial derivative of w with respect to x, consider y and z to be constant and differentiate with respect to x. A similar process is used to find the derivatives of w with respect to y and with respect to z. ⭸w f 共x ⫹ ⌬x, y, z兲 ⫺ f 共x, y, z兲 ⫽ fx共x, y, z兲 ⫽ lim ⌬x →0 ⭸x ⌬x ⭸w f 共x, y ⫹ ⌬y, z兲 ⫺ f 共x, y, z兲 ⫽ fy共x, y, z兲 ⫽ lim ⌬y→0 ⭸y ⌬y ⭸w f 共x, y, z ⫹ ⌬z兲 ⫺ f 共x, y, z兲 ⫽ fz共x, y, z兲 ⫽ lim ⌬z→0 ⭸z ⌬z In general, if w ⫽ f 共x1, x 2, . . . , xn兲, then there are n partial derivatives denoted by ⭸w ⫽ fxk共x1, x2, . . . , xn兲, k ⫽ 1, 2, . . . , n. ⭸xk To find the partial derivative with respect to one of the variables, hold the other variables constant and differentiate with respect to the given variable.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
894
Chapter 13
Functions of Several Variables
Finding Partial Derivatives a. To find the partial derivative of f 共x, y, z兲 ⫽ xy ⫹ yz 2 ⫹ xz with respect to z, consider x and y to be constant and obtain ⭸ 关xy ⫹ yz 2 ⫹ xz兴 ⫽ 2yz ⫹ x. ⭸z b. To find the partial derivative of f 共x, y, z兲 ⫽ z sin共xy 2 ⫹ 2z兲 with respect to z, consider x and y to be constant. Then, using the Product Rule, you obtain ⭸ ⭸ ⭸ 关z sin共xy 2 ⫹ 2z兲兴 ⫽ 共z兲 关sin共xy 2 ⫹ 2z兲兴 ⫹ sin共xy 2 ⫹ 2z兲 关z兴 ⭸z ⭸z ⭸z ⫽ 共z兲关cos共xy 2 ⫹ 2z兲兴共2兲 ⫹ sin共xy 2 ⫹ 2z兲 ⫽ 2z cos共xy 2 ⫹ 2z兲 ⫹ sin共xy 2 ⫹ 2z兲. c. To find the partial derivative of f 共x, y, z, w兲 ⫽
x⫹y⫹z w
with respect to w, consider x, y, and z to be constant and obtain ⭸ x⫹y⫹z x⫹y⫹z ⫽⫺ . ⭸w w w2
冤
冥
Higher-Order Partial Derivatives As is true for ordinary derivatives, it is possible to take second, third, and higher-order partial derivatives of a function of several variables, provided such derivatives exist. Higher-order derivatives are denoted by the order in which the differentiation occurs. For instance, the function z ⫽ f 共x, y兲 has the following second partial derivatives. 1. Differentiate twice with respect to x: ⭸ ⭸f ⭸ 2f ⫽ 2 ⫽ fxx. ⭸x ⭸x ⭸x
冢 冣
2. Differentiate twice with respect to y: ⭸ ⭸f ⭸ 2f ⫽ 2 ⫽ fyy. ⭸y ⭸y ⭸y
冢 冣
REMARK Note that the two types of notation for mixed partials have different conventions for indicating the order of differentiation. ⭸ ⭸f ⭸ 2f ⫽ ⭸y ⭸x ⭸y⭸x 共 fx兲y ⫽ fxy
冢 冣
Right-toleft order Left-toright order
You can remember the order by observing that in both notations you differentiate first with respect to the variable “nearest” f.
3. Differentiate first with respect to x and then with respect to y: ⭸ ⭸f ⭸ 2f ⫽ ⫽ fxy. ⭸y ⭸x ⭸y⭸x
冢 冣
4. Differentiate first with respect to y and then with respect to x: ⭸ ⭸f ⭸ 2f ⫽ ⫽ fyx. ⭸x ⭸y ⭸x⭸y
冢 冣
The third and fourth cases are called mixed partial derivatives.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.3
Partial Derivatives
895
Finding Second Partial Derivatives Find the second partial derivatives of f 共x, y兲 ⫽ 3xy 2 ⫺ 2y ⫹ 5x 2y 2 and determine the value of fxy共⫺1, 2兲. Solution
Begin by finding the first partial derivatives with respect to x and y.
fx共x, y兲 ⫽ 3y 2 ⫹ 10xy 2
and fy共x, y兲 ⫽ 6xy ⫺ 2 ⫹ 10x 2y
Then, differentiate each of these with respect to x and y. fxx共x, y兲 ⫽ 10y 2 and fyy共x, y兲 ⫽ 6x ⫹ 10x 2 fxy共x, y兲 ⫽ 6y ⫹ 20xy and fyx共x, y兲 ⫽ 6y ⫹ 20xy At 共⫺1, 2兲, the value of fxy is fxy共⫺1, 2兲 ⫽ 12 ⫺ 40 ⫽ ⫺28. Notice in Example 7 that the two mixed partials are equal. Sufficient conditions for this occurrence are given in Theorem 13.3. THEOREM 13.3 Equality of Mixed Partial Derivatives If f is a function of x and y such that fxy and fyx are continuous on an open disk R, then, for every 共x, y兲 in R, fxy共x, y兲 ⫽ fyx共x, y兲.
Theorem 13.3 also applies to a function f of three or more variables so long as all second partial derivatives are continuous. For example, if w ⫽ f 共x, y, z兲
Function of three variables
and all the second partial derivatives are continuous in an open region R, then at each point in R, the order of differentiation in the mixed second partial derivatives is irrelevant. If the third partial derivatives of f are also continuous, then the order of differentiation of the mixed third partial derivatives is irrelevant.
Finding Higher-Order Partial Derivatives Show that fxz ⫽ fzx and fxzz ⫽ fzxz ⫽ fzzx for the function f 共x, y, z兲 ⫽ ye x ⫹ x ln z. Solution First partials: fx共x, y, z兲 ⫽ ye x ⫹ ln z, fz共x, y, z兲 ⫽
x z
Second partials (note that the first two are equal): 1 fxz共x, y, z兲 ⫽ , z
1 fzx共x, y, z兲 ⫽ , z
fzz共x, y, z兲 ⫽ ⫺
x z2
Third partials (note that all three are equal): 1 fxzz共x, y, z兲 ⫽ ⫺ 2, z
1 fzxz共x, y, z兲 ⫽ ⫺ 2, z
fzzx共x, y, z兲 ⫽ ⫺
1 z2
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896
Chapter 13
Functions of Several Variables
13.3 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Examining a Partial Derivative In Exercises 1–6, explain whether the Quotient Rule should be used to find the partial derivative. Do not differentiate. ⭸ x2y 1. 2 ⭸x y ⫺ 3
⭸ x2y 2. 2 ⭸y y ⫺ 3
冢 冣 ⭸ x⫺y 3. ⭸y 冢 x ⫹ 1 冣 ⭸ xy 5. ⭸x 冢 x ⫹ 1 冣
冢 冣 ⭸ x⫺y 4. ⭸x 冢 x ⫹ 1 冣 ⭸ xy 6. ⭸y 冢 x ⫹ 1 冣
2
2
冢2, 4 冣
46. f 共x, y兲 ⫽ sin xy,
y 47. f 共x, y兲 ⫽ arctan , 共2, ⫺2兲 x 48. f 共x, y兲 ⫽ arccos xy, 共1, 1兲
2
49. f 共x, y兲 ⫽
2
50. f 共x, y兲 ⫽
xy , 共2, ⫺2兲 x⫺y 2xy 冪4x 2 ⫹ 5y 2
, 共1, 1兲
Finding Partial Derivatives In Exercises 7–38, find both
Finding the Slopes of a Surface In Exercises 51 and 52,
first partial derivatives.
find the slopes of the surface in the x- and y-directions at the given point.
7. f 共x, y兲 ⫽ 2x ⫺ 5y ⫹ 3 9. f 共x, y兲 ⫽
8. f 共x, y兲 ⫽ x 2 ⫺ 2y 2 ⫹ 4
51. g共x, y兲 ⫽ 4 ⫺ x 2 ⫺ y 2
10. f 共x, y兲 ⫽ 4x3y⫺2
x2y3
11. z ⫽ x冪y
共1, 1, 2兲
12. z ⫽ 2y 2冪x
13. z ⫽ x 2 ⫺ 4xy ⫹ 3y 2
14. z ⫽ y 3 ⫺ 2xy 2 ⫺ 1
15. z ⫽ e xy
16. z ⫽ e x兾y
17. z ⫽ x 2e 2y x 19. z ⫽ ln y
18. z ⫽ yey兾x
21. z ⫽ ln共x2 ⫹ y2兲
22. z ⫽ ln
23. z ⫽
5
24. z ⫽
x⫹y x⫺y
xy x2 ⫹ y 2
26. g共x, y兲 ⫽ ln
2 2 e⫺共x ⫹y 兲
4 3 2
冪x 2
x 2
⫹
29. z ⫽ cos xy
30. z ⫽ sin共x ⫹ 2y兲
31. z ⫽ tan共2x ⫺ y兲
32. z ⫽ sin 5x cos 5y
33. z ⫽ ey sin xy
34. z ⫽ cos共x 2 ⫹ y 2兲
35. z ⫽ sinh共2x ⫹ 3y兲
36. z ⫽ cosh xy2
冕 冕
y2
y
共t 2 ⫺ 1兲 dt
x y
冕
Finding Partial Derivatives In Exercises 39–42, use the limit definition of partial derivatives to find fx冇x, y冈 and fy冇x, y冈. 39. f 共x, y兲 ⫽ 3x ⫹ 2y
40. f 共x, y兲 ⫽ x 2 ⫺ 2xy ⫹ y 2
41. f 共x, y兲 ⫽ 冪x ⫹ y
42. f 共x, y兲 ⫽
1 x⫹y
Evaluating Partial Derivatives In Exercises 43–50, evaluate
e⫺x
first partial derivatives with respect to x, y, and z. 53. H共x, y, z兲 ⫽ sin共x ⫹ 2y ⫹ 3z兲 54. f 共x, y, z兲 ⫽ 3x 2 y ⫺ 5xyz ⫹ 10yz 2 55. w ⫽ 冪x 2 ⫹ y 2 ⫹ z 2 7xz x⫹y
45. f 共x, y兲 ⫽ cos共2x ⫺ y兲,
冢
, 4 3
58. G共x, y, z兲 ⫽
冣
1 冪1 ⫺ x ⫺ y 2 ⫺ z 2 2
Evaluating Partial Derivatives In Exercises 59–64, evaluate fx , fy , and fz at the given point. 59. f 共x, y, z兲 ⫽ x3yz2, 60. f 共x, y, z兲 ⫽
x2y3
共1, 1, 1兲
⫹ 2xyz ⫺ 3yz,
共⫺2, 1, 2兲
x 61. f 共x, y, z兲 ⫽ , 共1, ⫺1, ⫺1兲 yz 62. f 共x, y, z兲 ⫽
cos y, 共0, 0兲
y
57. F共x, y, z兲 ⫽ ln冪x 2 ⫹ y 2 ⫹ z 2
共2t ⫺ 1兲 dt
43. f 共x, y兲 ⫽ ey sin x, 共, 0兲
3
Finding Partial Derivatives In Exercises 53–58, find the
y
fx and fy at the given point.
3
y
2
x
56. w ⫽
x
共2t ⫹ 1兲 dt ⫹
x
44. f 共x, y兲 ⫽
7 6
28. f 共x, y兲 ⫽ 冪2x ⫹ y 3
38. f 共x, y兲 ⫽
z
4
27. f 共x, y兲 ⫽ 冪x 2 ⫹ y 2
37. f 共x, y兲 ⫽
共⫺2, 1, 3兲 z
20. z ⫽ ln冪xy
x2 3y 2 ⫹ 2y x
25. h共x, y兲 ⫽
52. h共x, y兲 ⫽ x 2 ⫺ y 2
xy , 共3, 1, ⫺1兲 x⫹y⫹z
63. f 共x, y, z兲 ⫽ z sin共x ⫹ y兲,
冢0, 2 , ⫺4冣
64. f 共x, y, z兲 ⫽ 冪3x2 ⫹ y2 ⫺ 2z2, 共1, ⫺2, 1兲
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.3
Using First Partial Derivatives In Exercises 65–72, for f 冇x, y冈, find all values of x and y such that fx冇x, y冈 ⴝ 0 and fy冇x, y冈 ⴝ 0 simultaneously. 66. f 共x, y兲 ⫽ x2 ⫺ xy ⫹ y2 ⫺ 5x ⫹ y
897
Heat Equation In Exercises 99 and 100, show that the function satisfies the heat equation z/t ⴝ c2冇 2z/x 2冈. 99. z ⫽ e⫺t cos
65. f 共x, y兲 ⫽ x2 ⫹ xy ⫹ y2 ⫺ 2x ⫹ 2y
Partial Derivatives
x c
100. z ⫽ e⫺t sin
x c
67. f 共x, y兲 ⫽ x 2 ⫹ 4xy ⫹ y 2 ⫺ 4x ⫹ 16y ⫹ 3
Using First Partial Derivatives In Exercises 101 and 102, determine whether there exists a function f 冇x, y冈 with the given
68. f 共x, y兲 ⫽ x2 ⫺ xy ⫹ y2
partial derivatives. Explain your reasoning. If such a function exists, give an example.
69. f 共x, y兲 ⫽
1 1 ⫹ ⫹ xy x y
70. f 共x, y兲 ⫽
3x3
⫺ 12xy ⫹
101. fx共x, y兲 ⫽ ⫺3 sin共3x ⫺ 2y兲, fy共x, y兲 ⫽ 2 sin共3x ⫺ 2y兲 102. fx共x, y兲 ⫽ 2x ⫹ y, fy共x, y兲 ⫽ x ⫺ 4y
y3
2 ⫹xy⫹y 2
71. f 共x, y兲 ⫽ e x
72. f 共x, y兲 ⫽ ln共x 2 ⫹ y 2 ⫹ 1兲
WRITING ABOUT CONCEPTS
Finding Second Partial Derivatives In Exercises 73–82,
103. First Partial Derivatives Let f be a function of two variables x and y. Describe the procedure for finding the first partial derivatives.
find the four second partial derivatives. Observe that the second mixed partials are equal. 73. z ⫽ 3xy2
74. z ⫽ x2 ⫹ 3y2
75. z ⫽ x 2 ⫺ 2xy ⫹ 3y 2
76. z ⫽ x 4 ⫺ 3x 2 y 2 ⫹ y 4
77. z ⫽
78. z ⫽ ln共x ⫺ y兲
冪x 2
⫹
y2
79. z ⫽ e x tan y
80. z ⫽ 2xe y ⫺ 3ye⫺ x
81. z ⫽ cos xy
82. z ⫽ arctan
y x
In Exercises 83–86, use a computer algebra system to find the first and second partial derivatives of the function. Determine whether there exist values of x and y such that fx冇x, y冈 ⴝ 0 and fy冇x, y冈 ⴝ 0 simultaneously.
85. f 共x, y兲 ⫽ ln
x x2 ⫹ y2
84. f 共x, y兲 ⫽ 冪25 ⫺ x 2 ⫺ y 2 86. f 共x, y兲 ⫽
105. Sketching a Graph Sketch the graph of a function z ⫽ f 共x, y兲 whose derivative fx is always negative and whose derivative fy is always positive. 106. Sketching a Graph Sketch the graph of a function z ⫽ f 共x, y兲 whose derivatives fx and fy are always positive.
Finding Partial Derivatives Using Technology
83. f 共x, y兲 ⫽ x sec y
104. First Partial Derivatives Sketch a surface representing a function f of two variables x and y. Use the sketch to give geometric interpretations of ⭸f兾⭸x and ⭸f兾⭸y.
107. Mixed Partial Derivatives If f is a function of x and y such that fxy and fyx are continuous, what is the relationship between the mixed partial derivatives? Explain.
HOW DO YOU SEE IT? Use the graph of the surface to determine the sign of each partial derivative. Explain your reasoning.
108.
xy x⫺y
Comparing Mixed Partial Derivatives In Exercises 87–90, show that the mixed partial derivatives fxyy , fyxy , and fyyx are equal. 87. f 共x, y, z兲 ⫽ xyz
z 2 −5
88. f 共x, y, z兲 ⫽ x2 ⫺ 3xy ⫹ 4yz ⫹ z3
5
89. f 共x, y, z兲 ⫽ e⫺x sin yz 90. f 共x, y, z兲 ⫽
y
5
2z x⫹y
x
Laplace’s Equation In Exercises 91–94, show that the function satisfies Laplace’s equation z/x ⴙ z/y ⴝ 0. 2
2
2
2
91. z ⫽ 5xy
92. z ⫽ 12共ey ⫺ e⫺y兲sin x
93. z ⫽ ex sin y
94. z ⫽ arctan
y x
Wave Equation In Exercises 95–98, show that the function satisfies the wave equation 2z/t 2 ⴝ c2冇 2z/x 2冈. 95. z ⫽ sin共x ⫺ ct兲
96. z ⫽ cos共4x ⫹ 4ct兲
97. z ⫽ ln共x ⫹ ct兲
98. z ⫽ sin ct sin x
(a) fx共4, 1兲
(b) fy共4, 1兲
(c) fx共⫺1, ⫺2兲
(d) fy共⫺1, ⫺2兲
109. Marginal Revenue A pharmaceutical corporation has two plants that produce the same over-the-counter medicine. If x1 and x2 are the numbers of units produced at plant 1 and plant 2, respectively, then the total revenue for the product is given by R ⫽ 200x1 ⫹ 200x2 ⫺ 4x12 ⫺ 8x1x2 ⫺ 4x22. When x1 ⫽ 4 and x2 ⫽ 12, find (a) the marginal revenue for plant 1, ⭸R兾⭸x1, and (b) the marginal revenue for plant 2, ⭸R兾⭸x2.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
898
Chapter 13
Functions of Several Variables 116. Apparent Temperature A measure of how hot weather feels to an average person is the Apparent Temperature Index. A model for this index is
110. Marginal Costs A company manufactures two types of woodburning stoves: a freestanding model and a fireplace-insert model. The cost function for producing x freestanding and y fireplace-insert stoves is
A ⫽ 0.885t ⫺ 22.4h ⫹ 1.20th ⫺ 0.544 where A is the apparent temperature in degrees Celsius, t is the air temperature, and h is the relative humidity in decimal form. (Source: The UMAP Journal) (a) Find
C ⫽ 32冪xy ⫹ 175x ⫹ 205y ⫹ 1050. (a) Find the marginal costs 共⭸C兾⭸x and ⭸C兾⭸y兲 when x ⫽ 80 and y ⫽ 20. (b) When additional production is required, which model of stove results in the cost increasing at a higher rate? How can this be determined from the cost model?
(b) Which has a greater effect on A, air temperature or humidity? Explain. 117. Ideal Gas Law The Ideal Gas Law states that PV ⫽ nRT, where P is pressure, V is volume, n is the number of moles of gas, R is a fixed constant (the gas constant), and T is absolute temperature. Show that ⭸T ⭸P
111. Psychology Early in the twentieth century, an intelligence test called the Stanford-Binet Test (more commonly known as the IQ test) was developed. In this test, an individual’s mental age M is divided by the individual’s chronological age C and the quotient is multiplied by 100. The result is the individual’s IQ. IQ共M, C兲 ⫽
M ⫻ 100 C
⭸V
118. Marginal Utility The utility function U ⫽ f 共x, y兲 is a measure of the utility (or satisfaction) derived by a person from the consumption of two products x and y. The utility function for two products is U ⫽ ⫺5x 2 ⫹ xy ⫺ 3y 2.
(b) Determine the marginal utility of product y.
112. Marginal Productivity Consider the Cobb-Douglas production function f 共x, y兲 ⫽ 200x 0.7 y 0.3. When x ⫽ 1000 and y ⫽ 500, find (a) the marginal productivity of labor, ⭸f兾⭸x, and (b) the marginal productivity of capital, ⭸f兾⭸y. 113. Think About It Let N be the number of applicants to a university, p the charge for food and housing at the university, and t the tuition. Suppose that N is a function of p and t such that ⭸N兾⭸p < 0 and ⭸N兾⭸t < 0. What information is gained by noticing that both partials are negative? 114. Investment The value of an investment of $1000 earning 6% compounded annually is
冤
⭸P
⭈ ⭸V ⭈ ⭸T ⫽ ⫺1.
(a) Determine the marginal utility of product x.
Find the partial derivatives of IQ with respect to M and with respect to C. Evaluate the partial derivatives at the point 共12, 10兲 and interpret the result. (Source: Adapted from Bernstein/Clark-Stewart/Roy/Wickens, Psychology, Fourth Edition)
1 ⫹ 0.06共1 ⫺ R兲 V共I, R兲 ⫽ 1000 1⫹I
⭸A ⭸A and when t ⫽ 30⬚ and h ⫽ 0.80. ⭸t ⭸h
冥
(c) When x ⫽ 2 and y ⫽ 3, should a person consume one more unit of product x or one more unit of product y? Explain your reasoning. (d) Use a computer algebra system to graph the function. Interpret the marginal utilities of products x and y graphically. 119. Modeling Data The expenditures (in billions of dollars) for different types of recreation in the United States from 2005 through 2010 are shown in the table. Expenditures on amusement parks and campgrounds, live entertainment (excluding sports), and spectator sports are represented by the variables x, y, and z. (Source: U.S. Bureau of Economic Analysis) Year
2005
2006
2007
2008
2009
2010
x
36.4
39.0
42.4
44.7
43.0
45.2
y
15.3
16.6
17.4
17.5
17.0
17.3
z
16.4
18.1
20.0
20.5
20.1
21.4
10
where I is the annual rate of inflation and R is the tax rate for the person making the investment. Calculate VI 共0.03, 0.28兲 and VR共0.03, 0.28兲. Determine whether the tax rate or the rate of inflation is the greater “negative” factor in the growth of the investment. 115. Temperature Distribution The temperature at any point 共x, y兲 in a steel plate is T ⫽ 500 ⫺ 0.6x 2 ⫺ 1.5y 2, where x and y are measured in meters. At the point 共2, 3兲, find the rates of change of the temperature with respect to the distances moved along the plate in the directions of the x- and y-axes.
A model for the data is given by z ⫽ 0.461x ⫹ 0.301y ⫺ 494. (a) Find
⭸z ⭸z and . ⭸x ⭸y
(b) Interpret the partial derivatives in the context of the problem. Amy Walters/Shutterstock.com
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.3 120. Modeling Data The table shows the public medical expenditures (in billions of dollars) for workers’ compensation x, Medicaid y, and Medicare z from 2005 through 2010. (Source: Centers for Medicare and Medicaid Services) Year
2005
2006
2007
2008
2009
2010
x
41.2
41.6
41.2
40.1
36.7
37.2
y
309.5
306.8
326.4
343.8
374.4
401.4
z
338.8
403.1
432.3
466.9
499.8
524.6
A model for the data is given by
Partial Derivatives
899
127. Using a Function Consider the function f 共x, y兲 ⫽ 共x2 ⫹ y2兲2兾3. Show that
冦
4x , fx共x, y兲 ⫽ 3共x2 ⫹ y2兲1兾3 0,
共x, y兲 ⫽ 共0, 0兲 . 共x, y兲 ⫽ 共0, 0兲
FOR FURTHER INFORMATION For more information about this problem, see the article “A Classroom Note on a Naturally Occurring Piecewise Defined Function” by Don Cohen in Mathematics and Computer Education.
z ⫽ 11.734x2 ⫺ 0.028y2 ⫺ 888.24x ⫹ 23.09y ⫹ 12,573.9. (a) Find
⭸2z ⭸2z and 2. ⭸x2 ⭸y
(b) Determine the concavity of traces parallel to the xz-plane. Interpret the result in the context of the problem. (c) Determine the concavity of traces parallel to the yz-plane. Interpret the result in the context of the problem.
True or False? In Exercises 121–124, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 121. If z ⫽ f 共x, y兲 and
⭸z ⭸z ⫽ , then z ⫽ c共x ⫹ y兲. ⭸x ⭸y
122. If z ⫽ f 共x兲g共y兲, then 123. If z ⫽ e xy, then
⭸z ⭸z ⫹ ⫽ f⬘共x兲g共 y兲 ⫹ f 共x兲g⬘共 y兲. ⭸x ⭸y
⭸ 2z ⫽ 共xy ⫹ 1兲e xy. ⭸y⭸x
Moiré Fringes Read the article “Moiré Fringes and the Conic Sections” by Mike Cullen in The College Mathematics Journal. The article describes how two families of level curves given by f 共x, y兲 ⫽ a and g共x, y兲 ⫽ b can form Moiré patterns. After reading the article, write a paper explaining how the expression ⭸f ⭸x
⭈
⭸g ⭸f ⫹ ⭸x ⭸y
⭸g
⭈ ⭸y
is related to the Moiré patterns formed by intersecting the two families of level curves. Use one of the following patterns as an example in your paper.
124. If a cylindrical surface z ⫽ f 共x, y兲 has rulings parallel to the y-axis, then ⭸z兾⭸y ⫽ 0. 125. Using a Function Consider the function defined by
冦
xy共x 2 ⫺ y 2兲 , f 共x, y兲 ⫽ x2 ⫹ y2 0,
共x, y兲 ⫽ 共0, 0兲 . 共x, y兲 ⫽ 共0, 0兲
(a) Find fx 共x, y兲 and fy 共x, y兲 for 共x, y兲 ⫽ 共0, 0兲. (b) Use the definition of partial derivatives to find fx共0, 0兲 and fy共0, 0兲.
冤Hint: f 共0, 0兲 ⫽ x
lim
⌬x→0
f 共⌬x, 0兲 ⫺ f 共0, 0兲 ⌬x
冥
(c) Use the definition of partial derivatives to find fxy共0, 0兲 and fyx共0, 0兲. (d) Using Theorem 13.3 and the result of part (c), what can be said about fxy or fyx? 126. Using a Function Consider the function f 共x, y兲 ⫽ 共x3 ⫹ y3兲1兾3. (a) Find fx共0, 0兲 and fy共0, 0兲. (b) Determine the points (if any) at which fx共x, y兲 or fy共x, y兲 fails to exist. Mike Cullen; Mike Cullen
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900
Chapter 13
Functions of Several Variables
13.4 Differentials Understand the concepts of increments and differentials. Extend the concept of differentiability to a function of two variables. Use a differential as an approximation.
Increments and Differentials In this section, the concepts of increments and differentials are generalized to functions of two or more variables. Recall from Section 3.9 that for y ⫽ f 共x兲, the differential of y was defined as dy ⫽ f ⬘共x兲 dx. Similar terminology is used for a function of two variables, z ⫽ f 共x, y兲. That is, ⌬x and ⌬y are the increments of x and y, and the increment of z is ⌬z ⫽ f 共x ⫹ ⌬x, y ⫹ ⌬y兲 ⫺ f 共x, y兲.
Increment of z
Definition of Total Differential If z ⫽ f 共x, y兲 and ⌬x and ⌬y are increments of x and y, then the differentials of the independent variables x and y are dx ⫽ ⌬x and dy ⫽ ⌬y and the total differential of the dependent variable z is dz ⫽
⭸z ⭸z dx ⫹ dy ⫽ fx 共x, y兲 dx ⫹ fy 共x, y兲 dy. ⭸x ⭸y
This definition can be extended to a function of three or more variables. For instance, if w ⫽ f 共x, y, z, u兲, then dx ⫽ ⌬x, dy ⫽ ⌬y, dz ⫽ ⌬z, du ⫽ ⌬u, and the total differential of w is dw ⫽
⭸w ⭸w ⭸w ⭸w dx ⫹ dy ⫹ dz ⫹ du. ⭸x ⭸y ⭸z ⭸u
Finding the Total Differential Find the total differential for each function. a. z ⫽ 2x sin y ⫺ 3x 2y 2
b. w ⫽ x 2 ⫹ y 2 ⫹ z 2
Solution a. The total differential dz for z ⫽ 2x sin y ⫺ 3x 2y 2 is ⭸z ⭸z dx ⫹ dy ⭸x ⭸y ⫽ 共2 sin y ⫺ 6xy 2兲 dx ⫹ 共2x cos y ⫺ 6x 2y兲 dy.
dz ⫽
Total differential dz
b. The total differential dw for w ⫽ x 2 ⫹ y 2 ⫹ z 2 is ⭸w ⭸w ⭸w dx ⫹ dy ⫹ dz ⭸x ⭸y ⭸z ⫽ 2x dx ⫹ 2y dy ⫹ 2z dz.
dw ⫽
Total differential dw
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13.4
Differentials
901
Differentiability In Section 3.9, you learned that for a differentiable function given by y ⫽ f 共x兲, you can use the differential dy ⫽ f⬘共x兲 dx as an approximation 共for small ⌬x兲 of the value ⌬y ⫽ f 共x ⫹ ⌬x兲 ⫺ f 共x兲. When a similar approximation is possible for a function of two variables, the function is said to be differentiable. This is stated explicitly in the next definition. Definition of Differentiability A function f given by z ⫽ f 共x, y兲 is differentiable at 共x 0, y0 兲 if ⌬z can be written in the form ⌬z ⫽ fx共x0, y0 兲 ⌬x ⫹ fy共x0, y0 兲 ⌬y ⫹ 1⌬x ⫹ 2 ⌬y where both 1 and 2 → 0 as
共⌬x, ⌬y兲 → 共0, 0兲. The function f is differentiable in a region R if it is differentiable at each point in R.
Showing that a Function Is Differentiable Show that the function f 共x, y兲 ⫽ x 2 ⫹ 3y is differentiable at every point in the plane. z
Solution plane is
4
1
4 x
−4
Figure 13.34
y
Letting z ⫽ f 共x, y兲, the increment of z at an arbitrary point 共x, y兲 in the
⌬z ⫽ f 共x ⫹ ⌬x, y ⫹ ⌬y兲 ⫺ f 共x, y兲 Increment of z 2 2 ⫽ 共x ⫹ 2x⌬x ⫹ ⌬x 兲 ⫹ 3共 y ⫹ ⌬y兲 ⫺ 共x 2 ⫹ 3y兲 ⫽ 2x⌬x ⫹ ⌬x 2 ⫹ 3⌬y ⫽ 2x共⌬x兲 ⫹ 3共⌬y兲 ⫹ ⌬x共⌬x兲 ⫹ 0共⌬y兲 ⫽ fx共x, y兲 ⌬x ⫹ fy共x, y兲 ⌬y ⫹ 1⌬x ⫹ 2⌬y where 1 ⫽ ⌬x and 2 ⫽ 0. Because 1 → 0 and 2 → 0 as 共⌬x, ⌬y兲 → 共0, 0兲, it follows that f is differentiable at every point in the plane. The graph of f is shown in Figure 13.34. Be sure you see that the term “differentiable” is used differently for functions of two variables than for functions of one variable. A function of one variable is differentiable at a point when its derivative exists at the point. For a function of two variables, however, the existence of the partial derivatives fx and fy does not guarantee that the function is differentiable (see Example 5). The next theorem gives a sufficient condition for differentiability of a function of two variables. THEOREM 13.4 Sufficient Condition for Differentiability If f is a function of x and y, where fx and fy are continuous in an open region R, then f is differentiable on R. A proof of Theorem 13.4 is given in Appendix A. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
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902
Chapter 13
Functions of Several Variables
Approximation by Differentials Theorem 13.4 tells you that you can choose 共x ⫹ ⌬x, y ⫹ ⌬y兲 close enough to 共x, y兲 to make 1⌬x and 2⌬y insignificant. In other words, for small ⌬x and ⌬y, you can use the approximation
z
∂z Δy ∂y
dz
⌬z ⬇ dz. ∂z Δx ∂x
Δz2 Δz1
This approximation is illustrated graphically in Figure 13.35. Recall that the partial derivatives ⭸z兾⭸x and ⭸z兾⭸y can be interpreted as the slopes of the surface in the Δz x- and y-directions. This means that dz ⫽
y x
(x + Δx, y) (x + Δx, y + Δy)
(x, y)
The exact change in z is ⌬z. This change can be approximated by the differential dz. Figure 13.35
⭸z ⭸z ⌬x ⫹ ⌬y ⭸x ⭸y
represents the change in height of a plane that is tangent to the surface at the point 共x, y, f 共x, y兲兲. Because a plane in space is represented by a linear equation in the variables x, y, and z, the approximation of ⌬z by dz is called a linear approximation. You will learn more about this geometric interpretation in Section 13.7.
Using a Differential as an Approximation See LarsonCalculus.com for an interactive version of this type of example.
Use the differential dz to approximate the change in z ⫽ 冪4 ⫺ x 2 ⫺ y 2 as 共x, y兲 moves from the point 共1, 1兲 to the point 共1.01, 0.97兲. Compare this approximation with the exact change in z. f(x + Δx, y + Δy) f (x, y)
z=
Solution
Letting 共x, y兲 ⫽ 共1, 1兲 and 共x ⫹ ⌬x, y ⫹ ⌬y兲 ⫽ 共1.01, 0.97兲 produces
dx ⫽ ⌬x ⫽ 0.01 and
4 − x2 − y2 z
dy ⫽ ⌬y ⫽ ⫺0.03.
So, the change in z can be approximated by ⌬z ⬇ dz ⫽
2
⭸z ⫺x ⫺y ⭸z dx ⫹ dy ⫽ ⌬x ⫹ ⌬y. ⭸x ⭸y 冪4 ⫺ x 2 ⫺ y 2 冪4 ⫺ x 2 ⫺ y 2
When x ⫽ 1 and y ⫽ 1, you have ⌬z ⬇ ⫺ 2
2 x
y
(1, 1) (1.01, 0.97)
As 共x, y兲 moves from 共1, 1兲 to the point 共1.01, 0.97兲, the value of f 共x, y兲 changes by about 0.0137. Figure 13.36
1 1 0.02 共0.01兲 ⫺ 共⫺0.03兲 ⫽ ⫽ 冪2 共0.01兲 ⬇ 0.0141. 冪2 冪2 冪2
In Figure 13.36, you can see that the exact change corresponds to the difference in the heights of two points on the surface of a hemisphere. This difference is given by ⌬z ⫽ f 共1.01, 0.97兲 ⫺ f 共1, 1兲 ⫽ 冪4 ⫺ 共1.01兲2 ⫺ 共0.97兲2 ⫺ 冪4 ⫺ 12 ⫺ 12 ⬇ 0.0137. A function of three variables w ⫽ f 共x, y, z兲 is differentiable at 共x, y, z兲 provided that ⌬w ⫽ f 共x ⫹ ⌬x, y ⫹ ⌬y, z ⫹ ⌬z兲 ⫺ f 共x, y, z兲 can be written in the form ⌬w ⫽ fx ⌬x ⫹ fy ⌬y ⫹ fz ⌬z ⫹ 1⌬x ⫹ 2⌬y ⫹ 3⌬z where 1, 2, and 3 → 0 as 共⌬x, ⌬y, ⌬z兲 → 共0, 0, 0兲. With this definition of differentiability, Theorem 13.4 has the following extension for functions of three variables: If f is a function of x, y, and z, where f, fx , fy , and fz are continuous in an open region R, then f is differentiable on R. In Section 3.9, you used differentials to approximate the propagated error introduced by an error in measurement. This application of differentials is further illustrated in Example 4.
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13.4
Differentials
903
Error Analysis z 20
The possible error involved in measuring each dimension of a rectangular box is ± 0.1 millimeter. The dimensions of the box are x ⫽ 50 centimeters, y ⫽ 20 centimeters, and z ⫽ 15 centimeters, as shown in Figure 13.37. Use dV to estimate the propagated error and the relative error in the calculated volume of the box.
y = 20
x = 50 z = 15
Solution 20 50
y
⭸V ⭸V ⭸V dx ⫹ dy ⫹ dz ⭸x ⭸y ⭸z ⫽ yz dx ⫹ xz dy ⫹ xy dz.
dV ⫽
x
Volume ⫽ xyz Figure 13.37
The volume of the box is V ⫽ xyz, and so
Using 0.1 millimeter ⫽ 0.01 centimeter, you have dx ⫽ dy ⫽ dz ⫽ ± 0.01 and the propagated error is approximately dV ⫽ 共20兲共15兲共± 0.01兲 ⫹ 共50兲共15兲共± 0.01兲 ⫹ 共50兲共20兲共± 0.01兲 ⫽ 300共± 0.01兲 ⫹ 750共± 0.01兲 ⫹ 1000共± 0.01兲 ⫽ 2050共± 0.01兲 ⫽ ± 20.5 cubic centimeters. Because the measured volume is V ⫽ 共50兲共20兲共15兲 ⫽ 15,000 cubic centimeters, the relative error, ⌬V兾V, is approximately ⌬V dV 20.5 ⬇ ⫽ ⬇ 0.14%. V V 15,000 As is true for a function of a single variable, when a function in two or more variables is differentiable at a point, it is also continuous there. THEOREM 13.5 Differentiability Implies Continuity If a function of x and y is differentiable at 共x0, y0 兲, then it is continuous at 共x0, y0 兲.
Proof
Let f be differentiable at 共x0, y0 兲, where z ⫽ f 共x, y兲. Then
⌬z ⫽ 关 fx 共x0, y0 兲 ⫹ 1兴 ⌬x ⫹ 关 fy共x0, y0 兲 ⫹ 2 兴 ⌬y where both 1 and 2 → 0 as 共⌬x, ⌬y兲 → 共0, 0兲. However, by definition, you know that ⌬z is ⌬z ⫽ f 共x0 ⫹ ⌬ x, y0 ⫹ ⌬y兲 ⫺ f 共x0 , y0 兲. Letting x ⫽ x0 ⫹ ⌬x and y ⫽ y0 ⫹ ⌬y produces f 共x, y兲 ⫺ f 共x0 , y0 兲 ⫽ 关 fx 共x0 , y0 兲 ⫹ 1兴 ⌬x ⫹ 关 fy 共x0 , y0 兲 ⫹ 2兴 ⌬y ⫽ 关 fx 共x0, y0 兲 ⫹ 1兴共x ⫺ x0 兲 ⫹ 关 fy 共x0 , y0 兲 ⫹ 2 兴共 y ⫺ y0 兲. Taking the limit as 共x, y兲 → 共x0 , y0 兲, you have lim
共x, y兲 → 共x0, y0兲
f 共x, y兲 ⫽ f 共x0, y0 兲
which means that f is continuous at 共x0 , y0 兲. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
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904
Chapter 13
Functions of Several Variables
Remember that the existence of fx and fy is not sufficient to guarantee differentiability, as illustrated in the next example.
A Function That Is Not Differentiable For the function
冦
⫺3xy , f 共x, y兲 ⫽ x2 ⫹ y2 0,
共x, y兲 ⫽ 共0, 0兲 共x, y兲 ⫽ 共0, 0兲
show that fx共0, 0兲 and fy共0, 0兲 both exist, but that f is not differentiable at 共0, 0兲. Solution You can show that f is not differentiable at 共0, 0兲 by showing that it is not continuous at this point. To see that f is not continuous at 共0, 0兲, look at the values of f 共x, y兲 along two different approaches to 共0, 0兲, as shown in Figure 13.38. Along the line y ⫽ x, the limit is lim
共x, x兲 → 共0, 0兲
f 共x, y兲 ⫽
⫺3x 2 3 ⫽⫺ 共x, x兲 → 共0, 0兲 2x 2 2 lim
whereas along y ⫽ ⫺x, you have lim
共x, ⫺x兲 → 共0, 0兲
f 共x, y兲 ⫽
3x 2 3 ⫽ . 共x, ⫺x兲 → 共0, 0兲 2x 2 2 lim
So, the limit of f 共x, y兲 as 共x, y兲 → 共0, 0兲 does not exist, and you can conclude that f is not continuous at 共0, 0兲. Therefore, by Theorem 13.5, you know that f is not differentiable at 共0, 0兲. On the other hand, by the definition of the partial derivatives fx and fy , you have fx共0, 0兲 ⫽ lim
f 共⌬x, 0兲 ⫺ f 共0, 0兲 0⫺0 ⫽ lim ⫽0 ⌬x→0 ⌬x ⌬x
fy 共0, 0兲 ⫽ lim
f 共0, ⌬y兲 ⫺ f 共0, 0兲 0⫺0 ⫽ lim ⫽ 0. ⌬y→0 ⌬y ⌬y
⌬x→0
and ⌬y→0
So, the partial derivatives at 共0, 0兲 exist.
f (x, y) =
−3xy , (x, y) ≠ (0, 0) x2 + y2 0,
(x, y) = (0, 0) z
Along the line y = −x, f(x, y) approaches 3/2.
(0, 0, 0)
z y
x y
Along the line y = x, f(x, y) approaches −3/2.
Figure 13.38
x
Generated by Mathematica
TECHNOLOGY A graphing utility can be used to graph piecewise-defined functions like the one given in Example 5. For instance, the graph shown at the left was generated by Mathematica.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.4
13.4 Exercises total differential.
3. z ⫽
2. z ⫽ 2x 4y ⫺ 8x2y3
⫺1 x2 ⫹ y2
4. w ⫽
x⫹y z ⫺ 3y 2 ⫹y2
⫺ e⫺x
2 ⫺y2
5. z ⫽ x cos y ⫺ y cos x
1 6. z ⫽ 2共ex
7. z ⫽ ex sin y
8. w ⫽ e y cos x ⫹ z2
9. w ⫽ 2z3y sin x
兲
10. w ⫽ x 2yz 2 ⫹ sin yz
25. Area The area of the shaded rectangle in the figure is A ⫽ lh. The possible errors in the length and height are ⌬l and ⌬h, respectively. Find dA and identify the regions in the figure whose areas are given by the terms of dA. What region represents the difference between ⌬A and dA? Δh
Δh
h
Using a Differential as an Approximation In Exercises 11–16, (a) evaluate f 冇2, 1冈 and f 冇2.1, 1.05冈 and calculate ⌬z, and (b) use the total differential dz to approximate ⌬z. 11. f 共x, y兲 ⫽ 2x ⫺ 3y
12. f 共x, y兲 ⫽ x2 ⫹ y2
13. f 共x, y兲 ⫽ 16 ⫺ x2 ⫺ y2
14. f 共x, y兲 ⫽
15. f 共x, y兲 ⫽ ye x
16. f 共x, y兲 ⫽ x cos y
y x
Approximating an Expression In Exercises 17–20, find z ⴝ f 冇x, y冈 and use the total differential to approximate the quantity. 17. 共2.01兲2共9.02兲 ⫺ 22 ⭈ 9
18.
1 ⫺ 共3.05兲2 1 ⫺ 32 ⫺ 共5.95兲 2 62
l
Δl Δr
Figure for 25
Figure for 26
26. Volume The volume of the red right circular cylinder in the figure is V ⫽ r 2h. The possible errors in the radius and the height are ⌬r and ⌬h, respectively. Find dV and identify the solids in the figure whose volumes are given by the terms of dV. What solid represents the difference between ⌬V and dV ? 27. Numerical Analysis A right circular cone of height h ⫽ 8 and radius r ⫽ 4 is constructed, and in the process, errors ⌬r and ⌬h are made in the radius and height, respectively. Complete the table to show the relationship between ⌬V and dV for the indicated errors.
19. 冪共5.05兲2 ⫹ 共3.1兲2 ⫺ 冪52 ⫹ 32 20. sin关共1.05兲2 ⫹ 共0.95兲 2兴 ⫺ sin共12 ⫹ 12兲
WRITING ABOUT CONCEPTS 21. Approximation Describe the change in accuracy of dz as an approximation of ⌬z as ⌬x and ⌬y increase. 22. Linear Approximation What is meant by a linear approximation of z ⫽ f 共x, y兲 at the point P共x0, y0兲? 23. Using Differentials When using differentials, what is meant by the terms propagated error and relative error?
24.
HOW DO YOU SEE IT? Which point has a greater differential, 共2, 2兲 or 共12, 12 兲? Explain. (Assume that dx and dy are the same for both points.) z
3 3
⌬r
⌬h
0.1
0.1
0.1
⫺0.1
0.001
0.002
⫺0.0001
0.0002
dV or dS
⌬V or ⌬S
⌬V ⫺ dV or ⌬S ⫺ dS
Table for Exercises 27 and 28
28. Numerical Analysis The height and radius of a right circular cone are measured as h ⫽ 16 meters and r ⫽ 6 meters. In the process of measuring, errors ⌬r and ⌬h are made. Let S be the lateral surface area of the cone. Complete the table above to show the relationship between ⌬S and dS for the indicated errors. 29. Volume The possible error involved in measuring each dimension of a rectangular box is ± 0.02 inch. The dimensions of the box are 8 inches by 5 inches by 12 inches. Approximate the propagated error and the relative error in the calculated volume of the box.
3
x
905
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding a Total Differential In Exercises 1–10, find the 1. z ⫽ 2x 2y 3
Differentials
y
30. Volume The possible error involved in measuring each dimension of a right circular cylinder is ± 0.05 centimeter. The radius is 3 centimeters and the height is 10 centimeters. Approximate the propagated error and the relative error in the calculated volume of the cylinder.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
906
Chapter 13
Functions of Several Variables
31. Wind Chill The formula for wind chill C (in degrees Fahrenheit) is given by C ⫽ 35.74 ⫹ 0.6215T ⫺ 35.75v0.16 ⫹ 0.4275Tv0.16 where v is the wind speed in miles per hour and T is the temperature in degrees Fahrenheit. The wind speed is 23 ± 3 miles per hour and the temperature is 8⬚ ± 1⬚. Use dC to estimate the maximum possible propagated error and relative error in calculating the wind chill. (Source: National Oceanic and Atmospheric Administration)
32. Resistance The total resistance R (in ohms) of two resistors connected in parallel is given by 1 1 1 ⫽ ⫹ . R R1 R 2 Approximate the change in R as R1 is increased from 10 ohms to 10.5 ohms and R2 is decreased from 15 ohms to 13 ohms. 33. Power Electrical power P is given by E2 P⫽ R where E is voltage and R is resistance. Approximate the maximum percent error in calculating power when 120 volts is applied to a 2000-ohm resistor and the possible percent errors in measuring E and R are 3% and 4%, respectively. 34. Acceleration The centripetal acceleration of a particle moving in a circle is a ⫽ v 2兾r, where v is the velocity and r is the radius of the circle. Approximate the maximum percent error in measuring the acceleration due to errors of 3% in v and 2% in r. 35. Volume A trough is 16 feet long (see figure). Its cross sections are isosceles triangles with each of the two equal sides measuring 18 inches. The angle between the two equal sides is .
16 ft
θ
18 in.
18 in. Not drawn to scale
36. Sports A baseball player in center field is playing approximately 330 feet from a television camera that is behind home plate. A batter hits a fly ball that goes to the wall 420 feet from the camera (see figure).
330 ft
9° 420 ft
(a) The camera turns 9⬚ to follow the play. Approximate the number of feet that the center fielder has to run to make the catch. (b) The position of the center fielder could be in error by as much as 6 feet and the maximum error in measuring the rotation of the camera is 1⬚. Approximate the maximum possible error in the result of part (a). 37. Inductance The inductance L (in microhenrys) of a straight nonmagnetic wire in free space is
冢
L ⫽ 0.00021 ln
2h ⫺ 0.75 r
冣
where h is the length of the wire in millimeters and r is the radius of a circular cross section. Approximate L when 1 1 millimeters and h ⫽ 100 ± 100 millimeters. r ⫽ 2 ± 16 38. Pendulum T⫽
The period T of a pendulum of length L is
2冪L 冪g
where g is the acceleration due to gravity. A pendulum is moved from the Canal Zone, where g ⫽ 32.09 feet per second per second, to Greenland, where g ⫽ 32.23 feet per second per second. Because of the change in temperature, the length of the pendulum changes from 2.5 feet to 2.48 feet. Approximate the change in the period of the pendulum.
Differentiability In Exercises 39– 42, show that the function is differentiable by finding values of 1 and 2 as designated in the definition of differentiability, and verify that both 1 and 2 approach 0 as 冇⌬x, ⌬y冈 → 冇0, 0冈. 39. f 共x, y兲 ⫽ x 2 ⫺ 2x ⫹ y
40. f 共x, y兲 ⫽ x 2 ⫹ y 2
41. f 共x, y兲 ⫽
42. f 共x, y兲 ⫽ 5x ⫺ 10y ⫹ y 3
x 2y
Differentiability In Exercises 43 and 44, use the function to show that fx冇0, 0冈 and fy冇0, 0冈 both exist, but that f is not differentiable at 冇0, 0冈.
冦 冦
(a) Write the volume of the trough as a function of and determine the value of such that the volume is a maximum.
3x 2y , 共x, y兲 ⫽ 共0, 0兲 43. f 共x, y兲 ⫽ x 4 ⫹ y 2 0, 共x, y兲 ⫽ 共0, 0兲
(b) The maximum error in the linear measurements is one-half inch and the maximum error in the angle measure is 2⬚. Approximate the change in the maximum volume.
5x 2y , 共x, y兲 ⫽ 共0, 0兲 ⫹ y3 44. f 共x, y兲 ⫽ 共x, y兲 ⫽ 共0, 0兲 0, x3
Roca/Shutterstock.com
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13.5
Chain Rules for Functions of Several Variables
907
13.5 Chain Rules for Functions of Several Variables Use the Chain Rules for functions of several variables. Find partial derivatives implicitly.
Chain Rules for Functions of Several Variables Your work with differentials in the preceding section provides the basis for the extension of the Chain Rule to functions of two variables. There are two cases—the first case involves w as a function of x and y, where x and y are functions of a single independent variable t, as shown in Theorem 13.6.
w
∂w ∂x x
∂w ∂y y
dx dt
dy dt t
t
Chain Rule: one independent variable w is a function of x and y, which are each functions of t. This diagram represents the derivative of w with respect to t. Figure 13.39
THEOREM 13.6 Chain Rule: One Independent Variable Let w ⫽ f 共x, y兲, where f is a differentiable function of x and y. If x ⫽ g共t兲 and y ⫽ h 共t兲, where g and h are differentiable functions of t, then w is a differentiable function of t, and dw ⭸w dx ⭸w dy ⫽ ⫹ . dt ⭸x dt ⭸y dt The Chain Rule is shown schematically in Figure 13.39. A proof of Theorem 13.6 is given in Appendix A. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Chain Rule: One Independent Variable Let w ⫽ x 2 y ⫺ y 2, where x ⫽ sin t and y ⫽ e t. Find dw兾dt when t ⫽ 0. Solution
By the Chain Rule for one independent variable, you have
dw ⭸w dx ⭸w dy ⫽ ⫹ dt ⭸x dt ⭸y dt ⫽ 2xy共cos t兲 ⫹ 共x 2 ⫺ 2y兲e t ⫽ 2共sin t兲共e t兲共cos t兲 ⫹ 共sin2 t ⫺ 2et兲et ⫽ 2e t sin t cos t ⫹ e t sin2 t ⫺ 2e2t. When t ⫽ 0, it follows that dw ⫽ ⫺2. dt The Chain Rules presented in this section provide alternative techniques for solving many problems in single-variable calculus. For instance, in Example 1, you could have used single-variable techniques to find dw兾dt by first writing w as a function of t, w ⫽ x 2y ⫺ y 2 ⫽ 共sin t兲 2 共e t兲 ⫺ 共e t兲 2 ⫽ e t sin 2 t ⫺ e 2t and then differentiating as usual. dw ⫽ 2e t sin t cos t ⫹ e t sin2 t ⫺ 2e2t dt
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908
Chapter 13
Functions of Several Variables
The Chain Rule in Theorem 13.6 can be extended to any number of variables. For example, if each xi is a differentiable function of a single variable t, then for w ⫽ f 共x1, x2, . . . , xn 兲 you have dw ⭸w dx1 ⭸w dx2 . . . ⭸w dxn ⫽ ⫹ ⫹ ⫹ . dt ⭸x1 dt ⭸x2 dt ⭸xn dt
An Application of a Chain Rule to Related Rates Two objects are traveling in elliptical paths given by the following parametric equations. x1 ⫽ 4 cos t and y1 ⫽ 2 sin t x2 ⫽ 2 sin 2t and y2 ⫽ 3 cos 2t y
t=
4
Solution From Figure 13.40, you can see that the distance s between the two objects is given by s ⫽ 冪共x2 ⫺ x1兲 2 ⫹ 共 y2 ⫺ y1兲 2
s
x
−2
4
When t ⫽ , the partial derivatives of s are as follows.
−4
y
t=
4
π 2
2
s x
−2
4
⫺ 共x2 ⫺ x1兲 ⫺ x1兲 2 ⫹ 共 y2 ⫺ 共 y2 ⫺ y1兲 ⫺ x1兲 2 ⫹ 共 y2 共x2 ⫺ x1兲 ⫺ x1兲 2 ⫹ 共 y2 共 y2 ⫺ y1兲 ⫺ x1兲 2 ⫹ 共 y2
1 4 ⫽ ⫺ 共0 ⫹ 4兲 ⫽ ⫺ 5 5 1 3 ⫽ ⫺ 共3 ⫺ 0兲 ⫽ ⫺ 5 5 ⫺ y1兲 2 1 4 ⫽ 共0 ⫹ 4兲 ⫽ 5 ⫺ y1兲 2 5 1 3 ⫽ 共3 ⫺ 0兲 ⫽ 5 ⫺ y1兲 2 5 ⫺ y1兲 2
dx1 ⫽ ⫺4 sin t ⫽ 0 dt dy1 ⫽ 2 cos t ⫽ ⫺2 dt
−4
y
dx2 ⫽ 4 cos 2t ⫽ 4 dt dy2 ⫽ ⫺6 sin 2t ⫽ 0. dt
t=π
s
x
−4
⭸s ⫽ ⭸x1 冪共x2 ⭸s ⫽ ⭸y1 冪共x2 ⭸s ⫽ ⭸x2 冪共x2 ⭸s ⫽ ⭸y2 冪共x2
When t ⫽ , the derivatives of x1, y1, x2, and y2 are
−2
4
and that when t ⫽ , you have x1 ⫽ ⫺4, y1 ⫽ 0, x 2 ⫽ 0, y2 ⫽ 3, and s ⫽ 冪共0 ⫹ 4兲 2 ⫹ 共3 ⫺ 0兲 2 ⫽ 5.
−2
−4
Second object
At what rate is the distance between the two objects changing when t ⫽ ?
π 3
2
−4
First object
4
−2 −4
Paths of two objects traveling in elliptical orbits Figure 13.40
So, using the appropriate Chain Rule, you know that the distance is changing at a rate of ⭸s dx1 ⭸s dy1 ⭸s dx2 ⭸s dy2 ds ⫽ ⫹ ⫹ ⫹ dt ⭸x1 dt ⭸y1 dt ⭸x2 dt ⭸y2 dt 4 3 4 3 ⫽ ⫺ 共0兲 ⫹ ⫺ 共⫺2兲 ⫹ 共4兲 ⫹ 共0兲 5 5 5 5 22 . ⫽ 5
冢 冣
冢 冣
冢冣
冢冣
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13.5
Chain Rules for Functions of Several Variables
909
In Example 2, note that s is the function of four intermediate variables, x1, y1, x2, and y2, each of which is a function of a single variable t. Another type of composite function is one in which the intermediate variables are themselves functions of more than one variable. For instance, for w ⫽ f 共x, y兲, where x ⫽ g 共s, t兲 and y ⫽ h 共s, t兲, it follows that w is a function of s and t, and you can consider the partial derivatives of w with respect to s and t. One way to find these partial derivatives is to write w as a function of s and t explicitly by substituting the equations x ⫽ g 共s, t兲 and y ⫽ h 共s, t兲 into the equation w ⫽ f 共x, y兲. Then you can find the partial derivatives in the usual way, as demonstrated in the next example.
Finding Partial Derivatives by Substitution Find ⭸w兾⭸s and ⭸w兾⭸t for w ⫽ 2xy, where x ⫽ s 2 ⫹ t 2 and y ⫽ s兾t. Solution to obtain
Begin by substituting x ⫽ s 2 ⫹ t 2 and y ⫽ s兾t into the equation w ⫽ 2xy
w ⫽ 2xy ⫽ 2共s 2 ⫹ t 2兲
冢st冣 ⫽ 2冢st ⫹ st冣. 3
Then, to find ⭸w兾⭸s, hold t constant and differentiate with respect to s. ⭸w 3s 2 ⫽2 ⫹t ⭸s t 6s 2 ⫹ 2t 2 ⫽ t
冢
冣
Similarly, to find ⭸w兾⭸t, hold s constant and differentiate with respect to t to obtain ⭸w s3 ⫽2 ⫺ 2⫹s ⭸t t ⫺s 3 ⫹ st 2 ⫽2 t2 2st 2 ⫺ 2s 3 ⫽ . t2
冢 冢
冣
冣
Theorem 13.7 gives an alternative method for finding the partial derivatives in Example 3, without explicitly writing w as a function of s and t.
∂w ∂x ∂x ∂t t
x
w
THEOREM 13.7 Chain Rule: Two Independent Variables Let w ⫽ f 共x, y兲, where f is a differentiable function of x and y. If x ⫽ g 共s, t兲 and y ⫽ h 共s, t兲 such that the first partials ⭸x兾⭸s, ⭸x兾⭸t, ⭸y兾⭸s, and ⭸y兾⭸t all exist, then ⭸w兾⭸s and ⭸w兾⭸t exist and are given by
∂w ∂y
∂x ∂s
∂y ∂t
s
t
y
∂y ∂s s
Chain Rule: two independent variables Figure 13.41
⭸w ⭸w ⭸x ⭸w ⭸y ⫽ ⫹ ⭸s ⭸x ⭸s ⭸y ⭸s and ⭸w ⭸w ⭸x ⭸w ⭸y ⫽ ⫹ . ⭸t ⭸x ⭸t ⭸y ⭸t The Chain Rule is shown schematically in Figure 13.41.
Proof To obtain ⭸w兾⭸s, hold t constant and apply Theorem 13.6 to obtain the desired result. Similarly, for ⭸w兾⭸t, hold s constant and apply Theorem 13.6. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
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910
Chapter 13
Functions of Several Variables
The Chain Rule with Two Independent Variables See LarsonCalculus.com for an interactive version of this type of example.
Use the Chain Rule to find ⭸w兾⭸s and ⭸w兾⭸t for w ⫽ 2xy where x ⫽ s 2 ⫹ t 2 and y ⫽ s兾t. Solution Note that these same partials were found in Example 3. This time, using Theorem 13.7, you can hold t constant and differentiate with respect to s to obtain ⭸w ⭸w ⭸x ⭸w ⭸y ⫽ ⫹ ⭸s ⭸x ⭸s ⭸y ⭸s 1 ⫽ 2y 共2s兲 ⫹ 2x t s 1 ⫽ 2 共2s兲 ⫹ 2共s2 ⫹ t2兲 t t 2 2 2 4s 2s ⫹ 2t ⫽ ⫹ t t 2 2 6s ⫹ 2t . ⫽ t
冢冣
冢冣
冢冣
s Substitute for y and s 2 ⫹ t 2 for x. t
Similarly, holding s constant gives ⭸w ⭸w ⭸x ⭸w ⭸y ⫽ ⫹ ⭸t ⭸x ⭸t ⭸y ⭸t ⫺s ⫽ 2y 共2t兲 ⫹ 2x 2 t s ⫺s ⫽ 2 共2t兲 ⫹ 2共s 2 ⫹ t 2兲 2 t t 3 2 2s ⫹ 2st ⫽ 4s ⫺ t2 4st 2 ⫺ 2s 3 ⫺ 2st 2 ⫽ t2 2st 2 ⫺ 2s 3 . ⫽ t2
冢 冣
冢冣
冢 冣
s Substitute for y and s 2 ⫹ t 2 for x. t
The Chain Rule in Theorem 13.7 can also be extended to any number of variables. For example, if w is a differentiable function of the n variables x1, x2, . . . , xn where each xi is a differentiable function of the m variables t1, t2, . . . , tm , then for w ⫽ f 共x1, x2, . . . , xn 兲 you obtain the following. ⭸w ⭸w ⭸x1 ⭸w ⫽ ⫹ ⭸t1 ⭸x1 ⭸t1 ⭸x2 ⭸w ⭸w ⭸x1 ⭸w ⫽ ⫹ ⭸t2 ⭸x1 ⭸t2 ⭸x2
⯗
⭸x2 . . . ⭸w ⭸xn ⫹ ⫹ ⭸t1 ⭸xn ⭸t1 ⭸x2 . . . ⭸w ⭸xn ⫹ ⫹ ⭸t2 ⭸xn ⭸t2
⭸w ⭸x1 ⭸w ⭸x2 . . . ⭸w ⭸xn ⭸w ⫽ ⫹ ⫹ ⫹ ⭸tm ⭸x1 ⭸tm ⭸x2 ⭸tm ⭸xn ⭸tm
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13.5
Chain Rules for Functions of Several Variables
911
The Chain Rule for a Function of Three Variables Find ⭸w兾⭸s and ⭸w兾⭸t when s ⫽ 1 and t ⫽ 2 for w ⫽ xy ⫹ yz ⫹ xz where x ⫽ s cos t, y ⫽ s sin t, and z ⫽ t. Solution
By extending the result of Theorem 13.7, you have
⭸w ⭸w ⭸x ⭸w ⭸y ⭸w ⭸z ⫽ ⫹ ⫹ ⭸s ⭸x ⭸s ⭸y ⭸s ⭸z ⭸s ⫽ 共 y ⫹ z兲共cos t兲 ⫹ 共x ⫹ z兲共sin t兲 ⫹ 共 y ⫹ x兲共0兲 ⫽ 共 y ⫹ z兲共cos t兲 ⫹ 共x ⫹ z兲共sin t兲. When s ⫽ 1 and t ⫽ 2, you have x ⫽ 1, y ⫽ 0, and z ⫽ 2. So, ⭸w ⫽ 共0 ⫹ 2兲共1兲 ⫹ 共1 ⫹ 2兲共0兲 ⫽ 2. ⭸s Furthermore, ⭸w ⭸w ⭸x ⭸w ⭸y ⭸w ⭸z ⫽ ⫹ ⫹ ⭸t ⭸x ⭸t ⭸y ⭸t ⭸z ⭸t ⫽ 共 y ⫹ z兲共⫺s sin t兲 ⫹ 共x ⫹ z兲共s cos t兲 ⫹ 共 y ⫹ x兲共1兲 and for s ⫽ 1 and t ⫽ 2, it follows that ⭸w ⫽ 共0 ⫹ 2兲共0兲 ⫹ 共1 ⫹ 2兲共1兲 ⫹ 共0 ⫹ 1兲共1兲 ⭸t ⫽ 2 ⫹ 2.
Implicit Partial Differentiation This section concludes with an application of the Chain Rule to determine the derivative of a function defined implicitly. Let x and y be related by the equation F共x, y兲 ⫽ 0, where y ⫽ f 共x兲 is a differentiable function of x. To find dy兾dx, you could use the techniques discussed in Section 2.5. You will see, however, that the Chain Rule provides a convenient alternative. Consider the function w ⫽ F共x, y兲 ⫽ F共x, f 共x兲兲. You can apply Theorem 13.6 to obtain dx dy dw ⫽ Fx 共x, y兲 ⫹ Fy 共x, y兲 . dx dx dx Because w ⫽ F共x, y兲 ⫽ 0 for all x in the domain of f, you know that dw ⫽0 dx and you have Fx 共x, y兲
dx dy ⫹ Fy 共x, y兲 ⫽ 0. dx dx
Now, if Fy共x, y兲 ⫽ 0, you can use the fact that dx兾dx ⫽ 1 to conclude that dy Fx 共x, y兲 . ⫽⫺ dx Fy 共x, y兲 A similar procedure can be used to find the partial derivatives of functions of several variables that are defined implicitly.
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912
Chapter 13
Functions of Several Variables
THEOREM 13.8 Chain Rule: Implicit Differentiation If the equation F共x, y兲 ⫽ 0 defines y implicitly as a differentiable function of x, then dy F 共x, y兲 ⫽⫺ x , dx Fy 共x, y兲
Fy共x, y兲 ⫽ 0.
If the equation F共x, y, z兲 ⫽ 0 defines z implicitly as a differentiable function of x and y, then ⭸z F 共x, y, z兲 ⫽⫺ x ⭸x Fz 共x, y, z兲
and
⭸z Fy 共x, y, z兲 ⫽⫺ , ⭸y Fz 共x, y, z兲
Fz共x, y, z兲 ⫽ 0.
This theorem can be extended to differentiable functions defined implicitly with any number of variables.
Finding a Derivative Implicitly Find dy兾dx for y 3 ⫹ y 2 ⫺ 5y ⫺ x 2 ⫹ 4 ⫽ 0. Solution
Begin by letting
F共x, y兲 ⫽ y 3 ⫹ y 2 ⫺ 5y ⫺ x 2 ⫹ 4. Then
REMARK Compare the solution to Example 6 with the solution to Example 2 in Section 2.5.
Fx 共x, y兲 ⫽ ⫺2x and
Fy 共x, y兲 ⫽ 3y 2 ⫹ 2y ⫺ 5.
Using Theorem 13.8, you have dy F 共x, y兲 ⫺ 共⫺2x兲 2x ⫽⫺ x ⫽ 2 ⫽ 2 . dx Fy 共x, y兲 3y ⫹ 2y ⫺ 5 3y ⫹ 2y ⫺ 5
Finding Partial Derivatives Implicitly Find ⭸z兾⭸x and ⭸z兾⭸y for 3x 2z ⫺ x 2 y 2 ⫹ 2z 3 ⫹ 3yz ⫺ 5 ⫽ 0. Solution
Begin by letting
F共x, y, z兲 ⫽ 3x 2z ⫺ x 2y 2 ⫹ 2z 3 ⫹ 3yz ⫺ 5. Then Fx 共x, y, z兲 ⫽ 6xz ⫺ 2xy 2 Fy 共x, y, z兲 ⫽ ⫺2x 2 y ⫹ 3z and Fz 共x, y, z兲 ⫽ 3x 2 ⫹ 6z 2 ⫹ 3y. Using Theorem 13.8, you have ⭸z 2xy 2 ⫺ 6xz Fx共x, y, z兲 ⫽ 2 ⫽⫺ ⭸x Fz共x, y, z兲 3x ⫹ 6z 2 ⫹ 3y and Fy共x, y, z兲 2x 2 y ⫺ 3z ⭸z ⫽⫺ ⫽ 2 . ⭸y Fz 共x, y, z兲 3x ⫹ 6z 2 ⫹ 3y
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.5
13.5 Exercises
Chain Rules for Functions of Several Variables
913
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Function
Using the Chain Rule In Exercises 1–4, find dw/dt using the appropriate Chain Rule. Evaluate dw/dt at the given value of t.
Values
16. w ⫽ x ⫺ 2
s ⫽ 3,
y2
x ⫽ s cos t,
t⫽
y ⫽ s sin t
4
Function
Value
1. w ⫽ x 2 ⫹ y 2
t⫽2
Using Different Methods In Exercises 17–20, find w/s and w/t (a) by using the appropriate Chain Rule and (b) by converting w to a function of s and t before differentiating.
t⫽0
17. w ⫽ xyz,
x ⫽ 2t, y ⫽ 3t 2. w ⫽ 冪x2 ⫹ y2 x ⫽ cos t, y ⫽
et
3. w ⫽ x sin y
18. w ⫽ x 2 ⫹ y 2 ⫹ z 2,
t⫽0
19. w ⫽ ze xy,
x ⫽ et, y ⫽ ⫺ t
(a) by using the appropriate Chain Rule and (b) by converting w to a function of t before differentiating. 6. w ⫽ cos共x ⫺ y兲, 7. w ⫽
x2
⫹
y2
⫹
8. w ⫽ xy cos z,
y ⫽ e⫺2t x ⫽ t 2,
z 2,
y ⫽ s ⫹ t,
s2,
y⫽
t 2,
z ⫽ st2
z ⫽ st
z ⫽ s ⫺ 2t
x ⫽ t,
22. sec xy ⫹ tan xy ⫹ 5 ⫽ 0 23. ln冪x 2 ⫹ y 2 ⫹ x ⫹ y ⫽ 4
y⫽1
y⫽
t 2,
y ⫽ sin t,
z ⫽ et
z ⫽ arccos t
x ⫽ t ⫺ 1,
10. w ⫽ xy2 ⫹ x2z ⫹ yz2,
21. x2 ⫺ xy ⫹ y2 ⫺ x ⫹ y ⫽ 0
24.
x ⫽ cos t,
9. w ⫽ xy ⫹ xz ⫹ yz,
y ⫽ t 2 ⫺ 1,
x ⫽ t 2, y ⫽ 2t,
z⫽t
z⫽2
Projectile Motion In Exercises 11 and 12, the parametric equations for the paths of two projectiles are given. At what rate is the distance between the two objects changing at the given value of t ? 11. x1 ⫽ 10 cos 2t, y1 ⫽ 6 sin 2t
First object
x2 ⫽ 7 cos t, y2 ⫽ 4 sin t
Second object
t ⫽ 兾2 12. x1 ⫽ 48冪2 t, y1 ⫽ 48冪2 t ⫺ 16t 2
First object
x2 ⫽ 48冪3 t, y2 ⫽ 48t ⫺ 16t 2
Second object
x ⫺ y2 ⫽ 6 x2 ⫹ y2
Finding Partial Derivatives Implicitly In Exercises 25 –32, differentiate implicitly to find the first partial derivatives of z. 25. x 2 ⫹ y 2 ⫹ z 2 ⫽ 1 27.
x2
⫹ 2yz ⫹
z2
⫽1
26. xz ⫹ yz ⫹ xy ⫽ 0 28. x ⫹ sin共 y ⫹ z兲 ⫽ 0
29. tan共x ⫹ y兲 ⫹ tan共 y ⫹ z兲 ⫽ 1 30. z ⫽ e x sin共 y ⫹ z兲 31. e x z ⫹ xy ⫽ 0 32. x ln y ⫹ y 2z ⫹ z 2 ⫽ 8
Finding Partial Derivatives Implicitly In Exercises 33 –36, differentiate implicitly to find the first partial derivatives of w. 33. xy ⫹ yz ⫺ wz ⫹ wx ⫽ 5 34. x 2 ⫹ y 2 ⫹ z 2 ⫺ 5yw ⫹ 10w 2 ⫽ 2
t⫽1
Finding Partial Derivatives In Exercises 13 –16, find w/s and w/t using the appropriate Chain Rule. Evaluate each partial derivative at the given values of s and t. Function
Values
13. w ⫽ x 2 ⫹ y 2
s ⫽ 1,
x ⫽ s ⫹ t, 14. w ⫽
y ⫽ t cos s,
differentiate implicitly to find dy/dx.
Using Different Methods In Exercises 5–10, find dw/ dt
x ⫽ e t,
z ⫽ st2
Finding a Derivative Implicitly In Exercises 21– 24,
x ⫽ cos t, y ⫽ sin t
5. w ⫽ xy,
x⫽
y ⫽ s ⫺ t,
x ⫽ t sin s,
x ⫽ s ⫺ t,
20. w ⫽ x cos yz,
t⫽ 4
y 4. w ⫽ ln x
x ⫽ s ⫹ t,
y⫽s⫺t
y3
⫺ 3x 2y
x ⫽ es,
y ⫽ et
15. w ⫽ sin共2x ⫹ 3y兲 x ⫽ s ⫹ t,
t⫽0
y⫽s⫺t
s ⫽ ⫺1, s ⫽ 0,
t⫽2 t⫽
2
35. cos xy ⫹ sin yz ⫹ wz ⫽ 20 36. w ⫺ 冪x ⫺ y ⫺ 冪y ⫺ z ⫽ 0
Homogeneous Functions A function f is homogeneous of degree n when f 冇tx, ty冈 ⴝ t nf 冇x, y冈. In Exercises 37– 40, (a) show that the function is homogeneous and determine n, and (b) show that xfx冇x, y冈 ⴙ yfy冇x, y冈 ⴝ nf 冇x, y冈. 37. f 共x, y兲 ⫽
xy 冪x 2 ⫹ y 2
38. f 共x, y兲 ⫽ x3 ⫺ 3xy 2 ⫹ y 3 39. f 共x, y兲 ⫽ e x兾y 40. f 共x, y兲 ⫽
x2 ⫹ y2
冪x 2
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
914
Chapter 13
Functions of Several Variables
41. Using a Table of Values Let w ⫽ f 共x, y兲, x ⫽ g共t兲, and y ⫽ h共t兲, where f, g, and h are differentiable. Use the appropriate Chain Rule to find dw兾dt when t ⫽ 2, given the following table of values. g共2兲
h共2兲
g⬘共2兲
h⬘共2兲
fx共4, 3兲
fy共4, 3兲
4
3
⫺1
6
⫺5
7
42. Using a Table of Values Let w ⫽ f 共x, y兲, x ⫽ g共s, t兲, and y ⫽ h共s, t兲, where f, g, and h are differentiable. Use the appropriate Chain Rule to find ws共1, 2兲 and wt共1, 2兲, given the following table of values.
48. Ideal Gas Law The Ideal Gas Law is pV ⫽ mRT, where p is the pressure, V is the volume, m is the constant mass, R is a constant, T is the temperature, and p and V are functions of time. Find dT兾dt, the rate at which the temperature changes with respect to time. 49. Moment of Inertia An annular cylinder has an inside radius of r1 and an outside radius of r2 (see figure). Its moment of inertia is I ⫽ 12m共r12 ⫹ r22 兲, where m is the mass. The two radii are increasing at a rate of 2 centimeters per second. Find the rate at which I is changing at the instant the radii are 6 centimeters and 8 centimeters. (Assume mass is a constant.) r r2
g共1, 2兲
h共1, 2兲
gs共1, 2兲
hs共1, 2兲
4
3
⫺3
5
h R
r1
gt共1, 2兲
ht共1, 2兲
fx共4, 3兲
fy共4, 3兲
⫺2
8
⫺5
7 Figure for 49
WRITING ABOUT CONCEPTS 43. Chain Rule Let w ⫽ f 共x, y兲 be a function in which x and y are functions of a single variable t. Give the Chain Rule for finding dw兾dt. 44. Chain Rule Let w ⫽ f 共x, y兲 be a function in which x and y are functions of two variables s and t. Give the Chain Rule for finding ⭸w兾⭸s and ⭸w兾⭸t. 45. Implicit Differentiation For f 共x, y兲 ⫽ 0, give the rule for finding dy兾dx implicitly. For f 共x, y, z兲 ⫽ 0, give the rule for finding ⭸z兾⭸x and ⭸z兾⭸y implicitly.
Figure for 50
50. Volume and Surface Area The two radii of the frustum of a right circular cone are increasing at a rate of 4 centimeters per minute, and the height is increasing at a rate of 12 centimeters per minute (see figure). Find the rates at which the volume and surface area are changing when the two radii are 15 centimeters and 25 centimeters, and the height is 10 centimeters. 51. Using the Chain Rule Show that ⭸w ⭸w ⫹ ⫽0 ⭸u ⭸v for w ⫽ f 共x, y兲, x ⫽ u ⫺ v, and y ⫽ v ⫺ u. 52. Using the Chain Rule Exercise 51 for
46.
HOW DO YOU SEE IT? The graph of the function w ⫽ f 共x, y兲 is shown below. z
2
w ⫽ 共x ⫺ y兲 sin共 y ⫺ x兲. 53. Cauchy-Riemann Equations Given the functions u共x, y兲 and v共x, y兲, verify that the Cauchy-Riemann differential equations ⭸u ⭸v ⫽ ⭸x ⭸y
−2 x
2
2
y
(a) Assume that x and y are functions of a single variable r. Give the chain rule for finding dw兾dr. (b) Assume that x and y are functions of two variables r and . Give the chain rule for finding ⭸w兾⭸r and ⭸w兾⭸.
47. Volume and Surface Area The radius of a right circular cylinder is increasing at a rate of 6 inches per minute, and the height is decreasing at a rate of 4 inches per minute. What are the rates of change of the volume and surface area when the radius is 12 inches and the height is 36 inches?
Demonstrate the result of
and
⭸u ⭸v ⫽⫺ ⭸y ⭸x
can be written in polar coordinate form as ⭸u 1 ⫽ ⭸r r
⭸v
⭈ ⭸
and
⭸v 1 ⫽⫺ ⭸r r
54. Cauchy-Riemann Equations Exercise 53 for the functions u ⫽ ln冪x 2 ⫹ y 2 and
⭸u
⭈ ⭸. Demonstrate the result of
y v ⫽ arctan . x
55. Homogeneous Function homogeneous of degree n, then
Show that if f 共x, y兲 is
x fx共x, y兲 ⫹ yfy共x, y兲 ⫽ nf 共x, y兲. [Hint: Let g共t兲 ⫽ f 共tx, ty兲 ⫽ t n f 共x, y兲. Find g⬘ 共t兲 and then let t ⫽ 1.]
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13.6
Directional Derivatives and Gradients
915
13.6 Directional Derivatives and Gradients Find and use directional derivatives of a function of two variables. Find the gradient of a function of two variables. Use the gradient of a function of two variables in applications. Find directional derivatives and gradients of functions of three variables.
Directional Derivative z You are standing on the hillside represented by z f x, y in Figure 13.42 and want to determine the hill’s incline toward the z-axis. You already know how to determine the slopes in two different directions—the slope in the y-direction is given by the partial derivative fy x, y, and the slope in the x-direction is given by the partial derivative fx x, y. In this y section, you will see that these two partial derivatives can be used to find the slope in x Surface: any direction. z = f(x, y) To determine the slope at a point on a surface, you will define a new type of Figure 13.42 derivative called a directional derivative. Begin by letting z f x, y be a surface and Px0, y0 be a point in the domain of f, as shown in Figure 13.43. The “direction” of the directional derivative is given by a unit vector
z
z = f(x, y)
u cos i sin j P u
θ
y
L x
Figure 13.43
Surface: z = f(x, y)
z
x x0 t cos
(x 0, y0 , f (x0 , y0 ))
Curve: C
where is the angle the vector makes with the positive x-axis. To find the desired slope, reduce the problem to two dimensions by intersecting the surface with a vertical plane passing through the point P and parallel to u, as shown in Figure 13.44. This vertical plane intersects the surface to form a curve C. The slope of the surface at x0, y0, f x0, y0 in the direction of u is defined as the slope of the curve C at that point. Informally, you can write the slope of the curve C as a limit that looks much like those used in single-variable calculus. The vertical plane used to form C intersects the xy-plane in a line L, represented by the parametric equations
and
(x, y, f (x, y))
y y0 t sin so that for any value of t, the point Qx, y lies on the line L. For each of the points P and Q, there is a corresponding point on the surface.
P
Q t
x
Figure 13.44
y
x0, y0, f x0, y0 x, y, f x, y
Point above P Point above Q
Moreover, because the distance between P and Q is x x02 y y02 t cos 2 t sin 2
t
you can write the slope of the secant line through x0, y0, f x0, y0 and x, y, f x, y as f x0 t cos , y0 t sin f x0, y0 f x, y f x0, y0 . t t Finally, by letting t approach 0, you arrive at the definition on the next page.
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916
Chapter 13
Functions of Several Variables
Definition of Directional Derivative Let f be a function of two variables x and y and let u cos i sin j be a unit vector. Then the directional derivative of f in the direction of u, denoted by Du f, is Du f x, y lim t→0
f x t cos , y t sin f x, y t
provided this limit exists.
Calculating directional derivatives by this definition is similar to finding the derivative of a function of one variable by the limit process (given in Section 2.1). A simpler “working” formula for finding directional derivatives involves the partial derivatives fx and fy. THEOREM 13.9 Directional Derivative If f is a differentiable function of x and y, then the directional derivative of f in the direction of the unit vector u cos i sin j is Du f x, y fxx, y cos fyx, y sin .
Proof
For a fixed point x0, y0, let
x x0 t cos
and
y y0 t sin .
Then, let gt f x, y. Because f is differentiable, you can apply the Chain Rule given in Theorem 13.6 to obtain gt fxx, y xt fyx, y yt fxx, y cos fyx, y sin . If t 0, then x x0 and y y0, so g0 fxx0, y0 cos fyx0, y0 sin . By the definition of gt, it is also true that gt g0 t→0 t f x0 t cos , y0 t sin f x0, y0 . lim t→0 t
g0 lim
Consequently, Du f x0, y0 fxx0, y0 cos fyx0, y0 sin . See LarsonCalculus.com for Bruce Edwards’s video of this proof.
There are infinitely many directional derivatives of a surface at a given point—one for each direction specified by u, as shown in Figure 13.45. Two of these are the partial derivatives fx and fy.
z
1. Direction of positive x-axis 0: u cos 0 i sin 0 j i
Surface: z = f(x, y)
Di f x, y fxx, y cos 0 fyx, y sin 0 fxx, y y
(x, y) x
The vector u
Figure 13.45
2. Direction of positive y-axis Dj f x, y fxx, y cos
: u cos i sin j j 2 2 2
fyx, y sin fyx, y 2 2
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13.6
Directional Derivatives and Gradients
917
Finding a Directional Derivative Find the directional derivative of 1 f x, y 4 x 2 y 2 4
Surface
at 1, 2 in the direction of
u cos
i sin j. 3 3
Direction
Solution Because fxx, y 2x and fyx, y y 2 are continuous, f is differentiable, and you can apply Theorem 13.9.
2y sin
Du f x, y fxx, y cos fyx, y sin 2x cos
Surface: f (x, y) = 4 − x 2 −
1 2 y 4
Evaluating at 3, x 1, and y 2 produces
z
Du f 1, 2 2
4
1
3
π 3
u
y
5
3
2
1.866.
(1, 2) x
12 1 23 See Figure 13.46.
Note in Figure 13.46 that you can interpret the directional derivative as giving the slope of the surface at the point 1, 2, 2 in the direction of the unit vector u.
Figure 13.46
You have been specifying direction by a unit vector u. When the direction is given by a vector whose length is not 1, you must normalize the vector before applying the formula in Theorem 13.9.
Finding a Directional Derivative z
See LarsonCalculus.com for an interactive version of this type of example. Surface: 25 f (x, y) = x 2 sin 2y
Find the directional derivative of f x, y x 2 sin 2y
20
Surface
at 1, 2 in the direction of
15
v 3i 4j. 10 5
u
Solution Because fxx, y 2x sin 2y and fyx, y 2x2 cos 2y are continuous, f is differentiable, and you can apply Theorem 13.9. Begin by finding a unit vector in the direction of v.
(1, π2 ( π /2
3
x
Direction
u
π y
v 3 4 i j cos i sin j v 5 5
Using this unit vector, you have Du f x, y 2x sin 2ycos 2x 2 cos 2ysin 3 4 Du f 1, 2 sin 2 cos 2 5 5 3 4 0 2 5 5 8 . See Figure 13.47. 5
− 25
Figure 13.47
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918
Chapter 13
Functions of Several Variables
The Gradient of a Function of Two Variables z
The gradient of a function of two variables is a vector-valued function of two variables. This function has many important uses, some of which are described later in this section. (x, y, f(x, y))
Definition of Gradient of a Function of Two Variables Let z f(x, y be a function of x and y such that fx and fy exist. Then the gradient of f, denoted by f x, y, is the vector f x, y fxx, y i fyx, y j. ∇f (x, y)
y
(x, y)
x
The gradient of f is a vector in the xy-plane. Figure 13.48
(The symbol f is read as “del f.”) Another notation for the gradient is grad f x, y. In Figure 13.48, note that for each x, y, the gradient f x, y is a vector in the plane (not a vector in space).
Notice that no value is assigned to the symbol by itself. It is an operator in the same sense that d dx is an operator. When operates on f x, y, it produces the vector f x, y.
Finding the Gradient of a Function Find the gradient of f x, y y ln x xy 2 at the point 1, 2. Solution
Using
fxx, y
y y2 x
and
fyx, y ln x 2xy
you have f x, y fxx, yi fyx, yj
yx y i ln x 2xyj. 2
At the point 1, 2, the gradient is f 1, 2
21 2 i ln 1 212 j 2
6i 4j. Because the gradient of f is a vector, you can write the directional derivative of f in the direction of u as Du f x, y fxx, y i fyx, yj cos i sin j . In other words, the directional derivative is the dot product of the gradient and the direction vector. This useful result is summarized in the next theorem. THEOREM 13.10 Alternative Form of the Directional Derivative If f is a differentiable function of x and y, then the directional derivative of f in the direction of the unit vector u is Du f x, y f(x, y u.
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13.6
Directional Derivatives and Gradients
919
Using f x, y to Find a Directional Derivative Find the directional derivative of f x, y 3x 2 2y 2 at 34, 0 in the direction from P 34, 0 to Q0, 1. Solution Because the partials of f are continuous, f is differentiable and you can apply Theorem 13.10. A vector in the specified direction is
\
PQ 0
Surface: f (x, y) = 3x 2 − 2y 2
z
3 i 1 0j 4
3 ij 4
and a unit vector in this direction is 3
\
u
PQ 3 4 i j. PQ 5 5
\
Unit vector in direction of PQ
\
Because
2
f x, y fxx, yi fyx, yj 6xi 4yj the gradient at 34, 0 is
1
3 9 f , 0 i 0j. 4 2 x
P
Gradient at 34 , 0
Consequently, at 34, 0, the directional derivative is
1
Q 2
y
Figure 13.49
3 3 Du f , 0 f , 0 u 4 4 9 3 4 i 0j i j 2 5 5 27 . Directional derivative at 34 , 0 10
See Figure 13.49.
Applications of the Gradient You have already seen that there are many directional derivatives at the point x, y on a surface. In many applications, you may want to know in which direction to move so that f x, y increases most rapidly. This direction is called the direction of steepest ascent, and it is given by the gradient, as stated in the next theorem. THEOREM 13.11 Properties of the Gradient Let f be differentiable at the point x, y.
REMARK Property 2 of Theorem 13.11 says that at the point x, y, f increases most rapidly in the direction of the gradient, f x, y.
1. If f x, y 0, then Du f x, y 0 for all u. 2. The direction of maximum increase of f is given by f x, y. The maximum value of Du f x, y is f x, y.
Maximum value of Du f x, y
3. The direction of minimum increase of f is given by f x, y. The minimum value of Du f x, y is f x, y.
Minimum value of Du f x, y
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920
Chapter 13
Functions of Several Variables
Proof
z
If f x, y 0, then for any direction (any u), you have
Du f x, y f x, y u 0i 0j cos i sin j 0. Maximum increase
If f x, y 0, then let be the angle between f x, y and a unit vector u. Using the dot product, you can apply Theorem 11.5 to conclude that Du f x, y f x, y u
(x, y, f (x, y))
f x, y u cos f x, y cos and it follows that the maximum value of Du f x, y will occur when
∇f (x, y)
x
y
(x, y)
The gradient of f is a vector in the xy-plane that points in the direction of maximum increase on the surface given by z f x, y. Figure 13.50
cos 1. So, 0, and the maximum value of the directional derivative occurs when u has the same direction as f x, y. Moreover, this largest value of Du f x, y is precisely f x, y cos f x, y. Similarly, the minimum value of Du f x, y can be obtained by letting
so that u points in the direction opposite that of f x, y, as shown in Figure 13.50. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
To visualize one of the properties of the gradient, imagine a skier coming down a mountainside. If f x, y denotes the altitude of the skier, then f x, y indicates the compass direction the skier should take to ski the path of steepest descent. (Remember that the gradient indicates direction in the xy-plane and does not itself point up or down the mountainside.) As another illustration of the gradient, consider the temperature Tx, y at any point x, y on a flat metal plate. In this case, Tx, y gives the direction of greatest temperature increase at the point x, y, as illustrated in the next example.
Level curves: T(x, y) = 20 − 4x 2 − y 2 y 5
Finding the Direction of Maximum Increase The temperature in degrees Celsius on the surface of a metal plate is Tx, y 20 4x 2 y 2 where x and y are measured in centimeters. In what direction from 2, 3 does the temperature increase most rapidly? What is this rate of increase? x −3
3
Solution
The gradient is
Tx, y Txx, yi Tyx, yj 8x i 2y j. (2, −3)
It follows that the direction of maximum increase is given by T2, 3 16i 6j
−5
The direction of most rapid increase in temperature at 2, 3 is given by 16i 6j. Figure 13.51
as shown in Figure 13.51, and the rate of increase is T2, 3 256 36 292
17.09 per centimeter.
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13.6
Directional Derivatives and Gradients
921
The solution presented in Example 5 can be misleading. Although the gradient points in the direction of maximum temperature increase, it does not necessarily point toward the hottest spot on the plate. In other words, the gradient provides a local solution to finding an increase relative to the temperature at the point 2, 3. Once you leave that position, the direction of maximum increase may change.
Finding the Path of a Heat-Seeking Particle A heat-seeking particle is located at the point 2, 3 on a metal plate whose temperature at x, y is Tx, y 20 4x 2 y 2. Find the path of the particle as it continuously moves in the direction of maximum temperature increase. Solution
Let the path be represented by the position vector
rt xti ytj.
Level curves: T(x, y) = 20 − 4x 2 − y 2
A tangent vector at each point xt, yt is given by
y
rt
5
dx dy i j. dt dt
Because the particle seeks maximum temperature increase, the directions of rt and Tx, y 8xi 2yj are the same at each point on the path. So, 8x k x −3
3
dx dt
and
2y k
dy dt
where k depends on t. By solving each equation for dt k and equating the results, you obtain dx dy . 8x 2y
(2, −3)
−5
Path followed by a heat-seeking particle Figure 13.52
The solution of this differential equation is x Cy 4. Because the particle starts at the point 2, 3, you can determine that C 2 81. So, the path of the heat-seeking particle is x
2 4 y . 81
The path is shown in Figure 13.52. In Figure 13.52, the path of the particle (determined by the gradient at each point) appears to be orthogonal to each of the level curves. This becomes clear when you consider that the temperature Tx, y is constant along a given level curve. So, at any point x, y on the curve, the rate of change of T in the direction of a unit tangent vector u is 0, and you can write f x, y u Du Tx, y 0.
u is a unit tangent vector.
Because the dot product of f x, y and u is 0, you can conclude that they must be orthogonal. This result is stated in the next theorem. THEOREM 13.12 Gradient Is Normal to Level Curves If f is differentiable at x0, y0 and f x0, y0 0, then f x0, y0 is normal to the level curve through x0, y0.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
922
Chapter 13
Functions of Several Variables
Finding a Normal Vector to a Level Curve Sketch the level curve corresponding to c 0 for the function given by f x, y y sin x and find a normal vector at several points on the curve. The level curve for c 0 is given by
Solution
0 y sin x or y sin x as shown in Figure 13.53(a). Because the gradient vector of f at x, y is f x, y fxx, yi fyx, yj cos xi j you can use Theorem 13.12 to conclude that f x, y is normal to the level curve at the point x, y. Some gradient vectors are f , 0 i j
f
2 3 1 , ij 3 2 2
f , 1 j 2
3 1 f , ij 3 2 2 f 0, 0 i j f
3 , 23 21 i j
f
2 , 1 j
f
23, 23 21 i j
and f , 0 i j. These are shown in Figure 13.53(b). z
y
4
3
Gradient is normal to the level curve.
2 −4 x
1
−π
π
π 2
−π
4
−4
(a) The surface is given by f x, y y sin x.
y
−2
π
x
y − sin x = 0
−3
(b) The level curve is given by f x, y 0.
Figure 13.53
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.6
Directional Derivatives and Gradients
923
Functions of Three Variables The definitions of the directional derivative and the gradient can be extended naturally to functions of three or more variables. As often happens, some of the geometric interpretation is lost in the generalization from functions of two variables to those of three variables. For example, you cannot interpret the directional derivative of a function of three variables as representing slope. The definitions and properties of the directional derivative and the gradient of a function of three variables are listed below. Directional Derivative and Gradient for Three Variables Let f be a function of x, y, and z, with continuous first partial derivatives. The directional derivative of f in the direction of a unit vector u ai bj ck is given by Du f x, y, z afxx, y, z bfyx, y, z cfzx, y, z. The gradient of f is defined as f x, y, z fxx, y, zi fyx, y, zj fzx, y, zk. Properties of the gradient are as follows. 1. Du f x, y, z f x, y, z u 2. If f x, y, z 0, then Du f x, y, z 0 for all u. 3. The direction of maximum increase of f is given by f x, y, z. The maximum value of Du f x, y, z is f x, y, z.
Maximum value of Du f x, y, z
4. The direction of minimum increase of f is given by f x, y, z. The minimum value of Du f x, y, z is f x, y, z.
Minimum value of Du f x, y, z
You can generalize Theorem 13.12 to functions of three variables. Under suitable hypotheses, f x0, y0, z0 z
is normal to the level surface through x0, y0, z0.
8
Finding the Gradient of a Function
6
Find f x, y, z for the function
4
−6
−4
2 x
f x, y, z x 2 y 2 4z
2
(2, −1, 1)
2
4
6
y
−2 −4
∇f (2, − 1, 1) = 4i − 2j − 4k
Level surface and gradient vector at 2, 1, 1 for f x, y, z x2 y2 4z Figure 13.54
and find the direction of maximum increase of f at the point 2, 1, 1. Solution
The gradient vector is
f x, y, z fxx, y, zi fyx, y, zj fzx, y, zk 2xi 2yj 4k. So, it follows that the direction of maximum increase at 2, 1, 1 is f 2, 1, 1 4i 2j 4k.
See Figure 13.54.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
924
Chapter 13
Functions of Several Variables
13.6 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding a Directional Derivative In Exercises 1–4, use Theorem 13.9 to find the directional derivative of the function at P in the direction of the unit vector u ⴝ cos i ⴙ sin j. 1. f x, y
x2
2. f x, y
y , xy
P3, 0,
3. f x, y sin2x y, P0, 0, 4. gx, y
23. gx, y x 2 y 2 1, P1, 2,
6
25. gx, y, z xyez, P2, 4, 0,
3
26. hx, y, z lnx y z,
Theorem 13.9 to find the directional derivative of the function at P in the direction of v. 5. f x, y 3x 4xy 9y, P1, 2, v 35 i 45 j
Function 27. f x, y
29. hx, y x tan y
P0, 0,
,
2
2
vij
Theorem 13.9 to find the directional derivative of the function at P in the direction of Q. 9. f x, y
P1, 1,
3y 2,
10. f x, y cosx y,
Q4, 5
P0, ,
Q
2 , 0
11. f x, y ey sin x, P0, 0, Q2, 1 12. f x, y sin 2x cos y, P, 0, Q
13. f x, y 3x 5y 2 1, 2, 1 15. z ln
y,
x2
17. w
3x 2
5y2
2, 0
18. w x tan y z, 4, 3, 1 the gradient to find the directional derivative of the function at P in the direction of v.
2 ,
P 1,
1, 4, 2 0, 0, 0
35. w xy 2z 2
2, 1, 1 2, 0, 4
xeyz
f x, y ⴝ 3 ⴚ
x y ⴚ . 3 2
4
(b)
2 3
39. Find Du f 3, 2, where u
Finding a Directional Derivative In Exercises 19–22, use
20. hx, y ex sin y,
1, 2
1 34. w 1 x 2 y 2 z2
(a)
1, 1, 2
19. f x, y xy, P0, 2, v
33. f x, y, z x 2 y 2 z 2
38. Find Du f 3, 2, where u cos i sin j, using each given value of .
3, 4
2z 2,
0, 5 y2
37. Sketch the graph of f in the first octant and plot the point 3, 2, 1 on the surface.
2, 3
16. z cosx 2 y 2,
32. gx, y
3 x2 ln
Using a Function In Exercises 37–42, consider the function
find the gradient of the function at the given point. 14. gx, y
2, 4 0, 3
31. gx, y yex
36. f x, y, z
2 ,
Finding the Gradient of a Function In Exercises 13–18,
2xe y x,
0, 1
30. hx, y y cosx y
Finding a Directional Derivative In Exercises 9–12, use
x2
1, 0
2xy
7. gx, y x 2 y 2, P3, 4, v 3i 4j 2 y 2
Q4, 3, 1
Point x2
i j
8. hx, y ex
P1, 0, 0,
find the gradient of the function and the maximum value of the directional derivative at the given point.
xy 28. f x, y y1
P4, 3, v
Q0, 0, 0
Using Properties of the Gradient In Exercises 27–36,
Finding a Directional Derivative In Exercises 5–8, use
6. f x, y x3 y 3,
Q2, 3
24. f x, y 3x 2 y 2 4, P1, 4, Q3, 6
2 P0, 2, 3
xey ,
Finding a Directional Derivative Using the Gradient In Exercises 23–26, use the gradient to find the directional derivative of the function at P in the direction of Q.
P1, 2, 4
y 2,
22. f x, y, z xy yz xz, P1, 2, 1, v 2i j k
1 2
i 3 j
4 3
(d)
6
v , using each given vector v. v (b) v 3i 4j
(c) v is the vector from 1, 2 to 2, 6. (d) v is the vector from 3, 2 to 4, 5. 40. Find f x, y.
v i
21. f x, y, z x 2 y 2 z 2, P1, 1, 1,
(a) v i j
(c)
41. Find the maximum value of the directional derivative at 3, 2. v
3
3
i j k
42. Find a unit vector u orthogonal to f 3, 2 and calculate Du f 3, 2. Discuss the geometric meaning of the result.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.6
Investigation In Exercises 43 and 44, (a) use the graph to estimate the components of the vector in the direction of the maximum rate of increase in the function at the given point. (b) Find the gradient at the point and compare it with your estimate in part (a). (c) In what direction would the function be decreasing at the greatest rate? Explain. 1 2 43. f x, y 10 x 3xy y 2
44. f x, y 12yx
1, 2
z
1
1
x
3
y
c 25, P3, 4
49. f x, y xy
50. f x, y
P1, 3
c 12,
x x2 y 2
P1, 1
x
Generated by Maple
Consider the function
f x, y x 2 y 2 at the point 4, 3, 7. (a) Use a computer algebra system to graph the surface represented by the function. (b) Determine the directional derivative Du f 4, 3 as a function of , where u cos i sin j. Use a computer algebra system to graph the function on the interval 0, 2. (c) Approximate the zeros of the function in part (b) and interpret each in the context of the problem. (d) Approximate the critical numbers of the function in part (b) and interpret each in the context of the problem. (e) Find f 4, 3 and explain its relationship to your answers in part (d). (f) Use a computer algebra system to graph the level curve of the function f at the level c 7. On this curve, graph the vector in the direction of f 4, 3, and state its relationship to the level curve. Consider the function
8y . 1 x2 y 2
(a) Analytically verify that the level curve of f x, y at the level c 2 is a circle. (b) At the point 3, 2 on the level curve for which c 2, sketch the vector showing the direction of the greatest rate of increase of the function. (To print an enlarged copy of the graph, go to MathGraphs.com.) (c) At the point 3, 2 on the level curve, sketch a vector such that the directional derivative is 0. (d) Use a computer algebra system to graph the surface to verify your answers in parts (a)–(c).
c 1,
52. f x, y x y2
P2, 10
53. f x, y
3 Generated by Maple
51. f x, y 4x2 y c 6,
2
f x, y
P0, 0
y
3
46. Investigation
c 6,
48. f x, y x 2 y 2
of the function at P, (b) find a unit normal vector to the level curve f x, y ⴝ c at P, (c) find the tangent line to the level curve f x, y ⴝ c at P, and (d) sketch the level curve, the unit normal vector, and the tangent line in the xy-plane.
1
45. Investigation
47. f x, y 6 2x 3y
Using a Function In Exercises 51–54, (a) find the gradient
2
3
925
Finding a Normal Vector In Exercises 47–50, find a normal vector to the level curve f x, y ⴝ c at P.
c 3,
1, 2 z
Directional Derivatives and Gradients
3x2
c 3, P4, 1
2y2
54. f x, y 9x2 4y2
P1, 1
c 40, P2, 1
WRITING ABOUT CONCEPTS 55. Directional Derivative Define the derivative of the function z f x, y in the direction u cos i sin j. 56. Directional Derivative Write a paragraph describing the directional derivative of the function f in the direction u cos i sin j when (a) 0 and (b) 90 . 57. Gradient Define the gradient of a function of two variables. State the properties of the gradient. 58. Sketching a Graph and a Vector Sketch the graph of a surface and select a point P on the surface. Sketch a vector in the xy-plane giving the direction of steepest ascent on the surface at P. 59. Gradient and Level Curves Describe the relationship of the gradient to the level curves of a surface given by z f x, y. 60. Using a Function
Consider the function
f x, y 9 x 2 y 2. (a) Sketch the graph of f in the first octant and plot the point 1, 2, 4 on the surface. (b) Find Du f 1, 2, where u cos i sin j, for 4. (c) Repeat part (b) for 3. (d) Find f 1, 2 and f 1, 2. (e) Find a unit vector u orthogonal to f 1, 2 and calculate Du f 1, 2. Discuss the geometric meaning of the result. 61. Topography equation
The surface of a mountain is modeled by the
hx, y 5000 0.001x 2 0.004y 2. A mountain climber is at the point 500, 300, 4390. In what direction should the climber move in order to ascend at the greatest rate?
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
926
Chapter 13
Functions of Several Variables 70. If f x, y x y, then 1 Du f x, y 1.
62.
HOW DO YOU SEE IT? The figure shows a topographic map carried by a group of hikers. Sketch the paths of steepest descent when the hikers start at point A and when they start at point B. (To print an enlarged copy of the graph, go to MathGraphs.com.)
71. If Du f x, y exists, then Du f x, y Du f x, y. 72. If Du f x0, y0 c for any unit vector u, then c 0. 73. Finding a Function
Find a function f such that
f e x cos y i e x sin y j z k. 74. Ocean Floor A team of oceanographers is mapping the ocean floor to assist in the recovery of a sunken ship. Using sonar, they develop the model
18 00
1671
B
D 250 30x 2 50 sin
1994
y , 0 x 2, 0 y 2 2
A 00 18
63. Temperature Distribution x, y on a metal plate is T
The temperature at the point
x . x2 y2
Find the direction of greatest increase in heat from the point 3, 4. 64. Temperature The temperature at the point x, y on a metal 2 plate is modeled by Tx, y 400ex y 2, x 0, y 0. (a) Use a computer algebra system to graph the temperature distribution function. (b) Find the directions of no change in heat on the plate from the point 3, 5. (c) Find the direction of greatest increase in heat from the point 3, 5.
Finding the Direction of Maximum Increase
In Exercises 65 and 66, the temperature in degrees Celsius on the surface of a metal plate is given by T x, y, where x and y are measured in centimeters. Find the direction from point P where the temperature increases most rapidly and the rate of increase. 65. Tx, y 80 3x2 y2,
P1, 5
66. Tx, y 50 x2 4y2,
P2, 1
Heat-Seeking Path In Exercises 67 and 68, find the path of a heat-seeking particle placed at point P on a metal plate with a temperature field T x, y.
where D is the depth in meters, and x and y are the distances in kilometers. (a) Use a computer algebra system to graph the surface. (b) Because the graph in part (a) is showing depth, it is not a map of the ocean floor. How could the model be changed so that the graph of the ocean floor could be obtained? (c) What is the depth of the ship if it is located at the coordinates x 1 and y 0.5? (d) Determine the steepness of the ocean floor in the positive x-direction from the position of the ship. (e) Determine the steepness of the ocean floor in the positive y-direction from the position of the ship. (f) Determine the direction of the greatest rate of change of depth from the position of the ship. 75. Using a Function Consider the function 3 xy. f x, y
(a) Show that f is continuous at the origin. (b) Show that fx and fy exist at the origin, but that the directional derivatives at the origin in all other directions do not exist. (c) Use a computer algebra system to graph f near the origin to verify your answers in parts (a) and (b). Explain. 76. Directional Derivative
67. Tx, y 400 2x 2 y 2,
P10, 10
4xy , f x, y x 2 y 2 0,
68. Tx, y 100 x 2 2y 2,
P4, 3
and the unit vector
True or False? In Exercises 69–72, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 69. If f x, y 1 x 2 y 2, then Du f 0, 0 0 for any unit vector u.
u
1 2
Consider the function
x, y 0, 0 x, y 0, 0
i j.
Does the directional derivative of f at P0, 0 in the direction of u exist? If f 0, 0 were defined as 2 instead of 0, would the directional derivative exist?
Brandelet/Shutterstock.com
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13.7
Tangent Planes and Normal Lines
927
13.7 Tangent Planes and Normal Lines Find equations of tangent planes and normal lines to surfaces. Find the angle of inclination of a plane in space. Compare the gradients f 冇x, y 冈 and F 冇x, y, z 冈.
Tangent Plane and Normal Line to a Surface Exploration
So far, you have represented surfaces in space primarily by equations of the form
Billiard Balls and Normal Lines In each of the three figures below, the cue ball is about to strike a stationary ball at point P. Explain how you can use the normal line to the stationary ball at point P to describe the resulting motion of each of the two balls. Assuming that each cue ball has the same speed, which stationary ball will acquire the greatest speed? Which will acquire the least? Explain your reasoning.
z f 共x, y兲.
Equation of a surface S
In the development to follow, however, it is convenient to use the more general representation F共x, y, z兲 0. For a surface S given by z f 共x, y兲, you can convert to the general form by defining F as F共x, y, z兲 f 共x, y兲 z. Because f 共x, y兲 z 0, you can consider S to be the level surface of F given by F共x, y, z兲 0.
Alternative equation of surface S
Writing an Equation of a Surface For the function F共x, y, z兲 x 2 y 2 z 2 4 describe the level surface given by
Normal line to stationary ball at point P
F共x, y, z兲 0. Moving cue ball
P
Solution x2
y2
The level surface given by F共x, y, z兲 0 can be written as z2 4
which is a sphere of radius 2 whose center is at the origin. Stationary ball
Normal line to stationary ball at point P
P
Moving cue ball
Stationary ball Normal line to stationary ball at point P
P
Stationary ball
Moving cue ball
You have seen many examples of the usefulness of normal lines in applications involving curves. Normal lines are equally important in analyzing surfaces and solids. For example, consider the collision of two billiard balls. When a stationary ball is struck at a point P on its surface, it moves along the line of impact determined by P and the center of the ball. The impact can occur in two ways. When the cue ball is moving along the line of impact, it stops dead and imparts all of its momentum to the stationary ball, as shown in Figure 13.55. When the cue ball is not moving along the line of impact, it is deflected to one side or the other and retains part of its momentum. The part of the momentum that is transferred to the stationary ball occurs along the line of impact, regardless of the direction of the cue ball, as shown in Figure 13.56. This line of impact is called the normal line to the surface of the ball at the point P.
Line of impact Line of impact
Figure 13.55
Line of impact
Figure 13.56
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
928
Chapter 13
Functions of Several Variables
In the process of finding a normal line to a surface, you are also able to solve the problem of finding a tangent plane to the surface. Let S be a surface given by F共x, y, z兲 0 and let P共x0, y0, z0兲 be a point on S. Let C be a curve on S through P that is defined by the vector-valued function r 共t兲 x共t兲i y 共t兲j z共t兲k. Then, for all t, F共x共t兲, y共t兲, z共t兲兲 0. If F is differentiable and x 共t兲, y 共t兲, and z 共t兲 all exist, then it follows from the Chain Rule that
Surface S: F(x, y, z) = 0 F
P(x0 , y0 , z0 )
0 F 共t兲 Fx 共x, y, z兲x 共t兲 Fy 共x, y, z兲y 共t兲 Fz 共x, y, z兲z 共t兲. At 共x0, y0, z0兲, the equivalent vector form is 0 F共x0, y0, z0兲 r 共t0兲. Gradient
Tangent plane to surface S at P Figure 13.57
Tangent vector
This result means that the gradient at P is orthogonal to the tangent vector of every curve on S through P. So, all tangent lines on S lie in a plane that is normal to F共x0, y0, z0 兲 and contains P, as shown in Figure 13.57. Definitions of Tangent Plane and Normal Line Let F be differentiable at the point P共x0, y0, z0兲 on the surface S given by F共x, y, z兲 0 such that F共x0, y0, z0兲 0.
REMARK In the remainder of this section, assume F共x0, y0, z0兲 to be nonzero unless stated otherwise.
1. The plane through P that is normal to F共x0, y0, z0兲 is called the tangent plane to S at P. 2. The line through P having the direction of F共x0, y0, z0兲 is called the normal line to S at P.
To find an equation for the tangent plane to S at 共x0, y0, z0兲, let 共x, y, z兲 be an arbitrary point in the tangent plane. Then the vector v 共x x0兲i 共 y y0兲j 共z z0兲k lies in the tangent plane. Because F共x0, y0, z0兲 is normal to the tangent plane at 共x0, y0, z0兲, it must be orthogonal to every vector in the tangent plane, and you have F共x0, y0, z0兲 v 0 which leads to the next theorem.
THEOREM 13.13 Equation of Tangent Plane If F is differentiable at 共x0, y0, z0兲, then an equation of the tangent plane to the surface given by F共x, y, z兲 0 at 共x0, y0, z0兲 is Fx 共x0, y0, z0兲共x x0兲 Fy 共x0, y0, z0兲共y y0兲 Fz 共x0, y0, z0兲共z z0) 0.
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13.7
TECHNOLOGY Some three-dimensional graphing utilities are capable of graphing tangent planes to surfaces. An example is shown below.
Tangent Planes and Normal Lines
929
Finding an Equation of a Tangent Plane Find an equation of the tangent plane to the hyperboloid z 2 2x 2 2y 2 12 at the point 共1, 1, 4兲.
z
Solution
Begin by writing the equation of the surface as
z 2x 2y 2 12 0. 2
2
Then, considering y x Generated by Mathematica
Sphere: x2 y2 z2 1
F共x, y, z兲 z 2 2x 2 2y 2 12 you have Fx 共x, y, z兲 4x,
Fy 共x, y, z兲 4y,
and Fz 共x, y, z兲 2z.
At the point 共1, 1, 4兲, the partial derivatives are Fx 共1, 1, 4兲 4,
Fy 共1, 1, 4兲 4,
and Fz 共1, 1, 4兲 8.
So, an equation of the tangent plane at 共1, 1, 4兲 is 4 共x 1兲 4 共 y 1兲 8共z 4兲 0 4x 4 4y 4 8z 32 0 4x 4y 8z 24 0 x y 2z 6 0. Figure 13.58 shows a portion of the hyperboloid and the tangent plane. Surface: z 2 − 2x 2 − 2y 2 − 12 = 0 z 6 5
F(1, − 1, 4) y 3 3 x
Tangent plane to surface Figure 13.58
To find the equation of the tangent plane at a point on a surface given by z f 共x, y兲, you can define the function F by F共x, y, z兲 f 共x, y兲 z. Then S is given by the level surface F共x, y, z兲 0, and by Theorem 13.13, an equation of the tangent plane to S at the point 共x0, y0, z0兲 is fx共x0, y0 兲共x x0兲 fy共x0, y0兲共y y0兲 共z z0兲 0.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
930
Chapter 13
Functions of Several Variables
Finding an Equation of the Tangent Plane Find the equation of the tangent plane to the paraboloid 1 2 共x 4y 2兲 10
z1
at the point 共1, 1, 2 兲. 1
Solution
1 From z f 共x, y兲 1 10共x 2 4y 2兲, you obtain
fx 共x, y兲
fx 共1, 1兲
1 5
and
Surface: z = 1 − 1 (x 2 + 4y 2 ) 10 z
fy 共x, y兲
)1, 1, 12 )
2
4y 5
4 fy 共1, 1兲 . 5
So, an equation of the tangent plane at 共1, 1, 2 兲 is 1
−6
冢 21冣 0 1 4 1 共x 1兲 共 y 1兲 冢z 冣 0 5 5 2
fx 共1, 1兲共x 1兲 fy 共1, 1兲共 y 1兲 z
−3 2 6
x 5
5
3
y
x
1 4 3 x y z 0. 5 5 2 This tangent plane is shown in Figure 13.59.
Figure 13.59
The gradient F共x, y, z兲 provides a convenient way to find equations of normal lines, as shown in Example 4.
Finding an Equation of a Normal Line to a Surface See LarsonCalculus.com for an interactive version of this type of example.
Find a set of symmetric equations for the normal line to the surface xyz 12 at the point 共2, 2, 3兲. Solution
Begin by letting
F共x, y, z兲 xyz 12.
Surface: xyz = 12
Then, the gradient is given by z y 2 −4
−2
and at the point 共2, 2, 3兲, you have
2 −2
4 x
−4 −6
∇F(2, − 2, −3)
Figure 13.60
4
F共x, y, z兲 Fx 共x, y, z兲i Fy 共x, y, z兲j Fz 共x, y, z兲k yz i xz j xyk F共2, 2, 3兲 共2兲共3兲i 共2兲(3兲j 共2兲共2兲k 6i 6j 4k. The normal line at 共2, 2, 3兲 has direction numbers 6, 6, and 4, and the corresponding set of symmetric equations is x2 y2 z3 . 6 6 4 See Figure 13.60.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.7
Tangent Planes and Normal Lines
931
Knowing that the gradient F共x, y, z兲 is normal to the surface given by F共x, y, z兲 0 allows you to solve a variety of problems dealing with surfaces and curves in space.
Finding the Equation of a Tangent Line to a Curve Describe the tangent line to the curve of intersection of the ellipsoid
Ellipsoid: x 2 + 2y 2 + 2z 2 = 20
x 2 2y 2 2z 2 20
z (0, 1, 3) 4
Tangent line
Ellipsoid
and the paraboloid x2 y2 z 4
Paraboloid
at the point (0, 1, 3), as shown in Figure 13.61. x
5
2 3 4 5
Paraboloid: x 2 + y 2 + z = 4
Figure 13.61
Solution y
Begin by finding the gradients to both surfaces at the point (0, 1, 3).
Ellipsoid F共x, y, z兲 x 2 2y 2 2z 2 20 F共x, y, z兲 2xi 4yj 4zk F共0, 1, 3兲 4j 12k
Paraboloid G共x, y, z兲 x 2 y 2 z 4 G共x, y, z兲 2xi 2yj k G共0, 1, 3兲 2j k
The cross product of these two gradients is a vector that is tangent to both surfaces at the point 共0, 1, 3兲.
ⱍ ⱍ
i F共0, 1, 3兲 G共0, 1, 3兲 0 0
j 4 2
k 12 20i 1
So, the tangent line to the curve of intersection of the two surfaces at the point 共0, 1, 3兲 is a line that is parallel to the x-axis and passes through the point 共0, 1, 3兲.
The Angle of Inclination of a Plane Another use of the gradient F共x, y, z兲 is to determine the angle of inclination of the tangent plane to a surface. The angle of inclination of a plane is defined as the angle 共0 兾2兲 between the given plane and the xy-plane, as shown in Figure 13.62. (The angle of inclination of a horizontal plane is defined as zero.) Because the vector k is normal to the xy-plane, you can use the formula for the cosine of the angle between two planes (given in Section 11.5) to conclude that the angle of inclination of a plane with normal vector n is
cos
ⱍn kⱍ ⱍn kⱍ. 储n储 储k储
Angle of inclination of a plane
储n储
z
n θ
k
y x
θ
The angle of inclination Figure 13.62
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932
Chapter 13
Functions of Several Variables
Finding the Angle of Inclination of a Tangent Plane Find the angle of inclination of the tangent plane to the ellipsoid x2 y2 z2 1 12 12 3 at the point 共2, 2, 1兲. Solution
Begin by letting
F共x, y, z兲
x2 y2 z2 1. 12 12 3
Then, the gradient of F at the point 共2, 2, 1兲 is
z 3
k θ
x y 2z F共x, y, z兲 i j k 6 6 3 1 1 2 F共2, 2, 1兲 i j k. 3 3 3
∇F(2, 2, 1)
Because F共2, 2, 1兲 is normal to the tangent plane and k is normal to the xy-plane, it follows that the angle of inclination of the tangent plane is
6 6 x
Ellipsoid: x2 y2 z2 + + =1 12 12 3
Figure 13.63
y
cos
ⱍF共2, 2, 1兲 kⱍ 储F共2, 2, 1兲储
2兾3 冪共1兾3兲 2 共1兾3兲 2 共2兾3兲 2
冪23
which implies that
冪23 ⬇ 35.3 ,
arccos
as shown in Figure 13.63. A special case of the procedure shown in Example 6 is worth noting. The angle of inclination of the tangent plane to the surface z f 共x, y兲 at 共x0, y0, z0兲 is cos
1 冪关 fx共x0, y0兲兴2 关 fy共x0, y0兲兴2 1
.
Alternative formula for angle of inclination 共See Exercise 67.兲
A Comparison of the Gradients f 冇x, y冈 and F 冇x, y, z冈 This section concludes with a comparison of the gradients f 共x, y兲 and F共x, y, z兲. In the preceding section, you saw that the gradient of a function f of two variables is normal to the level curves of f. Specifically, Theorem 13.12 states that if f is differentiable at 共x0, y0兲 and f 共x0, y0兲 0, then f 共x0, y0兲 is normal to the level curve through 共x0, y0兲. Having developed normal lines to surfaces, you can now extend this result to a function of three variables. The proof of Theorem 13.14 is left as an exercise (see Exercise 68). THEOREM 13.14 Gradient Is Normal to Level Surfaces If F is differentiable at 共x0, y0, z0兲 and F共x0, y0, z0兲 0 then F共x0, y0, z0兲 is normal to the level surface through 共x0, y0, z0兲. When working with the gradients f 共x, y兲 and F共x, y, z兲, be sure you remember that f 共x, y兲 is a vector in the xy-plane and F共x, y, z兲 is a vector in space.
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13.7
13.7 Exercises
Tangent Planes and Normal Lines
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Describing a Surface In Exercises 1–4, describe the level surface F冇x, y, z冈 ⴝ 0.
Finding an Equation of a Tangent Plane and a Normal Line In Exercises 21–30, find an equation of the tangent plane and find a set of symmetric equations for the normal line to the surface at the given point.
1. F共x, y, z兲 3x 5y 3z 15 2. F共x, y, z兲 x 2 y 2 z 2 25
21. x y z 9, 共3, 3, 3兲
3. F共x, y, z兲 4x 2 9y 2 4z 2
22. x 2 y 2 z 2 9, 共1, 2, 2兲
4. F共x, y, z兲 16x 2 9y 2 36z
23. x 2 y 2 z 9, 共1, 2, 4兲
Finding a Unit Normal Vector In Exercises 5–8, find a
24. z 16 x2 y2,
unit normal vector to the surface at the given point. [Hint: Normalize the gradient vector F冇x, y, z冈.]
25. z
Surface
Point
x2
共2, 2, 8兲
共3, 2, 5兲
y2,
26. xy z 0, 共2, 3, 6兲 27. xyz 10, 共1, 2, 5兲
5. 3x 4y 12z 0
共0, 0, 0兲
6. x 2 y 2 z 2 6
共1, 1, 2兲
7. x 2 3y z 3 9
共2, 1, 2兲
y 29. z arctan , x
8. x2y3 y2z 2xz3 4
共1, 1, 1兲
30. y ln xz2 2, 共e, 2, 1兲
10. f 共x, y兲
共2, 1, 8兲
y x
共1, 2, 2兲 z
z 10 8
6 4 2
x
6 x
y
2
共3, 4, 5兲 y 12. g共x, y兲 arctan , 共1, 0, 0兲 x 11. z 冪x 2 y 2,
13. g共x, y兲 x y , 2
共1, 1, 2兲
2
14. f 共x, y兲 x 2xy y 2,
共1, 2, 1兲
2
15. h 共x, y兲 ln 冪x 2 y 2,
冢
16. h 共x, y兲 cos y,
共3, 4, ln 5兲
冪2 5, , 4 2
冣
17. x 2 4y 2 z 2 36, 共2, 2, 4兲 18. x 2 2z 2 y 2,
共1, 3, 2兲
19. xy 2 3x z 2 8, 共1, 3, 2兲 20. z ex共sin y 1兲,
冢0, 2 , 2冣
31. x 2 y 2 2,
z x, 共1, 1, 1兲
32. z
z 4 y, 共2, 1, 5兲
34. z 35.
4
x2
y 2,
36. z
x2
冪x 2
y2
x2
y 2 z 2 25, 共3, 3, 4兲 5x 2y 3z 22, 共3, 4, 5兲
y 2,
z2
y 2,
14,
x y z 0, 共3, 1, 2兲
x y 6z 33, 共1, 2, 5兲
(1, 2, 2)
2 4
2
Finding the Equation of a Tangent Line to a Curve In Exercises 31–36, (a) find a set of symmetric equations for the tangent line to the curve of intersection of the surfaces at the given point, and (b) find the cosine of the angle between the gradient vectors at this point. State whether the surfaces are orthogonal at the point of intersection.
6
2
冢1, 1, 4 冣
33. x 2 z 2 25,
10
(2, 1, 8)
共0, 2, 2兲
28. z ye2xy,
Finding an Equation of a Tangent Plane In Exercises 9–20, find an equation of the tangent plane to the surface at the given point. 9. z x2 y2 3
933
6
y
Finding the Angle of Inclination of a Tangent Plane In Exercises 37–40, find the angle of inclination of the tangent plane to the surface at the given point. 37. 3x 2 2y 2 z 15, 共2, 2, 5兲 38. 2xy z3 0, 共2, 2, 2兲 39. x 2 y2 z 0, 共1, 2, 3兲 40. x 2 y 2 5, 共2, 1, 3兲
Horizontal Tangent Plane In Exercises 41–46, find the point(s) on the surface at which the tangent plane is horizontal. 41. z 3 x 2 y 2 6y 42. z 3x 2 2y 2 3x 4y 5 43. z x2 xy y2 2x 2y 44. z 4x2 4xy 2y2 8x 5y 4 45. z 5xy 46. z xy
1 1 x y
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934
Chapter 13
Functions of Several Variables
Tangent Surfaces In Exercises 47 and 48, show that the surfaces are tangent to each other at the given point by showing that the surfaces have the same tangent plane at this point. 47. x2 2y2 3z2 3, x2 y2 z2 6x 10y 14 0,
共1, 1, 0兲 48. x2 y2 z2 8x 12y 4z 42 0, x2 y2 2z 7, 共2, 3, 3兲
Perpendicular Tangent Planes In Exercises 49 and 50, (a) show that the surfaces intersect at the given point, and (b) show that the surfaces have perpendicular tangent planes at this point. 49. z 2xy2, 8x2 5y2 8z 13, 共1, 1, 2兲 50. x2 y2 z2 2x 4y 4z 12 0, 4x2 y2 16z2 24, 共1, 2, 1兲
57. Investigation f 共x, y兲
Consider the function
4xy 共x 2 1兲共 y 2 1兲
on the intervals 2 x 2 and 0 y 3. (a) Find a set of parametric equations of the normal line and an equation of the tangent plane to the surface at the point 共1, 1, 1兲. (b) Repeat part (a) for the point 共1, 2, 45 兲.
(c) Use a computer algebra system to graph the surface, the normal lines, and the tangent planes found in parts (a) and (b). 58. Investigation f 共x, y兲
Consider the function
sin y x
on the intervals 3 x 3 and 0 y 2 .
51. Using an Ellipsoid Find a point on the ellipsoid x2 4y2 z2 9 where the tangent plane is perpendicular to the line with parametric equations
(a) Find a set of parametric equations of the normal line and an equation of the tangent plane to the surface at the point
冢2, 2 , 12冣.
x 2 4t, y 1 8t, and z 3 2t. 52. Using a Hyperboloid Find a point on the hyperboloid x2 4y2 z2 1 where the tangent plane is parallel to the plane x 4y z 0.
WRITING ABOUT CONCEPTS
2 3 3 (b) Repeat part (a) for the point , , . 3 2 2
冢
冣
(c) Use a computer algebra system to graph the surface, the normal lines, and the tangent planes found in parts (a) and (b). 59. Using Functions Consider the functions
53. Tangent Plane Give the standard form of the equation of the tangent plane to a surface given by F共x, y, z兲 0 at 共x0, y0, z0兲. 54. Normal Lines For some surfaces, the normal lines at any point pass through the same geometric object. What is the common geometric object for a sphere? What is the common geometric object for a right circular cylinder? Explain. 55. Tangent Plane Discuss the relationship between the tangent plane to a surface and approximation by differentials.
f 共x, y兲 6 x 2
y2 4
and
g共x, y兲 2x y.
(a) Find a set of parametric equations of the tangent line to the curve of intersection of the surfaces at the point 共1, 2, 4兲, and find the angle between the gradient vectors. (b) Use a computer algebra system to graph the surfaces. Graph the tangent line found in part (a). 60. Using Functions Consider the functions f 共x, y兲 冪16 x 2 y 2 2x 4y and
56.
HOW DO YOU SEE IT? The graph shows the ellipsoid x2 4y2 z2 16. Use the graph to determine the equation of the tangent plane at each of the given points. z
y
(a) 共4, 0, 0兲
(b) 共0, 2, 0兲
冪2
2
冪1 3x 2 y 2 6x 4y.
(a) Use a computer algebra system to graph the first-octant portion of the surfaces represented by f and g. (b) Find one first-octant point on the curve of intersection and show that the surfaces are orthogonal at this point.
4
−3
g共x, y兲
3 45
x
(c) 共0, 0, 4兲
(c) These surfaces are orthogonal along the curve of intersection. Does part (b) prove this fact? Explain.
Writing a Tangent Plane In Exercises 61 and 62, show that the tangent plane to the quadric surface at the point 冇x0, y0, z0冈 can be written in the given form. x2 y2 z2 1 a2 b2 c2 x0 x y0 y z0 z Plane: 2 2 2 1 a b c
61. Ellipsoid:
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13.7 y2 z2 x2 1 a2 b2 c2 x0 x y0 y z0 z Plane: 2 2 2 1 a b c
Tangent Planes and Normal Lines
935
62. Hyperboloid:
Wildflowers
z 2 a 2x 2 b 2 y 2
The diversity of wildflowers in a meadow can be measured by counting the numbers of daisies, buttercups, shooting stars, and so on. When there are n types of wildflowers, each with a proportion pi of the total population, it follows that
passes through the origin.
p1 p2 . . . pn 1.
63. Tangent Planes of a Cone Show that any tangent plane to the cone
64. Tangent Planes Let f be a differentiable function and consider the surface
The measure of diversity of the population is defined as H
冢冣
y z xf . x
n
兺 p log p . i
i1
Show that the tangent plane at any point P共x0, y0, z0兲 on the surface passes through the origin. 65. Approximation Consider the following approximations for a function f 共x, y兲 centered at 共0, 0兲.
2 i
In this definition, it is understood that pi log2 pi 0 when pi 0. The tables show proportions of wildflowers in a meadow in May, June, August, and September. May
Linear Approximation:
Flower type
1
2
3
4
P1共x, y兲 f 共0, 0兲 fx 共0, 0兲 x fy 共0, 0兲 y
Proportion
5 16
5 16
5 16
1 16
Flower type
1
2
3
4
Proportion
1 4
1 4
1 4
1 4
Flower type
1
2
3
4
Proportion
1 4
0
1 4
1 2
Quadratic Approximation: P2共x, y兲 f 共0, 0兲 fx 共0, 0兲 x fy 共0, 0兲 y 1 2 fxx 共0,
0兲
x2
fxy 共0, 0兲 xy
1 2 fyy 共0,
June 0兲
y2
[Note that the linear approximation is the tangent plane to the surface at 共0, 0, f 共0, 0兲兲.] (a) Find the linear approximation of f 共x, y兲 e共xy兲 centered at 共0, 0兲. (b) Find the quadratic approximation of f 共x, y兲 e共xy兲 centered at 共0, 0兲. (c) When x 0 in the quadratic approximation, you obtain the second-degree Taylor polynomial for what function? Answer the same question for y 0. (d) Complete the table. x
y
0
0
0
0.1
0.2
0.1
0.2
0.5
1
0.5
f 共x, y兲
P1共x, y兲
P2共x, y兲
66. Approximation Repeat Exercise 65 for the function f 共x, y兲 cos共x y兲. 67. Proof Prove that the angle of inclination of the tangent plane to the surface z f 共x, y兲 at the point 共x0, y0, z0兲 is given by
68. Proof
September Flower type
1
2
3
4
Proportion
0
0
0
1
(a) Determine the wildflower diversity for each month. How would you interpret September’s diversity? Which month had the greatest diversity?
(e) Use a computer algebra system to graph the surfaces z f 共x, y兲, z P1共x, y兲, and z P2共x, y兲.
cos
August
1 冪[ fx 共x0, y0兲] 2 [ fy 共x0, y0兲] 2 1
.
(b) When the meadow contains 10 types of wildflowers in roughly equal proportions, is the diversity of the population greater than or less than the diversity of a similar distribution of 4 types of flowers? What type of distribution (of 10 types of wildflowers) would produce maximum diversity? (c) Let Hn represent the maximum diversity of n types of wildflowers. Does Hn approach a limit as n approaches ? FOR FURTHER INFORMATION Biologists use the concept of diversity to measure the proportions of different types of organisms within an environment. For more information on this technique, see the article “Information Theory and Biological Diversity” by Steven Kolmes and Kevin Mitchell in the UMAP Modules.
Prove Theorem 13.14.
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936
Chapter 13
Functions of Several Variables
13.8 Extrema of Functions of Two Variables Find absolute and relative extrema of a function of two variables. Use the Second Partials Test to find relative extrema of a function of two variables.
Absolute Extrema and Relative Extrema Surface: z = f(x, y) Minimum
In Chapter 3, you studied techniques for finding the extreme values of a function of a single variable. In this section, you will extend these techniques to functions of two variables. For example, in Theorem 13.15 below, the Extreme Value Theorem for a function of a single variable is extended to a function of two variables. Consider the continuous function f of two variables, defined on a closed bounded region R. The values f (a, b兲 and f 共c, d 兲 such that
z
Maximum
f 共a, b兲 ⱕ f 共x, y) ⱕ f 共c, d 兲 x
y
Closed bounded region R
R contains point(s) at which f 共x, y兲 is a minimum and point(s) at which f 共x, y兲 is a maximum. Figure 13.64
共a, b兲 and 共c, d 兲 are in R.
for all 共x, y兲 in R are called the minimum and maximum of f in the region R, as shown in Figure 13.64. Recall from Section 13.2 that a region in the plane is closed when it contains all of its boundary points. The Extreme Value Theorem deals with a region in the plane that is both closed and bounded. A region in the plane is bounded when it is a subregion of a closed disk in the plane. THEOREM 13.15 Extreme Value Theorem Let f be a continuous function of two variables x and y defined on a closed bounded region R in the xy-plane. 1. There is at least one point in R at which f takes on a minimum value. 2. There is at least one point in R at which f takes on a maximum value.
A minimum is also called an absolute minimum and a maximum is also called an absolute maximum. As in single-variable calculus, there is a distinction made between absolute extrema and relative extrema. Definition of Relative Extrema Let f be a function defined on a region R containing 共x0, y0兲. 1. The function f has a relative minimum at 共x0, y0兲 if f 共x, y兲 ⱖ f 共x0, y0 兲 for all 共x, y兲 in an open disk containing 共x0, y0兲. 2. The function f has a relative maximum at 共x0, y0兲 if f 共x, y兲 ⱕ f 共x0, y0 兲 z
for all 共x, y兲 in an open disk containing 共x0, y0兲.
5 5
y
To say that f has a relative maximum at 共x0, y0兲 means that the point 共x0, y0, z0兲 is at least as high as all nearby points on the graph of z ⫽ f 共x, y兲.
x
Relative extrema Figure 13.65
Similarly, f has a relative minimum at 共x0, y0兲 when 共x0, y0, z0兲 is at least as low as all nearby points on the graph. (See Figure 13.65.)
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13.8
Extrema of Functions of Two Variables
937
To locate relative extrema of f, you can investigate the points at which the gradient of f is 0 or the points at which one of the partial derivatives does not exist. Such points are called critical points of f. Definition of Critical Point Let f be defined on an open region R containing 共x0, y0兲. The point 共x0, y0兲 is a critical point of f if one of the following is true. 1. fx 共x0, y0兲 ⫽ 0 and fy 共x0, y0 兲 ⫽ 0 2. fx 共x0, y0兲 or fy 共x0, y0兲 does not exist.
Recall from Theorem 13.11 that if f is differentiable and
KARL WEIERSTRASS (1815–1897)
Although the Extreme Value Theorem had been used by earlier mathematicians, the first to provide a rigorous proof was the German mathematician Karl Weierstrass.Weierstrass also provided rigorous justifications for many other mathematical results already in common use.We are indebted to him for much of the logical foundation on which modern calculus is built.
ⵜf 共x0, y0兲 ⫽ fx (x0, y0兲i ⫹ fy 共x0, y0兲j ⫽ 0i ⫹ 0j then every directional derivative at 共x0, y0兲 must be 0. This implies that the function has a horizontal tangent plane at the point 共x0, y0兲, as shown in Figure 13.66. It appears that such a point is a likely location of a relative extremum. This is confirmed by Theorem 13.16. Surface: z = f(x, y)
z
Surface: z = f(x, y)
z
(x0 , y0 , z 0 )
See LarsonCalculus.com to read more of this biography.
(x 0 , y0 , z 0 ) y
y x
(x 0 , y0)
(x 0 , y0 )
x
Relative maximum Figure 13.66
Relative minimum
THEOREM 13.16 Relative Extrema Occur Only at Critical Points If f has a relative extremum at 共x0, y0兲 on an open region R, then 共x0, y0兲 is a critical point of f.
Exploration Use a graphing utility to graph z ⫽ x 3 ⫺ 3xy ⫹ y 3 using the bounds 0 ⱕ x ⱕ 3, 0 ⱕ y ⱕ 3, and ⫺3 ⱕ z ⱕ 3. This view makes it appear as though the surface has an absolute minimum. But does it? z 3
3
3 x
y
−3
Jacques Boyer/Roger-Viollet/The Image Works
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938
Chapter 13
Functions of Several Variables
Finding a Relative Extremum See LarsonCalculus.com for an interactive version of this type of example.
Determine the relative extrema of Surface: f(x, y) = 2x 2 + y 2 + 8x − 6y + 20
f 共x, y兲 ⫽ 2x 2 ⫹ y 2 ⫹ 8x ⫺ 6y ⫹ 20. Solution
z 6
fx 共x, y兲 ⫽ 4x ⫹ 8
Partial with respect to x
fy 共x, y兲 ⫽ 2y ⫺ 6
Partial with respect to y
and
5 4
are defined for all x and y, the only critical points are those for which both first partial derivatives are 0. To locate these points, set fx 共x, y兲 and fy 共x, y兲 equal to 0, and solve the equations
(− 2, 3, 3)
3 2 1
Begin by finding the critical points of f. Because
−2
−3
1
x
−4
4x ⫹ 8 ⫽ 0 and
2
3
4
to obtain the critical point 共⫺2, 3兲. By completing the square for f, you can see that for all 共x, y兲 ⫽ 共⫺2, 3兲
y
5
2y ⫺ 6 ⫽ 0
The function z ⫽ f 共x, y兲 has a relative minimum at 共⫺2, 3兲. Figure 13.67
f 共x, y兲 ⫽ 2共x ⫹ 2兲 2 ⫹ 共 y ⫺ 3兲 2 ⫹ 3 > 3. So, a relative minimum of f occurs at 共⫺2, 3兲. The value of the relative minimum is f 共⫺2, 3兲 ⫽ 3, as shown in Figure 13.67. Example 1 shows a relative minimum occurring at one type of critical point—the type for which both fx 共x, y兲 and fy 共x, y兲 are 0. The next example concerns a relative maximum that occurs at the other type of critical point—the type for which either fx 共x, y兲 or fy 共x, y兲 does not exist.
Finding a Relative Extremum Determine the relative extrema of f 共x, y兲 ⫽ 1 ⫺ 共x 2 ⫹ y 2兲 1兾3. Solution
Surface: f(x, y) = 1 − (x 2 + y 2)1/3 z
(0, 0, 1)
3
fx 共x, y兲 ⫽ ⫺
2x 3共x 2 ⫹ y 2兲 2兾3
Partial with respect to x
fy 共x, y兲 ⫽ ⫺
2y 3共x 2 ⫹ y 2兲 2兾3
Partial with respect to y
and
1
4
Because
2 4
x
fx共x, y兲 and fy共x, y兲 are undefined at 共0, 0兲. Figure 13.68
y
it follows that both partial derivatives exist for all points in the xy-plane except for 共0, 0兲. Moreover, because the partial derivatives cannot both be 0 unless both x and y are 0, you can conclude that 共0, 0兲 is the only critical point. In Figure 13.68, note that f 共0, 0兲 is 1. For all other 共x, y兲, it is clear that f 共x, y兲 ⫽ 1 ⫺ 共x 2 ⫹ y 2兲 1兾3 < 1. So, f has a relative maximum at 共0, 0兲. In Example 2, fx 共x, y兲 ⫽ 0 for every point on the y-axis other than 共0, 0兲. However, because fy 共x, y兲 is nonzero, these are not critical points. Remember that one of the partials must not exist or both must be 0 in order to yield a critical point.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.8
Extrema of Functions of Two Variables
939
The Second Partials Test Theorem 13.16 tells you that to find relative extrema, you need only examine values of f 共x, y兲 at critical points. However, as is true for a function of one variable, the critical points of a function of two variables do not always yield relative maxima or minima. Some critical points yield saddle points, which are neither relative maxima nor relative minima. As an example of a critical point that does not f (x, y) = y 2 − x 2 z yield a relative extremum, consider the hyperbolic paraboloid f 共x, y兲 ⫽ y 2 ⫺ x 2 as shown in Figure 13.69. At the point 共0, 0兲, both partial derivatives fx共x, y兲 ⫽ ⫺2x and
fy共x, y兲 ⫽ 2y
y
are 0. The function f does not, however, have a relative extremum at this point because in any x open disk centered at 共0, 0兲, the function takes on both negative values (along the x-axis) and positive values (along the y-axis). So, the point 共0, 0, 0兲 is a saddle point of the surface. (The Saddle point at 共0, 0, 0兲: term “saddle point” comes from the fact that fx共0, 0兲 ⫽ fy共0, 0兲 ⫽ 0 surfaces such as the one shown in Figure 13.69 Figure 13.69 resemble saddles.) For the functions in Examples 1 and 2, it was relatively easy to determine the relative extrema, because each function was either given, or able to be written, in completed square form. For more complicated functions, algebraic arguments are less convenient and it is better to rely on the analytic means presented in the following Second Partials Test. This is the two-variable counterpart of the Second Derivative Test for functions of one variable. The proof of this theorem is best left to a course in advanced calculus. THEOREM 13.17 Second Partials Test Let f have continuous second partial derivatives on an open region containing a point 共a, b兲 for which fx 共a, b兲 ⫽ 0 and fy 共a, b兲 ⫽ 0. To test for relative extrema of f, consider the quantity d ⫽ fxx 共a, b兲 fyy 共a, b兲 ⫺ 关 fxy 共a, b兲兴 2.
REMARK If d > 0, then fxx 共a, b兲 and fyy共a, b兲 must have the same sign. This means that fxx共a, b兲 can be replaced by fyy共a, b兲 in the first two parts of the test.
1. 2. 3. 4.
If d > 0 and fxx 共a, b兲 > 0, then f has a relative minimum at 共a, b兲. If d > 0 and fxx 共a, b兲 < 0, then f has a relative maximum at 共a, b兲. If d < 0, then 共a, b, f 共a, b兲兲 is a saddle point. The test is inconclusive if d ⫽ 0.
A convenient device for remembering the formula for d in the Second Partials Test is given by the 2 ⫻ 2 determinant d⫽
ⱍ
fxx 共a, b兲 fyx 共a, b兲
fxy 共a, b兲 fyy 共a, b兲
ⱍ
where fxy 共a, b兲 ⫽ fyx 共a, b兲 by Theorem 13.3.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
940
Chapter 13
Functions of Several Variables
Using the Second Partials Test Find the relative extrema of f 共x, y兲 ⫽ ⫺x 3 ⫹ 4xy ⫺ 2y 2 ⫹ 1. z
Solution
fx 共x, y兲 ⫽ ⫺3x 2 ⫹ 4y and fy 共x, y兲 ⫽ 4x ⫺ 4y
9
exist for all x and y, the only critical points are those for which both first partial derivatives are 0. To locate these points, set fx 共x, y兲 and fy 共x, y兲 equal to 0 to obtain
8 7
⫺3x 2 ⫹ 4y ⫽ 0 and
6
4x ⫺ 4y ⫽ 0.
From the second equation, you know that x ⫽ y, and, by substitution into the first 4 equation, you obtain two solutions: y ⫽ x ⫽ 0 and y ⫽ x ⫽ 3. Because
5 4
Relative maximum
Begin by finding the critical points of f. Because
fxx 共x, y兲 ⫽ ⫺6x, fyy 共x, y兲 ⫽ ⫺4, and fxy 共x, y兲 ⫽ 4
Saddle point (0, 0, 1)
3
it follows that, for the critical point 共0, 0兲, d ⫽ fxx 共0, 0兲 fyy 共0, 0兲 ⫺ 关 fxy 共0, 0兲兴 2 ⫽ 0 ⫺ 16 < 0
3
and, by the Second Partials Test, you can conclude that 共0, 0, 1兲 is a saddle point of f. 4 4 Furthermore, for the critical point 共3, 3 兲,
2
x
( 43 , 43 (
4
d ⫽ fxx
y
冢43, 43冣 f 冢43, 43冣 ⫺ 冤f 冢43, 43冣冥 yy
2
xy
⫽ ⫺8共⫺4兲 ⫺ 16 ⫽ 16 > 0 f(x, y) = −x 3 + 4xy − 2y 2 + 1
4 4 and because fxx共3, 3 兲 ⫽ ⫺8 < 0, you can conclude that f has a relative maximum at 4 4 共3, 3 兲, as shown in Figure 13.70.
Figure 13.70
The Second Partials Test can fail to find relative extrema in two ways. If either of the first partial derivatives does not exist, you cannot use the test. Also, if d ⫽ fxx 共a, b兲 fyy 共a, b兲 ⫺ 关 fxy 共a, b兲兴 ⫽ 0 2
the test fails. In such cases, you can try a sketch or some other approach, as demonstrated in the next example.
Failure of the Second Partials Test Find the relative extrema of f 共x, y兲 ⫽ x 2 y 2. Solution Because fx 共x, y兲 ⫽ 2xy 2 and fy 共x, y兲 ⫽ 2x 2y, you know that both partial derivatives are 0 when x ⫽ 0 or y ⫽ 0. That is, every point along the x- or y-axis is a critical point. Moreover, because
f(x, y) = x 2 y 2
fxx 共x, y兲 ⫽ 2y 2,
z
fyy 共x, y兲 ⫽ 2x 2, and
fxy 共x, y兲 ⫽ 4xy
you know that 1
2 x
If y = 0, then f(x, y) = 0.
Figure 13.71
2
y
If x = 0, then f (x, y) = 0.
d ⫽ fxx 共x, y兲 fyy 共x, y兲 ⫺ 关 fxy 共x, y兲兴 ⫽ 4x 2 y 2 ⫺ 16x 2 y 2 ⫽ ⫺12x2y2
2
which is 0 when either x ⫽ 0 or y ⫽ 0. So, the Second Partials Test fails. However, because f 共x, y兲 ⫽ 0 for every point along the x- or y-axis and f 共x, y兲 ⫽ x 2 y 2 > 0 for all other points, you can conclude that each of these critical points yields an absolute minimum, as shown in Figure 13.71.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.8
941
Extrema of Functions of Two Variables
Absolute extrema of a function can occur in two ways. First, some relative extrema also happen to be absolute extrema. For instance, in Example 1, f 共⫺2, 3兲 is an absolute minimum of the function. (On the other hand, the relative maximum found in Example 3 is not an absolute maximum of the function.) Second, absolute extrema can occur at a boundary point of the domain. This is illustrated in Example 5.
Finding Absolute Extrema Find the absolute extrema of the function f 共x, y兲 ⫽ sin xy on the closed region given by 0 ⱕ x ⱕ Solution
and 0 ⱕ y ⱕ 1.
From the partial derivatives
fx 共x, y兲 ⫽ y cos xy and fy 共x, y兲 ⫽ x cos xy you can see that each point lying on the hyperbola xy ⫽ 兾2 is a critical point. These points each yield the value f 共x, y兲 ⫽ sin
⫽1 2
which you know is the absolute maximum, as shown in Figure 13.72. The only other critical point of f lying in the given region is 共0, 0兲. It yields an absolute minimum of 0, because
z
Surface: f(x, y) = sin xy
Absolute minima
1
0 ⱕ xy ⱕ
Absolute maxima
implies that
y
1
0 ⱕ sin xy ⱕ 1. To locate other absolute extrema, you should consider the four boundaries of the region formed by taking traces with the vertical planes x ⫽ 0, x ⫽ , y ⫽ 0, and y ⫽ 1. In doing this, you will find that sin xy ⫽ 0 at all points on the x-axis, at all points on the y-axis, and at the point 共, 1兲. Each of these points yields an absolute minimum for the surface, as shown in Figure 13.72.
xy = 3 x
(π , 1) Absolute minima
π 2
Domain: 0≤x≤π 0≤y≤1
Figure 13.72
The concepts of relative extrema and critical points can be extended to functions of three or more variables. When all first partial derivatives of w ⫽ f 共x1, x2, x3, . . . , xn兲 exist, it can be shown that a relative maximum or minimum can occur at 共x1, x2, x3, . . . , xn兲 only when every first partial derivative is 0 at that point. This means that the critical points are obtained by solving the following system of equations. fx1 共x1, x2, x3, . . . , xn兲 ⫽ 0 fx2 共x1, x2, x3, . . . , xn兲 ⫽ 0
⯗
fxn 共x1, x2, x3, . . . , xn兲 ⫽ 0 The extension of Theorem 13.17 to three or more variables is also possible, although you will not consider such an extension in this text.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
942
Chapter 13
Functions of Several Variables
13.8 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding Relative Extrema In Exercises 1–6, identify any
19. z ⫽ e⫺x sin y
extrema of the function by recognizing its given form or its form after completing the square. Verify your results by using the partial derivatives to locate any critical points and test for relative extrema.
z 8 6
1. g 共x, y兲 ⫽ 共x ⫺ 1兲 2 ⫹ 共 y ⫺ 3兲 2
4
2. g 共x, y兲 ⫽ 5 ⫺ 共x ⫺ 3兲 2 ⫺ 共 y ⫹ 2兲 2
2
3. f 共x, y兲 ⫽ 冪x 2 ⫹ y 2 ⫹ 1 4. f 共x, y兲 ⫽ 冪 25 ⫺ 共x ⫺ 2兲 2 ⫺ y 2
3π
6 x
5. f 共x, y兲 ⫽ x 2 ⫹ y 2 ⫹ 2x ⫺ 6y ⫹ 6 6. f 共x, y兲 ⫽ ⫺x 2 ⫺ y 2 ⫹ 10x ⫹ 12y ⫺ 64
20. z ⫽
Using the Second Partials Test In Exercises 7–20, examine the function for relative extrema and saddle points.
冢12 ⫺ x
2
冣
2 ⫺y2
⫹ y 2 e1⫺x
z
7. h 共x, y兲 ⫽ 80x ⫹ 80y ⫺ x 2 ⫺ y 2 8. g 共x, y兲 ⫽
x2
⫺
y2
y
2
⫺x⫺y
9. g 共x, y兲 ⫽ xy 10. h 共x, y兲 ⫽ x 2 ⫺ 3xy ⫺ y 2 11. f 共x, y兲 ⫽ ⫺3x 2 ⫺ 2y 2 ⫹ 3x ⫺ 4y ⫹ 5 12. f 共x, y兲 ⫽
2x 2
⫹ 2xy ⫹
y2
4
⫹ 2x ⫺ 3
y
4
1 13. z ⫽ x2 ⫹ xy ⫹ 2 y2 ⫺ 2x ⫹ y
x
14. z ⫽ ⫺5x 2 ⫹ 4xy ⫺ y 2 ⫹ 16x ⫹ 10 15. f 共x, y兲 ⫽ 冪x 2 ⫹ y 2
Finding Relative Extrema and Saddle Points Using Technology In Exercises 21–24, use a computer algebra system to graph the surface and locate any relative extrema and saddle points.
16. h共x, y兲 ⫽ 共x 2 ⫹ y 2兲1兾3 ⫹ 2 17. f 共x, y兲 ⫽ x2 ⫺ xy ⫺ y2 ⫺ 3x ⫺ y
21. z ⫽
z 4
⫺4x x2 ⫹ y2 ⫹ 1
22. f 共x, y兲 ⫽ y 3 ⫺ 3yx 2 ⫺ 3y 2 ⫺ 3x 2 ⫹ 1 23. z ⫽ 共x 2 ⫹ 4y 2兲e1⫺x
2 ⫺y 2
24. z ⫽ exy
y 3 3 x
Finding Relative Extrema In Exercises 25 and 26, examine the function for extrema without using the derivative tests, and use a computer algebra system to graph the surface. (Hint: By observation, determine if it is possible for z to be negative. When is z equal to 0?)
1 18. f 共x, y兲 ⫽ 2xy ⫺ 2共x 4 ⫹ y 4兲 ⫹ 1
25. z ⫽
z 2
y −2
共x ⫺ y兲4 x2 ⫹ y2
x
共x 2 ⫺ y 2兲2 x2 ⫹ y2
Think About It In Exercises 27–30, determine whether there is a relative maximum, a relative minimum, a saddle point, or insufficient information to determine the nature of the function f 冇x, y冈 at the critical point 冇x0 , y0冈. 27. fxx共 x0, y0兲 ⫽ 9,
3
26. z ⫽
fyy 共x0, y0兲 ⫽ 4,
fxy 共x0, y0兲 ⫽ 6
28. fxx 共x0, y0兲 ⫽ ⫺3,
fyy 共x0, y0兲 ⫽ ⫺8,
29. fxx 共x0, y0兲 ⫽ ⫺9,
fyy 共x0, y0兲 ⫽ 6,
30. fxx 共x0, y0兲 ⫽ 25,
fyy 共x0, y0兲 ⫽ 8,
fxy 共x0, y0兲 ⫽ 2 fxy 共x0, y0兲 ⫽ 10
fxy 共x0, y0兲 ⫽ 10
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.8 31. Using the Second Partials Test A function f has continuous second partial derivatives on an open region containing the critical point 共3, 7兲. The function has a minimum at 共3, 7兲, and d > 0 for the Second Partials Test. Determine the interval for fxy共3, 7兲 when fxx 共3, 7兲 ⫽ 2 and fyy 共3, 7兲 ⫽ 8. 32. Using the Second Partials Test A function f has continuous second partial derivatives on an open region containing the critical point 共a, b兲. If fxx共a, b兲 and fyy共a, b兲 have opposite signs, what is implied? Explain.
Finding Relative Extrema and Saddle Points In Exercises 33–38, (a) find the critical points, (b) test for relative extrema, (c) list the critical points for which the Second Partials Test fails, and (d) use a computer algebra system to graph the function, labeling any extrema and saddle points. 33. f 共x, y兲 ⫽
x3
⫹
y3
34. f 共x, y兲 ⫽
x3
⫹
y3
⫺
6x 2
⫹
9y 2
Extrema of Functions of Two Variables
943
WRITING ABOUT CONCEPTS 49. Defining Terms Define each of the following for a function of two variables. (a) Relative minimum
(b) Relative maximum
(c) Critical point
(d) Saddle point
Sketching a Graph In Exercises 50–52, sketch the graph of an arbitrary function f satisfying the given conditions. State whether the function has any extrema or saddle points. (There are many correct answers.) 50. All of the first and second partial derivatives of f are 0. 51. fx 共x, y兲 > 0 and fy 共x, y兲 < 0 for all 共x, y兲. 52. fx 共0, 0兲 ⫽ 0, fy 共0, 0兲 ⫽ 0
冦 > 0,
fx 共x, y兲
⫹ 12x ⫹ 27y ⫹ 19
35. f 共x, y兲 ⫽ 共x ⫺ 1兲 2共 y ⫹ 4兲 2
< 0,
冦 < 0,
x < 0 , x > 0
fy 共x, y兲
> 0,
y < 0 y > 0
fxx 共x, y兲 > 0, fyy 共x, y兲 < 0, and fxy 共x, y兲 ⫽ 0 for all 共x, y兲.
36. f 共x, y兲 ⫽ 冪共x ⫺ 1兲 2 ⫹ 共 y ⫹ 2兲 2 37. f 共x, y兲 ⫽ x 2兾3 ⫹ y 2兾3
53. Comparing Functions
38. f 共x, y兲 ⫽ 共x 2 ⫹ y 2兲2兾3
f 共x, y兲 ⫽ x2 ⫺ y2 and
Examining a Function In Exercises 39 and 40, find the critical points of the function and, from the form of the function, determine whether a relative maximum or a relative minimum occurs at each point.
Consider the functions
g共x, y兲 ⫽ x2 ⫹ y2.
(a) Show that both functions have a critical point at 共0, 0兲. (b) Explain how f and g behave differently at this critical point.
39. f 共x, y, z兲 ⫽ x 2 ⫹ 共 y ⫺ 3兲 2 ⫹ 共z ⫹ 1兲 2 40. f 共x, y, z兲 ⫽ 9 ⫺ 关x共 y ⫺ 1兲共z ⫹ 2兲兴 2
54.
Finding Absolute Extrema In Exercises 41– 48, find the absolute extrema of the function over the region R. (In each case, R contains the boundaries.) Use a computer algebra system to confirm your results. 41. f 共x, y兲 ⫽ x 2 ⫺ 4xy ⫹ 5
y
D
R ⫽ 再共x, y兲 : 1 ⱕ x ⱕ 4, 0 ⱕ y ⱕ 2冎 42. f 共x, y兲 ⫽ x 2 ⫹ xy,
HOW DO YOU SEE IT? The figure shows the level curves for an unknown function f 共x, y兲. What, if any, information can be given about f at the points A, B, C, and D? Explain your reasoning.
ⱍⱍ
ⱍⱍ
A
x
R ⫽ 再共x, y兲 : x ⱕ 2, y ⱕ 1冎
43. f 共x, y兲 ⫽ 12 ⫺ 3x ⫺ 2y
B
C
R: The triangular region in the xy-plane with vertices 共2, 0兲, 共0, 1兲, and 共1, 2兲 44. f 共x, y兲 ⫽ 共2x ⫺ y兲2 R: The triangular region in the xy-plane with vertices 共2, 0兲, 共0, 1兲, and 共1, 2兲 45. f 共x, y兲 ⫽ 3x 2 ⫹ 2y 2 ⫺ 4y R: The region in the xy-plane bounded by the graphs of y ⫽ x 2 and y ⫽ 4 46. f 共x, y兲 ⫽ 2x ⫺ 2xy ⫹ y 2 R: The region in the xy-plane bounded by the graphs of y ⫽ x2 and y ⫽ 1 47. f 共x, y兲 ⫽ x 2 ⫹ 2xy ⫹ y 2, 48. f 共x, y兲 ⫽
ⱍⱍ
ⱍⱍ
R ⫽ 再共x, y兲 : x ⱕ 2, y ⱕ 1冎
4xy 共x 2 ⫹ 1兲共 y 2 ⫹ 1兲
R ⫽ 再共x, y兲 : 0 ⱕ x ⱕ 1, 0 ⱕ y ⱕ 1冎
True or False? In Exercises 55–58, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 55. If f has a relative maximum at 共x0, y0, z0兲, then fx 共x0, y0兲 ⫽ fy 共x0, y0兲 ⫽ 0. 56. If fx共x0, y0兲 ⫽ fy共x0, y0兲 ⫽ 0, then f has a relative maximum at 共x0, y0, z0兲. 57. Between any two relative minima of f, there must be at least one relative maximum of f. 58. If f is continuous for all x and y and has two relative minima, then f must have at least one relative maximum.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
944
Chapter 13
Functions of Several Variables
13.9 Applications of Extrema Solve optimization problems involving functions of several variables. Use the method of least squares.
Applied Optimization Problems In this section, you will survey a few of the many applications of extrema of functions of two (or more) variables.
Finding Maximum Volume See LarsonCalculus.com for an interactive version of this type of example. z
(0, 0, 8)
A rectangular box is resting on the xy-plane with one vertex at the origin. The opposite vertex lies in the plane
Plane: 6x + 4y + 3z = 24
6x ⫹ 4y ⫹ 3z ⫽ 24 as shown in Figure 13.73. Find the maximum volume of such a box. Solution Let x, y, and z represent the length, width, and height of the box. Because one vertex of the box lies in the plane 6x ⫹ 4y ⫹ 3z ⫽ 24, you know that z ⫽ 1324 ⫺ 6x ⫺ 4y, and you can write the volume xyz of the box as a function of two variables. Vx, y ⫽ xy 1324 ⫺ 6x ⫺ 4y ⫽ 1324xy ⫺ 6x 2y ⫺ 4xy 2
x
(4, 0, 0)
(0, 6, 0)
y
Next, find the first partial derivatives of V. 1 y Vxx, y ⫽ 24y ⫺ 12xy ⫺ 4y2 ⫽ 24 ⫺ 12x ⫺ 4y 3 3 1 x Vyx, y ⫽ 24x ⫺ 6x 2 ⫺ 8xy ⫽ 24 ⫺ 6x ⫺ 8y 3 3
Figure 13.73
Note that the first partial derivatives are defined for all x and y. So, by setting Vxx, y 1 and Vyx, y equal to 0 and solving the equations 3y24 ⫺ 12x ⫺ 4y ⫽ 0 and 1 4 3 x24 ⫺ 6x ⫺ 8y ⫽ 0, you obtain the critical points 0, 0 and 3 , 2. At 0, 0, the 4 volume is 0, so that point does not yield a maximum volume. At the point 3, 2, you can apply the Second Partials Test. Vxxx, y ⫽ ⫺4y ⫺8x Vyyx, y ⫽ 3 1 Vxyx, y ⫽ 24 ⫺ 12x ⫺ 8y 3
REMARK In many applied problems, the domain of the function to be optimized is a closed bounded region. To find minimum or maximum points, you must not only test critical points, but also consider the values of the function at points on the boundary.
Because 8 64 Vxx43, 2Vyy43, 2 ⫺ Vxy43, 2 ⫽ ⫺8⫺ 32 9 ⫺ ⫺ 3 ⫽ 3 > 0 2
2
and Vxx43, 2 ⫽ ⫺8 < 0 you can conclude from the Second Partials Test that the maximum volume is V43, 2 ⫽ 132443 2 ⫺ 643 2 ⫺ 443 22 ⫽ 64 9 cubic units. 2
Note that the volume is 0 at the boundary points of the triangular domain of V.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13.9
Applications of Extrema
945
Applications of extrema in economics and business often involve more than one independent variable. For instance, a company may produce several models of one type of product. The price per unit and profit per unit are usually different for each model. Moreover, the demand for each model is often a function of the prices of the other models (as well as its own price). The next example illustrates an application involving two products.
Finding the Maximum Profit An electronics manufacturer determines that the profit P (in dollars) obtained by producing and selling x units of an LCD television and y units of a plasma television is approximated by the model Px, y ⫽ 8x ⫹ 10y ⫺ 0.001x 2 ⫹ xy ⫹ y 2 ⫺ 10,000. Find the production level that produces a maximum profit. What is the maximum profit? Solution
The partial derivatives of the profit function are
Pxx, y ⫽ 8 ⫺ 0.0012x ⫹ y and Pyx, y ⫽ 10 ⫺ 0.001x ⫹ 2y. By setting these partial derivatives equal to 0, you obtain the following system of equations. 8 ⫺ 0.0012x ⫹ y ⫽ 0 10 ⫺ 0.001x ⫹ 2y ⫽ 0 After simplifying, this system of linear equations can be written as 2x ⫹ y ⫽ 8000 x ⫹ 2y ⫽ 10,000. Solving this system produces x ⫽ 2000 and y ⫽ 4000. The second partial derivatives of P are Pxx2000, 4000 ⫽ ⫺0.002 Pyy2000, 4000 ⫽ ⫺0.002 Pxy2000, 4000 ⫽ ⫺0.001. Because Pxx < 0 and Pxx2000, 4000Pyy2000, 4000 ⫺ Pxy2000, 4000 2 ⫽ ⫺0.0022 ⫺ ⫺0.0012 is greater than 0, you can conclude that the production level of x ⫽ 2000 units and y ⫽ 4000 units yields a maximum profit. The maximum profit is P2000, 4000 ⫽ 82000 ⫹ 104000 ⫺ 0.00120002 ⫹ 20004000 ⫹ 40002 ⫺ 10,000 ⫽ $18,000. In Example 2, it was assumed that the manufacturing plant is able to produce the required number of units to yield a maximum profit. In actual practice, the production would be bounded by physical constraints. You will study such constrained optimization problems in the next section. FOR FURTHER INFORMATION For more information on the use of mathematics in economics, see the article “Mathematical Methods of Economics” by Joel Franklin in The American Mathematical Monthly. To view this article, go to MathArticles.com.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
946
Chapter 13
Functions of Several Variables
The Method of Least Squares Many of the examples in this text have involved mathematical models. For instance, Example 2 involves a quadratic model for profit. There are several ways to develop such models; one is called the method of least squares. In constructing a model to represent a particular phenomenon, the goals are simplicity and accuracy. Of course, these goals often conflict. For instance, a simple linear model for the points in Figure 13.74 is y ⫽ 1.9x ⫺ 5. However, Figure 13.75 shows that by choosing the slightly more complicated quadratic model y ⫽ 0.20x2 ⫺ 0.7x ⫹ 1 you can achieve greater accuracy. y = 1.9x − 5
y
y = 0.20x 2 − 0.7x + 1 y
(11, 17) (11, 17)
15 15
(9, 12)
(9, 12)
10 10
(7, 6)
5
(7, 6)
5
(2, 1)
(5, 2)
(2, 1)
x 5
(5, 2) x
10 5
Figure 13.74
10
Figure 13.75
As a measure of how well the model y ⫽ f x fits the collection of points
x1, y1, x2, y2, x3, y3, . . . , xn, yn you can add the squares of the differences between the actual y-values and the values given by the model to obtain the sum of the squared errors
S⫽
n
f x ⫺ y . i
i
2
Sum of the squared errors
i⫽1
REMARK A method for finding the least squares regression quadratic for a collection of data is described in Exercise 31.
Graphically, S can be interpreted as the sum of the squares of the vertical distances between the graph of f and the given points in the plane, as shown in Figure 13.76. If the model is perfect, then S ⫽ 0. However, when perfection is not feasible, you can settle for a model that minimizes S. For instance, the sum of the squared errors for the linear model in Figure 13.74 is
y
(x1, y1) d1
y = f(x) d2 (x2 , y2 )
S ⫽ 17.6. Statisticians call the linear model that minimizes S the least squares regression line. The proof that this line actually minimizes S involves the minimizing of a function of two variables.
(x3, y3 ) d3 x
Sum of the squared errors: S ⫽ d12 ⫹ d22 ⫹ d32 Figure 13.76
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13.9
ADRIEN-MARIE LEGENDRE (1752–1833)
The method of least squares was introduced by the French mathematician Adrien-Marie Legendre. Legendre is best known for his work in geometry. In fact, his text Elements of Geometry was so popular in the United States that it continued to be used for 33 editions, spanning a period of more than 100 years. See LarsonCalculus.com to read more of this biography.
Applications of Extrema
947
THEOREM 13.18 Least Squares Regression Line The least squares regression line for x1, y1, x2, y2, . . . , xn, yn is given by f x ⫽ ax ⫹ b, where n
n a⫽
x i yi ⫺
i⫽1 n
x
2 i
n
⫺
i⫽1
Proof
n
n
y xi
i⫽1 n
i
i⫽1 2
and
x
b⫽
1 n
y ⫺ a x . n
n
i
i⫽1
i
i⫽1
i
i⫽1
Let Sa, b represent the sum of the squared errors for the model
f x ⫽ ax ⫹ b and the given set of points. That is, n
f x ⫺ y
Sa, b ⫽ ⫽
i
i
2
i⫽1 n
ax ⫹ b ⫺ y i
i
2
i⫽1
where the points xi, yi represent constants. Because S is a function of a and b, you can use the methods discussed in the preceding section to find the minimum value of S. Specifically, the first partial derivatives of S are n
2x ax ⫹ b ⫺ y
Saa, b ⫽
i
i
i
i⫽1
⫽ 2a
n
x2i ⫹ 2b
i⫽1
n
xi ⫺ 2
i⫽1
n
xy
i i
i⫽1
and n
2ax ⫹ b ⫺ y
Sba, b ⫽
i
i
i⫽1
⫽ 2a
n
n
x ⫹ 2nb ⫺ 2 y .
i⫽1
i
i
i⫽1
By setting these two partial derivatives equal to 0, you obtain the values of a and b that are listed in the theorem. It is left to you to apply the Second Partials Test (see Exercise 41) to verify that these values of a and b yield a minimum. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
If the x-values are symmetrically spaced about the y-axis, then xi ⫽ 0 and the formulas for a and b simplify to n
xy
i i
a⫽
i⫽1 n
x
2 i
i⫽1
and b⫽
1 n y. n i⫽1 i
This simplification is often possible with a translation of the x-values. For instance, given that the x-values in a data collection consist of the years 2009, 2010, 2011, 2012, and 2013, you could let 2011 be represented by 0.
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948
Chapter 13
Functions of Several Variables
Finding the Least Squares Regression Line Find the least squares regression line for the points
⫺3, 0, ⫺1, 1, 0, 2, and 2, 3. Solution The table shows the calculations involved in finding the least squares regression line using n ⫽ 4. x
y
xy
x2
⫺3
0
0
9
⫺1
1
⫺1
1
TECHNOLOGY Many
0
2
0
0
calculators have “built-in” least squares regression programs. If your calculator has such a program, use it to duplicate the results of Example 3.
2
3
6
4
n
n
n
x ⫽ ⫺2 y ⫽ 6 i
i⫽1
n
xy ⫽5 x
i
2 i
i i
i⫽1
i⫽1
⫽ 14
i⫽1
Applying Theorem 13.18 produces n
n a⫽
xi yi ⫺
i⫽1 n
x
2 i
n
i⫽1
⫺
n
n
y xi
i⫽1 n
i
i⫽1 2
x
i
i⫽1
45 ⫺ ⫺26 414 ⫺ ⫺22 8 ⫽ 13
⫽
and
n 1 n yi ⫺ a xi n i⫽1 i⫽1 1 8 ⫽ 6 ⫺ ⫺2 4 13
b⫽
⫽
47 . 26
The least squares regression line is f x ⫽
8 47 x⫹ 13 26
as shown in Figure 13.77. y
(2, 3) f(x) =
8 x 13
+
47 26
3 2
(0, 2)
1
(− 1, 1)
(−3, 0)
x −3
−2
−1
1
2
Least squares regression line Figure 13.77
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13.9
13.9 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding Minimum Distance In Exercises 1 and 2, find the minimum distance from the point to the plane x ⴚ y ⴙ z ⴝ 3. (Hint: To simplify the computations, minimize the square of the distance.) 1. 0, 0, 0
949
Applications of Extrema
2. 1, 2, 3
Finding Minimum Distance In Exercises 3 and 4, find the minimum distance from the point to the surface z ⴝ 1 ⴚ 2x ⴚ 2y. (Hint: To simplify the computations, minimize the square of the distance.)
15. Hardy-Weinberg Law Common blood types are determined genetically by three alleles A, B, and O. (An allele is any of a group of possible mutational forms of a gene.) A person whose blood type is AA, BB, or OO is homozygous. A person whose blood type is AB, AO, or BO is heterozygous. The Hardy-Weinberg Law states that the proportion P of heterozygous individuals in any given population is P p, q, r ⫽ 2pq ⫹ 2pr ⫹ 2qr
5. The product is 27, and the sum is a minimum.
where p represents the percent of allele A in the population, q represents the percent of allele B in the population, and r represents the percent of allele O in the population. Use the fact that
6. The sum is 32, and P ⫽ xy 2z is a maximum.
p⫹q⫹r⫽1
3. ⫺2, ⫺2, 0
4. ⫺4, 1, 0
Finding Positive Numbers In Exercises 5–8, find three positive integers x, y, and z that satisfy the given conditions.
7. The sum is 30, and the sum of the squares is a minimum. 8. The product is 1, and the sum of the squares is a minimum. 9. Cost A home improvement contractor is painting the walls and ceiling of a rectangular room. The volume of the room is 668.25 cubic feet. The cost of wall paint is $0.06 per square foot and the cost of ceiling paint is $0.11 per square foot. Find the room dimensions that result in a minimum cost for the paint. What is the minimum cost for the paint? 10. Maximum Volume The material for constructing the base of an open box costs 1.5 times as much per unit area as the material for constructing the sides. For a fixed amount of money C, find the dimensions of the box of largest volume that can be made.
to show that the maximum proportion of heterozygous individuals in any population is 23. 16. Shannon Diversity Index One way to measure species diversity is to use the Shannon diversity index H. If a habitat consists of three species, A, B, and C, then its Shannon diversity index is H ⫽ ⫺x ln x ⫺ y ln y ⫺ z ln z where x is the percent of species A in the habitat, y is the percent of species B in the habitat, and z is the percent of species C in the habitat. Use the fact that
11. Volume and Surface Area Show that a rectangular box of given volume and minimum surface area is a cube.
x⫹y⫹z⫽1
12. Maximum Volume Show that the rectangular box of maximum volume inscribed in a sphere of radius r is a cube.
to show that the maximum value of H occurs when 1 x ⫽ y ⫽ z ⫽ 3. What is the maximum value of H?
13. Maximum Revenue A company manufactures running shoes and basketball shoes. The total revenue from x1 units of running shoes and x2 units of basketball shoes is
17. Minimum Cost A water line is to be built from point P to point S and must pass through regions where construction costs differ (see figure). The cost per kilometer (in dollars) is 3k from P to Q, 2k from Q to R, and k from R to S. Find x and y such that the total cost C will be minimized.
R ⫽ ⫺5x12 ⫺ 8x22 ⫺ 2x1x 2 ⫹ 42x1 ⫹ 102x 2 where x1 and x2 are in thousands of units. Find x1 and x2 so as to maximize the revenue. 14. Maximum Profit A corporation manufactures candles at two locations. The cost of producing x1 units at location 1 is C1 ⫽ 0.02x12 ⫹ 4x1 ⫹ 500
P x 2 km
The candles sell for $15 per unit. Find the quantity that should be produced at each location to maximize the profit P ⫽ 15x1 ⫹ x2 ⫺ C1 ⫺ C2.
θ
30 − 2x
R y
and the cost of producing x2 units at location 2 is C2 ⫽ 0.05x22 ⫹ 4x 2 ⫹ 275.
θ
x Q
1 km
x
S 10 km
Figure for 17
Figure for 18
18. Area A trough with trapezoidal cross sections is formed by turning up the edges of a 30-inch-wide sheet of aluminum (see figure). Find the cross section of maximum area.
Sashkin/Shutterstock.com
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950
Chapter 13
Functions of Several Variables
WRITING ABOUT CONCEPTS 19. Applied Optimization Problems In your own words, state the problem-solving strategy for applied minimum and maximum problems.
30. Modeling Data The table shows the gross income tax collections (in billions of dollars) by the Internal Revenue Service for individuals x and businesses y. (Source: U.S. Internal Revenue Service)
20. Method of Least Squares In your own words, describe the method of least squares for finding mathematical models.
Finding the Least Squares Regression Line In Exercises 21–24, (a) find the least squares regression line, and (b) calculate S, the sum of the squared errors. Use the regression capabilities of a graphing utility to verify your results. y
21.
y
22. (2, 3)
3
4 3
2 1
2
(0, 1)
(−2, 0) −1
1
2
4
−3 −2 −1
1
−1
2 1
(3, 1)
1
(1, 1)
1
2
(6, 2)
x
3
1
x
2
156
288
397
540
Business, y
46
72
77
110
Year
1995
2000
2005
2010
Individual, x
676
1137
1108
1164
Business, y
174
236
307
278
3
4
5
y ⫽ ax 2 ⫹ bx ⫹ c for the points x1, y1, x2, y2, . . . , xn, yn by minimizing the sum
(4, 1)
(1, 0) (3, 0)
(2, 0)
Individual, x
31. Method of Least Squares Find a system of equations whose solution yields the coefficients a, b, and c for the least squares regression quadratic (4, 2)
(1, 3)
3
1990
3
2
(5, 2)
2
1985
(b) Use the model to estimate the business income taxes collected when the individual income taxes collected is $1300 billion.
y
24. (0, 4)
1980
(3, 2)
−2
y
23.
(1, 1)
1975
(a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data.
x
x
−2
(−1, 1) (−3, 0) 1
Year
6
Sa, b, c ⫽
n
y ⫺ ax
2 i
i
⫺ bxi ⫺ c2.
i⫽1
(2, 0)
4
Finding the Least Squares Regression Line In Exercises 25–28, find the least squares regression line for the points. Use the regression capabilities of a graphing utility to verify your results. Use the graphing utility to plot the points and graph the regression line. 25. 0, 0, 1, 1, 3, 4, 4, 2, 5, 5
HOW DO YOU SEE IT? Match the regression equation with the appropriate graph. Explain your reasoning. (Note that the x- and y-axes are broken.)
32.
(a) y ⫽ 0.22x ⫺ 7.5
(b) y ⫽ ⫺0.35x ⫹ 11.5
(c) y ⫽ 0.09x ⫹ 19.8
(d) y ⫽ ⫺1.29x ⫹ 89.8
y
(i)
26. 1, 0, 3, 3, 5, 6 27. 0, 6, 4, 3, 5, 0, 8, ⫺4, 10, ⫺5 28. 6, 4, 1, 2, 3, 3, 8, 6, 11, 8, 13, 8 29. Modeling Data The ages x (in years) and systolic blood pressures y of seven men are shown in the table.
y
(ii) 65
9 8 7 6 5 4 3
55 45 35 25 x
x
Age, x
16
Systolic Blood Pressure, y
25
39
45
49
64
10
70 (iii)
109
122
150
165
159
183
199
15
20
20
25
y
(iv)
(b) Use the model to approximate the change in systolic blood pressure for each one-year increase in age.
40
50
y
10 9 8 7 6 5 4
240 210 180
(a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data.
30
150 120 x 1200
1800
2400
x 50 55 60 65 70 75
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13.9
Finding the Least Squares Regression Quadratic In Exercises 33–36, use the result of Exercise 31 to find the least squares regression quadratic for the given points. Use the regression capabilities of a graphing utility to confirm your results. Use the graphing utility to plot the points and graph the least squares regression quadratic. 33. ⫺2, 0, ⫺1, 0, 0, 1, 1, 2, 2, 5 34. ⫺4, 5, ⫺2, 6, 2, 6, 4, 2 35. 0, 0, 2, 2, 3, 6, 4, 12
36. 0, 10, 1, 9, 2, 6, 3, 0
37. Modeling Data After a new turbocharger for an automobile engine was developed, the following experimental data were obtained for speed y in miles per hour at two-second time intervals x.
951
Applications of Extrema
40. Modeling Data The endpoints of the interval over which distinct vision is possible are called the near point and far point of the eye. With increasing age, these points normally change. The table shows the approximate near points y (in inches) for various ages x (in years). (Source: Ophthalmology & Physiological Optics) Age, x
16
32
44
50
60
Near Point, y
3.0
4.7
9.8
19.7
39.4
(a) Find a rational model for the data by taking the reciprocals of the near points to generate the points x, 1y. Use the regression capabilities of a graphing utility to find a least squares regression line for the revised data. The resulting line has the form 1y ⫽ ax ⫹ b. Solve for y.
Time, x
0
2
4
6
8
10
(b) Use a graphing utility to plot the data and graph the model.
Speed, y
0
15
30
50
65
70
(c) Do you think the model can be used to predict the near point for a person who is 70 years old? Explain.
(a) Find a least squares regression quadratic for the data. Use a graphing utility to confirm your results. (b) Use a graphing utility to plot the points and graph the model. 38. Modeling Data The table shows the world populations y (in billions) for five different years. Let x ⫽ 3 represent the year 2003. (Source: U.S. Census Bureau, International Data Base) Year, x Population, y
2003
2005
2007
2009
2011
6.3
6.5
6.6
6.8
6.9
(a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data. (b) Use the regression capabilities of a graphing utility to find the least squares regression quadratic for the data.
41. Using the Second Partials Test Use the Second Partials Test to verify that the formulas for a and b given in Theorem 13.18 yield a minimum.
Hint: Use the fact that n x ⱖ x . n
n
2 i
2
i
i⫽1
i⫽1
Building a Pipeline An oil company wishes to construct a pipeline from its offshore facility A to its refinery B. The offshore facility is 2 miles from shore, and the refinery is 1 mile inland. Furthermore, A and B are 5 miles apart, as shown in the figure. A
(c) Use a graphing utility to plot the data and graph the models. (d) Use both models to forecast the world population for the year 2020. How do the two models differ as you extrapolate into the future? 39. Modeling Data A meteorologist measures the atmospheric pressure P (in kilograms per square meter) at altitude h (in kilometers). The data are shown below. Altitude, h
0
5
10
15
20
Pressure, P
10,332
5583
2376
1240
517
(a) Use the regression capabilities of a graphing utility to find a least squares regression line for the points h, ln P. (b) The result in part (a) is an equation of the form ln P ⫽ ah ⫹ b. Write this logarithmic form in exponential form. (c) Use a graphing utility to plot the original data and graph the exponential model in part (b).
2 mi
5 mi P
x 1 mi B
The cost of building the pipeline is $3 million per mile in the water and $4 million per mile on land. So, the cost of the pipeline depends on the location of point P, where it meets the shore. What would be the most economical route of the pipeline? Imagine that you are to write a report to the oil company about this problem. Let x be the distance shown in the figure. Determine the cost of building the pipeline from A to P, and the cost of building it from P to B. Analyze some sample pipeline routes and their corresponding costs. For instance, what is the cost of the most direct route? Then use calculus to determine the route of the pipeline that minimizes the cost. Explain all steps of your development and include any relevant graphs.
(d) If your graphing utility can fit logarithmic models to data, then use it to verify the result in part (b).
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
952
Chapter 13
13.10
Functions of Several Variables
Lagrange Multipliers Understand the Method of Lagrange Multipliers. Use Lagrange multipliers to solve constrained optimization problems. Use the Method of Lagrange Multipliers with two constraints.
Lagrange Multipliers LAGRANGE MULTIPLIERS
The Method of Lagrange Multipliers is named after the French mathematician Joseph-Louis Lagrange. Lagrange first introduced the method in his famous paper on mechanics, written when he was just 19 years old.
Many optimization problems have restrictions, or constraints, on the values that can be used to produce the optimal solution. Such constraints tend to complicate optimization problems because the optimal solution can occur at a boundary point of the domain. In this section, you will study an ingenious technique for solving such problems. It is called the Method of Lagrange Multipliers. To see how this technique works, consider the problem of finding the rectangle of maximum area that can be inscribed in the ellipse x2 y2 ⫹ 2 ⫽ 1. 2 3 4 Let 共x, y兲 be the vertex of the rectangle in the first quadrant, as shown in Figure 13.78. Because the rectangle has sides of lengths 2x and 2y, its area is given by f 共x, y兲 ⫽ 4xy.
Objective function
You want to find x and y such that f 共x, y兲 is a maximum. Your choice of 共x, y兲 is restricted to first-quadrant points that lie on the ellipse x2 y2 ⫹ 2 ⫽ 1. 2 3 4
Constraint
Now, consider the constraint equation to be a fixed level curve of g共x, y兲 ⫽
x2 y2 ⫹ . 32 4 2
The level curves of f represent a family of hyperbolas f 共x, y兲 ⫽ 4xy ⫽ k. In this family, the level curves that meet the constraint correspond to the hyperbolas that intersect the ellipse. Moreover, to maximize f 共x, y兲, you want to find the hyperbola that just barely satisfies the constraint. The level curve that does this is the one that is tangent to the ellipse, as shown in Figure 13.79. Ellipse: x2 y2 + =1 32 42
y
Level curves of f: 4xy = k
y
5
(x, y) 3 2
3
1
2 x
−4
−2 −1 −1
1
2
−2 −3
k = 72 k = 56 k = 40 k = 24
1
4
x
−2 −1 −1
1
2
4
5
6
−2 −3
Objective function: f 共x, y兲 ⫽ 4xy
Constraint: g共x, y兲 ⫽
Figure 13.78
Figure 13.79
x2 y2 ⫹ 2⫽1 2 3 4
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13.10
Lagrange Multipliers
953
To find the appropriate hyperbola, use the fact that two curves are tangent at a point if and only if their gradient vectors are parallel. This means that ⵜf 共x, y兲 must be a scalar multiple of ⵜg共x, y兲 at the point of tangency. In the context of constrained optimization problems, this scalar is denoted by (the lowercase Greek letter lambda). ⵜf 共x, y兲 ⫽ ⵜg共x, y兲 The scalar is called a Lagrange multiplier. Theorem 13.19 gives the necessary conditions for the existence of such multipliers.
REMARK Lagrange’s Theorem can be shown to be true for functions of three variables, using a similar argument with level surfaces and Theorem 13.14.
THEOREM 13.19 Lagrange’s Theorem Let f and g have continuous first partial derivatives such that f has an extremum at a point 共x0, y0 兲 on the smooth constraint curve g共x, y兲 ⫽ c. If ⵜg共x0, y0 兲 ⫽ 0, then there is a real number such that ⵜf 共x0, y0 兲 ⫽ ⵜg共x0, y0 兲. Proof To begin, represent the smooth curve given by g共x, y兲 ⫽ c by the vectorvalued function r共t兲 ⫽ x共t兲i ⫹ y共t兲j, r⬘ 共t兲 ⫽ 0 where x⬘ and y⬘ are continuous on an open interval I. Define the function h as h 共t兲 ⫽ f 共x 共t兲, y 共t兲兲. Then, because f 共x0, y0 兲 is an extreme value of f, you know that h 共t0 兲 ⫽ f 共x共t0 兲, y 共t0 兲兲 ⫽ f 共x0, y0 兲 is an extreme value of h. This implies that h⬘共t0 兲 ⫽ 0, and, by the Chain Rule, h⬘ 共t0 兲 ⫽ fx共x0, y0 兲 x⬘ 共t0 兲 ⫹ fy 共x0, y0 兲y⬘ 共t0 兲 ⫽ ⵜf 共x0, y0 兲 ⭈ r⬘ 共t0 兲 ⫽ 0. So, ⵜf 共x0, y0 兲 is orthogonal to r⬘ 共t0 兲. Moreover, by Theorem 13.12, ⵜg共x0, y0 兲 is also orthogonal to r⬘ 共t0 兲. Consequently, the gradients ⵜf 共x0, y0 兲 and ⵜg共x0, y0 兲 are parallel, and there must exist a scalar such that ⵜf 共x0, y0 兲 ⫽ ⵜg共x0, y0 兲. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
The Method of Lagrange Multipliers uses Theorem 13.19 to find the extreme values of a function f subject to a constraint.
REMARK As you will see in Examples 1 and 2, the Method of Lagrange Multipliers requires solving systems of nonlinear equations. This often can require some tricky algebraic manipulation.
Method of Lagrange Multipliers Let f and g satisfy the hypothesis of Lagrange’s Theorem, and let f have a minimum or maximum subject to the constraint g共x, y兲 ⫽ c. To find the minimum or maximum of f, use these steps. 1. Simultaneously solve the equations ⵜf 共x, y兲 ⫽ ⵜg共x, y兲 and g共x, y兲 ⫽ c by solving the following system of equations. fx共x, y兲 ⫽ gx共x, y兲 fy共x, y兲 ⫽ gy共x, y兲 g共x, y兲 ⫽ c 2. Evaluate f at each solution point obtained in the first step. The greatest value yields the maximum of f subject to the constraint g共x, y兲 ⫽ c, and the least value yields the minimum of f subject to the constraint g共x, y兲 ⫽ c.
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954
Chapter 13
Functions of Several Variables
Constrained Optimization Problems In the problem at the beginning of this section, you wanted to maximize the area of a rectangle that is inscribed in an ellipse. Example 1 shows how to use Lagrange multipliers to solve this problem.
Using a Lagrange Multiplier with One Constraint Find the maximum value of f 共x, y兲 ⫽ 4xy, where x > 0 and y > 0, subject to the constraint 共x 2兾32兲 ⫹ 共 y 2兾42兲 ⫽ 1. Solution
REMARK Example 1 can also be solved using the techniques you learned in Chapter 3. To see how, try to find the maximum value of A ⫽ 4xy given that x2 32
⫹
y2 42
⫽ 1.
To begin, solve the second equation for y to obtain y ⫽ 43冪9 ⫺ x2. Then substitute into the first equation to obtain A ⫽ 4x共43冪9 ⫺ x2兲. Finally, use the techniques of Chapter 3 to maximize A.
To begin, let
g共x, y兲 ⫽
x2 y2 ⫹ 2 ⫽ 1. 2 3 4
By equating ⵜf 共x, y兲 ⫽ 4yi ⫹ 4xj and ⵜg共x, y兲 ⫽ 共2 x兾9兲 i ⫹ 共y兾8兲 j, you can obtain the following system of equations. 2 4y ⫽ x 9 1 4x ⫽ y 8 2 2 x y ⫹ ⫽1 32 42
fx共x, y兲 ⫽ gx共x, y兲 fy共x, y兲 ⫽ gy共x, y兲 Constraint
From the first equation, you obtain ⫽ 18y兾x, and substitution into the second equation produces 4x ⫽
冢 冣
1 18y y 8 x
x2 ⫽
9 2 y. 16
Substituting this value for x2 into the third equation produces
冢
冣
1 9 2 1 2 y ⫹ y ⫽1 9 16 16
y 2 ⫽ 8.
So, y ⫽ ± 2冪2. Because it is required that y > 0, choose the positive value and find that 9 2 y 16 9 ⫽ 共8兲 16 9 ⫽ 2 3 x⫽ . 冪2
x2 ⫽
So, the maximum value of f is f
冢冪32 , 2冪2冣 ⫽ 4xy ⫽ 4 冢冪32冣共2冪2 兲 ⫽ 24.
Note that writing the constraint as g共x, y兲 ⫽
x2 y2 ⫹ 2⫽1 2 3 4
or g共x, y兲 ⫽
x2 y2 ⫹ 2⫺1⫽0 2 3 4
does not affect the solution—the constant is eliminated when you form ⵜg.
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13.10
Lagrange Multipliers
955
A Business Application The Cobb-Douglas production function (see Example 5, Section 13.1) for a software manufacturer is given by f 共x, y兲 ⫽ 100x 3兾4 y1兾4
Objective function
where x represents the units of labor (at $150 per unit) and y represents the units of capital (at $250 per unit). The total cost of labor and capital is limited to $50,000. Find the maximum production level for this manufacturer. FOR FURTHER INFORMATION
For more information on the use of Lagrange multipliers in economics, see the article “Lagrange Multiplier Problems in Economics” by John V. Baxley and John C. Moorhouse in The American Mathematical Monthly. To view this article, go to MathArticles.com.
Solution
The gradient of f is
ⵜf 共x, y兲 ⫽ 75x⫺1兾4 y 1兾4 i ⫹ 25x 3兾4 y⫺3兾4 j. The limit on the cost of labor and capital produces the constraint g共x, y兲 ⫽ 150x ⫹ 250y ⫽ 50,000.
Constraint
So, ⵜg共x, y兲 ⫽ 150 i ⫹ 250 j. This gives rise to the following system of equations. 75x⫺1兾4 y 1兾4 ⫽ 150 25x 3兾4 y⫺3兾4 ⫽ 250 150x ⫹ 250y ⫽ 50,000
fx共x, y兲 ⫽ gx共x, y兲 fy共x, y兲 ⫽ gy共x, y兲 Constraint
By solving for in the first equation
⫽
75x⫺1兾4 y1兾4 x⫺1兾4 y1兾4 ⫽ 150 2
and substituting into the second equation, you obtain 25x 3兾4 y⫺3兾4 ⫽ 250
⫺1兾4 y1兾4
冢x
2
冣
25x ⫽ 125y
Multiply by x 1兾4 y 3兾4.
x ⫽ 5y. By substituting this value for x in the third equation, you have 150共5y兲 ⫹ 250y ⫽ 50,000 1000y ⫽ 50,000 y ⫽ 50 units of capital. This means that the value of x is x ⫽ 5共50兲 ⫽ 250 units of labor. So, the maximum production level is f 共250, 50兲 ⫽ 100共250兲3兾4共50兲1兾4 ⬇ 16,719 product units. Economists call the Lagrange multiplier obtained in a production function the marginal productivity of money. For instance, in Example 2, the marginal productivity of money at x ⫽ 250 and y ⫽ 50 is
⫽
x⫺1兾4 y 1兾4 共250兲⫺1兾4 共50兲1兾4 ⫽ ⬇ 0.334 2 2
which means that for each additional dollar spent on production, an additional 0.334 unit of the product can be produced.
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956
Chapter 13
Functions of Several Variables
Lagrange Multipliers and Three Variables See LarsonCalculus.com for an interactive version of this type of example.
Find the minimum value of f 共x, y, z兲 ⫽ 2x 2 ⫹ y 2 ⫹ 3z 2
Objective function
subject to the constraint 2x ⫺ 3y ⫺ 4z ⫽ 49. Solution
Let g共x, y, z兲 ⫽ 2x ⫺ 3y ⫺ 4z ⫽ 49. Then, because
ⵜf 共x, y, z兲 ⫽ 4xi ⫹ 2yj ⫹ 6zk and
ⵜg共x, y, z兲 ⫽ 2 i ⫺ 3 j ⫺ 4 k you obtain the following system of equations. 4x ⫽ 2 2y ⫽ ⫺3 6z ⫽ ⫺4 2x ⫺ 3y ⫺ 4z ⫽ 49
fx共x, y, z兲 ⫽ gx共x, y, z兲 fy共x, y, z兲 ⫽ gy共x, y, z兲 fz共x, y, z兲 ⫽ gz共x, y, z兲 Constraint
The solution of this system is x ⫽ 3, y ⫽ ⫺9, and z ⫽ ⫺4. So, the optimum value of f is f 共3, ⫺9, ⫺4兲 ⫽ 2共3兲 2 ⫹ 共⫺9兲2 ⫹ 3共⫺4兲 2 ⫽ 147.
Ellipsoid: 2x 2 + y 2 + 3z 2 = 147
From the original function and constraint, it is clear that f 共x, y, z兲 has no maximum. So, the optimum value of f determined above is a minimum.
z 8 y 16 −16 24
Point of tangency (3, − 9, − 4)
x
A graphical interpretation of constrained optimization problems was given at the beginning of this section. In three variables, the similar, except that level surfaces are used instead of level curves. Example 3, the level surfaces of f are ellipsoids centered at the constraint
in two variables interpretation is For instance, in origin, and the
2x ⫺ 3y ⫺ 4z ⫽ 49
Plane: 2x − 3y − 4z = 49
is a plane. The minimum value of f is represented by the ellipsoid that is tangent to the constraint plane, as shown in Figure 13.80.
Figure 13.80
Optimization Inside a Region Find the extreme values of f 共x, y兲 ⫽ x 2 ⫹ 2y 2 ⫺ 2x ⫹ 3 subject to the constraint x 2 ⫹ y 2 ⱕ 10.
z
(− 1, − 3, 24)
40
Relative maxima
32
Solution
(− 1, 3, 24)
24 16 8
Relative minimum (1, 0, 2)
2
3
4
( 10, 0, 6.675 (
x
Figure 13.81
Objective function
4
y
To solve this problem, you can break the constraint into two cases.
a. For points on the circle x 2 ⫹ y 2 ⫽ 10, you can use Lagrange multipliers to find that the maximum value of f 共x, y兲 is 24—this value occurs at 共⫺1, 3兲 and at 共⫺1, ⫺3兲. In a similar way, you can determine that the minimum value of f 共x, y兲 is approximately 6.675—this value occurs at 共冪10, 0兲. b. For points inside the circle, you can use the techniques discussed in Section 13.8 to conclude that the function has a relative minimum of 2 at the point 共1, 0兲. By combining these two results, you can conclude that f has a maximum of 24 at 共⫺1, ± 3兲 and a minimum of 2 at 共1, 0兲, as shown in Figure 13.81.
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13.10
Lagrange Multipliers
957
The Method of Lagrange Multipliers with Two Constraints For optimization problems involving two constraint functions g and h, you can introduce a second Lagrange multiplier, (the lowercase Greek letter mu), and then solve the equation ⵜf ⫽ ⵜg ⫹ ⵜh where the gradient vectors are not parallel, as illustrated in Example 5.
Optimization with Two Constraints Let T 共x, y, z兲 ⫽ 20 ⫹ 2x ⫹ 2y ⫹ z 2 represent the temperature at each point on the sphere x 2 ⫹ y 2 ⫹ z 2 ⫽ 11. Find the extreme temperatures on the curve formed by the intersection of the plane x ⫹ y ⫹ z ⫽ 3 and the sphere. Solution
The two constraints are
g共x, y, z兲 ⫽ x 2 ⫹ y 2 ⫹ z 2 ⫽ 11 and
h共x, y, z兲 ⫽ x ⫹ y ⫹ z ⫽ 3.
Using ⵜT 共x, y, z兲 ⫽ 2i ⫹ 2j ⫹ 2zk ⵜg共x, y, z兲 ⫽ 2 x i ⫹ 2y j ⫹ 2 z k and
ⵜh 共x, y, z兲 ⫽ i ⫹ j ⫹ k you can write the following system of equations. 2 ⫽ 2 x ⫹ 2 ⫽ 2y ⫹ 2z ⫽ 2z ⫹ 2 2 x ⫹ y ⫹ z 2 ⫽ 11 x⫹y⫹z⫽3
Tx共x, y, z兲 ⫽ gx共x, y, z兲 ⫹ hx共x, y, z兲 Ty共x, y, z兲 ⫽ gy共x, y, z兲 ⫹ hy共x, y, z兲 Tz共x, y, z兲 ⫽ gz共x, y, z兲 ⫹ hz共x, y, z兲 Constraint 1 Constraint 2
By subtracting the second equation from the first, you can obtain the following system.
REMARK The systems of equations that arise when the Method of Lagrange Multipliers is used are not, in general, linear systems, and finding the solutions often requires ingenuity.
共x ⫺ y兲 ⫽ 0 2z共1 ⫺ 兲 ⫺ ⫽ 0 x 2 ⫹ y 2 ⫹ z 2 ⫽ 11 x⫹y⫹z⫽3 From the first equation, you can conclude that ⫽ 0 or x ⫽ y. For ⫽ 0, you can show that the critical points are 共3, ⫺1, 1兲 and 共⫺1, 3, 1兲. 共Try doing this—it takes a little work.兲 For ⫽ 0, then x ⫽ y and you can show that the critical points occur when x ⫽ y ⫽ 共3 ± 2冪3 兲兾3 and z ⫽ 共3 ⫿ 4冪3 兲兾3. Finally, to find the optimal solutions, compare the temperatures at the four critical points. T 共3, ⫺1, 1兲 ⫽ T 共⫺1, 3, 1兲 ⫽ 25
冢 3 ⫺ 32 3, 3 ⫺ 32 3, 3 ⫹ 34 3 冣 ⫽ 913 ⬇ 30.33 3 ⫹ 2冪3 3 ⫹ 2冪3 3 ⫺ 4冪3 , , T冢 冣 ⫽ 913 ⬇ 30.33 3 3 3
T
冪
冪
冪
91 So, T ⫽ 25 is the minimum temperature and T ⫽ 3 is the maximum temperature on the curve.
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958
Chapter 13
Functions of Several Variables
13.10 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Using Lagrange Multipliers
In Exercises 1– 8, use Lagrange multipliers to find the indicated extrema, assuming that x and y are positive. 1. Maximize: f 共x, y兲 ⫽ xy 2. Minimize: f 共x, y兲 ⫽ 2x ⫹ y Constraint: xy ⫽ 32 3. Minimize f 共x, y兲 ⫽ x 2 ⫹ y 2 Constraint: x ⫹ 2y ⫺ 5 ⫽ 0 4. Maximize f 共x, y兲 ⫽ x 2 ⫺ y 2 5. Maximize f 共x, y兲 ⫽ 2x ⫹ 2xy ⫹ y
Constraint: x 2y ⫽ 6 7. Maximize f 共x, y兲 ⫽ 冪6 ⫺ x 2 ⫺ y 2 Constraint: x ⫹ y ⫺ 2 ⫽ 0
19. Line: x ⫺ y ⫽ 4
共0, 2兲
20. Line: x ⫹ 4y ⫽ 3
共1, 0兲
21. Parabola: y ⫽
共0, 3兲
x2
23. Circle:
x2
共⫺3, 0兲
⫹ 共y ⫺ 1兲 ⫽ 9
共4, 4兲
2
24. Circle: 共x ⫺ 4兲 ⫹
y2
共0, 10兲
⫽4
Point
25. Plane: x ⫹ y ⫹ z ⫽ 1
共2, 1, 1兲
26. Cone: z ⫽
共4, 0, 0兲
冪x 2
⫹
y2
highest point on the curve of intersection of the surfaces. 27. Cone: x 2 ⫹ y 2 ⫺ z 2 ⫽ 0
Using Lagrange Multipliers In Exercises 9 – 12, use Lagrange multipliers to find the indicated extrema, assuming that x, y, and z are positive. ⫹
共0, 0兲
Intersection of Surfaces In Exercises 27 and 28, find the y2
Constraint: 2x ⫹ 4y ⫺ 15 ⫽ 0
9. Minimize f 共x, y, z兲 ⫽
共0, 0兲
18. Line: 2x ⫹ 3y ⫽ ⫺1
Surface
6. Minimize f 共x, y兲 ⫽ 3x ⫹ y ⫹ 10
x2
17. Line: x ⫹ y ⫽ 1
2
Constraint: 2x ⫹ y ⫽ 100
⫹
Point
22. Parabola: y ⫽ x2
Constraint: 2y ⫺ x 2 ⫽ 0
8. Minimize f 共x, y兲 ⫽
Lagrange multipliers to find the minimum distance from the curve or surface to the indicated point. (Hint: To simplify the computations, minimize the square of the distance.) Curve
Constraint: x ⫹ y ⫽ 10
冪x 2
Finding Minimum Distance In Exercises 17–26, use
y2
⫹
Plane: x ⫹ 2z ⫽ 4 28. Sphere: x 2 ⫹ y 2 ⫹ z 2 ⫽ 36 Plane: 2x ⫹ y ⫺ z ⫽ 2
z2
Constraint: x ⫹ y ⫹ z ⫺ 9 ⫽ 0 10. Maximize f 共x, y, z兲 ⫽ xyz Constraint: x ⫹ y ⫹ z ⫺ 3 ⫽ 0 11. Minimize f 共x, y, z兲 ⫽ x 2 ⫹ y 2 ⫹ z 2 Constraint: x ⫹ y ⫹ z ⫽ 1
WRITING ABOUT CONCEPTS 29. Constrained Optimization Problems Explain what is meant by constrained optimization problems. 30. Method of Lagrange Multipliers Explain the Method of Lagrange Multipliers for solving constrained optimization problems.
12. Maximize f 共x, y, z兲 ⫽ x ⫹ y ⫹ z Constraint: x2 ⫹ y2 ⫹ z2 ⫽ 1
Using Lagrange Multipliers In Exercises 13 and 14, use Lagrange multipliers to find any extrema of the function subject to the constraint x 2 ⴙ y 2 ⱕ 1. 13. f 共x, y兲 ⫽ x 2 ⫹ 3xy ⫹ y 2
14. f 共x, y兲 ⫽ e⫺xy兾4
Using Lagrange Multipliers In Exercises 15 and 16, use Lagrange multipliers to find the indicated extrema of f subject to two constraints. In each case, assume that x, y, and z are nonnegative. 15. Maximize f 共x, y, z兲 ⫽ xyz Constraints: x ⫹ y ⫹ z ⫽ 32, x ⫺ y ⫹ z ⫽ 0 16. Minimize f 共x, y, z兲 ⫽ x 2 ⫹ y 2 ⫹ z 2 Constraints: x ⫹ 2z ⫽ 6, x ⫹ y ⫽ 12
Using Lagrange Multipliers In Exercises 31– 38, use Lagrange multipliers to solve the indicated exercise in Section 13.9. 31. Exercise 1
32. Exercise 2
33. Exercise 5
34. Exercise 6
35. Exercise 9
36. Exercise 10
37. Exercise 15
38. Exercise 16
39. Maximum Volume Use Lagrange multipliers to find the dimensions of a rectangular box of maximum volume that can be inscribed (with edges parallel to the coordinate axes) in the ellipsoid x2 y2 z2 ⫹ 2 ⫹ 2 ⫽ 1. 2 a b c
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13.10
HOW DO YOU SEE IT? The graphs show the
40.
constraint and several level curves of the objective function. Use the graph to approximate the indicated extrema. (b) Minimize z ⫽ x 2 ⫹ y 2
(a) Maximize z ⫽ xy Constraint: 2x ⫹ y ⫽ 4 y
Constraint: x ⫹ y ⫺ 4 ⫽ 0 y
c=2 c=4 c=6
4
c=8 c=6 c=4 c=2
2
−4
6
4
45. Refraction of Light When light waves traveling in a transparent medium strike the surface of a second transparent medium, they tend to “bend” in order to follow the path of minimum time. This tendency is called refraction and is described by Snell’s Law of Refraction, sin 1 sin 2 ⫽ v1 v2
P Medium 1
d1
x
4
6
959
where 1 and 2 are the magnitudes of the angles shown in the figure, and v1 and v2 are the velocities of light in the two media. Use Lagrange multipliers to derive this law using x ⫹ y ⫽ a.
4
x
2
Lagrange Multipliers
−4
θ1 y
x Medium 2 a
41. Minimum Cost A cargo container (in the shape of a rectangular solid) must have a volume of 480 cubic feet. The bottom will cost $5 per square foot to construct and the sides and the top will cost $3 per square foot to construct. Use Lagrange multipliers to find the dimensions of the container of this size that has minimum cost.
d2
θ2
Q
46. Area and Perimeter A semicircle is on top of a rectangle (see figure). When the area is fixed and the perimeter is a minimum, or when the perimeter is fixed and the area is a maximum, use Lagrange multipliers to verify that the length of the rectangle is twice its height.
42. Geometric and Arithmetic Means (a) Use Lagrange multipliers to prove that the product of three positive numbers x, y, and z, whose sum has the constant value S, is a maximum when the three numbers are equal. Use this result to prove that 3 xyz 冪
ⱕ
h
x⫹y⫹z . 3
l
(b) Generalize the result of part (a) to prove that the product x1 x2 x3 . . . xn is a maximum when x1 ⫽ x2 ⫽ x3 ⫽ . . . ⫽ xn,
n
兺 x ⫽ S, and all x i
i
ⱖ 0.
Production Level In Exercises 47 and 48, find the maximum production level P when the total cost of labor (at $72 per unit) and capital (at $60 per unit) is limited to $250,000, where x is the number of units of labor and y is the number of units of capital. 47. P共x, y兲 ⫽ 100x 0.25 y 0.75
48. P共x, y兲 ⫽ 100x 0.4y 0.6
i⫽1
Then prove that x ⫹ x 2 ⫹ x 3 ⫹ . . . ⫹ xn n 冪 x1 x 2 x 3 . . . xn ⱕ 1 . n This shows that the geometric mean is never greater than the arithmetic mean. 43. Minimum Surface Area Use Lagrange multipliers to find the dimensions of a right circular cylinder with volume V0 cubic units and minimum surface area. 44. Temperature Let T 共x, y, z兲 ⫽ 100 ⫹ x 2 ⫹ y 2 represent the temperature at each point on the sphere x 2 ⫹ y 2 ⫹ z 2 ⫽ 50.
Cost In Exercises 49 and 50, find the minimum cost of producing 50,000 units of a product, where x is the number of units of labor (at $72 per unit) and y is the number of units of capital (at $60 per unit). 49. P共x, y兲 ⫽ 100x 0.25 y 0.75
50. P共x, y兲 ⫽ 100x 0.6y 0.4
PUTNAM EXAM CHALLENGE 51. A can buoy is to be made of three pieces, namely, a cylinder and two equal cones, the altitude of each cone being equal to the altitude of the cylinder. For a given area of surface, what shape will have the greatest volume? This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
Find the maximum temperature on the curve formed by the intersection of the sphere and the plane x ⫺ z ⫽ 0.
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960
Chapter 13
Functions of Several Variables
Review Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Evaluating a Function In Exercises 1 and 2, find and
Limit and Continuity In Exercises 11–14, find the limit (if
simplify the function values.
it exists) and discuss the continuity of the function.
1. f 共x, y兲 3x2y (a) 共1, 3兲
(b) 共1, 1兲
2. f 共x, y兲 6 4x (a) 共0, 2兲
(c) 共4, 0兲
(d) 共x, 2兲
2y2
11. 13.
(b) 共5, 0兲
(c) 共1, 2兲
(d) 共3, y兲
lim
共x, y兲 → 共1, 1兲
xy x2 y2
y xey 共x, y兲 → 共0, 0兲 1 x 2
12.
xy x2 y2
lim
共x, y兲 → 共1, 1兲
2
lim
14.
lim
x4
共x, y兲 → 共0, 0兲
x2y y2
Finding Partial Derivatives In Exercises 15–22, find all Finding the Domain and Range of a Function In Exercises 3 and 4, find the domain and range of the function.
first partial derivatives. 15. f 共x, y兲 5x3 7y 3
16. f 共x, y兲 4x2 2xy y2
17. f 共x, y兲 e x cos y
18. f 共x, y兲
Sketching a Contour Map In Exercises 5 and 6, describe
19. f 共x, y兲 y3e4x
20. z ln共x 2 y 2 1兲
the level curves of the function. Sketch a contour map of the surface using level curves for the given c-values.
21. f 共x, y, z兲 2xz2 6xyz 5xy3
3. f 共x, y兲
冪x
4. f 共x, y兲 冪36 x2 y2
y
5. z 3 2x y, c 0, 2, 4, 6, 8
xy xy
22. w 冪x2 y2 z2
Finding Second Partial Derivatives In Exercises 23–26,
6. z 2x2 y2,
c 1, 2, 3, 4, 5
7. Conjecture
Consider the function f 共x, y兲 x2 y2.
find the four second partial derivatives. Observe that the second mixed partials are equal. x xy
(a) Sketch the graph of the surface given by f.
23. f 共x, y兲 3x2 xy 2y3
24. h共x, y兲
(b) Make a conjecture about the relationship between the graphs of f and g共x, y兲 f 共x, y兲 2. Explain your reasoning.
25. h共x, y兲 x sin y y cos x
26. g共x, y兲 cos共x 2y兲
(c) Make a conjecture about the relationship between the graphs of f and g共x, y兲 f 共x, y 2兲. Explain your reasoning. (d) On the surface in part (a), sketch the graphs of z f 共1, y兲 and z f 共x, 1兲. 8. Investment A principal of $2000 is deposited in a savings account that earns interest at a rate of r (written as a decimal) compounded continuously. The amount A共r, t兲 after t years is A共r, t兲 2000ert. Use this function of two variables to complete the table. Number of Years
27. Finding the Slopes of a Surface Find the slopes of the surface z x2 ln共y 1兲 in the x- and y-directions at the point 共2, 0, 0兲. 28. Marginal Revenue A company has two plants that produce the same lawn mower. If x1 and x2 are the numbers of units produced at plant 1 and plant 2, respectively, then the total revenue for the product is given by R 300x1 300x2 5x21 10x1x2 5x22. When x1 5 and x2 8, find (a) the marginal revenue for plant 1, R兾x1, and (b) the marginal revenue for plant 2, R兾x2.
Finding a Total Differential In Exercises 29–32, find the total differential.
Rate
5
10
15
20
0.02 0.04 0.06 0.07
Sketching a Level Surface In Exercises 9 and 10, sketch the graph of the level surface f 冇x, y, z冈 ⴝ c at the given value of c. 9. f 共x, y, z兲 x 2 y z 2, c 2 10. f 共x, y, z兲 4x 2 y 2 4z 2, c 0
29. z x sin xy
30. z 5x4y3
31. w 3xy2 2x3yz2
32. w
3x 4y y 3z
Using a Differential as an Approximation In Exercises 33 and 34, (a) evaluate f 冇2, 1冈 and f 冇2.1, 1.05冈 and calculate z, and (b) use the total differential dz to approximate z. 33. f 共x, y兲 4x 2y
34. f 共x, y兲 36 x2 y2
35. Volume The possible error involved in measuring each dimension of a right circular cone is ± 18 inch. The radius is 2 inches and the height is 5 inches. Approximate the propagated error and the relative error in the calculated volume of the cone.
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Review Exercises
961
36. Lateral Surface Area Approximate the propagated error and the relative error in the computation of the lateral surface area of the cone in Exercise 35. 共The lateral surface area is given by A r冪r2 h2.兲
Finding an Equation of a Tangent Plane In Exercises 53–56, find an equation of the tangent plane to the surface at the given point.
Using Different Methods In Exercises 37 and 38, find
54. 9x2 y2 4z2 25, 共0, 3, 2兲
dw兾dt (a) by using the appropriate Chain Rule, and (b) by converting w to a function of t before differentiating.
55. z 9 4x 6y x 2 y 2,
37. w ln共x 2 y兲, 38. w y2 x,
x 2t, y 4 t
w/r and w/t (a) by using the appropriate Chain Rule and (b) by converting w to a function of r and t before differentiating. xy , z
56. f 共x, y兲 冪25 y 2,
共2, 3, 4兲
共2, 3, 4兲
Finding an Equation of a Tangent Plane and a Normal Line In Exercises 57 and 58, find an equation of the tangent
x cos t, y sin t
Using Different Methods In Exercises 39 and 40, find
39. w
53. z x2 y2 2, 共1, 3, 12兲
x 2r t, y rt, z 2r t
plane and find a set of symmetric equations for the normal line to the surface at the given point. 57. f 共x, y兲 x 2 y, 共2, 1, 4兲 58. z 冪9 x 2 y 2,
共1, 2, 2兲
Finding Partial Derivatives Implicitly In Exercises 41
59. Angle of Inclination Find the angle of inclination of the tangent plane to the surface x 2 y 2 z 2 14 at the point 共2, 1, 3兲.
and 42, differentiate implicitly to find the first partial derivatives of z.
60. Approximation Consider the following approximations for a function f 共x, y兲 centered at 共0, 0兲.
40. w x 2 y 2 z 2,
x r cos t, y r sin t, z t
41. x2 xy y2 yz z2 0
42. xz 2 y sin z 0
Finding a Directional Derivative In Exercises 43 and 44, use Theorem 13.9 to find the directional derivative of the function at P in the direction of v. 43. f 共x, y兲 x 2y,
P共5, 5兲,
44. f 共x, y兲 14 y 2 x 2,
v 2i j
Finding a Directional Derivative In Exercises 45 and 46, use the gradient to find the directional derivative of the function at P in the direction of v. 45. w y 2 xz,
P共1, 2, 2兲, v 2i j 2k
46. w 5x 2 2xy 3y 2z,
P共1, 0, 1兲,
vijk
Using Properties of the Gradient In Exercises 47–50, find the gradient of the function and the maximum value of the directional derivative at the given point. 47. z x 2 y, 共2, 1兲 48. z ex cos y, 49. z
y , 共1, 1兲 x2 y2
x2 , 共2, 1兲 50. z xy
Using a Function In Exercises 51 and 52, (a) find the gradient of the function at P, (b) find a unit normal vector to the level curve f 冇x, y冈 ⴝ c at P, (c) find the tangent line to the level curve f 冇x, y冈 ⴝ c at P, and (d) sketch the level curve, the unit normal vector, and the tangent line in the xy-plane. 51. f 共x, y兲 9x2 4y2 c 65, P共3, 2兲
Quadratic approximation: P2共x, y兲 f 共0, 0兲 fx 共0, 0兲x fy 共0, 0兲y 0兲x 2 fxy共0, 0兲xy 12 fyy共0, 0兲y 2
[Note that the linear approximation is the tangent plane to the surface at 共0, 0, f 共0, 0兲兲.兴 (a) Find the linear approximation of f 共x, y兲 cos x sin y centered at 共0, 0兲. (b) Find the quadratic approximation of f 共x, y兲 cos x sin y centered at 共0, 0兲. (c) When y 0 in the quadratic approximation, you obtain the second-degree Taylor polynomial for what function? (d) Complete the table.
4
冢 冣 0,
P1共x, y兲 f 共0, 0兲 fx 共0, 0兲x fy 共0, 0兲y
1 2 fxx共0,
v 3i 4j
P共1, 4兲,
Linear approximation:
52. f 共x, y兲 4y sin x y c 3, P
冢2 , 1冣
x
y
0
0
0
0.1
0.2
0.1
0.5
0.3
1
0.5
f 共x, y兲
P1共x, y兲
P2共x, y兲
(e) Use a computer algebra system to graph the surfaces z f 共x, y兲, z P1共x, y兲, and z P2共x, y兲. How does the accuracy of the approximations change as the distance from 共0, 0兲 increases?
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
962
Chapter 13
Functions of Several Variables
Using the Second Partials Test In Exercises 61–66, examine the function for relative extrema and saddle points. 61. f 共x, y兲 x2 4y2 8x 8y 11
74. Modeling Data The data in the table show the yield y (in milligrams) of a chemical reaction after t minutes. Minutes, t
62. f 共x, y兲 x2 y2 16x 16y
Yield, y
63. f 共x, y兲 2x 2 6xy 9y 2 8x 14 64. f 共x, y兲 x2 3xy y2 5x
Minutes, t
1 1 65. f 共x, y兲 xy x y
Yield, y
66. f 共x, y兲 8x2 4xy y2 12x 7 67. Finding Minimum Distance Find the minimum distance from the point 共2, 1, 4兲 to the surface x y z 4. 共Hint: To simplify the computations, minimize the square of the distance.兲 68. Finding Positive Numbers Find three positive integers, x, y, and z, such that the product is 64 and the sum is a minimum. 69. Maximum Revenue A company manufactures two types of bicycles, a racing bicycle and a mountain bicycle. The total revenue from x1 units of racing bicycles and x2 units of mountain bicycles is R 6x12 10x22 2x1x2 32x1 84x2 where x1 and x2 are in thousands of units. Find x1 and x2 so as to maximize the revenue. 70. Maximum Profit A corporation manufactures digital cameras at two locations. The cost of producing x1 units at location 1 is C1 0.05x12 15x1 5400 and the cost of producing x2 units at location 2 is C2 0.03x 22 15x 2 6100. The digital cameras sell for $180 per unit. Find the quantity that should be produced at each location to maximize the profit P 180共x1 x2兲 C1 C2.
Finding the Least Squares Regression Line In Exercises 71 and 72, find the least squares regression line for the points. Use the regression capabilities of a graphing utility to verify your results. Use the graphing utility to plot the points and graph the regression line.
1
2
3
4
1.2
7.1
9.9
13.1
5
6
7
8
15.5
16.0
17.9
18.0
(a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data. Then use the graphing utility to plot the data and graph the model. (b) Use a graphing utility to plot the points 共ln t, y兲. Do these points appear to follow a linear pattern more closely than the plot of the given data in part (a)? (c) Use the regression capabilities of a graphing utility to find the least squares regression line for the points 共ln t, y兲 and obtain the logarithmic model y a b ln t. (d) Use a graphing utility to plot the original data and graph the linear and logarithmic models. Which is a better model? Explain.
Using Lagrange Multipliers In Exercises 75 – 80, use Lagrange multipliers to find the indicated extrema, assuming that x and y are positive. 75. Minimize: f 共x, y兲 x2 y2 Constraint: x y 8 0 76. Maximize: f 共x, y兲 xy Constraint: x 3y 6 0 77. Maximize: f 共x, y兲 2x 3xy y Constraint: x 2y 29 78. Minimize: f 共x, y兲 x2 y2 Constraint: x 2y 6 0 79. Maximize: f 共x, y兲 2xy Constraint: 2x y 12
71. 共0, 4兲, 共1, 5兲, 共3, 6兲, 共6, 8兲, 共8, 10兲
80. Minimize: f 共x, y兲 3x 2 y 2
72. 共0, 10兲, 共2, 8兲, 共4, 7兲, 共7, 5兲, 共9, 3兲, 共12, 0兲 73. Modeling Data An agronomist used four test plots to determine the relationship between the wheat yield y (in bushels per acre) and the amount of fertilizer x (in pounds per acre). The results are shown in the table. Fertilizer, x
100
150
200
250
Yield, y
35
44
50
56
Constraint: 2x 2y 5 0 81. Minimum Cost A water line is to be built from point P to point S and must pass through regions where construction costs differ (see figure). The cost per kilometer in dollars is 3k from P to Q, 2k from Q to R, and k from R to S. For simplicity, let k 1. Use Lagrange multipliers to find x, y, and z such that the total cost C will be minimized. P
(a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data.
2 km
(b) Use the model to approximate the wheat yield for a fertilizer application of 175 pounds per acre.
1 km
Q R x
y
S z
10 km
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
P.S. Problem Solving
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
P.S. Problem Solving 1. Area Heron’s Formula states that the area of a triangle with sides of lengths a, b, and c is given by A 冪s共s a兲共s b兲共s c兲 where s
abc , as shown in the figure. 2
(a) Use Heron’s Formula to find the area of the triangle with vertices 共0, 0兲, 共3, 4兲, and 共6, 0兲. (b) Show that among all triangles having a fixed perimeter, the triangle with the largest area is an equilateral triangle. (c) Show that among all triangles having a fixed area, the triangle with the smallest perimeter is an equilateral triangle. r b
a
963
c
5. Finding Maximum and Minimum Values (a) Let f 共x, y兲 x y and g共x, y兲 x2 y2 4. Graph various level curves of f and the constraint g in the xy-plane. Use the graph to determine the maximum value of f subject to the constraint g 4. Then verify your answer using Lagrange multipliers. (b) Let f 共x, y兲 x y and g共x, y兲 x2 y2 0. Find the maximum and minimum values of f subject to the constraint g 0. Does the Method of Lagrange Multipliers work in this case? Explain. 6. Minimizing Costs A heated storage room has the shape of a rectangular prism and has a volume of 1000 cubic feet, as shown in the figure. Because warm air rises, the heat loss per unit of area through the ceiling is five times as great as the heat loss through the floor. The heat loss through the four walls is three times as great as the heat loss through the floor. Determine the room dimensions that will minimize heat loss and therefore minimize heating costs.
h Figure for 1
V = xyz = 1000
Figure for 2
2. Minimizing Material An industrial container is in the shape of a cylinder with hemispherical ends, as shown in the figure. The container must hold 1000 liters of fluid. Determine the radius r and length h that minimize the amount of material used in the construction of the tank. 3. Tangent Plane Let P共x0, y0, z0兲 be a point in the first octant on the surface xyz 1. (a) Find the equation of the tangent plane to the surface at the point P. (b) Show that the volume of the tetrahedron formed by the three coordinate planes and the tangent plane is constant, independent of the point of tangency (see figure).
z y
x
7. Minimizing Costs Repeat Exercise 6 assuming that the heat loss through the walls and ceiling remain the same, but the floor is insulated so that there is no heat loss through the floor. 8. Temperature Consider a circular plate of radius 1 given by x 2 y 2 1, as shown in the figure. The temperature at any point P共x, y兲 on the plate is T 共x, y兲 2x 2 y 2 y 10. y 1
z
x2 + y2 ≤ 1
3
3
3 x
y
4. Using Functions Use a graphing utility to graph the functions 3 x3 1 f 共x兲 冪 and
−1
(a) Sketch the isotherm T 共x, y兲 10. To print an enlarged copy of the graph, go to MathGraphs.com. 9. Cobb-Douglas Production Function Cobb-Douglas production function
in the same viewing window. (a) Show that x→
1
(b) Find the hottest and coldest points on the plate.
g共x兲 x
lim 关 f 共x兲 g共x兲兴 0 and
x
−1
P
lim 关 f 共x兲 g共x兲兴 0.
x→
(b) Find the point on the graph of f that is farthest from the graph of g.
f 共x, y兲 Cxay1a,
Consider the
0 < a < 1.
(a) Show that f satisfies the equation x
f f y f. x y
(b) Show that f 共tx, ty兲 t f 共x, y兲.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
964
Chapter 13
10. Minimizing Area
Functions of Several Variables 15. Area The figure shows a rectangle that is approximately l 6 centimeters long and h 1 centimeter high.
Consider the ellipse
x2 y2 21 2 a b
l = 6 cm
that encloses the circle x2 y2 2x. Find values of a and b that minimize the area of the ellipse. 11. Projectile Motion A projectile is launched at an angle of 45 with the horizontal and with an initial velocity of 64 feet per second. A television camera is located in the plane of the path of the projectile 50 feet behind the launch site (see figure). y
h = 1 cm
(a) Draw a rectangular strip along the rectangular region showing a small increase in length. (b) Draw a rectangular strip along the rectangular region showing a small increase in height. (c) Use the results in parts (a) and (b) to identify the measurement that has more effect on the area A of the rectangle. (d) Verify your answer in part (c) analytically by comparing the value of dA when dl 0.01 and when dh 0.01.
(x, y)
(− 50, 0)
α
45°
x
16. Tangent Planes Let f be a differentiable function of one variable. Show that all tangent planes to the surface z y f 共x兾y兲 intersect in a common point. 17. Wave Equation
Show that
(a) Find parametric equations for the path of the projectile in terms of the parameter t representing time.
1 u共x, t兲 关sin共x t兲 sin共x t兲兴 2
(b) Write the angle that the camera makes with the horizontal in terms of x and y and in terms of t.
is a solution to the one-dimensional wave equation
(c) Use the results of part (b) to find
d . dt
(d) Use a graphing utility to graph in terms of t. Is the graph symmetric to the axis of the parabolic arch of the projectile? At what time is the rate of change of greatest? (e) At what time is the angle maximum? Does this occur when the projectile is at its greatest height? 12. Distance Consider the distance d between the launch site and the projectile in Exercise 11. (a) Write the distance d in terms of x and y and in terms of the parameter t. (b) Use the results of part (a) to find the rate of change of d. (c) Find the rate of change of the distance when t 2. (d) When is the rate of change of d minimum during the flight of the projectile? Does this occur at the time when the projectile reaches its maximum height?
2u 2u 2. t 2 x 18. Wave Equation
1 u共x, t兲 关 f 共x ct兲 f 共x ct兲兴 2 is a solution to the one-dimensional wave equation 2u 2u c2 2. 2 t x (This equation describes the small transverse vibration of an elastic string such as those on certain musical instruments.) 19. Verifying Equations Consider the function w f 共x, y兲, where x r cos and y r sin . Verify each of the following. (a)
2y 2
兲,
ⱍⱍ
0 < < .
(a) Use a computer algebra system to graph the function for
1 and 2, and identify any extrema or saddle points. (b) Use a computer algebra system to graph the function for
1 and 2, and identify any extrema or saddle points. (c) Generalize the results in parts (a) and (b) for the function f. 14. Proof Prove that if f is a differentiable function such that f 共x0, y0兲 0, then the tangent plane at 共x0, y0兲 is horizontal.
w w w sin cos x r r w w w cos sin y r r
13. Finding Extrema and Saddle Points Using Technology Consider the function f 共x, y兲 共 x 2 y 2兲e共x
Show that
w w 1 w 冢 冣 冢 冣 冢 冣冢 冣 冢w x 冣 y r r 2
(b)
2
2
2
2
20. Using a Function Demonstrate the result of Exercise 19(b) for y w arctan . x 21. Laplace’s Equation Rewrite Laplace’s equation 2u 2u 2u 20 x 2 y 2 z in cylindrical coordinates.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8
Multiple Integration Iterated Integrals and Area in the Plane Double Integrals and Volume Change of Variables: Polar Coordinates Center of Mass and Moments of Inertia Surface Area Triple Integrals and Applications Triple Integrals in Other Coordinates Change of Variables: Jacobians
Modeling Data (Exercise 34, p. 1008)
Center of Pressure on a Sail (Section Project, p. 1001)
Glacier (Exercise 60, p. 993)
Population (Exercise 57, p. 992) Average Production (Exercise 57, p. 984)
965 Clockwise from top left, AlexKZ/Shutterstock.com; Martynova Anna/Shutterstock.com; ValeStock/Shutterstock.com; Nataliya Hora/Shutterstock.com; Volodymyr Goinyk/Shutterstock.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
966
Chapter 14
Multiple Integration
14.1 Iterated Integrals and Area in the Plane Evaluate an iterated integral. Use an iterated integral to find the area of a plane region.
Iterated Integrals In Chapters 14 and 15, you will study several applications of integration involving functions of several variables. Chapter 14 is like Chapter 7 in that it surveys the use of integration to find plane areas, volumes, surface areas, moments, and centers of mass. In Chapter 13, you saw that it is meaningful to differentiate functions of several variables with respect to one variable while holding the other variables constant. You can integrate functions of several variables by a similar procedure. For example, consider the partial derivative fx共x, y兲 2xy. By considering y constant, you can integrate with respect to x to obtain f 共x, y兲
冕 冕 冕
y
fx共x, y兲 dx
Integrate with respect to x.
2xy dx
Hold y constant.
2x dx
Factor out constant y.
y共x 2兲 C共 y兲 x 2 y C共 y兲.
Antiderivative of 2x is x 2. C共 y兲 is a function of y.
The “constant” of integration, C共 y兲, is a function of y. In other words, by integrating with respect to x, you are able to recover f 共x, y兲 only partially. The total recovery of a function of x and y from its partial derivatives is a topic you will study in Chapter 15. For now, you will focus on extending definite integrals to functions of several variables. For instance, by considering y constant, you can apply the Fundamental Theorem of Calculus to evaluate
冕
2y
2y
冥
2xy dx x 2y
1
x is the variable of integration and y is fixed.
1
共2y兲2 y 共1兲2y 4y 3 y.
Replace x by the limits of integration.
The result is a function of y.
Similarly, you can integrate with respect to y by holding x fixed. Both procedures are summarized as follows.
冕 冕
h 2共 y兲
h1共 y兲
冥
g 共x兲 2
g1共x兲
h2共 y兲
fx共x, y兲 dx f 共x, y兲
h1共 y兲
f 共h2共 y兲, y兲 f 共h1共 y兲, y兲
With respect to x
g 共x兲
冥
fy共x, y兲 dy f 共x, y兲
2
g1共x兲
f 共x, g2共x兲兲 f 共x, g1共x兲兲
With respect to y
Note that the variable of integration cannot appear in either limit of integration. For instance, it makes no sense to write
冕
x
y dx.
0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.1
967
Iterated Integrals and Area in the Plane
Integrating with Respect to y
冕
x
Evaluate
共2x 2y2 2y兲 dy.
1
Solution
冕
Considering x to be constant and integrating with respect to y produces
x
共2x 2y2 2y兲 dy
1
冤
2x 2 y2 y
冥
x
Integrate with respect to y. 1
2x 2 2x 2 x2 1 x 1 3x 2 2x 1.
冢
冣 冢
冣
Notice in Example 1 that the integral defines a function of x and can itself be integrated, as shown in the next example.
The Integral of an Integral
冕 冤冕 2
Evaluate
1
冕 冤冕
共2x 2y2 2y兲 dy dx.
1
Solution 2
Using the result of Example 1, you have
冥 冕
x
1
冥
x
2
共2x 2y2 2y兲 dy dx
1
共3x 2 2x 1兲 dx
1
冤
冥
x3 x 2 x
2
Integrate with respect to x. 1
2 共1兲 3. The integral in Example 2 is an iterated integral. The brackets used in Example 2 are normally not written. Instead, iterated integrals are usually written simply as
冕冕 b
a
g2共x兲
g1(x兲
冕冕 d
f 共x, y兲 dy dx
and
c
h2共 y兲
h1共 y兲
f 共x, y兲 dx dy.
The inside limits of integration can be variable with respect to the outer variable of integration. However, the outside limits of integration must be constant with respect to both variables of integration. After performing the inside integration, you obtain a “standard” definite integral, and the second integration produces a real number. The limits of integration for an iterated integral identify two sets of boundary intervals for the variables. For instance, in Example 2, the outside limits indicate that x lies in the interval 1 x 2 and the inside limits indicate that y lies in the interval 1 y x. Together, these two intervals determine the region of integration R of the iterated integral, as shown in Figure 14.1. Because an iterated integral is just a special type of definite integral—one in which the integrand is also an integral—you can use the properties of definite integrals to evaluate iterated integrals.
y
y=x R: 1 ≤ x ≤ 2 1≤y≤x
2
1
x 1
2
The region of integration for
冕冕 2
1
x
f 冇x, y冈 dy dx
1
Figure 14.1
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
968
Chapter 14
Multiple Integration
Area of a Plane Region y
In the remainder of this section, you will take a new look at an old problem—that of finding the area of a plane region. Consider the plane region R bounded by a x b and g1共x兲 y g2共x兲, as shown in Figure 14.2. The area of R is
Region is bounded by a ≤ x ≤ b and g1(x) ≤ y ≤ g2(x)
冕
g2
b
关 g2共x兲 g1共x兲兴 dx.
Area of R
a
Using the Fundamental Theorem of Calculus, you can rewrite the integrand g2共x兲 g1共x兲 as a definite integral. Specifically, consider x to be fixed and let y vary from g1共x兲 to g2共x兲, and you can write
R g1
Δx
冕
g2共x兲
x
a
b Area =
b
g2(x)
a
g1(x)
dy dx
Vertically simple region Figure 14.2
g1共x兲
g2共x兲
冥
dy y
g1共x兲
g2共x兲 g1共x兲.
Combining these two integrals, you can write the area of the region R as an iterated integral
冕冕 b
a
g2共x兲
g1共x兲
冕
b
dy dx
a
g2共x兲
冥
y
g1共x兲
冕
b
dx
关g2共x兲 g1共x兲兴 dx.
Placing a representative rectangle in the region R helps determine both the order and the limits of integration. A vertical rectangle implies the order dy dx, with the inside limits corresponding to the upper and lower bounds of the rectangle, as shown in Figure 14.2. This type of region is vertically simple, because the outside limits of integration represent the vertical lines xa
Area of R
a
Region is bounded by c ≤ y ≤ d and h1(y) ≤ x ≤ h2(y) y
d R Δy c
and x b.
h1
h2
x
Similarly, a horizontal rectangle implies the order d h (y) 2 Area = dx dy dx dy, with the inside limits determined by the left c h (y) 1 and right bounds of the rectangle, as shown in Horizontally simple region Figure 14.3. This type of region is horizontally Figure 14.3 simple, because the outside limits represent the horizontal lines yc and y d. The iterated integrals used for these two types of simple regions are summarized as follows.
REMARK Be sure you see that the orders of integration of these two integrals are different––the order dy dx corresponds to a vertically simple region, and the order dx dy corresponds to a horizontally simple region.
Area of a Region in the Plane 1. If R is defined by a x b and g1共x兲 y g2共x兲, where g1 and g2 are continuous on 关a, b兴, then the area of R is
冕冕 b
A
a
g2共x兲
dy dx.
Figure 14.2 (vertically simple)
g1共x兲
2. If R is defined by c y d and h1共 y兲 x h2共 y兲, where h1 and h2 are continuous on 关c, d兴, then the area of R is
冕冕 d
A
c
h2共y兲
dx dy.
Figure 14.3 (horizontally simple)
h1共y兲
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.1
Iterated Integrals and Area in the Plane
969
If all four limits of integration happen to be constants, then the region of integration is rectangular, as shown in Example 3.
The Area of a Rectangular Region Use an iterated integral to represent the area of the rectangle shown in Figure 14.4. y
Solution The region shown in Figure 14.4 is both vertically simple and horizontally simple, so you can use either order of integration. By choosing the order dy dx, you obtain the following.
Rectangular region d
冕冕 b
d−c
R
a
冕 冕
d
b
dy dx
c
d
冥 dx
y
Integrate with respect to y.
c
a b
c
共d c兲 dx
a
b
a
b
冤
冥
共d c兲x
x
Integrate with respect to x. a
共d c兲共b a兲
b−a
Figure 14.4
Notice that this answer is consistent with what you know from geometry.
Finding Area by an Iterated Integral Use an iterated integral to find the area of the region bounded by the graphs of f 共x兲 sin x
Sine curve forms upper boundary.
g共x兲 cos x
Cosine curve forms lower boundary.
and between x 兾4 and x 5兾4. R: y
5π π ≤x≤ 4 4 cos x ≤ y ≤ sin x
Solution Because f and g are given as functions of x, a vertical representative rectangle is convenient, and you can choose dy dx as the order of integration, as shown in Figure 14.5. The outside limits of integration are
y = cos x x
−1
π 4
π 2
3π 2
π
Δx
5 . x 4 4 Moreover, because the rectangle is bounded above by f 共x兲 sin x and below by g共x兲 cos x, you have
y = sin x Area =
Figure 14.5
5π /4 sin x π /4
dy dx
Area of R
cos x
冕 冕 冕 冥 冕 5兾4
兾4
sin x
dy dx
cos x
5兾4
sin x
y
兾4
5兾4
兾4
冤
dx
Integrate with respect to y.
cos x
共sin x cos x兲 dx 5兾4
冥
cos x sin x
兾4
Integrate with respect to x.
2冪2. The region of integration of an iterated integral need not have any straight lines as boundaries. For instance, the region of integration shown in Figure 14.5 is vertically simple even though it has no vertical lines as left and right boundaries. The quality that makes the region vertically simple is that it is bounded above and below by the graphs of functions of x.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
970
Chapter 14
Multiple Integration
One order of integration will often produce a simpler integration problem than the other order. For instance, try reworking Example 4 with the order dx dy—you may be surprised to see that the task is formidable. However, if you succeed, you will see that the answer is the same. In other words, the order of integration affects the ease of integration, but not the value of the integral.
Comparing Different Orders of Integration See LarsonCalculus.com for an interactive version of this type of example.
Sketch the region whose area is represented by the integral
冕冕 2
0
4
dx dy.
y2
Then find another iterated integral using the order dy dx to represent the same area and show that both integrals yield the same value. y 3
R: 0 ≤ y ≤ 2 y2 ≤ x ≤ 4
2
x = y2
Solution y2
x
3
2
4
Area =
0 y2
2
dx dy
Outer limits of integration
you know that R is bounded below by the x-axis, as shown in Figure 14.6(a). The value of this integral is
冕冕
2 4
−1
Inner limits of integration
0 y 2
Δy
1
x 4
which means that the region R is bounded on the left by the parabola x y 2 and on the right by the line x 4. Furthermore, because
(4, 2)
1
From the given limits of integration, you know that
0
4
dx dy
y2
冥
4
x
dy
Integrate with respect to x.
y2
2
共4 y 2兲 dy
0
y
R: 0 ≤ x ≤ 4 0≤y≤ x
3
y=
冤
(4, 2)
x
1
2 Δx 3 4
Area =
dy dx 0 0
(b)
Figure 14.6
x
y3 3
4y
1
−1
2
0
(a)
2
冕 冕
x
4
冥
2
Integrate with respect to y.
0
16 . 3
To change the order of integration to dy dx, place a vertical rectangle in the region, as shown in Figure 14.6(b). From this, you can see that the constant bounds 0 x 4 serve as the outer limits of integration. By solving for y in the equation x y 2, you can conclude that the inner bounds are 0 y 冪x. So, the area of the region can also be represented by
冕冕 4
0
冪x
dy dx.
0
By evaluating this integral, you can see that it has the same value as the original integral.
冕冕 4
0
冪x
冕 冕
4
dy dx
0
0
冥
冪x
y
dx
Integrate with respect to y.
0
4
冪x dx
0
冥
2 3兾2 x 3
16 3
4
Integrate with respect to x. 0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.1
Iterated Integrals and Area in the Plane
971
Sometimes it is not possible to calculate the area of a region with a single iterated integral. In these cases, you can divide the region into subregions such that the area of each subregion can be calculated by an iterated integral. The total area is then the sum of the iterated integrals.
TECHNOLOGY Some computer software can perform symbolic integration for integrals such as those in Example 6. If you have access to such software, use it to evaluate the integrals in the exercises and examples given in this section.
An Area Represented by Two Iterated Integrals Find the area of the region R that lies below the parabola y 4x x 2
Parabola forms upper boundary.
above the x-axis, and above the line y 3x 6. Solution
Line and x-axis form lower boundary.
Begin by dividing R into the two subregions R1 and R2 shown in Figure 14.7. y
y = − 3x + 6
4
3
y = 4x − x 2 (1, 3) R1
REMARK In Examples 3
2
through 6, be sure you see the benefit of sketching the region of integration. You should develop the habit of making sketches to help you determine the limits of integration for all iterated integrals in this chapter.
R2
Δx
1
1 4x − x 2
2
Area =
−3x + 6
1
x
Δx
2
4
4
4x − x 2
2
0
dy dx +
dy dx
Figure 14.7
In both regions, it is convenient to use vertical rectangles, and you have
冕冕 冕 2
Area
1
4xx 2
3x6
冕冕 4
dy dx
2
4xx 2
dy dx
0
2
冕
4
共4x x 2 3x 6兲 dx
1
共4x x 2兲 dx
2
2 7x 2 x 3 x3 4 2x 2 6x 2 3 3 2 1 7 1 8 8 64 8 14 12 6 32 3 2 3 3 3
冤 冢
冥
冤
冥 冣 冢
冣
15 . 2
The area of the region is 15兾2 square units. Try checking this using the procedure for finding the area between two curves, as presented in Section 7.1. At this point, you may be wondering why you would need iterated integrals. After all, you already know how to use conventional integration to find the area of a region in the plane. (For instance, compare the solution of Example 4 in this section with that given in Example 3 in Section 7.1.) The need for iterated integrals will become clear in the next section. In this section, primary attention is given to procedures for finding the limits of integration of the region of an iterated integral, and the following exercise set is designed to develop skill in this important procedure.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
972
Chapter 14
Multiple Integration
14.1 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Evaluating an Integral In Exercises 1–10, evaluate the integral.
冕 冕 冕 冕 冕
共x 2y兲 dy
2.
0 2y
3.
1
x
y dx, x
4. 6.
y
8.
冪1y2
yey兾x dy
10.
1
2
共x 2 y 2兲 dx
6
4
x
y
37.
x2
11.
冕冕 冕冕 冕 冕 冕冕 冕冕 冕冕冢 冕冕 冕冕 冕冕 冕 冕 冕 冕
4 3
2
4
1
0
0
0
1
19.
2
1
0
2
0
3y
39. 冪x 冪y 2,
e xy dy dx
40. y x3兾2,
0
2yex dy dx
1
42.
x2
冪64
x3
dy dx
冪4y2
0
2 dx dy 26. 冪4 y 2
2 cos
r dr d
兾2
sin
0
3
1
y
0
x2
r dr d
yx2
y 2x,
x2
cos
y
47. 49.
0
50.
冪y
f 共x, y兲 dy dx
f 共x, y兲 dy dx
0
ex
f 共x, y兲 dy dx
0
兾2
52.
f 共x, y兲 dx dy
4x 2
1
1
1 x 2
0
2
2
f 共x, y兲 dx dy
0
1
51.
48.
ln y
1
3r 2 sin dr d
2
f 共x, y兲 dy dx
0
10
r dr d
46.
0
冪4x2
2
冕冕 冕冕 冕冕 冕 冕 4
f 共x, y兲 dx dy
0
2
4 dx dy y2
冪3
0
0
0
冪3 cos
兾4
30.
冕冕 冕冕 冕冕 冕冕 4
45.
3y dx dy
3y 26y
0
0
y0
1
43. y 4 x2,
2yy 2
兾4
28.
b2
x y 5,
sketch the region R of integration and switch the order of integration.
冕冕 冕冕 冕 冕 冕 冕 0
0
y0
Switching the Order of Integration In Exercises 45–52,
2
24.
a2
y2
44. y x,
冣
共x y兲 dx dy
x2
x 0,
y 2x
41. 2x 3y 0,
x
1 3 x2 y2 dx dy 4
冪1y2
5
ln 3
4 0
y
4
iterated integral to find the area of the region bounded by the graphs of the equations.
2y
0
29.
20.
dy dx
0
兾2
27.
x2
3
Finding the Area of a Region In Exercises 39– 44, use an
冪
4
x
2
1
4
共x 2 y 2兲 dy dx
共10 2x 2 2y 2兲 dx dy
0
25.
1
x
1 0
23.
共1 cos x兲 dy dx 18.
0
5
22.
4
冪1
0
21.
0
sin x
3
共x y2兲 dx dy
ln 4
16.
y cos x dy dx
1 1
3
1 1
0
兾2
17.
2
14.
1
2
1 2
共x 2 2y 2兲 dx dy
1
15.
12.
0
2≤x≤5
2 x
冕冕 冕冕 冕冕 冕冕 冕冕 1
共x y兲 dy dx
0
13.
2
1 x−1
y= 5
3
Evaluating an Iterated Integral In Exercises 11–30,
3
2
y
38. y=4−
sin3 x cos y dx
1
8
2
1
(2, 1)
x
y
evaluate the iterated integral.
(1, 1)
2
兾2
0
(2, 3)
(8, 3)
2
冪1y2
y ln x dx, y > 0 x y e x3
9.
6
共x 2 3y 2兲 dy
(1, 3) 3
4
x3
y
36.
8
y dx
冪x
x 2y dy
y
35.
0
0
7.
y dy x
cos y
y > 0
冪4x2
5.
冕 冕 冕 冕 冕
x2
x
1.
Finding the Area of a Region In Exercises 35–38, use an iterated integral to find the area of the region.
cos x
兾2
f 共x, y兲 dy dx
0
Evaluating an Improper Iterated Integral In Exercises
Switching the Order of Integration In Exercises 53–62,
31–34, evaluate the improper iterated integral.
sketch the region R whose area is given by the iterated integral. Then switch the order of integration and show that both orders yield the same area.
冕冕 冕冕
31.
1
1兾x
33.
1
y dy dx
32.
0
1
冕冕 冕冕 3
0
1 dx dy xy
0
x2 dy dx 1 y2
34.
0
2 2 xye共x y 兲 dx
冕冕 冕冕 1
dy
53.
0
0
1
55.
0
2
冕冕 冕冕 2
54.
dy dx
0
1
冪1y2
冪1y2
dx dy
56.
4
dx dy
2
2
冪4x2
2
冪4x2
dy dx
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.1
冕冕 冕冕 冕冕 冕冕 2
57.
0
x
0
4
58.
0
59.
0
1
61.
0
4x
2
x兾2
6x
dy dx
4
dy dx 71.
冕冕 冕冕
1
9
60.
x兾2
0
3 y 冪
62.
0
3
dy dx
1
冪x
72.
0
4y 2
2
dx dy
4
dx dy
73.
2 0
y2
0
63. Think About It Give a geometric argument for the equality. Verify the equality analytically.
冕冕 5
0
冪50x2
x
冕冕 0
x y dx dy
0
冕 冕
0
冪50y2
5
x2y2
dx dy
0
y
(0, 5
0
75. 5冪2
2 2
a
74.
2
x 2y 2 dy dx
y
5
76. 77. 78.
y
HOW DO YOU SEE IT? Complete the iterated integrals so that each one represents the area of the region R (see figure). (b) Area
2 dx dy 共x 1兲共 y 1兲
0
ax
共x 2 y 2兲 dy dx
0
4x 2 xy
e dy dx
0
2
冪16 x3 y3 dy dx
x
1cos
6r 2 cos dr d
0
1sin
15r dr d
0
Comparing Different Orders of Integration Using Technology In Exercises 79 and 80, (a) sketch the region of
x
dx dy
sin共x y兲 dx dy
y
0
5
冕冕
2y
兾2
(5, 5)
共x3 3y 2兲 dy dx
x2
0
y=x
(a) Area
2x
2
50 − x 2
y=
2)
2
0
5
64.
冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 冕 冕 2
0
dy dx
973
Evaluating an Iterated Integral Using Technology In Exercises 71–78, use a computer algebra system to evaluate the iterated integral.
dy dx
0
6
0
2
冕冕 冕冕 4
dy dx
Iterated Integrals and Area in the Plane
冕冕
dy dx
integration, (b) switch the order of integration, and (c) use a computer algebra system to show that both orders yield the same value.
冕冕 冕冕 2
79.
0
2
80.
0
4冪2y
y3
共x 2 y xy 2兲 dx dy
4x 2兾4
冪4x2
xy dy dx x2 y 2 1
y 2
y=
1
WRITING ABOUT CONCEPTS
(4, 2)
x
81. Iterated Integral Explain what is meant by an iterated integral. How is it evaluated?
x y= 2
R
x 1
2
3
4
Switching the Order of Integration In Exercises 65–70, sketch the region of integration. Then evaluate the iterated integral. (Hint: Note that it is necessary to switch the order of integration.)
冕冕 冕冕 冕冕 2
65.
0
1
67.
0
1
69.
0
2
66.
x
0
2
2
2
4ey dy dx
68.
2x
0
1
y
冕冕 冕冕 冕冕 4
x冪1 y3 dy dx
2
sin
x2
dx dy
70.
0
2
冪x
3 dy dx 2 y3
2
ey dy dx 2
x
83. Region of Integration Give a geometric description of the region of integration when the inside and outside limits of integration are constants. 84. Order of Integration Explain why it is sometimes an advantage to change the order of integration.
True or False? In Exercises 85 and 86, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.
冕冕 冕冕 b
85.
4
y2
82. Vertically Simple and Horizontally Simple Describe regions that are vertically simple and regions that are horizontally simple.
a
冪x sin x dx dy
1
86.
0
d
c
c
x
0
冕冕 冕冕 d
f 共x, y兲 dy dx
1
f 共x, y兲 dy dx
0
b
f 共x, y兲 dx dy
a y
f 共x, y兲 dx dy
0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
974
Chapter 14
Multiple Integration
14.2 Double Integrals and Volume Use a double integral to represent the volume of a solid region and use properties of double integrals. Evaluate a double integral as an iterated integral. Find the average value of a function over a region.
Double Integrals and Volume of a Solid Region Surface: z = f(x, y)
You already know that a definite integral over an interval uses a limit process to assign measures to quantities such as area, volume, arc length, and mass. In this section, you will use a similar process to define the double integral of a function of two variables over a region in the plane. Consider a continuous function f such that f 共x, y兲 0 for all 共x, y兲 in a region R in the xy-plane. The goal is to find the volume of the solid region lying between the surface given by
z
z f 共x, y兲 y
R
x
Figure 14.8
Surface lying above the xy-plane
and the xy-plane, as shown in Figure 14.8. You can begin by superimposing a rectangular grid over the region, as shown in Figure 14.9. The rectangles lying entirely within R form an inner partition , whose norm 储储 is defined as the length of the longest diagonal of the n rectangles. Next, choose a point 共xi, yi兲 in each rectangle and form the rectangular prism whose height is f 共xi, yi 兲
Height of ith prism
as shown in Figure 14.10. Because the area of the ith rectangle is Ai
Area of ith rectangle
it follows that the volume of the ith prism is f 共xi , yi 兲 Ai
Volume of ith prism
and you can approximate the volume of the solid region by the Riemann sum of the volumes of all n prisms, n
兺 f 共x , y 兲 A i
i
Riemann sum
i
i1
as shown in Figure 14.11. This approximation can be improved by tightening the mesh of the grid to form smaller and smaller rectangles, as shown in Example 1. Surface: z = f(x, y)
z
z
z
f (xi , yi ) (xi , yi) y
y x
R
The rectangles lying within R form an inner partition of R. Figure 14.9
y
x
x
Rectangular prism whose base has an area of Ai and whose height is f 共xi, yi兲 Figure 14.10
Volume approximated by rectangular prisms Figure 14.11
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14.2
Double Integrals and Volume
975
Approximating the Volume of a Solid Approximate the volume of the solid lying between the paraboloid 1 1 f 共x, y兲 1 x2 y 2 2 2 and the square region R given by 0 x 1, 0 y 1. Use a partition made up of 1 squares whose sides have a length of 4. Solution Begin by forming the specified partition of R. For this partition, it is convenient to choose the centers of the subregions as the points at which to evaluate f 共x, y兲.
z
共18, 18 兲 共38, 18 兲 共58, 18 兲 共78, 18 兲
1
共18, 38 兲 共38, 38 兲 共58, 38 兲 共78, 38 兲
共18, 58 兲 共38, 58 兲 共58, 58 兲 共78, 58 兲
共18, 78 兲 共38, 78 兲 共58, 78 兲 共78, 78 兲
1 Because the area of each square is Ai 16, you can approximate the volume by the sum 1
1 x
y
兺 f 共x , y 兲 A 兺 冢1 2 x 16
16
i
i
i
i1
Surface: f (x, y) = 1 − 1 x 2 − 1 y 2 2 2
Figure 14.12
1
i
2
i1
冣冢 冣
1 1 yi 2 ⬇ 0.672. 2 16
This approximation is shown graphically in Figure 14.12. The exact volume of the solid 2 is 3 (see Example 2). You can obtain a better approximation by using a finer partition. 1 For example, with a partition of squares with sides of length 10, the approximation is 0.668. z
TECHNOLOGY Some three-dimensional graphing utilities are capable of graphing figures such as that shown in Figure 14.12. For instance, the graph shown at the right was drawn with a computer program. In this graph, note that each of the rectangular prisms lies within the solid region.
y x
In Example 1, note that by using finer partitions, you obtain better approximations of the volume. This observation suggests that you could obtain the exact volume by taking a limit. That is, Volume lim
n
兺 f 共x , y 兲 A . i
储储→0 i1
i
i
The precise meaning of this limit is that the limit is equal to L if for every > 0, there exists a > 0 such that
ⱍ
L
n
兺 f 共x , y 兲 A i
i1
i
ⱍ
i
<
for all partitions of the plane region R (that satisfy 储储 < ) and for all possible choices of xi and yi in the ith region. Using the limit of a Riemann sum to define volume is a special case of using the limit to define a double integral. The general case, however, does not require that the function be positive or continuous.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
976
Chapter 14
Multiple Integration
Exploration The entries in the table represent the depths (in 10-yard units) of earth at the centers of the squares in the figure below. y x 1
1
2
3
10
9
7
2
7
7
4
3
5
5
4
4
4
5
3
Definition of Double Integral If f is defined on a closed, bounded region R in the xy-plane, then the double integral of f over R is
冕冕
n
兺 f 共x , y 兲 A
f 共x, y兲 d A lim
储储→0 i1
R
i
i
i
provided the limit exists. If the limit exists, then f is integrable over R.
z 20
30
40 x
Approximate the number of cubic yards of earth in the first octant. (This exploration was submitted by Robert Vojack, Ridgewood High School, Ridgewood, NJ.)
y
y Having defined a double integral, you will see that a definite integral is occasionally referred to R = R1 ∪ R2 as a single integral. Sufficient conditions for the double integral of f on the region R to exist are that R can be written as a union of a finite number of nonoverlapping R2 R1 subregions (see figure at the right) that are vertically or horizontally simple and that f is continuous on the region R. This means that the intersection of two nonoverlapping regions is a set that has an area of 0. In Figure 14.13, the area of the line segment common to R1 and R2 is 0. The two regions R1 and R2 are A double integral can be used to find the nonoverlapping. volume of a solid region that lies between the Figure 14.13 xy-plane and the surface given by z f 共x, y兲.
x
Volume of a Solid Region If f is integrable over a plane region R and f 共x, y兲 0 for all 共x, y兲 in R, then the volume of the solid region that lies above R and below the graph of f is V
冕冕
f 共x, y兲 dA.
R
Double integrals share many properties of single integrals. THEOREM 14.1 Properties of Double Integrals Let f and g be continuous over a closed, bounded plane region R, and let c be a constant. 1.
冕冕 冕冕 冕冕 冕冕 冕冕
冕冕
cf 共x, y兲 dA c
R
2.
关 f 共x, y兲 ± g共x, y兲兴 dA
R
3.
R
f 共x, y兲 d A ±
冕冕
g共x, y兲 dA
R
f 共x, y兲 dA 0, if f 共x, y兲 0 f 共x, y兲 dA
R
5.
冕冕 R
R
4.
f 共x, y兲 dA
R
冕冕 冕冕
g共x, y兲 dA,
if f 共x, y兲 g共x, y兲
R
f 共x, y兲 dA
R1
f 共x, y兲 dA
冕冕
f 共x, y兲 dA, where R is the union
R2
of two nonoverlapping subregions R1 and R2.
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14.2
Double Integrals and Volume
977
Evaluation of Double Integrals z
(0, 0, 2) 2
Plane: z = 2 − x − 2y
Height: z=2−x
1
(0, 1, 0) (2, 0, 0)
1 2
x
Triangular 2 cross section
Base: y = 2 − x 2
冕
y
1 1 2x 共2 x兲2 A共x兲 共base兲共height兲 . 共2 x兲 2 2 2 4
冢
冣
By the formula for the volume of a solid with known cross sections (Section 7.2), the volume of the solid is
2
Volume:
Normally, the first step in evaluating a double integral is to rewrite it as an iterated integral. To show how this is done, a geometric model of a double integral is used as the volume of a solid. Consider the solid region bounded by the plane z f 共x, y兲 2 x 2y and the three coordinate planes, as shown in Figure 14.14. Each vertical cross section taken parallel to the yz-plane is a triangular region whose base has a length of y 共2 x兲兾2 and whose height is z 2 x. This implies that for a fixed value of x, the area of the triangular cross section is
A共x兲 dx
0
冕 冕
b
Figure 14.14
Volume
A共x兲 dx
a 2
0
共2 x兲2 dx 4
共2 x兲3 12
2
冥
0
2 . 3 This procedure works no matter how A共x兲 is obtained. In particular, you can find A共x兲 by integration, as shown in Figure 14.15. That is, you consider x to be constant, and integrate z 2 x 2y from 0 to 共2 x兲兾2 to obtain z = 2 − x − 2y
A共x兲
冕
共2x兲兾2
共2 x 2y兲 dy
0
冤
y=
y=0
2−x 2
共2x兲兾2
冥
共2 x兲y y2
0
共2 x兲2 . 4
Triangular cross section Figure 14.15
Combining these results, you have the iterated integral Volume
冕冕
冕冕 2
f 共x, y兲 dA
0
R
共2x兲兾2
共2 x 2y兲 dy dx.
0
To understand this procedure better, it helps to imagine the integration as two sweeping motions. For the inner integration, a vertical line sweeps out the area of a cross section. For the outer integration, the triangular cross section sweeps out the volume, as shown in Figure 14.16. z
z
y x
z
y x
Integrate with respect to y to obtain the area of the cross section. Figure 14.16
x
z
y
x
Integrate with respect to x to obtain the volume of the solid.
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y
978
Chapter 14
Multiple Integration
The next theorem was proved by the Italian mathematician Guido Fubini (1879–1943). The theorem states that if R is a vertically or horizontally simple region and f is continuous on R, then the double integral of f on R is equal to an iterated integral. THEOREM 14.2 Fubini’s Theorem Let f be continuous on a plane region R. 1. If R is defined by a x b and g1共x兲 y g2共x兲, where g1 and g2 are continuous on 关a, b兴, then
冕冕
冕冕
g 共x兲
b
f 共x, y兲 dA
a
R
2
g1共x兲
f 共x, y兲 dy dx.
2. If R is defined by c y d and h 1共 y兲 x h 2共 y兲, where h 1 and h 2 are continuous on 关c, d兴, then
冕冕
冕冕 d
f 共x, y兲 dA
c
R
h2共 y兲
h1共 y兲
f 共x, y兲 dx dy.
Evaluating a Double Integral as an Iterated Integral y
Evaluate
冕 冕冢 R
R: 0 ≤ x ≤ 1 0≤y≤1
冣
1 1 1 x2 y 2 dA 2 2 1
where R is the region given by 0 x 1,
0 y 1.
Solution Because the region R is a square, it is both vertically and horizontally simple, and you can use either order of integration. Choose dy dx by placing a vertical representative rectangle in the region (see the figure at the right). This produces the following.
Δx
1 1
f (x, y) dA = R
冕 冕冢 R
冣
1 1 1 x2 y 2 dA 2 2
冕冕 冢 冕 冤冢 冕冢 1
0
0
1
0
1
1
0
f (x, y) dy dx 0 0
冣
1 1 1 x2 y 2 dy dx 2 2
冣
冥
1 y3 1 x2 y 2 6
1
dx
0
冣
5 1 2 x dx 6 2
冤 6 x x6 冥
2 3
5
x 1
3 1 0
The double integral evaluated in Example 2 represents the volume of the solid region approximated in Example 1. Note that the approximation obtained in Example 1 is quite good 共0.672 vs. 23 兲, even though you used a partition consisting of only 16 squares. The error resulted because the centers of the square subregions were used as the points in the approximation. This is comparable to the Midpoint Rule approximation of a single integral.
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14.2
979
Double Integrals and Volume
The difficulty of evaluating a single integral 兰a f 共x兲 dx usually depends on the function f, not on the interval 关a, b兴. This is a major difference between single and double integrals. In the next example, you will integrate a function similar to the one in Examples 1 and 2. Notice that a change in the region R produces a much more difficult integration problem. b
Exploration Volume of a Paraboloid Sector The solid in Example 3 has an elliptical (not a circular) base. Consider the region bounded by the circular paraboloid z a2 x2 y 2,
a > 0
and the xy-plane. How many ways of finding the volume of this solid do you now know? For instance, you could use the disk method to find the volume as a solid of revolution. Does each method involve integration?
Finding Volume by a Double Integral Find the volume of the solid region bounded by the paraboloid z 4 x2 2y 2 and the xy-plane, as shown in Figure 14.17(a). Solution By letting z 0, you can see that the base of the region in the xy-plane is the ellipse x2 2y 2 4, as shown in Figure 14.17(b). This plane region is both vertically and horizontally simple, so the order dy dx is appropriate. Variable bounds for y:
冪共4 2 x 兲 y 冪共4 2 x 兲 2
Constant bounds for x: 2 x 2
z
The volume is
a2
冕冕 冕冤
冪共4x 2兲兾2
2
V
2 冪共4x 2兲兾2
共4 x 2 2y 2兲 dy dx
2
−a a
2
a
y
x
REMARK In Example 3,
2
共4 x 2 兲y
4 3冪2 4 3冪2
冕 冕
2y 3 3
冥
冪共4x 2兲兾2
冪共4x 2兲兾2
See Figure 14.17(b).
dx
2
共4 x2兲3兾2 dx
2
兾2
兾2
冕
16 cos 4 d
兾2
64 共2兲 3冪2
x 2 sin
cos4 d
0
128 3
3冪2 16 4冪2 .
note the usefulness of Wallis’s Formula to evaluate
兾2 n 兰0 cos d . You may want to review this formula in Section 8.3.
冢 冣 z
Wallis’s Formula
Surface: f (x, y) = 4 − x 2 − 2y 2
Base: −2 ≤ x ≤ 2 (4 − x 2)/2 ≤ y ≤
−
(4 − x 2)/2
y
4
2 1 x −1
Δx
1
−1 −2
2 3
y
Volume: 2
(4 − x 2)/2
x −2 −
(a)
(4 − x 2)/2
(4 − x 2 − 2y 2)dy dx
(b)
Figure 14.17
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980
Chapter 14
Multiple Integration
In Examples 2 and 3, the problems could be solved with either order of integration because the regions were both vertically and horizontally simple. Moreover, had you used the order dx dy, you would have obtained integrals of comparable difficulty. There are, however, some occasions in which one order of integration is much more convenient than the other. Example 4 shows such a case.
Comparing Different Orders of Integration See LarsonCalculus.com for an interactive version of this type of example. Surface: 2 f(x, y) = e − x
z
Find the volume of the solid region bounded by the surface f 共x, y兲 ex
1
Solution The base of the solid region in the xy-plane is bounded by the lines y 0, x 1, and y x. The two possible orders of integration are shown in Figure 14.19.
z=0 1
1
y
y
x=1
Surface
and the planes z 0, y 0, y x, and x 1, as shown in Figure 14.18.
y=0
x
2
y
y=x
R: 0 ≤ x ≤ 1 0≤y≤x
Base is bounded by y 0, y x, and x 1. Figure 14.18
R: 0 ≤ y ≤ 1 y≤x≤1 (1, 1)
1
(1, 1)
1
Δy (1, 0) Δx
1 x
(1, 0)
x
x
1
1
1 1
e− x dy dx 2
0 0
0
e− x dx dy 2
y
Figure 14.19
By setting up the corresponding iterated integrals, you can see that the order dx dy requires the antiderivative
冕
ex dx 2
which is not an elementary function. On the other hand, the order dy dx produces
冕冕 1
0
x
冕 冕
1
ex dy dx 2
0
x 2
e
x
冥
y dx 0
0 1
xex dx 2
0
1 1 2 ex 2 0 1 1 1 2 e e1 2e ⬇ 0.316.
冥
冢
冣
TECHNOLOGY Try using a symbolic integration utility to evaluate the integral in Example 4.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.2
Double Integrals and Volume
981
Volume of a Region Bounded by Two Surfaces Find the volume of the solid region bounded above by the paraboloid z 1 x2 y2
Paraboloid
and below by the plane z1y
Plane
as shown in Figure 14.20. Paraboloid: z = 1 − x2 − y2
Plane: z=1−y
z
1
y
1
1 x
Figure 14.20 y
Solution Equating z-values, you can determine that the intersection of the two surfaces occurs on the right circular cylinder given by 1 y 1 x2 y2
x2 y y2.
So, the region R in the xy-plane is a circle, as shown in Figure 14.21. Because the volume of the solid region is the difference between the volume under the paraboloid and the volume under the plane, you have
1 2
Volume 共volume under paraboloid兲 共volume under plane兲
冕冕 冕冕 冕冤 冕
冪yy 2
1
x −1
1 2
2
y − y2 ≤ x ≤
Figure 14.21
冪yy 2
0
冪yy 2
1
R: 0 ≤ y ≤ 1 −
y − y2
冪yy 2
0
共1
共 y y 2兲x
4 3
1 6
共1 y兲 dx dy
x3 3
冪yy 2
冥
dy
冪yy 2
共 y y2兲3兾2 dy
0
冢 冣冢 冣冕 1 6
冪yy 2
1
4 3
兲 dx dy
冪yy 2
共 y y 2 x 2兲 dx dy
0
y2
0
1
冕冕 1
x2
冕 冕
1 8
1
关1 共2y 1兲2兴 3兾2 dy
0
兾2
cos 4 d 2 兾2
兾2
2y 1 sin
cos 4 d
0
冢16冣冢316 冣
. 32
Wallis’s Formula
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
982
Chapter 14
Multiple Integration
Average Value of a Function Recall from Section 4.4 that for a function f in one variable, the average value of f on the interval 关a, b兴 is 1 ba
冕
b
f 共x兲 dx.
a
Given a function f in two variables, you can find the average value of f over the plane region R as shown in the following definition. Definition of the Average Value of a Function Over a Region If f is integrable over the plane region R, then the average value of f over R is 1 A
Average value
冕冕
f 共x, y兲 dA
R
where A is the area of R.
Finding the Average Value of a Function Find the average value of 1 f 共x, y兲 xy 2 over the plane region R, where R is a rectangle with vertices
共0, 0兲, 共4, 0兲, 共4, 3兲, and 共0, 3兲. Solution
The area of the rectangular region R is
A 共4兲共3兲 12 z
as shown in Figure 14.22. The bounds for x are 0 x 4
6
and the bounds for y are 5
0 y 3. So, the average value is 1 Average value A
冕冕 冕冕 冕
1 12 1 12
4
f (x, y) = 12 xy
f 共x, y兲 dA
3
R
4
0
4
0
3
0
1 xy dy dx 2
冥
1
3
1 2 xy 4
dx
4
9 4
0
冤 冥
3 1 2 x 16 2
冢163 冣共8兲
4
y
1
0
冢 冣冢 冣冕 x dx 1 12
2
(0, 3)
(0, 0) 1
R
2 3 4
(4, 0)
(4, 3)
x
0
Figure 14.22
3 . 2
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14.2
14.2 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Approximation In Exercises 1–4, approximate the integral 兰R兰 f 冇x, y冈 d A by dividing the rectangle R with vertices 冇0, 0冈, 冇4, 0冈, 冇4, 2冈, and 冇0, 2冈 into eight equal squares and finding
冕冕
共x2 y2兲 d A
18.
R
R: semicircle bounded by y 冪4 x2, y 0
8
兺 f 冇x , y 冈 A , where 冇x , y 冈 is the center of the ith
the sum
i
i
i
i
i
iⴝ1
square. Evaluate the iterated integral and compare it with the approximation.
冕冕 冕冕 4
1.
0
2
4
3.
0
冕冕 冕冕 4
1 2. 2
共x y兲 dy dx
0 2
4.
0
0
Finding Volume In Exercises 19–26, use a double integral to find the volume of the indicated solid. 19.
2
0
4
共x 2 y 2兲 dy dx
x2y
5.
冕冕 冕冕 冕冕 冕冕
2
0
0
0
0
3
4
共x y兲 dx dy
0
冪a2 x2
a 冪a2 x2 1
10.
0
8.
y兾2
a
9.
6.
0
6
7.
共1 2x 2y兲 dy dx
冕冕 冕冕
1
z
1 dy dx 共x 1兲共 y 1兲
1 2
y
2
4 x
0≤x≤4 0≤y≤2
0
z
21.
冪y
x2 y2 dx dy
z
z=4−x−y
3
3
0
y1
2
冕冕 1
e x y dx dy
0
1
1y
e x y dx dy
4
1
0
integrals for both orders of integration. Use the more convenient order to evaluate the integral over the region R.
冕冕
6
y
2
y=x
x
x
y=2
23.
24. z
z
xy d A
z = 1 − xy
R
R: rectangle with vertices 共0, 0兲, 共0, 5兲, 共3, 5兲, 共3, 0兲
冕冕
1
3
sin x sin y d A 2
R
冕冕 R
y dA x2 y2
R: trapezoid bounded by y x, y 2x, x 1, x 2 14.
冕冕
xe y
y=x
x
y=1
冕冕
z=
2y dA
z
R
R: region bounded by y 4 x 2, y 4 x
冕冕 R
1
y dA 1 x2
R: region bounded by y 0, y 冪x, x 4 17.
2 x
25. Improper integral
R: triangle bounded by y 4 x, y 0, x 0
16.
1
y
1
1
1
2
y=x
y
y=2
dA
R
15.
z = 4 − y2
4
R: rectangle with vertices 共 , 0兲, 共 , 0兲, 共 , 兾2兲, 共 , 兾2兲 13.
0≤x≤4 0≤y≤2
22. 2x + 3y + 4z = 12
4
共x y兲 dy dx
y
2
4 x
sin2 x cos2 y dy dx
2
12.
z = 6 − 2y
6
3
Evaluating a Double Integral In Exercises 11–18, set up
11.
z
20.
y 2
1
兾2
1 2y
z=
dy dx
0
Evaluating a Double Integral In Exercises 5–10, sketch the region R and evaluate the iterated integral 兰R兰 f 冇x, y冈 dA. 2
983
Double Integrals and Volume
冕冕
x dA
26. Improper integral z
1 (x +
1)2(y
+
z = e − (x + y)/2
1
0≤x 2 there does not exist a real-valued function u such that for all x in the closed interval 1 0 x 1, u共x兲 1 兰x u共 y兲u共 y x兲 dy.
These problems were composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
986
Chapter 14
Multiple Integration
14.3 Change of Variables: Polar Coordinates Write and evaluate double integrals in polar coordinates.
Double Integrals in Polar Coordinates Some double integrals are much easier to evaluate in polar form than in rectangular form. This is especially true for regions such as circles, cardioids, and rose curves, and for integrands that involve x 2 y 2. In Section 10.4, you learned that the polar coordinates 共r, 兲 of a point are related to the rectangular coordinates 共x, y兲 of the point as follows. x r cos r 2 x2 y 2
y r sin y and tan x
and
Using Polar Coordinates to Describe a Region Use polar coordinates to describe each region shown in Figure 14.23. y
y
y 4
2
5 2
4
R
3 x
1
−4
R
−2
2
R
2
4 1
−2
x − 4 −3
x
1
2
−4
(a)
3
−1
4
−2
(b)
(c)
Figure 14.23
Solution a. The region R is a quarter circle of radius 2. It can be described in polar coordinates as R 再共r, 兲: 0 r 2,
0 兾2冎.
b. The region R consists of all points between concentric circles of radii 1 and 3. It can be described in polar coordinates as
π 2
θ2
Δr
R 再共r, 兲: 1 r 3, 0 2冎. θ1
R (ri, θi) r2
Δθ
c. The region R is a cardioid with a b 3. It can be described in polar coordinates as R 再共r, 兲: 0 r 3 3 sin , 0 2冎. The regions in Example 1 are special cases of polar sectors
r1 0
R 再共r, 兲: r1 r r2, Polar sector Figure 14.24
1 2冎
Polar sector
as shown in Figure 14.24.
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14.3 π 2
987
Change of Variables: Polar Coordinates
To define a double integral of a continuous function z f 共x, y兲 in polar coordinates, consider a region R bounded by the graphs of Δθ i (ri, θi)
r g1共兲 and
Ri
g2 Δri
β α
g1 0
Polar grid superimposed over region R Figure 14.25
r g2共兲
and the lines and . Instead of partitioning R into small rectangles, use a partition of small polar sectors. On R, superimpose a polar grid made of rays and circular arcs, as shown in Figure 14.25. The polar sectors Ri lying entirely within R form an inner polar partition , whose norm 储 储 is the length of the longest diagonal of the n polar sectors. Consider a specific polar sector Ri, as shown in Figure 14.26. It can be shown (see Exercise 70) that the area of Ri is Ai ri ri i
Area of R i
where ri r2 r1 and i 2 1. This implies that the volume of the solid of height f 共ri cos i, ri sin i 兲 above Ri is approximately f 共ri cos i, ri sin i 兲ri ri i and you have
冕冕
f 共x, y兲 dA ⬇
R
n
兺 f 共r cos , r sin 兲r r . i
i
i
i
i
i
i
i1
The sum on the right can be interpreted as a Riemann sum for f 共r cos , r sin 兲r. The region R corresponds to a horizontally simple region S in the r-plane, as shown in Figure 14.27. The polar sectors Ri correspond to rectangles Si, and the area Ai of Si is ri i. So, the right-hand side of the equation corresponds to the double integral
冕冕
f 共r cos , r sin 兲r dA.
S
From this, you can apply Theorem 14.2 to write
冕冕
f 共x, y兲 dA
R
冕冕 冕冕
f 共r cos , r sin 兲r dA
S
g2共兲
g1 共兲
f 共r cos , r sin 兲r dr d.
This suggests the theorem on the next page, the proof of which is discussed in Section 14.8. π 2
θ
r = g2(θ )
r = g1(θ )
θ2
β
θ1 Ri
Si r1
r2 (ri, θi)
α
(ri, θi) 0
The polar sector Ri is the set of all points 共r, 兲 such that r1 r r2 and 1 2. Figure 14.26
Horizontally simple region S Figure 14.27
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
r
988
Chapter 14
Multiple Integration
THEOREM 14.3 Change of Variables to Polar Form Let R be a plane region consisting of all points 共x, y兲 共r cos , r sin 兲 satisfying the conditions 0 g1共兲 r g2共兲, , where 0 共 兲 2. If g1 and g2 are continuous on 关, 兴 and f is continuous on R, then
冕冕
f 共x, y兲 dA
R
Exploration Volume of a Paraboloid Sector In the Exploration on page 979, you were asked to summarize the different ways you know of finding the volume of the solid bounded by the paraboloid
冕冕
g2共兲
g1 共兲
f 共r cos , r sin 兲r dr d.
If z f 共x, y兲 is nonnegative on R, then the integral in Theorem 14.3 can be interpreted as the volume of the solid region between the graph of f and the region R. When using the integral in Theorem 14.3, be certain not to omit the extra factor of r in the integrand. The region R is restricted to two basic types, r-simple regions and -simple regions, as shown in Figure 14.28. π 2
g2
θ =β
z a2 x2 y2, a > 0 Δθ
and the xy-plane. You now know another way. Use it to find the volume of the solid.
π 2
Fixed bounds for θ : α≤θ ≤β Variable bounds for r: 0 ≤ g1(θ ) ≤ r ≤ g2(θ )
Variable bounds for θ : 0 ≤ h1(r) ≤ θ ≤ h 2(r) h2
Fixed bounds for r: r1 ≤ r ≤ r2
g1
h1
θ =α
Δr 0
r = r1
r = r2
0
-Simple region
r -Simple region Figure 14.28
Evaluating a Double Polar Integral Let R be the annular region lying between the two circles x 2 y 2 1 and x 2 y 2 5. Evaluate the integral
冕冕
共x y兲 dA. 2
R
R: 1 ≤ r ≤ 5 0 ≤ θ ≤ 2π
Solution The polar boundaries are 1 r 冪5 and 0 2, as shown in Figure 14.29. Furthermore, x 2 共r cos 兲2 and y r sin . So, you have
π 2
冕冕
R
共x 2 y兲 dA
R
0 2
冕冕 冕冕 冕冢 冕冢 冕冢 2
0
2
0
3
2
0
r-Simple region Figure 14.29
2
冪5
冪5
2
r4 r3 cos2 sin 4 3 6 cos2
3
冪5
冣冥
1
d
冣
5冪5 1 sin d 3
3 3 cos 2
0
冢
共r 3 cos2 r 2 sin 兲 dr d
1
0
共r 2 cos2 r sin 兲r dr d
1
冣
5冪5 1 sin d 3
3 sin 2 5冪5 1
cos 2 3
2
冣冥
0
6.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.3
Change of Variables: Polar Coordinates
989
In Example 2, be sure to notice the extra factor of r in the integrand. This comes from the formula for the area of a polar sector. In differential notation, you can write dA r dr d which indicates that the area of a polar sector increases as you move away from the origin. 16 − x 2 − y 2
Surface: z =
Change of Variables to Polar Coordinates
z
Use polar coordinates to find the volume of the solid region bounded above by the hemisphere
4
z 冪16 x 2 y 2
Hemisphere forms upper surface.
and below by the circular region R given by x2 y 2 4 4 4
x
y
R: x 2 + y 2 ≤ 4
Circular region forms lower surface.
as shown in Figure 14.30. Solution
In Figure 14.30, you can see that R has the bounds
冪4 y 2 x 冪4 y 2,
Figure 14.30
2 y 2
and that 0 z 冪16 x 2 y 2. In polar coordinates, the bounds are and 0 2
0 r 2
with height z 冪16 x 2 y 2 冪16 r 2. Consequently, the volume V is
冕冕 冕冕 冕 冕
f 共x, y兲 dA
V
R
REMARK To see the benefit of polar coordinates in Example 3, you should try to evaluate the corresponding rectangular iterated integral
冕冕 2
冪4 y2
2
冪4 y2
2
2
冪16 x 2 y 2 dx dy.
冪16 r 2 r dr d
0
0
2
1 3
2
冥
共16 r 2兲3兾2
0 2
1 3
0
共24冪3 64兲 d
8 共3冪3 8兲 3 16 共8 3冪3 兲 3 ⬇ 46.979.
d 0
2
冥
0
TECHNOLOGY Any computer algebra system that can evaluate double integrals in rectangular coordinates can also evaluate double integrals in polar coordinates. The reason this is true is that once you have formed the iterated integral, its value is not changed by using different variables. In other words, if you use a computer algebra system to evaluate
冕冕 2
2
0
0
冪16 x2 x dx dy
you should obtain the same value as that obtained in Example 3. Just as with rectangular coordinates, the double integral
冕冕
dA
R
can be used to find the area of a region in the plane.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
990
Chapter 14
Multiple Integration
Finding Areas of Polar Regions See LarsonCalculus.com for an interactive version of this type of example. π 2
R: −
r = 3 cos 3θ
π π ≤θ≤ 6 6 0 ≤ r ≤ 3 cos 3θ π θ= 6
To use a double integral to find the area enclosed by the graph of r 3 cos 3, let R be one petal of the curve shown in Figure 14.31. This region is r-simple, and the boundaries are 兾6 兾6 and 0 r 3 cos 3. So, the area of one petal is 1 A 3
0
dA
R
3
θ=−
冕冕 冕 冕 冕 冥 冕 冕
π 6
Figure 14.31
兾6
3 cos 3
兾6 0
兾6
3 cos 3
r2
兾6 2 兾6
r dr d
9 2
9 4
9 1 sin 6 4 6
3 . 4
兾6
兾6
兾6
d
0
cos2 3 d
共1 cos 6兲 d
冤
兾6
冥
兾6
So, the total area is A 9兾4. As illustrated in Example 4, the area of a region in the plane can be represented by
冕冕
g2共兲
A
g1 共兲
r dr d.
For g1共兲 0, you obtain
冕冕
g2共兲
A
r dr d
0
冕
r2 2
g2共兲
冥
d
0
冕
1 共g 共兲兲2 d 2 2
which agrees with Theorem 10.13. So far in this section, all of the examples of iterated integrals in polar form have been of the form
冕冕
g2共兲
π 2
π 3θ
π 6
Changing the Order of Integration Find the area of the region bounded above by the spiral r 兾共3兲 and below by the polar axis, between r 1 and r 2. Solution
0 1
π 3r 1≤r≤2
R: 0 ≤ θ ≤
-Simple region Figure 14.32
f 共r cos , r sin 兲r dr d
in which the order of integration is with respect to r first. Sometimes you can obtain a simpler integration problem by switching the order of integration.
π θ= 3
θ=
r=
g1 共兲
2
The region is shown in Figure 14.32. The polar boundaries for the region are
1 r 2
and 0
. 3r
So, the area of the region can be evaluated as follows.
冕冕 2
A
1
兾共3r兲
0
冕
2
r d dr
1
r
兾共3r兲
冥
dr
0
冕
2
1
r dr 3 3
2
冥
1
3
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14.3
14.3 Exercises
1.
冕冕 冕冕 冕 冕 冕 冕 冕 冕 冕 冕 2
11.
6
兾4
12.
4
0
13.
3
0
3
2
R R
x −6
−2
2
x 2
1
3
y
3.
R
冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 冕冕
冪a2 y2
a
x −4
2
1
4
17.
R
0
x 1
−4
2
3
4
−1
冪a2 x2
a
18.
0
Describing a Region In Exercises 5–8, use polar coordinates
19.
y
冪4 x2
0
6
12
0
2
2
4
22.
x −8
−4
4
−2
2
4
0
4
y
8.
2
23.
−2
y
0
x −4
8
−4
7.
24.
0
4
10
25.
2
6 x −4
2
−2
4
2
26.
0
x −2 −2
2
4
6
8 10
−4
Evaluating a Double Integral In Exercises 9–16, evaluate the double integral 兰R兰 f 冇r, 冈 dA, and sketch the region R.
冕冕 冕冕
0
10.
0
冪9 x2
cos
冪8 y2
0
冪x 2 y2 dx dy
y
冪2x x2
xy dy dx
0
冪4y y2
x 2 dx dy
0
冪1 x2
cos共x2 y2兲 dy dx
0
冪4 x2
sin冪x2 y2 dy dx
0
Converting to Polar Coordinates In Exercises 27 and 28, combine the sum of the two iterated integrals into a single iterated integral by converting to polar coordinates. Evaluate the resulting iterated integral.
冕冕 冕 冕 2
r dr d
27.
r 2 dr d
28.
冕 冕 冕 冕 2冪2
x
冪8 x2
冪x 2 y 2 dy dx
0
0 sin
共x 2 y 2兲3兾2 dy dx
0
1
4
共x2 y2兲 dy dx
冪x x2
1
8
9.
冪x x2
3
21.
共x2 y2兲 dy dx
0
1
20.
y
6.
x dy dx
0
2
to describe the region shown.
y dx dy
0
2
5.
共sin 兲r dr d
0
Converting to Polar Coordinates In Exercises 17–26, evaluate the iterated integral by converting to polar coordinates.
2
2
1 cos
0
3 4
r dr d
0
兾2
y
4.
2
1sin
0
16.
re r dr d
0
兾2
15.
−4
4
3
0
−2
1
冪9 r 2 r dr d
2
兾2
14.
r 2 sin cos dr d
0
兾2
4
4
3r 2 sin dr d
0
0
y
2.
2
991
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Choosing a Coordinate System In Exercises 1–4, the region R for the integral 兰R兰 f 冇x, y冈 dA is shown. State whether you would use rectangular or polar coordinates to evaluate the integral. y
Change of Variables: Polar Coordinates
0
5冪2兾2
0
2
x
0
5
xy dy dx
5冪2兾2
冪x 2 y 2 dy dx
0
冪25 x2
xy dy dx
0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
992
Chapter 14
Multiple Integration
Converting to Polar Coordinates In Exercises 29–32, use polar coordinates to set up and evaluate the double integral 兰R兰 f 冇x, y冈 dA.
45.
π 2
r = 2 sin 3θ
29. f 共x, y兲 x y
0
0
R: x 2 y 2 4, x 0, y 0 30. f 共x, y兲 e 共x
π 2 r = 3 cos 2θ
46.
1
2
3
2 y 2 兲兾2
R: x 2 y 2 25, x 0 31. f 共x, y兲 arctan
y x
Area In Exercises 47–52, sketch a graph of the region bounded by the graphs of the equations. Then use a double integral to find the area of the region.
R: x 2 y 2 1, x 2 y 2 4, 0 y x 32. f 共x, y兲 9 x 2 y 2
47. Inside the circle r 2 cos and outside the circle r 1
R: x 2 y 2 9, x 0, y 0
Volume In Exercises 33–38, use a double integral in polar coordinates to find the volume of the solid bounded by the graphs of the equations. 34. z x 2 y 2 3, z 0, x 2 y 2 1
51. Inside the rose curve r 4 sin 3 and outside the circle r 2
35. z 冪x 2 y 2, z 0, x 2 y 2 25 37. Inside the hemisphere z 冪16 x 2 y 2 and inside the cylinder x 2 y 2 4x 0
39. Volume
y2
and
outside
the
cardioid
WRITING ABOUT CONCEPTS
and outside the
53. Polar Coordinates Describe the partition of the region R of integration in the xy-plane when polar coordinates are used to evaluate a double integral.
Find a such that the volume inside the hemisphere
54. Converting Coordinates Explain how to change from rectangular coordinates to polar coordinates in a double integral.
z 冪16 x 2 y 2 and outside the cylinder
55. Describing Regions In your own words, describe r-simple regions and -simple regions.
x2 y 2 a2 is one-half the volume of the hemisphere. 40. Volume Use a double integral in polar coordinates to find the volume of a sphere of radius a.
Area In Exercises 41–46, use a double integral to find the area of the shaded region. 41.
r2
52. Inside the circle r 2 2 cos
36. z ln共x 2 y 2兲, z 0, x 2 y 2 1, x 2 y 2 4
38. Inside the hemisphere z 冪16
cylinder x 2 y 2 1
49. Inside the circle r 3 cos and outside the cardioid r 1 cos 50. Inside the cardioid r 1 cos and outside the circle r 3 cos
33. z xy, x 2 y 2 1, first octant
x2
48. Inside the cardioid r 2 2 cos and outside the circle r1
π 2
42. r=2
r = 6 cos θ
π 2
56. Comparing Integrals Let R be the region bounded by the circle x2 y2 9. (a) Set up the integral
冕 冕 f 共x, y兲 dA. R
(b) Convert the integral in part (a) to polar coordinates. (c) Which integral would you choose to evaluate? Why?
r=4
57. Population The population density of a city is approximated by the model 0
0
1 2 3 4 5
43.
π 2
1
7
44.
r = 1 + cos θ
3
f 共x, y兲 4000e 0.01共x
2
y 2兲
for the region x 2 y 2 49, where x and y are measured in miles. Integrate the density function over the indicated circular region to approximate the population of the city.
π 2
0 0
1 2 3 4
r = 2 + sin θ ValeStock/Shutterstock.com
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.3
the statement is true or false. If it is false, explain why or give an example that shows it is false.
a region of integration for the double integral 兰R兰 f 共x, y兲 dA. For each region, state whether horizontal representative elements, vertical representative elements, or polar sectors would yield the easiest method for obtaining the limits of integration. Explain your reasoning. (a)
(b)
(c)
y
y
y
63. If 兰R兰 f 共r, 兲 dA > 0, then f 共r, 兲 > 0 for all 共r, 兲 in R. 64. If f 共r, 兲 is a constant function and the area of the region S is twice that of the region R, then 2
冕冕
f 共r, 兲 dA
R
I
R x
x
x
冕冕
f 共r, 兲 dA.
S
65. Probability R R
993
True or False? In Exercises 63 and 64, determine whether
HOW DO YOU SEE IT? Each figure shows
58.
Change of Variables: Polar Coordinates
冕
The value of the integral
e x 兾2 dx 2
is required in the development of the normal probability density function. (a) Use polar coordinates to evaluate the improper integral. 59. Volume Determine the diameter of a hole that is drilled vertically through the center of the solid bounded by the graphs 2 2 of the equations z 25e 共x y 兲兾4, z 0, and x2 y2 16 when one-tenth of the volume of the solid is removed.
e x
冕 冕
60. Glacier Horizontal cross sections of a piece of ice that broke from a glacier are in the shape of a quarter of a circle with a radius of approximately 50 feet. The base is divided into 20 subregions, as shown in the figure. At the center of each subregion, the height of the ice is measured, yielding the following points in cylindrical coordinates.
冢冕
I2
2兾2
冣冢冕
dx
e 共x
2
y 2兲兾2
冣
dy
dA
(b) Use the result of part (a) to determine I. FOR FURTHER INFORMATION For more information on 2 this problem, see the article “Integrating e x Without Polar Coordinates” by William Dunham in Mathematics Teacher. To view this article, go to MathArticles.com. 66. Evaluating Integrals Use the result of Exercise 65 and a change of variables to evaluate each integral. No integration is required.
(a) Approximate the volume of the solid.
67. Think About It Consider the region bounded by the graphs of y 2, y 4, y x, and y 冪3 x and the double integral 兰R兰 f dA. Determine the limits of integration when the region R is divided into (a) horizontal representative elements, (b) vertical representative elements, and (c) polar sectors.
(c) There are 7.48 gallons of water per cubic foot. Approximate the number of gallons of water in the solid. π 2
3π 8
冕
(a)
2
69. Probability
π 4
10 20 30 40 50
(b)
兾2
兾4
兾4
f 共x, y兲
冦0,ke
共x2 y2兲
,
0
2
is a probability density function. 0
r冪1 r 3 sin 冪 dr d
0 4
5re
e 4x dx
x 0, y 0 elsewhere
70. Area Show that the area A of the polar sector R (see figure) is A r r , where r 共r1 r2兲兾2 is the average radius of R.
R Δr
5
冪r
Find k such that the function
Approximation In Exercises 61 and 62, use a computer algebra system to approximate the iterated integral.
冕 冕 冕 冕
冕
e x dx
68. Think About It Repeat Exercise 67 for a region R bounded by the graph of the equation 共x 2兲2 y 2 4.
π 8
62.
2兾2
共5, 16 , 7兲, 共15, 16 , 8兲, 共25, 16 , 10兲, 共35, 16 , 12兲, 共45, 16 , 9兲, 共5, 316, 9兲, 共15, 316, 10兲, 共25, 316, 14兲, 共35, 316, 15兲, 共45, 316, 10兲, 共5, 516, 9兲, 共15, 516, 11兲, 共25, 516, 15兲, 共35, 516, 18兲, 共45, 516, 14兲, 共5, 716, 5兲, 共15, 716, 8兲, 共25, 716, 11兲, 共35, 716, 16兲, 共45, 716, 12兲 (b) Ice weighs approximately 57 pounds per cubic foot. Approximate the weight of the solid.
61.
e y
dr d
Δθ r1
r2
0
Volodymyr Goinyk/Shutterstock.com
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994
Chapter 14
Multiple Integration
14.4 Center of Mass and Moments of Inertia Find the mass of a planar lamina using a double integral. Find the center of mass of a planar lamina using double integrals. Find moments of inertia using double integrals.
Mass y
Section 7.6 discussed several applications of integration involving a lamina of constant density . For example, if the lamina corresponding to the region R, as shown in Figure 14.33, has a constant density , then the mass of the lamina is given by
g2
Mass A
R
冕冕 冕冕
g1 x=a
x
x=b
Lamina of constant density Figure 14.33
dA.
dA
R
Constant density
R
If not otherwise stated, a lamina is assumed to have a constant density. In this section, however, you will extend the definition of the term lamina to include thin plates of variable density. Double integrals can be used to find the mass of a lamina of variable density, where the density at 共x, y兲 is given by the density function . Definition of Mass of a Planar Lamina of Variable Density If is a continuous density function on the lamina corresponding to a plane region R, then the mass m of the lamina is given by m
冕冕
共x, y兲 dA.
Variable density
R
Density is normally expressed as mass per unit volume. For a planar lamina, however, density is mass per unit surface area.
Finding the Mass of a Planar Lamina Find the mass of the triangular lamina with vertices 共0, 0兲, 共0, 3兲, and 共2, 3兲, given that the density at 共x, y兲 is 共x, y兲 2x y. Solution As shown in Figure 14.34, region R has the boundaries x 0, y 3, and y 3x兾2 共or x 2y兾3兲. Therefore, the mass of the lamina is
y
y=3 3
2
(0, 3)
(2, 3)
m
R
冕冕 冕冕 冕冤 冕
共2x y兲 dA
R
3
x = 23 y
1
0
2y兾3
共2x y兲 dx dy
0
3
(0, 0) 1
x
2
3
Lamina of variable density 共x, y兲 2x y Figure 14.34
冥
x2 xy
0
10 9
dy 0
3
y 2 dy
0
3
冤 冥
10 y3 9 3 10.
2y兾3
0
In Figure 14.34, note that the planar lamina is shaded so that the darkest shading corresponds to the densest part.
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14.4
Center of Mass and Moments of Inertia
995
Finding Mass by Polar Coordinates y
Find the mass of the lamina corresponding to the first-quadrant portion of the circle x2 y2 4
x2 + y2 = 4 2
where the density at the point 共x, y兲 is proportional to the distance between the point and the origin, as shown in Figure 14.35.
(x, y) 1
共x, y兲 k冪共x 0兲2 共 y 0兲2 k冪x2 y2. x
1
At any point 共x, y兲, the density of the lamina is
Solution
R
2
Density at 共x, y兲: 共x, y兲 k冪x2 y2 Figure 14.35
Because 0 x 2 and 0 y 冪4 x2, the mass is given by m
冕冕 冕冕
k冪x2 y 2 dA
R
冪4x2
2
0
k冪x2 y 2 dy dx.
0
To simplify the integration, you can convert to polar coordinates, using the bounds 0 兾2
and
0 r 2.
So, the mass is m
冕冕 冕 冕 冕 冕 冕 冥 冕
k冪x2 y 2 dA
R
兾2
0
兾2
兾2
8k 3
2
kr 2 dr d
0
0
k冪r 2 r dr d
0
0
2
kr 3 3
兾2
2 0
d
d
0
兾2
冤冥
8k 3
4k . 3
0
TECHNOLOGY On many occasions, this text has mentioned the benefits of computer programs that perform symbolic integration. Even if you use such a program regularly, you should remember that its greatest benefit comes only in the hands of a knowledgeable user. For instance, notice how much simpler the integral in Example 2 becomes when it is converted to polar form. Rectangular Form
冕冕 2
0
冪4x2
0
k冪x2 y2 dy dx
Polar Form
冕 冕 兾2
0
2
kr2 dr d
0
If you have access to software that performs symbolic integration, use it to evaluate both integrals. Some software programs cannot handle the first integral, but any program that can handle double integrals can evaluate the second integral.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
996
Chapter 14
Multiple Integration
Moments and Center of Mass y
For a lamina of variable density, moments of mass are defined in a manner similar to that used for the uniform density case. For a partition of a lamina corresponding to a plane region R, consider the ith rectangle Ri of one area Ai , as shown in Figure 14.36. Assume that the mass of Ri is concentrated at one of its interior points 共xi , yi 兲. The moment of mass of Ri with respect to the x-axis can be approximated by
xi
Ri
(xi, yi)
共Mass兲共 yi 兲 ⬇ 关 共xi , yi 兲 Ai兴共 yi 兲. yi
Similarly, the moment of mass with respect to the y-axis can be approximated by x
Mx 共mass兲共yi兲 My 共mass兲共xi兲 Figure 14.36
共Mass兲共xi 兲 ⬇ 关 共xi , yi 兲 Ai兴共xi 兲. By forming the Riemann sum of all such products and taking the limits as the norm of approaches 0, you obtain the following definitions of moments of mass with respect to the x- and y-axes. Moments and Center of Mass of a Variable Density Planar Lamina Let be a continuous density function on the planar lamina R. The moments of mass with respect to the x- and y-axes are Mx
冕冕
y共x, y兲 dA
R
and My
冕冕
x共x, y兲 dA.
R
If m is the mass of the lamina, then the center of mass is
共x, y兲
冢Mm , Mm 冣. y
x
If R represents a simple plane region rather than a lamina, then the point 共x, y兲 is called the centroid of the region.
For some planar laminas with a constant density , you can determine the center of mass (or one of its coordinates) using symmetry rather than using integration. For instance, consider the laminas of constant density shown in Figure 14.37. Using symmetry, you can see that y 0 for the first lamina and x 0 for the second lamina. R: 0 ≤ x ≤ 1 − 1 − x2 ≤ y ≤
R: − 1 − y 2 ≤ x ≤ 0≤y≤1
1 − x2
z
1 − y2
z
1
1 −1
−1 −1
−1 1
1 x
y
−1
Lamina of constant density that is symmetric with respect to the x-axis Figure 14.37
1 x
1
y
−1
Lamina of constant density that is symmetric with respect to the y-axis
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14.4
Center of Mass and Moments of Inertia
997
Finding the Center of Mass See LarsonCalculus.com for an interactive version of this type of example.
Find the center of mass of the lamina corresponding to the parabolic region
Variable density: y ρ (x, y) = ky
0 y 4 x2
where the density at the point 共x, y兲 is proportional to the distance between 共x, y兲 and the x-axis, as shown in Figure 14.38.
y = 4 − x2 3
Solution The lamina is symmetric with respect to the y-axis and 共x, y兲 ky. So, the center of mass lies on the y-axis and x 0. To find y, first find the mass of the lamina.
(x, y) 2
冕冕 冕 冥 冕
4x2
2
Mass
1
x
−2
−1
Parabolic region
1
2
Parabolic region of variable density Figure 14.38
ky dy dx
2 0
k 2 k 2
2
4x2
y2
2
dx 0
2
2
共16 8x2 x 4兲 dx
冤 冢
k 8x3 x5 16x 2 3 5 64 32 k 32 3 5 256k 15
冥
2
2
冣
Next, find the moment about the x-axis.
冕冕 冕 冥 冕
4x 2
2
Mx
2 0
k 3 k 3
2
y3
2
共 y兲共ky兲 dy dx
4 x2 0
dx
2
2
共64 48x2 12x 4 x 6 兲 dx
冤
k 12x5 x7 64x 16x3 3 5 7
4096k 105
冥
2
2
So, Variable density: ρ (x, y) = ky z
y
R: −2 ≤ x ≤ 2 0 ≤ y ≤ 4 − x2
and the center of mass is 共0,
Center of mass: −2
x
Figure 14.39
16 7
兲.
)0, ) 16 7
1 2
Mx 4096k兾105 16 m 256k兾15 7
4
y
Although you can think of the moments Mx and My as measuring the tendency to rotate about the x- or y-axis, the calculation of moments is usually an intermediate step toward a more tangible goal. The use of the moments Mx and My is typical—to find the center of mass. Determination of the center of mass is useful in a variety of applications that allow you to treat a lamina as if its mass were concentrated at just one point. Intuitively, you can think of the center of mass as the balancing point of the lamina. For instance, the lamina in Example 3 should balance on the point of a pencil placed at 共0, 16 兲, as shown in Figure 14.39. 7
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998
Chapter 14
Multiple Integration
Moments of Inertia The moments of Mx and My used in determining the center of mass of a lamina are sometimes called the first moments about the x- and y-axes. In each case, the moment is the product of a mass times a distance. Mx
冕冕
共 y兲 共x, y兲 dA
My
R
冕冕
共x兲 共x, y兲 dA
R
Distance to x-axis
Distance to y-axis
Mass
Mass
You will now look at another type of moment—the second moment, or the moment of inertia of a lamina about a line. In the same way that mass is a measure of the tendency of matter to resist a change in straight-line motion, the moment of inertia about a line is a measure of the tendency of matter to resist a change in rotational motion. For example, when a particle of mass m is a distance d from a fixed line, its moment of inertia about the line is defined as I md 2 共mass兲共distance兲2. As with moments of mass, you can generalize this concept to obtain the moments of inertia about the x- and y-axes of a lamina of variable density. These second moments are denoted by Ix and Iy , and in each case the moment is the product of a mass times the square of a distance. Ix
冕冕
共 y 2兲 共x, y兲 dA
Iy
R
冕冕
共x2兲 共x, y兲 dA
R
Square of distance to x-axis
Mass
Square of distance to y-axis
Mass
The sum of the moments Ix and Iy is called the polar moment of inertia and is denoted by I0. For a lamina in the xy-plane, I0 represents the moment of inertia of the lamina about the z-axis. The term “polar moment of inertia” stems from the fact that the square of the polar distance r is used in the calculation. I0
冕冕
共x2 y 2兲 共x, y兲 d A
R
冕冕
r 2 共x, y兲 d A
R
Finding the Moment of Inertia Find the moment of inertia about the x-axis of the lamina in Example 3. Solution
From the definition of moment of inertia, you have
冕冕 冕 冥 冕
4x2
2
Ix
2 0
k 4 k 4
2
4x2
y4
2
y2共ky兲 dy dx dx
0
2
2
冤
共256 256x2 96x4 16x6 x8兲 dx
k 256x3 96x5 16x7 x9 256x 4 3 5 7 9 32,768k . 315
2
冥
2
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14.4
Center of Mass and Moments of Inertia
999
The moment of inertia I of a revolving lamina can be used to measure its kinetic energy. For example, suppose a planar lamina is revolving about a line with an angular speed of radians per second, as shown in Figure 14.40. The kinetic energy E of the revolving lamina is E
1 2 I . 2
Kinetic energy for rotational motion
On the other hand, the kinetic energy E of a mass m moving in a straight line at a velocity v is E Planar lamina revolving at radians per second Figure 14.40
1 mv 2. 2
Kinetic energy for linear motion
So, the kinetic energy of a mass moving in a straight line is proportional to its mass, but the kinetic energy of a mass revolving about an axis is proportional to its moment of inertia. The radius of gyration r of a revolving mass m with moment of inertia I is defined as r
冪mI .
Radius of gyration
If the entire mass were located at a distance r from its axis of revolution, it would have the same moment of inertia and, consequently, the same kinetic energy. For instance, the radius of gyration of the lamina in Example 4 about the x-axis is y
128 冪 ⬇ 2.469. 冪mI 冪32,768k兾315 256k兾15 21 x
Finding the Radius of Gyration Find the radius of gyration about the y-axis for the lamina corresponding to the region R: 0 y sin x, 0 x , where the density at 共x, y兲 is given by 共x, y兲 x. Solution The region R is shown in Figure 14.41. By integrating 共x, y兲 x over the region R, you can determine that the mass of the region is . The moment of inertia about the y-axis is
y
2
1
Variable density: ρ (x, y) = x
R: 0 ≤ x ≤ π 0 ≤ y ≤ sin x
Iy
(x, y) π 2
Figure 14.41
冕冕 冕 冥 冕
0
π
x
sin x
sin x
3
x y
0
x3 dy dx
0
dx 0
x3 sin x dx
0
冤
冥
共3x2 6兲共sin x兲 共x3 6x兲共cos x兲
0
3 6. So, the radius of gyration about the y-axis is
冪mI 6 冪
x
y
3
冪 2 6 ⬇ 1.967.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1000
Chapter 14
Multiple Integration
14.4 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding the Mass of a Lamina In Exercises 1–4, find the mass of the lamina described by the inequalities, given that its density is 冇x, y冈 ⴝ xy. (Hint: Some of the integrals are simpler in polar coordinates.)
23. y ex, y 0, x 0, x 2, kxy
1. 0 x 2, 0 y 2 2. 0 x 3, 0 y 9 x 2 3. 0 x 1, 0 y 冪1 x
2
4. x 0, 3 y 3 冪9 x 2
Finding the Center of Mass In Exercises 5–8, find the mass and center of mass of the lamina for each density. 5. R: square with vertices 共0, 0兲, 共a, 0兲, 共0, a兲, 共a, a兲 (a) k
(b) ky (c) kx
6. R: rectangle with vertices 共0, 0兲, 共a, 0兲, 共0, b兲, 共a, b兲 (a) kxy
(b) k 共x2 y 2兲
7. R: triangle with vertices 共0, 0兲, 共0, a兲, 共a, a兲 (a) k
(b) ky
Finding the Center of Mass Using Technology In Exercises 23–26, use a computer algebra system to find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density.
(c) kx
24. y ln x, y 0, x 1, x e, 25. r 2 cos 3,
(b) kxy
Finding the Radius of Gyration About Each Axis In Exercises 27–32, verify the given moment(s) of inertia and find x and y. Assume that each lamina has a density of ⴝ 1 gram per square centimeter. (These regions are common shapes used in engineering.) 27. Rectangle
28. Right triangle
y
Ix = Iy =
y
1 bh 3 3 1 3 b h 3
1 Ix = 12 bh 3 1 3 Iy = 12 b h
h
9. Translations in the Plane Translate the lamina in Exercise 5 to the right five units and determine the resulting center of mass. 10. Conjecture Use the result of Exercise 9 to make a conjecture about the change in the center of mass when a lamina of constant density is translated c units horizontally or d units vertically. Is the conjecture true when the density is not constant? Explain.
, k 6 6
26. r 1 cos , k
8. R: triangle with vertices 共0, 0兲, 共a兾2, a兲, 共a, 0兲 (a) k
k x
h b
b
x
29. Circle
x
30. Semicircle y
y
a
x
Finding the Center of Mass In Exercises 11–22, find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density. (Hint: Some of the integrals are simpler in polar coordinates.) 11. y 冪x, y 0, x 1, ky 12. y x2, y 0, x 2, kxy
I0 = 12 π a 4
31. Quarter circle
I0 = 14 π a 4
32. Ellipse y
y
I0 = 18 π a 4
13. y 4兾x, y 0, x 1, x 4, kx2
b
1 , y 0, x 1, x 1, k 14. y 1 x2 15. y ex, y 0, x 0, x 1, k 16. y ex, y 0, x 0, x 1, ky2 17. y 4 x2, y 0, ky 18. x 9 y2, x 0, kx 19. y sin
x , y 0, x 0, x L, k L
L x 20. y cos , y 0, x 0, x , ky L 2 21. y 冪a2 x2, 0 y x, k 22. x2 y2 a2, x 0, y 0, k共x2 y2兲
x
a
a
a
x
x
I0 = 14 π ab(a 2 + b 2)
Finding Moments of Inertia and Radii of Gyration In Exercises 33–36, find Ix, Iy, I0, x, and y for the lamina bounded by the graphs of the equations. 33. y 4 x2, y 0, x > 0, kx 34. y x, y x2, kxy 35. y 冪x, y 0, x 4, kxy 36. y x2, y2 x, kx
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14.4
Finding a Moment of Inertia Using Technology In Exercises 37–40, set up the double integral required to find the moment of inertia I, about the given line, of the lamina bounded by the graphs of the equations. Use a computer algebra system to evaluate the double integral.
Center of Mass and Moments of Inertia
WRITING ABOUT CONCEPTS
1001
( continued )
47. Radius of Gyration In your own words, describe what the radius of gyration measures.
37. x2 y2 b2, k, line: x a 共a > b兲 38. y 冪x, y 0, x 4, kx, line: x 6
HOW DO YOU SEE IT? The center of mass of the lamina of constant density shown in the figure is 共2, 85 兲. Make a conjecture about how the center of mass 共x, y兲 changes for each given nonconstant density 共x, y兲. Explain. 共Make your conjecture without performing any calculations.兲
48.
39. y 冪a2 x2, y 0, ky, line: y a 40. y 4 x2, y 0, k, line: y 2
Hydraulics In Exercises 41–44, determine the location of the horizontal axis ya at which a vertical gate in a dam is to be hinged so that there is no moment causing rotation under the indicated loading (see figure). The model for ya is
y 4
ya ⴝ y ⴚ
Iy hA
3 2
where y is the y-coordinate of the centroid of the gate, I y is the moment of inertia of the gate about the line y ⴝ y, h is the depth of the centroid below the surface, and A is the area of the gate.
(2, 85 (
1 x
y
1
y=L h y=y Iy ya = y − hA x
41.
y
y
42.
y=L
y=L
d
2
3
4
ⱍ
ⱍ
(a) 共x, y兲 ky
(b) 共x, y兲 k 2 x
(c) 共x, y兲 kxy
(d) 共x, y兲 k共4 x兲共4 y兲
49. Proof Prove the following Theorem of Pappus: Let R be a region in a plane and let L be a line in the same plane such that L does not intersect the interior of R. If r is the distance between the centroid of R and the line, then the volume V of the solid of revolution formed by revolving R about the line is given by V 2rA, where A is the area of R.
a b
43.
x
y
b
44. b
y=L
x
y
y=L
Center of Pressure on a Sail The center of pressure on a sail is the point 共xp, yp兲 at which the total aerodynamic force may be assumed to act. If the sail is represented by a plane region R, then the center of pressure is
d
xp x
a x
兰R 兰 xy dA 兰R 兰 y dA
and
yp
兰R 兰 y 2 dA . 兰R 兰 y dA
Consider a triangular sail with vertices at 共0, 0兲, 共2, 1兲, and 共0, 5兲. Verify the value of each integral. (a)
冕冕 冕冕 冕冕
y dA 10
R
WRITING ABOUT CONCEPTS
(b)
xy dA
35 6
y2 dA
155 6
R
45. Moments and Center of Mass Give the formulas for finding the moments and center of mass of a variable density planar lamina.
(c)
46. Moments of Inertia Give the formulas for finding the moments of inertia about the x- and y-axes for a variable density planar lamina.
Calculate the coordinates 共xp, yp兲 of the center of pressure. Sketch a graph of the sail and indicate the location of the center of pressure.
R
Martynova Anna/Shutterstock.com
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1002
Chapter 14
Multiple Integration
14.5 Surface Area Use a double integral to find the area of a surface.
Surface Area At this point, you know a great deal about the solid region lying between a surface and a closed and bounded region R in the xy-plane, as shown in Figure 14.42. For example, you know how to find the extrema of f on R (Section 13.8), the area of the base R of the solid (Section 14.1), the volume of the solid (Section 14.2), and the centroid of the base R (Section 14.4). In this section, you will learn how to find the upper surface area of the solid. Later, you will learn how to find the centroid of the solid (Section 14.6) and the lateral surface area (Section 15.2). To begin, consider a surface S given by z f 共x, y兲 Surface: z = f(x, y)
ΔTi
z
ΔSi ≈ ΔTi
R
Surface: z = f(x, y)
z
y
x
Region R in xy-plane
Figure 14.42
Surface defined over a region R
defined over a region R. Assume that R is closed and bounded and that f has continuous first partial derivatives. To find the surface area, construct an inner partition of R consisting of n rectangles, where the area of the ith rectangle Ri is Ai xi yi, as shown in Figure 14.43. In each Ri , let 共xi, yi兲 be the point that is closest to the origin. At the point 共xi, yi, zi兲 共xi, yi, f 共xi, yi 兲兲 on the surface S, construct a tangent plane Ti. The area of the portion of the tangent plane that lies directly above Ri is approximately equal to the area of the surface lying directly above Ri. That is, Ti ⬇ Si. So, the surface area of S is approximated by
y
n
n
兺 S ⬇ 兺 T .
x
i
i
i1
ΔAi
i1
To find the area of the parallelogram Ti, note that its sides are given by the vectors u xi i fx共xi , yi 兲 xi k
Figure 14.43
and v yi j fy共xi , yi 兲 yi k. From Theorem 11.8, the area of Ti is given by 储 u v储, where
ⱍ
i u v xi 0
j 0 yi
k fx共xi, yi 兲 xi fy共xi, yi 兲 yi
ⱍ
fx共xi, yi 兲 xi yi i fy共xi, yi 兲 xi yi j xi yi k 共fx共xi, yi 兲i fy共xi, yi 兲j k兲 Ai. So, the area of Ti is 储u v储 冪关 fx共xi, yi 兲兴 2 关 fy共xi, yi 兲兴 2 1 Ai, and Surface area of S ⬇ ⬇
n
兺 S
i
i1 n
兺 冪1 关 f 共x , y 兲兴 x
i
i
2
关 fy共xi, yi 兲兴 2 Ai.
i1
This suggests the definition of surface area on the next page.
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14.5
Surface Area
1003
Definition of Surface Area If f and its first partial derivatives are continuous on the closed region R in the xy-plane, then the area of the surface S given by z f 共x, y兲 over R is defined as Surface area
冕冕 冕冕
dS
R
冪1 关 fx共x, y兲兴 2 关 fy共x, y兲兴 2 dA.
R
As an aid to remembering the double integral for surface area, it is helpful to note its similarity to the integral for arc length.
冕 冕 冕 冕冕 冕冕 冕冕 b
Length on x-axis:
dx
a
b
Arc length in xy-plane:
b
a
Area in xy-plane:
冪1 关 f 共x兲兴 2 dx
ds
a
dA
R
Surface area in space:
dS
R
冪1 关 fx共x, y兲兴 2 关 fy共x, y兲兴 2 dA
R
Like integrals for arc length, integrals for surface area are often very difficult to evaluate. However, one type that is easily evaluated is demonstrated in the next example.
The Surface Area of a Plane Region Plane: z=2−x−y
Find the surface area of the portion of the plane
z
z2xy 2
that lies above the circle x 2 y 2 1 in the first quadrant, as shown in Figure 14.44. Because fx共x, y兲 1 and fy共x, y兲 1, the surface area is given by
Solution S
冕冕 冕冕 冕冕
冪1 关 fx共x, y兲兴 2 关 fy共x, y兲兴 2 dA
Formula for surface area
R
2 x
Figure 14.44
2
y
冪1 共1兲2 共1兲2 dA
Substitute.
R
R: x 2 + y 2 ≤ 1
R
冪3
冪3 dA
冕冕
dA.
R
Note that the last integral is 冪3 times the area of the region R. R is a quarter circle of radius 1, with an area of 14 共12兲 or 兾4. So, the area of S is S 冪3 共area of R兲 冪3 4 冪3 . 4
冢冣
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1004
Chapter 14
Multiple Integration
Finding Surface Area See LarsonCalculus.com for an interactive version of this type of example. z
Find the area of the portion of the surface f 共x, y兲 1 x 2 y that lies above the triangular region with vertices 共1, 0, 0兲, 共0, 1, 0兲, and 共0, 1, 0兲, as shown in Figure 14.45.
Surface: f(x, y) = 1 − x 2 + y
Because fx共x, y兲 2x and fy共x, y兲 1, you have
Solution
(0, 1, 2)
2
S
冕冕
冪1 关 fx共x, y兲兴 2 关 fy共x, y兲兴 2 dA
R
冕冕 冕 冕冤 冕共 1
S
−1
1x
0
y
1
1
冪2 4x 2 dy dx
x1
1
x
y冪2 4x2
0
冥
1x
dx x1
1
Figure 14.45
冥
共1 x兲冪2 4x2 共x 1兲冪2 4x2 dx
0
1
y
2冪2 4x 2 2x冪2 4x 2 兲 dx
0
y=1−x
共2 4x 2兲3兾2 6 1 冪6 ln共2 冪6 兲 冪6 ln 冪2 冪2 3 ⬇ 1.618.
冤
x
1
2
y=x−1
S
0
Because fx共x, y兲 2x and fy共x, y兲 2y, you have
冕冕
冪1 关 fx共x, y兲兴 2 关 fy共x, y兲兴 2 dA
R
Paraboloid: z = 1 + x2 + y2
冕冕
冪1 4x 2 4y 2 dA.
R
You can convert to polar coordinates by letting x r cos and y r sin . Then, because the region R is bounded by 0 r 1 and 0 2, you have
z
S
冕冕 冕 冕 2
0
2
2
0
R: x 2 + y 2 ≤ 1
2
0
R
Figure 14.47
冥
Find the surface area of the paraboloid z 1 x 2 y 2 that lies above the unit circle, as shown in Figure 14.47. Solution
1
1
Change of Variables to Polar Coordinates
Figure 14.46
x
Integration tables 共Appendix B兲, Formula 26 and Power Rule
x冪2 4x 2 ln共2x 冪2 4x 2 兲
R: 0 ≤ x ≤ 1 x−1≤y≤1−x
−1
冪1 4x 2 1 dA.
R
In Figure 14.46, you can see that the bounds for R are 0 x 1 and x 1 y 1 x. So, the integral becomes
1
1
冕冕
1
y
1
冪1 4r 2 r dr d
0 1
冥
1 共1 4r 2兲3兾2 12
0
d
5冪5 1 d
12
5冪5 1
12
冥
2 0
共5冪5 1兲 6 ⬇ 5.33.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.5
Surface Area
1005
Finding Surface Area Find the surface area S of the portion of the hemisphere
Hemisphere: 25 −
f(x, y) =
x2
−
y2
f 共x, y兲 冪25 x 2 y 2
z
that lies above the region R bounded by the circle x 2 y 2 9, as shown in Figure 14.48.
5
Solution
4 3 2
−4
1
−4
−2
Hemisphere
The first partial derivatives of f are x
fx共x, y兲
−6
冪25 x 2 y 2
and 2
1
4
2
3
6
R:
x
4
x2
+
y
5
y2
y 冪25 x 2 y 2
fy共x, y兲
≤9
Figure 14.48
and, from the formula for surface area, you have dS 冪1 关 fx共x, y兲兴 2 关 fy共x, y兲兴 2 dA
冪1 冢
x 冪25 x 2 y 2
5
y
冣 冢冪25 x 2
2
y 冣 2
2
dA
dA.
冪25 x 2 y 2
So, the surface area is S
冕冕 R
5 冪25 x 2 y 2
dA.
You can convert to polar coordinates by letting x r cos and y r sin . Then, because the region R is bounded by 0 r 3 and 0 2, you obtain S
冕冕 冕 冕 2
3
0
0
2
5
5
冥
冪25 r2
0
2
5
r dr d
冪25 r 2
3
0
d
d
0
10. The procedure used in Example 4 can be extended to find the surface area of a sphere by using the region R bounded by the circle x 2 y 2 a 2, where 0 < a < 5, as shown in Figure 14.49. The surface area of the portion of the hemisphere
Hemisphere:
25 − x 2 − y 2
f(x, y) =
f 共x, y兲 冪25 x 2 y 2
z
lying above the circular region can be shown to be
5
S
冕冕 冕冕 R
2
0
a
a
x
Figure 14.49
y
5
5
R:
x2
+
y2
≤
a2
5
冪25 x 2 y 2 a
0
5 冪25 r 2
dA
r dr d
10 共5 冪25 a2 兲. By taking the limit as a approaches 5 and doubling the result, you obtain a total area of 100. (The surface area of a sphere of radius r is S 4r 2.)
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1006
Chapter 14
Multiple Integration
You can use Simpson’s Rule or the Trapezoidal Rule to approximate the value of a double integral, provided you can get through the first integration. This is demonstrated in the next example.
Approximating Surface Area by Simpson’s Rule Find the area of the surface of the paraboloid
Paraboloid: f (x, y) = 2 − x 2 − y 2
f 共x, y兲 2 x 2 y 2
z
Paraboloid
that lies above the square region bounded by 1 x 1 and
2
1 y 1
as shown in Figure 14.50. Solution
Using the partial derivatives
fx共x, y兲 2x and
fy共x, y兲 2y
you have a surface area of S
R: −1 ≤ x ≤ 1 −1 ≤ y ≤ 1
x
冪1 关 fx共x, y兲兴 2 关 fy共x, y兲兴 2 dA
R
y
1 2
冕冕 冕冕 冕冕
冪1 共2x兲2 共2y兲2 dA
R
Figure 14.50
冪1 4x 2 4y 2 dA.
R
In polar coordinates, the line x 1 is given by r cos 1 or y
r sec
and you can determine from Figure 14.51 that one-fourth of the region R is bounded by
r = sec θ
θ= 1
π 4
0 r sec
and
. 4 4
Letting x r cos and y r sin produces x −1
1
−1
θ=−
π 4
One-fourth of the region R is bounded by 0 r sec and . 4 4 Figure 14.51
冕冕 冕 冕 冕 冕
1 1 S 4 4
冪1 4x 2 4y 2 dA
R
兾4
sec
兾4 0
兾4
冪1 4r 2 r dr d
1 12
兾4
兾4
sec
冥
1 共1 4r2兲3兾2 12 兾4
0
d
关共1 4 sec2 兲 3兾2 1兴 d .
Finally, using Simpson’s Rule with n 10, you can approximate this single integral to be S
1 3
冕
兾4
兾4
关共1 4 sec2 兲3兾2 1兴 d ⬇ 7.450.
TECHNOLOGY Most computer programs that are capable of performing symbolic integration for multiple integrals are also capable of performing numerical approximation techniques. If you have access to such software, use it to approximate the value of the integral in Example 5.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.5
14.5 Exercises
1. f 共x, y兲 2x 2y R: triangle with vertices 共0, 0兲, 共4, 0兲, 共0, 4兲 2. f 共x, y兲 15 2x 3y R: square with vertices 共0, 0兲, 共3, 0兲, 共0, 3兲, 共3, 3兲 3. f 共x, y兲 7 2x 2y, R 再共x, y兲: x 2 y 2 4冎 4. f 共x, y兲 12 2x 3y, R 再共x, y兲: x 2 y 2 9冎 5. f 共x, y兲 9 x 2 R: square with vertices 共0, 0兲, 共2, 0兲, 共0, 2兲, 共2, 2兲 6. f 共x, y兲 y 2 R: square with vertices 共0, 0兲, 共3, 0兲, 共0, 3兲, 共3, 3兲 7. f 共x, y兲 3 x3兾2 R: rectangle with vertices 共0, 0兲, 共0, 4兲, 共3, 4兲, 共3, 0兲 8. f 共x, y兲 2 23 y 3兾2 R 再共x, y兲: 0 x 2, 0 y 2 x冎
ⱍ
9. f 共x, y兲 ln sec x
冦
冧
, 0 y tan x 4 10. f 共x, y兲 13 x 2 y 2, R 再共x, y兲: x 2 y 2 4冎 11. f 共x, y兲 冪x 2 y 2, R 再共x, y兲: 0 f 共x, y兲 1冎 R 共x, y兲: 0 x
1007
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding Surface Area In Exercises 1–14, find the area of the surface given by z ⴝ f 冇x, y冈 over the region R. (Hint: Some of the integrals are simpler in polar coordinates.)
ⱍ
Surface Area
12. f 共x, y兲 xy, R 再共x, y兲: x 2 y 2 16冎 13. f 共x, y兲 冪a 2 x 2 y 2 R 再共x, y兲: x 2 y 2 b2, 0 < b < a冎
23. f 共x, y兲 4 x 2 y 2 R 再共x, y兲: 0 x 1, 0 y 1冎 24. f 共x, y兲 23x3兾2 cos x R 再共x, y兲: 0 x 1, 0 y 1冎
Setting Up a Double Integral In Exercises 25–28, set up a double integral that gives the area of the surface on the graph of f over the region R. 25. f 共x, y兲 exy, R 再共x, y兲: 0 x 4, 0 y 10冎 26. f 共x, y兲 x 2 3xy y 2 R 再共x, y兲: 0 x 4, 0 y x冎 27. f 共x, y兲 ex sin y, R 再共x, y兲: x 2 y 2 4冎
冦
28. f 共x, y兲 cos共x 2 y 2兲, R 共x, y兲: x 2 y 2
2
冧
WRITING ABOUT CONCEPTS 29. Surface Area State the double integral definition of the area of a surface S given by z f 共x, y兲 over the region R in the xy-plane. 30. Surface Area Answer each question about the surface area S on a surface given by a positive function z f 共x, y兲 over a region R in the xy-plane. Explain each answer. (a) Is it possible for S to equal the area of R? (b) Can S be greater than the area of R? (c) Can S be less than the area of R? 31. Surface Area Will the surface area of the graph of a function z f 共x, y兲 over a region R increase when the graph is shifted k units vertically? Why or why not?
14. f 共x, y兲 冪a 2 x 2 y 2 R 再共x, y兲: x 2 y 2 a 2冎
Finding Surface Area In Exercises 15–18, find the area of the surface. 15. The portion of the plane z 24 3x 2y in the first octant
32.
HOW DO YOU SEE IT? Consider the surface
f 共x, y兲 x2 y2 (see figure) and the surface area of f over each region R. Without integrating, order the surface areas from least to greatest. Explain.
16. The portion of the paraboloid z 16 x 2 y 2 in the first octant
z
17. The portion of the sphere x 2 y 2 z 2 25 inside the cylinder x 2 y 2 9
4
18. The portion of the cone z 2冪x 2 y 2 inside the cylinder x2 y 2 4
Finding Surface Area Using Technology In Exercises 19–24, write a double integral that represents the surface area of z ⴝ f 冇x, y冈 over the region R. Use a computer algebra system to evaluate the double integral.
−2
−2 x
2
2
y
19. f 共x, y兲 2y x 2, R: triangle with vertices 共0, 0兲, 共1, 0兲, 共1, 1兲
(a) R: rectangle with vertices 共0, 0兲, 共2, 0兲, 共2, 2兲, 共0, 2兲
20. f 共x, y兲 2x y 2, R: triangle with vertices 共0, 0兲, 共2, 0兲, 共2, 2兲
(b) R: triangle with vertices 共0, 0兲, 共2, 0兲, 共0, 2兲
21. f 共x, y兲 9
x2
y 2,
R 再共x, y兲: 0 f 共x, y兲冎
(c) R 再共x, y兲: x2 y2 4, first quadrant only冎
22. f 共x, y兲 x 2 y 2, R 再共x, y兲: 0 f 共x, y兲 16冎
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1008
Chapter 14
Multiple Integration
33. Product Design A company produces a spherical object of radius 25 centimeters. A hole of radius 4 centimeters is drilled through the center of the object.
36. Surface Area Show that the surface area of the cone z k冪x 2 y 2, k > 0, over the circular region x 2 y 2 r 2 in the xy-plane is r 2冪k 2 1 (see figure).
(a) Find the volume of the object.
z
(b) Find the outer surface area of the object. 34. Modeling Data z=k
A company builds a warehouse with dimensions 30 feet by 50 feet. The symmetrical shape and selected heights of the roof are shown in the figure.
R: x 2 + y 2 ≤ r 2
z 25
x2 + y2
(0, 0, 25) (0, 5, 22)
r
r
y
x
(0, 10, 17) (0, 15, 0) 20
y
50 x
(a) Use the regression capabilities of a graphing utility to find a model of the form
Capillary Action A well-known property of liquids is that they will rise in narrow vertical channels––this property is called “capillary action.” The figure shows two plates, that form a narrow wedge in a container of liquid. The upper surface of the liquid follows a hyperbolic shape given by z
z ay 3 by 2 cy d for the roof line.
k 冪x2 y2
where x, y, and z are measured in inches. The constant k depends on the angle of the wedge, the type of liquid, and the material that comprises the flat plates.
(b) Use the numerical integration capabilities of a graphing utility and the model in part (a) to approximate the volume of storage space in the warehouse.
z
θ = 2 arctan(0.01) 9 in.
(c) Use the numerical integration capabilities of a graphing utility and the model in part (a) to approximate the surface area of the roof. 13 in.
(d) Approximate the arc length of the roof line and find the surface area of the roof by multiplying the arc length by the length of the warehouse. Compare the results and the integrations with those found in part (c). 35. Surface Area Find the surface area of the solid of intersection of the cylinders x 2 z 2 1 and y 2 z 2 1 (see figure).
y x
z
y2 + z2 = 1
(a) Find the volume of the liquid that has risen in the wedge. 共Assume k 1.兲
2
(b) Find the horizontal surface area of the liquid that has risen in the wedge.
−3
3
3 x
−2
x2 + z2 = 1
y
Adaptation of Capillary Action problem from “Capillary Phenomena” by Thomas B. Greenslade, Jr., Physics Teacher, May 1992. By permission of the author. AlexKZ/Shutterstock.com
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.6
Triple Integrals and Applications
1009
14.6 Triple Integrals and Applications Use a triple integral to find the volume of a solid region. Find the center of mass and moments of inertia of a solid region.
Triple Integrals z
The procedure used to define a triple integral follows that used for double integrals. Consider a function f of three variables that is continuous over a bounded solid region Q. Then, encompass Q with a network of boxes and form the inner partition consisting of all boxes lying entirely within Q, as shown in Figure 14.52. The volume of the ith box is Vi xi yi zi .
Volume of ith box
The norm 储储 of the partition is the length of the longest diagonal of the n boxes in the partition. Choose a point 共xi, yi , z i 兲 in each box and form the Riemann sum y
n
兺 f 共x , y , z 兲 V . i
x
i
i
i
i1
Taking the limit as 储储 → 0 leads to the following definition. Solid region Q
Definition of Triple Integral If f is continuous over a bounded solid region Q, then the triple integral of f over Q is defined as
z
冕冕冕
n
兺 f 共x , y , z 兲 V
f 共x, y, z兲 dV lim
储储→0 i1
i
i
i
i
Q
provided the limit exists. The volume of the solid region Q is given by Volume of Q
冕冕冕
dV.
Q
y x
Volume of Q ⬇
n
兺 V
i
i1
Some of the properties of double integrals in Theorem 14.1 can be restated in terms of triple integrals.
Figure 14.52
1.
冕冕冕
cf 共x, y, z兲 dV c
冕冕冕
冕冕冕
关 f 共x, y, z兲 ± g共x, y, z兲兴 dV
冕冕冕
f 共x, y, z兲 dV
Q
2.
f 共x, y, z兲 dV
Q
Q
3.
Q
冕冕冕
f 共x, y, z兲 dV ±
Q
冕冕冕
f 共x, y, z兲 dV
Q1
冕冕冕
g共x, y, z兲 dV
Q
冕冕冕
f 共x, y, z兲 dV
Q2
In the properties above, Q is the union of two nonoverlapping solid subregions Q1 and Q 2. If the solid region Q is simple, then the triple integral 兰兰兰 f 共x, y, z兲 dV can be evaluated with an iterated integral using one of the six possible orders of integration: dx dy dz dy dx dz dz dx dy dx dz dy dy dz dx dz dy dx.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1010
Chapter 14
Multiple Integration
Exploration Volume of a Paraboloid Sector In the Explorations on pages 979 and 998, you were asked to summarize the different ways you know of finding the volume of the solid bounded by the paraboloid z a2 x2 y2, a > 0
The following version of Fubini’s Theorem describes a region that is considered simple with respect to the order dz dy dx. Similar descriptions can be given for the other five orders. THEOREM 14.4 Evaluation by Iterated Integrals Let f be continuous on a solid region Q defined by a x b, h 1共x兲 y h 2共x兲, g1共x, y兲 z g2共x, y兲 where h 1, h 2, g1, and g2 are continuous functions. Then,
冕冕冕
and the xy-plane. You now know one more way. Use it to find the volume of the solid.
冕冕 冕 b
f 共x, y, z兲 dV
h1共x兲
a
Q
h2共x兲
g2共x, y兲
g1共x, y兲
f 共x, y, z兲 dz dy dx.
z
To evaluate a triple iterated integral in the order dz dy dx, hold both x and y constant for the innermost integration. Then, hold x constant for the second integration.
a2
Evaluating a Triple Iterated Integral Evaluate the triple iterated integral
−a a x
a
y
冕冕冕 2
0
x
0
xy
Solution to z.
For the first integration, hold x and y constant and integrate with respect
冕冕冕 2
0
x
0
e x共 y 2z兲 dz dy dx.
0
xy
冕冕 冕冕 2
e x共 y 2z兲 dz dy dx
0
0
0
xy
冥
e x 共 yz z 2兲
0
2
x
dy dx 0
x
e x 共x 2 3xy 2y 2兲 dy dx
0
For the second integration, hold x constant and integrate with respect to y.
冕冕 2
0
x
冕冤 冢 冕 2
e x共x 2 3xy 2y 2兲 dy dx
0
e x x 2y
0
19 6
3xy 2 2y 3 2 3
冣冥
x
dx 0
2
x 3e x dx
0
Finally, integrate with respect to x. 19 6
冕
2
0
冤 冢
冥
19 x 3 e 共x 3x 2 6x 6兲 6 e2 19 1 3 ⬇ 65.797
x3e x dx
2 0
冣
Example 1 demonstrates the integration order dz dy dx. For other orders, you can follow a similar procedure. For instance, to evaluate a triple iterated integral in the order dx dy dz, hold both y and z constant for the innermost integration and integrate with respect to x. Then, for the second integration, hold z constant and integrate with respect to y. Finally, for the third integration, integrate with respect to z.
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14.6
Triple Integrals and Applications
1011
To find the limits for a particular order of integration, it is generally advisable first to determine the innermost limits, which may be functions of the outer two variables. Then, by projecting the solid Q onto the coordinate plane of the outer two variables, you can determine their limits of integration by the methods used for double integrals. For instance, to evaluate
z
z = g2(x, y)
冕冕冕
Q
f 共x, y, z兲 dz dy dx
Q
z = g1(x, y)
first determine the limits for z; the integral then has the form y
x
Projection onto xy-plane
冕冕 冤冕
g2共x, y兲
g1 共x, y兲
Solid region Q lies between two surfaces. Figure 14.53
冥
f 共x, y, z兲 dz dy dx.
By projecting the solid Q onto the xy-plane, you can determine the limits for x and y as you did for double integrals, as shown in Figure 14.53.
Using a Triple Integral to Find Volume Find the volume of the ellipsoid given by 4x 2 4y 2 z 2 16.
0 ≤ z ≤ 2 4 − x2 − y2 z
Solution Because x, y, and z play similar roles in the equation, the order of integration is probably immaterial, and you can arbitrarily choose dz dy dx. Moreover, you can simplify the calculation by considering only the portion of the ellipsoid lying in the first octant, as shown in Figure 14.54. From the order dz dy dx, you first determine the bounds for z.
4
0 z 2冪4 x 2 y 2 In Figure 14.55, you can see that the boundaries for x and y are and 0 y 冪4 x 2.
0 x 2 2
So, the volume of the ellipsoid is
1
x
2
V y
冕冕冕 冕冕 冕 冕冕 冥 冕冕 冕冤 冕 冕 冢冣 dV
Q
冪4x2
2
2冪4x2 y2
8
0
0
冪4x2
2
Ellipsoid: 4x 2 + 4y 2 + z 2 = 16
8
Figure 14.54
16
0
0≤x≤2 0 ≤ y ≤ 4 − x2
冪4x2
Integration tables 共Appendix B兲 Formula 37
冪共4 x 2兲 y 2 dy dx
0
2
y冪4 x 2 y 2 共4 x 2兲 arcsin
8
0
冢
y 冪4 x 2
冣冥
冪4x2
dx
0
2
2
关0 共4 x 2兲 arcsin共1兲 0 0兴 dx
8
x2 + y2 = 4
0
2
dx 2
共4 x 2兲
8
1
0
x
1
Figure 14.55
dy dx
0
0
2
y
2冪4x2 y2
z
0
dz dy dx
0
2
冤
4 4x
x3 3
2
冥
0
64 . 3
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1012
Chapter 14
Multiple Integration
Example 2 is unusual in that all six possible orders of integration produce integrals of comparable difficulty. Try setting up some other possible orders of integration to find the volume of the elipsoid. For instance, the order dx dy dz yields the integral
冕冕
冕
冪164y 2 z2兾2
冪16z2兾2
4
V8
0
0
dx dy dz.
0
The evaluation of this integral yields the same volume obtained in Example 2. This is always the case—the order of integration does not affect the value of the integral. However, the order of integration often does affect the complexity of the integral. In Example 3, the given order of integration is not convenient, so you can change the order to simplify the problem.
Changing the Order of Integration
冕 冕 冕 冪兾2
Evaluate
冪兾2
0
3
sin共 y 2兲 dz dy dx.
Q: 0 ≤ x ≤
1
x
Solution Note that after one integration in the given order, you would encounter the integral 2兰 sin共 y 2兲 dy, which is not an elementary function. To avoid this problem, change the order of integration to dz dx dy, so that y is the outer variable. From Figure 14.56, you can see that the solid region Q is
z
1≤z≤3
冪2 x y 冪 2
2
1 z 3
1
and the projection of Q in the xy-plane yields the bounds
冪
)
2
π 2
and
x
y=x
0 x y.
冪兾2
0
y
0
3
1
sin共 y 2兲 dz dx dy
冕 冕 冕 冕 冕 冕 冪兾2
y
冪兾2
2
π ,1 2 π 2
)
) y
3
冪兾2
冪兾2
dx dy
1
sin共 y 2兲 dx dy
0 y
冥
x sin共 y 2兲
0
2
π ,3 2
y
0
2
冥
z sin共 y2兲
0
0
π , 2
π , 2
Figure 14.56
So, evaluating the triple integral using the order dz dx dy produces
冕 冕冕
)
3
0 x
0 y
x≤y≤
π 2 π 2
dy 0
y sin共 y 2兲 dy
0
冥
cos共 y 2兲
冪兾2
0
1.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.6
Triple Integrals and Applications
1013
z
Determining the Limits of Integration 1
Set up a triple integral for the volume of each solid region.
z = 1 − y2
1 y
x=1−y 3
Δy
x=3−y
x
a. The region in the first octant bounded above by the cylinder z 1 y 2 and lying between the vertical planes x y 1 and x y 3 b. The upper hemisphere z 冪1 x 2 y 2 c. The region bounded below by the paraboloid z x 2 y 2 and above by the sphere x2 y 2 z2 6 Solution
Q: 0 ≤ z ≤ 1 − y 2 1−y≤x≤3−y 0≤y≤1
a. In Figure 14.57, note that the solid is bounded below by the xy-plane 共z 0兲 and above by the cylinder z 1 y 2. So, 0 z 1 y 2.
Figure 14.57
Bounds for z
Projecting the region onto the xy-plane produces a parallelogram. Because two sides of the parallelogram are parallel to the x-axis, you have the following bounds: z
1 y x 3 y and 0 y 1.
Hemisphere:
So, the volume of the region is given by
1 − x2 − y2
z=
冕冕冕 冕 冕 冕 1
V
1
3y
dV
0
1y
1y 2
dz dx dy.
0
Q
b. For the upper hemisphere z 冪1 x 2 y 2, you have 0 z 冪1 x 2 y 2. 1
1
x
y
Circular base: x2 + y2 = 1 Q: 0 ≤ z ≤ −
1 − x2 − y2
1 − y2 ≤ x ≤
Bounds for z
In Figure 14.58, note that the projection of the hemisphere onto the xy-plane is the circle x2 y 2 1 and you can use either order dx dy or dy dx. Choosing the first produces
1 − y2
冪1 y 2 x 冪1 y 2 and
−1 ≤ y ≤ 1
1 y 1
which implies that the volume of the region is given by
Figure 14.58
冕冕冕 冕 冕 冕 1
V
dV
Q
z
Sphere: x2 + y2 + z2 = 6
3
冪1y2
冪1x2 y2
dz dx dy.
1 冪1y 2 0
c. For the region bounded below by the paraboloid z x 2 y 2 and above by the sphere x 2 y 2 z 2 6, you have x 2 y 2 z 冪6 x 2 y 2.
Bounds for z
The sphere and the paraboloid intersect at z 2. Moreover, you can see in Figure 14.59 that the projection of the solid region onto the xy-plane is the circle Paraboloid: z = x2 + y2 −2 2
2
y
x 2 y 2 2. Using the order dy dx produces 冪2 x 2 y 冪2 x 2
and 冪2 x 冪2
x
Q: x 2 + y 2 ≤ z ≤ 6 − x 2 − y 2 − 2 − x2 ≤ y ≤ 2 − x2 − 2≤x≤ 2
Figure 14.59
which implies that the volume of the region is given by
冕冕冕 冕 冕 冕 冪2
V
dV
Q
冪2
冪2x2
冪6x2 y2
冪2x2 x2 y2
dz dy dx.
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1014
Chapter 14
Multiple Integration
Center of Mass and Moments of Inertia In the remainder of this section, two important engineering applications of triple integrals are discussed. Consider a solid region Q whose density is given by the density function . The center of mass of a solid region Q of mass m is given by 共x, y, z兲, where m
冕冕冕 冕冕冕 冕冕冕 冕冕冕
共x, y, z兲 dV
Mass of the solid
x 共x, y, z兲 dV
First moment about yz-plane
y 共x, y, z兲 dV
First moment about xz-plane
z 共x, y, z兲 dV
First moment about xy-plane
Q
Myz
Q
Mxz
Q
Mxy
Q
and x
Myz Mxz , y , m m
z
Mxy . m
The quantities Myz, Mxz, and Mxy are called the first moments of the region Q about the yz-, xz-, and xy-planes, respectively. The first moments for solid regions are taken about a plane, whereas the second moments for solids are taken about a line. The second moments (or moments of inertia) about the x-, y-, and z-axes are
REMARK In engineering and physics, the moment of inertia of a mass is used to find the time required for the mass to reach a given speed of rotation about an axis, as shown in Figure 14.60. The greater the moment of inertia, the longer a force must be applied for the mass to reach the given speed.
冕冕冕 冕冕冕
Ix
共 y 2 z 2兲 共x, y, z兲 dV
Moment of inertia about x-axis
共x 2 z 2兲 共x, y, z兲 dV
Moment of inertia about y-axis
共x 2 y 2兲 共x, y, z兲 dV.
Moment of inertia about z-axis
Q
Iy
Q
and Iz
z
冕冕冕 Q
For problems requiring the calculation of all three moments, considerable effort can be saved by applying the additive property of triple integrals and writing y
Ix Ixz Ixy,
Iy Iyz Ixy,
and Iz Iyz Ixz
where Ixy, Ixz, and Iyz are Ixy
x
冕冕冕 冕冕冕
z 2 共x, y, z兲 dV
Q
Ixz
Figure 14.60
y 2 共x, y, z兲 dV
Q
and Iyz
冕冕冕
x 2 共x, y, z兲 dV.
Q
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14.6
Triple Integrals and Applications
1015
Finding the Center of Mass of a Solid Region See LarsonCalculus.com for an interactive version of this type of example. z
Find the center of mass of the unit cube shown in Figure 14.61, given that the density at the point 共x, y, z兲 is proportional to the square of its distance from the origin. Solution Because the density at 共x, y, z兲 is proportional to the square of the distance between 共0, 0, 0兲 and 共x, y, z兲, you have
1
共x, y, z兲 k共x 2 y 2 z2兲. You can use this density function to find the mass of the cube. Because of the symmetry of the region, any order of integration will produce an integral of comparable difficulty.
(x, y, z)
1
1
冕冕冕 冕冕 冤 冕冕 冢 冕 冤冢 冕冢 1
y
m
1
0
x
1
0
1
1
k
Variable density: 共x, y, z兲 k共x2 y2 z2兲 Figure 14.61
k共x 2 y 2 z 2兲 dz dy dx
0
0
z3 3
共x 2 y 2兲z
0
1
1
k
0
x2 y 2
0
1
k
冣
0
1
k
x2
0
dy dx
0
冣
1 dy dx 3
1 y3 y 3 3
x2
1
冥
1
冥
dx
0
冣
2 dx 3 1
冤 x3 2x3 冥 3
k
0
k The first moment about the yz-plane is
冕冕冕 冕 冤冕 冕 1
1
1
x共x 2 y 2 z 2兲 dz dy dx
Myz k
0
0
1
k
0
1
冥
1
共x 2 y 2 z 2兲 dz dy dx.
x
0
0
0
Note that x can be factored out of the two inner integrals, because it is constant with respect to y and z. After factoring, the two inner integrals are the same as for the mass m. Therefore, you have
冕
1
Myz k
0
冤
k
冢
x x2
x 4 x2 4 3
冣
2 dx 3
1
冥
0
7k . 12
So, x
Myz 7k兾12 7 . m k 12
Finally, from the nature of and the symmetry of x, y, and z in this solid region, you 7 7 7 have x y z, and the center of mass is 共12, 12, 12 兲.
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1016
Chapter 14
Multiple Integration
Moments of Inertia for a Solid Region Find the moments of inertia about the x- and y-axes for the solid region lying between the hemisphere z 冪4 x 2 y 2 and the xy-plane, given that the density at 共x, y, z兲 is proportional to the distance between (x, y, z兲 and the xy-plane. 0 ≤ z ≤ 4 − x2 − y2 − 4 − x2 ≤ y ≤ 4 − x2 −2 ≤ x ≤ 2
Solution
共x, y, z兲 kz. Considering the symmetry of this problem, you know that Ix Iy, and you need to compute only one moment, say Ix. From Figure 14.62, choose the order dz dy dx and write
Hemisphere: z=
4−
x2 −
The density of the region is given by
y2
z
Ix
冕冕冕 冕冕 冕 冕冕 冤 冕冕 冤 冕冕 冕冤 冕 冕 冕
共 y 2 z 2兲 共x, y, z兲 dV
Q
2
冪4x2
2
2 冪4x2 0 冪4x2
2
k 2
y
x
Circular base: x2 + y2 = 4
Variable density: 共x, y, z兲 kz Figure 14.62
2 冪4x2 冪4x2
2
k
2
冪4x2 y 2
2 冪4x2
k 4 k 4 k 4
4k 5 4k 5
共 y 2 z 2兲共kz兲 dz dy dx
y 2z 2 z 4 2 4
冪4x2 y 2
冥
dy dx 0
y 2共4 x 2 y 2兲 共4 x 2 y 2兲 2 dy dx 2 4
冥
冪4x2
2
2 冪4x2
关共4 x2兲2 y 4兴 dy dx
2
共4 x 2兲2 y
2
y5 5
冥
冪4x2
冪4x2
dx
2
8 共4 x 2兲5兾2 dx 2 5 2
共4 x 2兲5兾2 dx
x 2 sin
0
兾2
64 cos6 d
0
5 冢256k 5 冣冢 32 冣
Wallis’s Formula
8k. So, Ix 8k Iy. In Example 6, notice that the moments of inertia about the x- and y-axes are equal to each other. The moment about the z-axis, however, is different. Does it seem that the moment of inertia about the z-axis should be less than or greater than the moments calculated in Example 6? By performing the calculations, you can determine that Iz
16 k. 3
This tells you that the solid shown in Figure 14.62 has a greater resistance to rotation about the x- or y-axis than about the z-axis.
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14.6
14.6 Exercises
Triple Integrals and Applications
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Evaluating a Triple Iterated Integral In Exercises 1–8,
z
19.
z
20.
evaluate the triple iterated integral.
冕冕冕 冕冕冕 冕冕冕 冕冕冕 冕冕 冕 冕 冕冕 3
1.
2
a
1
0
1
0
1
1
x 2y 2z 2 dx dy dz
1
0
4
1
0
8.
1
2
6.
0
兾2
y兾2
z=0 x 2 + y2 + z2 = a2
1
x
12
12
y
ln z dy dz dx
0
1
Volume In Exercises 21–24, use a triple integral to find the volume of the solid bounded by the graphs of the equations.
1x
0
0
z dz dx dy
0
e 2 1兾xz
4
2zex dy dx dz
0
0
0
y
x
冪y2 9x2
y兾3
0
x
兾2
冕冕 冕 冕冕 冕 9
4.
0
0
4
7.
xy
x dz dy dx
0
5.
x
a
a
1 1 1
3.
z = 36 − x 2 − y 2
36
共x y z兲 dx dz dy
0
2.
1017
x cos y dz dy dx
21. z 4 x2, y 4 x2, first octant
1兾y
22. z 9 x3, y x2 2, y 0, z 0, x 0
sin y dz dx dy
23. z 2 y, z 4 y2, x 0, x 3, y 0
0
Approximating a Triple Iterated Integral Using Technology In Exercises 9 and 10, use a computer algebra
24. z x, y x 2, y x2, first octant
system to approximate the triple iterated integral.
Changing the Order of Integration In Exercises 25–30, sketch the solid whose volume is given by the iterated integral and rewrite the integral using the indicated order of integration.
冕冕 冕冕 3
9.
冕 冕
y2
冪9y2 0
0
3
10.
冪9y2
0
2 共2y兾3兲
y dz dx dy x2y2
ze
25.
dx dz dy
0
0
0
0
y2
dz dy dx
1 0
Rewrite using the order dy dz dx.
Setting Up a Triple Integral In Exercises 11–16, set up a triple integral for the volume of the solid.
12. The solid bounded by z 9
x 2,
z 0, y 0, and y 2x
16. The solid bounded above by the cylinder z 4 below by the paraboloid z x2 3y2
0
x2
and
0
0
共123x6y兲兾4
dz dy dx
0
6xy
dz dy dx
0
0
冕冕冕 1
29.
1
y
冪1y2
dz dx dy
0
Rewrite using the order dz dy dx.
冕冕 冕 2
30.
0
z
冪9x2
Rewrite using the order dz dx dy.
0
volume of the solid shown in the figure.
4
2x
冪y2 4x2
dz dy dx
0
Rewrite using the order dx dy dz.
4
8
3
6
z = 2xy 0≤x≤2 0≤y≤2
4 x 4
2 y
x = 4 − y2 2
共4x兲兾2
冕冕 冕 3
28.
Volume In Exercises 17–20, use a triple integral to find the
z=0 y
dz dx dy
冕冕 冕 4
27.
15. The solid that is the common interior below the sphere x 2 y 2 z 2 80 and above the paraboloid z 21共x 2 y 2兲
18.
1x
Rewrite using the order dy dx dz.
14. The solid bounded by z 冪16 x2 y2 and z 0
z=x
1
Rewrite using the order dx dz dy.
13. The solid bounded by z 6 x2 y2 and z 0
z
冕 冕冕 1
26.
1 y 2 0
11. The solid in the first octant bounded by the coordinate planes and the plane z 5 x y
17.
冕冕 冕 1
62y3z
3 1
2
x
Orders of Integration In Exercises 31–34, list the six possible orders of integration for the triple integral over the solid region Q, 兰兰 兰 xyz dV. Q
31. Q 再共x, y, z兲: 0 x 1, 0 y x, 0 z 3冎 32. Q 再共x, y, z兲: 0 x 2, x 2 y 4, 0 z 2 x冎 33. Q 再共x, y, z兲: x 2 y 2 9, 0 z 4冎 34. Q 再共x, y, z兲: 0 x 1, y 1 x 2, 0 z 6冎
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1018
Chapter 14
Multiple Integration
Orders of Integration In Exercises 35 and 36, the figure shows the region of integration for the given integral. Rewrite the integral as an equivalent iterated integral in the five other orders.
冕冕 冕 1y2
1
35.
0
0
1y
冕冕冕 3
36.
dz dx dy
0
0
x
0
9x2
dz dy dx
0
z
x≥0 y≥0 z≥0
1
z=1−y x≥0 y≥0 z≥0
region bounded by the graphs of the equations or described by the figure. Use a computer algebra system to evaluate the triple integrals. (Assume uniform density and find the center of mass.) h 47. z 冪x 2 y 2, z h r 48. y 冪9 x 2, z y, z 0
z
z = 9 − x2
Centroid In Exercises 47–52, find the centroid of the solid
49. z 冪16 x 2 y 2, z 0
9
50. z
6
y=x
1 , z 0, x 2, x 2, y 0, y 1 y2 1 z
51.
3
1 1
x
x = 1 − y2
(0, 0, 4)
12 cm
y
x
z
52.
20 cm
3
5 cm y
3 y x
x
Mass and Center of Mass In Exercises 37–40, find the mass and the indicated coordinates of the center of mass of the solid region Q of density bounded by the graphs of the equations. 37. Find x using 共x, y, z兲 k. Q: 2x 3y 6z 12, x 0, y 0, z 0 38. Find y using 共x, y, z兲 ky. Q: 3x 3y 5z 15, x 0, y 0, z 0
y
(0, 3, 0)
(5, 0, 0)
Moments of Inertia In Exercises 53–56, find Ix , Iy, and Iz for the solid of given density. Use a computer algebra system to evaluate the triple integrals. 54. (a) 共x, y, z兲 k
53. (a) k
(b) 共x, y, z兲 k共x 2 y 2兲
(b) kxyz
z
z
39. Find z using 共x, y, z兲 kx.
a a 2
Q: z 4 x, z 0, y 0, y 4, x 0 40. Find y using 共x, y, z兲 k. Q:
x y z 1 共a, b, c > 0兲, x 0, y 0, z 0 a b c
Mass and Center of Mass In Exercises 41 and 42, set up the triple integrals for finding the mass and the center of mass of the solid of density bounded by the graphs of the equations. 41. x 0, x b, y 0, y b, z 0, z b
y
a
a
a 2
a 2
x
y
x
55. (a) 共x, y, z兲 k
56. (a) kz (b) k共4 z兲
(b) ky
共x, y, z兲 kxy
z
42. x 0, x a, y 0, y b, z 0, z c
z
z=4−x
4
4
z = 4 − y2
共x, y, z兲 kz
Think About It The center of mass of a solid of constant density is shown in the figure. In Exercises 43–46, make a conjecture about how the center of mass 冇x, y, z冈 will change for the nonconstant density 冇x, y, z冈. Explain.
4 x
4
2
y
y
4 x
z
) 2, ) 0, 85
Moments of Inertia In Exercises 57 and 58, verify the moments of inertia for the solid of uniform density. Use a computer algebra system to evaluate the triple integrals.
4 3 2
1 57. Ix 12 m共3a 2 L2兲
4 x
3
2
1
2
y
43. 共x, y, z兲 kx
44. 共x, y, z兲 kz
45. 共x, y, z兲 k共 y 2兲
46. 共x, y, z兲 kxz 2共 y 2兲 2
z
Iy 12ma 2
L
1 Iz 12 m共3a 2 L2兲
a x
a
a L 2
y
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14.6
Triple Integrals and Applications
1019
z
1 58. Ix 12 m共a 2 b2兲
WRITING ABOUT CONCEPTS
1 Iy 12 m共b 2 c 2兲
a
c
69. Think About It
b 2
1 Iz 12 m共a 2 c 2兲
to
冕冕冕 3
2
1
冕冕冕 冕 冕冕 冕冕冕
x
c 2
a 2
(a)
1
y
2
1
1
0
1
(b)
2
1 0 2
(c)
0
Moments of Inertia In Exercises 59 and 60, set up a triple integral that gives the moment of inertia about the z-axis of the solid region Q of density . 59. Q 再共x, y, z兲: 1 x 1, 1 y 1, 0 z 1 x冎
Which of the integrals below is equal
f 共x, y, z兲 dz dy dx? Explain.
1
0
3
b
1
( continued )
f 共x, y, z兲 dz dx dy
3
f 共x, y, z兲 dx dy dz
1
3
1
1
1
f 共x, y, z兲 dy dx dz
HOW DO YOU SEE IT? Consider two solids, solid A and solid B, of equal weight as shown below.
70.
冪x 2 y 2 z 2 60. Q 再共x, y, z兲: x 2 y 2 1, 0 z 4 x 2 y 2冎
kx 2
Setting Up Triple Integrals In Exercises 61 and 62, using
Axis of revolution
the description of the solid region, set up the integral for (a) the mass, (b) the center of mass, and (c) the moment of inertia about the z-axis.
Axis of revolution
61. The solid bounded by z 4 x2 y2 and z 0 with density function kz
Solid A
62. The solid in the first octant bounded by the coordinate planes and x2 y2 z2 25 with density function kxy
(a) Because the solids have the same weight, which has the greater density? Explain. (b) Which solid has the greater moment of inertia?
Average Value In Exercises 63–66, find the average value
(c) The solids are rolled down an inclined plane. They are started at the same time and at the same height. Which will reach the bottom first? Explain.
of the function over the given solid. The average value of a continuous function f 冇x, y, z冈 over a solid region Q is 1 V
冕冕冕
Solid B
f 冇x, y, z冈 dV
Q
where V is the volume of the solid region Q. 63. f 共x, y, z兲 z2 4 over the cube in the first octant bounded by the coordinate planes and the planes x 1, y 1, and z 1 64. f 共x, y, z兲 xyz over the cube in the first octant bounded by the coordinate planes and the planes x 4, y 4, and z 4 65. f 共x, y, z兲 x y z over the tetrahedron in the first octant with vertices 共0, 0, 0兲, 共2, 0, 0兲, 共0, 2, 0兲, and 共0, 0, 2兲 66. f 共x, y, z兲 x y over the solid bounded by the sphere x2 y2 z2 3
WRITING ABOUT CONCEPTS 67. Triple Integral Define a triple integral and describe a method of evaluating a triple integral. 68. Moment of Inertia Determine whether the moment of inertia about the y-axis of the cylinder in Exercise 57 will increase or decrease for the nonconstant density 共x, y, z兲 冪x 2 z 2 and a 4.
71. Maximizing a Triple Integral where the triple integral
Find the solid region Q
冕冕冕
共1 2x2 y2 3z2兲 dV
Q
is a maximum. Use a computer algebra system to approximate the maximum value. What is the exact maximum value? 72. Finding a Value
冕冕 1
0
冕
2
3ay
0
Solve for a in the triple integral. 2
4xy
dz dx dy
a
14 15
PUTNAM EXAM CHALLENGE 73. Evaluate
冕 冕 . . .冕 cos 冦2n 共x 1
lim
n→
1
1
2
0
0
0
1
冧
x2 . . . xn兲 dx1 dx2 . . . dxn.
This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1020
Chapter 14
Multiple Integration
14.7 Triple Integrals in Other Coordinates Write and evaluate a triple integral in cylindrical coordinates. Write and evaluate a triple integral in spherical coordinates.
Triple Integrals in Cylindrical Coordinates Many common solid regions, such as spheres, ellipsoids, cones, and paraboloids, can yield difficult triple integrals in rectangular coordinates. In fact, it is precisely this difficulty that led to the introduction of nonrectangular coordinate systems. In this section, you will learn how to use cylindrical and spherical coordinates to evaluate triple integrals. Recall from Section 11.7 that the rectangular conversion equations for cylindrical coordinates are x r cos y r sin z z. An easy way to remember these conversions is to note that the equations for x and y are the same as in polar coordinates and z is unchanged. In this coordinate system, the simplest solid region is a cylindrical block determined by
PIERRE SIMON DE LAPLACE (1749–1827)
r1 r r2 1 2
One of the first to use a cylindrical coordinate system was the French mathematician Pierre Simon de Laplace. Laplace has been called the “Newton of France,” and he published many important works in mechanics, differential equations, and probability.
and z1 z z2 as shown in Figure 14.63. To obtain the cylindrical coordinate form of a triple integral, consider a solid region Q whose projection R onto the xy-plane can be described in polar coordinates. That is,
See LarsonCalculus.com to read more of this biography.
Q 再共x, y, z兲: 共x, y兲 is in R,
h1共x, y兲 z h2共x, y兲冎
and R 再共r, 兲: 1 2, g1共兲 r g2共兲冎. If f is a continuous function on the solid Q, then you can write the triple integral of f over Q as
冕冕冕
z
f 共x, y, z兲 dV
θ =0
ri Δ θ
Volume of cylindrical block: Vi ri ri i zi Figure 14.63
冥
f 共x, y, z兲 dz dA
where the double integral over R is evaluated in polar coordinates. That is, R is a plane region that is either r-simple or -simple. If R is r-simple, then the iterated form of the triple integral in cylindrical form is
Δzi
Δri
h1共x, y兲
R
Q
θ=
冕 冕 冤冕
h2共x, y兲
π 2
冕冕冕
f 共x, y, z兲 dV
Q
冕冕 冕 2
1
g2共兲
g1共兲
h2共r cos , r sin 兲
h1共r cos , r sin 兲
f 共r cos , r sin , z兲r dz dr d.
This is only one of six possible orders of integration. The other five are dz d dr, dr dz d, dr d dz, d dz dr, and d dr dz. The Granger Collection, New York
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14.7 z
θ= θ =0
Triple Integrals in Other Coordinates
1021
To visualize a particular order of integration, it helps to view the iterated integral in terms of three sweeping motions—each adding another dimension to the solid. For instance, in the order dr d dz, the first integration occurs in the r-direction as a point sweeps out a ray. Then, as increases, the line sweeps out a sector. Finally, as z increases, the sector sweeps out a solid wedge, as shown in Figure 14.64.
π 2
Integrate with respect to r.
Exploration
z
θ= θ =0
π 2
Integrate with respect to .
z a 2 x 2 y 2,
θ=
π 2
Integrate with respect to z. Figure 14.64 Sphere: x2 + y2 + z2 = 4
a > 0
a2
−a
and the xy-plane. You now know one more way. Use it to find the volume of the solid. Compare the different methods. What are the advantages and disadvantages of each?
z
θ =0
z
Volume of a Paraboloid Sector In the Explorations on pages 979, 998, and 1010, you were asked to summarize the different ways you know of finding the volume of the solid bounded by the paraboloid
a
a
y
x
Finding Volume in Cylindrical Coordinates Find the volume of the solid region Q cut from the sphere x 2 y 2 z 2 4 by the cylinder r 2 sin , as shown in Figure 14.65.
z 2
Because x 2 y 2 z 2 r 2 z 2 4, the bounds on z are
Solution
冪4 r 2 z 冪4 r 2. R x
3 y
Cylinder: r = 2 sin θ
Figure 14.65
Let R be the circular projection of the solid onto the r-plane. Then the bounds on R are 0 r 2 sin
3
and 0 .
So, the volume of Q is V
冕冕 冕 冕 冕 冕 冕 冕 冕 冕 冕
2 sin
0
冪4r2
0
兾2
2
0
0
2
4 3
r dz dr d
冪4r 2
冪4r 2
2 sin
r dz dr d
2r冪4 r 2 dr d
0
兾2
0
2 sin
0
兾2
2
冪4r2
兾2
冥
2 共4 r 2兲3兾2 3
2 sin 0
d
共8 8 cos3 兲 d
0
兾2
32 3
关1 共cos 兲共1 sin2 兲兴 d
0
32 sin3 sin 3 3
16 共3 4兲 9
冤
兾2
冥
0
⬇ 9.644.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1022
Chapter 14
Multiple Integration
Finding Mass in Cylindrical Coordinates 0≤z≤
Find the mass of the ellipsoid Q given by 4x 2 4y 2 z 2 16, lying above the xy-plane. The density at a point in the solid is proportional to the distance between the point and the xy-plane.
16 − 4r 2 z
The density function is 共r, , z兲 kz. The bounds on z are
Solution
4
0 z 冪16 4x 2 4y 2 冪16 4r 2 where 0 r 2 and 0 2, as shown in Figure 14.66. The mass of the solid is m
冕 冕冕 冕冕 冕冕 冕冤 冕 2
0
0
2
2
0
x
2 2 y
Ellipsoid: 4x 2 + 4y 2 + z 2 = 16
k 2
2
冪164r 2
0
dr d
共16r 4r 3兲 dr d
0
2
冥
8r 2 r 4
0
2
8k
Figure 14.66
kzr dz dr d
2
0
k 2
冥
z2r
0
0
k 2
冪164r 2
2
2 0
d
d
0
16k. Integration in cylindrical coordinates is useful when factors involving x 2 y 2 appear in the integrand, as illustrated in Example 3.
Finding a Moment of Inertia z
Find the moment of inertia about the axis of symmetry of the solid Q bounded by the paraboloid z x 2 y 2 and the plane z 4, as shown in Figure 14.67. The density at each point is proportional to the distance between the point and the z-axis.
5
Solution Because the z-axis is the axis of symmetry and 共x, y, z兲 k冪x 2 y 2, it follows that Iz
冕冕冕
k共x 2 y 2兲冪x 2 y 2 dV.
Q
In cylindrical coordinates, 0 r 冪x 2 y 2 冪z. So, you have −2
冕冕 冕 冕冕 冥 冕冕 冕 4
1 2 x
Figure 14.67
1
2
Q: Bounded by z = x2 + y2 z=4
y
Iz k
0
0
4
k
0
0
k 5
2
0
4
k
2
2
0
冪z
冪z
r5 5
0
d dz
z5兾2 d dz 5
4
z5兾2 共2兲 dz
0
2k 2 7兾2 z 5 7 512k . 35
r 2共r兲r dr d dz
0
冤
冥
4 0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.7
1023
Triple Integrals in Other Coordinates
Triple Integrals in Spherical Coordinates Triple integrals involving spheres or cones are often easier to evaluate by converting to spherical coordinates. Recall from Section 11.7 that the rectangular conversion equations for spherical coordinates are x sin cos y sin sin z cos .
REMARK The Greek letter used in spherical coordinates is not related to density. Rather, it is the three-dimensional analog of the r used in polar coordinates. For problems involving spherical coordinates and a density function, this text uses a different symbol to denote density.
In this coordinate system, the simplest region is a spherical block determined by
再共 , , 兲: 1 2, 1 2, 1 2冎 where 1 0, 2 1 2, and 0 1 2 , as shown in Figure 14.68. If 共 , , 兲 is a point in the interior of such a block, then the volume of the block can be approximated by V ⬇ 2 sin . (See Exercise 8 in the Problem Solving exercises at the end of this chapter.) Using the usual process involving an inner partition, summation, and a limit, you can develop a triple integral in spherical coordinates for a continuous function f defined on the solid region Q. This formula, shown below, can be modified for different orders of integration and generalized to include regions with variable boundaries.
冕冕冕
f 共x, y, z兲 dV
Q
冕冕冕 2
2
1
1
2
1
z
ρi sin φi Δθi
Δ ρi
ρi Δ φi
y x
Spherical block: Vi ⬇ 2i sin i i i i Figure 14.68
f 共 sin cos , sin sin , cos 兲 2 sin d d d
Like triple integrals in cylindrical coordinates, triple integrals in spherical coordinates are evaluated with iterated integrals. As with cylindrical coordinates, you can visualize a particular order of integration by viewing the iterated integral in terms of three sweeping motions—each adding another dimension to the solid. For instance, the iterated integral
冕冕 冕 2
兾4
0
0
3
2 sin d d d
0
(which is used in Example 4) is illustrated in Figure 14.69. Cone: x2 + y2 = z2
z
Sphere: x2 + y2 + z2 = 9 ρ =3
z
θ φ
ρ 1 −2 2
−2 1
2
y
z
2 x
1
2
y
−2 2
1
2
x
x
varies from 0 to 3 with and held constant. Figure 14.69
varies from 0 to 兾4 with held constant.
varies from 0 to 2.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
y
1024
Chapter 14
Multiple Integration
Finding Volume in Spherical Coordinates Upper nappe of cone: z2 = x2 + y2
Find the volume of the solid region Q bounded below by the upper nappe of the cone z 2 x 2 y 2 and above by the sphere x 2 y 2 z 2 9, as shown in Figure 14.70.
z
Solution 3
In spherical coordinates, the equation of the sphere is
x y2 z2 9 2
3.
2
Furthermore, the sphere and cone intersect when −3
−2 3
2
1
共x 2 y 2兲 z 2 共z 2兲 z 2 9 1
2
3
x
y
3 冪2
and, because z cos , it follows that
. 4
冢冪32冣冢13冣 cos
Sphere: x2 + y2 + z2 = 9
z
Consequently, you can use the integration order d d d, where 0 3, 0 兾4, and 0 2. The volume is
冕冕冕 冕 冕 冕 冕冕 冕 冥 冕冢 冣
Figure 14.70
2
dV
兾4
0
0
3
2 sin d d d
0
Q
2
9 sin d d
0
0
9
兾4
2
兾4
cos
9
2
1
冪2
2
0
d
0
0
d
9 共2 冪2 兲 ⬇ 16.563.
Finding the Center of Mass of a Solid Region See LarsonCalculus.com for an interactive version of this type of example.
Find the center of mass of the solid region Q of uniform density, bounded below by the upper nappe of the cone z 2 x 2 y 2 and above by the sphere x 2 y 2 z 2 9. Solution Because the density is uniform, you can consider the density at the point 共x, y, z兲 to be k. By symmetry, the center of mass lies on the z-axis, and you need only calculate z Mxy兾m, where m kV 9k 共2 冪2 兲 from Example 4. Because z cos , it follows that Mxy
冕冕冕
冕冕 冕 冕冕 冕冕 2
3
kz dV k
Q
k
0
0
3
2
0
k 4
兾4
3
0
3
0
共 cos 兲 2 sin d d d
0
2
0
sin2
2
兾4
冥
d d
0
3 d d
k 2
冕
3
3 d
0
81k . 8
So, z
Mxy 9共2 冪2 兲 81k兾8 ⬇ 1.920 m 16 冪 9k 共2 2 兲
and the center of mass is approximately 共0, 0, 1.92兲.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.7
14.7 Exercises
Triple Integrals in Other Coordinates
1025
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Evaluating an Iterated Integral In Exercises 1–6, evaluate
18. Solid inside x 2 y 2 z 2 16 and outside z 冪x 2 y 2
the iterated integral.
19. Solid bounded above by z 2x and below by z 2x2 2y2
冕冕 冕 冕 冕 冕 冕 冕冕 冕冕 冕 冕 冕 冕 兾2
5
1. 3.
0
兾2
2 cos 2
0
5.
rz dz dr d
0
3
22. Solid inside the sphere x 2 y 2 z 2 4 and above the upper nappe of the cone z 2 x 2 y 2
2 sin d d d
Mass In Exercises 23 and 24, use cylindrical coordinates to find the mass of the solid Q of density .
0
兾4
兾4
cos
0
23. Q 再共x, y, z兲: 0 z 9 x 2y, x 2 y 2 4冎
2 sin cos d d d
共x, y, z兲 k冪x 2 y 2
0
Approximating an Iterated Integral Using Technology In Exercises 7 and 8, use a computer algebra system to approximate the iterated integral.
冕冕冕 冕 冕冕 4
7.
0
8.
兾2
z
0
0
兾2
0
re r
sin
0
20. Solid bounded above by z 2 x2 y2 and below by z x2 y2 21. Solid bounded by the graphs of the sphere r 2 z 2 a 2 and the cylinder r a cos
r sin dz dr d
e 2 d d d
cos
0
0
0
6r
0
兾4
0
6.
2
0
2
6
0
0
2.
0
4r 2
0
兾2
4.
r cos dr d dz
1 0
冕 冕冕 兾4
3
d dr dz
24. Q 再共x, y, z兲: 0 z 12e共x 共x, y, z兲 k
z
共2 cos 兲 2 d d d
( (
z=h 1−
0
9.
冕 冕冕 冕冕 冕 冕 冕冕 0 2
11.
0 0 兾2 4
兾6
0
2
12.
0
0
冕冕冕 2
er 2
3
r dz dr d
10.
0
0
r dz dr d
2 sin d d d
0
15.
25. Find the volume of the cone.
2 sin d d d
26. Find the centroid of the cone.
2
冪4x2
4
x2 y2
16.
0
a
冪a2 x2
0
冪x 2 y 2 dz dy dx
0
冪a2 x2
冪9x2
0
a 冪a2 x2 y2
冪x 2
y2
z2
29. Assume that the cone has uniform density and show that the moment of inertia about the z-axis is 3 Iz 10 mr02.
x dz dy dx
a
冪9x2 y2
27. Find the center of mass of the cone, assuming that its density at any point is proportional to the distance between the point and the axis of the cone. Use a computer algebra system to evaluate the triple integral. 28. Find the center of mass of the cone, assuming that its density at any point is proportional to the distance between the point and the base. Use a computer algebra system to evaluate the triple integral.
x dz dy dx
冪16x2 y2
冪4x2
a
3
y
x
5
冕冕 冕 冕冕 冕 冕冕 冕 冕冕 冕 2
r0
0
2 冪4x2
14.
h
0
integral from rectangular coordinates to both cylindrical and spherical coordinates, and evaluate the simplest iterated integral. 2
r r0
5r 2
冪5
Converting Coordinates In Exercises 13–16, convert the
13.
4, x 0, y 0冎
Using Cylindrical Coordinates In Exercises 25–30, use cylindrical coordinates to find the indicated characteristic of the cone shown in the figure.
Volume In Exercises 9–12, sketch the solid region whose volume is given by the iterated integral, and evaluate the iterated integral. 兾2
兲, x 2 y 2
2 y 2
dz dy dx
30. Assume that the density of the cone is 共x, y, z兲 k冪x 2 y 2 and find the moment of inertia about the z-axis.
0
Volume In Exercises 17–22, use cylindrical coordinates to find the volume of the solid. 17. Solid inside both x 2 y 2 z 2 a 2 and 共x a兾2兲 2
y2
共a兾2兲
2
Moment of Inertia In Exercises 31 and 32, use cylindrical coordinates to verify the given formula for the moment of inertia of the solid of uniform density. 31. Cylindrical shell: Iz 12 m共a 2 b2兲 0 < a r b,
0 z h
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1026
Chapter 14
Multiple Integration 3
32. Right circular cylinder: Iz 2ma 2 r 2a sin , 0 z h Use a computer algebra system to evaluate the triple integral.
Volume In Exercises 33 –36, use spherical coordinates to find the volume of the solid. 33. Solid inside x2 y2 z2 9, outside z 冪x2 y2, and above the xy-plane 34. Solid bounded above by x2 y2 z2 z and below by z 冪x2 y2
WRITING ABOUT CONCEPTS
47. Using Coordinates Describe the surface whose equation is a coordinate equal to a constant for each of the coordinates in (a) the cylindrical coordinate system and (b) the spherical coordinate system.
48.
35. The torus given by 4 sin (Use a computer algebra system to evaluate the triple integral.) 36. The solid between the spheres x 2 y 2 z 2 a 2 and
( continued )
HOW DO YOU SEE IT? The solid is bounded below by the upper nappe of a cone and above by a sphere (see figure). Would it be easier to use cylindrical coordinates or spherical coordinates to find the volume of the solid? Explain. z
Upper nappe of cone: z2 = x2 + y2
x 2 y 2 z 2 b2, b > a,
2
and inside the cone z 2 x 2 y 2
Mass In Exercises 37 and 38, use spherical coordinates to find the mass of the sphere x 2 1 y 2 1 z 2 ⴝ a 2 with the given density.
2
y
2
x
37. The density at any point is proportional to the distance between the point and the origin.
Sphere: x2 + y2 + z2 = 4
38. The density at any point is proportional to the distance of the point from the z-axis.
Center of Mass In Exercises 39 and 40, use spherical coordinates to find the center of mass of the solid of uniform density. 39. Hemispherical solid of radius r 40. Solid lying between two concentric hemispheres of radii r and R, where r < R
PUTNAM EXAM CHALLENGE 49. Find the volume of the region of points 共x, y, z兲 such that 共x2 y2 z2 8兲2 36共x2 y2兲. This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
Moment of Inertia In Exercises 41 and 42, use spherical coordinates to find the moment of inertia about the z-axis of the solid of uniform density.
41. Solid bounded by the hemisphere cos , , 4 2 and the cone 4 42. Solid lying between two concentric hemispheres of radii r and R, where r < R
Wrinkled and Bumpy Spheres In parts (a) and (b), find the volume of the wrinkled sphere or bumpy sphere. These solids are used as models for tumors. (a) Wrinkled sphere
(b) Bumpy sphere
1 0.2 sin 8 sin
1 0.2 sin 8 sin 4
0 2, 0
0 2, 0
z
z
WRITING ABOUT CONCEPTS 43. Coordinate Conversion Give the equations for conversion from rectangular to cylindrical coordinates and vice versa. 44. Coordinate Conversion Give the equations for conversion from rectangular to spherical coordinates and vice versa. 45. Cylindrical Form Give the iterated form of the triple integral 兰兰兰 f 共x, y, z兲 dV in cylindrical form. Q
46. Spherical Form Give the iterated form of the triple integral 兰兰兰 f 共x, y, z兲 dV in spherical form. Q
y
y x
x Generated by Maple
Generated by Maple
FOR FURTHER INFORMATION For more information on these types of spheres, see the article “Heat Therapy for Tumors” by Leah Edelstein-Keshet in The UMAP Journal.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.8
Change of Variables: Jacobians
1027
14.8 Change of Variables: Jacobians Understand the concept of a Jacobian. Use a Jacobian to change variables in a double integral.
Jacobians For the single integral
冕
b
f 共x兲 dx
a
you can change variables by letting x g共u兲, so that dx g 共u兲 du, and obtain
冕
冕
b
d
f 共x兲 dx
a
f 共 g 共u兲兲g 共u兲 du
c
where a g共c兲 and b g共d 兲. Note that the change of variables process introduces an additional factor g 共u兲 into the integrand. This also occurs in the case of double integrals
冕冕
f 共x, y兲 dA
R
冕冕
ⱍ
f 共 g共u, v兲, h 共u, v兲兲
S
Jacobian
CARL GUSTAV JACOBI (1804 –1851)
The Jacobian is named after the German mathematician Carl Gustav Jacobi. Jacobi is known for his work in many areas of mathematics, but his interest in integration stemmed from the problem of finding the circumference of an ellipse. See LarsonCalculus.com to read more of this biography.
ⱍ
x y y x du dv u v u v
where the change of variables x g共u, v兲 and
y h共u, v兲
introduces a factor called the Jacobian of x and y with respect to u and v. In defining the Jacobian, it is convenient to use the determinant notation shown below. Definition of the Jacobian If x g共u, v兲 and y h共u, v兲, then the Jacobian of x and y with respect to u and v, denoted by 共x, y兲兾共u, v兲, is
ⱍ ⱍ
x v x y y x . y u v u v v
x u 共x, y兲 y 共u, v兲 u
The Jacobian for Rectangular-to-Polar Conversion Find the Jacobian for the change of variables defined by x r cos Solution
and
ⱍⱍ ⱍ
y r sin .
From the definition of the Jacobian, you obtain
x 共x, y兲 r 共r, 兲 y r
x y
r sin r cos r cos 2 r sin2 r.
cos sin
ⱍ
Interfoto/Personalities/Alamy
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1028
Chapter 14
θ
Multiple Integration
Example 1 points out that the change of variables from rectangular to polar coordinates for a double integral can be written as
T(r, θ ) = (r cos θ, r sin θ)
冕冕
θ =β
β
S
r=a α
f 共x, y兲 dA
R
r=b
ⱍ ⱍ
f 共r cos , r sin 兲
S
a
r
b
y
θ =β
r=b
共x, y兲 dr d 共r, 兲
where S is the region in the r-plane that corresponds to the region R in the xy-plane, as shown in Figure 14.71. This formula is similar to that found in Theorem 14.3 on page 988. In general, a change of variables is given by a one-to-one transformation T from a region S in the uv-plane to a region R in the xy-plane, to be given by T 共u, v兲 共x, y兲 共 g共u, v兲, h 共u, v兲兲
R
r=a
f 共r cos , r sin 兲 r dr d, r > 0
S
θ =α
冕冕 冕冕
θ =α x
S in the region in the r-plane that corresponds to R in the xy-plane. Figure 14.71
where g and h have continuous first partial derivatives in the region S. Note that the point 共u, v兲 lies in S and the point 共x, y兲 lies in R. In most cases, you are hunting for a transformation in which the region S is simpler than the region R.
Finding a Change of Variables to Simplify a Region Let R be the region bounded by the lines x 2y 0, x 2y 4, x y 4, and
y
x+
y=
y=
y=
4
x+
3
2 x−
1
)
− 23 , 53
)
) 43, 83 ) x−
R
as shown in Figure 14.72. Find a transformation T from a region S to R such that S is a rectangular region (with sides parallel to the u- or v-axis).
−4
2y
=0
) 83, 43 )
1
x −2
1 −1
2
3
Solution To begin, let u x y and v x 2y. Solving this system of equations for x and y produces T 共u, v兲 共x, y兲, where 1 x 共2u v兲 and 3
1 y 共u v兲. 3
The four boundaries for R in the xy-plane give rise to the following bounds for S in the uv-plane.
) ) 2 1 , 3 3
−2
Bounds in the xy-Plane Region R in the xy-plane Figure 14.72 v
v=0 −1
u=1 (1 , 0)
u=4 (4, 0) u
2
3
−1 −2
xy1
x x x x
y1 y4 2y 0 2y 4
−3
u1 u4 v0 v 4
The region S is shown in Figure 14.73. Note that the transformation T共u, v兲 共x, y兲
S
Bounds in the uv-Plane
冢13 关2u v兴, 13 关u v兴冣
maps the vertices of the region S onto the vertices of the region R. For instance,
v = −4 (1, − 4) −5
Region S in the uv-plane Figure 14.73
(4, − 4)
冢13 关2共1兲 0兴 , 13 关1 0兴冣 冢23, 13冣 1 1 8 4 T 共4, 0兲 冢 关2共4兲 0兴 , 关4 0兴冣 冢 , 冣 3 3 3 3 1 1 4 8 T 共4, 4兲 冢 关2共4兲 4兴 , 关4 共4兲兴冣 冢 , 冣 3 3 3 3 1 1 2 5 T 共1, 4兲 冢 关2共1兲 4兴 , 关1 共4兲兴冣 冢 , 冣. 3 3 3 3 T 共1, 0兲
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.8
Change of Variables: Jacobians
1029
Change of Variables for Double Integrals THEOREM 14.5 Change of Variables for Double Integrals Let R be a vertically or horizontally simple region in the xy-plane, and let S be a vertically or horizontally simple region in the uv-plane. Let T from S to R be given by T共u, v兲 共x, y兲 共g共u, v兲, h共u, v兲兲, where g and h have continuous first partial derivatives. Assume that T is one-to-one except possibly on the boundary of S. If f is continuous on R, and 共x, y兲兾共u, v兲 is nonzero on S, then
冕冕
冕冕
f 共x, y兲 dx dy
R
v
(u, v + Δv)
S
共x, y兲 du dv. 共u, v兲
Proof Consider the case in which S is a rectangular region in the uv-plane with vertices 共u, v兲, 共u u, v兲, 共u u, v v兲, and 共u, v v兲, as shown in Figure 14.74. The images of these vertices in the xy-plane are shown in Figure 14.75. If u and v are small, then the continuity of g and h implies that R is approximately a parallelogram determined by the vectors MN and MQ . So, the area of R is
(u + Δu, v + Δv)
S
\
\
A ⬇ 储 MN
(u + Δu, v)
(u, v)
ⱍ ⱍ
f 共 g 共u, v兲, h共u, v兲兲
u
\
\
MQ 储 .
Moreover, for small u and v, the partial derivatives of g and h with respect to u can be approximated by
Area of S u v u > 0, v > 0 Figure 14.74
gu共u, v兲 ⬇
g共u u, v兲 g共u, v兲 u
hu共u, v兲 ⬇
and
h共u u, v兲 h共u, v兲 . u
Consequently, \
y
Q
P R
MN 关 g共u u, v兲 g共u, v兲兴 i 关h共u u, v兲 h共u, v兲兴 j ⬇ 关 gu共u, v兲 u兴 i 关hu共u, v兲 u兴 j x y ui uj. u u x y vi vj, which implies that Similarly, you can approximate MQ by v v k i j x y
ⱍ ⱍⱍ \
M = (x, y)
N x
x = g(u, v) y = h(u, v)
The vertices in the xy-plane are M共g共u, v兲, h共u, v兲兲, N共g共u u, v兲, h共u u, v兲兲, P共g共u u, v v兲, h共u u, v v兲兲, and Q共g共u, v v兲, h共u, v v兲兲. Figure 14.75
\
MN
y u u y v v
x MQ ⬇ u u x v v \
u x v
0 0
ⱍ
u u vk. y v
It follows that, in Jacobian notation, \
A ⬇ 储 MN
\
MQ 储 ⬇
ⱍ ⱍ
共x, y兲 u v. 共u, v兲
Because this approximation improves as u and v approach 0, the limiting case can be written as \
dA ⬇ 储 MN So,
冕冕
\
MQ 储 ⬇
f 共x, y兲 dx dy
R
ⱍ ⱍ
共x, y兲 du dv. 共u, v兲
冕冕
ⱍ ⱍ
f 共g共u, v兲, h共u, v兲兲
S
共x, y兲 du dv. 共u, v兲
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1030
Chapter 14
Multiple Integration
The next two examples show how a change of variables can simplify the integration process. The simplification can occur in various ways. You can make a change of variables to simplify either the region R or the integrand f 共x, y兲, or both.
Using a Change of Variables to Simplify a Region See LarsonCalculus.com for an interactive version of this type of example. y
x+ y=
y=
−4
x−
2y
Let R be the region bounded by the lines x 2y 0, x 2y 4, x y 4, and
4
x+
3
= 2y x−
1
R
=0
1 x −2
1 −1 −2
Figure 14.76
2
3
xy1
as shown in Figure 14.76. Evaluate the double integral
冕冕
3xy dA.
R
Solution
From Example 2, you can use the following change of variables.
1 x 共2u v兲 3
and y
1 共u v兲 3
The partial derivatives of x and y are x 2 , u 3
x 1 , v 3
y 1 , and u 3
y 1 v 3
ⱍ ⱍ ⱍⱍ
which implies that the Jacobian is x 共x, y兲 u 共u, v兲 y u
x v y v
2 1 3 3 1 1 3 3
2 1 9 9 1 . 3
So, by Theorem 14.5, you obtain
冕冕
3xy dA
R
冕冕 冤 冕冕 冕冤 冕冢 3
S
4
1
冥
0
1 共2u 2 uv v 2兲 dv du 4 9
4
1 9
ⱍ ⱍ
1 1 共x, y兲 共2u v兲 共u v兲 dv du 3 3 共u, v兲
2u 2v
1
4
1 9
8u 2 8u
1
冤
冥
uv 2 v 3 2 3
4
du
冣
64 du 3
冥
1 8u 3 64 4u 2 u 9 3 3
0
4 1
164 . 9
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14.8
Change of Variables: Jacobians
1031
Change of Variables: Simplifying an Integrand Let R be the region bounded by the square with vertices 共0, 1兲, 共1, 2兲, 共2, 1兲, and 共1, 0兲. Evaluate the integral
冕冕
共x y兲 2 sin2共x y兲 dA.
R
Solution Note that the sides of R lie on the lines x y 1, x y 1, x y 3, and x y 1, as shown in Figure 14.77. Letting u x y and v x y, you can determine the bounds for region S in the uv-plane to be
−1
y
1
x−
y=
3
1 u 3
x−
y=
(1, 2)
2
as shown in Figure 14.78. Solving for x and y in terms of u and v produces
x+
R
(0, 1)
y=
(2, 1)
3 y=
(1, 0)
1 x 共u v兲 2
x
3
x+
−1
1 and y 共u v兲. 2
The partial derivatives of x and y are
1
−1
and 1 v 1
x 1 , u 2
Region R in the xy-plane Figure 14.77
x 1 , v 2
y 1 , and u 2
ⱍ ⱍⱍ ⱍ
y 1 v 2
which implies that the Jacobian is v
1
u=1
(3, 1)
(1, 1) v=1
x 共x, y兲 u 共u, v兲 y u
u=3
S u 1
−1
2
3
v = −1 (1, − 1)
(3, − 1)
x v y v
1 2 1 2
1 2 1 2
By Theorem 14.5, it follows that
冕冕
1 1 1 . 4 4 2
冕冕 冕 冕 冕 1
共x y兲 2 sin2共x y兲 dA
R
Region S in the uv-plane Figure 14.78
3
u 2 sin2 v
1 1
1 2
1
1
共sin2 v兲
冢12冣 du dv
u3 3
冥
3
dv 1
1
13 3
sin2v dv
1 1
13 6
1
冤 冤
共1 cos 2v兲 dv
冥
1 13 1 v sin 2v 6 2 1 13 1 1 2 sin 2 sin共2兲 6 2 2 13 共2 sin 2兲 6 ⬇ 2.363.
冥
In each of the change of variables examples in this section, the region S has been a rectangle with sides parallel to the u- or v-axis. Occasionally, a change of variables can be used for other types of regions. For instance, letting T 共u, v兲 共x, 12 y兲 changes the circular region u 2 v 2 1 to the elliptical region x2
y2 1. 4
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1032
Chapter 14
Multiple Integration
14.8 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
(− 1, 0)
2. x au bv, y cu dv
−1
17.
冕冕
Using a Transformation In Exercises 9–12, sketch the
y
18.
冕冕
y
y
(− 1, 1)
6 4
(3, 3)
2
R
6
−2
5 4
2
(6, 3)
3
R
2
(3, 0)
19.
(2, 2)
R
1
x
(4, 1) x
3
(0, 0) (4, 0) 6
冕冕
8
−1
exy兾2 dA
x
(0, 0) 2 3 4 5 6
20.
冕冕
11. x y
v兲
12. x
v兲
y
冪uv, y
u x , yv v
冪uv
y
3
y= 2
y
2
3
) 32, 32 )
R
)
− 13 ,
1
) ) 1 1 , 2 2
1
4 3
2
)
R
− 2 −1 −1
x 2
) 23, 103) ) 43, 83 )
) 13, 23 ) 1
2
2
y=
x 3
1 x 4
3
y=
1 x
13. Example 3
22. f 共x, y兲 共3x 2y兲2冪2y x
14. Example 4
Evaluating a Double Integral Using a Change of Variables In Exercises 15–20, use the indicated change of
冕冕 R
x y
16.
冕冕
60xy dA
R
1 2 共u 1 2 共u
v兲
x 12共u v兲
v兲
y 12共u v兲
x 1
2
3
4
Exercises 21–28, use a change of variables to find the volume of the solid region lying below the surface z ⴝ f 冇x, y冈 and above the plane region R. 21. f 共x, y兲 48xy
4共x2 y 2兲 dA
y=1
Finding Volume Using a Change of Variables In
verify the result of the indicated example by setting up the integral using dy dx or dx dy for dA. Then use a computer algebra system to evaluate the integral.
15.
R
x
Verifying a Change of Variables In Exercises 13 and 14,
variables to evaluate the double integral.
xy = 4
3
R
1 4
(0, 1) (1, 2)
4 x
y=4
xy = 1
y = 2x
1 3 共v u兲 1 3 共2v u兲
y
y sin xy dA
R
y 1 2 共u 1 2 共u
x
(0, 0) 1
−1
R
R
(1, 1)
(7, 3)
x
(2, 3)
4共x y兲exy dA
R
y 12共u v兲
y
2
y 共x y兲 dA
yu
y 13 共u v兲
(0, 0)
Figure for 16
x 12共u v兲
10. x 13 共4u v兲
y 3v
2
xuv
image S in the uv-plane of the region R in the xy-plane using the given transformations. 9. x 3u 2v
x
(1, 0) 1
R
8. x u兾v, y u v
R
(0, −1)
Figure for 15
7. x eu sin v, y eu cos v
1
1
4. x uv 2u, y uv 6. x u a, y v a
(0, 1) (1, 2) (2, 1)
x
5. x u cos v sin , y u sin v cos
2
(1, 0)
R
−1
3. x u v 2, y u v
1
(0, 1)
1
1. x 12共u v兲, y 12共u v兲
3
y
y
Finding a Jacobian In Exercises 1–8, find the Jacobian 冇x, y冈 / 冇u, v冈 for the indicated change of variables.
R: region bounded by the square with vertices 共1, 0兲, 共0, 1兲, 共1, 2兲, 共2, 1兲 R: region bounded by the parallelogram with vertices 共0, 0兲, 共2, 3兲, 共2, 5兲, 共4, 2兲 23. f 共x, y兲 共x y兲exy R: region bounded by the square with vertices 共4, 0兲, 共6, 2兲, 共4, 4兲, 共2, 2兲 24. f 共x, y兲 共x y兲2 sin2共x y兲 R: region bounded by the square with vertices 共 , 0兲, 共3 兾2, 兾2兲, 共 , 兲, 共 兾2, 兾2兲
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.8 25. f 共x, y兲 冪共x y兲共x 4y兲 R: region bounded by the parallelogram with vertices 共0, 0兲, 共1, 1兲, 共5, 0兲, 共4, 1兲 26. f 共x, y兲 共3x 2y兲共2y x兲3兾2 R: region bounded by the parallelogram with vertices 共0, 0兲, 共2, 3兲, 共2, 5兲, 共4, 2兲 27. f 共x, y兲 冪x y
xy 28. f 共x, y兲 1 x 2y 2
29. Using a Transformation The substitutions u 2x y and v x y make the region R (see figure) into a simpler region S in the uv-plane. Determine the total number of sides of S that are parallel to either the u-axis or the v-axis. y
(2, 7)
4
R
2
4
6
(0, 6)
(− 2, 6)
(6, 4)
S (− 2, 2)
(4, 2) x 5
1
33. Jacobian
State the definition of the Jacobian.
34. Change of Variables Describe how to use the Jacobian to change variables in double integrals.
6
y ⴝ g冇u, v, w冈, and z ⴝ h冇u, v, w冈
u −5 − 4 − 3 −2 −1
x v y v z v
x w y w . z w
1 1 37. x 2共u v兲, y 2共u v兲, z 2uvw
(0, 2) 1
(1, 1) 4
冣
36. x 4u v, y 4v w, z u w
3
R
3
y2 b2
35. x u共1 v兲, y uv共1 w兲, z uvw
5
(3, 3)
2
WRITING ABOUT CONCEPTS
x u y 冇x, y, z冈 ⴝ u 冇u, v, w冈 z u
v
y
1
b2
2
then the Jacobian of x, y, and z with respect to u, v, and w is
transformed into a simpler region S (see figure). Which substitution can be used to make the transformation?
1
y2
2
ⱍ ⱍ
8
HOW DO YOU SEE IT? The region R is
2
a2
x ⴝ f 冇u, v, w冈 ,
(0, 0) 2
3
x2
2
for the indicated change of variables. If
(6, 3) x
4
冢 冪ax
(b) f 共x, y兲 A cos
冇x, y, z冈 冇u, v, w冈
6
5
x2 y2
1 16 9
Finding a Jacobian In Exercises 35–40, find the Jacobian
8
30.
(a) f 共x, y兲 16 x 2 y 2
R:
R: region bounded by the graphs of xy 1, xy 4, x 1, x 4 共Hint: Let x u, y v兾u.兲
1033
32. Volume Use the result of Exercise 31 to find the volume of each dome-shaped solid lying below the surface z f 共x, y兲 and above the elliptical region R. (Hint: After making the change of variables given by the results in Exercise 31, make a second change of variables to polar coordinates.)
R:
R: region bounded by the triangle with vertices 共0, 0兲, 共a, 0兲, 共0, a兲, where a > 0
Change of Variables: Jacobians
38. x u v w, y 2uv, z u v w
1 2
(a) u 3y x, v y x (b) u y x, v 3y x
39. Spherical Coordinates x sin cos , y sin sin , z cos 40. Cylindrical Coordinates
31. Using an Ellipse Consider the region R in the xy-plane bounded by the ellipse x2 y2 21 2 a b and the transformations x au and y bv. (a) Sketch the graph of the region R and its image S under the given transformation. 共x, y兲 (b) Find . 共u, v兲
x r cos , y r sin , z z
PUTNAM EXAM CHALLENGE 41. Let A be the area of the region in the first quadrant bounded by the line y 12 x, the x-axis, and the ellipse 1 2 2 9 x y 1. Find the positive number m such that A is equal to the area of the region in the first quadrant bounded by the line y mx, the y-axis, and the ellipse 19 x2 y2 1. This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
(c) Find the area of the ellipse.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1034
Chapter 14
Multiple Integration
Review Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Evaluating an Integral In Exercises 1 and 2, evaluate the
Finding Volume In Exercises 17–20, use a double integral
integral.
to find the volume of the indicated solid.
冕
2x
1.
冕
2y
xy3 dy
2.
共x 2 y 2兲 dx
z
17.
y
0
z
18. z=4
5
4
z =5−x
Evaluating an Iterated Integral In Exercises 3–6, evaluate
y=x
the iterated integral.
冕冕 冕冕 冕冕 冕冕 1
3.
0
1x
2
4.
0
0
0
y
3
共x 2 2y兲 dy dx
冪9x2
z
19.
4x dy dx
4
0
共9 3x2 3y2兲 dx dy
9. y x, y 2x 2, x 0, x 4 10. x y 2 1, x 0, y 0, y 2
2
2 x
sketch the region R whose area is given by the iterated integral. Then switch the order of integration and show that both orders yield the same area.
冕冕 冕冕 冕冕 冕冕 4
5
−1 ≤ x ≤ 1 −1 ≤ y ≤ 1
3
x
0
4
13.
3
R: rectangle with vertices 共2, 2兲, 共2, 2兲, 共2, 2兲, 共2, 2兲 22. f 共x兲 2x2 y2 R: square with vertices 共0, 0兲, 共3, 0兲, 共3, 3兲, 共0, 3兲
6x
T共x, y兲 40 6x2 y2
dy dx
0
where x and y are measured in centimeters. Estimate the average temperature when x varies between 0 and 3 centimeters and y varies between 0 and 5 centimeters.
8
2x
3
9y2
dx dy
3 0
24. Average Profit A firm’s profit P from marketing two soft drinks is P 192x 576y x2 5y2 2xy 5000
Evaluating a Double Integral In Exercises 15 and 16, set up integrals for both orders of integration. Use the more convenient order to evaluate the integral over the region R. 15.
冕冕
where x and y represent the numbers of units of the two soft drinks. Estimate the average weekly profit when x varies between 40 and 50 units and y varies between 45 and 60 units.
4xy dA
R
R: rectangle with vertices 共0, 0兲, 共0, 4兲, 共2, 4兲, 共2, 0兲 16.
x
21. f 共x兲 16 x2 y2
dy dx
0
14.
冕冕 6
dy dx
y
2
23. Average Temperature The temperature in degrees Celsius on the surface of a metal plate is
dx dy
1
2
y
Average Value In Exercises 21 and 22, find the average value of f 冇x, y冈 over the plane region R.
Switching the Order of Integration In Exercises 11–14,
0
x+y+z=2
0
8. y 6x x 2, y x 2 2x
12.
z 2
7. x 3y 3, x 0, y 0
2
x=2
20.
2y
Finding the Area of a Region In Exercises 7–10, use an iterated integral to find the area of the region bounded by the graphs of the equations.
11.
z = 4 − x2 − y2
x
y
2
2
0≤x≤3 0≤y≤2
x
x2
1
6.
2
2x
3
5.
共3x 2y兲 dy dx
0
冕冕
Converting to Polar Coordinates In Exercises 25 and 26, evaluate the iterated integral by converting to polar coordinates.
6x dA
R
冕冕 冕冕 h
2
25.
0
R: region bounded by y 0, y 冪x, x 1
4
26.
0
x
冪x 2 y 2 dy dx
0 冪16y2
共x 2 y 2兲 dx dy
0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Review Exercises
1035
polar coordinates to find the volume of the solid bounded by the graphs of the equations.
Finding Moments of Inertia and Radii of Gyration In Exercises 39 and 40, find Ix, Iy, I0, x, and y for the lamina bounded by the graphs of the equations.
27. z xy2, x2 y2 9, first octant
39. y 0, y b, x 0, x a, kx
Volume In Exercises 27 and 28, use a double integral in
28. z 冪25
x2
y2,
z 0,
x2
y2
40. y 4 x 2, y 0, x > 0, ky
16
Area In Exercises 29 and 30, use a double integral to find the area of the shaded region. π 2
29.
π 2
30. r = 2 + cos θ
r = 2 sin 2θ
2
41. f 共x, y兲 25 x 2 y 2 R 再共x, y兲: x 2 y 2 25冎
0
0 1
Finding Surface Area In Exercises 41– 44, find the area of the surface given by z ⴝ f 冇x, y冈 over the region R. 冇Hint: Some of the integrals are simpler in polar coordinates.冈
4
42. f 共x, y兲 8 4x 5y R 再共x, y兲: x2 y2 1冎
2
43. f 共x, y兲 9 y2
Area In Exercises 31 and 32, sketch a graph of the region bounded by the graphs of the equations. Then use a double integral to find the area of the region.
R: triangle bounded by the graphs of the equations y x, y x, and y 3 44. f 共x, y兲 4 x2 R: triangle bounded by the graphs of the equations y x, y x, and y 2
31. Inside the cardioid r 2 2 cos and outside the circle r 3 32. Inside the circle r 3 sin and outside the cardioid r 1 sin
45. Building Design A new auditorium is built with a foundation in the shape of one-fourth of a circle of radius 50 feet. So, it forms a region R bounded by the graph of
33. Area and Volume Consider the region R in the xy-plane bounded by the graph of the equation
x2 y2 502
共x 2 y 2兲2 9共x 2 y 2兲.
with x 0 and y 0. The following equations are models for the floor and ceiling.
(a) Convert the equation to polar coordinates. Use a graphing utility to graph the equation.
Floor: z
(b) Use a double integral to find the area of the region R.
Ceiling: z 20
(c) Use a computer algebra system to determine the volume of the solid over the region R and beneath the hemisphere z 冪9 x 2 y 2 . 34. Converting to Polar Coordinates Combine the sum of the two iterated integrals into a single iterated integral by converting to polar coordinates. Evaluate the resulting iterated integral.
冕 冕 8兾冪13
0
3x兾2
0
xy dy dx
冕 冕 4
8兾冪13
xy 5 xy 100
(a) Calculate the volume of the room, which is needed to determine the heating and cooling requirements. (b) Find the surface area of the ceiling. 46. Surface Area The roof over the stage of an open air theater at a theme park is modeled by
冤
冪16x2
f 共x, y兲 25 1 e共x
xy dy dx
2 y 2兲兾1000
y 冢x 1000 冣冥 2
cos 2
2
0
where the stage is a semicircle bounded by the graphs of y 冪50 2 x 2 and y 0.
Finding the Center of Mass In Exercises 35–38, find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density. 共Hint: Some of the integrals are simpler in polar coordinates.兲
(a) Use a computer algebra system to graph the surface. (b) Use a computer algebra system to approximate the number of square feet of roofing required to cover the surface.
35. y x3, y 0, x 2, kx
Evaluating a Triple Iterated Integral In Exercises 47–50,
2 36. y , y 0, x 1, x 2, ky x
evaluate the triple iterated integral.
37. y 2x, y 2x3, x 0, y 0, kxy
47.
38. y 6 x, y 0, x 0, kx2
冕冕冕 冕冕冕 4
0
2
48.
0
1
0
y
0
2
共2x y 4z兲 dy dz dx
0 xy
y dz dx dy
0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1036
Chapter 14
冕冕冕 冕冕 冕 a
49.
0
3
50.
0
b
Multiple Integration
c
0
共x 2 y 2 z 2兲 dx dy dz
0
5
Approximating an Iterated Integral Using Technology In Exercises 63 and 64, use a computer algebra system to approximate the iterated integral.
冕 冕冕 冕 冕 冕
z sin x dy dx dz
兾2 2
63.
2
0
0
Approximating a Triple Iterated Integral Using Technology In Exercises 51 and 52, use a computer algebra
冕冕 冕 冕冕 冕
1 冪1x2 冪1x2 y2 2
52.
冪1x2 y2
冪1x2
1
0
共x 2 y 2兲 dz dy dx
冪4x2 y2
冪4x2
0
xyz dz dy dx
0
Volume In Exercises 53 and 54, use a triple integral to find the volume of the solid bounded by the graphs of the equations. 53. z xy, z 0, 0 x 3, 0 y 4
0
兾2
64.
0
system to approximate the iterated integral. 51.
3
冪z 2 4 dz dr d
兾2
0
cos
2 cos d d d
0
65. Volume Use cylindrical coordinates to find the volume of the solid bounded above by z 8 x2 y2 and below by z x2 y2. 66. Volume Use spherical coordinates to find the volume of the solid bounded above by x2 y2 z2 36 and below by z 冪x2 y2.
Finding a Jacobian In Exercises 67–70, find the Jacobian 冇x, y冈 / 冇u, v冈 for the indicated change of variables.
54. z 8 x y, z 0, y x, y 3, x 0
67. x u 3v, y 2u 3v
Changing the Order of Integration In Exercises 55 and
69. x u sin v cos , y u cos v sin
56, sketch the solid whose volume is given by the iterated integral and rewrite the integral using the indicated order of integration.
冕冕冕 1
55.
0
冪1x2
y
0
dz dx dy
0
Rewrite using the order dz dy dx.
冕冕 冕 6
56.
0
6x
0
68. x u2 v2, y u2 v2
70. x uv, y
Evaluating a Double Integral Using a Change of Variables In Exercises 71–74, use the indicated change of variables to evaluate the double integral.
6xy
dz dy dx
71.
冕冕
y 5
R
2
0
兾2
60.
0
兾2
61.
0
兾4
62.
0
0
1
兾2
0
兾4
0
73.
2
3
共xy x2兲 dA
R
0 cos
6
(4, 4)
0
v u
x=1
4 3
R
2 1
4
5
(1, 3)
3
cos d d d
3
x dA 1 x2y2
x u, y
5
d d d
2
y
2
4
1
冕冕
y
z dr dz d 2
x
74.
1 x u, y 共u v兲 3
(2, 2) (1, 0)
−1
4
冕冕
4z
0
R
1
(2, 1)
R
r cos dr d dz
0
3
(0, 2)
x
Evaluating an Iterated Integral In Exercises 59–62, evaluate
0
3
(3, 2)
(1, 2) 1
(1, 4)
4
3
Q: z 5 y, z 0, y 0, x 0, x 5
4
1 1 x 共u v兲, y 共v u兲 4 2
(2, 3)
58. Find y using 共x, y, z兲 kx.
冕冕 冕 冕 冕冕 冕 冕 冕 冕 冕 冕
16xy dA
R
4
Q: x y z 10, x 0, y 0, z 0
the iterated integral.
冕冕
y
57. Find x using 共x, y, z兲 k.
兾3
72.
1 1 x 共u v兲, y 共u v兲 2 2
Mass and Center of Mass In Exercises 57 and 58, find the mass and the indicated coordinates of the center of mass of the solid region Q of density bounded by the graphs of the equations.
3
ln共x y兲 dA
R
0
Rewrite using the order dy dx dz.
59.
v u
xy = 5
2
(4, 2)
1
(1, 1)
R
x=5 x
x
1
2
3
4
5
1 xy = 1 4
5
6
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P.S. Problem Solving
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
P.S. Problem Solving 1. Volume Find the volume of the solid of intersection of the three cylinders x 2 z 2 1, y 2 z 2 1, and x 2 y 2 1 (see figure). z
z
5. Deriving a Sum Derive Euler’s famous result that was mentioned in Section 9.3,
1
兺n
2
n1
3
3
2 6
by completing each step.
冕
1 dv v C. arctan 2 u 2 v 2 冪2 u 2 冪2 u 2
(a) Prove that −3
−3 x
3
y
3
3
3
x
y
(b) Prove that
冕 冕 冪2兾2
I1
u
2 2 dv du 2 2 18 u 2 u v
0
−3
1037
−3
by using the substitution u 冪2 sin . (c) Prove that
2. Surface Area Let a, b, c, and d be positive real numbers. The first octant of the plane ax by cz d is shown in the figure. Show that the surface area of this portion of the plane is equal to
冕 冕 冕
u 冪2
冪2
I2
4
A共R兲 冪a 2 b 2 c 2 c
u 冪2
冪2兾2
兾2
arctan
兾6
2 dv du 2 u2 v2
1 sin d cos
by using the substitution u 冪2 sin .
where A共R兲 is the area of the triangular region R in the xy-plane, as shown in the figure.
(d) Prove the trigonometric identity 1 sin 共 兾2兲 tan . cos 2
冢
z
冕 冕 冪2
(e) Prove that I2
冪2兾2
冣
u 冪2
u 冪2
2 2 dv du . 2 u2 v2 9
(f) Use the formula for the sum of an infinite geometric series to verify that
R
1 2 n n1
兺
y
x
冕冕 1
0
1
0
1 dx dy. 1 xy
(g) Use the change of variables
3. Using a Change of Variables region R bounded by the curves y 冪x, y 冪2x, y
The figure shows the
u
x2 x2 , and y . 3 4
xy 冪2
and
v
yx 冪2
to prove that
1 2 n n1
兺
Use the change of variables x u1兾3 v2兾3 and y u2兾3 v1兾3 to find the area of the region R.
冕冕 1
0
1
0
1 2 dx dy I1 I2 . 1 xy 6
6. Evaluating a Double Integral Evaluate the integral
y
y=
y = 14 x 2
1 2 x 3
冕冕
0
y=
2x
0
1 dx dy. 共1 x2 y2兲2
7. Evaluating Double Integrals Evaluate the integrals R
y=
x
x
冕冕 1
4. Proof
Prove that lim
n→
0
1
0
1
0
xy dx dy and 共x y兲3
冕冕 1
0
1
0
xy dy dx. 共x y兲3
Are the results the same? Why or why not?
1
0
冕冕
x n y n dx dy 0.
8. Volume Show that the volume of a spherical block can be approximated by V ⬇ 2 sin .
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1038
Chapter 14
Multiple Integration
Evaluating an Integral In Exercises 9 and 10, evaluate the integral. (Hint: See Exercise 65 in Section 14.3.)
冕 冕冪
9.
x 2ex dx 2
0
1
10.
(a) z1 2 x
1 ln dx x
0
(b) z2 5 (c) z3 10 5x 9y
11. Joint Density Function Consider the function f 共x, y兲
ke共xy兲兾a,
冦0,
x 0, y 0 elsewhere.
Find the relationship between the positive constants a and k such that f is a joint density function of the continuous random variables x and y. 12. Volume Find the volume of the solid generated by revolving 2 the region in the first quadrant bounded by y ex about the y-axis. Use this result to find
冕
15. Surface Area Use the result of Exercise 14 to order the planes in ascending order of their surface areas for a fixed region R in the xy-plane. Explain your ordering without doing any calculations.
ex dx. 2
(d) z4 3 x 2y 16. Sprinkler Consider a circular lawn with a radius of 10 feet, as shown in the figure. Assume that a sprinkler distributes water in a radial fashion according to the formula f 共r兲
r r2 16 160
(measured in cubic feet of water per hour per square foot of lawn), where r is the distance in feet from the sprinkler. Find the amount of water that is distributed in 1 hour in the following two annular regions. A 再共r, 兲: 4 r 5, 0 2 }
13. Volume and Surface Area From 1963 to 1986, the volume of the Great Salt Lake approximately tripled while its top surface area approximately doubled. Read the article “Relations between Surface Area and Volume in Lakes” by Daniel Cass and Gerald Wildenberg in The College Mathematics Journal. Then give examples of solids that have “water levels” a and b such that V共b兲 3V共a兲 and A共b兲 2A共a兲 (see figure), where V is volume and A is area.
B 再共r, 兲: 9 r 10, 0 2 } Is the distribution of water uniform? Determine the amount of water the entire lawn receives in 1 hour.
1 ft B
A
A(b) 4 ft
V(b)
A(a) V(a)
17. Changing the Order of Integration Sketch the solid whose volume is given by the sum of the iterated integrals
冕冕 冕 6
0
14. Proof The angle between a plane P and the xy-plane is , where 0 < 兾2. The projection of a rectangular region in P onto the xy-plane is a rectangle whose sides have lengths x and y, as shown in the figure. Prove that the area of the rectangular region in P is sec x y.
3
y
冕冕 冕 6
dx dy dz
z兾2 z兾2
0
共12z兲兾2 6y
3
dx dy dz.
z兾2
Then write the volume as a single iterated integral in the order dy dz dx. 18. Volume The figure shows a solid bounded below by the plane z 2 and above by the sphere x 2 y 2 z 2 8. z 4
Area: sec θ Δx Δy
x2 + y2 + z2 = 8
P −2
θ
Δy
2
2
y
x
Δx
(a) Find the volume of the solid using cylindrical coordinates.
Area in xy-plane: ΔxΔy
(b) Find the volume of the solid using spherical coordinates.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8
Vector Analysis Vector Fields Line Integrals Conservative Vector Fields and Independence of Path Green’s Theorem Parametric Surfaces Surface Integrals Divergence Theorem Stokes’s Theorem
Work (Exercise 39, p. 1073) An Application of Curl (Example 3, p. 1118)
Finding the Mass of a Spring (Example 5, p. 1055)
Building Design (Exercise 72, p. 1064)
Earth’s Magnetic Field (Exercise 83, p. 1050)
1039 Clockwise from top left, Caroline Warren/Photodisc/Getty Images; Elaine Davis/Shutterstock.com; nui7711/Shutterstock.com; Thufir/Big Stock Photo; David Stockman/iStockphoto.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1040
Chapter 15
Vector Analysis
15.1 Vector Fields Understand the concept of a vector field. Determine whether a vector field is conservative. Find the curl of a vector field. Find the divergence of a vector field.
Vector Fields In Chapter 12, you studied vector-valued functions—functions that assign a vector to a real number. There you saw that vector-valued functions of real numbers are useful in representing curves and motion along a curve. In this chapter, you will study two other types of vector-valued functions—functions that assign a vector to a point in the plane or a point in space. Such functions are called vector fields, and they are useful in representing various types of force fields and velocity fields.
Definition of Vector Field A vector field over a plane region R is a function F that assigns a vector Fx, y to each point in R. A vector field over a solid region Q in space is a function F that assigns a vector Fx, y, z to each point in Q.
Although a vector field consists of infinitely many vectors, you can get a good idea of what the vector field looks like by sketching several representative vectors Fx, y whose initial points are x, y. The gradient is one example of a vector field. For instance, if f x, y ⫽ x 2y ⫹ 3xy3 then the gradient of f ⵜf x, y ⫽ fxx, y i ⫹ fyx, y j ⫽ 2xy ⫹ 3y3 i ⫹ x2 ⫹ 9xy2 j
Vector field in the plane
is a vector field in the plane. From Chapter 13, the graphical interpretation of this field is a family of vectors, each of which points in the direction of maximum increase along the surface given by z ⫽ f x, y. Similarly, if f x, y, z ⫽ x 2 ⫹ y 2 ⫹ z 2 then the gradient of f ⵜf x, y, z ⫽ fxx, y, zi ⫹ fyx, y, zj ⫹ fzx, y, zk ⫽ 2xi ⫹ 2yj ⫹ 2zk
Vector field in space
is a vector field in space. Note that the component functions for this particular vector field are 2x, 2y, and 2z. A vector field Fx, y, z ⫽ Mx, y, zi ⫹ Nx, y, zj ⫹ Px, y, zk is continuous at a point if and only if each of its component functions M, N, and P is continuous at that point.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.1
Vector Fields
1041
Some common physical examples of vector fields are velocity fields, gravitational fields, and electric force fields.
Velocity field
1. Velocity fields describe the motions of systems of particles in the plane or in space. For instance, Figure 15.1 shows the vector field determined by a wheel rotating on an axle. Notice that the velocity vectors are determined by the locations of their initial points—the farther a point is from the axle, the greater its velocity. Velocity fields are also determined by the flow of liquids through a container or by the flow of air currents around a moving object, as shown in Figure 15.2. 2. Gravitational fields are defined by Newton’s Law of Gravitation, which states that the force of attraction exerted on a particle of mass m 1 located at x, y, z by a particle of mass m 2 located at 0, 0, 0 is
Rotating wheel Figure 15.1
Fx, y, z ⫽
⫺Gm 1m 2 u x2 ⫹ y 2 ⫹ z2
where G is the gravitational constant and u is the unit vector in the direction from the origin to x, y, z. In Figure 15.3, you can see that the gravitational field F has the properties that Fx, y, z always points toward the origin, and that the magnitude of Fx, y, z is the same at all points equidistant from the origin. A vector field with these two properties is called a central force field. Using the position vector r ⫽ xi ⫹ yj ⫹ zk for the point x, y, z, you can write the gravitational field F as
Air flow vector field Figure 15.2
Fx, y, z ⫽
z
⫺Gm 1m 2 r ⫺Gm 1m 2 ⫽ u. r2 r r 2
3. Electric force fields are defined by Coulomb’s Law, which states that the force exerted on a particle with electric charge q1 located at x, y, z by a particle with electric charge q2 located at 0, 0, 0 is
(x, y, z)
y
Fx, y, z ⫽
cq1q2 u r2
where r ⫽ x i ⫹ yj ⫹ zk, u ⫽ rr, and c is a constant that depends on the choice of units for r, q1, and q2. x
m 1 is located at (x, y, z). m 2 is located at (0, 0, 0).
Gravitational force field Figure 15.3
Note that an electric force field has the same form as a gravitational field. That is, Fx, y, z ⫽
k u. r2
Such a force field is called an inverse square field. Definition of Inverse Square Field Let rt ⫽ xti ⫹ ytj ⫹ ztk be a position vector. The vector field F is an inverse square field if Fx, y, z ⫽
k u r2
where k is a real number and u⫽
r r
is a unit vector in the direction of r.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1042
Chapter 15
Vector Analysis
Because vector fields consist of infinitely many vectors, it is not possible to create a sketch of the entire field. Instead, when you sketch a vector field, your goal is to sketch representative vectors that help you visualize the field.
Sketching a Vector Field Sketch some vectors in the vector field Fx, y ⫽ ⫺yi ⫹ xj. Solution You could plot vectors at several random points in the plane. It is more enlightening, however, to plot vectors of equal magnitude. This corresponds to finding level curves in scalar fields. In this case, vectors of equal magnitude lie on circles. F ⫽ c
y
Vectors of length c
x 2 ⫹ y 2 ⫽ c
3
x2 ⫹ y 2 ⫽ c2
2 1
Equation of circle
To begin making the sketch, choose a value for c and plot several vectors on the resulting circle. For instance, the following vectors occur on the unit circle. x 1
Vector field: F(x, y) = − yi + xj
Figure 15.4
3
Point
Vector
1, 0 0, 1 ⫺1, 0 0, ⫺1
F1, 0 ⫽ j F0, 1 ⫽ ⫺i F⫺1, 0 ⫽ ⫺j F0, ⫺1 ⫽ i
These and several other vectors in the vector field are shown in Figure 15.4. Note in the figure that this vector field is similar to that given by the rotating wheel shown in Figure 15.1.
Sketching a Vector Field Sketch some vectors in the vector field Fx, y ⫽ 2xi ⫹ yj. Solution For this vector field, vectors of equal length lie on ellipses given by
y 4
F ⫽ c
3
2x2 ⫹ y2 ⫽ c
c=2
c=1
which implies that 4x 2 ⫹ y 2 ⫽ c 2.
Equation of ellipse
−4
−3
x
−2
2
3
For c ⫽ 1, sketch several vectors 2xi ⫹ yj of magnitude 1 at points on the ellipse given by −3
4x 2 ⫹ y 2 ⫽ 1.
−4
For c ⫽ 2, sketch several vectors 2xi ⫹ yj of magnitude 2 at points on the ellipse given by 4x 2
⫹
y2
Vector field: F(x, y) = 2xi + yj
⫽ 4.
These vectors are shown in Figure 15.5.
Figure 15.5
TECHNOLOGY A computer algebra system can be used to graph vectors in a vector field. If you have access to a computer algebra system, use it to graph several representative vectors for the vector field in Example 2.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.1
Vector Fields
1043
Sketching a Velocity Field z
Sketch some vectors in the velocity field vx, y, z ⫽ 16 ⫺ x 2 ⫺ y 2k
16
where x 2 ⫹ y 2 ⱕ 16. Solution You can imagine that v describes the velocity of a liquid flowing through a tube of radius 4. Vectors near the z-axis are longer than those near the edge of the tube. For instance, at the point 0, 0, 0, the velocity vector is v0, 0, 0 ⫽ 16k, whereas at the point 0, 3, 0, the velocity vector is v0, 3, 0 ⫽ 7k. Figure 15.6 shows these and several other vectors for the velocity field. From the figure, you can see that the speed of the liquid is greater near the center of the tube than near the edges of the tube.
Conservative Vector Fields Notice in Figure 15.5 that all the vectors appear to be normal to the level curve from which they emanate. Because this is a property of gradients, it is natural to ask whether the vector field Fx, y ⫽ 2xi ⫹ yj
4
4
x
Velocity field: v(x, y, z) = (16 − x 2 − y 2)k
Figure 15.6
y
is the gradient of some differentiable function f. The answer is that some vector fields can be represented as the gradients of differentiable functions and some cannot—those that can are called conservative vector fields. Definition of Conservative Vector Field A vector field F is called conservative when there exists a differentiable function f such that F ⫽ f. The function f is called the potential function for F.
Conservative Vector Fields a. The vector field given by Fx, y ⫽ 2xi ⫹ yj is conservative. To see this, consider 1 the potential function f x, y ⫽ x 2 ⫹ 2 y 2. Because ⵜf ⫽ 2xi ⫹ yj ⫽ F it follows that F is conservative. b. Every inverse square field is conservative. To see this, let Fx, y, z ⫽
k u r2
and f x, y, z ⫽
⫺k x 2 ⫹ y 2 ⫹ z 2
where u ⫽ rr. Because kx ky kz i⫹ 2 j⫹ 2 k 2 32 2 2 32 2 x ⫹ y ⫹ z x ⫹ y ⫹ z x ⫹ y ⫹ z232 k x i ⫹ yj ⫹ zk ⫽ 2 2 2 x ⫹ y ⫹ z x 2 ⫹ y 2 ⫹ z 2 k r ⫽ r2 r k ⫽ u r2
ⵜf ⫽
2
2
it follows that F is conservative.
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1044
Chapter 15
Vector Analysis
As can be seen in Example 4(b), many important vector fields, including gravitational fields and electric force fields, are conservative. Most of the terminology in this chapter comes from physics. For example, the term “conservative” is derived from the classic physical law regarding the conservation of energy. This law states that the sum of the kinetic energy and the potential energy of a particle moving in a conservative force field is constant. (The kinetic energy of a particle is the energy due to its motion, and the potential energy is the energy due to its position in the force field.) The next theorem gives a necessary and sufficient condition for a vector field in the plane to be conservative.
REMARK Theorem 15.1 is valid on simply connected domains. A plane region R is simply connected when every simple closed curve in R encloses only points that are in R. (See Figure 15.26 in Section 15.4.)
THEOREM 15.1 Test for Conservative Vector Field in the Plane Let M and N have continuous first partial derivatives on an open disk R. The vector field Fx, y ⫽ Mi ⫹ Nj is conservative if and only if ⭸N ⭸M ⫽ . ⭸x ⭸y
Proof To prove that the given condition is necessary for F to be conservative, suppose there exists a potential function f such that Fx, y ⫽ ⵜf x, y ⫽ Mi ⫹ Nj. Then you have ⭸M ⭸y ⭸N fyx x, y ⫽ ⭸x
fxx, y ⫽ M
fxyx, y ⫽
fy x, y ⫽ N
and, by the equivalence of the mixed partials fxy and fyx, you can conclude that ⭸N⭸x ⫽ ⭸M⭸y for all x, y in R. The sufficiency of this condition is proved in Section 15.4. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Testing for Conservative Vector Fields in the Plane Decide whether the vector field given by F is conservative. a. Fx, y ⫽ x 2yi ⫹ xyj
b. Fx, y ⫽ 2xi ⫹ yj
Solution a. The vector field Fx, y ⫽ x 2yi ⫹ xyj is not conservative because ⭸M ⭸ ⫽ x 2y ⫽ x 2 and ⭸y ⭸y
⭸N ⭸ ⫽ xy ⫽ y. ⭸x ⭸x
b. The vector field Fx, y ⫽ 2xi ⫹ yj is conservative because ⭸ ⭸M ⫽ 2x ⫽ 0 and ⭸y ⭸y
⭸ ⭸N ⫽ y ⫽ 0. ⭸x ⭸x
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.1
Vector Fields
1045
Theorem 15.1 tells you whether a vector field is conservative. It does not tell you how to find a potential function of F. The problem is comparable to antidifferentiation. Sometimes you will be able to find a potential function by simple inspection. For instance, in Example 4, you observed that 1 f x, y ⫽ x 2 ⫹ y 2 2 has the property that ⵜf x, y ⫽ 2x i ⫹ yj.
Finding a Potential Function for F x, y Find a potential function for Fx, y ⫽ 2xyi ⫹ x 2 ⫺ yj. Solution
From Theorem 15.1, it follows that F is conservative because
⭸ 2xy ⫽ 2x and ⭸y
⭸ 2 x ⫺ y ⫽ 2x. ⭸x
If f is a function whose gradient is equal to Fx, y, then ⵜf x, y ⫽ 2xyi ⫹ x 2 ⫺ yj which implies that fxx, y ⫽ 2xy and fyx, y ⫽ x 2 ⫺ y. To reconstruct the function f from these two partial derivatives, integrate fxx, y with respect to x f x, y ⫽
fxx, y dx ⫽
2xy dx ⫽ x 2y ⫹ g y
and integrate fyx, y with respect to y f x, y ⫽
fyx, y dy ⫽
x 2 ⫺ y dy ⫽ x 2 y ⫺
y2 ⫹ hx. 2
Notice that g y is constant with respect to x and hx) is constant with respect to y. To find a single expression that represents f x, y, let g y ⫽ ⫺
y2 2
and hx ⫽ K.
Then, you can write f x, y ⫽ x 2 y ⫹ g y ⫹ K y2 ⫽ x2 y ⫺ ⫹ K. 2 You can check this result by forming the gradient of f. You will see that it is equal to the original function F. Notice that the solution to Example 6 is comparable to that given by an indefinite integral. That is, the solution represents a family of potential functions, any two of which differ by a constant. To find a unique solution, you would have to be given an initial condition that is satisfied by the potential function.
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1046
Chapter 15
Vector Analysis
Curl of a Vector Field Theorem 15.1 has a counterpart for vector fields in space. Before stating that result, the definition of the curl of a vector field in space is given. Definition of Curl of a Vector Field The curl of Fx, y, z ⫽ Mi ⫹ Nj ⫹ Pk is curl Fx, y, z ⫽ ⫻ Fx, y, z ⭸P ⭸N ⭸P ⭸M ⭸N ⭸M ⫽ ⫺ i⫺ ⫺ j⫹ ⫺ k. ⭸y ⭸z ⭸x ⭸z ⭸x ⭸y
If curl F ⫽ 0, then F is said to be irrotational. The cross product notation used for curl comes from viewing the gradient ⵜf as the result of the differential operator ⵜ acting on the function f. In this context, you can use the following determinant form as an aid in remembering the formula for curl.
curl Fx, y, z ⫽ ⵜ
⫻
Fx, y, z
i
j
k
⭸ ⭸ ⫽ ⭸x ⭸y
⭸ ⭸z
M
P
⫽
N
⭸N ⭸P ⭸M ⭸N ⭸M ⫺ i⫺ ⫺ j⫹ ⫺ k ⭸P ⭸y ⭸z ⭸x ⭸z ⭸x ⭸y
Finding the Curl of a Vector Field See LarsonCalculus.com for an interactive version of this type of example.
Find curl F of the vector field Fx, y, z ⫽ 2xyi ⫹ x 2 ⫹ z 2j ⫹ 2yzk. Is F irrotational? Solution
The curl of F is
curl Fx, y, z ⫽ ⵜ
⫽
⫽
⫻
Fx, y, z
i
j
k
⭸ ⭸x 2xy
⭸ ⭸y x2 ⫹ z2
⭸ ⭸z 2yz
⭸ ⭸y
x2 ⫹ z2
⭸ ⭸ ⭸z i ⫺ ⭸x 2yz 2xy
⭸ ⭸ ⭸z j ⫹ ⭸x 2yz 2xy
⫽ 2z ⫺ 2zi ⫺ 0 ⫺ 0j ⫹ 2x ⫺ 2xk ⫽ 0.
⭸ ⭸y
k
x2 ⫹ z2
Because curl F ⫽ 0, F is irrotational.
TECHNOLOGY Some computer algebra systems have a command that can be used to find the curl of a vector field. If you have access to a computer algebra system that has such a command, use it to find the curl of the vector field in Example 7. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.1
Vector Fields
1047
Later in this chapter, you will assign a physical interpretation to the curl of a vector field. But for now, the primary use of curl is shown in the following test for conservative vector fields in space. The test states that for a vector field in space, the curl is 0 at every point in its domain if and only if F is conservative. The proof is similar to that given for Theorem 15.1.
REMARK Theorem 15.2 is valid for simply connected domains in space. A simply connected domain in space is a domain D for which every simple closed curve in D can be shrunk to a point in D without leaving D. (See Section 15.4.)
THEOREM 15.2 Test for Conservative Vector Field in Space Suppose that M, N, and P have continuous first partial derivatives in an open sphere Q in space. The vector field Fx, y, z ⫽ Mi ⫹ Nj ⫹ Pk is conservative if and only if curl Fx, y, z ⫽ 0. That is, F is conservative if and only if ⭸P ⭸N ⫽ , ⭸y ⭸z
⭸P ⭸M ⫽ , and ⭸x ⭸z
⭸N ⭸M ⫽ . ⭸x ⭸y
From Theorem 15.2, you can see that the vector field given in Example 7 is conservative because curl Fx, y, z ⫽ 0. Try showing that the vector field Fx, y, z ⫽ x 3y 2zi ⫹ x 2zj ⫹ x 2yk is not conservative—you can do this by showing that its curl is curl Fx, y, z ⫽ x3y 2 ⫺ 2xyj ⫹ 2xz ⫺ 2x 3yzk ⫽ 0. For vector fields in space that pass the test for being conservative, you can find a potential function by following the same pattern used in the plane (as demonstrated in Example 6).
Finding a Potential Function for F x, y, z REMARK Examples 6 and 8 are illustrations of a type of problem called recovering a function from its gradient. If you go on to take a course in differential equations, you will study other methods for solving this type of problem. One popular method gives an interplay between successive “partial integrations” and partial differentiations.
Find a potential function for Fx, y, z ⫽ 2xyi ⫹ x 2 ⫹ z 2j ⫹ 2yzk. Solution From Example 7, you know that the vector field given by F is conservative. If f is a function such that Fx, y, z ⫽ ⵜf x, y, z, then fxx, y, z ⫽ 2xy, fyx, y, z ⫽ x 2 ⫹ z 2, and fzx, y, z ⫽ 2yz and integrating with respect to x, y, and z separately produces f x, y, z ⫽ f x, y, z ⫽ f x, y, z ⫽
M dx ⫽ N dy ⫽ P dz ⫽
2xy dx ⫽ x 2 y ⫹ g y, z
x 2 ⫹ z 2 dy ⫽ x 2 y ⫹ yz 2 ⫹ hx, z
2yz dz ⫽ yz 2 ⫹ kx, y.
Comparing these three versions of f x, y, z, you can conclude that g y, z ⫽ yz 2 ⫹ K,
hx, z ⫽ K,
and kx, y ⫽ x 2 y ⫹ K.
So, f x, y, z is given by f x, y, z ⫽ x 2 y ⫹ yz 2 ⫹ K.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1048
Chapter 15
Vector Analysis
Divergence of a Vector Field You have seen that the curl of a vector field F is itself a vector field. Another important function defined on a vector field is divergence, which is a scalar function. Definition of Divergence of a Vector Field The divergence of Fx, y ⫽ Mi ⫹ Nj is div Fx, y ⫽ ⵜ
⭈ Fx, y ⫽
⭸M ⭸N ⫹ . ⭸x ⭸y
Plane
The divergence of Fx, y, z ⫽ Mi ⫹ Nj ⫹ Pk is div Fx, y, z ⫽ ⵜ
⭈ Fx, y, z ⫽
⭸M ⭸N ⭸P ⫹ ⫹ . ⭸x ⭸y ⭸z
Space
If div F ⫽ 0, then F is said to be divergence free. The dot product notation used for divergence comes from considering ⵜ as a differential operator, as follows. ⵜ ⭈ Fx, y, z ⫽ ⫽
TECHNOLOGY Some computer algebra systems have a command that can be used to find the divergence of a vector field. If you have access to a computer algebra system that has such a command, use it to find the divergence of the vector field in Example 9.
⭸
⭸
⭸
⭸x i ⫹ ⭸y j ⫹ ⭸zk ⭈ M i ⫹ Nj ⫹ Pk ⭸M ⭸N ⭸P ⫹ ⫹ ⭸x ⭸y ⭸z
Finding the Divergence of a Vector Field Find the divergence at 2, 1, ⫺1 for the vector field Fx, y, z ⫽ x3y 2zi ⫹ x 2zj ⫹ x 2yk. Solution
The divergence of F is
div Fx, y, z ⫽
⭸ 3 2 ⭸ ⭸ x y z ⫹ x 2z ⫹ x 2y ⫽ 3x 2y 2z. ⭸x ⭸y ⭸z
At the point 2, 1, ⫺1, the divergence is div F2, 1, ⫺1 ⫽ 32212⫺1 ⫽ ⫺12. Divergence can be viewed as a type of derivative of F in that, for vector fields representing velocities of moving particles, the divergence measures the rate of particle flow per unit volume at a point. In hydrodynamics (the study of fluid motion), a velocity field that is divergence free is called incompressible. In the study of electricity and magnetism, a vector field that is divergence free is called solenoidal. There are many important properties of the divergence and curl of a vector field F see Exercise 77(a)–(g) . One that is used often is described in Theorem 15.3. You are asked to prove this theorem in Exercise 77(h). THEOREM 15.3 Divergence and Curl If Fx, y, z ⫽ Mi ⫹ Nj ⫹ Pk is a vector field and M, N, and P have continuous second partial derivatives, then divcurl F ⫽ 0.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.1
15.1 Exercises
graph. [The graphs are labeled (a), (b), (c), and (d).] y
27. Fx, y ⫽ sin y i ⫹ x cos yj 29. Fx, y ⫽
y
(b) 6
31. Fx, y ⫽
x
x
−6
5
1049
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Matching In Exercises 1–4, match the vector field with its (a)
Vector Fields
32. Fx, y ⫽
6
1 yi ⫺ xj xy 1 x 2 ⫹ y 2
1 1 ⫹ xy
28. Fx, y ⫽ 5y 2 yi ⫹ 3xj 30. Fx, y ⫽
2 2xy e yi ⫺ xj y2
i ⫹ j
yi ⫹ xj
Finding a Potential Function In Exercises 33– 42, determine y
(c)
whether the vector field is conservative. If it is, find a potential function for the vector field.
−6
−5
y
(d)
5
5
x
x
5
33. Fx, y ⫽ yi ⫹ xj
34. Fx, y ⫽ 3x 2y 2 i ⫹ 2x 3 yj
35. Fx, y ⫽ 2xyi ⫹ x 2j
36. Fx, y ⫽ xe x y2yi ⫹ xj
37. Fx, y ⫽ 15y 3 i ⫺ 5xy 2 j
38. Fx, y ⫽
1 yi ⫺ 2xj y2
40. Fx, y ⫽
xi ⫹ yj x2 ⫹ y 2
5
39. Fx, y ⫽ −5
2. Fx, y ⫽ x j
3. Fx, y ⫽ y i ⫺ xj
4. Fx, y ⫽ x i ⫹ 3yj
Sketching a Vector Field In Exercises 5–10, find F and sketch several representative vectors in the vector field. 5. Fx, y ⫽ i ⫹ j
6. Fx, y ⫽ y i ⫺ 2x j
7. Fx, y, z ⫽ 3yj
8. Fx, y ⫽ y i ⫹ xj 10. Fx, y, z ⫽ x i ⫹ yj ⫹ zk
Graphing a Vector Field In Exercises 11–14, use a computer algebra system to graph several representative vectors in the vector field. 11. Fx, y ⫽
1 8 2xyi
13. Fx, y, z ⫽
⫹
y 2j
x2 2y i⫺ 2j x y
41. Fx, y ⫽ e x cos yi ⫺ sin yj
1. Fx, y ⫽ y i
9. Fx, y, z ⫽ i ⫹ j ⫹ k
2
12. Fx, y ⫽ 2y ⫺ x, 2y ⫹ x
x i ⫹ yj ⫹ zk x 2 ⫹ y 2 ⫹ z 2
42. Fx, y ⫽
2xi ⫹ 2yj x 2 ⫹ y 22
Finding the Curl of a Vector Field In Exercises 43–46, find curl F for the vector field at the given point. 43. Fx, y, z ⫽ xyzi ⫹ xyz j ⫹ xyzk; 2, 1, 3 44. Fx, y, z ⫽ x 2zi ⫺ 2xzj ⫹ yzk; 2, ⫺1, 3 45. Fx, y, z ⫽ ex sin yi ⫺ e x cos yj; 0, 0, 1 46. Fx, y, z ⫽ e⫺xyz i ⫹ j ⫹ k; 3, 2, 0
Finding the Curl of a Vector Field In Exercises 47–50, use a computer algebra system to find the curl F for the vector field. 47. Fx, y, z ⫽ arctan
xy i ⫹ lnx
2
⫹ y2 j ⫹ k
yz xz xy i⫹ j⫹ k y⫺z x⫺z x⫺y
14. Fx, y, z ⫽ x, ⫺y, z
48. Fx, y, z ⫽
Finding a Conservative Vector Field In Exercises 15–24,
49. Fx, y, z ⫽ sinx ⫺ yi ⫹ sin y ⫺ zj ⫹ sinz ⫺ xk
find the conservative vector field for the potential function by finding its gradient.
50. Fx, y, z ⫽ x 2 ⫹ y 2 ⫹ z 2 i ⫹ j ⫹ k
15. f x, y ⫽ x2 ⫹ 2y 2
16. f x, y ⫽ x2 ⫺ 14 y 2
17. gx, y ⫽ 5x 2 ⫹ 3xy ⫹ y 2
18. gx, y ⫽ sin 3x cos 4y
Finding a Potential Function In Exercises 51–56, determine whether the vector field F is conservative. If it is, find a potential function for the vector field.
19. f x, y, z ⫽ 6xyz
20. f x, y, z ⫽ x2 ⫹ 4y2 ⫹ z2
21. gx, y, z ⫽ z ⫹ ye x
2
23. hx, y, z ⫽ xy lnx ⫹ y
22. gx, y, z ⫽
y z xz ⫹ ⫺ z x y
24. hx, y, z ⫽ x arcsin yz
Testing for a Conservative Vector Field In Exercises 25–32, determine whether the vector field is conservative. 25. Fx, y ⫽ x y 2 i ⫹ x 2 yj
26. Fx, y ⫽
1 yi ⫺ xj x2
51. Fx, y, z ⫽ xy 2 z 2 i ⫹ x 2 yz 2j ⫹ x 2 y 2zk 52. Fx, y, z ⫽ y 2z3i ⫹ 2xyz 3j ⫹ 3xy 2z 2 k 53. Fx, y, z ⫽ sin zi ⫹ sin xj ⫹ sin yk 54. Fx, y, z ⫽ ye z i ⫹ ze x j ⫹ xe y k z xz 55. Fx, y, z ⫽ i ⫺ 2 j ⫹ y y x i⫹ 56. Fx, y, z ⫽ 2 x ⫹ y2
x k y y j⫹k x2 ⫹ y 2
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1050
Chapter 15
Vector Analysis
Finding the Divergence of a Vector Field In Exercises
(e) ⵜ ⫻ ⵜf ⫹ ⵜ ⫻ F ⫽ ⵜ ⫻ ⵜ ⫻ F
57–60, find the divergence of the vector field F.
(f) ⵜ ⫻ f F ⫽ f ⵜ ⫻ F ⫹ ⵜf ⫻ F
57. Fx, y ⫽ x 2 i ⫹ 2y 2j
(g) div f F ⫽ f div F ⫹ ⵜf ⭈ F
58. Fx, y ⫽ xe x i ⫹ ye y j
(h) divcurl F ⫽ 0 (Theorem 15.3)
59. Fx, y, z ⫽ sin x i ⫹ cos yj ⫹ z 2k 60. Fx, y, z ⫽ lnx 2 ⫹ y 2i ⫹ xyj ⫹ ln y 2 ⫹ z 2k
Finding the Divergence of a Vector Field In Exercises 61–64, find the divergence of the vector field F at the given point. 61. Fx, y, z ⫽ xyzi ⫹ xyj ⫹ zk; 2, 1, 1 62. Fx, y, z ⫽ x 2z i ⫺ 2xzj ⫹ yzk; 2, ⫺1, 3 63. Fx, y, z ⫽ e x sin yi ⫺ e x cos yj ⫹ z2 k; 3, 0, 0
78.
HOW DO YOU SEE IT? Several representative vectors in the vector fields Fx, y ⫽
67. Curl
Define the curl of a vector field.
68. Divergence Define the divergence of a vector field in the plane and in space.
Curl of a Cross Product In Exercises 69 and 70, find curl F ⴛ G ⴝ ⵜ ⴛ F ⴛ G . 69. Fx, y, z ⫽ i ⫹ 3xj ⫹ 2yk
70. Fx, y, z ⫽ x i ⫺ zk
Gx, y, z ⫽ x i ⫺ yj ⫹ zk
Gx, y, z ⫽ x 2 i ⫹ yj ⫹ z 2k
xi ⫺ yj x2 ⫹ y2
y
y
4 3 2
4 3 2
WRITING ABOUT CONCEPTS
66. Conservative Vector Field What is a conservative vector field? How do you test for it in the plane and in space?
and Gx, y ⫽
are shown below. Explain any similarities or differences in the vector fields.
64. Fx, y, z ⫽ lnxyzi ⫹ j ⫹ k; 3, 2, 1
65. Vector Field Define a vector field in the plane and in space. Give some physical examples of vector fields.
xi ⫹ yj x2 ⫹ y2
x
x
−4 − 3 −2
2 3 4 −2 −3 −4
Vector field: xi + yj F(x, y) = x2 + y2
−4
−2
2
4
−2 −3 −4
Vector field: xi − yj G(x, y) = x2 + y2
True or False? In Exercises 79–82, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 79. If Fx, y ⫽ 4xi ⫺ y2j, then Fx, y → 0 as x, y → 0, 0.
Curl of the Curl of a Vector Field In Exercises 71 and 72, find curl curl F ⴝ ⵜ ⴛ ⵜ ⴛ F .
80. If Fx, y ⫽ 4xi ⫺ y2j and x, y is on the positive y-axis, then the vector points in the negative y-direction.
71. Fx, y, z ⫽ xyzi ⫹ yj ⫹ zk
81. If f is a scalar field, then curl f is a meaningful expression. 82. If F is a vector field and curl F ⫽ 0, then F is irrotational but not conservative.
72. Fx, y, z ⫽ x 2zi ⫺ 2xz j ⫹ yzk
Divergence of a Cross Product In Exercises 73 and 74,
find div F ⴛ G ⴝ ⵜ ⭈ F ⴛ G . 73. Fx, y, z ⫽ i ⫹ 3xj ⫹ 2yk
74. Fx, y, z ⫽ x i ⫺ zk
Gx, y, z ⫽ x i ⫺ yj ⫹ zk
Gx, y, z ⫽ x 2i ⫹ yj ⫹ z 2k
Divergence of the Curl of a Vector Field In Exercises 75 and 76, find div curl F ⴝ ⵜ ⭈ ⵜ ⴛ F . 75. Fx, y, z ⫽ xyzi ⫹ yj ⫹ zk 76. Fx, y, z ⫽ x 2zi ⫺ 2xz j ⫹ yzk 77. Proof In parts (a)–(h), prove the property for vector fields F and G and scalar function f. (Assume that the required partial derivatives are continuous.) (a) curlF ⫹ G ⫽ curl F ⫹ curl G (b) curlⵜf ⫽ ⵜ ⫻ ⵜf ⫽ 0 (c) divF ⫹ G ⫽ div F ⫹ div G
83. Earth’s Magnetic Field A cross section of Earth’s magnetic field can be represented as a vector field in which the center of Earth is located at the origin and the positive y-axis points in the direction of the magnetic north pole. The equation for this field is Fx, y ⫽ Mx, yi ⫹ Nx, yj m ⫽ 2 3xyi ⫹ 2y2 ⫺ x2)j
x ⫹ y252 where m is the magnetic moment of Earth. Show that this vector field is conservative.
(d) divF ⫻ G ⫽ curl F ⭈ G ⫺ F ⭈ curl G Thufir/Big Stock Photo
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15.2
Line Integrals
1051
15.2 Line Integrals Understand and use the concept of a piecewise smooth curve. Write and evaluate a line integral. Write and evaluate a line integral of a vector field. Write and evaluate a line integral in differential form.
Piecewise Smooth Curves A classic property of gravitational fields is that, subject to certain physical constraints, the work done by gravity on an object moving between two points in the field is independent of the path taken by the object. One of the constraints is that the path must be a piecewise smooth curve. Recall that a plane curve C given by rt xti ytj,
a t b
is smooth when dx dt
and
dy dt
are continuous on [a, b] and not simultaneously 0 on a, b. Similarly, a space curve C given by rt xti ytj ztk, JOSIAH WILLARD GIBBS (1839–1903)
Many physicists and mathematicians have contributed to the theory and applications described in this chapter––Newton, Gauss, Laplace, Hamilton, and Maxwell, among others. However, the use of vector analysis to describe these results is attributed primarily to the American mathematical physicist Josiah Willard Gibbs. See LarsonCalculus.com to read more of this biography.
C = C1 + C2 + C3
(0, 0, 0) C1 (1, 2, 0)
Figure 15.7
dx , dt
dy , dt
and
dz dt
are continuous on [a, b] and not simultaneously 0 on a, b. A curve C is piecewise smooth when the interval [a, b] can be partitioned into a finite number of subintervals, on each of which C is smooth.
Finding a Piecewise Smooth Parametrization
Solution Because C consists of three line segments C1, C2, and C3, you can construct a smooth parametrization for each segment and piece them together by making the last t-value in Ci correspond to the first t-value in Ci1.
1
1
is smooth when
Find a piecewise smooth parametrization of the graph of C shown in Figure 15.7.
z
x
a t b
(1, 2, 1) C3 C2
C1: xt 0, C2: xt t 1, C3: xt 1,
(0, 2, 0) y
yt 2t, yt 2, yt 2,
zt 0, zt 0, zt t 2,
0 t 1 1 t 2 2 t 3
So, C is given by
2tj, rt t 1i 2j, i 2j t 2k,
0 t 1 1 t 2. 2 t 3
Because C1, C2, and C3 are smooth, it follows that C is piecewise smooth. Recall that parametrization of a curve induces an orientation to the curve. For instance, in Example 1, the curve is oriented such that the positive direction is from 0, 0, 0, following the curve to 1, 2, 1. Try finding a parametrization that induces the opposite orientation. The Granger Collection
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1052
Chapter 15
Vector Analysis
Line Integrals Up to this point in the text, you have studied various types of integrals. For a single integral
b
f x dx
Integrate over interval [a, b].
a
you integrated over the interval [a, b]. Similarly, for a double integral
f x, y dA
Integrate over region R.
R
you integrated over the region R in the plane. In this section, you will study a new type of integral called a line integral
f x, y ds
Integrate over curve C.
C
for which you integrate over a piecewise smooth curve C. (The terminology is somewhat unfortunate—this type of integral might be better described as a “curve integral.”) To introduce the concept of a line integral, consider the mass of a wire of finite length, given by a curve C in space. The density (mass per unit length) of the wire at the point x, y, z is given by f x, y, z. Partition the curve C by the points P0, P1, . . . , Pn z
producing n subarcs, as shown in Figure 15.8. The length of the ith subarc is given by si. Next, choose a point xi, yi, zi in each subarc. If the length of each subarc is small, then the total mass of the wire can be approximated by the sum Mass of wire
P0
P1 P2 C
(xi , yi , zi ) Pi − 1 Pi P Δsi
x
n
f x , y , z s . i
i
i
i
n−1
Pn y
Partitioning of curve C Figure 15.8
i1
By letting denote the length of the longest subarc and letting approach 0, it seems reasonable that the limit of this sum approaches the mass of the wire. This leads to the next definition. Definition of Line Integral If f is defined in a region containing a smooth curve C of finite length, then the line integral of f along C is given by
f x, y ds lim
f x, y, z ds lim
C
or
C
n
f x , y s
→0 i1
i
i
Plane
i
n
f x , y , z s
→0 i1
i
i
i
i
Space
provided this limit exists.
As with the integrals discussed in Chapter 14, evaluation of a line integral is best accomplished by converting it to a definite integral. It can be shown that if f is continuous, then the limit given above exists and is the same for all smooth parametrizations of C.
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15.2
Line Integrals
1053
To evaluate a line integral over a plane curve C given by rt xti ytj, use the fact that ds rt dt xt 2 yt 2 dt. A similar formula holds for a space curve, as indicated in Theorem 15.4. THEOREM 15.4 Evaluation of a Line Integral as a Definite Integral Let f be continuous in a region containing a smooth curve C. If C is given by rt xti ytj, where a t b, then
C
b
f x, y ds
f xt, yt xt 2 yt 2 dt.
a
If C is given by rt xti ytj ztk, where a t b, then
b
f x, y, z ds
f xt, yt, zt xt 2 yt 2 zt 2 dt.
a
C
Note that if f x, y, z 1, then the line integral gives the arc length of the curve C, as defined in Section 12.5. That is,
b
1 ds
rt dt length of curve C.
a
C
z
Evaluating a Line Integral 1
(0, 0, 0)
Evaluate C
1 x
x2 y 3z ds
C
1 2
Figure 15.9
(1, 2, 1)
y
where C is the line segment shown in Figure 15.9. Solution x t,
Begin by writing a parametric form of the equation of the line segment: y 2t, and z t,
0 t 1.
Therefore, xt 1, yt 2, and zt 1, which implies that xt 2 yt 2 zt 2 12 22 12 6.
So, the line integral takes the following form.
C
1
x2 y 3z ds
t 2 2t 3t 6 dt
0
1
6
t 2 t dt
0
6
t3 t2 3 2
1 0
5 6 6 The value of the line integral in Example 2 does not depend on the parametrization of the line segment C; any smooth parametrization will produce the same value. To convince yourself of this, try some other parametrizations, such as x 1 2t, y 2 4t, and z 1 2t, 12 t 0, or x t, y 2t, and z t, 1 t 0.
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1054
Chapter 15
Vector Analysis
Let C be a path composed of smooth curves C1, C2, . . . , Cn. If f is continuous on C, then it can be shown that
f x, y ds
f x, y ds
C1
C
f x, y ds . . .
C2
f x, y ds.
Cn
This property is used in Example 3.
Evaluating a Line Integral Over a Path y
Evaluate
C = C1 + C2
(1, 1)
1
x ds
C
where C is the piecewise smooth curve shown in Figure 15.10.
y=x
Solution Begin by integrating up the line y x, using the following parametrization. y = x2
C1
C1: x t, y t,
C2
(0, 0)
x 1
Figure 15.10
0 t 1
For this curve, rt ti tj, which implies that xt 1 and yt 1. So, xt 2 yt 2 2
and you have
1
x ds
C1
t 2 dt
0
2
2
1
t2
0
2
2
.
Next, integrate down the parabola y x2, using the parametrization C2: x 1 t,
y 1 t2,
0 t 1.
For this curve, rt 1 ti 1 t2j which implies that xt 1 and yt 21 t. So, xt 2 yt 2 1 41 t2
and you have
1
x ds
C2
Consequently,
C
1 t 1 4 1 t2 dt
0
x ds
1
1 2 1 41 t2 32 8 3
0
1 32 5 1. 12
x ds
C1
x ds
C2
2
2
1 32 5 1 1.56. 12
For parametrizations given by rt xti ytj ztk, it is helpful to remember the form of ds as ds rt dt xt 2 yt 2 zt 2 dt. This is demonstrated in Example 4.
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15.2
Line Integrals
1055
Evaluating a Line Integral Evaluate
x 2 ds, where C is the curve represented by
C
4 1 rt ti t 32j t 2 k, 0 t 2. 3 2 Because rt i 2t12j tk and
Solution
rt xt 2 yt 2 zt 2 1 4t t 2 it follows that
2
x 2 ds
t 2 1 4t t 2 dt
0
C
2
1 2t 21 4t t 212 dt 2 0 2 1 1 4t t 232 3 0 1 13 13 1 3 15.29.
The next example shows how a line integral can be used to find the mass of a spring whose density varies. In Figure 15.11, note that the density of this spring increases as the spring spirals up the z-axis.
Finding the Mass of a Spring z
Find the mass of a spring in the shape of the circular helix rt
1 2
Density: ρ (x, y, z) = 1 + z
cos t i sin tj t k
where 0 t 6 and the density of the spring is
x, y, z 1 z as shown in Figure 15.11. Solution
Because
rt
1 2
sin t2 cos t2 12 1
it follows that the mass of the spring is Mass
1 z ds
C 6
1 t 2 dt t t 2 2 3 6 1 2
2 x
2
y
r(t) = 1 (cos ti + sin tj + tk) 2
0
2
6
Figure 15.11
0
144.47. David Stockman/iStockphoto.com
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1056
Chapter 15
Vector Analysis
Line Integrals of Vector Fields One of the most important physical applications of line integrals is that of finding the work done on an object moving in a force field. For example, Figure 15.12 shows an inverse square force field similar to the gravitational field of the sun. Note that the magnitude of the force along a circular path about the center is constant, whereas the magnitude of the force along a parabolic path varies from point to point. To see how a line integral can be used to find work done in a force field F, consider an object moving along a path C in the field, as shown in Figure 15.13. To determine the work done by the force, you need consider only that part of the force that is acting in the same direction as that in which the object is moving (or the opposite direction). This means that at each point on C, you can consider the projection F T of the force vector F onto the unit tangent vector T. On a small subarc of length si, the increment of work is Wi forcedistance Fxi, yi, zi Txi, yi, zi si
Inverse square force field F
where xi, yi, z i is a point in the ith subarc. Consequently, the total work done is given by the integral W
C
Fx, y, z Tx, y, z ds.
z
Vectors along a parabolic path in the force field F Figure 15.12
z
z
C
T has the direction of F.
F
T
y
T
(F • T)T
C
T
F
(F • T)T C y
y
(F • T)T x
x
At each point on C, the force in the direction of motion is F Figure 15.13
x
TT.
This line integral appears in other contexts and is the basis of the definition of the line integral of a vector field shown below. Note in the definition that F T ds F
F
rt rt dt rt
rt dt
F dr.
Definition of the Line Integral of a Vector Field Let F be a continuous vector field defined on a smooth curve C given by rt, a t b. The line integral of F on C is given by
C
F dr
F T ds
C b
a
Fxt, yt, zt rt dt.
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15.2
Line Integrals
1057
Work Done by a Force See LarsonCalculus.com for an interactive version of this type of example. z
(− 1, 0, 3π )
Find the work done by the force field 1 1 1 Fx, y, z xi yj k 2 2 4
3π
Force field F
on a particle as it moves along the helix given by rt cos t i sin tj tk
−2
π
−2
−1
−1
from the point 1, 0, 0 to 1, 0, 3, as shown in Figure 15.14. Solution
(1, 0, 0) 1
2
2
x
Space curve C
Because
rt xti ytj ztk cos t i sin tj tk
y
Figure 15.14
it follows that xt cos t,
yt sin t, and
zt t.
So, the force field can be written as 1 1 1 Fxt, yt, zt cos t i sin tj k. 2 2 4 To find the work done by the force field in moving a particle along the curve C, use the fact that rt sin t i cos tj k and write the following. W
F
dr
C
b
a
Fxt, yt, zt rt dt
3
1 1 1 cos t i sin tj k 2 2 4
0
3
0
z
3
0
sin t i cos tj k dt
1 1 1 sin t cos t sin t cos t dt 2 2 4
1 dt 4
1 3 t 4 0 3 4
In Example 6, note that the x- and y-components of the force field end up contributing nothing to the total work. This occurs because in this particular example, the z-component of the force field is the only portion of the force that is acting in the same (or opposite) direction in which the particle is moving (see Figure 15.15). y x Generated by Mathematica
Figure 15.15
TECHNOLOGY The computer-generated view of the force field in Example 6 shown in Figure 15.15 indicates that each vector in the force field points toward the z-axis.
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1058
Chapter 15
Vector Analysis
For line integrals of vector functions, the orientation of the curve C is important. If the orientation of the curve is reversed, the unit tangent vector Tt is changed to Tt, and you obtain
C
F dr
C
F dr.
Orientation and Parametrization of a Curve Let Fx, y yi x2j and evaluate the line integral
C1: r1(t) = (4 − t)i + (4t − t 2)j C2: r2(t) = ti + (4t − t 2)j
y
C
F dr
for each parabolic curve shown in Figure 15.16.
4
a. C1: r1t 4 ti 4t t 2 j, 0 t 3 b. C2: r2t t i 4t t 2j, 1 t 4
(1, 3) 3
2
C2
C1
Solution a. Because r1 t i 4 2tj and
1
Fxt, yt 4t t 2i 4 t2j
(4, 0) x
1
2
3
Figure 15.16
the line integral is
C1
F dr
3
0
4t t 2i 4 t2j i 4 2tj dt
3
4t t 2 64 64t 20t 2 2t 3 dt
0 3
2t 3 21t 2 68t 64 dt
0
REMARK Although the value of the line integral in Example 7 depends on the orientation of C, it does not depend on the parametrization of C. To see this, let C3 be represented by r3 t 2i 4 t2j where 1 t 2. The graph of this curve is the same parabolic segment shown in Figure 15.16. Does the value of the line integral over C3 agree with the value over C1 or C2? Why or why not?
t4 7t 3 34t 2 64t 2
3 0
69 . 2
b. Because r2 t i 4 2tj and Fxt, yt 4t t 2i t 2j the line integral is
C2
F dr
4
1
4t t 2 i t 2j i 4 2tj dt
4
4t t 2 4t 2 2t 3 dt
1 4
2t 3 3t 2 4t dt
1
t4 t 3 2t 2 2 69 . 2
4 1
The answer in part (b) is the negative of that in part (a) because C1 and C2 represent opposite orientations of the same parabolic segment.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.2
Line Integrals
1059
Line Integrals in Differential Form A second commonly used form of line integrals is derived from the vector field notation used in Section 15.1. If F is a vector field of the form Fx, y Mi Nj, and C is given by rt xti ytj, then F dr is often written as M dx N dy.
F dr
C
F
C
b
a
Mi Nj xti ytj dt
b
M
a
REMARK The parentheses are often omitted from this differential form, as shown below.
dx dy N dt dt dt
M dx N dy
C
This differential form can be extended to three variables.
Evaluating a Line Integral in Differential Form
M dx N dy
Let C be the circle of radius 3 given by
C
rt 3 cos t i 3 sin tj, 0 t 2
In three variables, the differential form is
dr dt dt
as shown in Figure 15.17. Evaluate the line integral
M dx N dy P dz.
C
y3 dx x3 3xy2 dy.
C
Solution Because x 3 cos t and y 3 sin t, you have dx 3 sin t dt and dy 3 cos t dt. So, the line integral is
M dx N dy
C
y
4
2
27 sin3 t3 sin t 27 cos3 t 81 cos t sin2 t3 cos t dt
81
2
cos4 t sin4 t 3 cos2 t sin2 t dt
0 2
cos t sin t 43 sin 2t dt 3 1 cos 4t 81
cos 2t 4 2 dt sin 2t 3 3 sin 4t 81
t 2 8 32 81
x −2
y3 dx x3 3xy2 dy
C 2 0
2
−4
2
2
2
0 2
4
−2
0
2
−4
0
r(t) = 3 cos ti + 3 sin tj
Figure 15.17
243 . 4
The orientation of C affects the value of the differential form of a line integral. Specifically, if C has the orientation opposite to that of C, then
C
M dx N dy
M dx N dy.
C
So, of the three line integral forms presented in this section, the orientation of C does not affect the form C f x, y ds, but it does affect the vector form and the differential form.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1060
Chapter 15
Vector Analysis
For curves represented by y gx, a x b, you can let x t and obtain the parametric form y gt,
x t and
a t b.
Because dx dt for this form, you have the option of evaluating the line integral in the variable x or the variable t. This is demonstrated in Example 9.
Evaluating a Line Integral in Differential Form y
Evaluate
C: y = 4x − x 2 4
y dx x2 dy
C
3
(1, 3)
where C is the parabolic arc given by y 4x x2 from 4, 0 to 1, 3, as shown in Figure 15.18.
2
Solution Rather than converting to the parameter t, you can simply retain the variable x and write
1
x 1
2
Figure 15.18
3
4
dy 4 2x dx.
y 4x x2
(4, 0)
Then, in the direction from 4, 0 to 1, 3, the line integral is
1
y dx x2 dy
4x x2 dx x24 2x dx
4
C
1
4x 3x2 2x3 dx
4
2x2 x3
69 . 2
x4 2
1 4
See Example 7.
Exploration Finding Lateral Surface Area The figure below shows a piece of tin that has been cut from a circular cylinder. The base of the circular cylinder is modeled by x2 y2 9. At any point x, y on the base, the height of the object is f x, y 1 cos
x . 4
Explain how to use a line integral to find the surface area of the piece of tin. z
2
1 −2
x
1 + cos
−1
πx 4
3 3
y
(x, y)
x2 + y2 = 9
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.2
15.2 Exercises
Exercises 1–6, find a piecewise smooth parametrization of the path C. (There is more than one correct answer.) y
y=
parametrization of the path C, and (b) evaluate
C
x
(2, 4)
4
(1, 1)
1
3
C
C y = x2
2
y=x
15. C: x-axis from x 0 to x 1
1
x
1 y
3.
(3, 3)
3
2
3
4
5
18. C: counterclockwise around the square with vertices 0, 0, 2, 0, 2, 2, and 0, 2
(5, 4)
4 2
3
C
C
2
1
Finding a Parametrization and Evaluating a Line Integral In Exercises 19 and 20, (a) find a piecewise smooth
1 x
x
1
2 y
5.
1
3
x2 + y2 = 9
2 y
6. 4
2
3
4
5
x2 y2 + =1 16 9
x 1
x −2
2
C
8.
xy ds
C
C
3x y ds
20.
z
0 t 1
x2
y2
z2
ds
10.
0 t 2 2xyz ds
C
C: rt sin ti cos tj 2k 0 t 2
C: rt 12ti 5tj 84tk 0 t 1
Evaluating a Line Integral In Exercises 11–14, (a) find a parametrization of the path C, and (b) evaluate
x2 ⴙ y2 ds
C
along C. 11. C: line segment from 0, 0 to 1, 1 12. C: line segment from 0, 0 to 2, 4 13. C: counterclockwise around the circle x2 y2 1 from 1, 0 to 0, 1
z
1
C
C (0, 0, 0)
(0, 0, 0)
(0, 1, 0) y
(1, 0, 0) x
(0, 1, 1)
1
(1, 0, 1)
1
(1, 1, 1)
1 y x
C: rt ti 2 tj
C
19.
C
C: rt 4ti 3tj
9.
2x ⴙ y2 ⴚ z ds
C
−4
Evaluating a Line Integral In Exercises 7–10, evaluate the line integral along the given path. 7.
2
−2
−2
parametrization of the path C shown in the figure, and (b) evaluate
along C.
2
1 −2 − 1
16. C: y-axis from y 1 to y 9 17. C: counterclockwise around the triangle with vertices 0, 0, 1, 0, and 0, 1
y
4.
x ⴙ 4 y ds
along C.
1
x
14. C: counterclockwise around the circle x2 y2 4 from 2, 0 to 0, 2
Evaluating a Line Integral In Exercises 15–18, (a) find a
y
2.
1061
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Finding a Piecewise Smooth Parametrization In
1.
Line Integrals
Mass In Exercises 21 and 22, find the total mass of two turns of a spring with density in the shape of the circular helix rt ⴝ 2 cos t i ⴙ 2 sin t j ⴙ t k, 0 t 4. 1 21. x, y, z 2x2 y2 z2
22. x, y, z z
Mass In Exercises 23–26, find the total mass of the wire with density . 23. rt cos ti sin tj, x, y x y 2, 0 t 3 24. rt t2 i 2tj, x, y 4 y, 0 t 1
25. rt t2 i 2tj tk, x, y, z kz k > 0, 1 t 3 26. rt 2 cos ti 2 sin tj 3tk, x, y, z k z
k > 0, 0 t 2
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1062
Chapter 15
Vector Analysis
Evaluating a Line Integral of a Vector Field In
37. Fx, y xi yj
Exercises 27–32, evaluate
C
C: counterclockwise around the triangle with vertices 0, 0, 1, 0, and 0, 1 (Hint: See Exercise 17a.)
F dr
y
y
where C is represented by rt. 27. Fx, y xi yj
3
(0, 1)
1
C
C: rt ti tj, 0 t 1
1
C
28. Fx, y xyi yj
x
C: rt 4 cos ti 4 sin tj, 0 t 2
1
x −2
−1
Figure for 37
30. Fx, y 3x i 4yj C: rt ti 4 t2 j,
C: counterclockwise along the semicircle y 4 x2 from 2, 0 to 2, 0
31. Fx, y, z xyi xz j yzk 32. Fx, y, z
x2i
y2j
39. Fx, y, z x i yj 5zk
0 t 1
C: rt 2 cos ti 2 sin tj tk, 0 t 2
z2k
C: rt 2 sin ti 2 cos tj
1 2 2 t k,
0 t 2π
Exercises 33 and 34, use a computer algebra system to evaluate the integral
C
z
z
Evaluating a Line Integral of a Vector Field In
Figure for 38
38. Fx, y yi xj
2 t 2
C: rt ti t 2j 2t k,
2
−1
29. Fx, y 3x i 4yj C: rt cos ti sin tj, 0 t 2
1
3
C
2
π
F dr
x
−3
C
5
1
3
−3
y
where C is represented by rt.
x
33. Fx, y, z
Figure for 39
x2zi
C: rt ti
6yj
t2j
yz2k
ln tk, 1 t 3
3
y
Figure for 40
40. Fx, y, z yzi xzj xyk
xi yj zk 34. Fx, y, z x2 y2 z2 C: rt ti tj et k,
3
C: line from 0, 0, 0 to 5, 3, 2
0 t 2
Work In Exercises 41–44, determine whether the work done along the path C is positive, negative, or zero. Explain.
Work In Exercises 35–40, find the work done by the force field F on a particle moving along the given path.
y
41.
35. Fx, y x i 2yj C: x t, y t3 from 0, 0 to 2, 8 C
y
y
x
(2, 8)
8
1
6 4
y
42.
C
C
2
x 1
x 2
4
6
Figure for 35
x
8
Figure for 36
C
36. Fx, y x2 i xyj C: x cos3 t, y sin3 t from 1, 0 to 0, 1
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.2 y
43.
Line Integrals
1063
Evaluating a Line Integral in Differential Form In Exercises 55–62, evaluate the integral
C
2x ⴚ y dx ⴙ x ⴙ 3y dy
C
along the path C. x
55. C: x-axis from x 0 to x 5 56. C: y-axis from y 0 to y 2 57. C: line segments from 0, 0 to 3, 0 and 3, 0 to 3, 3
y
44.
58. C: line segments from 0, 0 to 0, 3 and 0, 3 to 2, 3 59. C: arc on y 1 x 2 from 0, 1 to 1, 0 60. C: arc on y x32 from 0, 0 to 4, 8
C
61. C: parabolic path x t, y 2t 2 from 0, 0 to 2, 8 62. C: elliptic path x 4 sin t, y 3 cos t from 0, 3 to 4, 0 x
Lateral Surface Area In Exercises 63 –70, find the area of the lateral surface (see figure) over the curve C in the xy-plane and under the surface z ⴝ f x, y, where Evaluating a Line Integral of a Vector Field In Exercises 45 and 46, evaluate C F dr for each curve. Discuss the orientation of the curve and its effect on the value of the integral.
Lateral surface area ⴝ
f x, y ds.
C
z
Surface: z = f (x, y)
45. Fx, y x 2 i xyj (a) r1t 2ti t 1j,
1 t 3
(b) r2t 23 ti 2 tj, 0 t 2 46. Fx, y x 2 yi xy32j (a) r1t t 1 i t 2j,
Lateral surface
0 t 2
(b) r2t 1 2 cos ti 4 cos2 tj,
0 t 2
Demonstrate a Property In Exercises 47–50, demonstrate
(xi, yi) x
P
the property that
C
Q
C: Curve in xy-plane
F dr ⴝ 0
63. f x, y h, C: line from 0, 0 to 3, 4
regardless of the initial and terminal points of C, where the tangent vector rⴕ t is orthogonal to the force field F.
64. f x, y y, C: line from 0, 0 to 4, 4)
47. Fx, y yi xj
66. f x, y x y, C: x2 y2 1 from 1, 0 to 0, 1
48. Fx, y 3yi xj
C: rt t i 2tj
C: rt t i
49. Fx, y x3 2x2i x
t 3j
Evaluating a Line Integral in Differential Form In Exercises 51–54, evaluate the line integral along the path C given by x ⴝ 2t, y ⴝ 10t, where 0 t 1. 52.
C
53.
C
x 3y2 dx
C
xy dx y dy
C: y 1 x2 from 1, 0 to 0, 1
70. f x, y x2 y2 4,
C: rt 3 sin ti 3 cos tj
x 3y2 dy
C: x2 y2 1 from 1, 0 to 0, 1
67. f x, y h, C: y 1 x2 from 1, 0 to 0, 1 69. f x, y xy,
50. Fx, y xi yj
65. f x, y xy,
68. f x, y y 1, C: y 1 x2 from 1, 0 to 0, 1
y j 2
C: rt t i t2j
51.
y
Δsi
54.
C
3y x dx y2 dy
C: x2 y2 4
71. Engine Design A tractor engine has a steel component with a circular base modeled by the vector-valued function rt 2 cos t i 2 sin tj. Its height is given by z 1 y2. (All measurements of the component are in centimeters.) (a) Find the lateral surface area of the component. (b) The component is in the form of a shell of thickness 0.2 centimeter. Use the result of part (a) to approximate the amount of steel used in its manufacture. (c) Draw a sketch of the component.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1064
Chapter 15
Vector Analysis
72. Building Design
WRITING ABOUT CONCEPTS
The ceiling of a building has a height above the floor given by z 20 14x. One of the walls follows a path modeled by y x 32. Find the surface area of the wall for 0 x 40. (All measurements are in feet.)
79. Line Integral Define a line integral of a function f along a smooth curve C in the plane and in space. How do you evaluate the line integral as a definite integral? 80. Line Integral of a Vector Field Define a line integral of a continuous vector field F on a smooth curve C. How do you evaluate the line integral as a definite integral? 81. Ordering Surfaces Order the surfaces in ascending order of the lateral surface area under the surface and over the curve y x from 0, 0 to 4, 2 in the xy-plane. Explain your ordering without doing any calculations.
Moments of Inertia Consider a wire of density x, y given by the space curve
(a) z1 2 x
(b) z2 5 x
(c) z3 2
(d) z4 10 x 2y
C: rt ⴝ xti ⴙ ytj, 0 t b. The moments of inertia about the x- and y-axes are given by Ix ⴝ
y x, y ds 2
and Iy ⴝ
C
HOW DO YOU SEE IT? For each of the following, determine whether the work done in moving an object from the first to the second point through the force field shown in the figure is positive, negative, or zero. Explain your answer.
82. x x, y ds. 2
C
In Exercises 73 and 74, find the moments of inertia for the wire of density .
y
73. A wire lies along rt a cos ti a sin tj, where 0 t 2 and a > 0, with density x, y 1. 74. A wire lies along rt a cos ti a sin tj, where 0 t 2 and a > 0, with density x, y y.
x
75. Investigation The top outer edge of a solid with vertical sides and resting on the xy-plane is modeled by rt 3 cos t i 3 sin tj 1 sin2 2tk, where all measurements are in centimeters. The intersection of the plane y b 3 < b < 3 with the top of the solid is a horizontal line. (a) Use a computer algebra system to graph the solid. (b) Use a computer algebra system to approximate the lateral surface area of the solid. (c) Find (if possible) the volume of the solid.
(a) From 3, 3 to 3, 3 (b) From 3, 0 to 0, 3 (c) From 5, 0 to 0, 3
True or False? In Exercises 83–86, determine whether the
76. Work A particle moves along the path y x2 from the point 0, 0 to the point 1, 1. The force field F is measured at five points along the path, and the results are shown in the table. Use Simpson’s Rule or a graphing utility to approximate the work done by the force field.
x, y
0, 0
14, 161
12, 14
34, 169
1, 1
Fx, y
5, 0
3.5, 1
2, 2
1.5, 3
1, 5
77. Work Find the work done by a person weighing 175 pounds walking exactly one revolution up a circular helical staircase of radius 3 feet when the person rises 10 feet. 78. Investigation Determine the value of c such that the work done by the force field Fx, y 15 4 x2yi xyj on an object moving along the parabolic path y c1 x2 between the points 1, 0 and 1, 0 is a minimum. Compare the result with the work required to move the object along the straight-line path connecting the points.
statement is true or false. If it is false, explain why or give an example that shows it is false. 83. If C is given by xt t, yt t, where 0 t 1, then
1
xy ds
t 2 dt.
0
C
84. If C2 C1, then
C1
f x, y ds
f x, y ds 0.
C2
85. The vector functions r1 t i t 2j, where 0 t 1, and r2 1 ti 1 t2j, where 0 t 1, define the same curve. 86. If
F T ds 0, then F and T are orthogonal.
C
87. Work Consider a particle that moves through the force field Fx, y y xi xyj from the point 0, 0 to the point 0, 1 along the curve x kt1 t, y t. Find the value of k such that the work done by the force field is 1. nui7711/Shutterstock.com
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15.3
Conservative Vector Fields and Independence of Path
1065
15.3 Conservative Vector Fields and Independence of Path Understand and use the Fundamental Theorem of Line Integrals. Understand the concept of independence of path. Understand the concept of conservation of energy.
Fundamental Theorem of Line Integrals The discussion at the beginning of Section 15.2 pointed out that in a gravitational field the work done by gravity on an object moving between two points in the field is independent of the path taken by the object. In this section, you will study an important generalization of this result—it is called the Fundamental Theorem of Line Integrals. To begin, an example is presented in which the line integral of a conservative vector field is evaluated over three different paths.
y
(1, 1)
1
Line Integral of a Conservative Vector Field
C1
Find the work done by the force field
x
(0, 0)
1
1 1 F共x, y兲 ⫽ xyi ⫹ x 2j 2 4 on a particle that moves from 共0, 0兲 to 共1, 1兲 along each path, as shown in Figure 15.19.
C1: y = x
a. C1: y ⫽ x
(a)
Solution
b. C2: x ⫽ y 2
Note that F is conservative because the first partial derivatives are equal.
⭸ 1 1 xy ⫽ x and ⭸y 2 2
冤 冥
y
⭸ 1 2 1 x ⫽ x ⭸x 4 2
冤 冥
a. Let r共t兲 ⫽ ti ⫹ tj for 0 ⱕ t ⱕ 1, so that
(1, 1)
1
dr ⫽ 共i ⫹ j兲 dt and
C2
Then, the work done is x
(0, 0)
c. C3: y ⫽ x 3
1
W⫽
冕
C1
C2: x = y 2
F ⭈ dr ⫽
冕
1
0
1 1 F共x, y兲 ⫽ t 2 i ⫹ t 2j. 2 4 1
冥
3 2 1 t dt ⫽ t 3 4 4
0
1 ⫽ . 4
b. Let r共t兲 ⫽ ti ⫹ 冪t j for 0 ⱕ t ⱕ 1, so that
(b)
冢
dr ⫽ i ⫹
冣
1 1 1 j dt and F共x, y兲 ⫽ t 3兾2 i ⫹ t 2j. 2 4 2冪t
y
Then, the work done is
1
W⫽
(1, 1)
冕
C2
c. Let r共t兲 ⫽ C3
dr ⫽ x
(0, 0)
1
C3: y = (c)
Figure 15.19
x3
1 2 ti
F ⭈ dr ⫽ ⫹
1 3 8t j
0
冢12 i ⫹ 83 t j冣 dt 2
冕
C3
1
冥
5 3兾2 1 t dt ⫽ t 5兾2 8 4
1 ⫽ . 4
0
for 0 ⱕ t ⱕ 2, so that
Then, the work done is W⫽
冕
1
F ⭈ dr ⫽
冕
2
0
and F共x, y兲 ⫽
1 4 1 t i ⫹ t 2j. 32 16 2
冥
5 4 1 5 t dt ⫽ t 128 128
0
1 ⫽ . 4
So, the work done by the conservative vector field F is the same for each path.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1066
Chapter 15
Vector Analysis
In Example 1, note that the vector field F共x, y兲 ⫽ 12xyi ⫹ 14x 2j is conservative because F共x, y兲 ⫽ ⵜf 共x, y兲, where f 共x, y兲 ⫽ 14x 2y. In such cases, the next theorem states that the value of 兰C F ⭈ dr is given by
冕
C
F ⭈ dr ⫽ f 共x共1兲, y共1兲兲 ⫺ f 共x共0兲, y共0兲兲 1 ⫺0 4 1 ⫽ . 4
⫽
REMARK Notice how the Fundamental Theorem of Line Integrals is similar to the Fundamental Theorem of Calculus (Section 4.4), which states that
冕
THEOREM 15.5 Fundamental Theorem of Line Integrals Let C be a piecewise smooth curve lying in an open region R and given by r共t兲 ⫽ x共t兲i ⫹ y共t兲j,
If F共x, y兲 ⫽ Mi ⫹ Nj is conservative in R, and M and N are continuous in R, then
冕
b
f 共x兲 dx ⫽ F 共b兲 ⫺ F 共a兲
a
where F⬘共x兲 ⫽ f 共x兲.
a ⱕ t ⱕ b.
C
F ⭈ dr ⫽
冕
C
ⵜf ⭈ dr ⫽ f 共x共b兲, y共b兲兲 ⫺ f 共x共a兲, y共a兲兲
where f is a potential function of F. That is, F共x, y兲 ⫽ ⵜf 共x, y兲.
Proof A proof is provided only for a smooth curve. For piecewise smooth curves, the procedure is carried out separately on each smooth portion. Because F共x, y兲 ⫽ ⵜf 共x, y兲 ⫽ fx共x, y兲i ⫹ fy共x, y兲j it follows that
冕
C
F ⭈ dr ⫽ ⫽
冕 ⭈ 冕冤 b
dr dt dt dx dy fx共x, y兲 ⫹ fy共x, y兲 dt dt dt
F
a b a
冥
and, by the Chain Rule (Theorem 13.6), you have
冕
C
冕
b
d 关 f 共x共t兲, y共t兲兲兴 dt a dt ⫽ f 共x共b兲, y共b兲兲 ⫺ f 共x共a兲, y 共a兲兲.
F ⭈ dr ⫽
The last step is an application of the Fundamental Theorem of Calculus. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
In space, the Fundamental Theorem of Line Integrals takes the following form. Let C be a piecewise smooth curve lying in an open region Q and given by r共t兲 ⫽ x共t兲i ⫹ y共t兲j ⫹ z共t兲k,
a ⱕ t ⱕ b.
If F共x, y, z兲 ⫽ Mi ⫹ Nj ⫹ Pk is conservative and M, N, and P are continuous, then
冕
C
F ⭈ dr ⫽
冕
C
ⵜf ⭈ dr ⫽ f 共x共b兲, y共b兲, z共b兲兲 ⫺ f 共x共a兲, y共a兲, z共a兲兲
where F共x, y, z兲 ⫽ ⵜf 共x, y, z兲. The Fundamental Theorem of Line Integrals states that if the vector field F is conservative, then the line integral between any two points is simply the difference in the values of the potential function f at these points.
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15.3
Conservative Vector Fields and Independence of Path
1067
Using the Fundamental Theorem of Line Integrals F(x, y) = 2xyi + (x 2 − y)j
Evaluate
(−1, 4)
4
as shown in Figure 15.20. Solution where
From Example 6 in Section 15.1, you know that F is the gradient of f,
(1, 2)
2
f 共x, y兲 ⫽ x 2y ⫺
1
x
−2
F ⭈ dr, where C is a piecewise smooth curve from 共⫺1, 4兲 to 共1, 2兲 and
F共x, y兲 ⫽ 2xyi ⫹ 共x 2 ⫺ y兲j
3
C
冕
C
y
−1
1
2
Using the Fundamental Theorem of Line Integrals, 兰C F ⭈ dr Figure 15.20
y2 ⫹ K. 2
Consequently, F is conservative, and by the Fundamental Theorem of Line Integrals, it follows that
冕
C
F ⭈ dr ⫽ f 共1, 2兲 ⫺ f 共⫺1, 4兲
冤
⫽ 12共2兲 ⫺
冥 冤
22 42 ⫺ 共⫺1兲2共4兲 ⫺ 2 2
冥
⫽ 4. Note that it is unnecessary to include a constant K as part of f, because it is canceled by subtraction.
Using the Fundamental Theorem of Line Integrals Evaluate
F(x, y, z) = 2xyi + (x 2 + z 2)j + 2yzk
冕
C
z
F ⭈ dr, where C is a piecewise smooth curve from 共1, 1, 0兲 to 共0, 2, 3兲 and
F共x, y, z兲 ⫽ 2xyi ⫹ 共x 2 ⫹ z 2兲j ⫹ 2yz k 3
as shown in Figure 15.21. (0, 2, 3)
2
1
C
From Example 8 in Section 15.1, you know that F is the gradient of f,
f 共x, y, z兲 ⫽ x 2y ⫹ yz 2 ⫹ K. Consequently, F is conservative, and by the Fundamental Theorem of Line Integrals, it follows that
1 2 x
Solution where
(1, 1, 0)
2
y
Using the Fundamental Theorem of Line Integrals, 兰C F ⭈ dr Figure 15.21
冕
C
F ⭈ dr ⫽ f 共0, 2, 3兲 ⫺ f 共1, 1, 0兲 ⫽ 关共0兲2共2兲 ⫹ 共2兲共3兲2兴 ⫺ 关共1兲2共1兲 ⫹ 共1兲共0兲 2兴 ⫽ 17.
In Examples 2 and 3, be sure you see that the value of the line integral is the same for any smooth curve C that has the given initial and terminal points. For instance, in Example 3, try evaluating the line integral for the curve given by r共t兲 ⫽ 共1 ⫺ t兲 i ⫹ 共1 ⫹ t兲 j ⫹ 3tk. You should obtain
冕
C
F ⭈ dr ⫽
冕
1
共30t 2 ⫹ 16t ⫺ 1兲 dt
0
⫽ 17.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1068
Chapter 15
Vector Analysis
Independence of Path A C R1
R2 B
R1 is connected.
R2 is not connected.
From the Fundamental Theorem of Line Integrals, it is clear that if F is continuous and conservative in an open region R, then the value of 兰C F ⭈ dr is the same for every piecewise smooth curve C from one fixed point in R to another fixed point in R. This result is described by saying that the line integral 兰C F ⭈ dr is independent of path in the region R. A region in the plane (or in space) is connected when any two points in the region can be joined by a piecewise smooth curve lying entirely within the region, as shown in Figure 15.22. In open regions that are connected, the path independence of 兰C F ⭈ dr is equivalent to the condition that F is conservative.
Figure 15.22
THEOREM 15.6
Independence of Path and Conservative Vector Fields If F is continuous on an open connected region, then the line integral
冕
C
F ⭈ dr
is independent of path if and only if F is conservative.
(x1, y)
(x, y)
C2 C4
C1 C3
(x, y1)
Proof If F is conservative, then, by the Fundamental Theorem of Line Integrals, the line integral is independent of path. Now establish the converse for a plane region R. Let F共x, y兲 ⫽ Mi ⫹ Nj, and let 共x0, y0兲 be a fixed point in R. For any point 共x, y兲 in R, choose a piecewise smooth curve C running from 共x0, y0兲 to 共x, y兲, and define f by f 共x, y兲 ⫽
(x0, y0)
冕
C
Figure 15.23
F ⭈ dr ⫽
冕
M dx ⫹ N dy.
C
The existence of C in R is guaranteed by the fact that R is connected. You can show that f is a potential function of F by considering two different paths between 共x0, y0兲 and 共x, y兲. For the first path, choose 共x1, y兲 in R such that x ⫽ x1. This is possible because R is open. Then choose C1 and C2, as shown in Figure 15.23. Using the independence of path, it follows that f 共x, y兲 ⫽
冕 冕
M dx ⫹ N dy
C
⫽
M dx ⫹ N dy ⫹
C1
冕
M dx ⫹ N dy.
C2
Because the first integral does not depend on x, and because dy ⫽ 0 in the second integral, you have f 共x, y兲 ⫽ g共 y兲 ⫹
冕
M dx
C2
and it follows that the partial derivative of f with respect to x is fx共x, y兲 ⫽ M. For the second path, choose a point 共x, y1 兲. Using reasoning similar to that used for the first path, you can conclude that fy共x, y兲 ⫽ N. Therefore, ⵜf 共x, y兲 ⫽ fx共x, y兲 i ⫹ fy共x, y兲 j ⫽ M i ⫹ Nj ⫽ F共x, y兲 and it follows that F is conservative. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
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15.3
Conservative Vector Fields and Independence of Path
1069
Finding Work in a Conservative Force Field For the force field given by F共x, y, z兲 ⫽ e x cos yi ⫺ e x sin yj ⫹ 2k show that 兰C F ⭈ dr is independent of path, and calculate the work done by F on an object moving along a curve C from 共0, 兾2, 1兲 to 共1, , 3兲. Solution Writing the force field in the form F共x, y, z兲 ⫽ Mi ⫹ Nj ⫹ Pk, you have M ⫽ e x cos y, N ⫽ ⫺e x sin y, and P ⫽ 2, and it follows that ⭸P ⭸N ⫽0⫽ ⭸y ⭸z ⭸P ⭸M ⫽0⫽ ⭸x ⭸z and ⭸N ⭸M ⫽ ⫺e x sin y ⫽ . ⭸x ⭸y So, F is conservative. If f is a potential function of F, then fx共x, y, z兲 ⫽ e x cos y fy共x, y, z兲 ⫽ ⫺e x sin y and fz共x, y, z兲 ⫽ 2. By integrating with respect to x, y, and z separately, you obtain f 共x, y, z兲 ⫽ f 共x, y, z兲 ⫽ and f 共x, y, z兲 ⫽
冕 冕 冕
fx共x, y, z兲 dx ⫽ fy共x, y, z兲 dy ⫽
fz共x, y, z兲 dz ⫽
冕 冕
冕
e x cos y dx ⫽ e x cos y ⫹ g共 y, z兲 ⫺e x sin y dy ⫽ e x cos y ⫹ h共x, z兲
2 dz ⫽ 2z ⫹ k共x, y兲.
By comparing these three versions of f 共x, y, z兲, you can conclude that f 共x, y, z兲 ⫽ e x cos y ⫹ 2z ⫹ K. Therefore, the work done by F along any curve C from 共0, 兾2, 1兲 to 共1, , 3兲 is W⫽
冕
C
冤
F ⭈ dr 共1, , 3兲
冥
⫽ e x cos y ⫹ 2z
共0, 兾2, 1兲
⫽ 共⫺e ⫹ 6兲 ⫺ 共0 ⫹ 2兲 ⫽ 4 ⫺ e. For the object in Example 4, how much work is done when the object moves on a curve from 共0, 兾2, 1兲 to 共1, , 3兲 and then back to the starting point 共0, 兾2, 1兲? The Fundamental Theorem of Line Integrals states that there is zero work done. Remember that, by definition, work can be negative. So, by the time the object gets back to its starting point, the amount of work that registers positively is canceled out by the amount of work that registers negatively.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1070
Chapter 15
Vector Analysis
A curve C given by r共t兲 for a ⱕ t ⱕ b is closed when r 共a兲 ⫽ r 共b兲. By the Fundamental Theorem of Line Integrals, you can conclude that if F is continuous and conservative on an open region R, then the line integral over every closed curve C is 0.
REMARK Theorem 15.7 gives you options for evaluating a line integral involving a conservative vector field. You can use a potential function, or it might be more convenient to choose a particularly simple path, such as a straight line.
THEOREM 15.7 Equivalent Conditions Let F共x, y, z兲 ⫽ Mi ⫹ Nj ⫹ Pk have continuous first partial derivatives in an open connected region R, and let C be a piecewise smooth curve in R. The conditions listed below are equivalent. 1. F is conservative. That is, F ⫽ ⵜf for some function f. 2.
冕 冕
C
3.
C
F ⭈ dr is independent of path. F ⭈ dr ⫽ 0 for every closed curve C in R.
Evaluating a Line Integral See LarsonCalculus.com for an interactive version of this type of example. C1: r(t) = (1 − cos t)i + sin tj
Evaluate
y
冕
C1
F ⭈ dr, where
F共x, y兲 ⫽ 共 y 3 ⫹ 1兲i ⫹ 共3xy 2 ⫹ 1兲j and C1 is the semicircular path from 共0, 0兲 to 共2, 0兲, as shown in Figure 15.24.
1
C1
Solution
C2 (0, 0)
(2, 0) 1
C2: r(t) = ti
2
x
You have the following three options.
a. You can use the method presented in Section 15.2 to evaluate the line integral along the given curve. To do this, you can use the parametrization r共t兲 ⫽ 共1 ⫺ cos t兲 i ⫹ sin t j, where 0 ⱕ t ⱕ . For this parametrization, it follows that dr ⫽ r⬘ 共t兲 dt ⫽ 共sin ti ⫹ cos tj兲 dt
Figure 15.24
and
冕
C1
F ⭈ dr ⫽
冕
共sin t ⫹ sin4 t ⫹ cos t ⫹ 3 sin2 t cos t ⫺ 3 sin2 t cos2 t兲 dt.
0
This integral should dampen your enthusiasm for this option. b. You can try to find a potential function and evaluate the line integral by the Fundamental Theorem of Line Integrals. Using the technique demonstrated in Example 4, you can find the potential function to be f 共x, y兲 ⫽ xy 3 ⫹ x ⫹ y ⫹ K, and, by the Fundamental Theorem, W⫽
冕
C1
F ⭈ dr ⫽ f 共2, 0兲 ⫺ f 共0, 0兲 ⫽ 2.
c. Knowing that F is conservative, you have a third option. Because the value of the line integral is independent of path, you can replace the semicircular path with a simpler path. Choose the straight-line path C2 from 共0, 0兲 to 共2, 0兲. Let r共t兲 ⫽ ti for 0 ⱕ t ⱕ 2, so that F共x, y兲 ⫽ i ⫹ j.
dr ⫽ i dt and Then, the integral is
冕
C1
F ⭈ dr ⫽
冕
C2
F ⭈ dr ⫽
冕
2
0
冥
1 dt ⫽ t
2 0
⫽ 2.
Of the three options, obviously the third one is the easiest.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.3
Conservative Vector Fields and Independence of Path
1071
Conservation of Energy In 1840, the English physicist Michael Faraday wrote, “Nowhere is there a pure creation or production of power without a corresponding exhaustion of something to supply it.” This statement represents the first formulation of one of the most important laws of physics—the Law of Conservation of Energy. In modern terminology, the law is stated as follows: In a conservative force field, the sum of the potential and kinetic energies of an object remains constant from point to point. You can use the Fundamental Theorem of Line Integrals to derive this law. From physics, the kinetic energy of a particle of mass m and speed v is 1 k ⫽ mv 2. 2
The potential energy p of a particle at point 共x, y, z兲 in a conservative vector field F is defined as p共x, y, z兲 ⫽ ⫺f 共x, y, z兲, where f is the potential function for F. Consequently, the work done by F along a smooth curve C from A to B is
MICHAEL FARADAY (1791–1867)
W⫽
Several philosophers of science have considered Faraday’s Law of Conservation of Energy to be the greatest generalization ever conceived by humankind. Many physicists have contributed to our knowledge of this law.Two early and influential ones were James Prescott Joule (1818 –1889) and Hermann Ludwig Helmholtz (1821–1894).
冕
冥
F ⭈ dr ⫽ f 共x, y, z兲
C
W⫽
冕 冕 冕 冕 冕
a b
⫽
a b
⫽
a b
⫽
a
⫽ ⫽
y
A
C
B x
C
F ⭈ dr ⫽ p共A兲 ⫺ p共B兲.
Figure 15.25
B A
⫽ p共A兲 ⫺ p共B兲
m 2 m 2
F ⭈ r⬘共t兲 dt F ⭈ v共t兲 dt
关mv⬘共t兲兴 ⭈ v共t兲 dt m关 v⬘共t兲 ⭈ v共t兲兴 dt
冕 冕
b
a b
a
冤 冤
d 关v共t兲 ⭈ v共t兲兴 dt dt d 关储v共t兲 储2兴 dt dt
冥 冥
b m 储v共t兲 储2 2 a b m ⫽ 关v共t兲兴 2 2 a 1 1 ⫽ m 关v共b兲兴 2 ⫺ m 关v共a兲兴 2 2 2 ⫽ k共B兲 ⫺ k共A兲.
⫽
F
冕
A
冥
⫽ ⫺p共x, y, z兲
F ⭈ dr
b
⫽
W⫽
B
as shown in Figure 15.25. In other words, work W is equal to the difference in the potential energies of A and B. Now, suppose that r共t兲 is the position vector for a particle moving along C from A ⫽ r共a兲 to B ⫽ r共b兲. At any time t, the particle’s velocity, acceleration, and speed are v共t兲 ⫽ r⬘共t兲, a共t兲 ⫽ r⬙ 共t兲, and v共t兲 ⫽ 储v共t兲 储, respectively. So, by Newton’s Second Law of Motion, F ⫽ ma共t兲 ⫽ m共v⬘共t兲兲, and the work done by F is
C
The work done by F along C is
Kinetic energy
Equating these two results for W produces p共A兲 ⫺ p共B兲 ⫽ k共B兲 ⫺ k共A兲 p共A兲 ⫹ k共A兲 ⫽ p共B兲 ⫹ k共B兲 which implies that the sum of the potential and kinetic energies remains constant from point to point. The Granger Collection, NYC — All rights reserved.
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1072
Chapter 15
Vector Analysis
15.3 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Evaluating a Line Integral for Different Parametrizations In Exercises 1–4, show that the value of 兰C F ⭈ d r is the same for each parametric representation of C.
14. F共x, y兲 ⫽ xy 2 i ⫹ 2x 2y j 1 (a) r1共t兲 ⫽ t i ⫹ j, 1 ⱕ t ⱕ 3 t
1. F共x, y兲 ⫽ x 2 i ⫹ xy j (a) r1共t兲 ⫽ t i ⫹
1 (b) r2共t兲 ⫽ 共t ⫹ 1兲 i ⫺ 3共t ⫺ 3兲 j,
0 ⱕ t ⱕ 1
t 2 j,
(b) r2共兲 ⫽ sin i ⫹
sin2
j, 0 ⱕ ⱕ 2
15.
冕
⫹ w j,
y
(a) 4
0 ⱕ w ⱕ 2
(a) r1共兲 ⫽ sec i ⫹ tan j, 0 ⱕ ⱕ
1 − x2
y= C2
(4, 4)
3
C1
2
3. F共x, y兲 ⫽ y i ⫺ x j
3
4
y
(c)
(b) r2共w兲 ⫽ 共2 ⫹ ln w兲 i ⫹ 共3 ⫺ ln w兲 j, 1 ⱕ w ⱕ e3
y
(d) C3
(− 1, 2)
(a) r1共t兲 ⫽ 共2 ⫹ t兲 i ⫹ 共3 ⫺ t兲 j, 0 ⱕ t ⱕ 3
C4
1
(−1, 0) 1
5–10, determine whether the vector field is conservative.
16.
冕
x
1
2 −1
(1, − 1)
(− 1, − 1)
(1, 0)
−1
x
5. F共x, y兲 ⫽ e x共sin y i ⫹ cos yj兲 6. F共x, y兲 ⫽ 15x 2y 2 i ⫹ 10x 3yj
1 − x2
y=
(2, 2)
Testing for Conservative Vector Fields In Exercises
1 共 y i ⫹ xj兲 y2
−1
x
3
2
x
1
(0, 0) 1
(b) r2共t兲 ⫽ 冪t ⫹ 1 i ⫹ 冪t j, 0 ⱕ t ⱕ 3
(1, 0)
(−1, 0) −1
1
4. F共x, y兲 ⫽ y i ⫹ x 2 j
7. F共x, y兲 ⫽
y
(b) (3, 4)
(a) r1共t兲 ⫽ t i ⫹ 冪t j, 0 ⱕ t ⱕ 4 (b) r2共w兲 ⫽
y 2 dx ⫹ 2xy dy
C
2. F共x, y兲 ⫽ 共x 2 ⫹ y 2兲 i ⫺ x j w2 i
0 ⱕ t ⱕ 2
共2x ⫺ 3y ⫹ 1兲 dx ⫺ 共3x ⫹ y ⫺ 5兲 dy
C
8. F共x, y, z兲 ⫽ y ln z i ⫺ x ln z j ⫹
xy k z
9. F共x, y, z兲 ⫽ y 2z i ⫹ 2xyz j ⫹ xy 2 k
(0, 1)
x
−1
(4, 1) 1
(0, 0) −1
x
1
F ⭈ dr.
6
x
0 ⱕ t ⱕ 1
(c) r3共t兲 ⫽ t i ⫹
t 3 j,
0 ⱕ t ⱕ 1
−1 x
1
17.
冕
1 − y2 C4
C3 2
−1
(0, −1)
2xy dx ⫹ 共x 2 ⫹ y 2兲 dy
C
(a) C: ellipse
(b) r2共t兲 ⫽ t i ⫹
x=
1
(0, 1)
(a) r1共t兲 ⫽ t i ⫺ 共t ⫺ 3兲 j, 0 ⱕ t ⱕ 3
t 2 j,
(0, 1)
y = ex
12. F共x, y兲 ⫽ ye xy i ⫹ xe xy j
(a) r1共t兲 ⫽ t i ⫹ t j, 0 ⱕ t ⱕ 1
y
(2, e 2)
2
(b) r2共t兲 ⫽ t i ⫹ t 3 j, 0 ⱕ t ⱕ 1
13. F共x, y兲 ⫽ y i ⫺ x j
(0, −1)
4
4
0 ⱕ t ⱕ 1
(b) The closed path consisting of line segments from 共0, 3兲 to 共0, 0兲, from 共0, 0兲 to 共3, 0兲, and then from 共3, 0兲 to 共0, 3兲
3
(d)
8
11. F共x, y兲 ⫽ 2xy i ⫹ x 2 j (a) r1共t兲 ⫽ t i ⫹ t 2 j,
2
y
(c)
(Hint: If F is conservative, the integration may be easier on an alternative path.)
C2
C1
2
Exercises 11–24, find the value of the line integral
1 − y2
x=
1
(2, 3)
3
Evaluating a Line Integral of a Vector Field In
C
y
(b)
4
10. F共x, y, z兲 ⫽ sin yz i ⫹ xz cos yz j ⫹ xy sin yz k
冕
y
(a)
y2 x2 ⫹ ⫽ 1 from 共5, 0兲 to 共0, 4兲 25 16
(b) C: parabola y ⫽ 4 ⫺ x 2 from 共2, 0兲 to 共0, 4兲
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15.3 18.
冕
共x 2 ⫹ y 2兲 dx ⫹ 2xy dy
Conservative Vector Fields and Independence of Path 31.
C
(a) C: line segment from 共0, 0, 0兲 to 共1, 1, 1兲
0 ⱕ t ⱕ 2
(b) r2共t兲 ⫽ 2 cos t i ⫹ 2 sin t j, 0 ⱕ t ⱕ
2
(b) C: line segments from 共0, 0, 0兲 to 共0, 0, 1兲 to 共1, 1, 1兲 (c) C: line segments from 共0, 0, 0兲 to 共1, 0, 0兲 to 共1, 1, 0兲 to 共1, 1, 1兲
19. F共x, y, z兲 ⫽ yz i ⫹ xz j ⫹ xy k (a) r1共t兲 ⫽ t i ⫹ 2 j ⫹ t k, 0 ⱕ t ⱕ 4 (b) r2共t兲 ⫽
t2i
⫹ tj ⫹
t 2 k,
32. Repeat Exercise 31 using the integral
0 ⱕ t ⱕ 2
20. F共x, y, z兲 ⫽ i ⫹ z j ⫹ y k (b) r2共t兲 ⫽ 共1 ⫺ 2t兲 i ⫹
2t k,
0 ⱕ t ⱕ
0 ⱕ t ⱕ 1
33.
(a) r1共t兲 ⫽ t i ⫹
(a) r1共t兲 ⫽ cos t i ⫹ sin t j ⫹ t k, 0 ⱕ t ⱕ (b) r2共t兲 ⫽ 共1 ⫺ 2t兲 i ⫹ t k, 0 ⱕ t ⱕ 1 23. F共x, y, z兲 ⫽ ez共 y i ⫹ x j ⫹ xy k兲
冕
6x dx ⫺ 4z dy ⫺ 共4y ⫺ 20z兲 dz
C: smooth curve from 共0, 0, 0兲 to 共3, 4, 0兲
Work In Exercises 35 and 36, find the work done by the force field F in moving an object from P to Q.
(a) r1共t兲 ⫽ 4 cos t i ⫹ 4 sin t j ⫹ 3k, 0 ⱕ t ⱕ
35. F共x, y兲 ⫽ 9x 2y 2 i ⫹ 共6x3y ⫺ 1兲 j; P共0, 0兲, Q共5, 9兲
(b) r2共t兲 ⫽ 共4 ⫺ 8t兲 i ⫹ 3k,
36. F共x, y兲 ⫽
0 ⱕ t ⱕ 1
24. F共x, y, z兲 ⫽ y sin z i ⫹ x sin z j ⫹ xy cos xk (a) r1共t兲 ⫽ t 2 i ⫹ t 2 j,
0 ⱕ t ⱕ 2
(b) r2共t兲 ⫽ 4t i ⫹ 4tj, 0 ⱕ t ⱕ 1
Using the Fundamental Theorem of Line Integrals In Exercises 25–34, evaluate the line integral using the Fundamental Theorem of Line Integrals. Use a computer algebra system to verify your results.
冕
C
共3yi ⫹ 3xj兲 ⭈ dr
冕
C
关2共x ⫹ y兲 i ⫹ 2共x ⫹ y兲 j兴 ⭈ dr
C: smooth curve from 共⫺1, 1兲 to 共3, 2兲
冕
cos x sin y dx ⫹ sin x cos y dy
C
C: line segment from 共0, ⫺ 兲 to
冕
C
冢32, 2 冣
y dx ⫺ x dy x2 ⫹ y 2
C: line segment from 共1, 1兲 to 共2冪3, 2兲
冕
2x x2 i ⫺ 2 j; P共⫺1, 1兲, Q共3, 2兲 y y
37. Work A stone weighing 1 pound is attached to the end of a two-foot string and is whirled horizontally with one end held fixed. It makes 1 revolution per second. Find the work done by the force F that keeps the stone moving in a circular path. [Hint: Use Force ⫽ (mass)(centripetal acceleration).] 38. Work Let F共x, y, z兲 ⫽ a1i ⫹ a 2 j ⫹ a3 k be a constant force vector field. Show that the work done in moving a particle along any path from P to Q is W ⫽ F ⭈ PQ . \
39. Work
C: smooth curve from 共0, 0兲 to 共3, 8兲
29.
34.
冢2 , 3, 4冣
C
22. F共x, y, z兲 ⫽ ⫺y i ⫹ x j ⫹ 3xz 2 k
28.
⫺sin x dx ⫹ z dy ⫹ y dz
C: smooth curve from 共0, 0, 0兲 to
⫹ k, 0 ⱕ t ⱕ 1
(b) r2共t兲 ⫽ t i ⫹ tj ⫹ 共 2t ⫺ 1兲2 k, 0 ⱕ t ⱕ 1
27.
zy dx ⫹ xz dy ⫹ xy dz.
C
21. F共x, y, z兲 ⫽ 共2y ⫹ x兲 i ⫹ 共x 2 ⫺ z兲 j ⫹ 共2y ⫺ 4z兲 k t 2j
冕 冕
C
(a) r1共t兲 ⫽ cos t i ⫹ sin t j ⫹ t 2 k,
26.
共z ⫹ 2y兲 dx ⫹ 共2x ⫺ z兲 dy ⫹ 共 x ⫺ y兲 dz
C
(a) r1共t兲 ⫽ t 3 i ⫹ t 2 j,
25.
冕
1073
A zip line is installed 50 meters above ground level. It runs to a point on the ground 50 meters away from the base of the installation. Show that the work done by the gravitational force field for a 175-pound person moving the length of the zip line is the same for each path. (a) r共t兲 ⫽ t i ⫹ 共50 ⫺ t兲 j 1 (b) r共t兲 ⫽ t i ⫹ 50 共50 ⫺ t兲2j
e x sin y dx ⫹ e x cos y dy
C
C: cycloid x ⫽ ⫺ sin , y ⫽ 1 ⫺ cos from 共0, 0兲 to 共2, 0兲
冕
2x 2y 30. 2 ⫹ y2兲 2 dx ⫹ 共x 2 ⫹ y 2兲 2 dy 共 x C C: circle 共x ⫺ 4兲2 ⫹ 共 y ⫺ 5兲 2 ⫽ 9 clockwise from 共7, 5兲 to 共1, 5兲
40. Work Can you find a path for the zip line in Exercise 39 such that the work done by the gravitational force field would differ from the amounts of work done for the two paths given? Explain why or why not.
Caroline Warren/Photodisc/Getty Images
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1074
Chapter 15
Vector Analysis
WRITING ABOUT CONCEPTS 41. Fundamental Theorem of Line Integrals the Fundamental Theorem of Line Integrals.
State
42. Independence of Path What does it mean that a line integral is independent of path? State the method for determining whether a line integral is independent of path.
Graphical Reasoning In Exercises 45 and 46, consider the force field shown in the figure. Is the force field conservative? Explain why or why not. y
45.
x
y x i⫺ 2 j. Find x2 ⫹ y2 x ⫹ y2 the value of the line integral 兰C F ⭈ dr.
43. Think About It
x
Let F共x, y兲 ⫽
y
(a)
y
46.
(b)
y
True or False? In Exercises 47–50, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.
C2
C1 x
x
y
(c)
(d)
47. If C1, C2, and C3 have the same initial and terminal points and 兰C1 F ⭈ dr1 ⫽ 兰C2 F ⭈ dr2, then 兰C1 F ⭈ dr1 ⫽ 兰C3 F ⭈ dr3. 48. If F ⫽ y i ⫹ x j and C is given by r共t兲 ⫽ 共4 sin t兲 i ⫹ 共3 cos t兲 j, for 0 ⱕ t ⱕ , then 兰C F ⭈ dr ⫽ 0.
y
49. If F is conservative in a region R bounded by a simple closed path and C lies within R, then 兰C F ⭈ dr is independent of path. C4
C3
50. If F ⫽ M i ⫹ N j and ⭸M兾⭸x ⫽ ⭸N兾⭸y, then F is conservative.
x x
51. Harmonic Function A function f is called harmonic ⭸ 2f ⭸ 2f when 2 ⫹ 2 ⫽ 0. Prove that if f is harmonic, then ⭸x ⭸y
冕冢 C
44.
HOW DO YOU SEE IT? Consider the force field shown in the figure. To print an enlarged copy of the graph, go to MathGraphs.com. y
⭸f ⭸f dx ⫺ dy ⫽ 0 ⭸y ⭸x
冣
where C is a smooth closed curve in the plane. 52. Kinetic and Potential Energy The kinetic energy of an object moving through a conservative force field is decreasing at a rate of 15 units per minute. At what rate is the potential energy changing? Let F共x, y兲 ⫽
53. Investigation
y x i⫺ 2 j. x2 ⫹ y 2 x ⫹ y2
(a) Show that x
−5
⭸N ⭸M ⫽ ⭸x ⭸y where
−5
(a) Give a verbal argument that the force field is not conservative because you can identify two paths that require different amounts of work to move an object from 共⫺4, 0兲 to 共3, 4兲. Of the two paths, which requires the greater amount of work? (b) Give a verbal argument that the force field is not conservative because you can find a closed curve C such that 兰C F ⭈ dr ⫽ 0.
M⫽
y x2 ⫹ y 2
and
N⫽
⫺x . x2 ⫹ y 2
(b) Let r共t兲 ⫽ cos t i ⫹ sin t j for 0 ⱕ t ⱕ . Find 兰C F ⭈ dr. (c) Let r共t兲 ⫽ cos t i ⫺ sin t j for 0 ⱕ t ⱕ . Find 兰C F ⭈ dr. (d) Let r共t兲 ⫽ cos t i ⫹ sin t j for 0 ⱕ t ⱕ 2. Find 兰C F ⭈ dr. Why doesn’t this contradict Theorem 15.7?
冢
(e) Show that ⵜ arctan
冣
x ⫽ F. y
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15.4
Green’s Theorem
1075
15.4 Green’s Theorem Use Green’s Theorem to evaluate a line integral. Use alternative forms of Green’s Theorem.
Green’s Theorem r(a) = r(b)
R1
Simply connected
In this section, you will study Green’s Theorem, named after the English mathematician George Green (1793–1841). This theorem states that the value of a double integral over a simply connected plane region R is determined by the value of a line integral around the boundary of R. A curve C given by r共t兲 x共t兲i y共t兲j, where a t b, is simple when it does not cross itself—that is, r共c兲 r共d兲 for all c and d in the open interval 共a, b兲. A connected plane region R is simply connected when every simple closed curve in R encloses only points that are in R (see Figure 15.26). Informally, a simply connected region cannot consist of separate parts or holes.
R3 R2
THEOREM 15.8 Green’s Theorem Let R be a simply connected region with a piecewise smooth boundary C, oriented counterclockwise (that is, C is traversed once so that the region R always lies to the left). If M and N have continuous first partial derivatives in an open region containing R, then
Not simply connected
Figure 15.26
冕
M dx N dy
C
y
冕冕 冢 R
N M dA. x y
冣
Proof A proof is given only for a region that is both vertically simple and horizontally simple, as shown in Figure 15.27.
C2: y = f2(x)
冕
M dx
冕 冕 冕
M dx
C1
C
R
冕
M dx
C2
b
C1: y = f1(x)
冕
a
M共x, f1共x兲兲 dx
a
M共x, f2共x兲兲 dx
b
b
a C = C1 + C2
b
关M共x, f1共x兲兲 M共x, f2共x兲兲兴 dx
a
On the other hand,
R is vertically simple. y
x
冕冕
C ′1: x = g1(y)
R
M dA y
冕冕 冕 冕 b
a
b
d
C ′2: x = g2(y)
f1 x
x
关M共x, f2共x兲兲 M共x, f1共x兲兲兴 dx.
a
Consequently,
R is horizontally simple. Figure 15.27
f2共x兲
冥 共 兲 dx
b
R
C ′ = C ′1 + C ′2
f1共x兲
M dy dx y
M共x, y兲
a
c
f2共x兲
冕
C
M dx
冕冕 R
M dA. y
Similarly, you can use g1共 y兲 and g2共 y兲 to show that 兰C N dy 兰R兰 N兾x dA. By adding the integrals 兰C M dx and 兰C N dy, you obtain the conclusion stated in the theorem. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1076
Chapter 15
Vector Analysis
An integral sign with a circle is sometimes used to indicate a line integral around a simple closed curve, as shown below. To indicate the orientation of the boundary, an arrow can be used. For instance, in the second integral, the arrow indicates that the boundary C is oriented counterclockwise. M dx N dy
1.
M dx N dy
2.
C
C
Using Green’s Theorem y
Use Green’s Theorem to evaluate the line integral
冕
C = C1 + C2 (1, 1)
y=x
1
where C is the path from 共0, 0兲 to 共1, 1兲 along the graph of y x3 and from 共1, 1兲 to 共0, 0兲 along the graph of y x, as shown in Figure 15.28.
C1 C2 (0, 0)
y 3 dx 共x3 3xy 2兲 dy
C
y = x3
Solution x
1
C is simple and closed, and the region R always lies to the left of C. Figure 15.28
Because M y 3 and N x 3 3xy 2, it follows that
N 3x 2 3y 2 and x
M 3y 2. y
Applying Green’s Theorem, you then have
冕
y 3 dx 共x 3 3xy 2兲 dy
C
冕冕 冢 冕冕 冕冕 冕 冥 冕 R 1
N M dA x y
冣
x
关共3x2 3y 2兲 3y 2兴 dy dx
0 x3 1 x
3x 2 dy dx
0 x3 1
3x 2y
0 1
x
dx x3
共3x 3 3x5兲 dx
0
冤
3x 4 x 6 4 2
1
冥
0
1 . 4 GEORGE GREEN (1793–1841)
Green, a self-educated miller’s son, first published the theorem that bears his name in 1828 in an essay on electricity and magnetism. At that time, there was almost no mathematical theory to explain electrical phenomena. “Considering how desirable it was that a power of universal agency, like electricity, should, as far as possible, be submitted to calculation, . . . I was induced to try whether it would be possible to discover any general relations existing between this function and the quantities of electricity in the bodies producing it.”
Green’s Theorem cannot be applied to every line integral. Among other restrictions stated in Theorem 15.8, the curve C must be simple and closed. When Green’s Theorem does apply, however, it can save time. To see this, try using the techniques described in Section 15.2 to evaluate the line integral in Example 1. To do this, you would need to write the line integral as
冕
y 3 dx 共x 3 3xy 2兲 dy
C
冕
y 3 dx 共x 3 3xy 2兲 dy
C1
冕
y 3 dx 共x 3 3xy 2兲 dy
C2
where C1 is the cubic path given by r共t兲 ti t 3j from t 0 to t 1, and C2 is the line segment given by r共t兲 共1 t兲i 共1 t兲j from t 0 to t 1.
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15.4
Green’s Theorem
1077
Using Green’s Theorem to Calculate Work While subject to the force
F(x, y) = y 3 i + (x 3 + 3xy 2)j
F共x, y兲 y 3i 共x3 3xy 2兲j
y
a particle travels once around the circle of radius 3 shown in Figure 15.29. Use Green’s Theorem to find the work done by F.
C 2
Solution
1
冕
x
−2
−1
1
2
From Example 1, you know by Green’s Theorem that
y 3 dx 共x 3 3xy 2兲 dy
C
−1
冕冕
3x 2 dA.
R
In polar coordinates, using x r cos and dA r dr d , the work done is
−2
W
r=3
冕冕 冕冕 冕冕 冕 冕 冕
3x 2 dA
R
Figure 15.29
2
2
3
2
r4 cos2 4
0
2
3
0
r 3 cos2 dr d
0
0
3
3共r cos 兲2r dr d
0
0
3
3
3
共1 cos 2 兲 d
0
243 sin 2 8 2 243 . 4
d
0
81 cos2 d 4
2
243 8
冥
冤
2
冥
0
When evaluating line integrals over closed curves, remember that for conservative vector fields (those for which N兾x M兾y), the value of the line integral is 0. This is easily seen from the statement of Green’s Theorem:
冕
M dx N dy
C
冕冕 冢 R
N M dA 0. x y
冣
y
Green’s Theorem and Conservative Vector Fields C
Evaluate the line integral
冕
y 3 dx 3xy 2 dy
C
x
where C is the path shown in Figure 15.30. Solution From this line integral, M y 3 and N 3xy 2. So, N兾x 3y 2 and M兾y 3y 2. This implies that the vector field F Mi Nj is conservative, and because C is closed, you can conclude that
C is closed. Figure 15.30
冕
y 3 dx 3xy 2 dy 0.
C
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1078
Chapter 15
Vector Analysis
Using Green’s Theorem See LarsonCalculus.com for an interactive version of this type of example. y
Evaluate
冕
(0, 3) C
共arctan x y 2兲 dx 共e y x2兲 dy
C
R
where C is the path enclosing the annular region shown in Figure 15.31. x
(−3, 0)
(−1, 0)
(1, 0)
C is piecewise smooth. Figure 15.31
(3, 0)
Solution
In polar coordinates, R is given by 1 r 3 for 0 . Moreover,
N M 2x 2y 2共r cos r sin 兲. x y So, by Green’s Theorem,
冕
共arctan x y 2兲 dx 共ey x 2兲 dy
C
冕冕 冕冕 冕 冕冢
2共x y兲 dA
R
0
3
2r 共cos sin 兲r dr d
1
2共cos sin 兲
0
0
冥
3 1
d
冣
52 共cos sin 兲 d 3
冤
52 sin cos 3 104 . 3
r3 3
冥
0
In Examples 1, 2, and 4, Green’s Theorem was used to evaluate line integrals as double integrals. You can also use the theorem to evaluate double integrals as line integrals. One useful application occurs when N兾x M兾y 1.
冕
M dx N dy
C
冕冕 冢 冕冕 R
N M dA x y
1 dA
R
冣
N M 1 x y
area of region R Among the many choices for M and N satisfying the stated condition, the choice of M
y 2
and
N
x 2
produces the following line integral for the area of region R. THEOREM 15.9 Line Integral for Area If R is a plane region bounded by a piecewise smooth simple closed curve C, oriented counterclockwise, then the area of R is given by A
1 2
冕
x dy y dx.
C
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15.4
Green’s Theorem
1079
Finding Area by a Line Integral Use a line integral to find the area of the ellipse 共x2兾a2兲 共 y2兾b2兲 1. y
Solution Using Figure 15.32, you can induce a counterclockwise orientation to the elliptical path by letting x a cos t and y b sin t, 0 t 2. So, the area is
x2 y2 + =1 a2 b2
A b
1 2
冕
冕 冕
2
1 关共a cos t兲共b cos t兲 dt 共b sin t兲共a sin t兲 dt兴 2 0 2 ab 共cos 2 t sin2 t兲 dt 2 0 ab 2 t 2 0 ab.
x dy y dx
C
a x
R
冤冥
Figure 15.32
Green’s Theorem can be extended to cover some regions that are not simply connected. This is demonstrated in the next example.
Green’s Theorem Extended to a Region with a Hole y 2
C1
R
−3
冕
C3 C2
−2
C3: y = 0, 1 ≤ x ≤ 3 C4: y = 0, 1 ≤ x ≤ 3
Figure 15.33
Let R be the region inside the ellipse 共x 2兾9兲 共 y 2兾4兲 1 and outside the circle x 2 y 2 1. Evaluate the line integral
C1: Ellipse C2: Circle
3
C4
2xy dx 共x 2 2x兲 dy
C
x
where C C1 C2 is the boundary of R, as shown in Figure 15.33. Solution To begin, introduce the line segments C3 and C4, as shown in Figure 15.33. Note that because the curves C3 and C4 have opposite orientations, the line integrals over them cancel. Furthermore, apply Green’s Theorem to the region R using the boundary C1 C4 C2 C3 to obtain
冕
2xy dx 共x 2 2x兲 dy
C
冕冕 冢 冕冕 冕冕 R
N M dA x y
冣
共2x 2 2x兲 dA
R
2
dA
R
2共area of R兲 2共ab r 2兲 2关 共3兲共2兲 共12兲兴 10. In Section 15.1, a necessary and sufficient condition for conservative vector fields was listed. There, only one direction of the proof was shown. You can now outline the other direction, using Green’s Theorem. Let F共x, y兲 Mi Nj be defined on an open disk R. You want to show that if M and N have continuous first partial derivatives and M兾y N兾x, then F is conservative. Let C be a closed path forming the boundary of a connected region lying in R. Then, using the fact that M兾y N兾x, apply Green’s Theorem to conclude that
冕
C
F dr
冕
C
M dx N dy
冕冕 冢 R
N M dA 0. x y
冣
This, in turn, is equivalent to showing that F is conservative (see Theorem 15.7).
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1080
Chapter 15
Vector Analysis
Alternative Forms of Green’s Theorem This section concludes with the derivation of two vector forms of Green’s Theorem for regions in the plane. The extension of these vector forms to three dimensions is the basis for the discussion in the remaining sections of this chapter. For a vector field F in the plane, you can write F共x, y, z兲 Mi Nj 0k
ⱍ ⱍ
so that the curl of F, as described in Section 15.1, is given by i curl F F x M Consequently,
冤
共curl F兲 k
j y N
k N i M j N M k. z z z x y 0
冢
冣
N M N M i j k z z x y
冢
M . 冣 冥 k N x y
With appropriate conditions on F, C, and R, you can write Green’s Theorem in the vector form
冕
C
F dr
冕冕 冢 冕冕 R
N M dA x y
冣
共curl F兲 k dA.
R
First alternative form
The extension of this vector form of Green’s Theorem to surfaces in space produces Stokes’s Theorem, discussed in Section 15.8. For the second vector form of Green’s Theorem, assume the same conditions for F, C, and R. Using the arc length parameter s for C, you have r共s兲 x共s兲i y共s兲j. So, a unit tangent vector T to curve C is given by r 共s兲 T x 共s兲i y 共s兲j. From Figure 15.34, you can see that the outward unit normal vector N can then be written as n C
T θ
N = −n
T cos i sin j n cos i sin j 2 2 sin i cos j N sin i cos j Figure 15.34
冢
冣
冢
N y 共s兲i x 共s兲j. Consequently, for F共x, y兲 Mi Nj, you can apply Green’s Theorem to obtain
冕
C
F N ds
冣
冕 冕冢 冕 冕 冕冕 冢 冕冕 b
a b
共Mi Nj兲 共 y 共s兲i x 共s兲j兲 ds M
a
冣
dy dx N ds ds ds
M dy N dx
C
N dx M dy
C
R
M N dA x y
冣
Green’s Theorem
div F dA.
R
Therefore,
冕
C
F N ds
冕冕
div F dA.
Second alternative form
R
The extension of this form to three dimensions is called the Divergence Theorem and will be discussed in Section 15.7. The physical interpretations of divergence and curl will be discussed in Sections 15.7 and 15.8.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.4
15.4 Exercises
y2 dx ⴙ x 2 dy ⴝ
C
冕冕 冸 R
N M ⴚ dA x y
冹
13.
3. C: square with vertices 共0, 0兲, 共1, 0兲, 共1, 1兲, and 共0, 1兲
15.
16.
Green’s Theorem by using a computer algebra system to evaluate both integrals
冕冕 冸 R
17.
C: r 1 cos
e x cos 2y dx 2e x sin 2y dy
18.
冇 y ⴚ x冈 dx ⴙ 冇2x ⴚ y冈 dy
C
for the given path. 7. C: boundary of the region lying between the graphs of y x and y x 2 2x 8. C: x 2 cos , y sin 9. C: boundary of the region lying inside the rectangle bounded by x 5, x 5, y 3, and y 3, and outside the square bounded by x 1, x 1, y 1, and y 1 10. C: boundary of the region lying inside the semicircle y 冪25 x 2 and outside the semicircle y 冪9 x 2
Evaluating a Line Integral Using Green’s Theorem In Exercises 11–20, use Green’s Theorem to evaluate the line integral. 2xy dx 共x y兲 dy
cos y dx 共xy x sin y兲 dy
冕
共ex 兾2 y兲 dx 共ey 2
兾2
2
x兲 dy
C: boundary of the region lying between the graphs of the circle x 6 cos , y 6 sin and the ellipse x 3 cos , y 2 sin 19.
冕
共x 3y兲 dx 共x y兲 dy
C
6. C: boundary of the region lying between the graphs of y x and y x3 in the first quadrant
In Exercises 7–10, use Green’s Theorem to evaluate the integral
冕
C
冹
Evaluating a Line Integral Using Green’s Theorem
y 2 arctan dx ln共x 2 y 2兲 dy x C
C: boundary of the region lying between the graphs of y x and y 冪x
N M ⴚ dA x y
5. C: circle given by x 2 y 2 4
冕
C
for the given path.
冕
共x 2 y 2兲 dx 2xy dy
C
C: x 4 2 cos , y 4 sin
Verifying Green’s Theorem In Exercises 5 and 6, verify
11.
冕
冕
C: x 2 y 2 a 2
4. C: rectangle with vertices 共0, 0兲, 共3, 0兲, 共3, 4兲, and 共0, 4兲
冕
14.
C: x 2 y 2 16
2. C: boundary of the region lying between the graphs of y x and y 冪x
C
共x 2 y 2兲 dx 2xy dy
C
1. C: boundary of the region lying between the graphs of y x and y x2
xe y dx ⴙ e x dy ⴝ
冕
C
for the given path.
冕
1081
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Verifying Green’s Theorem In Exercises 1– 4, verify Green’s Theorem by evaluating both integrals
冕
Green’s Theorem
C: boundary of the region lying between the graphs of x 2 y 2 1 and x 2 y 2 9 20.
冕
3x 2e y dx ey dy
C
C: boundary of the region lying between the squares with vertices 共1, 1兲, 共1, 1兲, 共1, 1兲, and 共1, 1兲, and 共2, 2兲, 共2, 2兲, 共2, 2兲, and 共2, 2兲
Work In Exercises 21–24, use Green’s Theorem to calculate the work done by the force F on a particle that is moving counterclockwise around the closed path C. 21. F共x, y兲 xyi 共x y兲j C: x 2 y 2 1 22. F共x, y兲 共e x 3y兲i 共ey 6x兲j C: r 2 cos 23. F共x, y兲 共x 3兾2 3y兲i 共6x 5冪y 兲j C: boundary of the triangle with vertices 共0, 0兲, 共5, 0兲, and 共0, 5兲 24. F共x, y兲 共3x 2 y兲i 4xy 2j C: boundary of the region lying between the graphs of y 冪x, y 0, and x 9
C
C: boundary of the region lying between the graphs of y 0 and y 1 x 2 12.
冕
y 2 dx xy dy
C
C: boundary of the region lying between the graphs of y 0, y 冪x, and x 9
Area In Exercises 25–28, use a line integral to find the area of the region R. 25. R: region bounded by the graph of x 2 y 2 a 2 26. R: triangle bounded by the graphs of x 0, 3x 2y 0, and x 2y 8
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1082
Chapter 15
Vector Analysis
27. R: region bounded by the graphs of y 5x 3 and y x2 1
42.
HOW DO YOU SEE IT? Use Green’s Theorem to explain why
28. R: region inside the loop of the folium of Descartes bounded by the graph of
冕
f 共x兲 dx g共 y兲 dy 0
C
3t 3t 2 x 3 , y 3 t 1 t 1
where f and g are differentiable functions and C is a piecewise smooth simple closed path (see figure).
WRITING ABOUT CONCEPTS 29. Green’s Theorem
y
State Green’s Theorem.
30. Area Give the line integral for the area of a region R bounded by a piecewise smooth simple curve C.
C x
Using Green’s Theorem to Verify a Formula
In Exercises 31 and 32, use Green’s Theorem to verify the line integral formulas. 31. The centroid of the region having area A bounded by the simple closed path C is 1 2A
x
冕
x 2 dy,
y
C
1 2A
冕
y 2 dx.
C
32. The area of a plane region bounded by the simple closed path C given in polar coordinates is A
1 2
冕
C
r 2 d .
43. Green’s Theorem: Region with a Hole Let R be the region inside the circle x 5 cos , y 5 sin and outside the ellipse x 2 cos , y sin . Evaluate the line integral
冕
共ex 兾2 y兲 dx 共ey 兾2 x兲 dy 2
2
C
where C C1 C2 is the boundary of R, as shown in the figure.
Centroid In Exercises 33–36, use the results of Exercise 31
y
to find the centroid of the region. 33. R: region bounded by the graphs of y 0 and y 4 x 2
4 3 2
34. R: region bounded by the graphs of y 冪a x and y 0 2
2
35. R: region bounded by the graphs of y x3 and y x, 0x1
37. r a共1 cos 兲 38. r a cos 3
40. r
2 3 4 −2 −3 −4
44. Green’s Theorem: Region with a Hole Let R be the region inside the ellipse x 4 cos , y 3 sin and outside the circle x 2 cos , y 2 sin . Evaluate the line integral
共3x 2y 1兲 dx 共x 3 4x兲 dy
C
3 2 cos
where C C1 C2 is the boundary of R, as shown in the figure. y
41. Maximum Value (a) Evaluate
冕
C1
C1
C2 x
冕
39. r 1 2 cos (inner loop)
R
−4 −3 −2
36. R: triangle with vertices 共a, 0兲, 共a, 0兲, and 共b, c兲, where a b a
Area In Exercises 37–40, use the results of Exercise 32 to find the area of the region bounded by the graph of the polar equation.
C1: Circle C2: Ellipse
y3 dx 共27x x3兲 dy,
where C1 is the unit circle given by r共t兲 cos t i sin t j, for 0 t 2. (b) Find the maximum value of
冕
C
y3 dx 共27x x3兲 dy,
C1: Ellipse C2: Circle
R C2
1
x −3
−1 −1
1
3
C1
where C is any closed curve in the xy-plane, oriented counterclockwise.
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15.4 45. Think About It
冕
I
Let
y dx x dy x2 y 2
C
where C is a circle oriented counterclockwise. Show that I 0 when C does not contain the origin. What is I when C does contain the origin? 46. Think About It For each given path, verify Green’s Theorem by showing that
冕
y2 dx x2 dy
C
冕 冕冢 R
冣
(b) C: circle given by x2 y2 1 (a) Let C be the line segment joining 共x1, y1兲 and 共x2, y2兲. Show that 兰C y dx x dy x1y2 x2 y1. (b) Let 共x1, y1兲, 共x2, y2兲, . . . , 共xn, yn 兲 be the vertices of a polygon. Prove that the area enclosed is 1 关共x y x y 兲 共x y x y 兲 . . . 2 3
共 f 2g f g兲 dA
50. Green’s second identity:
共 f 2g g 2f 兲 dA
R
47. Proof
2 1
冕冕 冕冕
(a) C: triangle with vertices 共0, 0兲, 共4, 0兲, and 共4, 4兲
1 2
49. Green’s first identity:
冕
f DNg ds
C
[Hint: Use the second alternative form of Green’s Theorem and the property div共 f G兲 f div G f G.兴
For each path, which integral is easier to evaluate? Explain.
2
1083
Proof In Exercises 49 and 50, prove the identity, where R is a simply connected region with boundary C. Assume that the required partial derivatives of the scalar functions f and g are continuous. The expressions DN f and DN g are the derivatives in the direction of the outward normal vector N of C, and are defined by DN f ⴝ f N, and DN g g N.
R
N M dA. x y
Green’s Theorem
冕
共 f DNg gDN f 兲 ds
C
(Hint: Use Green’s first identity from Exercise 49 twice.) 51. Proof Let F Mi Nj, where M and N have continuous first partial derivatives in a simply connected region R. Prove that if C is simple, smooth, and closed, and Nx My , then 兰C F dr 0.
3 2
PUTNAM EXAM CHALLENGE
共xn1 yn xnyn1兲 共xn y1 x1yn兲兴. 48. Area Use the result of Exercise 47(b) to find the area enclosed by the polygon with the given vertices. (a) Pentagon: 共0, 0兲, 共2, 0兲, 共3, 2兲, 共1, 4兲, and 共1, 1兲 (b) Hexagon: 共0, 0兲, 共2, 0兲, 共3, 2兲, 共2, 4兲, 共0, 3兲, and 共1, 1兲
52. Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola xy 1 and both branches of the hyperbola xy 1. 共A set S in the plane is called convex if for any two points in S the line segment connecting them is contained in S.兲 This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
Hyperbolic and Trigonometric Functions (a) Sketch the plane curve represented by the vector-valued function r共t兲 cosh ti sinh tj on the interval 0 t 5. Show that the rectangular equation corresponding to r共t兲 is the hyperbola x2 y2 1. Verify your sketch by using a graphing utility to graph the hyperbola. (b) Let P 共cosh , sinh 兲 be the point on the hyperbola corresponding to r共兲 for > 0. Use the formula for area A
1 2
冕
x dy y dx
(d) Consider the unit circle given by x2 y2 1. Let be the angle formed by the x-axis and the radius to 共x, y兲. The area of the corresponding sector is 12 . That is, the trigonometric functions f 共 兲 cos and g共 兲 sin could have been defined as the coordinates of that point 共cos , sin 兲 on the unit circle that determines a sector of area 12 . Write a short paragraph explaining how you could define the hyperbolic functions in a similar manner, using the “unit hyperbola” x2 y2 1.
C
y 1 2 .
to verify that the area of the region shown in the figure is (c) Show that the area of the indicated region is also given by the integral A
冕
sinh
0
(cosh φ, sinh φ)
关冪1 y2 共coth 兲y兴 dy.
Confirm your answer in part (b) by numerically approximating this integral for 1, 2, 4, and 10. (0, 0)
(1, 0)
x
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1084
Chapter 15
Vector Analysis
15.5 Parametric Surfaces Understand the definition of a parametric surface, and sketch the surface. Find a set of parametric equations to represent a surface. Find a normal vector and a tangent plane to a parametric surface. Find the area of a parametric surface.
Parametric Surfaces You already know how to represent a curve in the plane or in space by a set of parametric equations—or, equivalently, by a vector-valued function. rt ⫽ xti ⫹ ytj rt ⫽ xti ⫹ ytj ⫹ ztk
Plane curve Space curve
In this section, you will learn how to represent a surface in space by a set of parametric equations—or by a vector-valued function. For curves, note that the vector-valued function r is a function of a single parameter t. For surfaces, the vector-valued function is a function of two parameters u and v. Definition of Parametric Surface Let x, y, and z be functions of u and v that are continuous on a domain D in the uv-plane. The set of points x, y, z given by ru, v ⫽ xu, vi ⫹ yu, vj ⫹ zu, vk
Parametric surface
is called a parametric surface. The equations x ⫽ xu, v,
y ⫽ yu, v,
and
z ⫽ zu, v
Parametric equations
are the parametric equations for the surface.
If S is a parametric surface given by the vector-valued function r, then S is traced out by the position vector ru, v as the point u, v moves throughout the domain D, as shown in Figure 15.35. v
z
D
S (u, v)
r(u, v)
y u
x
Figure 15.35
TECHNOLOGY Some computer algebra systems are capable of graphing surfaces that are represented parametrically. If you have access to such software, use it to graph some of the surfaces in the examples and exercises in this section.
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15.5
Parametric Surfaces
1085
Sketching a Parametric Surface Identify and sketch the parametric surface S given by ru, v ⫽ 3 cos ui ⫹ 3 sin uj ⫹ vk where 0 ⱕ u ⱕ 2 and 0 ⱕ v ⱕ 4. z
Solution Because x ⫽ 3 cos u and y ⫽ 3 sin u, you know that for each point x, y, z on the surface, x and y are related by the equation
3
x2 ⫹ y2 ⫽ 32. 4
In other words, each cross section of S taken parallel to the xy-plane is a circle of radius 3, centered on the z-axis. Because z ⫽ v, where 0 ⱕ v ⱕ 4
y x
you can see that the surface is a right circular Figure 15.36 cylinder of height 4. The radius of the cylinder is 3, and the z-axis forms the axis of the cylinder, as shown in Figure 15.36. As with parametric representations of curves, parametric representations of surfaces are not unique. That is, there are many other sets of parametric equations that could be used to represent the surface shown in Figure 15.36.
Sketching a Parametric Surface z
c3
Identify and sketch the parametric surface S given by ru, v ⫽ sin u cos vi ⫹ sin u sin vj ⫹ cos uk
c2
where 0 ⱕ u ⱕ and 0 ⱕ v ⱕ 2.
d1 c4
Solution To identify the surface, you can try to use trigonometric identities to eliminate the parameters. After some experimentation, you can discover that
c1
d2
d3
x
d4
Figure 15.37
y
x2 ⫹ y2 ⫹ z2 ⫽ sin u cos v2 ⫹ sin u sin v2 ⫹ cos u2 ⫽ sin2 u cos2 v ⫹ sin2 u sin2 v ⫹ cos2 u ⫽ sin2 ucos2 v ⫹ sin2 v ⫹ cos2 u ⫽ sin2 u ⫹ cos2 u ⫽ 1. So, each point on S lies on the unit sphere, centered at the origin, as shown in Figure 15.37. For fixed u ⫽ di , ru, v traces out latitude circles x2 ⫹ y2 ⫽ sin2 di ,
0 ⱕ di ⱕ
that are parallel to the xy-plane, and for fixed v ⫽ ci , ru, v traces out longitude (or meridian) half-circles. To convince yourself further that ru, v traces out the entire unit sphere, recall that the parametric equations x ⫽ sin cos ,
y ⫽ sin sin , and
z ⫽ cos
where 0 ⱕ ⱕ 2 and 0 ⱕ ⱕ , describe the conversion from spherical to rectangular coordinates, as discussed in Section 11.7.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1086
Chapter 15
Vector Analysis
Finding Parametric Equations for Surfaces In Examples 1 and 2, you were asked to identify the surface described by a given set of parametric equations. The reverse problem—that of writing a set of parametric equations for a given surface—is generally more difficult. One type of surface for which this problem is straightforward, however, is a surface that is given by z ⫽ f x, y. You can parametrize such a surface as rx, y ⫽ x i ⫹ yj ⫹ f x, yk.
z
3
Representing a Surface Parametrically Write a set of parametric equations for the cone given by
2
z ⫽ x2 ⫹ y2 as shown in Figure 15.38. Solution Because this surface is given in the form z ⫽ f x, y, you can let x and y be the parameters. Then the cone is represented by the vector-valued function
−2 1 1
2
2
x
rx, y ⫽ x i ⫹ yj ⫹ x2 ⫹ y2 k
y
where x, y varies over the entire xy-plane.
Figure 15.38
A second type of surface that is easily represented parametrically is a surface of revolution. For instance, to represent the surface formed by revolving the graph of y ⫽ f x, a ⱕ x ⱕ b about the x-axis, use x ⫽ u,
y ⫽ f u cos v, and
z ⫽ f u sin v
where a ⱕ u ⱕ b and 0 ⱕ v ⱕ 2.
Representing a Surface of Revolution Parametrically See LarsonCalculus.com for an interactive version of this type of example. z
Write a set of parametric equations for the surface of revolution obtained by revolving 1 f x ⫽ , x
1 1
1 ⱕ x ⱕ 10
y
about the x-axis. Solution
Use the parameters u and v as described above to write
x ⫽ u, y ⫽ f u cos v ⫽
10 x
Figure 15.39
1 cos v, u
and
z ⫽ f u sin v ⫽
1 sin v u
where 1 ⱕ u ⱕ 10
and 0 ⱕ v ⱕ 2.
The resulting surface is a portion of Gabriel’s Horn, as shown in Figure 15.39. The surface of revolution in Example 4 is formed by revolving the graph of y ⫽ f x about the x-axis. For other types of surfaces of revolution, a similar parametrization can be used. For instance, to parametrize the surface formed by revolving the graph of x ⫽ f z about the z-axis, you can use z ⫽ u,
x ⫽ f u cos v, and
y ⫽ f u sin v.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.5
Parametric Surfaces
1087
Normal Vectors and Tangent Planes Let S be a parametric surface given by ru, v ⫽ xu, vi ⫹ yu, vj ⫹ zu, vk over an open region D such that x, y, and z have continuous partial derivatives on D. The partial derivatives of r with respect to u and v are defined as ru ⫽
⭸x ⭸y ⭸z u, vi ⫹ u, vj ⫹ u, vk ⭸u ⭸u ⭸u
rv ⫽
⭸x ⭸y ⭸z u, vi ⫹ u, vj ⫹ u, vk. ⭸v ⭸v ⭸v
and
Each of these partial derivatives is a vector-valued function that can be interpreted geometrically in terms of tangent vectors. For instance, if v ⫽ v0 is held constant, then ru, v0 is a vector-valued function of a single parameter and defines a curve C1 that lies on the surface S. The tangent vector to C1 at the point
xu0, v0 , yu0, v0 , zu0, v0 is given by ruu0, v0 ⫽ N
as shown in Figure 15.40. In a similar way, if u ⫽ u0 is held constant, then ru0, v is a vector-valued function of a single parameter and defines a curve C2 that lies on the surface S. The tangent vector to C2 at the point xu0, v0 , yu0, v0 , zu0, v0 is given by
z
(x0, y0, z 0 ) rv C2
⭸x ⭸y ⭸z u , v i ⫹ u0, v0 j ⫹ u0, v0 k ⭸u 0 0 ⭸u ⭸u
rvu0, v0 ⫽
ru C1 S
x y
⭸x ⭸y ⭸z u0, v0 i ⫹ u0, v0 j ⫹ u0, v0 k. ⭸v ⭸v ⭸v
If the normal vector ru ⫻ rv is not 0 for any u, v in D, then the surface S is called smooth and will have a tangent plane. Informally, a smooth surface is one that has no sharp points or cusps. For instance, spheres, ellipsoids, and paraboloids are smooth, whereas the cone given in Example 3 is not smooth.
Figure 15.40
Normal Vector to a Smooth Parametric Surface Let S be a smooth parametric surface ru, v ⫽ xu, vi ⫹ yu, vj ⫹ zu, vk defined over an open region D in the uv-plane. Let u0, v0 be a point in D. A normal vector at the point
x0, y0, z0 ⫽ xu0, v0 , yu0, v0 , zu0, v0 is given by
i
⭸x N ⫽ ruu0, v0 ⫻ rvu0, v0 ⫽ ⭸u ⭸x ⭸v
j
⭸y ⭸u ⭸y ⭸v
Figure 15.40 shows the normal vector ru S and points in the opposite direction.
k
⭸z ⭸u . ⭸z ⭸v
⫻ rv . The
vector rv ⫻ ru is also normal to
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1088
Chapter 15
Vector Analysis
Finding a Tangent Plane to a Parametric Surface Find an equation of the tangent plane to the paraboloid ru, v ⫽ ui ⫹ vj ⫹ u2 ⫹ v2k at the point 1, 2, 5. Solution The point in the uv-plane that is mapped to the point x, y, z ⫽ 1, 2, 5 is u, v ⫽ 1, 2. The partial derivatives of r are
z 7
ru ⫽ i ⫹ 2uk and
6
rv ⫽ j ⫹ 2vk.
The normal vector is given by
(1, 2, 5)
ru
⫻ rv
i ⫽ 1 0
j 0 1
k 2u ⫽ ⫺2ui ⫺ 2vj ⫹ k 2v
which implies that the normal vector at 1, 2, 5 is ru −3
−2
⫻ rv
⫽ ⫺2i ⫺ 4j ⫹ k.
So, an equation of the tangent plane at 1, 2, 5 is −1
1
2
2
y
3
3 x
⫺2x ⫺ 1 ⫺ 4 y ⫺ 2 ⫹ z ⫺ 5 ⫽ 0 ⫺2x ⫺ 4y ⫹ z ⫽ ⫺5. The tangent plane is shown in Figure 15.41.
Figure 15.41
Area of a Parametric Surface
v
To define the area of a parametric surface, you can use a development that is similar to that given in Section 14.5. Begin by constructing an inner partition of D consisting of n rectangles, where the area of the ith rectangle Di is ⌬Ai ⫽ ⌬ui ⌬vi , as shown in Figure 15.42. In each Di , let ui, vi be the point that is closest to the origin. At the point xi, yi, zi ⫽ xui, vi , yui, vi , zui, vi on the surface S, construct a tangent plane Ti . The area of the portion of S that corresponds to Di, ⌬Ti , can be approximated by a parallelogram in the tangent plane. That is, ⌬Ti ⌬Si . So, the surface of S is given by ⌬Si ⌬Ti . The area of the parallelogram in the tangent plane is
Di Δvi
⌬ui ru
Δui
⫻
⌬vi rv ⫽ ru
u
Area of a Parametric Surface Let S be a smooth parametric surface
z
ru, v ⫽ xu, vi ⫹ yu, vj ⫹ zu, vk
Δvi rv
defined over an open region D in the uv-plane. If each point on the surface S corresponds to exactly one point in the domain D, then the surface area of S is given by
S
Surface area ⫽
dS ⫽
S
Δui ru
ru
⫻ rv
dA
D
where y
Figure 15.42
⌬ui ⌬vi
which leads to the next definition.
(ui, vi)
x
⫻ rv
ru ⫽
⭸x ⭸y ⭸z i⫹ j⫹ k ⭸u ⭸u ⭸u
and
rv ⫽
⭸x ⭸y ⭸z i⫹ j⫹ k. ⭸v ⭸v ⭸v
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.5
Parametric Surfaces
1089
For a surface S given by z ⫽ f x, y, this formula for surface area corresponds to that given in Section 14.5. To see this, you can parametrize the surface using the vectorvalued function rx, y ⫽ xi ⫹ yj ⫹ f x, yk defined over the region R in the xy-plane. Using rx ⫽ i ⫹ fxx, yk
you have
i rx ⫻ ry ⫽ 1 0 and
and ry ⫽ j ⫹ fyx, yk
j 0 1
k fxx, y ⫽ ⫺fxx, yi ⫺ fyx, yj ⫹ k fyx, y
rx ⫻ ry ⫽ fxx, y 2 ⫹ fyx, y 2 ⫹ 1. This implies that the surface area of S is Surface area ⫽
rx ⫻ ry dA
R
⫽
1 ⫹ fxx, y 2 ⫹ fyx, y 2 dA.
R
Finding Surface Area REMARK The surface in Example 6 does not quite fulfill the hypothesis that each point on the surface corresponds to exactly one point in D. For this surface, ru, 0 ⫽ ru, 2 for any fixed value of u. However, because the overlap consists of only a semicircle (which has no area), you can still apply the formula for the area of a parametric surface.
Find the surface area of the unit sphere ru, v ⫽ sin u cos vi ⫹ sin u sin vj ⫹ cos uk where the domain D is 0 ⱕ u ⱕ and 0 ⱕ v ⱕ 2. Solution
Begin by calculating ru and rv.
ru ⫽ cos u cos vi ⫹ cos u sin vj ⫺ sin uk rv ⫽ ⫺sin u sin vi ⫹ sin u cos vj
The cross product of these two vectors is ru
i j k ⫽ cos u cos v cos u sin v ⫺sin u ⫺sin u sin v sin u cos v 0
⫻ rv
⫽ sin u cos vi ⫹ sin u sin vj ⫹ sin u cos uk 2
2
which implies that ru
⫻ rv
⫽ sin2 u cos v2 ⫹ sin2 u sin v2 ⫹ sin u cos u2 ⫽ sin4 u ⫹ sin2 u cos2 u ⫽ sin2 u ⫽ sin u. sin u > 0 for 0 ⱕ u ⱕ
Finally, the surface area of the sphere is A⫽ ⫽
ru
D 2 0
⫽
2
⫻ rv
dA
sin u du dv
0
2 dv
0
⫽ 4. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1090
Chapter 15
Vector Analysis
Finding Surface Area z
Find the surface area of the torus given by ru, v ⫽ 2 ⫹ cos u cos vi ⫹ 2 ⫹ cos u sin vj ⫹ sin uk where the domain D is given by 0 ⱕ u ⱕ 2 and 0 ⱕ v ⱕ 2. (See Figure 15.43.) Solution
Begin by calculating ru and rv .
ru ⫽ ⫺sin u cos vi ⫺ sin u sin vj ⫹ cos uk rv ⫽ ⫺ 2 ⫹ cos u sin vi ⫹ 2 ⫹ cos u cos vj y x
Figure 15.43
The cross product of these two vectors is ⫽
⫻ rv
ru
i j k ⫺sin u cos v ⫺sin u sin v cos u ⫺ 2 ⫹ cos u sin v 2 ⫹ cos u cos v 0
⫽ ⫺ 2 ⫹ cos ucos v cos ui ⫹ sin v cos uj ⫹ sin uk which implies that ru
⫽ 2 ⫹ cos ucos v cos u2 ⫹ sin v cos u2 ⫹ sin2 u ⫽ 2 ⫹ cos ucos2 ucos2 v ⫹ sin2 v ⫹ sin2 u ⫽ 2 ⫹ cos ucos2 u ⫹ sin2 u ⫽ 2 ⫹ cos u.
⫻ rv
Finally, the surface area of the torus is
Exploration For the torus in Example 7, describe the function ru, v for fixed u. Then describe the function ru, v for fixed v.
A⫽
ru
⫻ rv
dA
D
⫽
2
0
⫽
2
2 ⫹ cos u du dv
0
2
4 dv
0
⫽ 8 2. For a surface of revolution, you can show that the formula for surface area given in Section 7.4 is equivalent to the formula given in this section. For instance, suppose f is a nonnegative function such that f⬘ is continuous over the interval a, b . Let S be the surface of revolution formed by revolving the graph of f, where a ⱕ x ⱕ b, about the x-axis. From Section 7.4, you know that the surface area is given by
b
Surface area ⫽ 2
f x1 ⫹ f⬘x 2 dx.
a
To represent S parametrically, let x ⫽ u,
y ⫽ f u cos v, and
z ⫽ f u sin v
where a ⱕ u ⱕ b and 0 ⱕ v ⱕ 2. Then, ru, v ⫽ ui ⫹ f u cos vj ⫹ f u sin vk. Try showing that the formula Surface area ⫽
ru
⫻ rv
dA
D
is equivalent to the formula given above (see Exercise 58).
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.5
15.5 Exercises
with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] z
z
(b) 2
2
−2
2
0 ⱕ v ⱕ 2
14. ru, v ⫽ 2u cos vi ⫹ 2u sin vj ⫹ vk
− 2 −1
x
0 ⱕ u ⱕ 2,
0 ⱕ u ⱕ 2, 0 ⱕ v ⱕ 2
1
y
−2
12. ru, v ⫽ 2 cos v cos ui ⫹ 4 cos v sin uj ⫹ sin vk 13. ru, v ⫽ 2 sinh u cos vi ⫹ sinh u sin vj ⫹ cosh uk
2
1
1091
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Matching In Exercises 1– 6, match the vector-valued function (a)
Parametric Surfaces
1
2
y
2
0 ⱕ u ⱕ 1,
0 ⱕ v ⱕ 3
15. ru, v ⫽ u ⫺ sin u cos vi ⫹ 1 ⫺ cos u sin vj ⫹ uk
x
0 ⱕ u ⱕ , 0 ⱕ v ⱕ 2 z
(c)
2
2
y
4
4
x
16. ru, v ⫽ cos3 u cos vi ⫹ sin3 u sin vj ⫹ uk 0 ⱕ u ⱕ , 0 ⱕ v ⱕ 2 2
z
(d)
x
2
2
y
Think About It In Exercises 17–20, determine how the graph of the surface s u, v differs from the graph of r u, v ⴝ u cos vi ⴙ u sin vj ⴙ u2 k (see figure), where 0 ⱕ u ⱕ 2 and 0 ⱕ v ⱕ 2. (It is not necessary to graph s.) z
z
(e)
z
(f)
4 2
4
r(u, v) 2
−4
y
4
4
x
2 y
−2
−2
x x
1. ru, v ⫽ ui ⫹ vj ⫹ uvk 2. ru, v ⫽ u cos vi ⫹ u sin vj ⫹ uk 1 3. ru, v ⫽ ui ⫹ 2u ⫹ v j ⫹ vk
4. ru, v ⫽ ui ⫹
1 3 4v j
⫹ vk
5. ru, v ⫽ 2 cos v cos ui ⫹ 2 cos v sin uj ⫹ 2 sin vk 6. ru, v ⫽ 4 cos ui ⫹ 4 sin uj ⫹ vk
Sketching a Parametric Surface In Exercises 7–10, find the rectangular equation for the surface by eliminating the parameters from the vector-valued function. Identify the surface and sketch its graph. 7. ru, v ⫽ ui ⫹ vj ⫹
v k 2
8. ru, v ⫽ 2u cos vi ⫹ 2u sin vj ⫹ 12u2 k 9. ru, v ⫽ 2 cos ui ⫹ vj ⫹ 2 sin uk 10. ru, v ⫽ 3 cos v cos ui ⫹ 3 cos v sin uj ⫹ 5 sin vk
Graphing a Parametric Surface In Exercises 11–16, use a computer algebra system to graph the surface represented by the vector-valued function. 11. ru, v ⫽ 2u cos vi ⫹ 2u sin vj ⫹ u 4k 0 ⱕ u ⱕ 1,
0 ⱕ v ⱕ 2
2
2
y
17. su, v ⫽ u cos vi ⫹ u sin vj ⫺ u2k 0 ⱕ u ⱕ 2,
0 ⱕ v ⱕ 2
18. su, v ⫽ u cos vi ⫹ u2j ⫹ u sin vk 0 ⱕ u ⱕ 2,
0 ⱕ v ⱕ 2
19. su, v ⫽ u cos vi ⫹ u sin vj ⫹ u2k 0 ⱕ u ⱕ 3,
0 ⱕ v ⱕ 2
20. su, v ⫽ 4u cos vi ⫹ 4u sin vj ⫹ u2k 0 ⱕ u ⱕ 2,
0 ⱕ v ⱕ 2
Representing a Surface Parametrically In Exercises 21–30, find a vector-valued function whose graph is the indicated surface. 21. The plane z ⫽ y 22. The plane x ⫹ y ⫹ z ⫽ 6 23. The cone y ⫽ 4x2 ⫹ 9z2 24. The cone x ⫽ 16y2 ⫹ z2 25. The cylinder x2 ⫹ y2 ⫽ 25 26. The cylinder 4x2 ⫹ y2 ⫽ 16 27. The cylinder z ⫽ x2 28. The ellipsoid
x2 y2 z2 ⫹ ⫹ ⫽1 9 4 1
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1092
Chapter 15
Vector Analysis
29. The part of the plane z ⫽ 4 that lies inside the cylinder x2 ⫹ y2 ⫽ 9 30. The part of the paraboloid z ⫽ x2 ⫹ y2 that lies inside the cylinder x2 ⫹ y2 ⫽ 9
Surface of Revolution In Exercises 31–34, write a set of parametric equations for the surface of revolution obtained by revolving the graph of the function about the given axis. Function x 31. y ⫽ , 2
x-axis
33. x ⫽ sin z, 0 ⱕ z ⱕ
z-axis
34. z ⫽ y2 ⫹ 1,
y-axis
0 ⱕ y ⱕ 2
43. The part of the cone ru, v ⫽ au cos vi ⫹ au sin vj ⫹ uk, where 0 ⱕ u ⱕ b and 0 ⱕ v ⱕ 2
Tangent Plane In Exercises 35– 38, find an equation of the tangent plane to the surface represented by the vector-valued function at the given point. 35. ru, v ⫽ u ⫹ vi ⫹ u ⫺ vj ⫹ vk, 1, ⫺1, 1 z
40. The part of the paraboloid ru, v ⫽ 2u cos v i ⫹ 2u sin vj ⫹ u2 k, where 0 ⱕ u ⱕ 2 and 0 ⱕ v ⱕ 2
42. The sphere ru, v ⫽ a sin u cos vi ⫹ a sin u sin vj ⫹ a cos uk, where 0 ⱕ u ⱕ and 0 ⱕ v ⱕ 2
x-axis
32. y ⫽ x, 0 ⱕ x ⱕ 4
39. The part of the plane ru, v ⫽ 4ui ⫺ vj ⫹ vk, where 0 ⱕ u ⱕ 2 and 0 ⱕ v ⱕ 1
41. The part of the cylinder ru, v ⫽ a cos ui ⫹ a sin uj ⫹ vk, where 0 ⱕ u ⱕ 2 and 0 ⱕ v ⱕ b
Axis of Revolution
0 ⱕ x ⱕ 6
Area In Exercises 39–46, find the area of the surface over the given region. Use a computer algebra system to verify your results.
44. The torus ru, v ⫽ a ⫹ b cos v cos ui ⫹ a ⫹ b cos v sin uj ⫹ b sin vk, where a > b, 0 ⱕ u ⱕ 2, and 0 ⱕ v ⱕ 2 45. The surface of revolution ru, v ⫽ u cos vi ⫹ u sin vj ⫹ uk, where 0 ⱕ u ⱕ 4 and 0 ⱕ v ⱕ 2 46. The surface of revolution ru, v ⫽ sin u cos vi ⫹ uj ⫹ sin u sin vk, where 0 ⱕ u ⱕ and 0 ⱕ v ⱕ 2
z
(1, − 1, 1)
WRITING ABOUT CONCEPTS
2
2 −1
47. Parametric Surface
−2 1 (1, 1, 1)
2
y
x
Define a parametric surface.
48. Surface Area Give the double integral that yields the surface area of a parametric surface over an open region D.
2
2 −2
y
1 2 x
Figure for 35
Figure for 36
49. Representing a Cone Parametrically Show that the cone in Example 3 can be represented parametrically by ru, v ⫽ u cos vi ⫹ u sin vj ⫹ uk, where 0 ⱕ u and 0 ⱕ v ⱕ 2.
36. ru, v ⫽ ui ⫹ vj ⫹ uv k, 1, 1, 1 37. ru, v ⫽ 2u cos vi ⫹ 3u sin vj ⫹ u2 k, 0, 6, 4
HOW DO YOU SEE IT? The figures below are graphs of ru, v ⫽ ui ⫹ sin u cos vj ⫹ sin u sin vk, where 0 ⱕ u ⱕ 2 and 0 ⱕ v ⱕ 2. Match each of the graphs with the point in space from which the surface is viewed. The points are 10, 0, 0, ⫺10, 10, 0, 0, 10, 0, and 10, 10, 10.
50.
z 6 5
(0, 6, 4)
z
(a)
z
(b)
−6 2
4
2
x
y 4
6
y
x
y
1 38. ru, v ⫽ 2u cosh vi ⫹ 2u sinh vj ⫹ 2 u2 k, ⫺4, 0, 2
z
z
(c)
(d)
z
4 x
(−4, 0, 2)
y
2
x
6
4
2
−2
−4
−6
4 y
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15.5 51. Astroidal Sphere x, y, and z is
An equation of an astroidal sphere in
57. Area
Parametric Surfaces
1093
Graph and find the area of one turn of the spiral ramp
ru, v ⫽ u cos vi ⫹ u sin vj ⫹ 2vk
x23 ⫹ y23 ⫹ z23 ⫽ a23.
where 0 ⱕ u ⱕ 3 and 0 ⱕ v ⱕ 2.
A graph of an astroidal sphere is shown below. Show that this surface can be represented parametrically by ru, v ⫽ a sin3 u cos3 vi ⫹ a sin3 u sin3 vj ⫹ a cos3 uk where 0 ⱕ u ⱕ and 0 ⱕ v ⱕ 2. z
58. Surface Area Let f be a nonnegative function such that f⬘ is continuous over the interval a, b . Let S be the surface of revolution formed by revolving the graph of f, where a ⱕ x ⱕ b, about the x-axis. Let x ⫽ u, y ⫽ f u cos v, and z ⫽ f u sin v, where a ⱕ u ⱕ b and 0 ⱕ v ⱕ 2. Then, S is represented parametrically by ru, v ⫽ ui ⫹ f u cos vj ⫹ f u sin vk. Show that the following formulas are equivalent.
b
Surface area ⫽ 2
f x1 ⫹ f⬘x 2 dx
a
y x
Surface area ⫽
ru ⫻ rv dA
D
59. Open-Ended Project
The parametric equations
x ⫽ 3 ⫹ sin u 7 ⫺ cos3u ⫺ 2v ⫺ 2 cos3u ⫹ v y ⫽ 3 ⫹ cos u 7 ⫺ cos3u ⫺ 2v ⫺ 2 cos3u ⫹ v 52. Different Views of a Surface Use a computer algebra system to graph three views of the graph of the vector-valued function ru, v ⫽ u cos vi ⫹ u sin vj ⫹ vk, 0 ⱕ u ⱕ , 0 ⱕ v ⱕ
z ⫽ sin3u ⫺ 2v ⫹ 2 sin3u ⫹ v where ⫺ ⱕ u ⱕ and ⫺ ⱕ v ⱕ , represent the surface shown below. Try to create your own parametric surface using a computer algebra system.
from the points 10, 0, 0, 0, 0, 10, and 10, 10, 10. 53. Investigation torus
Use a computer algebra system to graph the
ru, v ⫽ a ⫹ b cos v cos ui ⫹
a ⫹ b cos v sin uj ⫹ b sin vk for each set of values of a and b, where 0 ⱕ u ⱕ 2 and 0 ⱕ v ⱕ 2. Use the results to describe the effects of a and b on the shape of the torus. (a) a ⫽ 4, b ⫽ 1
(b) a ⫽ 4,
b⫽2
(c) a ⫽ 8, b ⫽ 1
(d) a ⫽ 8,
b⫽3
54. Investigation
Consider the function in Exercise 14.
(a) Sketch a graph of the function where u is held constant at u ⫽ 1. Identify the graph.
60. Möbius Strip The surface shown in the figure is called a Möbius strip and can be represented by the parametric equations
(b) Sketch a graph of the function where v is held constant at v ⫽ 23. Identify the graph.
x ⫽ a ⫹ u cos
(c) Assume that a surface is represented by the vector-valued function r ⫽ ru, v. What generalization can you make about the graph of the function when one of the parameters is held constant?
where ⫺1 ⱕ u ⱕ 1, 0 ⱕ v ⱕ 2, and a ⫽ 3. Try to graph other Möbius strips for different values of a using a computer algebra system.
55. Surface Area is given by
z
The surface of the dome on a new museum 2
ru, v ⫽ 20 sin u cos vi ⫹ 20 sin u sin vj ⫹ 20 cos uk
−3
where 0 ⱕ u ⱕ 3, 0 ⱕ v ⱕ 2, and r is in meters. Find the surface area of the dome. 56. Hyperboloid hyperboloid
v v v cos v, y ⫽ a ⫹ u cos sin v, z ⫽ u sin 2 2 2
Find a vector-valued function for the
x2 ⫹ y2 ⫺ z2 ⫽ 1
x
4
2
−4
−1
1 3 −2
y
and determine the tangent plane at 1, 0, 0.
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1094
Chapter 15
Vector Analysis
15.6 Surface Integrals Evaluate a surface integral as a double integral. Evaluate a surface integral for a parametric surface. Determine the orientation of a surface. Understand the concept of a flux integral.
Surface Integrals The remainder of this chapter deals primarily with surface integrals. You will first consider surfaces given by z gx, y. Later in this section, you will consider more general surfaces given in parametric form. Let S be a surface given by z gx, y and let R be its projection onto the xy-plane, as shown in Figure 15.44. Let g, gx, and gy be continuous at all points in R and let f be a scalar function defined on S. Employing the procedure used to find surface area in Section 14.5, evaluate f at xi, yi , z i and form the sum
z
S: z = g(x, y)
n
f x , y , z S i
i
i
i
i1
(xi , yi , zi )
where Si 1 gxxi , yi 2 gyxi , yi 2 Ai.
x
(xi , yi )
R
Scalar function f assigns a number to each point of S. Figure 15.44
y
Provided the limit of this sum as approaches 0 exists, the surface integral of f over S is defined as
n
f x , y , z S .
f x, y, z dS lim
→0 i1
S
i
i
i
i
This integral can be evaluated by a double integral. THEOREM 15.10 Evaluating a Surface Integral Let S be a surface with equation z gx, y and let R be its projection onto the xy-plane. If g, gx , and gy are continuous on R and f is continuous on S, then the surface integral of f over S is
f x, y, z dS
S
f x, y, gx, y1 gxx, y 2 gyx, y 2 dA.
R
For surfaces described by functions of x and z (or y and z), you can make the following adjustments to Theorem 15.10. If S is the graph of y gx, z and R is its projection onto the xz-plane, then
f x, y, z dS
S
f x, gx, z, z1 gxx, z 2 gzx, z 2 dA.
R
If S is the graph of x g y, z and R is its projection onto the yz-plane, then
S
f x, y, z dS
f g y, z, y, z1 gy y, z 2 gz y, z 2 dA.
R
If f x, y, z 1, the surface integral over S yields the surface area of S. For instance, suppose the surface S is the plane given by z x, where 0 x 1 and 0 y 1. The surface area of S is 2 square units. Try verifying that
f x, y, z dS 2.
S
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15.6
Surface Integrals
1095
Evaluating a Surface Integral Evaluate the surface integral
y 2 2yz dS
S
where S is the first-octant portion of the plane 2x y 2z 6. Solution
Begin by writing S as
1 z 6 2x y 2 1 gx, y 6 2x y. 2 Using the partial derivatives gxx, y 1 and gyx, y 2, you can write 1
1 gxx, y 2 gyx, y 2 z
Using Figure 15.45 and Theorem 15.10, you obtain
z = 12 (6 − 2x − y)
(0, 0, 3)
y 2 2yz dS
S
S
f x, y, gx, y1 gxx, y 2 gyx, y 2 dA y 2 2y
R
23x
3
y
(3, 0, 0)
R
(0, 6, 0) x
1 1 41 23 .
3
y = 2(3 − x)
0
12 6 2x y 32 dA
y3 x dy dx
0
3
Figure 15.45
3 x3 dx
6
0
3 3 x4 2 243 . 2 z
(0, 0, 3) z=
6−y 2
1 gy y, z 2 gz y, z 2 (0, 6, 0) y
(3, 0, 0) x = 12 (6 − y − 2z)
Figure 15.46
0
An alternative solution to Example 1 would be to project S onto the yz-plane, as 1 shown in Figure 15.46. Then, x 26 y 2z, and
S
x
3
So, the surface integral is
y 2 2yz dS
S
1 41 1 23 .
f g y, z, y, z1 gy y, z 2 gz y, z 2 dA
R
6y 2
6
0
3 8
y 2 2yz
0
32 dz dy
6
36y y 3 dy
0
243 . 2
Try reworking Example 1 by projecting S onto the xz-plane.
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1096
Chapter 15
Vector Analysis
In Example 1, you could have projected the surface S onto any one of the three coordinate planes. In Example 2, S is a portion of a cylinder centered about the x-axis, and you can project it onto either the xz-plane or the xy-plane.
Evaluating a Surface Integral See LarsonCalculus.com for an interactive version of this type of example. z 3
Evaluate the surface integral
R: 0 ≤ x ≤ 4 0≤y≤3
x z dS
S
where S is the first-octant portion of the cylinder y 2 z2 9
4
3
2
between x 0 and x 4, as shown in Figure 15.47.
1 3
x
S: y 2 + z 2 = 9
y
Solution
Project S onto the xy-plane, so that
z gx, y 9 y 2 and obtain
Figure 15.47 1 gxx, y 2 gyx, y 2
1 3
y 9 y 2
2
.
9 y 2
Theorem 15.10 does not apply directly, because gy is not continuous when y 3. However, you can apply Theorem 15.10 for 0 b < 3 and then take the limit as b approaches 3, as follows.
b
x z dS lim b→3
S
4
0
0
b
lim 3 b→3
0
b
b→3
4
0
lim 3
x 9 y 2
0
b→3
8
9 y 2
0
b
b 3
36 24
dy
0
4 dy y 3
lim 3 4b 8 arcsin b→3
4
lim 3 4y 8 arcsin b→3
1 dx dy
x2 x 29 y 2
b
lim 3
x 9 y 2
3 dx dy 9 y 2
0
2
36 12
TECHNOLOGY Some computer algebra systems are capable of evaluating improper integrals. If you have access to such computer software, use it to evaluate the improper integral
3
0
4
0
x 9 y2
3 9 y2
dx dy.
Do you obtain the same result as in Example 2?
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15.6
Surface Integrals
1097
You have already seen that when the function f defined on the surface S is simply f x, y, z 1, the surface integral yields the surface area of S.
Area of surface
1 dS
S
On the other hand, when S is a lamina of variable density and x, y, z is the density at the point x, y, z, then the mass of the lamina is given by Mass of lamina
x, y, z dS.
S
Finding the Mass of a Surface Lamina z
A cone-shaped surface lamina S is given by
Cone: z=4−2
x2 + y2
4
z 4 2x 2 y 2,
0 z 4
as shown in Figure 15.48. At each point on S, the density is proportional to the distance between the point and the z-axis. Find the mass m of the lamina. 3
Solution
S: z 4 2x2 y2 gx, y, 0 z 4 R: x 2 y 2 4
2
with a density of x, y, z kx 2 y 2. Using a surface integral, you can find the mass to be
1
1
m
1
2
R:
+
y2
=4
x, y, z dS
S
2
x
x2
Projecting S onto the xy-plane produces
y
kx 2 y 21 gxx, y 2 gyx, y 2 dA
R
Figure 15.48
k
1 x 4x y
x 2 y 2
R
k
2
2
2
4y 2 dA x y2 2
5x 2 y 2 dA
R
k
2
5r r dr d
0
0
2
5k
3
2
2
r3
0
0
85k 3
Polar coordinates
2
d
d
0
2
85k 3
165k . 3
0
TECHNOLOGY Use a computer algebra system to confirm the result shown in Example 3. The computer algebra system Mathematica evaluated the integral as follows.
2
4y 2
k
2
4y 2
5x 2 y 2 dx dy k
2
0
2
0
5r r dr d
165k 3
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1098
Chapter 15
Vector Analysis
Parametric Surfaces and Surface Integrals For a surface S given by the vector-valued function ru, v xu, v i yu, vj zu, vk
Parametric surface
defined over a region D in the uv-plane, you can show that the surface integral of f x, y, z over S is given by
f x, y, z dS
S
f xu, v, yu, v, zu, vruu, v rvu, v dA.
D
Note the similarity to a line integral over a space curve C.
b
f x, y, z ds
f xt, yt, ztr t dt
Line integral
a
C
Also, notice that ds and dS can be written as ds r t dt
and dS ruu, v rvu, v dA.
Evaluating a Surface Integral z
Example 2 demonstrated an evaluation of the surface integral
3
x z dS
S
where S is the first-octant portion of the cylinder y 2 z2 9 1 2
3
3
y
between x 0 and x 4 (see Figure 15.49). Reevaluate this integral in parametric form. Solution
4 x Generated by Mathematica
Figure 15.49
In parametric form, the surface is given by
rx, xi 3 cos j 3 sin k where 0 x 4 and 0 2. To evaluate the surface integral in parametric form, begin by calculating the following. rx i r 3 sin j 3 cos k
i j k rx r 1 0 0 0 3 sin 3 cos
3 cos j 3 sin k
rx r 9 cos 2 9 sin 2 3 So, the surface integral can be evaluated as follows.
4
x 3 sin 3 dA
0
D
2
3x 9 sin d dx
0
4
3x 9 cos
0
4
0
dx 0
3 x 9 dx 2
3 2 x 9x 4 12 36
2
4 0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.6
Surface Integrals
1099
Orientation of a Surface Unit normal vectors are used to induce an orientation to a surface S in space. A surface is orientable when a unit normal vector N can be defined at every nonboundary point of S in such a way that the normal vectors vary continuously over the surface S. The surface S is called an oriented surface. An orientable surface S has two distinct sides. So, when you orient a surface, you are selecting one of the two possible unit normal vectors. For a closed surface such as a sphere, it is customary to choose the unit normal vector N to be the one that points outward from the sphere. Most common surfaces, such as spheres, paraboloids, ellipses, and planes, are orientable. (See Exercise 43 for an example of a surface that is not orientable.) Moreover, for an orientable surface, the gradient vector provides a convenient way to find a unit normal vector. That is, for an orientable surface S given by z gx, y
Orientable surface
let Gx, y, z z gx, y. Then, S can be oriented by either the unit normal vector N S: z = g(x, y) z
Gx, y, z Gx, y, z gxx, yi gyx, yj k 1 gxx, y 2 gyx, y 2
Upward unit normal vector
or the unit normal vector
N = ∇G ⎜⎜∇G ⎜⎜
N
S
Gx, y, z Gx, y, z gxx, yi gyx, yj k 1 gxx, y 2 gyx, y 2
Downward unit normal vector
as shown in Figure 15.50. If the smooth orientable surface S is given in parametric form by r u, v xu, v i yu, v j zu, v k
y
then the unit normal vectors are given by
Upward direction
x
S is oriented in an upward direction.
S: z = g(x, y) z
Parametric surface
N
ru rv ru rv
N
rv ru . rv ru
and
N = −∇G ⎜⎜∇G ⎜⎜
For an orientable surface given by y gx, z or x g y, z
S
you can use the gradient vector y x
Downward direction
S is oriented in a downward direction. Figure 15.50
Gx, y, z gxx, zi j gzx, zk
Gx, y, z y gx, z
Gx, y, z i gy y, zj gz y, zk
Gx, y, z x g y, z
or
to orient the surface.
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1100
Chapter 15
Vector Analysis
Flux Integrals z
N
One of the principal applications involving the vector form of a surface integral relates to the flow of a fluid through a surface. Consider an oriented surface S submerged in a fluid having a continuous velocity field F. Let S be the area of a small patch of the surface S over which F is nearly constant. Then the amount of fluid crossing this region per unit of time is approximated by the volume of the column of height F N, as shown in Figure 15.51. That is,
F
F·N ΔS
y x
The velocity field F indicates the direction of the fluid flow. Figure 15.51
V heightarea of base F N S. Consequently, the volume of fluid crossing the surface S per unit of time (called the flux of F across S) is given by the surface integral in the next definition. Definition of Flux Integral Let Fx, y, z M i Nj Pk, where M, N, and P have continuous first partial derivatives on the surface S oriented by a unit normal vector N. The flux integral of F across S is given by
S
F N dS.
Geometrically, a flux integral is the surface integral over S of the normal component of F. If x, y, z is the density of the fluid at x, y, z, then the flux integral
S
F N dS
represents the mass of the fluid flowing across S per unit of time. To evaluate a flux integral for a surface given by z gx, y, let Gx, y, z z gx, y. Then, N dS can be written as follows. Gx, y, z dS Gx, y, z Gx, y, z gx 2 gy2 1 dA gx 2 gy2 1
N dS
Gx, y, z dA THEOREM 15.11 Evaluating a Flux Integral Let S be an oriented surface given by z gx, y and let R be its projection onto the xy-plane.
S
S
F N dS F N dS
R
R
F gxx, yi gyx, yj k dA
Oriented upward
F gxx, yi gyx, yj k dA
Oriented downward
For the first integral, the surface is oriented upward, and for the second integral, the surface is oriented downward.
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15.6
1101
Surface Integrals
Using a Flux Integral to Find the Rate of Mass Flow z
Let S be the portion of the paraboloid z gx, y 4 x 2 y 2
8
lying above the xy-plane, oriented by an upward unit normal vector, as shown in Figure 15.52. A fluid of constant density is flowing through the surface S according to the vector field
6
Fx, y, z xi yj zk. Find the rate of mass flow through S. Solution Begin by computing the partial derivatives of g.
−4
gxx, y 2x
4
4
x
and
Figure 15.52
gyx, y 2y The rate of mass flow through the surface S is
S
F N dS
F
xi yj 4 x 2 y 2 k 2x i 2yj k dA
R
gxx, yi gyx, yj k dA
R
2x 2 2y 2 4 x 2 y 2 dA
R
4 x 2 y 2 dA
R
2
2
4 r 2r dr d
Polar coordinates
0
0
2
12 d
0
24 . For an oriented surface S given by the vector-valued function ru, v xu, vi yu, vj zu, vk
Parametric surface
defined over a region D in the uv-plane, you can define the flux integral of F across S as
S
F N dS
D
D
F
r
ru
F ru
rv
u rv
rv
r
u
rv
dA
dA.
Note the similarity of this integral to the line integral
C
F dr
C
F T ds.
A summary of formulas for line and surface integrals is presented on page 1103.
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y
1102
Chapter 15
Vector Analysis
Finding the Flux of an Inverse Square Field Find the flux over the sphere S given by
S: x 2 + y 2 + z 2 = a 2
x2 y 2 z2 a2
z
N
Sphere S
where F is an inverse square field given by a
Fx, y, z
N
N
kq r kqr r 2 r r 3
Inverse square field F
and a x
a
N
r xi y j zk.
y
Assume S is oriented outward, as shown in Figure 15.53. Solution
R: x 2 + y 2 ≤ a 2
The sphere is given by
r u, v xu, v i yu, v j zu, v k a sin u cos vi a sin u sin vj a cos uk
Figure 15.53
where 0 u and 0 v 2. The partial derivatives of r are ruu, v a cos u cos vi a cos u sin vj a sin uk and rvu, v a sin u sin vi a sin u cos vj
which implies that the normal vector ru ru
rv
i a cos u cos v a sin u sin v
a2
sin 2
u cos vi
rv
is
j a cos u sin v a sin u cos v sin 2
k a sin u 0
u sin vj sin u cos uk.
Now, using Fx, y, z
kqr r3
kq
xi yj zk xi yj zk3
kq a sin u cos vi a sin u sin vj a cos uk a3
it follows that F ru
rv
kq a sin u cos vi a sin u sin vj a cos uk a3 a2sin2 u cos vi sin2 u sin vj sin u cos uk kqsin3 u cos2 v sin3 u sin2 v sin u cos2 u
kq sin u. Finally, the flux over the sphere S is given by
S
F N dS
kq sin u dA
D
2
0
kq sin u du dv
0
4 kq.
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15.6
Surface Integrals
1103
The result in Example 6 shows that the flux across a sphere S in an inverse square field is independent of the radius of S. In particular, if E is an electric field, then the result in Example 6, along with Coulomb’s Law, yields one of the basic laws of electrostatics, known as Gauss’s Law:
S
E N dS 4 kq
Gauss’s Law
where q is a point charge located at the center of the sphere and k is the Coulomb constant. Gauss’s Law is valid for more general closed surfaces that enclose the origin, and relates the flux out of the surface to the total charge q inside the surface. Surface integrals are also used in the study of heat flow. Heat flows from areas of higher temperature to areas of lower temperature in the direction of greatest change. As a result, measuring heat flux involves the gradient of the temperature. The flux depends on the area of the surface. It is the normal direction to the surface that is important, because heat that flows in directions tangential to the surface will produce no heat loss. So, assume that the heat flux across a portion of the surface of area S is given by H kT N dS, where T is the temperature, N is the unit normal vector to the surface in the direction of the heat flow, and k is the thermal diffusivity of the material. The heat flux across the surface is given by H
S
kT N dS.
This section concludes with a summary of different forms of line integrals and surface integrals.
SUMMARY OF LINE AND SURFACE INTEGRALS Line Integrals ds r t dt x t 2 y t 2 z t 2 dt
C
C
b
f x, y, z ds F dr
C b
a
f xt, yt, zt ds
Scalar form
a
F T ds Fxt, yt, zt r t dt
Vector form
Surface Integrals [z ⴝ g x, y] dS 1 gxx, y 2 gyx, y 2 dA
S
f x, y, z dS
S
f x, y, gx, y1 gxx, y 2 gyx, y 2 dA
Scalar form
R
F N dS
F
gxx, y i gyx, y j k dA
Vector form (upward normal)
R
Surface Integrals (parametric form) dS ruu, v rvu, v dA
S
f x, y, z dS
S
f xu, v, yu, v, zu, v dS
Scalar form
D
F N dS
F
ru rv dA
Vector form
D
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1104
Chapter 15
Vector Analysis
15.6 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Evaluating a Surface Integral In Exercises 1–4, evaluate
15. f x, y x y S: ru, v 2 cos u i 2 sin u j v k
x ⴚ 2y ⴙ z dS.
S
1. S: z 4 x, 0 x 4,
0 y 3
2. S: z 15 2x 3y, 0 x 2, 3. S: z 2,
0 y 4
0 u
16. f x, y x y S: ru, v 4u cos v i 4u sin v j 3u k
x2 y2 1
2 4. S: z 3 x 3 2,
0 x 1,
0 u 4, 0 v
0 y x
Evaluating a Surface Integral In Exercises 5 and 6, evaluate
, 0 v 1 2
Evaluating a Surface Integral In Exercises 17–22, evaluate
S
5. S: z 3 x y, first octant
17. f x, y, z x 2 y 2 z 2
xy dS.
6. S: z h, 0 x 2,
S: z x y, x 2 y 2 1
0 y 4 x 2
Evaluating a Surface Integral In Exercises 7 and 8, use a
18. f x, y, z
computer algebra system to evaluate
19. f x, y, z x 2 y 2 z 2
xy dS.
7. S: z 9 x 2, 8. S: z 12 xy,
0 x 2, 0 y x
0 x 4,
0 y 4
Evaluating a Surface Integral In Exercises 9 and 10, use
S: z x 2 y 2,
21. f x, y, z x 2 y 2 z 2
S: x 2 y 2 9,
y 2,
0 x 2,
0 y 2
1 10. S: z cos x, 0 x , 0 y x 2 2
Mass In Exercises 11 and 12, find the mass of the surface lamina S of density . 11. S: 2x 3y 6z 12, first octant, x, y, z x 2 y 2 12. S: z a 2 x 2 y 2,
x, y, z kz
Evaluating a Surface Integral In Exercises 13–16, evaluate
0 z 9
0 x 3, 0 z x
Evaluating a Flux Integral In Exercises 23–28, find the flux of F through S,
S
F N dS
where N is the upward unit normal vector to S. 23. Fx, y, z 3z i 4j yk S: z 1 x y, first octant 24. Fx, y, z x i yj S: z 6 3x 2y, first octant
f x, y dS.
25. Fx, y, z x i yj zk
S
13. f x, y y 5 S: ru, v u i vj 2vk,
S: z 1 x 2 y 2, 0 u 1, 0 v 2
14. f x, y xy
z 0
26. Fx, y, z x i yj zk S: x 2 y 2 z 2 36, first octant
S: ru, v 2 cos u i 2 sin u j v k 0 u
0 y 3,
22. f x, y, z x 2 y 2 z 2
S
x 12 y 2 1
S: z x 2 y 2,
S: x 2 y 2 9, 0 x 3,
x 2 ⴚ 2xy dS. x2
x2 y 2 4
20. f x, y, z x 2 y 2 z 2
a computer algebra system to evaluate
9. S: z 10
xy z
S: z x 2 y 2, 4 x 2 y 2 16
S
f x, y, z dS.
S
, 0 v 1 2
27. Fx, y, z 4 i 3j 5k S: z x 2 y 2,
x2 y 2 4
28. Fx, y, z x i yj 2zk S: z a 2 x 2 y 2
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.6
Surface Integrals
1105
Evaluating a Flux Integral In Exercises 29 and 30, find the flux of F over the closed surface. (Let N be the outward unit normal vector of the surface.)
42.
HOW DO YOU SEE IT? Is the surface shown in the figure orientable? Explain why or why not.
29. Fx, y, z x y i yj zk S: z 16 x 2 y 2, 30. Fx, y, z 4xy i
z2j
z0 yzk
S: unit cube bounded by x 0, x 1, y 0, y 1, z 0, z1 31. Electrical Charge Let E yz i xz j xy k be an electrostatic field. Use Gauss’s Law to find the total charge enclosed by the closed surface consisting of the hemisphere z 1 x 2 y 2 and its circular base in the xy-plane. 32. Electrical Charge Let E x i y j 2z k be an electrostatic field. Use Gauss’s Law to find the total charge enclosed by the closed surface consisting of the hemisphere z 1 x 2 y 2 and its circular base in the xy-plane.
Moment of Inertia In Exercises 33 and 34, use the following formulas for the moments of inertia about the coordinate axes of a surface lamina of density . Ix ⴝ
y 2 1 z 2 x, y, z dS
Iz ⴝ
x 2 1 z 2 x, y, z dS
1
(a) Use a computer algebra system to graph the vector-valued function ru, v 4 v sin u cos2ui 4 v sin u sin2uj v cos uk,
0 u ,
1 v 1.
This surface is called a Möbius strip. (c) Use a computer algebra system to graph the space curve represented by ru, 0. Identify the curve. (d) Construct a Möbius strip by cutting a strip of paper, making a single twist, and pasting the ends together.
S
x2
43. Investigation
(b) Is the surface orientable? Explain why or why not.
S
Iy ⴝ
Double twist
x, y, z dS
y2
S
33. Verify that the moment of inertia of a conical shell of uniform density about its axis is 12ma 2, where m is the mass and a is the radius and height.
(e) Cut the Möbius strip along the space curve graphed in part (c), and describe the result.
34. Verify that the moment of inertia of a spherical shell of uniform density about its diameter is 23ma 2, where m is the mass and a is the radius.
Hyperboloid of One Sheet
Moment of Inertia In Exercises 35 and 36, find Iz for the
ru, v a cosh u cos vi a cosh u sin vj b sinh uk.
given lamina with uniform density of 1. Use a computer algebra system to verify your results. 35. x 2 y 2 a 2,
0 z h
36. z x 2 y 2,
0 z h
Flow Rate In Exercises 37 and 38, use a computer algebra system to find the rate of mass flow of a fluid of density through the surface S oriented upward when the velocity field is given by Fx, y, z ⴝ 0.5zk. 37. S: z 16 x 2 y 2, z 0 38. S: z 16 x 2 y 2
Consider the parametric surface given by the function
(a) Use a graphing utility to graph r for various values of the constants a and b. Describe the effect of the constants on the shape of the surface. (b) Show that the surface is a hyperboloid of one sheet given by x2 y2 z2 2 2 1. 2 a a b (c) For fixed values u u0, describe the curves given by ru0, v a cosh u0 cos vi a cosh u0 sin vj b sinh u0k. (d) For fixed values v v0, describe the curves given by
WRITING ABOUT CONCEPTS 39. Surface Integral Define a surface integral of the scalar function f over a surface z gx, y. Explain how to evaluate the surface integral.
ru, v0 a cosh u cos v0i a cosh u sin v0 j b sinh uk. (e) Find a normal vector to the surface at u, v 0, 0.
40. Orientable Surface Describe an orientable surface. 41. Flux Integral is evaluated.
Define a flux integral and explain how it
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1106
Chapter 15
Vector Analysis
15.7 Divergence Theorem Understand and use the Divergence Theorem. Use the Divergence Theorem to calculate flux.
Divergence Theorem Recall from Section 15.4 that an alternative form of Green’s Theorem is
冕
C
F N ds
冕冕 冢 冕冕 R
M N dA x y
冣
div F dA.
R
CARL FRIEDRICH GAUSS (1777–1855)
In an analogous way, the Divergence Theorem gives the relationship between a triple integral over a solid region Q and a surface integral over the surface of Q. In the statement of the theorem, the surface S is closed in the sense that it forms the complete boundary of the solid Q. Regions bounded by spheres, ellipsoids, cubes, tetrahedrons, or combinations of these surfaces are typical examples of closed surfaces. Let Q be a solid region on which a triple integral can be evaluated, and let S be a closed surface that is oriented by outward unit normal vectors, as shown in Figure 15.54. With these restrictions on S and Q, the Divergence Theorem can be stated as shown below the figure. z
The Divergence Theorem is also called Gauss’s Theorem, after the famous German mathematician Carl Friedrich Gauss. Gauss is recognized, with Newton and Archimedes, as one of the three greatest mathematicians in history. One of his many contributions to mathematics was made at the age of 22, when, as part of his doctoral dissertation, he proved the Fundamental Theorem of Algebra.
S1: Oriented by upward unit normal vector N S2: Oriented by downward unit normal vector
S1
S2 y
See LarsonCalculus.com to read more of this biography.
N x
Figure 15.54
THEOREM 15.12 The Divergence Theorem Let Q be a solid region bounded by a closed surface S oriented by a unit normal vector directed outward from Q. If F is a vector field whose component functions have continuous first partial derivatives in Q, then
冕冕 S
F N dS
冕冕冕
div F dV.
Q
REMARK As noted at the left above, the Divergence Theorem is sometimes called Gauss’s Theorem. It is also sometimes called Ostrogradsky’s Theorem, after the Russian mathematician Michel Ostrogradsky (1801–1861). akg-images/Newscom
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15.7
Proof
冕冕
REMARK This proof is restricted to a simple solid region. The general proof is best left to a course in advanced calculus.
S
Divergence Theorem
1107
For F共x, y, z兲 Mi Nj Pk, the theorem takes the form F N dS
冕冕 冕冕冕 冢
共Mi N Nj N Pk N兲 dS
S
M N P dV. x y z
冣
Q
You can prove this by verifying that the following three equations are valid.
冕冕 冕冕 冕冕 S
S
S
Mi N dS
冕冕冕 冕冕冕 冕冕冕 Q
Nj N dS
Q
Pk N dS
z g1共x, y兲
S3
冕冕
y
S
S1: z = g1(x, y)
Figure 15.55
Lower surface
whose projections onto the xy-plane coincide and form region R. If Q has a lateral surface like S3 in Figure 15.55, then a normal vector is horizontal, which implies that Pk N 0. Consequently, you have
N (downward) R
Upper surface
and lower surface
S2
S1
x
P dV z
Q
z g2共x, y兲
N (horizontal)
N dV y
Because the verifications of the three equations are similar, only the third is discussed. Restrict the proof to a simple solid region with upper surface
S2: z = g2(x, y) z
N (upward)
M dV x
Pk N dS
冕冕
Pk N dS
S1
冕冕 S2
Pk N dS 0.
On the upper surface S2, the outward normal vector is upward, whereas on the lower surface S1, the outward normal vector is downward. So, by Theorem 15.11, you have
冕冕 S1
Pk N dS
冕冕 冕冕
P共x, y, g1共x, y兲兲k
R
g1
g1
冢 x i y j k冣 dA
P共x, y, g1共x, y兲兲 dA
R
and
冕冕 S2
Pk N dS
冕冕 冕冕 R
冢
P共x, y, g2共x, y兲兲k
g2 g i 2 j k dA x y
冣
P共x, y, g2共x, y兲兲 dA.
R
Adding these results, you obtain
冕冕 S
Pk N dS
冕冕 冕 冕 冤冕 冕冕冕
关P共x, y, g2共x, y兲兲 P共x, y, g1共x, y兲兲兴 dA
R
g2共x, y兲
R
g1共x, y兲
冥
P dz dA z
P dV. z
Q
See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1108
Chapter 15
Vector Analysis
Using the Divergence Theorem Let Q be the solid region bounded by the coordinate planes and the plane 2x 2y z 6 and let F xi y 2j zk. Find
冕冕
F N dS
S
where S is the surface of Q. z
Solution From Figure 15.56, you can see that Q is bounded by four subsurfaces. So, you would need four surface integrals to evaluate
6
冕冕
S2: yz-plane S1: xz-plane
F N dS.
S
However, by the Divergence Theorem, you need only one triple integral. Because M N P x y z 1 2y 1 2 2y
div F S4
4
3
x
3
S4: 2x + 2y + z = 6
you have 4
S3: xy-plane
y
冕冕
F N dS
S
冕冕冕 冕冕 冕 冕冕 冕冕 冕冤 冕
div F dV
Q
3
Figure 15.56
0
3
0
62x2y
共2 2y兲 dz dx dy
0
3y
3
62x2y
冥
共2z 2yz兲
0
3y
0
dx dy 0
3y
共12 4x 8y 4xy 4y 2兲 dx dy
0
0
3
冥
12x 2x 2 8xy 2x 2y 4xy 2
0
3y
dy
0
3
共18 6y 10y 2 2y 3兲 dy
0
冤
18y 3y 2
10y 3 y 4 3 2
冥
3 0
63 . 2
TECHNOLOGY If you have access to a computer algebra system that can evaluate triple-iterated integrals, use it to verify the result in Example 1. When you are using such a utility, note that the first step is to convert the triple integral to an iterated integral—this step may be done by hand. To give yourself some practice with this important step, find the limits of integration for the following iterated integrals. Then use a computer to verify that the value is the same as that obtained in Example 1.
冕冕冕 ?
?
?
?
?
?
共2 2y兲 dy dz dx,
冕冕冕 ?
?
?
?
?
共2 2y兲 dx dy dz
?
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15.7
Divergence Theorem
1109
Verifying the Divergence Theorem z
S2: z = 4 − x 2 − y 2
Let Q be the solid region between the paraboloid z 4 x2 y 2
4
and the xy-plane. Verify the Divergence Theorem for
N2
F共x, y, z兲 2z i xj y 2k.
S1: z = 0
2
2
x
N1 = − k R: x 2 + y 2 ≤ 4
Figure 15.57
y
Solution From Figure 15.57, you can see that the outward normal vector for the surface S1 is N1 k, whereas the outward normal vector for the surface S2 is N2
2xi 2yj k 冪4x 2 4y 2 1
.
So, by Theorem 15.11, you have
冕冕 S
F N dS
冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 冕冤 冕 F
S1
F
y 2 dA
2
冪4y 2
2
冪4y 2 冪4y 2
2 冪4y 2 2
冪4y 2
2 冪4y 2
2
dS
共4xz 2xy y 2兲 dA
y 2 dx dy
冪4y 2
2 冪4y 2
共4xz 2xy y 2兲 dx dy
共4xz 2xy兲 dx dy
关4x共4 x 2 y 2兲 2xy兴 dx dy 共16x 4x 3 4xy 2 2xy兲 dx dy
2
冪4x2 4y2 1
2
2 冪4y 2
2
共2xi 2yj k兲
F
R
冪4y 2
2
N2 dS
S2
R
F
S2
共k兲 dS
S1
冕冕 冕冕 冕冕 冕冕
N1 dS
冥
8x2 x 4 2x 2y 2 x 2y
冪4y 2
冪4y 2
dy
2
0 dy
2
0. On the other hand, because div F
关2z兴 关x兴 关 y 2兴 0 0 0 0 x y z
you can apply the Divergence Theorem to obtain the equivalent result
冕冕 S
F N dS
冕冕冕 冕冕冕
div F dV
Q
0 dV
Q
0.
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1110
Chapter 15
Vector Analysis
Using the Divergence Theorem Let Q be the solid bounded by the cylinder x 2 y 2 4, the plane x z 6, and the xy-plane, as shown in Figure 15.58. Find
z 9
Plane: x+z=6
冕冕
8
S
7
F N dS
where S is the surface of Q and
6
F共x, y, z兲 共x 2 sin z兲i 共xy cos z兲j eyk. Solution Direct evaluation of this surface integral would be difficult. However, by the Divergence Theorem, you can evaluate the integral as follows.
冕冕 S
2
2
F N dS
冕冕冕 冕冕冕 冕冕冕 冕 冕冕 冕冕 冕
div F dV
Q
y
x
共2x x 0兲 dV
Q
Cylinder: x2 + y2 = 4
Figure 15.58
3x dV
Q
2
0
0
2
6r cos
共3r cos 兲r dz dr d
0
2
共18r 2 cos 3r 3 cos 2 兲 dr d
0
0
2
2
共48 cos 12 cos 2 兲 d
0
冢
冤
48 sin 6
1 sin 2 2
2
冣冥
0
12 Notice that cylindrical coordinates with x r cos
and
dV r dz dr d
were used to evaluate the triple integral. z
Even though the Divergence Theorem was stated for a simple solid region Q bounded by a closed surface, the theorem is also valid for regions that are the finite unions of simple solid regions. For example, let Q be the solid bounded by the closed surfaces S1 and S2, as shown in Figure 15.59. To apply the Divergence Theorem to this solid, let S S1 傼 S2. The normal vector N to S is given by N1 on S1 and by N2 on S2. So, you can write
N2
S2
−N1 S1
冕冕冕
div F dV
F
N dS
F
共N1兲 dS
S
Q
x
冕冕 冕冕 冕冕
y
S1
Figure 15.59
F
S1
N1 dS
冕冕 冕冕 F
S2
F
S2
N2 dS
N2 dS.
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15.7
Divergence Theorem
1111
Flux and the Divergence Theorem To help understand the Divergence Theorem, consider the two sides of the equation
冕冕 S
F N dS
冕冕冕
div F dV.
Q
You know from Section 15.6 that the flux integral on the left determines the total fluid flow across the surface S per unit of time. This can be approximated by summing the fluid flow across small patches of the surface. The triple integral on the right measures this same fluid flow across S, but from a very different perspective—namely, by calculating the flow of fluid into (or out of) small cubes of volume Vi. The flux of the ith cube is approximately div F共xi, yi, zi兲 Vi for some point 共xi, yi, zi兲 in the ith cube. Note that for a cube in the interior of Q, the gain (or loss) of fluid through any one of its six sides is offset by a corresponding loss (or gain) through one of the sides of an adjacent cube. After summing over all the cubes in Q, the only fluid flow that is not canceled by adjoining cubes is that on the outside edges of the cubes on the boundary. So, the sum n
兺 div F共x , y , z 兲 V i
i
i
i
i1
approximates the total flux into (or out of) Q, and therefore through the surface S. To see what is meant by the divergence of F at a point, consider V to be the volume of a small sphere S of radius and center 共x0, y0, z0兲 contained in region Q, as shown in Figure 15.60. Applying the Divergence Theorem to S produces
z
Sα
Solid region Q
(x0, y0, z 0 )
冕冕 冕 Q
div F dV ⬇ div F共x0, y0, z0,兲 V
where Q is the interior of S . Consequently, you have y
x
Flux of F across S
Figure 15.60
div F共x0, y0, z0兲 ⬇
flux of F across S
V
and, by taking the limit as → 0, you obtain the divergence of F at the point 共x0, y0, z0兲. div F共x0, y0, z0兲 lim
→0
flux of F across S flux per unit volume at 共x0, y0, z0兲
V
The point 共x0, y0, z0兲 in a vector field is classified as a source, a sink, or incompressible, as shown in the list below.
REMARK In hydrodynamics, a source is a point at which additional fluid is considered as being introduced to the region occupied by the fluid. A sink is a point at which fluid is considered as being removed.
1. Source, for div F > 0 2. Sink, for div F < 0 3. Incompressible, for div F 0
(a) Source: div F > 0
See Figure 15.61(a). See Figure 15.61(b). See Figure 15.61(c).
(b) Sink: div F < 0
(c) Incompressible: div F 0
Figure 15.61
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1112
Chapter 15
Vector Analysis
Calculating Flux by the Divergence Theorem See LarsonCalculus.com for an interactive version of this type of example.
Let Q be the region bounded by the sphere x 2 y 2 z 2 4. Find the outward flux of the vector field F共x, y, z兲 2x3 i 2y 3j 2z3 k through the sphere. Solution
By the Divergence Theorem, you have
Flux across S
冕冕 冕冕冕 冕冕冕 冕冕 冕 冕冕 冕 F
N dS
S
div F dV
Q
6共x 2 y 2 z 2兲 dV
Q 2
6
0
0
2
6
0
2
4 sin d d d
Spherical coordinates
0
2 4 sin d d
0
2
12
2 4 d
0
24
15.7 Exercises
冢325冣
768 . 5
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Verifying the Divergence Theorem In Exercises 1–6,
3. F共x, y, z兲 共2x y兲i 共2y z兲j zk
verify the Divergence Theorem by evaluating
冕冕 S
S: surface bounded by the plane 2x 4y 2z 12 and the coordinate planes
F N dS
4. F共x, y, z兲 xyi zj 共x y兲k S: surface bounded by the planes y 4 and z 4 x and the coordinate planes
as a surface integral and as a triple integral. 1. F共x, y, z兲 2x i 2yj z 2k
z
S: cube bounded by the planes x 0, x a, y 0, y a, z 0, z a 2. F共x, y, z兲 2x i 2yj
z
6
4
z 2k
S: cylinder x 2 y 2 4, 0 z h z
z 4
a
x
3
h
6
4 y
y
x
Figure for 3
Figure for 4
5. F共x, y, z兲 xzi zyj 2z2 k a
a
S: surface bounded by z 1 x2 y2 and z 0
y
x x
Figure for 1
2
Figure for 2
2
y
6. F共x, y, z兲 xy2 i yx2 j ek S: surface bounded by z 冪x2 y2 and z 4
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.7
Divergence Theorem
1113
Using the Divergence Theorem In Exercises 7–16, use the Divergence Theorem to evaluate
冕冕 S
HOW DO YOU SEE IT? The graph of a vector field F is shown. Does the graph suggest that the divergence of F at P is positive, negative, or zero?
22.
F N dS
y
and find the outward flux of F through the surface of the solid bounded by the graphs of the equations. Use a computer algebra system to verify your results.
4
7. F共x, y, z兲 x 2i y 2j z 2k
2
S: x 0, x a, y 0, y a, z 0, z a
P
8. F共x, y, z兲 x 2z2 i 2yj 3xyzk
x
−2
S: x 0, x a, y 0, y a, z 0, z a
2
9. F共x, y, z兲 x 2 i 2xyj xyz 2 k
4
−2
S: z 冪a 2 x 2 y 2, z 0 10. F共x, y, z兲 xy i yz j yzk S: z 冪a 2 x 2 y 2, z 0
23. Volume
11. F共x, y, z兲 x i yj zk
(a) Use the Divergence Theorem to verify that the volume of the solid bounded by a surface S is
S: x 2 y 2 z 2 9
冕冕
12. F共x, y, z兲 xyz j S: x 2 y 2 4, z 0, z 5
14. F共x, y, z兲 共xy 2 cos z兲i 共x 2y sin z兲j ezk S: z 12冪x 2 y 2, z 8
冕冕
S: z 4 y, z 0, x 0, x 6, y 0
S
冕冕
Using the Divergence Theorem In Exercises 17 and 18, evaluate
S
S
curl F N dS
where S is the closed surface of the solid bounded by the graphs of x ⴝ 4 and z ⴝ 9 ⴚ y 2, and the coordinate planes. 18. F共x, y, z兲 xy cos z i yz sin xj xyzk
WRITING ABOUT CONCEPTS State the Divergence Theorem.
20. Classifying a Point in a Vector Field How do you determine whether a point 共x0, y0, z0兲 in a vector field is a source, a sink, or incompressible?
1 储 F储
curl F N dS 0
S
F N dS
DN f ⴝ f N,
冕冕冕
3 储 F储
冕冕冕
vector
field
dV.
Q
DN g ⴝ g N.
共 f 2g f g兲 dV
Q
冕冕
f DNg dS
S
[Hint: Use div共 f G兲 f div G f G.] 28.
冕冕冕 Q
for any closed surface S.
冕冕
the
Q, S, and N meet the conditions of the Divergence Theorem and that the required partial derivatives of the scalar functions f and g are continuous. The expressions DN f and DN g are the derivatives in the direction of the vector N and are defined by
21. Closed Surface Verify that
S
F N dS 3V
Proof In Exercises 27 and 28, prove the identity, assuming that
27.
冕冕
F N dS 0
26. Verifying an Identity For F共x, y, z兲 x i yj zk, verify that
17. F共x, y, z兲 共4xy z 2兲i 共2x 2 6yz兲j 2xzk
19. Divergence Theorem
z dx dy.
S
25. Volume For the vector field F共x, y, z兲 x i yj zk, verify the following integral, where V is the volume of the solid bounded by the closed surface S.
S: x 2 y 2 z 2 16
冕冕
冕冕
y dz dx
S
24. Constant Vector Field For the constant vector field F共x, y, z兲 a1i a2 j a3k, verify the following integral for any closed surface S.
15. F共x, y, z兲 xezi yezj ezk 16. F共x, y, z兲 xyi 4yj xzk
冕冕
(b) Verify the result of part (a) for the cube bounded by x 0, x a, y 0, y a, z 0, and z a.
13. F共x, y, z兲 x i y 2j zk S: x2 y2 25, z 0, z 7
x dy dz
S
共 f 2g g 2f 兲 dV
冕冕
共 f DNg gDN f 兲 dS
S
(Hint: Use Exercise 27 twice.)
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1114
Chapter 15
Vector Analysis
15.8 Stokes’s Theorem Understand and use Stokes’s Theorem. Use curl to analyze the motion of a rotating liquid.
Stokes’s Theorem A second higher-dimension analog of Green’s Theorem is called Stokes’s Theorem, after the English mathematical physicist George Gabriel Stokes. Stokes was part of a group of English mathematical physicists referred to as the Cambridge School, which included William Thomson (Lord Kelvin) and James Clerk Maxwell. In addition to making contributions to physics, Stokes worked with infinite series and differential equations, as well as with the integration results presented in this section. Stokes’s Theorem gives the relationship between a surface integral over an oriented surface S and a line integral along a closed space curve C forming the boundary of S, as shown in Figure 15.62. The positive direction along C is counterclockwise relative to the normal vector N. That is, if you imagine grasping the normal vector N with your right hand, with your thumb pointing in the direction of N, then your fingers will point in the positive direction C, as shown in Figure 15.63. GEORGE GABRIEL STOKES (1819–1903)
z
Stokes became a Lucasian professor of mathematics at Cambridge in 1849. Five years later, he published the theorem that bears his name as a prize examination question there.
N Surface S
See LarsonCalculus.com to read more of this biography.
N
C y
S
R x
Figure 15.62
C
Direction along C is counterclockwise relative to N. Figure 15.63
THEOREM 15.13 Stokes’s Theorem Let S be an oriented surface with unit normal vector N, bounded by a piecewise smooth simple closed curve C with a positive orientation. If F is a vector field whose component functions have continuous first partial derivatives on an open region containing S and C, then
冕
C
F dr
冕冕 S
共curl F兲 N dS.
In Theorem 15.13, note that the line integral may be written in the differential form 兰C M dx N dy P dz or in the vector form 兰C F T ds. Bettmann/Corbis
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.8
Stokes’s Theorem
1115
Using Stokes’s Theorem z
Let C be the oriented triangle lying in the plane 2x 2y z 6
6
S: 2x + 2y + z = 6
as shown in Figure 15.64. Evaluate
冕
C
C2
C3
where F共x, y, z兲 y 2 i zj xk. N (upward)
R 3 x
C1 x+y=3
Figure 15.64
F dr
3
Solution
y
ⱍ ⱍ
Using Stokes’s Theorem, begin by finding the curl of F.
i curl F x y 2 Considering
j y z
k i j 2yk z x
z g共x, y兲 6 2x 2y you can use Theorem 15.11 for an upward normal vector to obtain
冕
C
F dr
冕冕 冕冕 冕冕 冕冕 冕
共curl F兲 N dS
S
R
R
3
共i j 2yk兲 关gx共x, y兲 i gy共x, y兲 j k兴 dA 共i j 2yk兲 共2i 2j k兲 dA
3y
0
共2y 4兲 dx dy
0
3
共2y 2 10y 12兲 dy
0
冤
冥
2y 3 5y 2 12y 3 9.
3 0
Try evaluating the line integral in Example 1 directly, without using Stokes’s Theorem. One way to do this would be to consider C as the union of C1, C2, and C3, as follows. C1: r1共t兲 共3 t兲 i tj, 0 t 3 C2: r2共t兲 共6 t兲 j 共2t 6兲 k, 3 t 6 C3: r3共t兲 共t 6兲 i 共18 2t兲 k, 6 t 9 The value of the line integral is
冕
C
F dr
冕 冕
C1
F r1 共t兲 dt
3
冕
冕
C2
F r2 共t兲 dt
6
t 2 dt
0
3
冕
冕
C3
F r3 共t兲 dt
9
共2t 6兲 dt
共2t 12兲 dt
6
999 9.
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1116
Chapter 15
Vector Analysis
Verifying Stokes’s Theorem See LarsonCalculus.com for an interactive version of this type of example. S: z = 4 − x 2 − y 2
Let S be the portion of the paraboloid
z
z 4 x2 y2
4
lying above the xy-plane, oriented upward (see Figure 15.65). Let C be its boundary curve in the xy-plane, oriented counterclockwise. Verify Stokes’s Theorem for
S N (upward)
F共x, y, z兲 2zi xj y2k by evaluating the surface integral and the equivalent line integral.
−3
R 3 x
C
3
R: x 2 + y 2 ≤ 4
Figure 15.65
y
Solution As a surface integral, you have z g共x, y兲 4 x 2 y 2, gx 2x, gy 2y, and
ⱍ ⱍ
i curl F x 2z
j y x
k 2yi 2j k. z y2
By Theorem 15.11, you obtain
冕冕 S
共curl F兲 N dS
冕冕 冕冕 冕冤 冕
共2yi 2j k兲 共2x i 2yj k兲 dA
R
冪4x 2
2
2 冪4x 2
共4xy 4y 1兲 dy dx
2
冥
2xy 2 2y2 y
2
冪4x 2
dx
冪4x 2
2
2
2冪4 x 2 dx
Area of circle of radius 2 4 . As a line integral, you can parametrize C as r共t兲 2 cos t i 2 sin t j 0k, 0 t 2 . For F共x, y, z兲 2zi xj y 2 k, you obtain
冕
C
F dr
冕 冕 冕 冕 冕
M dx N dy P dz
C
2z dx x dy y 2 dz
C
2
关0 2 cos t 共2 cos t兲 0兴 dt
0
2
4 cos 2 t dt
0
2
2
共1 cos 2t兲 dt
0
冤
2 t
1 sin 2t 2
冥
2 0
4 .
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
15.8
1117
Stokes’s Theorem
Physical Interpretation of Curl T F
α (x, y, z)
Cα
F T F N
N
Stokes’s Theorem provides insight into a physical interpretation of curl. In a vector field F, let S be a small circular disk of radius , centered at 共x, y, z兲 and with boundary C , as shown in Figure 15.66. At each point on the circle C , F has a normal component F N and a tangential component F T. The more closely F and T are aligned, the greater the value of F T. So, a fluid tends to move along the circle rather than across it. Consequently, you say that the line integral around C measures the circulation of F around C␣ . That is,
冕
Disk Sα
Figure 15.66
C
F T ds circulation of F around C .
Now consider a small disk S to be centered at some point 共x, y, z兲 on the surface S, as shown in Figure 15.67. On such a small disk, curl F is nearly constant, because it varies little from its value at 共x, y, z兲. Moreover, curl F N is also nearly constant on S because all unit normals to S are about the same. Consequently, Stokes’s Theorem yields
冕
C
F T ds
冕冕 S
共curl F兲 N dS
⬇ 共curl F兲 N
冕冕
curl F N
S (x, y, z)
Sα
Figure 15.67
dS
S
⬇ 共curl F兲 N 共 2兲. So,
共curl F兲 N ⬇
冕
C
F T ds
2
circulation of F around C
area of disk S
rate of circulation.
Assuming conditions are such that the approximation improves for smaller and smaller disks 共 → 0兲, it follows that 1
→0 2
共curl F兲 N lim
冕
C
F T ds
which is referred to as the rotation of F about N. That is, curl F共x, y, z兲 N rotation of F about N at 共x, y, z兲. In this case, the rotation of F is maximum when curl F and N have the same direction. Normally, this tendency to rotate will vary from point to point on the surface S, and Stokes’s Theorem
冕冕 S
共curl F兲 N dS
Surface integral
冕
C
F dr
Line integral
says that the collective measure of this rotational tendency taken over the entire surface S (surface integral) is equal to the tendency of a fluid to circulate around the boundary C (line integral).
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1118
Chapter 15
Vector Analysis
An Application of Curl z
A liquid is swirling around in a cylindrical container of radius 2, so that its motion is described by the velocity field F共x, y, z兲 y冪x 2 y 2 i x冪x 2 y 2 j as shown in Figure 15.68. Find
冕冕 S
2 x 2
y
共curl F兲 N dS
where S is the upper surface of the cylindrical container. Solution
Figure 15.68
ⱍ
The curl of F is given by
i curl F x y冪x 2 y 2 Letting N k, you have
冕冕 S
共curl F兲 N dS
j y x冪x 2 y 2
冕冕 冕冕 冕 冥 冕
3冪x 2 y 2 dA
R
2
2
共3r兲 r dr d
0
0
2
2
d
r3
0
ⱍ
k 3冪x 2 y 2 k. z 0
2
0
8 d
0
16 . If curl F 0 throughout region Q, then the rotation of F about each unit normal N is 0. That is, F is irrotational. From Section 15.1, you know that this is a characteristic of conservative vector fields.
SUMMARY OF INTEGRATION FORMULAS Fundamental Theorem of Calculus
冕
Fundamental Theorem of Line Integrals
b
F 共x兲 dx F共b兲 F共a兲
a
M dx N dy
C
C
F N ds
冕冕
冕 冕冢 R
N M dA x y
冕
冕冕
冣
冕
C
F T ds
冕
F dr
冕冕
C
C
f dr f 共x共b兲, y 共b兲兲 f 共x共a兲, y 共a兲兲
R
共curl F兲 k dA
div F dA
R
Divergence Theorem
S
F dr
C
Green’s Theorem
冕 冕
冕
F N dS
冕冕冕 Q
Stokes’s Theorem
div F dV
冕
C
F dr
冕冕 S
共curl F兲 N dS
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15.8
15.8 Exercises
Finding the Curl of a Vector Field In Exercises 1–4, find 1. F共x, y, z兲 共2y z兲 i e z j xyzk 2. F共x, y, z兲 x sin y i y cos x j yz 2 k 2 y 2
i ey
2 z 2
j xyz k
4. F共x, y, z兲 arcsin y i 冪1 x 2 j y 2 k
Verifying Stokes’s Theorem In Exercises 5–8, verify
Stokes’s Theorem by evaluating 兰C F T ds ⴝ 兰C F dr as a line integral and as a double integral. 5. F共x, y, z兲 共y z兲 i 共x z兲 j 共x y兲 k S: z 9 x 2 y 2,
1119
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
the curl of the vector field F.
3. F共x, y, z兲 e x
Stokes’s Theorem
z 0
6. F共x, y, z兲 共y z兲 i 共x z兲 j 共x y兲 k S: z 冪1 x 2 y 2 7. F共x, y, z兲 xyz i y j zk
Motion of a Liquid In Exercises 19 and 20, the motion of a liquid in a cylindrical container of radius 1 is described by the velocity field F冇x, y, z冈. Find 兰S兰 冇curl F冈 N dS, where S is the upper surface of the cylindrical container. 19. F共x, y, z兲 i j 2k
20. F共x, y, z兲 z i yk
WRITING ABOUT CONCEPTS 21. Stokes’s Theorem 22. Curl
State Stokes’s Theorem.
Give a physical interpretation of curl.
23. Proof Let C be a constant vector. Let S be an oriented surface with a unit normal vector N, bounded by a smooth curve C. Prove that
冕冕 S
C N dS
1 2
冕
C
共C r兲 dr.
S: 6x 6y z 12, first octant 8. F共x, y, z兲 z 2 i x 2 j y 2 k S: z
y 2,
0 x a,
HOW DO YOU SEE IT? Let S1 be the portion of the paraboloid lying above the xy-plane, and let S2 be the hemisphere, as shown in the figures. Both surfaces are oriented upward.
24.
0 y a
Using Stokes’s Theorem In Exercises 9–18, use Stokes’s
Theorem to evaluate 兰C F dr. In each case, C is oriented counterclockwise as viewed from above. S1
9. F共x, y, z兲 2y i 3z j x k
−a
x 10. F共x, y, z兲 arctan i ln冪x 2 y 2 j k y 11. F共x, y, z兲 z 2 i 2x j y 2 k S: z 1 x 2 y 2,
13. F共x, y, z兲
z2 i
2a
a
a
a x
a
y
For a vector field F共x, y, z兲 with continuous partial derivatives, does
S1
z ≥ 0
−a
y
a
a x
冕冕
z ≥ 0
12. F共x, y, z兲 4xz i y j 4xy k S: z 9 x 2 y 2,
2a
S2
C: triangle with vertices 共2, 0, 0兲, 共0, 2, 0兲, and 共0, 0, 2兲
C: triangle with vertices 共0, 0, 0兲, 共1, 1, 1兲, and 共0, 0, 2兲
z
z
共curl F兲 N dS1
冕冕 S2
共curl F兲 N dS2?
Explain your reasoning.
yj zk
S: z 冪4 x 2 y 2 14. F共x, y, z兲 x 2 i z 2 j xyz k
PUTNAM EXAM CHALLENGE
S: z 冪4 x 2 y 2 15. F共x, y, z兲 ln冪x 2 y 2 i arctan
x jk y
S: z 9 2x 3y over r 2 sin 2 in the first octant 16. F共x, y, z兲 yz i 共2 3y兲 j 共x 2 y 2兲 k, x2 y2 16 S: the first-octant portion of x 2 z 2 16 over x 2 y 2 16 17. F共x, y, z兲 xyz i y j z k S: z x 2,
0 x a,
0 y a
N is the downward unit normal to the surface. 18. F共x, y, z兲 xyz i y j z k, x2 y2 a2
25. Let G共x, y兲
冢x
2
y x , ,0 . 4y2 x2 4y2
冣
Prove or disprove that there is a vector-valued function F共x, y, z兲 共M共x, y, z兲, N共x, y, z兲, P共x, y, z兲兲 with the following properties: (i) M, N, P have continuous partial derivatives for all 共x, y, z兲 共0, 0, 0兲; (ii) Curl F 0 for all 共x, y, z兲 共0, 0, 0兲; (iii) F共x, y, 0兲 G共x, y兲. This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
S: the first-octant portion of z x 2 over x 2 y 2 a 2
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1120
Chapter 15
Vector Analysis
Review Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Sketching a Vector Field In Exercises 1 and 2, find 储F储 and sketch several representative vectors in the vector field. Use a computer algebra system to verify your results. 1. F共x, y, z兲 x i j 2 k
2. F共x, y兲 i 2y j
23.
3. f 共x, y, z兲 2x 2 xy z 2
C: r 共t兲 共1 sin t兲 i 共1 cos t兲 j, 24.
25.
4. f 共x, y, z兲 x 2e yz
6. F共x, y兲
y 1 i 2j y x
7. F共x, y兲 共 xy 2 x 2兲 i 共x 2 y y 2兲 j 8. F共x, y兲 共2y 3 sin 2x兲 i 3y 2 共1 cos 2x兲j 9. F共x, y, z兲 4x y2i 2x 2 j 2z k 10. F共x, y, z兲 共4xy z 2兲 i 共2x 2 6yz兲 j 2 xz k yz i xz j xy k 11. F共x, y, z兲 y 2z 2
共x 2 y 2兲 ds
C: r 共t兲 共cos t t sin t兲 i 共sin t t cos t兲 j, 0 t 2
冕
共2x y兲 dx 共x 2y兲 dy
C
(a) C: line segment from 共0, 0兲 to 共3, 3兲 (b) C: one revolution counterclockwise around the circle x 3 cos t, y 3 sin t
whether the vector field is conservative. If it is, find a potential function for the vector field. y 1 i j x2 x
冕
26.
冕
共2x y兲 dx 共x 3y兲 dy
C
C: r 共t兲 共cos t t sin t兲 i 共sin t t sin t兲 j, 0 t 兾2
Evaluating a Line Integral In Exercises 27 and 28, use a computer algebra system to evaluate the line integral over the given path. 27.
冕
共2x y兲 ds
28.
C
12. F共x, y, z兲 sin z 共 y i x j k兲
Divergence and Curl In Exercises 13–20, find (a) the
0 t 2
C
Finding a Potential Function In Exercises 5–12, determine
5. F共x, y兲
共x 2 y 2兲 ds
C
Finding a Conservative Vector Field In Exercises 3 and 4, find the conservative vector field for the potential function by finding its gradient.
冕
冕
共x 2 y 2 z 2兲 ds
C
r 共t兲 a cos 3 ti a sin 3 tj,
r 共t兲 t i t 2 j t 3兾2k,
0 t 兾2
0 t 4
divergence of the vector field F and (b) the curl of the vector field F.
Lateral Surface Area In Exercises 29 and 30, find the lateral surface area over the curve C in the xy-plane and under the surface z ⴝ f 冇x, y冈.
13. F共x, y, z兲 x 2 i xy 2 j x 2z k
29. f 共x, y兲 3 sin共x y兲; C: y 2x from 共0, 0兲 to 共2, 4兲
14. F共x, y, z兲 y 2 j z 2 k 15. F共x, y, z兲 共cos y y cos x兲 i 共sin x x sin y兲 j xyz k
30. f 共x, y兲 12 x y; C: y x 2 from 共0, 0兲 to 共2, 4兲
16. F共x, y, z兲 共3x y兲 i 共 y 2z兲 j 共z 3x兲 k
Evaluating a Line Integral of a Vector Field In Exercises
17. F共x, y, z兲 arcsin x i xy 2 j yz 2 k
31–36, evaluate
18. F共x, y, z兲 共
31. F共x, y兲 xy i 2xy j
x2
19. F共x, y, z兲 ln共
y兲 i 共x
x2
兲j
兲 i ln共
y2
x2
冕
C
sin2 y
y 2兲 j z k
F dr.
C: r 共t兲 t 2 i t 2 j,
0 t 1
z z 20. F共x, y, z兲 i j z 2 k x y
32. F共x, y兲 共x y兲 i 共x y兲 j
Evaluating a Line Integral In Exercises 21–26, evaluate
33. F共x, y, z兲 x i y j z k
C: r 共t兲 4 cos t i 3 sin t j,
the line integral along the given path(s). 21.
冕
共
x2
C: r 共t兲 2 cos t i 2 sin t j t k, 0 t 2 34. F共x, y, z兲 共2y z兲 i 共z x兲 j 共x y兲 k
兲 ds
y2
C: curve of intersection of x 2 z 2 4 and y 2 z 2 4 from 共2, 2, 0兲 to 共0, 0, 2兲
C
(a) C: line segment from 共0, 0) to 共3, 4兲 (b) C: x 2 y 2 1, one starting at 共1, 0兲 22.
冕
0 t 2
revolution
counterclockwise,
xy ds
C
(a) C: line segment from 共0, 0兲 to 共5, 4兲 (b) C: counterclockwise around the triangle with vertices 共0, 0兲, 共4, 0兲, and 共0, 2兲
35. F共x, y, z兲 共 y z兲 i 共x z兲 j 共x y兲 k C: curve of intersection of z x 2 y 2 and y x from 共0, 0, 0兲 to 共2, 2, 8兲 36. F共x, y, z兲 共x 2 z兲 i 共 y 2 z兲 j x k C: curve of intersection of z x 2 and x 2 y 2 4 from 共0, 2, 0兲 to 共0, 2, 0兲
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Review Exercises
Evaluating a Line Integral In Exercises 37 and 38, use a computer algebra system to evaluate the line integral. 37.
冕
C
冕
C
39. Work Find the work done by the force field F x i 冪y j along the path y x 3兾2 from 共0, 0兲 to 共4, 8兲. 40. Work A 20-ton aircraft climbs 2000 feet while making a 90 turn in a circular arc of radius 10 miles. Find the work done by the engines.
Using the Fundamental Theorem of Line Integrals In Exercises 41 and 42, evaluate the integral using the Fundamental Theorem of Line Integrals. 2xyz dx x 2z dy x 2y dz
C
C: smooth curve from 共0, 0, 0兲 to 共1, 3, 2兲 42.
冕
y dx x dy
C
1 dz z
C: smooth curve from 共0, 0, 1兲 to 共4, 4, 4兲 43. Evaluating a Line Integral Evaluate the line integral
冕
y 2 dx 2xy dy.
(a) C: r 共t兲 共1 3t兲 i 共1 t兲 j, 0 t 1 (b) C: r 共t兲 t i 冪t j, 1 t 4 (c) Use the Fundamental Theorem of Line Integrals, where C is a smooth curve from 共1, 1兲 to 共4, 2兲. 44. Area and Centroid Consider the region bounded by the x-axis and one arch of the cycloid with parametric equations x a共 sin 兲 and y a共1 cos 兲. Use line integrals to find (a) the area of the region and (b) the centroid of the region.
Evaluating a Line Integral In Exercises 45–50, use Green’s Theorem to evaluate the line integral.
冕
y dx 2x dy
C: boundary of the square with vertices 共0, 0兲, 共0, 1兲, 共1, 0兲, and 共1, 1兲
冕
xy dx 共x y 兲 dy 2
2
C
C: boundary of the square with vertices 共0, 0兲, 共0, 2兲, 共2, 0兲, and 共2, 2兲 47.
冕
xy dx x 2 dy
C
50.
冕
y 2 dx x 4兾3 dy
C
C: x 2兾3 y 2兾3 1
Graphing a Parametric Surface In Exercises 51 and 52, use a computer algebra system to graph the surface represented by the vector-valued function. 51. r共u, v兲 sec u cos vi 共1 2 tan u兲 sin vj 2uk 0 u
, 0 v 2 3
u 52. r共u, v兲 eu兾4 cos vi eu兾4 sin vj k 6 0 u 4,
0 v 2
53. Investigation Consider the surface represented by the vector-valued function r共u, v兲 3 cos v cos ui 3 cos v sin uj sin vk. Use a computer algebra system to do the following. (a) Graph the surface for 0 u 2 and
(c) Graph the surface for 0 u
v . 2 2
v . 4 2
and 0 v . 4 2
(d) Graph and identify the space curve for 0 u 2 and v
. 4
(e) Approximate the area of the surface graphed in part (b). (f) Approximate the area of the surface graphed in part (c). 54. Evaluating a Surface Integral integral 兰S兰 z dS over the surface S:
Evaluate the surface
r 共u, v兲 共u v兲 i 共u v兲 j sin v k where 0 u 2 and 0 v .
C
46.
冕
(b) Graph the surface for 0 u 2 and
C
45.
49.
C: boundary of the region between the graphs of y x 2 and y1
C: r 共t兲 共2 cos t 2t sin t兲 i 共2 sin t 2t cos t兲 j, 0 t
冕
共x 2 y 2兲 dx 2xy dy
C: x 2 y 2 a 2
F dr
F共x, y兲 共2x y兲 i 共2y x兲 j
41.
冕
C
xy dx 共x2 y2兲 dy
C: y x 2 from 共0, 0兲 to 共2, 4兲 and y 2x from 共2, 4兲 to 共0, 0兲 38.
48.
1121
xy 2 dx x 2 y dy
C
C: x 4 cos t, y 4 sin t
55. Approximating a Surface Integral Use a computer algebra system to graph the surface S and approximate the surface integral
冕冕
共x y兲 dS
S
where S is the surface S: r 共u, v兲 u cos v i u sin v j 共u 1兲共2 u兲 k over 0 u 2 and 0 v 2.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1122
Chapter 15
56. Mass
Vector Analysis
A cone-shaped surface lamina S is given by
z a共a 冪x 2 y 2兲,
Q: solid region bounded by the coordinate planes and the plane 2x 3y 4z 12
0 z a 2.
At each point on S, the density is proportional to the distance between the point and the z-axis.
Verifying Stokes’s Theorem In Exercises 59 and 60, verify
(a) Sketch the cone-shaped surface.
Stokes’s Theorem by evaluating
(b) Find the mass m of the lamina.
Verifying the Divergence Theorem In Exercises 57 and 58, verify the Divergence Theorem by evaluating
冕冕 S
58. F共x, y, z兲 x i y j z k
冕
C
F dr
as a line integral and as a double integral. 59. F共x, y, z兲 共cos y y cos x兲 i 共sin x x sin y兲 j xyz k
F N dS
S: portion of z y 2 over the square in the xy-plane with vertices 共0, 0兲, 共a, 0兲, 共a, a兲, and 共0, a兲
as a surface integral and as a triple integral.
N is the upward unit normal vector to the surface.
57. F共x, y, z兲 x 2 i xy j z k
60. F共x, y, z兲 共x z兲 i 共 y z兲 j x 2 k
Q: solid region bounded by the coordinate planes and the plane 2x 3y 4z 12
S: first-octant portion of the plane 3x y 2z 12 61. Proof Prove that it is not possible for a vector field with twicedifferentiable components to have a curl of xi yj zk.
The Planimeter You have learned many calculus techniques for finding the area of a planar region. Engineers use a mechanical device called a planimeter for measuring planar areas, which is based on the area formula given in Theorem 15.9 (page 1078). As you can see in the figure, the planimeter is fixed at point O (but free to pivot) and has a hinge at A. The end of the tracer arm AB moves counterclockwise around the region R. A small wheel at B is perpendicular to AB and is marked with a scale to measure how much it rolls as B traces out the boundary of region R. In this project, you will show that the area of R is given by the length L of the tracer arm AB multiplied by the distance D that the wheel rolls. Assume that point B traces out the boundary of R for a t b. Point A will move back and forth along a circular arc around the origin O. Let 共t兲 denote the angle in the figure and let 共x共t兲, y共t兲兲 denote the coordinates of A.
(d) Let N sin i cos j. Explain why the distance D that the wheel rolls is given by D
冕
C
N T ds.
(e) Show that the area of region R is given by I 1 I 2 I3 I 4 DL. B r(t) O
R
\
(a) Show that the vector OB is given by the vector-valued function
Wheel
L
A(x, y)
θ
r共t兲 关x共t兲 L cos 共t兲兴 i 关 y 共t兲 L sin 共t兲兴 j. (b) Show that the following two integrals are equal to zero.
冕
b
I1
a
冕
b
1 2 d L dt 2 dt
I2
冕
a
冢
冣
dx 1 dy y x dt 2 dt dt
b
(c) Use the integral
FOR FURTHER INFORMATION For more information about Green’s Theorem and planimeters, see the article “As the Planimeter’s Wheel Turns: Planimeter Proofs for Calculus Class” by Tanya Leise in The College Mathematics Journal. To view this article, go to MathArticles.com.
关x 共t兲 sin 共t兲 y 共t兲 cos 共t兲兴 dt to show
a
that the following two integrals are equal.
冕 冕
b
I3
a b
I4
a
1 d d L y sin x cos dt 2 dt dt
冢
冢
冣
冣
1 dx dy L sin cos dt 2 dt dt
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1123
P.S. Problem Solving
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
P.S. Problem Solving 1. Heat Flux Consider a single heat source located at the origin with temperature T共x, y, z兲
25 冪x 2 y 2 z 2
3. Moments of Inertia Consider a wire of density 共x, y, z兲 given by the space curve C: r共t兲 x共t兲i y共t兲j z共t兲k,
.
a t b.
The moments of inertia about the x-, y-, and z-axes are given by
(a) Calculate the heat flux across the surface
冦
S 共x, y, z兲: z 冪1 x 2,
冧
1 1 x , 0 y 1 2 2
as shown in the figure.
Iy 兰C 共x 2 z 2兲 共x, y, z兲 ds Iz 兰C 共x 2 y 2兲 共x, y, z兲 ds. Find the moments of inertia for a wire of uniform density 1 in the shape of the helix
z
2
Ix 兰C 共 y 2 z 2兲 共x, y, z兲 ds
r共t兲 3 cos ti 3 sin tj 2tk, 0 t 2 (see figure).
N
r(t) = 3 cos ti + 3 sin t j + 2t k
S
1
r(t) =
z
2 2t 3/2 t2 i + tj + k 3 2 z
12 2
10
1
1
y
8
x
1
6
(b) Repeat the calculation in part (a) using the parametrization
4
x cos u, y v, z sin u where
x
2
2
x
2 u 3 3
0 v 1.
and
25 冪x 2 y 2 z 2
.
S 再 共x, y, z兲: z 冪1 x 2 y 2, x 2 y 2 1冎 as shown in the figure.
of density
1 given by the curve 1t
t2 2冪2 t 3兾2 i tj k, 2 3
(a) Show that 共ln f 兲
F . f2
冢1f 冣 fF .
(b) Show that
where
(c) Show that f n nf n2F. and
3
(d) The Laplacian is the differential operator
0 v 2.
2
z
2 2 2 2 2 x y z2
and Laplace’s equation is
N
x
0 t 1 (see figure).
5. Laplace’s Equation Let F共x, y, z兲 xi yj zk, and let f 共x, y, z兲 储F共x, y, z兲储.
x sin u cos v, y sin u sin v, z cos u
2
y
Figure for 4
(b) Repeat the calculation in part (a) using the parametrization
0 u
2
4. Moments of Inertia Find the moments of inertia for a wire
C: r共t兲
(a) Calculate the heat flux across the surface
1
y
Figure for 3
2. Heat Flux Consider a single heat source located at the origin with temperature T共x, y, z兲
1
2
1
1
S
1
2w
y
2w 2w 2w 2 2 0. x 2 y z
Any function that satisfies this equation is called harmonic. Show that the function w 1兾f is harmonic.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1124
Chapter 15
Vector Analysis
6. Green’s Theorem
冕
Consider the line integral
y n dx x n dy
C
where C is the boundary of the region lying between the graphs of y 冪a 2 x 2 共a > 0兲 and y 0.
10. Work The force field F共x, y兲 共3x 2 y 2兲 i 共2x 3 y兲 j is shown in the figure below. Three particles move from the point 共1, 1兲 to the point 共2, 4兲 along different paths. Explain why the work done is the same for each particle, and find the value of the work. y
(a) Use a computer algebra system to verify Green’s Theorem for n, an odd integer from 1 through 7.
6 5
(b) Use a computer algebra system to verify Green’s Theorem for n, an even integer from 2 through 8.
4
(c) For n an odd integer, make a conjecture about the value of the integral.
3 2
7. Area Use a line integral to find the area bounded by one arch of the cycloid x共兲 a共 sin 兲, y共兲 a共1 cos 兲, 0 2, as shown in the figure. y
y
2a
x 1 −1
x
2π a
Figure for 7
Figure for 8
8. Area Use a line integral to find the area bounded by the two loops of the eight curve x共t兲
1 sin 2t, y共t兲 sin t, 2
x 1
2
3
4
5
6
11. Proof Let S be a smooth oriented surface with normal vector N, bounded by a smooth simple closed curve C. Let v be a constant vector, and prove that
1
−1
1
0 t 2
冕冕 S
共2v N兲 dS
冕
共v r兲 dr.
C
12. Area and Work How does the area of the ellipse x2 y2 2 1 compare with the magnitude of the work done by 2 a b the force field 1 1 F共x, y兲 yi xj 2 2 on a particle that moves once around the ellipse (see figure)? y
as shown in the figure. 9. Work The force field F共x, y兲 共x y兲i 共x 2 1兲j acts on an object moving from the point 共0, 0兲 to the point 共0, 1兲, as shown in the figure.
1
y
x −1
1
1 −1
13. Verifying Identities
x 1
(a) Let f and g be scalar functions with continuous partial derivatives, and let C and S satisfy the conditions of Stokes’s Theorem. Verify each identity.
(a) Find the work done when the object moves along the path x 0, 0 y 1.
(i)
(b) Find the work done when the object moves along the path x y y 2, 0 y 1.
(ii)
(c) The object moves along the path x c共 y y 2兲, 0 y 1, c > 0. Find the value of the constant c that minimizes the work.
冕 冕
共 f g兲 dr
C
冕冕 S
共 f f 兲 dr 0
C
共f g兲 N dS (iii)
冕
共 f g gf 兲 dr 0
C
(b) Demonstrate the results of part (a) for the functions f 共x, y, z兲 xyz and g共x, y, z兲 z. Let S be the hemisphere z 冪4 x2 y2.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
16 16.1 16.2 16.3 16.4
Additional Topics in Differential Equations Exact First-Order Equations Second-Order Homogeneous Linear Equations Second-Order Nonhomogeneous Linear Equations Series Solutions of Differential Equations
Electrical Circuits (Exercises 33 and 34, p. 1147)
Parachute Jump (Section Project, p. 1148)
Undamped or Damped Motion? (Exercise 47, p. 1140)
Motion of a Spring (Example 8, p. 1138) Cost (Exercise 45, p. 1132) 1125 Clockwise from top left, nsj-images/iStockphoto; Danshutter/Shutterstock.com; ICHIRO/Photodisc/Getty Images; Mircea BEZERGHEANU/Shutterstock.com; Monty Rakusen/Cultura/Getty Images
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1126
Chapter 16
Additional Topics in Differential Equations
16.1 Exact First-Order Equations Solve an exact differential equation. Use an integrating factor to make a differential equation exact.
Exact Differential Equations In Chapter 6, you studied applications of differential equations to growth and decay problems. You also learned more about the basic ideas of differential equations and studied the solution technique known as separation of variables. In this chapter, you will learn more about solving differential equations and using them in real-life applications. This section introduces you to a method for solving the first-order differential equation M共x, y兲 dx N共x, y兲 dy 0 for the special case in which this equation represents the exact differential of a function z f 共x, y兲. Definition of an Exact Differential Equation The equation M 共x, y兲 dx N共x, y兲 dy 0 is an exact differential equation when there exists a function f of two variables x and y having continuous partial derivatives such that fx 共x, y兲 M 共x, y兲
and
fy 共x, y兲 N共x, y兲.
The general solution of the equation is f 共x, y兲 C.
From Section 13.3, you know that if f has continuous second partials, then M 2f 2f N . y yx x y x This suggests the following test for exactness. THEOREM 16.1 Test for Exactness Let M and N have continuous partial derivatives on an open disk R. The differential equation M 共x, y兲 dx N共x, y兲 dy 0 is exact if and only if M N . y x
Every differential equation of the form M 共x兲 dx N共 y兲 dy 0 is exact. In other words, a separable differential equation is actually a special type of an exact equation. Exactness is a fragile condition in the sense that seemingly minor alterations in an exact equation can destroy its exactness. This is demonstrated in the next example.
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16.1
Exact First-Order Equations
1127
Testing for Exactness Determine whether each differential equation is exact. a. 共xy2 x兲 dx yx2 dy 0
b. cos y dx 共y2 x sin y兲 dy 0
Solution a. This differential equation is exact because M 关xy 2 x兴 2xy y y
and
N 关 yx 2兴 2xy. x x
Notice that the equation 共 y 2 1兲 dx xy dy 0 is not exact, even though it is obtained by dividing each side of the first equation by x. b. This differential equation is exact because M 关cos y兴 sin y and y y
N 2 关 y x sin y兴 sin y. x x
Notice that the equation cos y dx 共 y 2 x sin y兲 dy 0 is not exact, even though it differs from the first equation only by a single sign. Note that the test for exactness of M共x, y兲 dx N共x, y兲 dy 0 is the same as the test for determining whether F共x, y兲 M 共x, y兲 i N 共x, y兲 j is the gradient of a potential function (Theorem 15.1). This means that a general solution f 共x, y兲 C to an exact differential equation can be found by the method used to find a potential function for a conservative vector field.
Solving an Exact Differential Equation See LarsonCalculus.com for an interactive version of this type of example.
Solve the differential equation 共2xy 3x 2兲 dx 共 x 2 2y兲 dy 0. Solution
This differential equation is exact because
M 关2xy 3x 2兴 2x and y y
N 关x2 2y兴 2x. x x
The general solution, f 共x, y兲 C, is f 共x, y兲
冕
M 共x, y兲 dx
冕
共2xy 3x 2兲 dx x 2y x 3 g共 y兲.
In Section 15.1, you determined g共 y兲 by integrating N共x, y兲 with respect to y and reconciling the two expressions for f 共x, y兲. An alternative method is to partially differentiate this version of f 共x, y兲 with respect to y and compare the result with N共x, y兲. In other words, y
C = 1000
24 20
N共x, y兲
2 fy共x, y兲 关 x y x 3 g共 y兲兴 x 2 g 共 y兲 x 2 2y. y
C = 100
16
g 共 y兲 2y
12 8
So, g共 y兲 2y, and it follows that g共 y兲 y 2 C1. Therefore,
C = 10
C=1
x −12
−8
−4
Figure 16.1
4
8
12
f 共x, y兲 x 2 y x 3 y 2 C1 and the general solution is x 2 y x 3 y 2 C. Figure 16.1 shows the solution curves that correspond to C 1, 10, 100, and 1000.
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1128
Chapter 16
Additional Topics in Differential Equations
Solving an Exact Differential Equation Find the particular solution of
共cos x x sin x y 2兲 dx 2xy dy 0 that satisfies the initial condition y 1 when x . Solution
The differential equation is exact because M y
N x
关cos x x sin x y 2兴 2y 关2xy兴. y x Because N 共x, y兲 is simpler than M共x, y兲, it is better to begin by integrating N共x, y兲. f 共x, y兲
冕
N共x, y兲 dy
冕
2xy dy xy 2 g共x兲 M共x, y兲
TECHNOLOGY A graphing utility can be used to graph a particular solution that satisfies the initial condition of a differential equation. In Example 3, the differential equation and initial conditions are satisfied when xy2 x cos x 0, which implies that the particular solution can be written as x 0 or y ± 冪cos x. On a graphing utility screen, the solution would be represented by Figure 16.2 together with the y-axis.
So, g 共x兲 cos x x sin x and g共x兲
Figure 16.2
冕
共cos x x sin x兲 dx
x cos x C1 which implies that f 共x, y兲 xy 2 x cos x C1, and the general solution is xy 2 x cos x C.
General solution
Applying the given initial condition produces
共1兲 2 cos C which implies that C 0. So, the particular solution is 4
−4
关 xy 2 g共x兲兴 y 2 g 共x兲 cos x x sin x y 2 x g 共x兲 cos x x sin x
4
−4
fx共x, y兲
y
xy 2 x cos x 0. The graph of the particular solution is shown in Figure 16.3. Notice that the graph consists of two parts: the ovals are given by y 2 cos x 0, and the y-axis is given by x 0.
4 2
(π , 1) x
−3π
−2π
−π −2
π
2π
3π
−4
Figure 16.3
In Example 3, note that for z f 共x, y兲 xy 2 x cos x, the total differential of z is given by dz fx共x, y兲 dx fy共x, y兲 dy 共cos x x sin x y 2兲 dx 2xy dy M 共x, y兲 dx N 共x, y兲 dy. In other words, M dx N dy 0 is called an exact differential equation because M dx N dy is exactly the differential of f 共x, y兲.
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16.1
Exact First-Order Equations
1129
Integrating Factors When the differential equation M共x, y兲 dx N共x, y兲 dy 0 is not exact, it may be possible to make it exact by multiplying by an appropriate factor u共x, y兲, which is called an integrating factor for the differential equation.
Multiplying by an Integrating Factor a. When the differential equation 2y dx x dy 0
Not an exact equation
is multiplied by the integrating factor u共x, y兲 x, the resulting equation 2xy dx x 2 dy 0
Exact equation
is exact—the left side is the total differential of x 2 y. b. When the equation y dx x dy 0
Not an exact equation
is multiplied by the integrating factor u共x, y兲 1兾y 2, the resulting equation 1 x dx 2 dy 0 y y
Exact equation
is exact—the left side is the total differential of x兾y. Finding an integrating factor can be difficult. There are two classes of differential equations, however, whose integrating factors can be found routinely— namely, those that possess integrating factors that are functions of either x alone or y alone. The next theorem, which is presented without proof, outlines a procedure for finding these two special categories of integrating factors.
REMARK When either h共x兲 or k共x兲 is constant, Theorem 16.2 still applies. As an aid to remembering these formulas, note that the subtracted partial derivative identifies both the denominator and the variable for the integrating factor.
THEOREM 16.2 Integrating Factors Consider the differential equation M共x, y兲 dx N共x, y兲 dy 0. 1. If 1 关M 共x, y兲 Nx共x, y兲兴 h共x兲 N共x, y兲 y is a function of x alone, then e兰h共x兲 dx is an integrating factor. 2. If 1 关N 共x, y兲 My共x, y兲兴 k共 y兲 M共x, y兲 x is a function of y alone, then e兰k共 y兲 dy is an integrating factor.
Exploration In Chapter 6, you solved the first-order linear differential equation dy P共x兲y Q共x兲 dx by using the integrating factor u共x兲 e兰P共x兲 dx. Show that you can obtain this integrating factor by using the methods of this section.
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1130
Chapter 16
Additional Topics in Differential Equations
Finding an Integrating Factor Solve the differential equation
共 y 2 x兲 dx 2y dy 0. Solution This equation is not exact because My共x, y兲 2y and
Nx共x, y兲 0.
However, because My共x, y兲 Nx共x, y兲 2y 0 1 h共x兲 N共x, y兲 2y it follows that e兰h共x兲 dx e兰 dx e x is an integrating factor. Multiplying the differential equation by e x produces the exact differential equation
共 y 2e x xe x兲 dx 2ye x dy 0 whose solution is obtained as follows. f 共x, y兲
冕
fx共x, y兲
y 2e x
N共x, y兲 dy
冕
2ye x dy y 2e x g共x兲 M共x, y兲
g 共x兲
y 2e x
xe x
g 共x兲 xe x
Therefore, g 共x兲 xe x and g共x兲 xe x e x C1, which implies that f 共x, y兲 y 2e x xe x e x C1. The general solution is y 2e x xe x e x C, or y 2 x 1 Cex.
General solution
The next example shows how a differential equation can help in sketching a force field given by F共x, y兲 M共x, y兲 i N共x, y兲 j.
An Application to Force Fields
Force field:
y2 − x
2y
j i− x2 + y2 x2 + y2 Family of curves tangent to F: F(x, y) =
y2
=x−1+
Sketch the force field F共x, y兲
Ce − x y
by finding and sketching the family of curves tangent to F. Solution
−1 −2 −3
Figure 16.4
At the point 共x, y兲 in the plane, the vector F共x, y兲 has a slope of
dy 共 y 2 x兲兾冪x 2 y 2 共 y 2 x兲 dx 2y兾冪x 2 y 2 2y
2
−3
2y y2 x i j 冪x 2 y 2 冪x 2 y 2
x 3
which, in differential form, is 2y dy 共 y 2 x兲 dx 共 y 2 x兲 dx 2y dy 0. From Example 5, you know that the general solution of this differential equation is y 2 x 1 Cex. Figure 16.4 shows several representative curves from this family. Note that the force vector at 共x, y兲 is tangent to the curve passing through 共x, y兲.
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16.1
16.1 Exercises
Exact First-Order Equations
1131
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Testing for Exactness In Exercises 1– 4, determine whether the differential equation is exact. Explain your reasoning.
Finding a Particular Solution In Exercises 17–22, find the particular solution that satisfies the initial condition.
1. 共2x xy2兲 dx 共3 x2y兲 dy 0
Differential Equation
Initial Condition
17.
y dx 关ln共x 1兲 2y兴 dy 0 x1
y共2兲 4
18.
1 共x dx y dy兲 0 x2 y2
y共0兲 4
Solving an Exact Differential Equation In Exercises
19. e 3x 共sin 3y dx cos 3y dy兲 0
y共0兲
5–14, determine whether the differential equation is exact. If it is, find the general solution.
20. 共x y 兲 dx 2xy dy 0
y共3兲 1
21. 共2xy 9x 兲 dx 共2y x 1兲 dy 0
y共0兲 3
22. 共2xy2 4兲 dx 共2x2y 6兲 dy 0
y共1兲 8
2. 共1 xy兲 dx 共 y xy兲 dy 0 3. x sin y dx x cos y dy 0 4. ye xy dx xe xy dy 0
2
2
2
5. 共2x 3y兲 dx 共2y 3x兲 dy 0 6. ye x dx e x dy 0
Finding an Integrating Factor In Exercises 23–32, find
7. 共3y 2 10xy 2兲 dx 共6xy 2 10x 2y兲 dy 0
the integrating factor that is a function of x or y alone and use it to find the general solution of the differential equation.
8. 2 cos共2x y兲 dx cos共2x y兲 dy 0 9. 共4x 3 6xy 2兲 dx 共4y 3 6xy兲 dy 0
23. y dx 共x 6y 2兲 dy 0
10. 2y 2e xy dx 2xye xy dy 0 2
2
24. 共2x 3 y兲 dx x dy 0
1 共x dy y dx兲 0 11. 2 x y2 12. e共x 13.
25. 共5x 2 y兲 dx x dy 0 26. 共5x 2 y 2兲 dx 2y dy 0
共x dx y dy兲 0
2 y 2 兲
27. 共x y兲 dx tan x dy 0
1 共 y 2 dx x 2 dy兲 0 共x y兲2
28. 共2x 2 y 1兲 dx x 3 dy 0
14. e cos xy 关 y dx 共x tan xy兲 dy兴 0
29. y 2 dx 共xy 1兲 dy 0
y
Graphical and Analytic Analysis In Exercises 15 and 16, (a) sketch an approximate solution of the differential equation satisfying the initial condition on the slope field, (b) find the particular solution that satisfies the initial condition, and (c) use a graphing utility to graph the particular solution. Compare the graph with the sketch in part (a). Differential Equation
Initial Condition
15. 共2x tan y 5兲 dx 共x 2 sec 2 y兲 dy 0 16.
1 共x dx y dy兲 0 冪x y 2 2
4
4
2
2
2
4
−4
−4
Figure for 15
integrating factor to find the general solution of the differential equation.
共4, 3兲
34. u共x, y兲
共3y 2 5x 2 y兲 dx 共3xy 2x 3兲 dy 0
Figure for 16
x2y
35. u共x, y兲 x2 y3
共y 5 x 2 y兲 dx 共2xy 4 2x 3兲 dy 0
36. u共x, y兲 x2 y2
y 3 dx 共xy 2 x 2兲 dy 0
37. Integrating Factor Show that each expression is an integrating factor for the differential equation y dx x dy 0. 2
−2
Using an Integrating Factor In Exercises 33–36, use the
共4x 2 y 2y 2兲 dx 共3x 3 4xy兲 dy 0
−2
−2
32. 共2y 3 1兲 dx 共3xy 2 x 3兲 dy 0
Integrating Factor Differential Equation
x −4
31. 2y dx 共x sin 冪y 兲 dy 0
33. u共x, y兲 xy 2
x −2
30. 共x 2 2x y兲 dx 2 dy 0
冢12, 4 冣 y
y
−4
2
4
(a)
1 x2
(b)
1 y2
(c)
38. Integrating Factor
1 xy
(d)
1 x2 y2
Show that the differential equation
共axy 2 by兲 dx 共bx 2y ax兲 dy 0 is exact only when a b. For a b, show that x m y n is an integrating factor, where m
2b a 2a b , n . ab ab
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1132
Chapter 16
Additional Topics in Differential Equations
Tangent Curves In Exercises 39–42, use a graphing utility to graph the family of curves tangent to the force field. 39. F共x, y兲 40. F共x, y兲
y 冪x 2 y 2
x 冪x 2 y 2
i i
x 冪x 2 y 2
y 冪x 2 y 2
冢
41. F共x, y兲 4x 2 y i 2xy 2
j j
冣
x j y2
Euler’s Method In Exercises 47 and 48, (a) use Euler’s Method and a graphing utility to graph the particular solution of the initial value problem over the indicated interval with the specified value of h and initial condition, (b) find the exact solution of the differential equation analytically, and (c) use a graphing utility to graph the particular solution and compare the result with the graph in part (a). Differential Equation
42. F共x, y兲 共1 x 2兲 i 2xy j
Finding an Equation of a Curve In Exercises 43 and 44, find an equation of the curve with the specified slope passing through the given point. Slope
Point
dy yx 43. dx 3y x
共2, 1兲
dy 2xy 44. dx x 2 y 2
共0, 2兲
Interval
h
Initial Condition
47. y
xy x2 y2
关2, 4兴
0.05
y共2兲 1
48. y
6x y 2 y共3y 2x兲
关0, 5兴
0.2
y共0兲 1
49. Euler’s Method Repeat Exercise 47 for h 1 and discuss how the accuracy of the result changes. 50. Euler’s Method Repeat Exercise 48 for h 0.5 and discuss how the accuracy of the result changes.
45. Cost
WRITING ABOUT CONCEPTS
In a manufacturing process where y C共x兲 represents the cost of producing x units, the elasticity of cost is defined as
51. Testing for Exactness Explain how to determine whether a differential equation is exact.
marginal cost C 共x兲 x dy E共x兲 . average cost C共x兲兾x y dx
52. Finding an Integrating Factor Outline the procedure for finding an integrating factor for the differential equation M共x, y兲 dx N共x, y兲 dy 0.
Find the cost function when the elasticity function is
True or False? In Exercises 53–56, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.
20x y E共x兲 2y 10x
53. The differential equation 2xy dx 共 y 2 x 2兲 dy 0 is exact.
where
54. If M dx N dy 0 is exact, then xM dx xN dy 0 is also exact.
C共100兲 500
55. If M dx N dy 0 is exact, then 关 f 共x兲 M 兴 dx 关 g共 y兲 N 兴 dy 0 is also exact.
and x 100.
56. The differential equation f 共x兲 dx g共 y兲 dy 0 is exact.
Exact Differential Equation In Exercises 57 and 58, find 46.
HOW DO YOU SEE IT? The graph of the
all values of k such that the differential equation is exact.
cost function in Exercise 45 is shown below. Use the figure to estimate the limit of the cost function as x approaches 100 from the right.
57. 共xy2 kx2y x3兲 dx 共x3 x2y y2兲 dy 0
y
58. 共 ye2xy 2x兲 dx 共kxe2xy 2y兲 dy 0 59. Exact Differential Equation Find all nonzero functions f and g such that
5000 4000
g共 y兲 sin x dx y2 f 共x兲 dy 0
3000
is exact.
2000
60. Exact Differential Equation Find all nonzero functions g such that
1000 x 50 100 150 200 250 300 350 400 450 500
g共 y兲 ey dx xy dy 0 is exact. Mircea BEZERGHEANU/Shutterstock.com
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16.2
Second-Order Homogeneous Linear Equations
1133
16.2 Second-Order Homogeneous Linear Equations Solve a second-order linear differential equation. Solve a higher-order linear differential equation. Use a second-order linear differential equation to solve an applied problem.
Second-Order Linear Differential Equations In this section and the next section, you will learn methods for solving higher-order linear differential equations. Definition of Linear Differential Equation of Order n Let g1, g2, . . . , gn and f be functions of x with a common (interval) domain. An equation of the form
REMARK Notice that this use of the term homogeneous differs from that in Section 6.3.
y共n兲 g1共x兲y共n1兲 g2共x兲y共n2兲 . . . gn1共x兲y gn共x兲y f 共x兲 is a linear differential equation of order n. If f 共x兲 0, then the equation is homogeneous; otherwise, it is nonhomogeneous.
Homogeneous equations are discussed in this section, and the nonhomogeneous case is discussed in the next section. The functions y1, y2, . . . , yn are linearly independent when the only solution of the equation C1y1 C2y2 . . . Cn yn 0 is the trivial one C1 C2 . . . Cn 0. Otherwise, this set of functions is linearly dependent.
Linearly Independent and Dependent Functions a. The functions y1共x兲 sin x
and y2共x兲 x
are linearly independent because the only values of C1 and C2 for which C1 sin x C2x 0 for all x are C1 0 and C2 0. b. It can be shown that two functions form a linearly dependent set if and only if one is a constant multiple of the other. For example, y1共x兲 x and
y2共x兲 3x
are linearly dependent because C1x C2共3x兲 0 has the nonzero solutions C1 3 and
C2 1.
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1134
Chapter 16
Additional Topics in Differential Equations
The next theorem points out the importance of linear independence in constructing the general solution of a second-order linear homogeneous differential equation with constant coefficients. THEOREM 16.3 Linear Combinations of Solutions If y1 and y2 are linearly independent solutions of the differential equation y ay by 0, then the general solution is y C1 y1 C2 y2
General solution
where C1 and C2 are constants.
Proof
This theorem is proved in only one direction. Letting y1 and y2 be solutions of
y ay by 0 you obtain the following system of equations. y1 共x兲 ay1 共x兲 by1共x兲 0 y2 共x兲 ay2 共x兲 by2共x兲 0 Multiplying the first equation by C1, multiplying the second by C2, and adding the resulting equations together produces
关C1 y1 共x兲 C2 y2 共x兲兴 a关C1 y1共x兲 C2 y2共x兲兴 b关C1 y1共x兲 C2 y2共x兲兴 0 which means that y C1 y1 C2 y2 is a solution, as desired. The proof that all solutions are of this form is best left to a full course on differential equations. See LarsonCalculus.com for Bruce Edwards’s video of this proof.
Theorem 16.3 states that when you can find two linearly independent solutions, you can obtain the general solution by forming a linear combination of the two solutions. To find two linearly independent solutions, note that the nature of the equation y ay by 0 suggests that it may have solutions of the form y emx. If so, then y memx and
y m2emx.
So, by substitution, y emx is a solution if and only if y ay by 0 m e amemx bemx 0 emx共m2 am b兲 0. 2 mx
Because emx is never 0, y emx is a solution if and only if m2 am b 0.
Characteristic equation
This is the characteristic equation of the differential equation y ay by 0. Note that the characteristic equation can be determined from its differential equation simply by replacing y with m2, y with m, and y with 1.
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16.2
(a) y 9y 0 (b) y 6y 8y 0
1135
Characteristic Equation: Distinct Real Zeros
Exploration For each differential equation below, find the characteristic equation. Solve the characteristic equation for m, and use the values of m to find a general solution of the differential equation. Using your results, develop a general solution of differential equations with characteristic equations that have distinct real roots.
Second-Order Homogeneous Linear Equations
Solve the differential equation y 4y 0. Solution
In this case, the characteristic equation is
m2 4 0.
Characteristic equation
So, m ± 2. Therefore, y1 em1x e2x
and y2 em2x e2x
are particular solutions of the differential equation. Furthermore, because these two solutions are linearly independent, you can apply Theorem 16.3 to conclude that the general solution is y C1e2x C2e2x.
General solution
The characteristic equation in Example 2 has two distinct real zeros. From algebra, you know that this is only one of three possibilities for quadratic equations. In general, the quadratic equation m2 am b 0 has zeros m1
a 冪a2 4b 2
and
m2
a 冪a2 4b 2
which fall into one of three cases. 1. Two distinct real zeros, m1 m2 2. Two equal real zeros, m1 m2 3. Two complex conjugate zeros, m1 i and m2 i In terms of the differential equation y ay by 0 these three cases correspond to three different types of general solutions. THEOREM 16.4 The solutions of
Solutions of y ⴙ ay ⴙ by 0
y ay by 0 fall into one of the following three cases, depending on the solutions of the characteristic equation, m2 am b 0. 1. Distinct Real Zeros If m1 m2 are distinct real zeros of the characteristic equation, then the general solution is y C1em1x C2em2x. FOR FURTHER INFORMATION
For more information on Theorem 16.4, see the article “A Note on a Differential Equation” by Russell Euler in the 1989 winter issue of the Missouri Journal of Mathematical Sciences.
2. Equal Real Zeros If m1 m2 are equal real zeros of the characteristic equation, then the general solution is y C1em1x C2xem1x 共C1 C2x兲em1x. 3. Complex Zeros If m1 i and m2 i are complex zeros of the characteristic equation, then the general solution is y C1e x cos x C2e x sin x.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1136
Chapter 16
Additional Topics in Differential Equations
Characteristic Equation: Complex Zeros Find the general solution of the differential equation y 6y 12y 0. Solution
The characteristic equation
m2 6m 12 0
y
has two complex zeros, as follows.
3
f
6 ± 冪36 48 2 冪 6 ± 12 2 6 ± 2冪3 2 3 ± 冪3 3 ± 冪3i
m f+g g
x
g−f
1
2
3
4
The basic solutions in Example 3, f 共x兲 e3x cos 冪3x and g共x兲 e3x sin 冪3x, are shown in the graph along with other members of the family of solutions. Notice that as x → , all of these solutions approach 0. Figure 16.5
So, 3 and 冪3, and the general solution is y C1e3x cos 冪3x C2e3x sin 冪3x. The graphs of the basic solutions f 共x兲 e3x cos 冪3x
and g共x兲 e3x sin 冪3x
along with other members of the family of solutions, are shown in Figure 16.5. In Example 3, note that although the characteristic equation has two complex zeros, the solution of the differential equation is real.
Characteristic Equation: Repeated Zeros Solve the differential equation y 4y 4y 0 subject to the initial conditions y共0兲 2 and y共0兲 1. Solution m2
The characteristic equation
4m 4 0
共m 2兲2 0
has two equal zeros given by m 2. So, the general solution is y C1e2x C2xe2x.
General solution
Now, because y 2 when x 0, you have 2 C1共1兲 C2共0兲共1兲
2 C1.
Furthermore, because y 1 when x 0, you have y 2C1e2x C2共2xe2x e2x兲 1 2共2兲共1兲 C2关2共0兲共1兲 1兴 5 C2. Therefore, the solution is y 2e2x 5xe2x.
Particular solution
Try checking this solution in the original differential equation.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
16.2
Second-Order Homogeneous Linear Equations
1137
Higher-Order Linear Differential Equations For higher-order homogeneous linear differential equations, you can find the general solution in much the same way as you do for second-order equations. That is, you begin by determining the n zeros of the characteristic equation. Then, based on these n zeros, you form a linearly independent collection of n solutions. The major difference is that with equations of third or higher order, zeros of the characteristic equation may occur more than twice. When this happens, the linearly independent solutions are formed by multiplying by increasing powers of x, as demonstrated in Examples 6 and 7.
Solving a Third-Order Equation Find the general solution of y y 0. Solution
The characteristic equation and its zeros are
m3 m 0 m共m 1兲共m 1兲 0 m 0, 1, 1. Because the characteristic equation has three distinct zeros, the general solution is y C1 C2ex C3e x.
General solution
Solving a Third-Order Equation Find the general solution of y 3y 3y y 0. Solution m3
The characteristic equation and its zeros are
3m2 3m 1 0 共m 1兲3 0 m 1.
Because the zero m 1 occurs three times, the general solution is y C1ex C2 xex C3x2ex.
General solution
Solving a Fourth-Order Equation See LarsonCalculus.com for an interactive version of this type of example.
Find the general solution of y共4兲 2y y 0. Solution
The characteristic equation and its zeros are
m4 2m2 1 0 共m2 1兲2 0 m ± i. Because each of the zeros m1 i 0 i and
m2 i 0 i
occurs twice, the general solution is y C1 cos x C2 sin x C3x cos x C4x sin x.
General solution
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1138
Chapter 16
Additional Topics in Differential Equations
Application
l = natural length
y = displacement m
A rigid object of mass m attached to the end of the spring causes a displacement of y. Figure 16.6
One of the many applications of linear differential equations is describing the motion of an oscillating spring. According to Hooke’s Law, a spring that is stretched (or compressed) y units from its natural length l tends to restore itself to its natural length by a force F that is proportional to y. That is, F共 y兲 ky, where k is the spring constant and indicates the stiffness of the spring. A rigid object of mass m is attached to the end of a spring and causes a displacement, as shown in Figure 16.6. Assume that the mass of the spring is negligible compared with m. When the object is pulled downward and released, the resulting oscillations are a product of two opposing forces—the spring force F共y兲 ky and the weight mg of the object. Under such conditions, you can use a differential equation to find the position y of the object as a function of time t. According to Newton’s Second Law of Motion, the force acting on the weight is F ma, where a d 2 y兾dt2 is the acceleration. Assuming that the motion is undamped—that is, there are no other external forces acting on the object—it follows that m共d 2 y兾dt2兲 ky, and you have
冢 冣
d2y k y 0. dt2 m
Undamped motion of a spring
Undamped Motion of a Spring A 4-pound weight stretches a spring 8 inches from its natural length. The weight is pulled downward an additional 6 inches and released with an initial upward velocity of 8 feet per second. Find a formula for the position of the weight as a function of time t. 2 Solution By Hooke’s Law, 4 k共3 兲, so k 6. Moreover, because the weight w is given by mg, it follows that
m
w 4 1 . g 32 8
So, the resulting differential equation for this undamped motion is d2y 48y 0. dt2 The characteristic equation m2 48 0 has complex zeros m 0 ± 4冪3i, so the general solution is A common type of spring is a coil spring, also called a helical spring because the shape of the spring is a helix. A tension coil spring resists being stretched (see photo above and Example 8). A compression coil spring resists being compressed, such as the spring in a car suspension.
y C1e0 cos 4冪3 t C2e0 sin 4冪3 t C1 cos 4冪3 t C2 sin 4冪3 t. 1 When t 0 seconds, y 6 inches 2 foot. Using this initial condition, you have
1 C1共1兲 C2共0兲 2
1 C1 . 2
y 共0兲 2 1
To determine C2, note that y 8 feet per second when t 0 seconds. y共t兲 4冪3 C1 sin 4冪3 t 4冪3 C2 cos 4冪3 t 1 8 4冪3 共0兲 4冪3 C2共1兲 y共0兲 8 2
冢冣
2冪3 C2 3 Consequently, the position at time t is given by y
1 2冪3 sin 4冪3 t. cos 4冪3 t 3 2
ICHIRO/Photodisc/Getty Images
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16.2
Second-Order Homogeneous Linear Equations
1139
The object in Figure 16.7 undergoes an additional damping or frictional force that is proportional to its velocity. A case in point would be the damping force resulting from friction and movement through a fluid. Considering this damping force p
dy dt
Damping force
the differential equation for the oscillation is m
d2y dy ky p dt2 dt
or, in standard linear form, A damped vibration could be caused by friction and movement through a liquid. Figure 16.7
冢 冣
d2y p dy k y 0. 2 dt m dt m
16.2 Exercises
Damped motion of a spring
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Verifying a Solution In Exercises 1–4, verify the solution of the differential equation. Then use a graphing utility to graph the particular solutions for several different values of C1 and C2. What do you observe?
32. Finding a Particular Solution Determine C and such that y C sin冪3 t is a particular solution of the differential equation y y 0, where y共0兲 5.
1. y 共C1 C2x兲e3x
y 6y 9y 0
Finding a Particular Solution: Initial Conditions In Exercises 33–38, find the particular solution of the linear differential equation that satisfies the initial conditions.
2. y C1
y 4y 0
33. y y 30y 0
Solution e2x
Differential Equation C2
e2x
y 4y 0
3. y C1 cos 2x C2 sin 2x 4. y C1
ex
cos 3x
C2ex
sin 3x
y 2y 10y 0
Finding a General Solution In Exercises 5–30, find the general solution of the linear differential equation. 5. y y 0
6. y 2y 0
7. y y 6y 0
8. y 6y 5y 0
9. 2y 3y 2y 0
10. 16y 16y 3y 0
11. y 6y 9y 0
12. y 10y 25y 0
13. 16y 8y y 0
14. 9y 12y 4y 0
15. y y 0
16. y 4y 0
17. y 9y 0
18. y 2y 0
19. y 2y 4y 0
20. y 4y 21y 0
21. y 3y y 0
22. 3y 4y y 0
23. 9y 12y 11y 0
24. 2y 6y 7y 0
25.
y共4兲
y0
26. y共4兲 y 0
27. y 6y 11y 6y 0
28. y y y y 0
29. y 3y 7y 5y 0
30. y 3y 3y y 0
31. Finding a Particular Solution Consider the differential equation and the solution y 100y 0 y C1 cos 10x C2 sin 10x. Find the particular solution satisfying each initial condition. (a) y共0兲 2, y共0兲 0
(b) y共0兲 0,
y共0兲 2
y共0兲 1,
y共0兲 4
35. y 16y 0 y共0兲 0,
y共0兲 2
37. y 2y 3y 0 y共0兲 2,
y共0兲 1
34. y 7y 12y 0 y共0兲 3,
y共0兲 3
36. 9y 6y y 0 y共0兲 2,
y共0兲 1
38. 4y 4y y 0 y共0兲 3,
y共0兲 1
Finding a Particular Solution: Boundary Conditions In Exercises 39–44, find the particular solution of the linear differential equation that satisfies the boundary conditions, if possible. 39. y 4y 3y 0 y共0兲 1,
y共1兲 3
41. y 9y 0 y共0兲 3,
y共 兲 5
43. 4y 28y 49y 0 y共0兲 2,
y共1兲 1
40. 4y y 0 y共0兲 2,
y共 兲 5
42. 4y 20y 21y 0 y共0兲 3,
y共2兲 0
44. y 6y 45y 0 y共0兲 4,
y共 兲 8
WRITING ABOUT CONCEPTS 45. Characteristic Equation The solutions of the differential equation y ay by 0 fall into what three cases? What is the relationship of these solutions to the characteristic equation of the differential equation? 46. Linearly Independent Functions Two functions are said to be linearly independent provided what?
(c) y共0兲 1, y共0兲 3
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1140
Chapter 16
Additional Topics in Differential Equations
Vibrating Spring In Exercises 53–58, describe the motion of a 32-pound weight suspended on a spring. Assume that the weight stretches the spring 32 foot from its natural position.
47. Undamped or Damped Motion? Several shock absorbers are shown at the right. Do you think the motion of the spring in a shock absorber is undamped or damped?
53. The weight is pulled 21 foot below the equilibrium position and released. 54. The weight is raised 32 foot above the equilibrium position and released. 55. The weight is raised 32 foot above the equilibrium position and started off with a downward velocity of 21 foot per second. 56. The weight is pulled 21 foot below the equilibrium position and started off with an upward velocity of 12 foot per second. 1
57. The weight is pulled 2 foot below the equilibrium position and released. The motion takes place in a medium that furnishes a damping force of magnitude 18 speed at all times.
HOW DO YOU SEE IT? Give a geometric
48.
argument to explain why the graph cannot be a solution of the differential equation. (It is not necessary to solve the differential equation.)
58. The weight is pulled 12 foot below the equilibrium position and released. The motion takes place in a medium that furnishes a damping force of magnitude 14 v at all times.
(b) y 12 y
(a) y y y
ⱍⱍ
y
5
3
4
2
3
1
59. Real Zeros The characteristic equation of the differential equation y ay by 0 has two equal real zeros given by m r. Show that y C1erx C2 xerx is a solution. x
2
−3 −2
1
1
2
3
x −3 −2 −1
1
2
y ay by 0
−3
3
60. Complex Zeros The characteristic equation of the differential equation
has complex zeros given by m1 i and m2 i. Show that y C1e x cos x C2ex sin x is a solution.
Vibrating Spring In Exercises 49–52, match the differential equation with the graph of a particular solution. [The graphs are labeled (a), (b), (c), and (d).] The correct match can be made by comparing the frequency of the oscillations or the rate at which the oscillations are being damped with the appropriate coefficient in the differential equation. (a)
(b)
y
2
4
5
61. y C1e3x C2e3x is the general solution of y 6y 9 0. is
the
general
63. y x is a solution of an y 共n兲 an1y 共n1兲 . . . a1 y a0 y 0 if and only if a1 a0 0. x
x 1
statement is true or false. If it is false, explain why or give an example that shows it is false.
62. y 共C1 C2 x兲sin x 共C3 C4x兲cos x solution of y共4兲 2y y 0.
y 3
3
True or False? In Exercises 61–64, determine whether the
1
6
2
3
4
6
64. It is possible to choose a and b such that y x 2e x is a solution of y ay by 0.
Wronskian The Wronskian of two differentiable functions f and g, denoted by W冇 f, g冈, is defined as the function given by (c)
(d)
y
the determinant
y 3
3
W冇 f, g冈 ⴝ x
x 2
3
4
5
1
6
6
3
49. y 9y 0 51. y 2y 10y 0
50. y 25y 0 52. y y
37 4
y0
ⱍ ⱍ f f
g . g
The functions f and g are linearly independent when there exists at least one value of x for which W冇 f, g冈 ⴝ 0. In Exercises 65–68, use the Wronskian to verify the linear independence of the two functions. 65. y1 eax
66. y1 eax
y2 ebx, a b 67. y1
eax
sin bx
y2
eax
cos bx,
y2 xeax 68. y1 x b 0
y2 x2
Monty Rakusen/Cultura/Getty Images
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16.3
Second-Order Nonhomogeneous Linear Equations
1141
16.3 Second-Order Nonhomogeneous Linear Equations Recognize the general solution of a second-order nonhomogeneous linear differential equation. Use the method of undetermined coefficients to solve a second-order nonhomogeneous linear differential equation. Use the method of variation of parameters to solve a second-order nonhomogeneous linear differential equation.
Nonhomogeneous Equations In the preceding section, damped oscillations of a spring were represented by the homogeneous second-order linear equation
冢 冣
d 2y p dy k y 0. dt 2 m dt m
Free motion
This type of oscillation is called free because it is determined solely by the spring and gravity and is free of the action of other external forces. If such a system is also subject to an external periodic force, such as a sin bt, caused by vibrations at the opposite end of the spring, then the motion is called forced, and it is characterized by the nonhomogeneous equation
冢 冣
d 2y p dy k y a sin bt. dt 2 m dt m SOPHIE GERMAIN (1776–1831)
Many of the early contributors to calculus were interested in forming mathematical models for vibrating strings and membranes, oscillating springs, and elasticity. One of these was the French mathematician Sophie Germain, who in 1816 was awarded a prize by the French Academy for a paper entitled “Memoir on the Vibrations of Elastic Plates.” See LarsonCalculus.com to read more of this biography.
Forced motion
In this section, you will study two methods for finding the general solution of a nonhomogeneous linear differential equation. In both methods, the first step is to find the general solution of the corresponding homogeneous equation. y yh
General solution of homogeneous equation
Having done this, you try to find a particular solution of the nonhomogeneous equation. y yp
Particular solution of nonhomogeneous equation
By combining these two results, you can conclude that the general solution of the nonhomogeneous equation is y yh yp as stated in the next theorem. THEOREM 16.5 Let
Solution of Nonhomogeneous Linear Equation
y ay by F共x兲 be a second-order nonhomogeneous linear differential equation. If yp is a particular solution of this equation and yh is the general solution of the corresponding homogeneous equation, then y yh yp is the general solution of the nonhomogeneous equation. The Granger Collection, NYC
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1142
Chapter 16
Additional Topics in Differential Equations
Method of Undetermined Coefficients You already know how to find the solution yh of a linear homogeneous differential equation. The remainder of this section looks at ways to find the particular solution yp. When F共x兲 in y ay by F共x兲 consists of sums or products of xn, emx, cos x, or sin x, you can find a particular solution yp by the method of undetermined coefficients. The object of this method is to guess that the solution yp is a generalized form of F共x兲. Here are some examples. 1. For F共x兲 3x 2, choose yp Ax 2 Bx C. 2. For F共x兲 4xex, choose yp Axex Bex. 3. For F共x兲 x sin 2x, choose yp 共Ax B兲 C sin 2x D cos 2x. Then, by substitution, determine the coefficients for the generalized solution.
Method of Undetermined Coefficients Find the general solution of the equation y 2y 3y 2 sin x. Solution
To find yh, solve the characteristic equation.
m 2 2m 3 0 共m 1兲共m 3兲 0 m 1 or m 3 So, yh C1ex C2e3x. Next, let yp be a generalized form of 2 sin x. yp A cos x B sin x yp A sin x B cos x yp A cos x B sin x Substitution into the original differential equation yields y 2y 3y 2 sin x A cos x B sin x 2A sin x 2B cos x 3A cos x 3B sin x 2 sin x 共4A 2B兲cos x 共2A 4B兲sin x 2 sin x. By equating coefficients of like terms, you obtain 4A 2B 0 and 2A 4B 2 with solutions A
1 5
2 and B . 5
Therefore, yp
1 2 cos x sin x 5 5
and the general solution is y yh yp C1ex C2e3x
1 2 cos x sin x. 5 5
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16.3
Second-Order Nonhomogeneous Linear Equations
1143
In Example 1, the form of the homogeneous solution yh C1ex C2e3x has no overlap with the function F共x兲 in the equation y ay by F共x兲. However, suppose the given differential equation in Example 1 were of the form y 2y 3y ex. Now it would make no sense to guess that the particular solution was y Aex because this solution would yield 0. In such cases, you should alter your guess by multiplying by the lowest power of x that removes the duplication. For this particular problem, you would guess yp Axex.
Method of Undetermined Coefficients Find the general solution of y 2y x 2ex. Solution
The characteristic equation
m 2m 0 2
has solutions m 0 and m 2. So, yh C1 C2e 2x. Because F共x兲 x 2ex, your first choice for yp would be 共A Bx兲 Cex. However, because yh already contains a constant term C1, you should multiply the polynomial part by x and use yp Ax Bx 2 Cex yp A 2Bx Ce x yp 2B Cex. Substitution into the differential equation produces y 2y x 2ex 2B Cex 2共A 2Bx Cex兲 x 2ex 共2B 2A兲 4Bx Cex x 2ex. Equating coefficients of like terms yields the system 2B 2A 0, 4B 1, C 2 with solutions A B 14 and C 2. Therefore, 1 1 yp x x 2 2e x 4 4 and the general solution is y yh yp 1 1 C1 C2e 2x x x 2 2e x. 4 4
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1144
Chapter 16
Additional Topics in Differential Equations
In Example 2, the polynomial part of the initial guess 共A Bx兲 Cex for yp overlapped by a constant term with yh C1 C2e2x and it was necessary to multiply the polynomial part by a power of x that removed the overlap. The next example further illustrates some choices for yp that eliminate overlap with yh. Remember that in all cases, the first guess for yp should match the types of functions occurring in F共x兲.
Choosing the Form of the Particular Solution Determine a suitable choice for yp for each differential equation, given its general solution of the homogeneous equation. y ⴙ ay ⴙ by ⴝ F 冇x冈 a. y b. y 2y 10y 4 sin 3x c. y 4y 4 e 2x x2
yh C1 C2x C1ex cos 3x C2ex sin 3x C1e 2x C2 xe 2x
Solution a. Because F共x兲 x 2, the normal choice for yp would be A Bx Cx 2. However, because yh C1 C2x already contains a linear term, you should multiply by x 2 to obtain yp Ax 2 Bx3 Cx 4. b. Because F共x兲 4 sin 3x and each term in yh contains a factor of ex, you can simply let yp A cos 3x B sin 3x. c. Because F共x兲 e2x, the normal choice for yp would be Ae2x. However, because yh C1e2x C2xe2x already contains an xe2x term, you should multiply by x2 to get yp Ax2e2x.
Solving a Third-Order Equation See LarsonCalculus.com for an interactive version of this type of example.
Find the general solution of y 3y 3y y x. Solution From Example 6 in Section 16.2, you know that the homogeneous solution is yh C1ex C2 xex C3 x 2ex. Because F共x兲 x, let yp A Bx and obtain yp B and yp 0. So, by substitution into the general solution, you have 0 3共0兲 3共B兲 A Bx x 共3B A兲 Bx x. So, B 1 and A 3, which implies that yp 3 x. Therefore, the general solution is y yh yp C1ex C2xex C3x 2ex 3 x.
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16.3
Second-Order Nonhomogeneous Linear Equations
1145
Variation of Parameters The method of undetermined coefficients works well when F共x兲 is made up of polynomials or functions whose successive derivatives have a cyclical pattern. For functions such as 1兾x and tan x, which do not have such characteristics, it is better to use a more general method called variation of parameters. In this method, you assume that yp has the same form as yh, except that the constants in yh are replaced by variables. Variation of Parameters To find the general solution of the equation y ay by F共x兲, use these steps. 1. Find yh C1y1 C2 y2. 2. Replace the constants by variables to form yp u1y1 u2 y2. 3. Solve the following system for u1 and u2. u1 y1 u 2 y2 0 u1 y1 u 2 y2 F共x兲 4. Integrate to find u1 and u2. The general solution is y yh yp.
Variation of Parameters Solve the differential equation y 2y y Solution
ex , 2x
x > 0.
The characteristic equation
m2 2m 1 0
共m 1兲2 0
has one repeated solution, m 1. So, the homogeneous solution is yh C1y1 C2 y2 C1ex C2xex. Replacing C1 and C2 by u1 and u2 produces yp u1y1 u2 y2 u1e x u 2xe x. The resulting system of equations is u1 e x u2 xe x 0 ex u1 e x u2 共xe x e x兲 . 2x Subtracting the second equation from the first produces u2 1兾共2x兲. Then, by substitution 1 in the first equation, you have u1 2. Finally, integration yields u1
冕
1 x dx 2 2
and u2
1 2
冕
1 1 dx ln x ln 冪x. x 2
From this result, it follows that a particular solution is 1 yp xe x 共ln 冪x 兲xe x 2 and the general solution is y C1e x C2xe x
1 x xe xe x ln 冪x. 2
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1146
Chapter 16
Additional Topics in Differential Equations
Exploration Notice in Example 5 that the constants of integration were not introduced when finding u1 and u2. Show that for x u1 a1 2
and u2 ln 冪x a2
the general solution 1 y yh yp C1ex C2ex xex xex ln 冪x 2 yields the same result as the solution obtained in the example.
Variation of Parameters Solve the differential equation y y tan x. Solution Because the characteristic equation m2 1 0 has solutions m ± i, the homogeneous solution is yh C1 cos x C2 sin x. Replacing C1 and C2 by u1 and u2 produces yp u1 cos x u2 sin x. The resulting system of equations is u1 cos x u2 sin x 0 u1 sin x u2 cos x tan x. Multiplying the first equation by sin x and the second by cos x produces u1 sin x cos x u2 sin2 x 0 u1 sin x cos x u2 cos2 x sin x. Adding these two equations produces u2 sin x, which implies that sin2 x cos x cos2 x 1 cos x cos x sec x.
u1
Integration yields u1
冕
共cos x sec x兲 dx sin x ln ⱍsec x tan xⱍ
u2
冕
sin x dx cos x
and
so that
ⱍ
ⱍ
yp sin x cos x cos x ln sec x tan x sin x cos x cos x ln sec x tan x
ⱍ
ⱍ
and the general solution is y yh yp C1 cos x C2 sin x cos x ln sec x tan x .
ⱍ
ⱍ
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16.3
16.3 Exercises of the differential equation. Solution
Differential Equation
1. y 2共e2x cos x兲
y y 10e2x y y cos x
ⱍ
ⱍ
3. y 3 sin x cos x ln sec x tan x y y tan x 4. y 共5 ln sin x 兲cos x x sin x
ⱍ
ⱍ
1147
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Verifying a Solution In Exercises 1–4, verify the solution
1 2. y 共2 2x兲sin x
Second-Order Nonhomogeneous Linear Equations
y y csc x cot x
Finding a Particular Solution In Exercises 5–10, find a
WRITING ABOUT CONCEPTS 31. Choosing yp Using the method of undetermined coefficients, determine a suitable choice for yp for each differential equation. Explain your reasoning. (You do not need to solve the differential equations.) (a) y y 12y x2
(b) y y 12y e4x
32. Variation of Parameters Describe the steps for solving a differential equation by the method of variation of parameters.
particular solution of the differential equation. 5. y 7y 12y 3x 1
Electrical Circuits
6. y y 6y 4
In Exercises 33 and 34, use the electrical circuit differential equation
7. y 8y 16y e3x
冸冹
8. y y 3y e2x
冸 冹
冸冹
d 2q R dq 1 1 ⴙ ⴙ qⴝ E冇t冈 dt2 L dt LC L
9. y 2y 15y sin x 10. y 4y 5y ex cos x
where R is the resistance (in ohms), C is the capacitance (in farads), L is the inductance (in henrys), E冇t冈 is the electromotive force (in volts), and q is the charge on the capacitor (in coulombs). Find the charge q as a function of time for the electrical circuit described. Assume that q冇0冈 ⴝ 0 and q 冇0冈 ⴝ 0.
Method of Undetermined Coefficients In Exercises 11–18, solve the differential equation by the method of undetermined coefficients. 11. y 3y 2y 2x
12. y 2y 3y x 2 1
13. y 2y 2ex
14. y 9y 5e3x
15. y 10y 25y 5 6ex 16. 16y 8y y 4共x ex兲 17. y 9y sin 3x
33. R 20, C 0.02, L 2, E共t兲 12 sin 5t
18. y 3y 2y 2e2x
34. R 20, C 0.02, L 1, E共t兲 10 sin 5t
Method of Undetermined Coefficients In Exercises 19–24, solve the differential equation by the method of undetermined coefficients that satisfies the initial condition(s). 19. y y x3 y共0兲 1, y共0兲 0 21. y y 2 sin x y共0兲 0, y共0兲 3 23. y 4y xex xe4x y共0兲
1 3
20. y 4y 4 y共0兲 1, y共0兲 6 22. y y 2y 3 cos 2x y共0兲 1, y共0兲 2 24. y 2y sin x y
冢2 冣 52
Method of Variation of Parameters In Exercises 25–30, solve the differential equation by the method of variation of parameters. 25. y y sec x
26. y y sec x tan x
27. y 4y csc 2x
28. y 4y 4y x 2e2x
29. y 2y y ex ln x
30. y 4y 4y
e2x x
Vibrating Spring In Exercises 35–38, find the particular solution of the differential equation w w y 冇t冈 ⴙ by 冇t冈 ⴙ ky冇t冈 ⴝ F冇t冈 g g for the oscillating motion of an object on the end of a spring. Use a graphing utility to graph the solution. In the equation, y is the displacement from equilibrium (positive direction is downward), measured in feet, and t is time in seconds (see figure). The constant w is the weight of the object, g is the acceleration due to gravity, b is the magnitude of l = natural the resistance to the motion, k length is the spring constant from y = displacement Hooke’s Law, and F冇t冈 is the m acceleration imposed on the Spring displacement system. 35.
24 32 y
48y 24 32 共48 sin 4t兲
y共0兲 14, y共0兲 0
nsj-images/iStockphoto
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1148 36.
2 32 y
Chapter 16
Additional Topics in Differential Equations
True or False? In Exercises 43 and 44, determine whether
2 4y 32 共4 sin 8t兲
the statement is true or false. If it is false, explain why or give an example that shows it is false.
y共0兲 14, y共0兲 0 37.
2 32 y
2 y 4y 32 共4 sin 8t兲
43. yp e2x cos ex is a particular solution of the differential equation
y共0兲 14, y共0兲 3 38.
4 32 y
12y 25 2y 0
y 3y 2y cos ex.
y共0兲 12, y共0兲 4
44. yp 18e2x is a particular solution of the differential equation
39. Vibrating Spring Rewrite yh in the solution to Exercise 35 by using the identity a cos t b sin t 冪a2 b2 sin共 t 兲
PUTNAM EXAM CHALLENGE
where arctan a兾b.
40.
45. For all real x, the real-valued function y f 共x兲 satisfies y 2y y 2ex.
HOW DO YOU SEE IT? The figure shows the particular solution of the differential equation
that satisfies the initial conditions y共0兲 and y共0兲 4 for values of the resistance component b in the interval 关0, 1兴. (Note that when b 12, the problem is identical to that of Exercise 38.) According to the figure, is the motion damped or undamped when b 0? when b > 0? (You do not need to solve the differential equation.) 1 2
y
b=0 b=
Generated by Maple
1 2
b
41. Vibrating Spring Refer to the differential equation and the initial conditions given in Exercise 40. (a) When there is no resistance to the motion 共b 0兲, describe the motion. (b) For b > 0, what is the ultimate effect of the retarding force? (c) Is there a real number M such that there will be no oscillations of the spring for b > M ? Explain your answer. 42. Solving a Differential Equation Solve the differential equation given that y1 and y2 are solutions of the corresponding homogeneous equation. (a) x2y xy y 4x ln x y1 x, y2 x ln x (b) x2y xy 4y sin 共ln x兲 y1 sin 共ln x2兲, y2 cos 共ln x2兲
This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.
Parachute Jump The fall of a parachutist is described by the second-order linear differential equation dy w d 2y k w g dt 2 dt
b=1
t
(a) If f 共x兲 > 0 for all real x, must f 共x兲 > 0 for all real x? Explain. (b) If f 共x兲 > 0 for all real x, must f 共x兲 > 0 for all real x? Explain.
4 25 y by y0 32 2
Danshutter/Shutterstock.com
y 6y e2x.
where w is the weight of the parachutist, y is the height at time t, g is the acceleration due to gravity, and k is the drag factor of the parachute. (a) The parachute is opened at 2000 feet, so y共0兲 2000. At that time, the velocity is y共0兲 100 feet per second. For a 160-pound parachutist, using k 8, the differential equation is 5y 8y 160. Using the initial conditions, verify that the solution of the differential equation is y 1950 50e1.6t 20t. (b) Consider a 192-pound parachutist who has a parachute with a drag factor of k 9. Using the initial conditions given in part (a), write and solve a differential equation that describes the fall of the parachutist.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
16.4
Series Solutions of Differential Equations
1149
16.4 Series Solutions of Differential Equations Use a power series to solve a differential equation. Use a Taylor series to find the series solution of a differential equation.
Power Series Solution of a Differential Equation Power series can be used to solve certain types of differential equations. This section begins with the general power series solution method. Recall from Chapter 9 that a power series represents a function f on an interval of convergence, and that you can successively differentiate the power series to obtain a series for f ⬘, f ⬙, and so on. These properties are used in the power series solution method demonstrated in the first two examples.
Power Series Solution Use a power series to solve the differential equation y⬘ ⫺ 2y ⫽ 0. ⬁
兺a x
Assume that y ⫽
Solution
n
n
is a solution. Then,
n⫽0
y⬘ ⫽
⬁
兺 na x n
n⫺1.
n⫽1
Substituting for y⬘ and ⫺2y, you obtain the following series form of the differential equation. (Note that, from the third step to the fourth, the index of summation is changed to ensure that xn occurs in both sums.) y⬘ ⫺ 2y ⫽ 0 ⬁
兺 na x
n⫺1
n
⫺2
n⫽1
⬁
兺ax n
n
⫽0
n⫽0
⬁
兺 na x n
n⫺1
n⫽1
⬁
兺 共n ⫹ 1兲a
n
n
n⫽0
n n⫹1x
n⫽0
⬁
兺 2a x
⫽
⬁
⫽
兺 2a x n
n
n⫽0
Now, by equating coefficients of like terms, you obtain the recursion formula
共n ⫹ 1兲an⫹1 ⫽ 2an which implies that an⫹1 ⫽
Exploration In Example 1, the differential equation could be solved easily without using a series. Determine which method should be used to solve the differential equation y⬘ ⫺ 2y ⫽ 0 and show that the result is the same as that obtained in the example.
2an , n ≥ 0. n⫹1
This formula generates the following results. a0
a1
a0
2a0
a2 22a0 2
a3 23a0 3!
a4 24a0 4!
a5 25a0 5!
. . . . . .
Using these values as the coefficients for the solution series, you have ⬁
2na0 n x n⫽0 n! ⬁ 共2x兲n ⫽ a0 n⫽0 n! ⫽ a0e 2x.
y⫽
兺
兺
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1150
Chapter 16
Additional Topics in Differential Equations
In Example 1, the differential equation could be solved easily without using a series. The differential equation in Example 2 cannot be solved by any of the methods discussed in previous sections.
Power Series Solution Use a power series to solve the differential equation y⬙ ⫹ xy⬘ ⫹ y ⫽ 0. ⬁
兺a x
Assume that y ⫽
Solution
n
n
is a solution. Then you have
n⫽0
y⬘ ⫽
⬁
兺 na x n
⬁
兺 na x ,
xy⬘ ⫽
n⫺1,
n
n⫽1
⬁
兺 n共n ⫺ 1兲a x
y⬙ ⫽
n
n⫽1
n
n⫺2.
n⫽2
Substituting for y⬙, xy⬘, and y in the given differential equation, you obtain the following series. ⬁
兺 n共n ⫺ 1兲a x n
n⫺2
⫹
n⫽2
⬁
兺 na x n
⬁
⫹
n
n⫽0
兺ax n
n
⫽0
n⫽0
⬁
兺 n共n ⫺ 1兲a x n
n⫺2
⬁
⫽⫺
n⫽2
兺 共n ⫹ 1兲a x n
n
n⫽0
To obtain equal powers of x, adjust the summation indices by replacing n by n ⫹ 2 in the left-hand sum, to obtain ⬁
兺 共n ⫹ 2兲共n ⫹ 1兲a
n n⫹2 x
⫽⫺
n⫽0
⬁
兺 共n ⫹ 1兲a x . n
n
n⫽0
By equating coefficients, you have
共n ⫹ 2兲共n ⫹ 1兲an⫹2 ⫽ ⫺ 共n ⫹ 1兲an from which you obtain the recursion formula an⫹2 ⫽ ⫺
a 共n ⫹ 1兲 an ⫽ ⫺ n , n ⱖ 0, 共n ⫹ 2兲共n ⫹ 1兲 n⫹2
and the coefficients of the solution series are as follows. a0 2 a2 a a4 ⫽ ⫺ ⫽ 0 4 2⭈4 a a0 a6 ⫽ ⫺ 4 ⫽ ⫺ 6 2⭈4⭈6
a1 3 a3 a a5 ⫽ ⫺ ⫽ 1 5 3⭈5 a a1 a7 ⫽ ⫺ 5 ⫽ ⫺ 7 3⭈5⭈7
a2 ⫽ ⫺
a3 ⫽ ⫺
⯗
⯗
共⫺1兲k a0 共⫺1兲ka0 a2k ⫽ ⫽ 2 ⭈ 4 ⭈ 6 . . . 共2k兲 2k共k!兲
a2k⫹1 ⫽
共⫺1兲k a1 3 ⭈ 5 ⭈ 7 . . . 共2k ⫹ 1兲
So, you can represent the general solution as the sum of two series—one for the even-powered terms with coefficients in terms of a0 , and one for the odd-powered terms with coefficients in terms of a1.
冢
冣
冢
x4 x5 x2 x3 ⫹ ⫺ . . . ⫹ a1 x ⫺ ⫹ ⫺. . . 2 2⭈4 3 3⭈5 ⬁ 共⫺1兲kx 2k ⬁ 共⫺1兲kx 2k⫹1 ⫽ a0 ⫹ a 1 k . . . 共2k ⫹ 1兲 k⫽0 2 共k!兲 k⫽0 3 ⭈ 5 ⭈ 7
y ⫽ a0 1 ⫺
兺
冣
兺
The solution has two arbitrary constants, a0 and a1, as you would expect in the general solution of a second-order differential equation.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
16.4
Series Solutions of Differential Equations
1151
Approximation by Taylor Series A second type of series solution method involves a differential equation with initial conditions and makes use of Taylor series, as given in Section 9.10.
Approximation by Taylor Series See LarsonCalculus.com for an interactive version of this type of example.
Use a Taylor series to find the first six terms of the series solution of y⬘ ⫽ y 2 ⫺ x for the initial condition y ⫽ 1 when x ⫽ 0. Then, use this polynomial to approximate values of y for 0 ⱕ x ⱕ 1. Solution
Recall from Section 9.10 that, for c ⫽ 0,
y ⫽ y共0兲 ⫹ y⬘共0兲x ⫹
y⬙ 共0兲 2 y⬘⬘⬘共0兲 3 . . . x ⫹ x ⫹ . 2! 3!
Because y共0兲 ⫽ 1 and y⬘ ⫽ y 2 ⫺ x, you obtain the following. y共0兲 ⫽ 1 y⬘ ⫽
y2
y⬘共0兲 ⫽ 1
⫺x
y⬙ 共0兲 ⫽ 2 ⫺ 1 ⫽ 1
y⬙ ⫽ 2yy⬘ ⫺ 1 y⬘⬘⬘ ⫽ 2yy⬙ ⫹ 2共 y⬘兲
y⬘⬘⬘共0兲 ⫽ 2 ⫹ 2 ⫽ 4
y共4兲
y共4兲共0兲 ⫽ 8 ⫹ 6 ⫽ 14
2
⫽ 2yy⬘⬘⬘ ⫹ 6y⬘y⬙
y共5兲 ⫽ 2yy共4兲 ⫹ 8y⬘y⬘⬘⬘ ⫹ 6共 y⬙ 兲2
y共5兲共0兲 ⫽ 28 ⫹ 32 ⫹ 6 ⫽ 66
So, y can be approximated by the first six terms of the series solution shown below. y⬙ 共0兲 2 y⬘⬘⬘共0兲 3 y共4兲共0兲 4 y共5兲共0兲 5 x ⫹ x ⫹ x ⫹ x 2! 3! 4! 5! 1 4 14 4 66 5 ⫽ 1 ⫹ x ⫹ x 2 ⫹ x3 ⫹ x ⫹ x 2 3! 4! 5!
y ⬇ y共0兲 ⫹ y⬘共0兲x ⫹
Using this polynomial, you can approximate values for y in the interval 0 ⱕ x ⱕ 1, as shown in the table below. x
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
y
1.0000
1.1057
1.2264
1.3691
1.5432
1.7620
2.0424
2.4062
2.8805
3.4985
4.3000
In addition to approximating values of a function, you can also use a series solution to sketch a graph. In Figure 16.8, the series solutions of y⬘ ⫽ y2 ⫺ x using the first two, four, and six terms are shown, along with an approximation found using a computer algebra system. The approximations are nearly the same for values of x close to 0. As x approaches 1, however, there is a noticeable difference among the approximations. For a series solution that is more accurate near x ⫽ 1, repeat Example 3 using c ⫽ 1.
y
8
6
6 terms
4
4 terms
2
2 terms x 0.2
0.4
0.6
0.8
1.0
Figure 16.8
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1152
Chapter 16
Additional Topics in Differential Equations
16.4 Exercises
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Equivalent Solution Techniques In Exercises 1–6, verify that the power series solution of the differential equation is equivalent to the solution found using previously learned solution techniques. 1. y⬘ ⫺ y ⫽ 0
2. y⬘ ⫺ ky ⫽ 0
3. y⬙ ⫺ 9y ⫽ 0
4. y⬙ ⫺ k 2y ⫽ 0
5. y⬙ ⫹ 4y ⫽ 0
6. y⬙ ⫹ k 2y ⫽ 0
HOW DO YOU SEE IT? Consider the differential
18.
equation y⬙ ⫹ 9y ⫽ 0 with initial conditions y共0兲 ⫽ 2 and y⬘共0兲 ⫽ 6. The figure shows the graph of the solution of the differential equation and the third-degree and fifth-degree polynomial approximations of the solution. Identify each.
Power Series Solution In Exercises 7–10, use power series to solve the differential equation and find the interval of convergence of the series. 7. y⬘ ⫹ 3xy ⫽ 0
y 3
8. y⬘ ⫺ 2xy ⫽ 0
9. y⬙ ⫺ xy⬘ ⫽ 0
2
10. y⬙ ⫺ xy⬘ ⫺ y ⫽ 0
1 x 1 2
Finding Terms of a Power Series Solution In Exercises 11 and 12, find the first three terms of each of the power series representing independent solutions of the differential equation. 11. 共x 2 ⫹ 4兲y⬙ ⫹ y ⫽ 0
12. y⬙ ⫹ x 2y ⫽ 0
Approximation by Taylor Series In Exercises 13 and 14, use a Taylor series to find the first n terms of the series solution of the differential equation under the specified initial conditions. Use this polynomial to approximate y for the given value of x and compare the result with the approximation given by Euler’s Method for h ⴝ 0.1. 13. y⬘ ⫹ 共2x ⫺ 1兲y ⫽ 0, y共0兲 ⫽ 2, n ⫽ 5, x ⫽ 12 14. y⬘ ⫺ 2xy ⫽ 0, y共0兲 ⫽ 1, n ⫽ 4, x ⫽ 1
WRITING ABOUT CONCEPTS
Approximation by Taylor Series In Exercises 19–22, use a Taylor series to find the first n terms of the series solution of the differential equation under the specified initial conditions. Use this polynomial to approximate y for the given value of x. 19. y⬙ ⫺ 2xy ⫽ 0, y共0兲 ⫽ 1, y⬘ 共0兲 ⫽ ⫺3, n ⫽ 6, x ⫽ 14 20. y⬙ ⫺ 2xy⬘ ⫹ y ⫽ 0, y共0兲 ⫽ 1, y⬘ 共0兲 ⫽ 2, n ⫽ 8, x ⫽ 12 1 21. y⬙ ⫹ x2y⬘ ⫺ 共cos x兲y ⫽ 0, y共0兲 ⫽ 3, y⬘ 共0兲 ⫽ 2, n ⫽ 4, x ⫽ 3 1 22. y⬙ ⫹ e xy⬘ ⫺ 共sin x兲y ⫽ 0, y共0兲 ⫽ ⫺2, y⬘ 共0兲 ⫽ 1, n ⫽ 4, x ⫽ 5
15. Power Series Solution Method Describe how to use power series to solve a differential equation.
Verifying that a Series Converges In Exercises 23–26, verify that the series converges to the given function on the indicated interval. (Hint: Use the given differential equation.)
16. Recursion Formula Give an example.
23.
What is a recursion formula?
⬁
xn
兺 n! ⫽ e , 共⫺ ⬁, ⬁兲 x
n⫽0
Differential equation: y⬘ ⫺ y ⫽ 0 17. Investigation
Consider the differential equation
24.
y⬙ ⫺ xy⬘ ⫽ 0
Differential equation: y⬙ ⫹ y ⫽ 0
with the initial conditions y共0兲 ⫽ 0 and y⬘共0兲 ⫽ 2.
25.
共⫺1兲nx 2n⫹1 ⫽ arctan x, 共⫺1, 1兲 2n ⫹ 1 n⫽0 ⬁
兺
Differential equation: 共x 2 ⫹ 1兲y⬙ ⫹ 2xy⬘ ⫽ 0
(See Exercise 9.) (a) Find the series solution satisfying the initial conditions.
共⫺1兲nx2n ⫽ cos x, 共⫺ ⬁, ⬁兲 共2n兲! n⫽0 ⬁
兺
26.
⬁
n⫽0
(b) Use a graphing utility to graph the third-degree and fifthdegree series approximations of the solution. Identify the approximations. (c) Identify the symmetry of the solution.
共2n兲!x2n⫹1
兺 共2 n!兲 共2n ⫹ 1兲 ⫽ arcsin x, 共⫺1, 1兲 n
2
Differential equation: 共1 ⫺ x 2兲y⬙ ⫺ xy⬘ ⫽ 0 27. Airy’s Equation Find the first six terms in the series solution of Airy’s equation, y⬙ ⫺ xy ⫽ 0.
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Review Exercises
Review Exercises
1153
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
Testing for Exactness In Exercises 1 and 2, determine
Finding an Integrating Factor In Exercises 13–16, find
whether the differential equation is exact. Explain your reasoning.
the integrating factor that is a function of x or y alone and use it to find the general solution of the differential equation.
1. 共 y ⫹ x3 ⫹ xy2兲 dx ⫺ x dy ⫽ 0 2. 共5x ⫺ y兲 dx ⫹ 共5y ⫺ x兲 dy ⫽ 0
Solving an Exact Differential Equation In Exercises 3–8, determine whether the differential equation is exact. If it is, find the general solution.
13. 共3x2 ⫺ y2兲 dx ⫹ 2xy dy ⫽ 0 14. 2xy dx ⫹ 共 y2 ⫺ x2兲 dy ⫽ 0 15. dx ⫹ 共3x ⫺ e⫺2y兲 dy ⫽ 0 16. cos y dx ⫺ 关2共x ⫺ y兲 sin y ⫹ cos y兴 dy ⫽ 0
3. 共10x ⫹ 8y ⫹ 2兲 dx ⫹ 共8x ⫹ 5y ⫹ 2兲 dy ⫽ 0
Verifying a Solution In Exercises 17 and 18, verify the
4. 共2x ⫺ 2y3 ⫹ y兲 dx ⫹ 共x ⫺ 6xy2兲 dy ⫽ 0
solution of the differential equation. Then use a graphing utility to graph the particular solutions for several different values of C1 and C2. What do you observe?
5. 共x ⫺ y ⫺ 5兲 dx ⫺ 共x ⫹ 3y ⫺ 2兲 dy ⫽ 0 6. 共3x2 ⫺ 5xy2兲 dx ⫹ 共2y3 ⫺ 5xy2兲 dy ⫽ 0 x x 7. dx ⫺ 2 dy ⫽ 0 y y 8. y sin共xy兲 dx ⫹ 关x sin共xy兲 ⫹ y兴 dy ⫽ 0
Graphical and Analytic Analysis In Exercises 9 and 10, (a) sketch an approximate solution of the differential equation satisfying the initial condition on the slope field, (b) find the particular solution that satisfies the initial condition, and (c) use a graphing utility to graph the particular solution. Compare the graph with the sketch in part (a). 9. 共2x ⫺ y兲 dx ⫹ 共2y ⫺ x兲 dy ⫽ 0, y共2兲 ⫽ 2
Differential Equation ⫹ C2
y⬙ ⫺ 4y ⫽ 0
e⫺2x
18. y ⫽ C1 cos 2x ⫹ C2 sin 2x
y⬙ ⫹ 4y ⫽ 0
Finding a Particular Solution: Initial Conditions In Exercises 19–22, find the particular solution of the differential equation that satisfies the initial conditions. Use a graphing utility to graph the solution. Differential Equation
Initial Conditions y共0兲 ⫽ 0, y⬘ 共0兲 ⫽ 3
20. y⬙ ⫹ 4y⬘ ⫹ 5y ⫽ 0
y共0兲 ⫽ 2, y⬘ 共0兲 ⫽ ⫺7
4
21. y⬙ ⫹ 2y⬘ ⫺ 3y ⫽ 0
y共0兲 ⫽ 2, y⬘ 共0兲 ⫽ 0
2
22. y⬙ ⫹ 12y⬘ ⫹ 36y ⫽ 0
y共0兲 ⫽ 2, y⬘ 共0兲 ⫽ 1
x
−2
2
4
Finding a Particular Solution: Boundary Conditions In Exercises 23 and 24, find the particular solution of the differential equation that satisfies the boundary conditions. Use a graphing utility to graph the solution.
−2 −4
10. 共6xy ⫺ y3兲 dx ⫹ 共4y ⫹ 3x2 ⫺ 3xy2兲 dy ⫽ 0, y共0兲 ⫽ 1
Differential Equation
Boundary Conditions
23. y⬙ ⫹ 2y⬘ ⫹ 5y ⫽ 0
y共1兲 ⫽ 4, y共2兲 ⫽ 0
4
24. y⬙ ⫹ y ⫽ 0
y共0兲 ⫽ 2, y共兾2兲 ⫽ 1
2
25. Think About It
y
−4
17. y ⫽ C1
e2x
19. y⬙ ⫺ y⬘ ⫺ 2y ⫽ 0
y
−4
Solution
x
−2
2
4
−2 −4
Finding a Particular Solution In Exercises 11 and 12, find the particular solution that satisfies the initial condition. 11. 共2x ⫹ y ⫺ 3兲 dx ⫹ 共x ⫺ 3y ⫹ 1兲 dy ⫽ 0, y共2兲 ⫽ 0 12. 3x2y2 dx ⫹ 共2x3y ⫺ 3y2兲 dy ⫽ 0, y共1兲 ⫽ 2
Is the differential equation
y⬙ ⫺ y⬘ ⫺ 5y ⫽ sin x homogeneous? Why or why not? 26. Solutions of a Differential Equation Find all values of k for which the differential equation y⬙ ⫹ 2ky⬘ ⫹ ky ⫽ 0 has a general solution of the indicated form. (a) y ⫽ C1em3 x ⫹ C2em2x (b) y ⫽ C1em1x ⫹ C2xem1x (c) y ⫽ C1e␣x cos x ⫹ C2e␣xx
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1154
Chapter 16
Additional Topics in Differential Equations
Finding a General Solution In Exercises 27–32, find the general solution of the second-order differential equation.
yp ⫽
27. y⬙ ⫹ y ⫽ x3 ⫹ x 28. y⬙ ⫹ 2y ⫽ e2x ⫹ x
The function
1 cos x 4
is a particular solution of the differential equation
29. y⬙ ⫹ y ⫽ 2 cos x
y⬙ ⫹ 4y⬘ ⫹ 5y ⫽ sin x ⫹ cos x.
30. y⬙ ⫹ 5y⬘ ⫹ 4y ⫽ x2 ⫹ sin 2x
43. Think About It
31. y⬙ ⫺ 2y⬘ ⫹ y ⫽ 2xe x 32. y⬙ ⫹ 2y⬘ ⫹ y ⫽
42. True or False?
(a) Explain how, by observation, you know that a form of a particular solution of the differential equation
1 x2e x
Finding a Particular Solution In Exercises 33–38, find the particular solution of the differential equation that satisfies the initial conditions. Differential Equation
Initial Conditions
33. y⬙ ⫺ y⬘ ⫺ 6y ⫽ 54
y共0兲 ⫽ 2, y⬘ 共0兲 ⫽ 0
34. y⬙ ⫹ 25y ⫽ e x
y共0兲 ⫽ 0, y⬘ 共0兲 ⫽ 0
35. y⬙ ⫹ 4y ⫽ cos x
y共0兲 ⫽ 6, y⬘ 共0兲 ⫽ ⫺6
36. y⬙ ⫹ 3y⬘ ⫽ 6x
y共0兲 ⫽ 2, y⬘ 共0兲 ⫽ 10 3
37. y⬙ ⫺ y⬘ ⫺ 2y ⫽ 1 ⫹ xe⫺x
y共0兲 ⫽ 1, y⬘ 共0兲 ⫽ 3
38. y⬘⬘⬘ ⫺ y⬙ ⫽ 4x2
y共0兲 ⫽ 1, y⬘ 共0兲 ⫽ 1, y⬙ 共0兲 ⫽ 1
Vibrating Spring In Exercises 39 and 40, describe the motion of a 64-pound weight suspended on a spring. Assume that the weight stretches the spring 43 feet from its natural position. 1 2
39. The weight is pulled foot below the equilibrium position and released.
y⬙ ⫹ 3y ⫽ 12 sin x is yp ⫽ A sin x. (b) Use your explanation in part (a) to find a particular solution of the differential equation y⬙ ⫹ 5y ⫽ 10 cos x. (c) Compare the algebra required to find particular solutions in parts (a) and (b) with that required when the form of the particular solution is yp ⫽ A cos x ⫹ B sin x. 44. Think About It Explain how you can find a particular solution of the differential equation y⬙ ⫹ 4y⬘ ⫹ 6y ⫽ 30 by observation.
40. The weight is pulled 12 foot below the equilibrium position and released. The motion takes place in a medium that furnishes a damping force of magnitude 18 speed at all times.
Power Series Solution In Exercises 45 and 46, find the series solution of the differential equation.
41. Investigation
46. y⬙ ⫹ 3xy⬘ ⫺ 3y ⫽ 0
The differential equation
8 8 y⬙ ⫹ by⬘ ⫹ ky ⫽ F共t兲, 32 32
1 y共0兲 ⫽ , y⬘ 共0兲 ⫽ 0 2
models the motion of a weight suspended on a spring. (a) Solve the differential equation and use a graphing utility to graph the solution for each of the assigned quantities for b, k, and F共t兲. (i)
b ⫽ 0, k ⫽ 1, F共t兲 ⫽ 24 sin t
(ii) b ⫽ 0, k ⫽ 2, F共t兲 ⫽ 24 sin共2冪2 t兲
45. 共x ⫺ 4兲y⬘ ⫹ y ⫽ 0
Approximation by Taylor Series In Exercises 47 and 48, use a Taylor series to find the first n terms of the series solution of the differential equation under the specified initial conditions. Use this polynomial to approximate y for the given value of x. 47. y⬙ ⫹ y⬘ ⫺ e xy ⫽ 0, y共0兲 ⫽ 2, y⬘ 共0兲 ⫽ 0, n ⫽ 4, x ⫽ 48. y⬙ ⫹ xy ⫽ 0, y共0兲 ⫽ 1, y⬘共0兲 ⫽ 1, n ⫽ 6, x ⫽
1 4
1 2
(iii) b ⫽ 0.1, k ⫽ 2, F共t兲 ⫽ 0 (iv) b ⫽ 1, k ⫽ 2, F共t兲 ⫽ 0 (b) Describe the effect of increasing the resistance to motion b. (c) Explain how the motion of the object changes when a stiffer spring 共greater value of k兲 is used. (d) Matching the input and natural frequencies of a system is known as resonance. In which case of part (a) does this occur, and what is the result?
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
P.S. Problem Solving
See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.
P.S. Problem Solving 1. Finding a General Solution Find the value of k that makes the differential equation
1155
9. Pendulum Consider a pendulum of length L that swings by the force of gravity only.
共3x2 ⫹ kxy2兲 dx ⫺ 共5x2y ⫹ ky2兲 dy ⫽ 0 exact. Using this value of k, find the general solution. 2. Using an Integrating Factor The differential equation 共kx2 ⫹ y2兲 dx ⫺ kxy dy ⫽ 0 is not exact, but the integrating factor 1兾x2 makes it exact.
θ
L
(a) Use this information to find the value of k. (b) Using this value of k, find the general solution. 3. Finding a General Solution Find the general solution of the differential equation y⬙ ⫺ a2y ⫽ 0, a > 0. Show that the general solution can be written in the form y ⫽ C1 cosh ax ⫹ C2 sinh ax.
For small values of ⫽ 共t兲, the motion of the pendulum can be approximated by the differential equation d 2 g ⫹ ⫽0 dt 2 L where g is the acceleration due to gravity.
4. Finding a General Solution Find the general solution of the differential equation y⬙ ⫹  2 y ⫽ 0. Show that the general solution can be written in the form
(a) Find the general solution of the differential equation and show that it can be written in the form
冤冪
共t兲 ⫽ A cos
y ⫽ C sin共x ⫹ 兲, 0 ⱕ < 2. 5. Distinct Real Zeros Given that the characteristic equation of the differential equation y⬙ ⫹ ay⬘ ⫹ by ⫽ 0 has two distinct real zeros, m1 ⫽ r ⫹ s and m2 ⫽ r ⫺ s, where r and s are real, show that the general solution of the differential equation can be written in the form y ⫽ erx共C1 cosh sx ⫹ C2 sinh sx兲.
冥
g 共t ⫹ 兲 . L
(b) Find the particular solution for a pendulum of length 0.25 meter when the initial conditions are 共0兲 ⫽ 0.1 radian and ⬘ 共0兲 ⫽ 0.5 radian per second. (Use g ⫽ 9.8 meters per second per second.) (c) Determine the period of the pendulum. (d) Determine the maximum value of .
6. Limit of a Solution Given that a and b are positive and that y共x兲 is a solution of the differential equation
(e) How much time from t ⫽ 0 does it take for to be 0 the first time? the second time? (f) What is the angular velocity ⬘ when ⫽ 0 the first time? the second time?
y⬙ ⫹ ay⬘ ⫹ by ⫽ 0 show that lim y共x兲 ⫽ 0. x→ ⬁
7. Trivial and Nontrivial Solutions Consider the differential equation y⬙ ⫹ ay ⫽ 0 with boundary conditions y共0兲 ⫽ 0 and y共L兲 ⫽ 0 for some nonzero real number L.
10. Deflection of a Beam A horizontal beam with a length of 2 meters rests on supports located at the ends of the beam. 2 meters
(a) For a ⫽ 0, show that the differential equation has only the trivial solution y ⫽ 0. (b) For a < 0, show that the differential equation has only the trivial solution y ⫽ 0. (c) For a > 0, find the value(s) of a for which the solution is nontrivial. Then find the corresponding solution(s). 8. Euler’s Differential Equation equation is of the form
Euler’s differential
x2 y⬙ ⫹ axy⬘ ⫹ by ⫽ 0, x > 0 when a and b are constants. (a) Show that this equation can be transformed into a second-order linear equation with constant coefficients by using the substitution x ⫽ et. (b) Solve
x2
y⬙ ⫹ 6xy⬘ ⫹ 6y ⫽ 0.
The beam is supporting a load of W kilograms per meter. The resulting deflection y of the beam at a horizontal distance of x meters from the left end can be modeled by A
d 2y 1 ⫽ 2Wx ⫺ Wx 2 dx 2 2
where A is a positive constant. (a) Solve the differential equation to find the deflection y as a function of the horizontal distance x. (b) Use a graphing utility to determine the location and value of the maximum deflection.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1156
Chapter 16
Additional Topics in Differential Equations
Damped Motion In Exercises 11–14, consider a damped mass-spring system whose motion is described by the differential equation
17. Bessel’s Equation: Order Zero The differential equation x2y⬙ ⫹ xy⬘ ⫹ x 2 y ⫽ 0 is known as Bessel’s equation of order zero. (a) Use a power series of the form
d 2y dy ⴙ 2 ⴙ 2y ⴝ 0. dt 2 dt
y⫽
The zeros of its characteristic equation are
m2 ⴝ ⴚ ⴚ 冪2 ⴚ 2.
18. Bessel’s Equation: Order One The differential equation
For 2 ⴚ 2 > 0, the system is overdamped; for 2 ⴚ 2 ⴝ 0, it is critically damped; and for 2 ⴚ 2 < 0, it is underdamped. (a) Determine whether the differential equation represents an overdamped, critically damped, or underdamped system. (b) Find the particular solution corresponding to the given initial conditions. (c) Use a graphing utility to graph the particular solution found in part (b). Explain how the graph illustrates the type of damping in the system. 12.
y共0兲 ⫽ 1, y⬘ 共0兲 ⫽ 1 d 2y dy 13. ⫹ 20 ⫹ 64y ⫽ 0 dt 2 dt
d 2y dy ⫹ 2 ⫹ 26y ⫽ 0 dt 2 dt y共0兲 ⫽ 1, y⬘ 共0兲 ⫽ 4
d 2y dy 14. ⫹2 ⫹y⫽0 dt 2 dt
y共0兲 ⫽ 2, y⬘ 共0兲 ⫽ ⫺20
y共0兲 ⫽ 2, y⬘ 共0兲 ⫽ ⫺1
15. Airy’s Equation Consider Airy’s equation given in Section 16.4, Exercise 27. Rewrite the equation as y⬙ ⫺ 共x ⫺ 1兲y ⫺ y ⫽ 0. Then use a power series of the form y⫽
⬁
兺 a 共x ⫺ 1兲 n
n
(b) Compare your result with that of the function J0共x兲 given in Section 9.8, Exercise 65.
and
d 2y dy ⫹ 8 ⫹ 16y ⫽ 0 dt 2 dt
n
n⫽0
to find the solution.
m1 ⴝ ⴚ ⴙ 冪2 ⴚ 2
11.
⬁
兺a x
n
n⫽0
to find the first eight terms of the solution. Compare your result with that of Exercise 27 in Section 16.4. 16. Chebyshev’s Equation Consider Chebyshev’s equation
x 2 y⬙ ⫹ xy⬘ ⫹ 共x2 ⫺ 1兲y ⫽ 0 is known as Bessel’s equation of order one. (a) Use a power series of the form y⫽
⬁
兺a x n
n
n⫽0
to find the solution. (b) Compare your result with that of the function J1共x兲 given in Section 9.8, Exercise 66. 19. Hermite’s Equation
Consider Hermite’s equation
y⬙ ⫺ 2xy⬘ ⫹ 2ky ⫽ 0. (a) Use a power series of the form y⫽
⬁
兺ax n
n
n⫽0
to find the solution when k ⫽ 4. [Hint: Choose the arbitrary constants such that the leading term is 共2x兲k.] (b) Polynomial solutions of Hermite’s equation are called Hermite polynomials and are denoted by Hk(x). The general form for Hk共x兲 can be written as Hk共x兲 ⫽
共⫺1兲nk!共2x兲k⫺2n n⫽0 n!共k ⫺ 2n兲! P
兺
where P is the greatest integer less than or equal to k兾2. Use this formula to determine the Hermite polynomials H0共x兲, H1共x兲, H2共x兲, H3共x兲, and H4共x兲. 20. Laguerre’s Equation
Consider Laguerre’s equation
共1 ⫺ x2兲y⬙ ⫺ xy⬘ ⫹ k2 y ⫽ 0.
xy⬙ ⫹ 共1 ⫺ x兲y⬘ ⫹ ky ⫽ 0.
Polynomial solutions of this differential equation are called Chebyshev polynomials and are denoted by Tk共x兲. They satisfy the recursion equation
(a) Polynomial solutions of Laguerre’s equation are called Laguerre polynomials and are denoted by Lk共x兲. Use a power series of the form
Tn⫹1共x兲 ⫽ 2xTn共x兲 ⫺ Tn⫺1共x).
y⫽
⬁
兺
an x n
(a) Given that T0共x兲 ⫽ 1 and T1共x兲 ⫽ x, determine the Chebyshev polynomials T2共x兲, T3共x兲, and T4共x兲.
to show that
(b) Verify that T0共x兲, T1共x兲, T2共x兲, T3共x兲, and T4共x兲 are solutions of the given differential equation.
Lk共x兲 ⫽
(c) Show that T5共x兲 ⫽ 16x5 ⫺ 20x3 ⫹ 5x, T6共x兲 ⫽ 32x6 ⫺ 48x4 ⫹ 18x2 ⫺ 1, and T7共x兲 ⫽ 64x7 ⫺ 112x5 ⫹ 56x3 ⫺ 7x.
n⫽0
k
共⫺1兲nk!xn
兺 共k ⫺ n兲!共n!兲 .
n⫽0
2
Assume that a0 ⫽ 1. (b) Determine the Laguerre polynomials L0共x兲, L1共x兲, L2共x兲, L3共x兲, and L4共x兲.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Appendices Appendix A Appendix B Appendix C
Appendix D Appendix E Appendix F
Proofs of Selected Theorems A2 Integration Tables A3 Precalculus Review (Online) C.1 Real Numbers and the Real Number Line C.2 The Cartesian Plane C.3 Review of Trigonometric Functions Rotation and the General Second-Degree Equation (Online) Complex Numbers (Online) Business and Economic Applications (Online)
A1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A
Proofs of Selected Theorems
For this edition, we have made Appendix A, Proofs of Selected Theorems, available in video format at LarsonCalculus.com. When you navigate to that website, you will find a link to Bruce Edwards explaining each proof in the text, including those in this appendix. We hope these videos enhance your study of calculus. The text version of this appendix is available at CengageBrain.com.
Proofs of Selected Theorems sample at LarsonCalculus.com
Bruce Edwards’s Proof of the Power Rule at LarsonCalculus.com
A2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
B
Integration Tables
Forms Involving un 1.
冕
un du ⫽
un⫹1 ⫹ C, n ⫽ ⫺1 n⫹1
2.
冕
1 du ⫽ ln u ⫹ C u
4.
冕
u 1 a du ⫽ 2 ⫹ lnⱍa ⫹ buⱍ ⫹ C 共a ⫹ bu兲2 b a ⫹ bu
ⱍⱍ
Forms Involving a ⴙ bu 3. 5. 6. 7. 8. 9. 10. 12.
冕 冕 冕 冕 冕 冕 冕 冕
u 1 du ⫽ 2 共bu ⫺ a ln a ⫹ bu 兲 ⫹ C a ⫹ bu b
ⱍ
ⱍ
冢
冣
u 1 ⫺1 a du ⫽ 2 ⫹ ⫹ C, n ⫽ 1, 2 共a ⫹ bu兲n b 共n ⫺ 2兲共a ⫹ bu兲n⫺2 共n ⫺ 1兲共a ⫹ bu兲n⫺1
冤
冥
冤
ⱍ冥 ⫹ C
u2 1 bu du ⫽ 3 ⫺ 共2a ⫺ bu兲 ⫹ a2 ln a ⫹ bu a ⫹ bu b 2
ⱍ
冢
ⱍ冣 ⫹ C
u2 1 a2 ⫺ 2a ln a ⫹ bu 2 du ⫽ 3 bu ⫺ 共a ⫹ bu兲 b a ⫹ bu
ⱍ
冤
ⱍ冥 ⫹ C
u2 1 2a a2 du ⫽ 3 ⫺ ⫹ ln a ⫹ bu 3 共a ⫹ bu兲 b a ⫹ bu 2共a ⫹ bu兲2
ⱍ
1 ⫺1 2a a2 u2 du ⫽ 3 ⫹ ⫺ ⫹ C, n ⫽ 1, 2, 3 n n⫺3 n⫺2 共a ⫹ bu兲 b 共n ⫺ 3兲共a ⫹ bu兲 共n ⫺ 2兲共a ⫹ bu兲 共n ⫺ 1兲共a ⫹ bu兲n⫺1
冤
ⱍ ⱍ
1 1 u du ⫽ ln ⫹C u共a ⫹ bu兲 a a ⫹ bu
冢
11.
ⱍ ⱍ冣
1 1 1 b u du ⫽ ⫺ ⫹ ln u 2共a ⫹ bu兲 a u a a ⫹ bu
⫹C
13.
冕 冕
冥
冢
ⱍ ⱍ冣 ⱍ ⱍ冥
1 1 1 1 u du ⫽ ⫹ ln u共a ⫹ bu兲2 a a ⫹ bu a a ⫹ bu
冤
⫹C
1 1 a ⫹ 2bu 2b u du ⫽ ⫺ 2 ⫹ ln u 2共a ⫹ bu兲2 a u共a ⫹ bu兲 a a ⫹ bu
⫹C
Forms Involving a ⴙ bu ⴙ cu2, b2 ⴝ 4ac
14.
15.
冕 冕
1 du ⫽ a ⫹ bu ⫹ cu2
冦
2 冪4ac ⫺ b2
arctan
ⱍ
2cu ⫹ b 冪4ac ⫺ b2
⫹ C,
b2 < 4ac
ⱍ
1 2cu ⫹ b ⫺ 冪b2 ⫺ 4ac ln ⫹ C, b2 > 4ac 冪b ⫺ 4ac 2cu ⫹ b ⫹ 冪b2 ⫺ 4ac 2
冢 ⱍ
u 1 ln a ⫹ bu ⫹ cu 2 ⫺ b 2 du ⫽ a ⫹ bu ⫹ cu 2c
ⱍ
冕
1 du a ⫹ bu ⫹ cu 2
冣
Forms Involving 冪a ⴙ bu 16.
冕
17.
冕
18.
冕
un冪a ⫹ bu du ⫽
冤
2 un共a ⫹ bu兲3兾2 ⫺ na b共2n ⫹ 3兲
冦
ⱍ
ⱍ
冕
冥
un⫺1冪a ⫹ bu du
1 冪a ⫹ bu ⫺ 冪a ln ⫹ C, a > 0 1 冪a 冪a ⫹ bu ⫹ 冪a du ⫽ u冪a ⫹ bu 2 a ⫹ bu arctan ⫹ C, a < 0 冪⫺a ⫺a
冪
1 冪a ⫹ bu 共2n ⫺ 3兲b ⫺1 du ⫽ ⫹ ⫹ bu a共n ⫺ 1兲 un⫺1 2
un冪a
冤
冕
1 un⫺1冪a
⫹ bu
冥
du , n ⫽ 1 A3
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A4
19. 20. 21. 22.
Appendix B
冕 冕 冕 冕
Integration Tables
冕
冪a ⫹ bu
1 du u冪a ⫹ bu
du ⫽ 2冪a ⫹ bu ⫹ a
u 冪a ⫹ bu
un u 冪a ⫹ bu
du ⫽
⫺1 共a ⫹ bu兲3兾2 共2n ⫺ 5兲b ⫹ a共n ⫺ 1兲 un⫺1 2
du ⫽
⫺2共2a ⫺ bu兲 冪a ⫹ bu ⫹ C 3b 2
冤
冢
un 2 du ⫽ un冪a ⫹ bu ⫺ na 冪a ⫹ bu 共2n ⫹ 1兲b
冕
冕
冪a ⫹ bu
un⫺1
un⫺1 du 冪a ⫹ bu
冥
du , n ⫽ 1
冣
Forms Involving a2 ± u2, a > 0 23. 24. 25.
冕 冕 冕
1 1 u du ⫽ arctan ⫹ C a2 ⫹ u2 a a 1 du ⫽ ⫺ u2 ⫺ a2
冕
ⱍ ⱍ
1 1 u⫺a du ⫽ ln ⫹C a2 ⫺ u2 2a u ⫹ a
冤
冕
1 1 u du ⫽ 2 ⫹ 共2n ⫺ 3兲 共a2 ± u2兲n 2a 共n ⫺ 1兲 共a2 ± u2兲n⫺1
冥
1 du , n ⫽ 1 共a2 ± u2兲n⫺1
Forms Involving 冪u2 ± a2, a > 0 26. 27. 28. 29. 30. 31. 32. 34. 35.
冕 冕 冕 冕 冕 冕 冕 冕 冕
冪u2 ± a2 du ⫽
ⱍ
u2冪u2 ± a2 du ⫽ 冪u2 ⫹ a2
u ⫺ u
冪u2
a2
冪u2 ± a2
u2 1 冪u2 ± a2
ⱍ
⫹
± a2
ⱍ
ⱍuⱍ ⫹ C a
ⱍ
ⱍ
⫺ 冪u2 ± a2 ⫹ ln u ⫹ 冪u2 ± a2 ⫹ C u
ⱍ
ⱍ
du ⫽ ln u ⫹ 冪u2 ± a2 ⫹ C
a2
u2 冪u2
ⱍ
a ⫹ 冪u2 ⫹ a2 ⫹C u
du ⫽ 冪u2 ⫺ a2 ⫺ a arcsec
1 u冪u2
ⱍ
1 关u共2u2 ± a2兲冪u2 ± a2 ⫺ a4 ln u ⫹ 冪u2 ± a2 兴 ⫹ C 8
du ⫽ 冪u2 ⫹ a2 ⫺ a ln
du ⫽
ⱍ
1 共u冪u2 ± a2 ± a2 ln u ⫹ 冪u2 ± a2 兲 ⫹ C 2
du ⫽
du ⫽
ⱍ
ⱍ
⫺1 a ⫹ 冪u2 ⫹ a2 ln ⫹C a u
ⱍ
33.
ⱍ
1 共u冪u2 ± a2 ⫿ a2 ln u ⫹ 冪u2 ± a2 兲 ⫹ C 2
冪u2 ± a2 1 du ⫽ ⫹C ⫿ a2u u2冪u2 ± a2
36.
冕
1 1 u du ⫽ arcsec ⫹C u冪u2 ⫺ a2 a a
冕
1 ±u du ⫽ 2 2 ⫹C 共u2 ± a2兲3兾2 a 冪u ± a2
ⱍⱍ
Forms Involving 冪a2 ⴚ u2, a > 0 37. 38.
冕 冕
冪a2 ⫺ u2 du ⫽
冢
冣
1 u u冪a2 ⫺ u2 ⫹ a2 arcsin ⫹C 2 a
u2冪a2 ⫺ u2 du ⫽
冤
冥
1 u u共2u2 ⫺ a2兲冪a2 ⫺ u2 ⫹ a4 arcsin ⫹C 8 a
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Appendix B
39. 41. 43. 45.
冕 冕 冕 冕
冪a2 ⫺ u2
u
ⱍ
du ⫽ 冪a2 ⫺ u2 ⫺ a ln
1 u du ⫽ arcsin ⫹ C ⫺ u2 a
ⱍ
a ⫹ 冪a2 ⫺ u2 ⫹C u
42.
冪a2
u2 冪a2 ⫺ u2
du ⫽
40.
冢
冣
1 u ⫺u冪a2 ⫺ u2 ⫹ a2 arcsin ⫹C 2 a
44.
冕 冕 冕
冪a2 ⫺ u2
u2
du ⫽
1 u冪a2
⫺
u2
Integration Tables
A5
⫺ 冪a2 ⫺ u2 u ⫺ arcsin ⫹ C u a
du ⫽
ⱍ
ⱍ
⫺1 a ⫹ 冪a2 ⫺ u2 ln ⫹C a u
1 du ⫽ u2冪a2 ⫺ u2
⫺ 冪a2
⫺ a2u
u2
⫹C
u 1 du ⫽ 2 2 ⫹C 共a2 ⫺ u2兲3兾2 a 冪a ⫺ u2
Forms Involving sin u or cos u 46. 48. 50. 52. 54. 56. 58.
冕 冕 冕 冕 冕 冕 冕
sin u du ⫽ ⫺cos u ⫹ C
47.
1 sin2 u du ⫽ 共u ⫺ sin u cos u兲 ⫹ C 2
49.
sinn u du ⫽ ⫺
sinn⫺1 u cos u n ⫺ 1 ⫹ n n
冕
sinn⫺2 u du
u sin u du ⫽ sin u ⫺ u cos u ⫹ C
冕
un sin u du ⫽ ⫺un cos u ⫹ n
51. 53.
un⫺1 cos u du
1 du ⫽ tan u ⫿ sec u ⫹ C 1 ± sin u
55. 57.
冕 冕 冕 冕 冕 冕
cos u du ⫽ sin u ⫹ C 1 cos2 u du ⫽ 共u ⫹ sin u cos u兲 ⫹ C 2 cosn u du ⫽
cosn⫺1 u sin u n ⫺ 1 ⫹ n n
冕
cosn⫺2 u du
u cos u du ⫽ cos u ⫹ u sin u ⫹ C un cos u du ⫽ un sin u ⫺ n
冕
un⫺1 sin u du
1 du ⫽ ⫺cot u ± csc u ⫹ C 1 ± cos u
1 du ⫽ ln tan u ⫹ C sin u cos u
ⱍ
ⱍ
Forms Involving tan u, cot u, sec u, or csc u 59. 61. 62. 63. 65. 67. 69. 70.
冕 冕 冕 冕 冕 冕 冕 冕
ⱍ
ⱍ
tan u du ⫽ ⫺ln cos u ⫹ C
60.
ⱍ
ⱍ
ⱍ
ⱍ
冕
ⱍ
ⱍ
cot u du ⫽ ln sin u ⫹ C
sec u du ⫽ ln sec u ⫹ tan u ⫹ C csc u du ⫽ ln csc u ⫺ cot u ⫹ C
or
冕
ⱍ
tan2 u du ⫽ ⫺u ⫹ tan u ⫹ C
64.
sec2 u du ⫽ tan u ⫹ C
66.
冕
tann u du ⫽
tann⫺1 u ⫺ n⫺1
secn u du ⫽
secn⫺2 u tan u n ⫺ 2 ⫹ n⫺1 n⫺1
cscn u du ⫽ ⫺
tann⫺2 u du, n ⫽ 1
冕
cscn⫺2 u cot u n ⫺ 2 ⫹ n⫺1 n⫺1
ⱍ
csc u du ⫽ ⫺ln csc u ⫹ cot u ⫹ C
68.
冕 冕 冕
cot2 u du ⫽ ⫺u ⫺ cot u ⫹ C csc2 u du ⫽ ⫺cot u ⫹ C
cot n u du ⫽ ⫺
cot n⫺1 u ⫺ n⫺1
冕
共cot n⫺2 u兲 du, n ⫽ 1
secn⫺2 u du, n ⫽ 1
冕
cscn⫺2 u du, n ⫽ 1
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A6
71. 73.
Appendix B
冕 冕
Integration Tables
1 1 du ⫽ 共u ± ln cos u ± sin u 兲 ⫹ C 1 ± tan u 2
ⱍ
72.
1 du ⫽ u ⫹ cot u ⫿ csc u ⫹ C 1 ± sec u
74.
ⱍ
冕 冕
1 1 du ⫽ 共u ⫿ ln sin u ± cos u 兲 ⫹ C 1 ± cot u 2
ⱍ
ⱍ
1 du ⫽ u ⫺ tan u ± sec u ⫹ C 1 ± csc u
Forms Involving Inverse Trigonometric Functions 75. 77. 79.
冕 冕 冕
arcsin u du ⫽ u arcsin u ⫹ 冪1 ⫺ u2 ⫹ C
76.
arctan u du ⫽ u arctan u ⫺ ln冪1 ⫹ u2 ⫹ C
78.
ⱍ
ⱍ
arcsec u du ⫽ u arcsec u ⫺ ln u ⫹ 冪u2 ⫺ 1 ⫹ C
80.
Forms Involving eu 81. 83. 85.
冕 冕 冕
eu du ⫽ eu ⫹ C
82.
冕
uneu du ⫽ uneu ⫺ n eau sin bu du ⫽
a2
un⫺1eu du
84.
eau 共a sin bu ⫺ b cos bu兲 ⫹ C ⫹ b2
86.
冕 冕 冕
arccos u du ⫽ u arccos u ⫺ 冪1 ⫺ u2 ⫹ C arccot u du ⫽ u arccot u ⫹ ln冪1 ⫹ u2 ⫹ C
ⱍ
ⱍ
arccsc u du ⫽ u arccsc u ⫹ ln u ⫹ 冪u2 ⫺ 1 ⫹ C
冕 冕 冕
ueu du ⫽ 共u ⫺ 1兲eu ⫹ C 1 du ⫽ u ⫺ ln共1 ⫹ eu兲 ⫹ C 1 ⫹ eu eau cos bu du ⫽
a2
eau 共a cos bu ⫹ b sin bu兲 ⫹ C ⫹ b2
Forms Involving ln u 87. 89. 90.
冕 冕 冕
ln u du ⫽ u共⫺1 ⫹ ln u兲 ⫹ C un ln u du ⫽
88.
un⫹1 关⫺1 ⫹ 共n ⫹ 1兲 ln u兴 ⫹ C, n ⫽ ⫺1 共n ⫹ 1兲2
共ln u兲2 du ⫽ u 关2 ⫺ 2 ln u ⫹ 共ln u兲2兴 ⫹ C
91.
冕
u ln u du ⫽
冕
共ln u兲n du ⫽ u共ln u兲n ⫺ n
u2 共⫺1 ⫹ 2 ln u兲 ⫹ C 4
冕
共ln u兲n⫺1 du
Forms Involving Hyperbolic Functions 92. 94. 96.
冕 冕 冕
cosh u du ⫽ sinh u ⫹ C
93.
sech2 u du ⫽ tanh u ⫹ C
95.
sech u tanh u du ⫽ ⫺sech u ⫹ C
97.
冕 冕 冕
sinh u du ⫽ cosh u ⫹ C csch2 u du ⫽ ⫺coth u ⫹ C csch u coth u du ⫽ ⫺csch u ⫹ C
Forms Involving Inverse Hyperbolic Functions (in logarithmic form) 98. 100.
冕 冕
du 冪u2 ± a2
⫽ ln 共u ⫹ 冪u2 ± a2 兲 ⫹ C
du 1 a⫹ ⫽ ⫺ ln a u冪a2 ± u2
冪a2
ⱍⱍ u
±
u2
99.
冕
ⱍ ⱍ
du 1 a⫹u ⫽ ln ⫹C a2 ⫺ u2 2a a ⫺ u
⫹C
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A85
Answers to Odd-Numbered Exercises 23.
1. (a) 具4, 2典 (b) y
−u
3. (a) 具6, 0典 (b)
y
5
−v x
4
4
x
2
(−6, 0)
3
(4, 2)
−8
2
−6
v −4
x
−2
y
25.
u −4
x 1
2
3
4
5. u v 具2, 4典 9. (a) and (d)
5
u−v
7. u v 具6, 5典 11. (a) and (d)
y
−v
x
(5, 5)
6 4
4
v
v
2
2
4
(6, − 1)
(2, 0) x 1
−1
2
3
4
5
v
2
( 12 , 3(
(6, 6)
45.
3
6
(−1, ( 2 5 3
v
v
51.
( ( 3 4 , 2 3
2
−2 x
6
(b) 具0, 4典 (c) v 4j 17. (a) 具6, 10典
x
−1
1
2
(b) 具 1, 53典 (c) v i 53 j (b) 具9, 15典
y
(6, 10)
(3, 5)
6
v
3
8 6
(3, 5)
4
v
x −15 −12 −9 −6 −3
2v
3
55. 57.
59. 63. 67.
y
10
u 1
−1
y
4
6
2
−2
4
6
8
10
4
5
6
7
y
10 2
8
−3v −15
(a) (b)
6 4
具 212, 352典
(d) 具 2, 10 3典
y
(
18
21 35 , 2 2
9
(3, 5) 7 v 2
6
3
2
4
6
8
6
9
12 15 18
71. (a) ± 15 具4, 3典 (b) ± 15 具3, 4典
(2, 103 (
v
2 v 3
−1
1
2
3
4
(a)
5
4
(b)
(3, 4)
3
19. (a)
具
8 3,
6典
(b) 具6, 15典
1
2
(c) 具2, 14典
2
(d) 具18, 7典
1 −1
x 1
2
3
4
具 冪2兾2, 冪2兾2典 10.7, 584.6 lb 71.3, 228.5 lb (a) 0 (b) 180 (c) No, the resultant can only be less than or equal to the sum. 81. Horizontal: 1193.43 ft兾sec Vertical: 125.43 ft兾sec 83. 38.3 north of west 882.9 km兾h 73. 75. 77. 79.
y
x −1
x
10
1 x
−3 −3
(b)
(3, 5)
2
v
3
(1, 1)
x
4 3
1
(a)
(3, 9)
−2
5
15 12
2
y
(
冭
y
−12
(−9, −15)
3
冬
−9 2
x
2
储u储 储v储 冪5 冪41 and 储u v储 冪74 冪74 冪5 冪41 47. 具 冪5, 2冪5 典 49. 具3, 0典 具0, 6典 冪 冪 具 冪3, 1典 53. 2 23 2, 3 2 2 具2 cos 4 cos 2, 2 sin 4 sin 2典 Answers will vary. Example: A scalar is a single real number, such as 2. A vector is a line segment having both direction and magnitude. The vector 具 冪3, 1典, given in component form, has a direction of 兾6 and a magnitude of 2. 61. a 1, b 1 共4, 1兲, 共6, 5兲, 共10, 3兲 65. a 23, b 13 a 1, b 2 (a) ± 共1兾冪37兲具1, 6典 69. (a) ± 共1兾冪10兲 具1, 3典 (b) ± 共1兾冪37 兲具6, 1典 (b) ± 共1兾冪10兲 具3, 1典
−6 x
(f) 1
u+v
3
y
2
(e) 1
4
(−2, −4)
(b) 具2, 4典 (c) v 2i 4j 15. (a) and (d)
(6, 2)
(d) 1
5
1
(0, 4)
(c) 冪85兾2
6
8
−6
(b) 具3, 5典 (c) v 3i 5j 13. (a) and (d)
(b) 冪13
7 x
−4 −2
1
41. (a) 冪5兾2 y 43.
(8, 3)
2
3
(c)
29. 7 共3, 5兲 5 33. 冪61 具 冪17兾17, 4冪17兾17典 具 3冪34兾34, 5冪34兾34典 (a) 冪2 (b) 冪5 (c) 1 (d) 1 (e) 1 (f) 1
y
(3, 5)
5
−2
27. 31. 35. 37. 39.
−2
v
1
4
y
(page 755)
Section 11.1
−1
y
21.
Chapter 11
5
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A86
Answers to Odd-Numbered Exercises
ⱍⱍ
85. True 87. True 89. False. 储ai bj储 冪2 a 2 91–93. Proofs 95. x y 2 25
z
(c) 3
3
2
(page 763)
Section 11.2 z
1.
(2, 1, 3)
−3
6 5 4 3
3 2 1
(5, −2, 2) −3 4
3
−2 −3
1 4
3
1 2 3
y
y
5. 11. 13. 15. 17. 19. 21. 23. 29. 31. 35. 39. 41. 43. 45.
3
−3
5
5
4
4
〈− 2, 2, 2〉
3
\
1
−2 1
1
2
2
3
3
1
2
3
y
x
3
\
87.
具0, 冪3, ± 1典
z
2
−2
〈0,
1
−2
89. 共2, 1, 2兲
3, 1〉
−1 1 y
−1
〈0,
−2
3, − 1〉
(b) a 0, a b 0, b 0 (c) a 1, a b 2, b 1 (d) Not possible
z
1
x
v
49. v 具1, 1, 6典 储v储 冪38 1 具1, 1, 6典 u 冪38
51. (a) and (d) z 5 4
1
(3, 3, 4) (−1, 2, 3)
−2
v 2 4
y
4
z
z
(b)
5
2
3
〈2, 4, 4〉
2 −2 2 3 4
−3
1
1
2 3
2 y
x
−2
(c)
−3
30
18.4
11.5
10
9.3
40
45
50
9.0
8.7
8.6
(d) Proof
L = 18
97. 0
(e) 30 in.
−2
〈−1, −2, −2〉 1
T
3
4
y
99. 共冪3兾3兲具1, 1, 1典 101. (a) T 8L兾冪L2 182, L > 18 (b) L 20 25 30 35
x
(b) 具4, 1, 1典 (c) v 4i j k
53. 共3, 1, 8兲 55. (a)
1
93. x 0 is directed distance to yz-plane. y0 is directed distance to xz-plane. z 0 is directed distance to xy-plane. 95. 共x x 0 兲2 共 y y0 兲2 共z z 0 兲2 r 2
3
(4, 1, 1) 2
u
x
(0, 0, 0) 2
x
\
v
y
4
\
tv
u
91. (a)
2 4
−3
u + tv
su
−3
1
y
3
su + tv
2
1
−2
2
−2
\
\
x
2
〈0, 0, 0〉 1
57. 具7, 0, 4典 59. 具 3, 典 61. a and b 63. a 65. Collinear 67. Not collinear 69. AB 具1, 2, 3典, CD 具1, 2, 3典, BD 具2, 1, 1典, AC 具2, 1, 1典; Because AB CD and BD AC , the given points form the vertices of a parallelogram. 71. 0 73. 冪34 75. 冪14 77. (a) 13 具2, 1, 2典 (b) 13 具2, 1, 2典 冪2 冪2 2冪2 3冪2 2冪2 3冪2 79. (a) i j k (b) i j k 5 2 10 5 2 10 81. The terminal points of the 83. 具 0, 10兾冪2, 10兾冪2 典 vectors tu, u tv, and 85. 具 1, 1, 12典 su tv are collinear.
〈− 3, 0, 3〉
3
−3
2
−3
5 2
\
7. 共12, 0, 0兲 9. 0 共3, 4, 5兲 Six units above the xy-plane Three units behind the yz-plane To the left of the xz-plane Within three units of the xz-plane Three units below the xy-plane, and below either quadrant I or quadrant III Above the xy-plane and above quadrants II or IV, or below the xy-plane and below quadrants I or III 冪69 25. 冪61 27. 7, 7冪5, 14; Right triangle 冪41, 冪41, 冪14; Isosceles triangle 33. 共2, 6, 3兲 共0, 0, 9兲, 共2, 6, 12兲, 共6, 4, 3兲 共32, 3, 5兲 37. 共x 0兲2 共 y 2兲2 共z 5兲2 4 共x 1兲2 共 y 3兲2 共z 0兲2 10 共x 1兲2 共 y 3兲2 共z 4兲2 25 Center: 共1, 3, 4兲 Radius: 5 共x 13 兲2 共 y 1兲2 z 2 1 Center: 共13, 1, 0兲 Radius: 1 (a) 具2, 2, 2典 47. (a) 具3, 0, 3典 (b) v 2i 2j 2k (b) v 3i 3k z z (c) (c)
−1
x
\
(5, −2, −2)
x
−2 1
y
−2
1
2
7 2,
x
2 3 4
−3
1 −2
x
1
2
〈 32 , 3, 3〉
3
2
−3
−2 2
(−1, 2, 1)
2
−2
z
3.
z
(d)
2
3
y
T=8
−2 −3
0
100 0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A87
Answers to Odd-Numbered Exercises 103. Tension in cable AB: 202.919 N Tension in cable AC: 157.909 N Tension in cable AD: 226.521 N 2 2 105. 共x 43 兲 共 y 3兲2 共z 13 兲 44 9 4 1 2冪11 Sphere; center: , 3, , radius: 3 3 3
冢
Section 11.3 1. 3. 5. 7. 9. 13.
(a) (a) (a) (a) (a) (a)
17. 20
冣
(page 773)
17 (b) 25 (c) 25 (d) 具17, 85典 (e) 34 (e) 52 26 (b) 52 (c) 52 (d) 具78, 52典 2 (b) 29 (c) 29 (d) 具0, 12, 10典 (e) 4 1 (b) 6 (c) 6 (d) i k (e) 2 11. (a) 1.7127 (b) 98.1° 兾2 (b) 90° 1.0799 (b) 61.9° 15. (a) 2.0306 (b) 116.3° 19. Orthogonal
21. Neither
69. (a) 共1, 0兲, 共1, 0兲 (b) To y 1 x 2 at 共1, 0兲: 具 ± 冪5兾5, 2冪5兾5典 To y x 2 1 at 共1, 0兲: 具 ± 冪5兾5, ± 2冪5兾5典 To y 1 x 2 at 共1, 0兲: 具 ± 冪5兾5, ± 2冪5兾5典 To y x 2 1 at (1, 0兲: 具 ± 冪5兾5, 2冪5兾5典 (c) At 共1, 0兲: 53.13 At 共1, 0兲: 53.13 71. Proof z 73. (a) (b) k冪2 (c) 60 (d) 109.5 k (k, 0, k)
k
x
23. Orthogonal
(0, k, k)
k
y
(k, k, 0)
25. Right triangle; answers will vary. 27. Acute triangle; answers will vary. 1 3 29. cos , ⬇ 70.5 31. cos , ⬇ 43.3 3 冪17 2 2 cos , ⬇ 48.2 , ⬇ 61.0 cos 3 冪17 2 2 cos , ⬇ 48.2 , ⬇ 119.0 cos 3 冪17 33. cos 0, ⬇ 90 cos 3兾冪13, ⬇ 33.7 cos 2兾冪13, ⬇ 123.7 35. (a) 具2, 8典 (b) 具4, 1典 37. (a) 具 52, 12典 (b) 具 12, 52典 39. (a) 具2, 2, 2典 (b) 具2, 1, 1典
44 8 6 41. (a) 具 0, 33 (b) 具 2, 25 , 25典 25 , 25 典 43. See “Definition of Dot Product” page 766. 45. (a) and (b) are defined. (c) and (d) are not defined because it is not possible to find the dot product of a scalar and a vector or to add a scalar to a vector. 47. See Figure 11.29 on page 770. 49. Yes. 51. $17,490.25; Total revenue u v v u v u 储v储2 储u储2
储
储 储
储v储
储
储u储
ⱍu vⱍ 储v储2 ⱍv uⱍ 储u储2 1 1 储v储 储u储 储u储 储v储 Answers will vary. Example: 具12, 2典 and 具12, 2典 Answers will vary. Example: 具2, 0, 3典 and 具2, 0, 3典 arccos共1兾冪3 兲 ⬇ 54.7 (a) 8335.1 lb (b) 47,270.8 lb 425 ft-lb 63. 2900.2 km-N False. For example, 具1, 1典 具2, 3典 5 and 具1, 1典 具1, 4典 5, but 具2, 3典 具1, 4典. 67. (a) 共0, 0兲, 共1, 1兲 (b) To y x 2 at 共1, 1兲: 具 ± 冪5兾5, ± 2冪5兾5典 To y x 1兾3 at 共1, 1兲: 具 ± 3冪10兾10, ± 冪10兾10典 To y x 2 at 共0, 0兲: 具± 1, 0典 To y x 1兾3 at (0, 0兲: 具0, ± 1典 (c) At 共1, 1兲: 45 At 共0, 0兲: 90 53. 55. 57. 59. 61. 65.
75–77. Proofs
(page 781)
Section 11.4 1. k
z
z
3. i
1
1
k j 1 x
j 1
i
−k
1 y
−1
5. j
x
i
1 y
−1
z
1
−1
1 x
−j
k
i
1 y
−1
7. (a) 20i 10j 16k (b) 20i 10j 16k (c) 0 9. (a) 17i 33j 10k (b) 17i 33j 10k (c) 0 11. 具0, 0, 54典 13. 具1, 1, 1典 15. 具2, 3, 1典 7 5 13 7 5 13 17. or , , , , 冪 冪 冪 冪 冪 9 3 9 3 9 3 9 3 9 3 9冪3 7 1 3 7 1 3 19. or , , , , 冪59 冪59 冪59 冪59 冪59 冪59 21. 1 23. 6冪5 25. 9冪5 27. 11 2 29. 10 cos 40 ⬇ 7.66 ft-lb 31. (a) F 180共cos j sink兲 (b) 储AB F 储 225 sin 180 cos
冬 冬
冭 冬 冭 冬
\
\
ⱍ
冭 冭
ⱍ
(c) 储AB F 储 225共1兾2兲 180共冪3兾2兲 ⬇ 268.38 (d) 141.34 AB and F are perpendicular. (e) 400 From part (d), the zero is ⬇ 141.34, when the vectors are parallel. \
0
180 0
33. 1 35. 6 37. 2 39. 75 41. (a) (b) (c) (h) and (e) (f) (g) 43. See “Definition of Cross Product of Two Vectors in Space” on page 775.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A88
Answers to Odd-Numbered Exercises
45. The magnitude of the cross product will increase by a factor of 4. 47. False. The cross product of two vectors is not defined in a two-dimensional coordinate system. 49. False. Let u 具1, 0, 0典, v 具1, 0, 0典, and w 具1, 0, 0典. Then u v u w 0, but v w. 51– 59. Proofs
Section 11.5
(page 790)
1. (a) Yes (b) No Parametric Symmetric Direction Equations (a) Equations (b) Numbers z x 3. x 3t 3, 1, 5 y 3 5 yt z 5t y z3 x2 5. x 2 2t 2, 4, 2 2 4 2 y 4t z 3 2t y z1 x1 7. x 1 3t 3, 2, 1 3 2 1 y 2t z1t x5 y3 z2 9. x 5 17t 17, 11, 9 17 11 9 y 3 11t z 2 9t 11. x 7 10t Not possible 10, 2, 0 y 2 2t z6 13. x 2 15. x 2 3t 17. x 5 2t y3 y 3 2t y 3 t z4t z4t z 4 3t 19. x 2 t 21. P共3, 1, 2兲; 23. P共7, 6, 2兲; y1t v 具1, 2, 0典 v 具4, 2, 1典 z2t 25. L 1 L 2 and is parallel to L 3. 27. L1 and L 3 are identical. 29. 共2, 3, 1兲; cos 7冪17兾51 31. Not intersecting 33. (a) Yes (b) Yes 35. y 3 0 37. 2x 3y z 10 39. 2x y 2z 6 0 41. 3x 19y 2z 0 43. 4x 3y 4z 10 45. z 3 47. x y z 5 49. 7x y 11z 5 51. y z 1 53. x z 0 55. 9x 3y 2z 21 0 57. Orthogonal 59. Neither; 83.5 61. Parallel z z 63. 65.
71. P1 and P2 are parallel. 73. P1 P4 and is parallel to P2. 75. (a) ⬇ 65.91 (b) x 2 y1t z 1 2t 77. 共2, 3, 2兲; The line does not lie in the plane. 79. Not intersecting 81. 6冪14兾7 83. 11冪6兾6 85. 2冪26兾13 87. 27冪94兾188 89. 冪2533兾17 91. 7冪3兾3 93. 冪66兾3 95. Parametric equations: x x1 at, y y1 bt, and z z1 ct x x1 y y1 z z1 Symmetric equations: a b c You need a vector v 具a, b, c典 parallel to the line and a point P共x1, y1, z1兲 on the line. 97. Simultaneously solve the two linear equations representing the planes and substitute the values back into one of the original equations. Then choose a value for t and form the corresponding parametric equations for the line of intersection. 99. Yes. If v1 and v2 are the direction vectors for the lines L1 and L2, then v v1 v2 is perpendicular to both L1 and L2. 101. (a) Year
2005
2006
2007
2008
2009
2010
z (approx.)
16.39
17.98
19.78
20.87
19.94
21.04
The approximations are close to the actual values. (b) An increase 103. (a) 冪70 in. (b) 15 (c) The distance is never zero. (d) 5 in.
0
48 23 105. 共77 107. 共 12, 94, 14 兲 109. True 111. True 13 , 13 , 13 兲 113. False. Plane 7x y 11z 5 and plane 5x 2y 4z 1 are both perpendicular to plane 2x 3y z 3 but are not parallel.
4
1. c 2. e 7. Plane
(0, −4, 0)
(0, 0, 2)
(
3
4. b
5. d 6. a 9. Right circular cylinder
z
z 4
2
−2
1 2
−3
1 −1
1
7 6
2
−2
x
(
3. f
3
3
0, 0, 4 2
(page 802)
Section 11.6
3
6
15 0
4
−3
4
x
5
y
y
−4
(3, 0, 0)
x
−1
(0, 6, 0)
4
6
6
y
3
y
11. Elliptic cylinder
(2, 0, 0)
z
67. 8
z
69.
13. Ellipsoid
z
x
z
3
2
3
(0, 0, 6)
−3 2 3
(6, 0, 0)
5
x
8 x
x
5
8
y
y
2 2
3
y
2
x
y
−2
(5, 0, 0)
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A89
Answers to Odd-Numbered Exercises 15. Hyperboloid of one sheet
17. Hyperboloid of two sheets
z
25. x 2 y 2 z 2 5
z 3
3 2 −2
3
2 3
3
−2
−2
3
2
x
3
y
x
−2
−3
21. Hyperbolic paraboloid z
z
39.
2
−3
1
3
4
2 3
3 2
x
1
31. 共4冪2, 2兾3, 兾4兲
29. 共4, 0, 兾2兲 33. 共4, 兾6, 兾6兲
3
3
共
5 5 2, 2,
5冪2兾2
35. 共冪6, 冪2, 2冪2 兲
兲
43. 7
47. tan2 2
49. x 2 y 2 z 2 25
y
37. 共0, 0, 12兲
41. 2 csc csc
45. 4 csc
y
y
2
−1
−3
19. Elliptic paraboloid
x
1
2
y
y
−3
2
2 1
3
3
z
−3 x
x
27. x 2 y 2 2y 0
z
51. 3x 2 3y 2 z 2 0
z
−2
z
6 5
−3
2
23. Elliptic cone
−2
z
−1 6 5
1
x
x
5
−3
6
2
1
−1
y
−2
−1
1
2 y
1 x
−6 3
53. x 2 y 2 共z 2兲2 4
y
−1
55. x 2 y 2 1
z
25. Let C be a curve in a plane and let L be a line not in a parallel plane. The set of all lines parallel to L and intersecting C is called a cylinder. C is called the generating curve of the cylinder, and the parallel lines are called rulings. 27. See pages 796 and 797. 29. xy-plane: ellipse; three-space: hyperboloid of one sheet 31. x 2 z 2 4y 33. 4x 2 4y 2 z 2 2 2 2 35. y z 4兾x 37. y 冪2z 共or x 冪2z 兲 39. 128兾3 冪 41. (a) Major axis: 4 2 (b) Major axis: 8冪2 Minor axis: 4 Minor axis: 8 Foci: 共0, ± 2, 2兲 Foci: 共0, ± 4, 8兲 43. x 2 z 2 8y; Elliptic paraboloid 45. x 2兾39632 y 2兾39632 z 2兾39502 1 47. x at, y bt, z 0; x at, y bt ab 2, z 2abt a 2 b 2 49. True 51. False. A trace of an ellipsoid can be a single point. 53. The Klein bottle does not have both an “inside” and an “outside.” It is formed by inserting the small open end through the side of the bottle and making it contiguous with the top of the bottle.
(page 809)
Section 11.7 1. 7. 13. 19. 21.
3. 共3冪2兾2, 3冪2兾2, 1兲 5. 共2冪3, 2, 3兲 共7, 0, 5兲 9. 共2冪2, 兾4, 4兲 11. 共2, 兾3, 4兲 共5, 兾2, 1兲 15. r 2 z 2 17 17. r sec tan z4 r 2 sin2 10 z 2 23. x 冪3y 0 x2 y 2 9
5
3
2 −2 x
57. 63. 67. 69. 75. 79.
81.
83. 85. 87. 89. 91.
2 4
y
2
3
−2 1
2
1
−1
2
y
−2
y
a
3 −a
2
−2
−a
1
1
x
2
2
−2
1
−3
x
d 58. e 59. c 60. a 61. f 62. b 共4, 兾4, 兾2兲 65. 共4冪2, 兾2, 兾4兲 共2冪13, 兾6, arccos关3兾冪13 兴兲 71. 共10, 兾6, 0兲 73. 共36, , 0兲 共13, , arccos关5兾13兴兲 共3冪3, 兾6, 3兲 77. 共4, 7兾6, 4冪3兲 Rectangular to cylindrical: r 2 x 2 y 2, tan y兾x, z z Cylindrical to rectangular: x r cos , y r sin , z z Rectangular to spherical: 2 x 2 y 2 z 2, tan y兾x, arccos共z兾冪x 2 y 2 z 2兲 Spherical to rectangular: x sin cos , y sin sin , z cos (a) r 2 z 2 25 (b) 5 (a) r 2 共z 1兲2 1 (b) 2 cos (a) r 4 sin (b) 4 sin 兾sin 4 sin csc (a) r 2 9兾共cos2 sin2 兲 (b) 2 9 csc2 兾共cos2 sin2 兲 z z 93.
3
y −3
2
3
2
3
−2
5
1
4 3
1
3
2
x
2
4
z
z
z
2
3
y
x
a
a
x
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
y
A90
Answers to Odd-Numbered Exercises z
95.
z
97.
65. x2 z2 2y
63. Cylinder z
a
2
30°
2
2 x y
2
2 x
99. 101. 103. 105. 107.
y
x y
−2
67. (a) 共4, 3兾4, 2兲
Rectangular: 0 x 10; 0 y 10; 0 z 10 Spherical: 4 6 Cylindrical: r2 z2 9, r 3 cos , 0 False. r z represents a cone. False. See page 805. 109. Ellipse
(b) 共2冪5, 3兾4, arccos关冪5兾5兴兲
69. 共50冪5, 兾6, arccos关1兾冪5兴兲
71. 共25冪2兾2, 兾4, 25冪2兾2兲 73. (a) r 2 cos 2 2z (b) 2 sec 2 cos csc2 2 75. 共x 52 兲 y2 25 77. x y 4 z
(page 811)
Review Exercises for Chapter 11
1. (a) u 具3, 1典, v 具4, 2典 (b) u 3i j, v 4i 2j (c) 储u储 冪10, 储v储 2冪5 (d) 10i 3. v 具 4, 4冪3 典 5. 共5, 4, 0兲 7. 冪22 225 2 2 2 9. 共x 3兲 共 y 2兲 共z 6兲 4 11. 共x 2兲2 共 y 3兲2 z 2 9; Center: 共2, 3, 0兲; Radius: 3 13. (a) and (d) 15. Collinear z 17. 共1兾冪38兲 具2, 3, 5典 (2, −1, 3) 3 19. (a) u 具1, 4, 0典 2 1 v 具3, 0, 6典 1 2 (b) 3 (c) 45 3 3 y −2 4 5 5 21. (a) (b) 15° x 12 23. (a) (b) 180° 25. Orthogonal −8 27. 具2, 10典 −9 −10 29. 具1, 0, 1典
37. 39. 41. 43. 47. 53. 55.
y
2
4
3
2
x
1 3
−3
(page 813)
P.S. Problem Solving
1–3. Proofs 5. (a) 3冪2兾2 ⬇ 2.12 7. (a) 兾2 (b) 12 共abk兲k (c) V 12 共ab兲k 2
(b) 冪5 ⬇ 2.24
V 12 共area of base兲height 9. Proof z 11. (a) (b)
z
2 −3
−3
−2 1
冬
冭 冬
冭
x
1
3
2
y
3
−2
x
−2
13. (a) Tension: 2冪3兾3 ⬇ 1.1547 lb Magnitude of u: 冪3兾3 ⬇ 0.5774 lb (b) T sec ; 储u储 tan ; Domain: 0 90 (c) 0 10 20 30
(d)
2
(0, 0, 2)
y
3
3
z
z
y
4
2
(b) u 具2, 5, 10典 (c) u 2i 5j 10k Answers will vary. Example: 具6, 5, 0典, 具6, 5, 0典 (a) 9i 26j 7k (b) 9i 26j 7k (c) 0 (a) 8i 10j 6k (b) 8i 10j 6k (c) 0 8 12 13 8 12 13 or , , , , 冪377 冪377 冪377 冪377 冪377 冪377 100 sec 20 ⬇ 106.4 lb (a) x 3 6t, y 11t, z 2 4t (b) 共x 3兲兾6 y兾11 共z 2兲兾4 45. x t, y 1 t, z 1 x 1, y 2 t, z 3 49. x 2y 1 51. 87 27x 4y 32z 33 0 冪35兾7 Plane 57. Plane 3
3
3
x
(4, 4, − 7) (2, 5, −10)
31. 33. 35.
z
3
T
1
1.0154
1.0642
1.1547
储u储
0
0.1763
0.3640
0.5774
40
50
60
T
1.3054
1.5557
2
储u储
0.8391
1.1918
1.7321
(e) Both are increasing functions
2.5
T 3
(0, 3, 0)
6
0
61. Hyperboloid of two sheets z 2
2 −2
−4 4 −2
y
60 0
z
5
⎜⎜u ⎜⎜
x
59. Ellipsoid
x
y
6
(6, 0, 0)
x
2
y
5 x
5
y
(f)
lim
→ 兾2
T and lim 储u储 → 兾2
Yes. As increases, both T and 储u储 increase. 15. 具0, 0, cos sin cos sin 典; Proof PQ n w 共u v兲 共u v兲 w u 共v w兲 17. D 储n储 储u v储 储u v储 储u v储 19. Proof
ⱍ
\
ⱍ ⱍ
ⱍ ⱍ
ⱍ ⱍ
ⱍ
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A91
Answers to Odd-Numbered Exercises z
39.
Chapter 12
−3
(page 821)
Section 12.1
−2
3 2 1 x − 4 −3 −2 −1
1
2
3
2 3
−5
y
Parabola
Helix (a) The helix is translated two units back on the x-axis. 2π (b) The height of the helix increases at a greater rate. π (c) The orientation of the graph is −2 −2 reversed. 2 (d) The axis of the helix is the x-axis. 2 y x (e) The radius of the helix is increased from 2 to 6. 45–51. Answers will vary. z 53. z
43.
2, −
(
2, 4 ) 5
−3
(−
1
2
2, 4)
2,
2
y
3
3
1 2 3 4 5
x
r共t兲 t i tj 2t 2 k z
55.
y
29.
−2
−1 1
x
−4
−2 −3
27.
y
−3
x
−5 −4 −3 − 2 −1
y
2
−2
4
−2
2
−1
x
7 6 5 4 3 2
4
1
z
41.
−3
2
3
1. 共 , 1兲 傼 共1, 兲 3. 共0, 兲 5. 关0, 兲 7. 共 , 兲 9. (a) 12 i (b) j (c) 12 共s 1兲2 i sj (d) 12t共t 4兲i tj 1 11. (a) ln 2i j 6k (b) Not possible 2 1 (c) ln共t 4兲i j 3共t 4兲k t4 t (d) ln共1 t兲i j 3tk 1 t 13. r共t兲 3t i t j 2t k, 0 t 1 x 3t, y t, z 2t, 0 t 1 15. r共t兲 共2 t兲i 共5 t兲j 共3 12t兲k, 0 t 1 x 2 t, y 5 t, z 3 12t, 0 t 1 17. t 2共5t 1兲; No, the dot product is a scalar. 19. b 20. c 21. d 22. a y y 23. 25.
−2
4
12 9
2
6
1
3
x
−3 −2
2
x
−12 −9 −6
3
6
−3
9 12
−3
−6
3
−9
z
31.
z
57.
7
(0, 6, 5)
5
r共t兲 2 sin t i 2 cos tj 4 sin2 tk
z
33.
y
3
x
−12
3
4 3
(2, − 2, 1)
(1, 2, 3) −3
1
3
4 5
6
3
x
−3 3
3 x
x
z
35.
y −3
r共t兲 共1 sin t兲i 冪2 cos tj 共1 sin t兲k and r共t兲 共1 sin t兲i 冪2 cos tj 共1 sin t兲k
z
37. 6
6
)2, 4, 163 ) 59.
4
z
2
3
(0, 0, 2)
2 x −3 3 x
3
y
y
3
3
y
−2
5
y
−4 −6
)− 2, 4, − 163 )
4 x
3
2 4
y
(2, 2, 0)
r共t兲 t i tj 冪4 t 2 k
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A92
Answers to Odd-Numbered Exercises
61. Let x t, y 2t cos t, and z 2t sin t. Then y 2 z2 共2t cos t兲2 共2t sin t兲2 4t 2 cos2 t 4t 2 sin2 t 4t 2共cos2 t sin2 t兲 4t 2. Because x t, y 2 z2 4x 2.
5. r 共t兲 具et, 2e2t典 r共0兲 i j r 共0兲 i 2j
y
3
r′
2
1
(1, 1) r
z
x 1
2
3
16
r 共t0兲 is tangent to the curve at t0. z 7. r 共t兲 2 sin t i 2 cos t j k 3π 0, − 2, ) ) 2 2π 3
3
2j k r r′ 2 2 3
r π r 2i k 2
12 8 4
7
4
5
6
8
12
x
16
冢 冣 冢 冣
y
冢 冣
−2
63. 69. 73. 75. 77. 79.
i j 65. 0 67. i j k 共 , 0兲, 共0, 兲 71. 关1, 1兴 共 兾2 n , 兾2 n 兲, n is an integer. s共t兲 t 2 i 共t 3兲 j 共t 3兲k s共t兲 共t 2 2兲 i 共t 3兲 j tk A vector-valued function r is continuous at t a if the limit of r共t兲 exists as t → a and lim r共t兲 r共u兲. The function t→a
冦
i j, t 2 is not continuous at t 0. r共t兲 i j, t < 2 81. Answers will vary. Sample answer: z
3 2 −2
1
−2
−1
−1 1 2
−1
x
2
y
r共t兲 1.5 cos t i 1.5 sin tj 83– 85. Proofs 87. Yes; Yes 91. True 93. True
1 tk, 0 t 2
89. Not necessarily
1. r 共t兲 2t i j r共2兲 4i 2j r 共2兲 4i j
3. r 共t兲 sin t i cos t j r共 兾2兲 j r 共 兾2兲 i
y
y
4
(4, 2)
r′
r′
(0, 1)
2
r x 2
4
6
8
r x
1
−2
9. 13. 17. 19. 21. 23. 25.
1
2
y
11. 2 sin t i 5 cos t j 3t 3j 15. 3a sin t cos2 t i 3a sin2 t cos t j 6i 14tj 3t 2 k t i 共5te t 5e t兲k e 具sin t t cos t, cos t t sin t, 1典 (a) 3t 2 i t j (b) 6t i j (c) 18t 3 t (a) 4 sin t i 4 cos t j (b) 4 cos t i 4 sin tj (c) 0 (a) t i j 12 t 2 k (b) i tk (c) t 3兾2 t 2i
(d) t i 12 t 2j k 27. (a) 具t cos t, t sin t, 1典 (b) 具cos t t sin t, sin t t cos t, 0典 (c) t (d) 具sin t t cos t, cos t t sin t, t 2 典 29. 共 , 0兲, 共0, 兲 31. 共n 兾2, 共n 1兲 兾2兲 33. 共 , 兲 35. 共 , 0), 共0, 兲 37. 共 兾2 n , 兾2 n 兲, n is an integer. 39. (a) i 3j 2tk (b) i 共9 2t兲j 共6t 3t2兲k (c) 40ti 15t 2j 20t 3k (d) 8t 9t 2 5t 4 (e) 8t 3 i 共12t 2 4t 3兲j 共3t 2 24t兲k (f) 2i 6j 8tk 41. (a) 7t 6 (b) 12t 5 i 5t 4 j 43. t 2 i t j tk C 2 5兾2 45. ln t i t j 5 t k C 47. 共t 2 t兲i t 4 j 2t 3兾2 k C 49. tan t i arctan t j C 51. 4i 12 j k 53. ai aj 共 兾2兲 k 55. 2i 共e2 1兲j 共e2 1兲k 57. 2e2t i 3共e t 1兲 j 59. 600冪3 t i 共16t 2 600t兲 j 2 t t 61. 共共2 e 兲兾2 兲i 共e 2兲j 共t 1兲k 63. See “Definition of the Derivative of a Vector-Valued Function” and Figure 12.8 on page 824. 65. The three components of u are increasing functions of t at t t0. 67–73. Proofs 75. (a) 5 The curve is a cycloid.
ⱍⱍ
(page 830)
Section 12.2
2 x
−4
r 共t0兲 is tangent to the curve at t0.
r 共t0兲 is tangent to the curve at t0.
0
40 0
(b) The maximum of 储 r 储 is 2; the minimum of 储 r 储 is 0. The maximum and the minimum of 储 r 储 are 1. 77. Proof 79. True 81. False. Let r共t兲 cos t i sin tj k, then d兾dt 关储 r共t兲 储兴 0, but 储 r 共t兲 储 1.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Answers to Odd-Numbered Exercises
Section 12.3
23. v共t兲 sin t i cos t j k r共t兲 cos t i sin t j t k r共2兲 共cos 2兲i 共sin 2兲j 2k 25. Maximum height: 45.5 ft; The ball will clear the fence. 27. v0 40冪6 ft兾sec; 78 ft 29. Proof 440 440 31. (a) r共t兲 共 3 cos 0兲 ti 关3 共 3 sin 0兲t 16t2兴 j (b) 100
(page 838)
1. (a) v共t兲 3i j 储v共t兲储 冪10 a共t兲 0 (b) v共1兲 3i j a共1兲 0 y (c)
A93
3. (a) v共t兲 2ti j 储v共t兲储 冪4t2 1 a共t兲 2i (b) v共2兲 4i j a共2兲 2i (c) y
θ 0 = 20
θ 0 = 25
4
2
v
(3, 0)
(4, 2)
a
2
x 4
v 0
6
−2
4
6
8
−4 y
(c) 3
v
(
2)
2,
x 3 0
(π , 2)
1 1 3 r共t兲 共t 3兾6 92 t 14 3 兲 j 共t 兾6 2 t 3 兲k
300 0
40
(c)
4
2
0
Maximum height: 2.1 ft Range: 46.6 ft
y
(c)
50 0
Maximum height: 10.0 ft Range: 227.8 ft (d)
200
v
a π
2π
9. (a) v共t兲 i 5j 3k 11. (a) v共t兲 i 2t j tk 储v共t兲 储 冪35 储v共t兲 储 冪1 5t2 a共t兲 0 a共t兲 2j k (b) v共1兲 i 5j 3k (b) v共4兲 i 8j 4k a共1兲 0 a共4兲 2j k 13. (a) v共t兲 i j 共t兾冪9 t 2兲 k 储v共t兲 储 冪共18 t2兲兾共9 t2兲 a共t兲 共9兾共9 t 2兲3兾2兲 k (b) v共0兲 i j 1 a共0兲 3k 15. (a) v共t兲 4i 3 sin tj 3 cos tk 储v共t兲 储 5 a共t兲 3 cos tj 3 sin tk (b) v共 兲 具4, 0, 3典 a共 兲 具0, 3, 0典 17. (a) v共t兲 共et cos t et sin t兲i 共et sin t et cos t兲j et k 储v共t兲储 et冪3 a共t兲 2et sin t i 2et cos t j et k (b) v共0兲 具1, 1, 1典 a共0兲 具0, 2, 1典 19. v共t兲 t 共i j k兲 r共t兲 共t 2兾2兲共i j k兲 r共2兲 2共i j k兲 9 1 21. v共t兲 共t 2兾2 2 兲 j 共t 2兾2 2 兲k
θ 0 = 15
a −3
−3
7. (a) v共t兲 具1 cos t, sin t典 储v共t兲储 冪2 2 cos t a共t兲 具sin t, cos t典 (b) v共 兲 具2, 0典 a共 兲 具0, 1典
θ 0 = 10
The minimum angle appears to be 0 20. (c) 0 ⬇ 19.38 33. (a) v0 28.78 ft兾sec; 58.28 (b) v0 ⬇ 32 ft兾sec 35. 1.91 37. (a) 5 (b) 15
−2
−4
5. (a) v共t兲 2 sin t i 2 cos t j 储v共t兲储 2 a共t兲 2 cos t i 2 sin t j (b) v共 兾4兲 冪2 i 冪2 j a共 兾4兲 冪2 i 冪2 j
500 0
x 2
x
0
200
0
0
Maximum height: 34.0 ft Range: 136.1 ft (e)
140 0
39. 41. 43.
45. 47.
49. 51. 53.
Maximum height: 166.5 ft Range: 666.1 ft (f)
60
0
800 0
300
0
600 0
Maximum height: 51.0 ft Maximum height: 249.8 ft Range: 117.9 ft Range: 576.9 ft Maximum height: 129.1 m; Range: 886.3 m Proof v共t兲 b 关共1 cos t兲i sin tj兴 a共t兲 b 2共sin ti cos tj兲 (a) 储v共t兲 储 0 when t 0, 2 , 4 , . . . . (b) 储v共t兲 储 is maximum when t , 3 , . . . . v共t兲 b sin ti b cos tj v共t兲 r共t兲 0 a共t兲 b 2共cos ti sin tj兲 2r共t兲; a共t兲 is a negative multiple of a unit vector from 共0, 0兲 to 共cos t, sin t兲, so a共t兲 is directed toward the origin. 8冪10 ft兾sec The velocity of an object involves both magnitude and direction of motion, whereas speed involves only magnitude. Proof
2 r共2兲 17 3 j 3k
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A94
Answers to Odd-Numbered Exercises
55. (a) v共t兲 6 sin t i 3 cos tj 储v共t兲 储 3冪3 sin2 t 1 a共t兲 6 cos t i 3 sin t j (b) t 0
兾4 Speed
3冪10兾2
3
兾2
2 兾3
6
3冪13兾2
3
35. r共2兲 5i 4j i 2j T共2兲 冪5 2i j N共2兲 , perpendicular to T共2兲 冪5 y 2
y
(c) 8 6 4 2 −8
−4 −2 −2
x 2
4
6
8
(d) The speed is increasing when the angle between v and a is in the interval 关0, 兾2兲, and decreasing when the angle is in the interval 共 兾2, 兴.
−4 −6 −8
57. Proof 59. False. Acceleration is the derivative of the velocity. 61. True
(page 848)
Section 12.4
1. T共1兲 共冪2兾2兲 共i j兲 3. T共 兾4兲 共冪2兾2兲 共i j兲 5. T共e兲 共3ei j兲兾冪9e2 1 ⬇ 0.9926i 0.1217j 7. T共0兲 共冪2兾2兲 共i k兲 9. T共0兲 共冪10兾10兲 共3j k兲 xt x3 y0 y 3t zt zt 11. T共 兾4兲 12 具 冪2, 冪2, 0典 x 冪2 冪2 t y 冪2 冪2 t z4 13. N共2兲 共冪5兾5兲 共2i j兲 15. N共2兲 共 冪5兾5兲共2i j兲 17. N共1兲 共 冪14兾14兲共i 2j 3k兲 19. N共3 兾4兲 共冪2兾2兲共i j兲 21. T共1兲 共冪2兾2兲 共i j兲 23. T共1兲 共 冪5兾5兲 共i 2 j兲 N共1兲 共冪2兾2兲 共i j兲 N共1兲 共 冪5兾5兲 共2i j兲 a T 冪2 a T 14冪5兾5 a N 冪2 a N 8冪5兾5 25. T共0兲 共冪5兾5兲 共i 2 j兲 27. T共 兾2兲 共冪2兾2兲共i j兲 N共0兲 共冪5兾5兲共2i j兲 N共 兾2兲 共 冪2兾2兲共i j兲 a T 7冪5兾5 a T 冪2e 兾2 a N 6冪5兾5 a N 冪2e 兾2 29. T共t兲 sin共 t兲i cos共 t兲j N共t兲 cos共 t兲i sin共 t兲j aT 0 aN a 2 31. 储v共t兲 储 a ; The speed is constant because a T 0. 33. r共2兲 2i 12 j T共2兲 共冪17兾17兲 共4i j兲 N共2兲 共冪17兾17兲 共i 4j兲 y
3
2
N 1
)2, 12 ) 1
T 2
x
x
−6 −4 −2
2
−4 −6
4
6
(5, − 4) N
T
−8
37. T共1兲 共冪14兾14兲 共i 2j 3k兲 N共1兲 is undefined. a T is undefined. a N is undefined. 39. T共1兲 共冪6兾6兲 共i 2j k兲 N共1兲 共冪30兾30兲 共5i 2j k兲 a T 5冪6兾6 a N 冪30兾6 41. T共0兲 共冪3兾3兲共i j k兲 N共0兲 共冪2兾2兲共i j兲 aT 冪3 aN 冪2 43. Let C be a smooth curve represented by r on an open interval I. The unit tangent vector T共t兲 at t is defined as r共t兲 , r共t兲 0. T共t兲 储r共t兲储 The principal unit normal vector N共t兲 at t is defined as T共t兲 N共t兲 , T共t兲 0. 储T共t兲储 The tangential and normal components of acceleration are defined as a共t兲 aTT共t兲 aNN共t兲. 45. (a) The particle’s motion is in a straight line. (b) The particle’s speed is constant. 47. v共t兲 r 共t兲 3i 4j 储v共t兲储 冪9 16 5 a共t兲 v 共t兲 0 3 4 v共t兲 i j T共t兲 储v共t兲储 5 5 T 共t兲 0 ⇒ N共t兲 does not exist. The path is a line. The speed is constant 共5兲. 49. (a) t 12 : a T 冪2 2兾2, a N 冪2 2兾2 t 1: a T 0, a N 2 t 32 : aT 冪2 2兾2, aN 冪2 2兾2 (b) t 12 : Increasing because a T > 0. t 1: Maximum because a T 0. t 32 : Decreasing because a T < 0. 51. T共 兾2兲 共冪17兾17兲 共4i k兲 N共 兾2兲 j B共 兾2兲 共冪17兾17兲共i 4k兲 53. T共 兾4兲 共冪2兾2兲共j k兲 N共 兾4兲 共冪2兾2兲共 j k兲 B共 兾4兲 i
3
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A95
Answers to Odd-Numbered Exercises 55. T共 兾3兲 共冪5兾5兲共i 冪3 j k兲 N共 兾3兲 12 共冪3 i j兲 B共 兾3兲 共冪5兾10兲共i 冪3 j 4k兲 1 共4t i 3j兲 57. N共t兲 冪16t 2 9 1 共 t i 2tj 5k兲 59. N共t兲 冪5t 2 25 32共v0 sin 32t兲 61. a T 2 冪v0 cos2 共v0 sin 32t兲2 32v0 cos aN 冪v02 cos2 共v0 sin 32t兲2 At maximum height, a T 0 and a N 32. 63. (a) r共t兲 60冪3 t i 共5 60t 16t 2兲j (b) 70
y
5. a
x
−a
a −a
6a 7. (a) r共t兲 共50t冪2兲 i 共3 50t冪2 16t 2兲 j (b) 649 (c) 315.5 ft (d) 362.9 ft 8 ⬇ 81 ft z 9. 11. 2
−3
−1
1
21 2
−2
2.0
2.5
3.0
Speed
104
105.83
109.98
15. 17.
aN 4
aT −20
The speed is decreasing when aT and aN have opposite signs. 65. (a) 4冪625 2 1 ⬇ 314 mi兾h (b) a T 0, a N 1000 2 a T 0 because the speed is constant. 67. (a) The centripetal component is quadrupled. (b) The centripetal component is halved. 69. 4.74 mi兾sec 71. 4.67 mi兾sec 73. False; centripetal acceleration may occur with constant speed. 75. (a) Proof (b) Proof 77–79. Proofs
y
(a, 0, 0)
6
9
y
19. 31. 39. 43. 47. 51. 53. 55.
y
2 冪a 2 b 2 (a) 2冪21 ⬇ 9.165 (b) 9.529 (c) Increase the number of line segments. (d) 9.571 s s s (a) s 冪5 t (b) r共s兲 2 cos i 2 sin j k 冪5 冪5 冪5 (c) s 冪5: 共1.081, 1.683, 1.000兲 s 4: 共0.433, 1.953, 1.789兲 (d) Proof 0 21. 25 23. 0 25. 冪2兾2 27. 1 29. 14 12 3 33. 冪5兾共1 5t 2兲3兾2 35. 25 37. 125 1兾a 41. K 0, 1兾K is undefined. 7冪26兾676 45. K 4, 1兾K 1兾4 K 4兾17 3兾2, 1兾K 17 3兾2兾4 49. (a) 共1, 3兲 (b) 0 K 12兾1453兾2, 1兾K 1453兾2兾12 (a) K → as x → 0 (No maximum) (b) 0 (a) 共1兾冪2, ln 2兾2兲 (b) 0 57. 共 兾2 K , 0兲 共0, 1兲
冕
b
59. s
(0, 0) x 9
(9, −3)
(0, 0)
x
冕
b
冪关x 共t兲兴2 关 y 共t兲兴2 关z 共t兲兴2 dt
a
储r 共t兲储 dt
a
61. The curve is a line. 2 6x 2 1 63. (a) K 共16x 6 16x 4 4x 2 1兲 3兾2 1 2 1 (b) x 0: x 2 y 2 4 2 1 5 x 1: x 2 y −3 2 4
冢 冢
(1, 1) 1
3冪10
(6 π , 0, −1)
−3 −6 −9 −12
y
ⱍ
(page 860) 3.
−6
6
2π b
x
104.61
40
−3
9
1.5
t
6
12
πb
107.63
3
5
15
6 −6
3冪17 兾2
(a, 0, 2π b)
112.85
y
4
18
−9
z
13.
Speed
Section 12.5
3
x
冪26
Maximum height ⬇ 61.245 ft Range ⬇ 398.186 ft (c) v共t兲 60冪3 i 共60 32t兲j 储v共t兲储 8冪16t 2 60t 225 a共t兲 32j (d) t 0.5 1.0
0
(− 1, 4, 3)
1
400 0
1.
−2
(0, 0, 0)
x
(e)
−12
3
3
0
(0, − 1, 0)
4
2
z
ⱍ
冣 冣
2
f 3
−2
1
共13冪13 8兲兾27
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A96
Answers to Odd-Numbered Exercises 5
(c)
−3
3 −2
65. 73. 79.
83. 87.
The curvature tends to be greatest near the extrema of the function and decreases as x → ± . However, f and K do not have the same critical numbers. Critical numbers of f : x 0, ± 冪2兾2 ⬇ ± 0.7071 Critical numbers of K: x 0, ± 0.7647, ± 0.4082 Proof 67. (a) 12.25 units (b) 12 69–71. Proofs (a) 0 (b) 0 75. 14 77. Proof 81. 3327.5 lb K 关1兾共4a兲兴 csc 共兾2兲 Minimum: K 1兾共4a兲 There is no maximum. Proof 85. False. See Exploration on page 851. True 89–95. Proofs
ⱍ
ⱍ
(page 863)
Review Exercises for Chapter 12
4 3
1
−5 x
−4
−2 −1
−3
(2, 2, 2)
(1, − 1, 0) − 2
2
1
2
4
4
13. r共t兲 t i 共 34t 3兲j
4
y
−4
5
2
2 x − 10
1
2
3
z 12 10 8 6 4 2
y
3
x
x t, y t, z 2t 17. 4i k 19. (a) 共2t 4兲i 6tj (b) 2i 6j (c) 40t 8
2
2
21. (a) 6t 2 i 4j 2tk (b) 12t i 2k (c) 72t 3 4t (d) 8i 12t 2j 48tk 23. (a) 3i j (b) 5i 共2t 2兲j 2t 2 k (c) 18t i 共6t 3兲j (d) 4t 3t 2 (e) 共83 t 3 2t 2兲 i 8t 3 j 共9t 2 2t 1兲k (f) 2i 8tj 16t 2 k 25. t i 3tj 2t 2 k C 27. 2t 3兾2 i 2 ln t j tk C
ⱍⱍ
x
−2
(10, − 15)
2
10
− 10
5冪13 57.
−3
10
2 4 6 8 10 12 14
−4 −6 −8 −10 −12 −14 −16
−2
z
15.
(0, 0)
−4 −2
−3
x
−4
3
2 3
−2
2
x
−2
(0, −4, −2)
冢冣
冢冣
1. (a) All reals except 共 兾2兲 n , n is an integer. (b) Continuous except at t 共 兾2兲 n , n is an integer. 3. (a) 共0, 兲 (b) Continuous for all t > 0 5. (a) i 冪2 k (b) 3i 4j (c) 共2c 1兲i 共c 1兲2 j 冪c 1 k (d) 2t i t共t 2兲j 共冪t 3 冪3兲k 7. r共t兲 共3 t兲i 2tj 共5 2t兲k, 0 t 1 x 3 t, y 2t, z 5 2t, 0 t 1 z y 9. 11. 4
29. 32 31. 2共e 1兲 i 8j 2k 3j 33. r共t兲 共t 2 1兲i 共e t 2兲 j 共et 4兲k 35. (a) v共t兲 4i 3t 2 j k 储v共t兲储 冪17 9t 4 a共t兲 6tj (b) v共1兲 4i 3j k a共1兲 6j 37. (a) v共t兲 具3 cos2 t sin t, 3 sin2 t cos t, 3典 储v共t兲 储 3冪sin2 t cos2 t 1 a共t兲 具3 cos t 共2 sin2 t cos2 t兲, 3 sin t共2 cos2 t sin2 t兲, 0典 (b) v共 兲 具0, 0, 3典 a共 兲 具3, 0, 0典 39. About 191.0 ft 41. About 38.1 m兾sec 冪10 3冪10 43. T共1兲 i j 10 10 冪15 冪5 冪5
45. T i j k; 3 5 5 5
x 冪3t 1, y t 冪3, z t 3 冪10 3冪10
47. N共1兲 49. N j i j 4 10 10 冪13 18冪13 51. T共3兲 i j 65 65 冪13 18冪13 i j N共3兲 65 65 2冪13 aT 585 4冪13 aN 65 y y 53. 55.
60 z
59.
(− 9, 6, 12)
π 2
8 x
(0, 0, 0) 2 4 6 8
10
6
)0, 8, π2 )
4 4
(8, 0, 0)
6 8
y
y
冪65 兾2 3冪29 61. 0 63. 共2冪5兲兾共4 5t 2兲3兾2 65. 冪2兾3 67. K 冪17兾289; r 17冪17 69. K 冪2兾4; r 2冪2 71. 2016.7 lb
P.S. Problem Solving
(page 865)
1. (a) a (b) a (c) K a 3. Initial speed: 447.21 ft兾sec; ⬇ 63.43 5–7. Proofs
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A97
Answers to Odd-Numbered Exercises 4 3 9. Unit tangent: 具 5, 0, 5典 Unit normal: 具0, 1, 0典 Binormal: 具 35, 0, 45典
z
33.
4 3 2
B
T 3
N
3π
B
T
4
1
4
1
y
z
37. 4
2 3
8
2
6
y
2
x
2
4
(b) 6.766
−3
y
3
z
39.
1 −2
(b) Proof
2
4 x
y
x
11. (a) Proof 13. (a)
2
x
1
2
1
5
N 3
5
5
z
6π
z
35.
2
4
3
4
y
x
z
41.
z
43.
−2
(c) K 关 共 2 t 2 2兲兴兾共 2 t 2 1兲 3兾2 K共0兲 2 K共1兲 关 共 2 2兲兴兾共 2 1兲3兾2 ⬇ 1.04 K共2兲 ⬇ 0.51 (d) 5 (e) lim K 0
y
y x
x
t→
45. c 46. d 47. b 49. Lines: x y c y 0
5
4
0
(f) As t → , the graph spirals outward and the curvature decreases.
48. a 51. Ellipses: x2 4y2 c (except x2 4y2 0 is the point 共0, 0兲.兲 y
2
c=0 c=1 c=2 c=3 c=4
2 x
−2
Chapter 13 Section 13.1
2
4 c=4
−2
1. Not a function because for some values of x and y 共for example x y 0兲, there are two z-values. 3. z is a function of x and y. 5. z is not a function of x and y. 7. (a) 6 (b) 4 (c) 150 (d) 5y (e) 2x (f) 5t 9. (a) 5 (b) 3e2 (c) 2兾e (d) 5e y (e) xe 2 (f) te t 3 10 2 11. (a) 3 (b) 0 (c) 2 (d) 3 13. (a) 冪2 (b) 3 sin 1 (c) 3冪3兾2 (d) 4 15. (a) 4 (b) 6 (c) 25 (d) 94 4 17. (a) 2, x 0 (b) 2y y, y 0 19. Domain: 再共x, y兲: x is any real number, y is any real number冎 Range: z 0 21. Domain: 再共x, y兲: y 0冎 Range: all real numbers 23. Domain: 再共x, y兲: x 0, y 0冎 Range: all real numbers 25. Domain: 再共x, y兲: x 2 y 2 ≤ 4冎 Range: 0 ≤ z ≤ 2 27. Domain: 再共x, y兲: 1 ≤ x y ≤ 1冎 Range: 0 z 29. Domain: 再共x, y兲: y < x 4冎 Range: all real numbers 31. (a) 共20, 0, 0兲 (b) 共15, 10, 20兲 (c) 共20, 15, 25兲 (d) 共20, 20, 0兲
c = −1
x
−2
c=2
(page 876)
2
c=0 −2
53. Hyperbolas: xy c y
1
−1
1 −1
c=6 c=5 c=4 c=3 c=2 c=1 x c = −1 c = −2 c = −3 c = −4 c = −5 c = −6
55. Circles passing through 共0, 0兲 Centered at 共1兾共2c兲, 0兲 y
c = −1 2
2
c = −3
c=1
2
c=2 x
2
c= 3
c = −2
2
c= 1
c = −1
57.
59.
6
−9
9
−6
2
4
−6
6
−4
61. The graph of a function of two variables is the set of all points 共x, y, z兲 for which z f 共x, y兲 and 共x, y兲 is in the domain of f. The graph can be interpreted as a surface in space. Level curves are the scalar fields f 共x, y兲 c, where c is a constant. 63. f 共x, y兲 x兾y; the level curves are the lines y 共1兾c兲x.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A98
Answers to Odd-Numbered Exercises
65. The surface may be shaped like a saddle. For example, let f 共x, y兲 xy. The graph is not unique; any vertical translation will produce the same level curves. 67. Inflation Rate Tax Rate
0
0.03
0.05
0
$1790.85
$1332.56
$1099.43
0.28
$1526.43
$1135.80
$937.09
0.35
$1466.07
$1090.90
$900.04
z
69.
37.
f 共x, y兲
共x, y兲
共x, y兲
共0.01, 0兲
共0.001, 0兲
0
0
0
0
0
共1, 1兲
共0.5, 0.5兲
共0.1, 0.1兲
1 2
1 2
1 2
共0.01, 0.01兲
共0.001, 0.001兲
1 2
1 2
f 共x, y兲
y x: 12 Limit does not exist. Continuous except at 共0, 0兲
1 −4 −1
4
4
x 2
共0.1, 0兲
f 共x, y兲
4
1
共0.5, 0兲
z
71.
1
共1, 0兲
y 0: 0
2
−2
共x, y兲
y
39.
y
2
−4
x
共x, y兲 f 共x, y兲
共1, 0兲 共0.5, 0兲 共0.1, 0兲 共0.01, 0兲 共0.001, 0兲 0
0
0
0
0
z
73.
y 0; 0
2
共x, y兲
−2 −2 1
2
2
共1, 1兲
共0.5, 0.5兲
共0.1, 0.1兲
1 2
1
5
f 共x, y兲
y
x
共x, y兲 75. (a) 243 board-ft y 77. c = 600 c = 500 c = 400
30
41.
30
−30
81. (a) k 520 3 (b) P 520T兾共3V兲 The level curves are lines. 83. (a) C (b) A (c) B 85. C 1.20xy 1.50共xz yz兲 87. False. Let f 共x, y兲 4. 89. True 91. Putnam Problem A1, 2008
Section 13.2
43. 53. 57. 61. 65. 69. 71. 73.
(page 887)
1–3. Proofs 5. 1 7. 12 9. 9, continuous 11. e 2, continuous 13. 0, continuous for y 0 15. 12, continuous except at 共0, 0兲 17. 0, continuous 19. 0, continuous for xy 1, xy 1 21. 2冪2, continuous for x y z ≥ 0 23. 0 25. Limit does not exist. 27. Limit does not exist. 29. Limit does not exist. 31. 0 33. Limit does not exist. 35. Continuous, 1
ⱍ ⱍ
共0.001, 0.001兲
50
500
f 共x, y兲
(b) 507 board-ft 79. Proof
c = 300 c = 200 c = 100 c=0
x
−30
共0.01, 0.01兲
75. 81.
y x;
The limit does not exist. Continuous except at 共0, 0兲 f is continuous. g is continuous except at 共0, 0兲. g has a removable discontinuity at 共0, 0兲. 0 45. 0 47. 1 49. 1 51. 0 Continuous except at 共0, 0, 0兲 55. Continuous Continuous 59. Continuous Continuous for y 2x兾3 63. (a) 2x (b) 4 (a) 1兾y (b) x兾y2 67. (a) 3 y (b) x 2 True ln(x2 y2兲, x 0, y 0 False. Let f 共x, y兲 . 0, x 0, y 0 (a) 共1 a2兲兾a, a 0 (b) Limit does not exist. (c) No, the limit does not exist. Different paths result in different limits. 0 77. 兾2 79. Proof See “Definition of the Limit of a Function of Two Variables” on page 881; show that the value of f 共x, y兲 is not the lim
冦
共x, y兲 → 共x0 , y0兲
same for two different paths to 共x0, y0 兲. 83. (a) No. The existence of f 共2, 3兲 has no bearing on the existence of the limit as 共x, y兲 → 共2, 3兲. (b) No. f 共2, 3兲 can equal any number, or not even be defined.
Section 13.3
(page 896)
1. No. Because you are finding the partial derivative with respect to x, you consider y to be constant. So, the denominator is considered a constant and does not contain any variables.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Answers to Odd-Numbered Exercises 3. No. Because you are finding the partial derivative with respect to y, you consider x to be constant. So, the denominator is considered a constant and does not contain any variables. 5. Yes. Because you are finding the partial derivative with respect to x, you consider y to be constant. So, both the numerator and denominator contain variables. 7. fx 共x, y兲 2 9. fx共x, y兲 2xy3 fy 共x, y兲 5 fy共x, y兲 3x2y2 11. z兾 x 冪y 13. z兾 x 2x 4y z兾 y x兾共2冪y兲 z兾 y 4x 6y 15. z兾 x yexy 17. z兾 x 2xe 2y z兾 y xexy z兾 y 2x 2e2y 19. z兾 x 1兾x 21. z兾 x 2x兾共x 2 y 2兲 z兾 y 1兾y z兾 y 2y兾共x 2 y 2兲 23. z兾 x 共x 3 3y 3兲兾共x 2 y兲 z兾 y 共x 3 12y 3兲兾共2xy 2兲 2 2 25. hx共x, y兲 2xe共x y 兲 27. fx 共x, y兲 x兾冪x 2 y 2 2 2 hy共x, y兲 2ye共x y 兲 fy共x, y兲 y兾冪x 2 y 2 29. z兾 x y sin xy 31. z兾 x 2 sec2共2x y兲 z兾 y x sin xy z兾 y sec2共2x y兲 33. z兾 x yey cos xy z兾 y e y 共x cos xy sin xy兲 35. z兾 x 2 cosh共2x 3y兲 37. fx 共x, y兲 1 x 2 fy 共x, y兲 y 2 1 z兾 y 3 cosh共2x 3y兲 39. fx 共x, y兲 3 fy 共x, y兲 2 43. fx 1 fy 0 1 47. fx 4
fy
1 4
41. fx 共x, y兲 1兾共2冪x y 兲 fy 共x, y兲 1兾共2冪x y兲 45. fx 1
81.
83.
85.
87. 89. 91. 93. 95. 97. 99. 101. 103.
fy 12 1 49. fx 4
fy 14
51. gx 共1, 1兲 2 gy 共1, 1兲 2 53. Hx 共x, y, z兲 cos共x 2y 3z兲 Hy 共x, y, z兲 2 cos共x 2y 3z兲 Hz 共x, y, z兲 3 cos共x 2y 3z兲 w x x 55. 57. Fx 共x, y, z兲 2 x x y 2 z2 冪x 2 y 2 z2 w y y Fy 共x, y, z兲 2 y x y 2 z2 冪x 2 y 2 z2 w z z Fz 共x, y, z兲 2 z x y 2 z2 冪x 2 y 2 z2 59. fx 3; fy 1; fz 2 61. fx 1; fy 1; fz 1 63. fx 0; fy 0; fz 1 65. x 2, y 2 67. x 6, y 4 69. x 1, y 1 71. x 0, y 0 2z 2z 0 2 73. 75. x2 x 2 2z 2z 6x 6 2 y y 2 2z 2z 2z 2z 6y 2 y x x y y x x y 2z y2 2z e x tan y 77. 79. x 2 共x 2 y 2兲3兾2 x 2 2z x2 2z 2ex sec2 y tan y y 2 共x 2 y 2兲3兾2 y 2 2z 2z xy 2z 2z 2 ex sec2 y 2 3兾2 y x x y 共x y 兲 y x x y
A99
2z y2 cos xy x2 2z x2 cos xy y2 2z 2z xy cos xy sin xy y x x y z兾 x sec y z兾 y x sec y tan y 2z兾 x 2 0 2z兾 y 2 x sec y共sec2 y tan2 y兲 2z兾 y x 2z兾 x y sec y tan y No values of x and y exist such that fx共x, y兲 fy共x, y兲 0. z兾 x 共 y 2 x 2兲兾关x共x 2 y 2兲兴 z兾 y 2y兾共x 2 y 2兲 2z兾 x2 共x 4 4x 2 y 2 y4兲兾关x 2共x 2 y 2兲2兴 2z兾 y 2 2共 y 2 x 2兲兾共x 2 y 2兲2 2z兾 y x 2z兾 x y 4xy兾共x 2 y 2兲2 No values of x and y exist such that fx共x, y兲 fy共x, y兲 0. fxyy共x, y, z兲 fyxy共x, y, z兲 fyyx 共x, y, z兲 0 fxyy 共x, y, z兲 fyxy 共x, y, z兲 fyyx 共x, y, z兲 z 2ex sin yz 2z兾 x 2 2z兾 y2 0 0 0 2z兾 x2 2z兾 y2 e x sin y e x sin y 0 2z兾 t2 c 2 sin共x ct兲 c 2共 2z兾 x 2兲 2z兾 t2 c 2兾共x ct兲2 c 2共 2z兾 x 2兲 z兾 t et cos x兾c c 2共 2z兾 x 2兲 Yes, f 共x, y兲 cos共3x 2y兲. If z f 共x, y兲, then to find fx , you consider y constant and differentiate with respect to x. Similarly, to find fy, you consider x constant and differentiate with respect to y. z
105. 4 2
4
2 4
y
x
107. The mixed partial derivatives are equal. See Theorem 13.3. 109. (a) 72 (b) 72 100 111. IQM , IQM共12, 10兲 10 C IQ increases at a rate of 10 points per year of mental age when the mental age is 12 and the chronological age is 10. 100M IQC 2 , IQC共12, 10兲 12 C IQ decreases at a rate of 12 points per year of chronological age when the mental age is 12 and the chronological age is 10. 113. An increase in either the charge for food and housing or the tuition will cause a decrease in the number of applicants. 115. T兾 x 2.4 兾m, T兾 y 9 兾m 117. T PV兾共nR兲 ⇒ T兾 P v兾共nR兲 P nRT兾V ⇒ P兾 V nRT兾V 2 V nRT兾P ⇒ V兾 T nR兾P T兾 P P兾 V V兾 T nRT兾共VP兲 nRT兾共nRT兲 1
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A100
Answers to Odd-Numbered Exercises
z z 0.461; 0.301 x y (b) As the expenditures on amusement parks and campgrounds 共x兲 increase, the expenditures on spectator sports 共z兲 increase. As the expenditures on live entertainment 共y兲 increase, the expenditures on spectator sports 共z兲 also increase. 121. False. Let z x y 1. 123. True y共x 4 4x 2 y 2 y 4兲 125. (a) fx 共x, y兲 共x 2 y 2兲2 4 x共x 4x 2 y 2 y 4兲 fy 共x, y兲 共x 2 y 2兲2 (b) fx 共0, 0兲 0, fy 共0, 0兲 0 (c) fxy 共0, 0兲 1, fyx 共0, 0兲 1 (d) fxy or fyx or both are not continuous at 共0, 0兲. 127. Proof 119. (a)
Section 13.4 1. 3. 5. 7. 9. 11. 13. 15.
17. 21. 23.
25.
ΔA
dA
h
29. 31. 35. 37.
21. 25.
29.
33.
37.
t
冢冪x xy y 冣 t f 共x, y兲; n 1 2
2
(b) xfx共x, y兲 yfy共x, y兲
Δl
A dA dl dh 27.
17.
3. e t共sin t cos t兲; 1 5. (a) and (b) et 26t; 52 (a) and (b) 2e 2t 9. (a) and (b) 3共2t 2 1兲 11冪29兾29 ⬇ 2.04 15. w兾 s 5 cos共5s t兲, 0 w兾 s 4s, 4 w兾 t 4t, 0 w兾 t cos共5s t兲, 0 (a) and (b) 19. (a) and (b) w w 2 2 t 2共3s 2 t 2兲 te s t 共2s 2 1兲 s s w w 2 2 2st共s2 2t 2兲 se s t 共1 2t 2兲 t t y 2x 1 x2 y 2 x 23. 2 2y x 1 x y2 y z x x z 27. x z x yz z y z z y z y yz z sec2共x y兲 z 共zexz y兲 31. 2 x sec 共 y z兲 x xexz 2 z sec 共x y兲 z 1 exz y sec2共 y z兲 y w yw w y sin xy 35. x xz x z w xz w x sin xy z cos yz y xz y z w w w y y cos yz w z xz z z 共tx兲共ty兲 (a) f 共tx, ty兲 冪共tx兲2 共ty兲2
dA
dA l
1. 7. 11. 13.
41. Answers will vary. Example: 1 y x 2 2x x 共 x兲2
(page 913)
Section 13.5
(page 905)
dz 4xy 3 dx 6x 2 y 2 dy dz 2共x dx y dy兲兾共x 2 y2兲2 dz 共cos y y sin x兲 dx 共x sin y cos x兲 dy dz 共e x sin y兲 dx 共e x cos y兲 dy dw 2z 3 y cos x dx 2z 3 sin x dy 6z 2 y sin x dz (a) f 共2, 1兲 1, f 共2.1, 1.05兲 1.05, z 0.05 (b) dz 0.05 (a) f 共2, 1兲 11, f 共2.1, 1.05兲 10.4875, z 0.5125 (b) dz 0.5 (a) f 共2, 1兲 e2 ⬇ 7.3891, f 共2.1, 1.05兲 1.05e2.1 ⬇ 8.5745, z ⬇ 1.1854 (b) dz ⬇ 1.1084 0.44 19. 0.094 In general, the accuracy worsens as x and y increase. If z f 共x, y兲, then z ⬇ dz is the propagated error and z dz is the relative error. ⬇ z z dA h dl l dh Δh
39. Answers will vary. Example: 1 x 2 0 43. Proof
r
h
dV
V
V dV
0.1
0.1
8.3776
8.5462
0.1686
0.1
0.1
5.0265
5.0255
0.0010
0.001
0.002
0.1005
0.1006
0.0001
0.0001
0.0002
0.0034
0.0034
0.0000
± 3.92 cubic inches; 0.82% dC ± 2.4418; dC兾C 19%
33. 10% (a) V 18 sin ft3; 兾2 (b) 1.047 ft3 L ⬇ 8.096 104 ± 6.6 106 microhenrys
xy 冪x2 y2
1f 共x, y兲
39. (a) f 共tx, ty兲 etx兾ty e x兾y f 共x, y兲; n 0 xe x兾y xe x兾y (b) xfx共x, y兲 yfy共x, y兲 0 y y 41. 47 43. dw兾dt 共 w兾 x dx兾dt兲 共 w兾 y dy兾dt兲 dy fx共x, y兲 45. dx fy共x, y兲 z fx共x, y, z兲 x fz共x, y, z兲 z fy共x, y, z兲 y fz共x, y, z兲 47. 4608 in.3兾min; 624 in.2兾min 51–55. Proofs
Section 13.6 1. 冪2
3.
49. 28m cm2兾sec
(page 924) 2 冪3 2
5. 1
7.
7 25
9. 6
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Answers to Odd-Numbered Exercises 13. 3i 10j 15. 4i j 2冪5兾5 19. 1 21. 2冪3兾3 23. 3冪2 6i 10j 8k 27. 2关共x y兲i xj兴; 2冪2 8兾冪5 31. ex共yi j兲; 冪26 tan yi x sec2 yj; 冪17 x i yj zk 33. 35. yz共 yzi 2xzj 2xyk兲; 冪33 ;1 冪x 2 y 2 z 2 z 37. 39. (a) 5冪2兾12 (3, 2, 1) 3 (b) 3兾5 (c) 1兾5 6 y (d) 11冪10兾60 11. 17. 25. 29.
9
x
41. 冪13兾6 43. (a) Answers will vary. Example: 4i j 1 1 (b) 25 i 10 j (c) 25 i 10 j The direction opposite that of the gradient z 45. (a)
x y
53. (a) 6 i 4j (d)
y
(b) 共冪13兾13兲共3i 2j兲
A101
(c) y 32 x 12
3 2 1 x
−3 −2 −1
1
2
3
−2 −3
55. The directional derivative of z f 共x, y兲 in the direction of u cos i sin j is f 共x t cos , y t sin 兲 f 共x, y兲 Du f 共x, y兲 lim t→0 t if the limit exists. 57. See the definition on page 918. See the properties on page 919. 59. The gradient vector is normal to the level curves. 1 61. 5h 共5i 12j兲 63. 625 共7i 24 j兲 65. 6i 10j; 11.66 per centimeter 67. y 2 10x 69. True 71. True 73. f 共x, y, z兲 e x cos y 12 z 2 C 75. (a) Proof (b) Proof z (c)
(b) Du f 共4, 3兲 8 cos 6 sin
3
Du f 12 8 −2
4
π
−4
−1
θ
2π
2
y
2
−8 x
−12
Generated by Mathematica
Section 13.7
(c) ⬇ 2.21, ⬇ 5.36 Directions in which there is no change in f (d) ⬇ 0.64, ⬇ 3.79 Directions of greatest rate of change in f (e) 10; Magnitude of the greatest rate of change y (f) 6 4 2
x
−6 − 4
−2
2
4
6
−4 −6
Generated by Mathematica
Orthogonal to the level curve 47. 2i 3j 49. 3i j 51. (a) 16 i j (b) 共冪257兾257兲共16i j兲 (c) y 16x 22 y (d)
−15 −10 −5
x −5
−10
5
10
15
(page 933)
1. The level surface can be written as 3x 5y 3z 15, which is an equation of a plane in space. 3. The level surface can be written as 4x 2 9y2 4z2 0, which is an elliptic cone that lies on the z-axis. 1 1 5. 13 7. 13 共3i 4j 12k兲 共4i 3j 12k兲 9. 4x 2y z 2 11. 3x 4y 5z 0 13. 2x 2y z 2 15. 3x 4y 25z 25共1 ln 5兲 17. x 4y 2 z 18 19. 6x 3y 2z 11 21. x y z 9 23. 2x 4y z 14 x1 y2 z4 x3y3z3 2 4 1 25. 6x 4y z 5 27. 10x 5y 2z 30 x3 y2 z5 x1 y2 z5 6 4 1 10 5 2 29. x y 2z 兾2 共x 1兲 共 y 1兲 z 共兾4兲 1 1 2 x1 y1 z1 1 31. (a) (b) , not orthogonal 1 1 1 2 x3 y3 z4 16 33. (a) (b) , not orthogonal 4 4 3 25 x3 y1 z2 35. (a) (b) 0, orthogonal 1 5 4 37. 86.0 39. 77.4 41. 共0, 3, 12兲 43. 共2, 2, 4兲 45. 共0, 0, 0兲 47. Proof 49. (a) Proof (b) Proof
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A102
Answers to Odd-Numbered Exercises
51. 共2, 1, 1兲 or 共2, 1, 1兲 53. Fx共x0, y0, z0兲共x x0兲 Fy共x0, y0, z0兲共 y y0兲 Fz共x0, y0, z0兲共z z0兲 0 55. Answers will vary. 57. (a) Line: x 1, y 1, z 1 t Plane: z 1 6 (b) Line: x 1, y 2 25 t, z 45 t Plane: 6y 25z 32 0 z z (c) 1
1
2 x
y
−1
−1
2
−2
2 y
−4
67. Proof
3
4
−4 4
y
5
x
−2 1
x
z
(b)
4
−2
y
59. (a) x 1 t y 2 2t z4 ⬇ 48.2
z
P1
(page 942)
Section 13.8 2
x
f P2
−2
2
3
(e)
−4
x2 y2 z2 1 a2 b2 c2 2x Fx共x, y, z兲 2 a 2y Fy共x, y, z兲 2 b 2z Fz共x, y, z兲 2 c 2y0 2z 0 2x0 Plane: 2 共x x0 兲 2 共 y y0 兲 2 共z z 0 兲 0 a b c x 0 x y0 y z 0 z 2 2 1 a2 b c 63. F共x, y, z兲 a 2 x 2 b 2 y 2 z2 Fx共x, y, z兲 2a2x Fy共x, y, z兲 2b2 y Fz共x, y, z兲 2z Plane: 2a 2x 0共x x0 兲 2b 2 y0共 y y0 兲 2z 0共z z 0 兲 0 a2x0 x b2 y0 y z 0 z 0 Therefore, the plane passes through the origin. 65. (a) P1共x, y兲 1 x y (b) P2共x, y兲 1 x y 12 x 2 xy 12 y 2 (c) If x 0, P2共0, y兲 1 y 12 y 2. This is the second-degree Taylor polynomial for ey. If y 0, P2共x, 0兲 1 x 12 x 2. This is the second-degree Taylor polynomial for e x. (d) x y f 共x, y兲 P1共x, y兲 P2共x, y兲 61. F共x, y, z兲
0
0
1
1
1
0
0.1
0.9048
0.9000
0.9050
1. Relative minimum: 共1, 3, 0兲 5. Relative minimum: 共1, 3, 4兲 9. Saddle point: 共0, 0, 0兲 13. Relative minimum: 共3, 4, 5兲 17. Saddle point: 共1, 1, 1兲 z 21.
6 5
−4 4
1.1052
1.1000
1.1050
0.2
0.5
0.7408
0.7000
0.7450
1
0.5
1.6487
1.5000
1.6250
−4
y
−4
5
x
−4 x
4
4
y
Relative maximum: 共1, 0, 2兲 Relative minimum: 共0, 0, 0兲 Relative minimum: 共1, 0, 2兲 Relative maxima: 共0, ± 1, 4兲 Saddle points: 共± 1, 0, 1兲 25. z is never negative. Minimum: z 0 when x y 0. z 60 40
3 3
x
y
27. Insufficient information 29. Saddle point 31. 4 < fxy 共3, 7兲 < 4 33. (a) 共0, 0兲 (b) Saddle point: 共0, 0, 0兲 (c) 共0, 0兲 z (d) 2
2
y
0.1
z
23.
4
x
0.2
3. Relative minimum: 共0, 0, 1兲 7. Relative maximum: 共40, 40, 3200兲 11. Relative maximum: 共12, 1, 314 兲 15. Relative minimum: 共0, 0, 0兲 19. There are no critical numbers.
−2
1 2
−2
Saddle point (0, 0, 0) −2
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A103
Answers to Odd-Numbered Exercises 35. (a) 共1, a兲, 共b, 4兲 (b) Absolute minima: 共1, a, 0兲, 共b, 4, 0兲 (c) 共1, a兲, 共b, 4兲 (d) z
1. 冪3 3. 冪7 5. x y z 3 7. 10, 10, 10 9. 9 ft 9 ft 8.25 ft; $26.73 11. Let x, y, and z be the length, width, and height, respectively, and let V0 be the given volume. Then V0 xyz and z V0兾xy. The surface area is S 2xy 2yz 2xz 2共xy V0兾x V0兾y兲. Sx 2共 y V0兾x2兲 0 x 2 y V0 0 Sy 2共x V0兾y2兲 0 xy 2 V0 0
6
4
y
−2
x
冧
4
2
6
−4
Absolute minimum (b, −4, 0)
Absolute minimum (1, a, 0)
13. 17. 19.
6
21. 25. 4
2
2
4
3 So, x V0, y V0, and z 冪 V0. 15. Proof x 1 3; x 2 6 x 冪2兾2 ⬇ 0.707 km y 共3冪2 2冪3兲兾6 ⬇ 1.284 km Write the equation to be maximized or minimized as a function of two variables. Take the partial derivatives and set them equal to zero or undefined to obtain the critical points. Use the Second Partials Test to test for relative extrema using the critical points. Check the boundary points. (a) y 34 x 43 (b) 16 23. (a) y 2x 4 (b) 2 37 7 945 27. y 175 y 43 x 43 x 148 148 3 冪
37. (a) 共0, 0兲 (b) Absolute minimum: 共0, 0, 0兲 (c) 共0, 0兲 z (d)
6
6
y
x
(page 949)
Section 13.9
3 冪
8
7 y = 37 x + 43 43
7
Absolute minimum (0, 0, 0)
(0, 6)
(5, 5)
39. Relative minimum: 共0, 3, 1兲 41. Absolute maximum: 43. Absolute maximum: 共0, 1, 10兲 共4, 0, 21兲 Absolute minimum: Absolute minimum: 共1, 2, 5兲 共4, 2, 11兲 45. Absolute maxima: 47. Absolute maxima: 共± 2, 4, 28兲 共2, 1, 9兲, 共2, 1, 9兲 Absolute minimum: Absolute minima: 共0, 1, 2兲 共x, x, 0兲, x 1 49. (a) See the definition on page 936. (b) See the definition on page 936. (c) See the definition on page 937. (d) See the definition on page 939. 51. Answers will vary. Sample answer:
ⱍⱍ
z
(4, 3)
(3, 4)
−4
(5, 0) (8, − 4)
(4, 2) (1, 1) (0, 0)
−2
10
−6
n
兺x a兺 x a兺 x i1 n
3 i
33. y
n
2 i
2 i
i1 n
i
i
i1
i
6 5x
i
i1
n
i
i1
3 2 7x
2 i i
i1 n
i1 n
2 i
i1
n
3 i
b
i1 n
i1 n
(b) 1.6
兺 x c兺 x 兺 x y b兺 x c兺 x 兺 x y b 兺 x cn 兺 y
4 i
31. a
n
(10, − 5)
y = − 175 x + 945 148 148
−1
29. (a) y 1.6x 84
18
i1
26 35
35. y x 2 x
8
14
(4, 12) (−1, 0)
(2, 5) (3, 6)
(1, 2) −9
(0, 1)
(−2, 0)
6
−5
(0, 0)
7
−2
−2
75
(2, 2)
37. (a) y 0.22x 2 9.66x 1.79 120 (b)
60 45 30
2 x 2
−1
y
No extrema 53. (a) fx 2x 0, fy 2y 0 ⇒ 共0, 0兲 is a critical point. gx 2x 0, gy 2y 0 ⇒ 共0, 0兲 is a critical point. (b) d 2共2兲 0 < 0 ⇒ 共0, 0兲 is a saddle point. d 2共2兲 0 > 0 ⇒ 共0, 0兲 is a relative minimum. 55. False. Let f 共x, y兲 1 x y at the point 共0, 0, 1兲. 57. False. Let f 共x, y兲 x2 y2 (see Example 4 on page 940).
ⱍⱍ ⱍⱍ
14 −20
39. (a) ln P 0.1499h 9.3018 (b) P 10,957.7e0.1499h (c) 14,000 (d) Proof
−2
24
−2,000
41. Proof
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A104
Answers to Odd-Numbered Exercises
(page 958)
Section 13.10
1. f 共5, 5兲 25 3. f 共1, 2兲 5 5. f 共25, 50兲 2600 7. f 共1, 1兲 2 9. f 共3, 3, 3兲 27 11. f 共13, 13, 13 兲 13 13. Maxima: f 共冪2兾2, 冪2兾2兲 5兾2 f 共 冪2兾2, 冪2兾2兲 5兾2 Minima: f 共 冪2兾2, 冪2兾2兲 1兾2 f 共冪2兾2, 冪2兾2兲 1兾2 15. f 共8, 16, 8兲 1024 17. 冪2兾2 19. 3冪2 21. 冪11兾2 23. 2 25. 冪3 27. 共4, 0, 4兲 29. Optimization problems that have restrictions or constraints on the values that can be used to produce the optimal solutions are called constrained optimization problems. 31. 冪3 33. x y z 3 35. 9 ft 9 ft 8.25 ft; $26.73 37. Proof 39. 2冪3a兾3 2冪3b兾3 2冪3c兾3 3 3 3 41. 冪 360 冪 360 43 冪 360 ft 43. r
冪2v and h 2冪2v 0
3
3
0
45. Proof
47. P共15,625兾18,3125兲 ⬇ 226,869 49. x ⬇ 191.3 y ⬇ 688.7 Cost ⬇ $55,095.60 51. Putnam Problem 2, morning session, 1938
(page 960)
Review Exercises for Chapter 13 1. (a) 9 (b) 3 (c) 0 (d) 6x2 3. Domain: 再共x, y兲: x 0 and y 0冎 Range: all real numbers 5. Lines: y 2x 3 c yc =
c=6 c=4 8 c=2 c=0
6
x − 6 −4
2
4
6
z
7. (a)
(b) g is a vertical translation of f two units upward. (c) g is a horizontal translation of f two units to the right.
5 4
z
9. 2
−2 2
y
3
x −2
11. Limit: 12 13. Limit: 0 Continuous except at 共0, 0兲 Continuous 15. fx共x, y兲 15x2 17. fx 共x, y兲 e x cos y fy共x, y兲 7 fy 共x, y兲 e x sin y 3 4x 19. fx共x, y兲 4y e fy共x, y兲 3y2e4x 21. fx共x, y, z兲 2z2 6yz 5y3 fy共x, y, z兲 6xz 15xy2 fz共x, y, z兲 4xz 6xy 23. fxx 共x, y兲 6 fyy 共x, y兲 12y fxy 共x, y兲 fyx 共x, y兲 1 25. hxx 共x, y兲 y cos x hyy共x, y兲 x sin y hxy 共x, y兲 hyx 共x, y兲 cos y sin x 27. Slope in x-direction: 0 Slope in y-direction: 4 29. 共xy cos xy sin xy兲 dx 共x2 cos xy兲 dy 31. dw 共3y2 6x2yz2兲 dx 共6xy 2x 3z 2兲 dy 共4x3yz兲 dz 33. (a) f 共2, 1兲 10 (b) dz 0.5 f 共2.1, 1.05兲 10.5 z 0.5 35. ± cubic inches; 15% 37. dw兾dt 共8t 1兲兾共4t 2 t 4兲 39. w兾 r 共4r 2 t 4rt 2 t 3兲兾共2r t兲2 w兾 t 共4r 2 t rt 2 4r 3兲兾共2r t兲2 41. z兾 x 共2x y兲兾共 y 2z兲 z兾 y 共x 2y z兲兾共 y 2z兲 43. 50 45. 23 47. 具4, 4典, 4冪2 49. 具 12, 0典, 12 27 65 8 27 51. (a) 54i 16j (b) i j (c) y x 8 8 冪793 冪793 y (d) 6 4
Unit normal vector
2 −6 −4
x −2
4
6
−4 −2
1
2
2
−6
y
Tangent line
x
z
z
(d)
53. 2x 6y z 8 55. z 4 57. Tangent plane: 4x 4y z 8 Normal line: x 2 4t, y 1 4t, z 4 t 59. ⬇ 36.7 61. Relative maximum: 共4, 1, 9兲 63. Relative minimum: 共4, 43, 2兲 65. Relative minimum: 共1, 1, 3兲
5
5
4
4
z = f (1, y)
z = f (x, 1) 2
2 x
2
y
x 2 y
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A105
Answers to Odd-Numbered Exercises 67. 73. 75. 81.
456 69. x1 2, x2 4 71. y 161 226 x 113 (a) y 0.138x 22.1 (b) 46.25 bushels per acre 77. f 共15, 7兲 352 79. f 共3, 6兲 36 f 共4, 4兲 32 x 冪2兾2 ⬇ 0.707 km; y 冪3兾3 ⬇ 0.577 km; z 共60 3冪2 2冪3兲6 ⬇ 8.716 km
冪3
P.S. Problem Solving
(page 963)
1. (a) 12 square units (b) Proof (c) Proof 3. (a) y0 z0共x x0兲 x0z 0共 y y0兲 x0 y0共z z0兲 0 (b) x0 y0 z0 1 ⇒ z 0 1兾x0 y 0 Then the tangent plane is 1 1 1 y0 共x x0兲 x0 共 y y0兲 x0 y0 z 0. x0 y0 x0 y0 x0 y0
冢 冣
冢 冣
冢
冢
Intercepts: 共3x0, 0, 0兲, 共0, 3y0, 0兲, 0, 0, 5. (a)
y
k=0 k=1 k=2
(b)
y
g(x, y)
冣
z
(b) 1
1
1
y
2
2
x
x
Minimum: 共0, 0, 0兲 Maxima: 共0, ± 1, 2e1兲 Saddle points: 共± 1, 0, e1兲 (c) > 0 Minimum: 共0, 0, 0兲 Maxima: 共0, ± 1, e1兲 Saddle points: 共± 1, 0, e1兲 6 cm 15. (a)
y
2 −1
Minima: 共± 1, 0, e1兲 Maxima: 共0, ± 1, 2e1兲 Saddle point: 共0, 0, 0兲 < 0 Minima: 共± 1, 0, e1兲 Maxima: 共0, ± 1, e1兲 Saddle point: 共0, 0, 0兲
1 cm
k=0 k=1 k=2
2
g(x, y)
1 −1
3 x0 y0
冣
z
13. (a)
x −1
1
3
4
−2 −1
1
−1 −2
k=3 −3
−3
−4
−4
Maximum value: 2冪2
1
2
3
4
冣
Maximum and minimum value: 0 The method of Lagrange multipliers does not work because g共x0, y0兲 0.
冢
1 cm
k=3
3 150 2 冪 3 150 5 冪 3 150兾3 7. 2冪 f f 9. (a) x y xCy1aax a1 yCx a共1 a兲y1a1 x y ax aCy1a 共1 a兲x aC共 y1a兲 Cx ay1a关a 共1 a兲兴 Cx ay1a f 共x, y兲 (b) f 共tx, ty兲 C共tx兲a共ty兲1a Ctx ay1a tCx ay1a t f 共x, y兲 11. (a) x 32冪2t y 32冪2t 16t2 y 32冪2t 16t2 (b) arctan arctan x 50 32冪2t 50
冢
6 cm
(b) x
(c) Height (d) dl 0.01, dh 0: dA 0.01 dl 0, dh 0.01: dA 0.06 17–21. Proofs
Chapter 14 1. 7. 13. 25. 33. 45.
3. y ln共2y兲 5. 共4x 2 x 4兲兾2 2x 2 2 2 2 2 9. x2共1 ex x 2ex 兲 11. 3 共 y兾2兲 关共ln y兲 y 兴 8 1 1 15. 2 17. 2 19. 3 21. 1629 23. 23 3 4 27. 兾2 29. 共 2兾32兲 共1兾8兲 31. 12 16 8 Diverges 35. 24 37. 3 39. 3 41. 5 43. 92 y y 47. 3 3 1
2 1
−2
冕冕 4
0
1
2
3
1
2
−1
4
冕冕
4
冪4y2
2
f 共x, y兲 dy dx
0
x
y
49.
x
−1
x
冣
16共8冪2t2 25t 25冪2兲 d (c) dt 64t 4 256冪2t 3 1024t2 800冪2t 625 (d) 30 No; The rate of change of is greatest when the projectile is closest to the camera.
(page 972)
Section 14.1
冪4y2
f 共x, y兲 dx dy
y
51. 4
8
3
6 2
4 0
4
2 −5
(e) is maximum when t 0.98 second. No; the projectile is at its maximum height when t 冪2 ⬇ 1.41 seconds.
x 1
冕 冕 ln 10
0
2
3
10
ex
−2
冕冕 1
f 共x, y兲 dy dx
−1
0
x 1
2
冪y
冪y
f 共x, y兲 dx dy
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A106
Answers to Odd-Numbered Exercises
y
53.
冕冕 1
3
2
0
dy dx
0
y
67.
冕冕 2
1
0
dx dy 2
3
0 2
2
1
1
x
x
1
2
3
冕冕 1
y
55. 1
x
冪1y 2
0
2
1
冕冕
冪1x 2
1
dx dy
4e y dy dx e4 1 ⬇ 53.598 y
1
冪1y 2
3
2x
69.
冕冕 1
2
2
0
−1
1
2
dy dx
1 0
x
1
y
57.
冕冕
3
1
2
0
1
2
3
y
1 sin x2 dx dy 共1 cos 1兲 ⬇ 0.230 2 73. 共ln 5兲2
71. 1664 105 79. (a)
x 1
1
4
77. 15兾2
75. 20.5648
y
−1
冕冕 2
0
x
冕冕 4
dy dx
0
2
4x
dy dx
0
0
y
59.
冕冕 2
2
0
4
冕冕 2
x = y3
4y
dx dy 4
y
1
x 2
冕冕 1
dy dx
x兾2
−2
2y
0
dx dy 1
0
1
冕冕 0
x
61.
2
x= 3 y
冕冕
y
1
x = y2
2
0
y
3 冪 y
2
dx dy
冕冕 1
0
冪x
x3
dy dx
5 12
4
6
3 x 冪
x 2兾32
共x 2 y xy 2兲 dy dx
(1, 1)
(page 983)
1. 24 (approximation is exact) 3. Approximation: 52; Exact: 160 3 5. y 7.
x
1
2
63. The first integral arises using vertical representative rectangles. The second two integrals arise using horizontal representative rectangles. Value of the integrals: 15,625兾24 65. y 2 2 26 3 x冪1 y3 dy dx 9 0 x
4
2
1
2 x
1
2
x
3
2
8
4
6
36 y
9. a
x
2
(3, 6) 6
2
1
y
3
冕冕
1
(c) 67,520兾693
81. An iterated integral is an integral of a function of several variables. Integrate with respect to one variable while holding the other variables constant. 83. If all four limits of integration are constant, the region of integration is rectangular. 85. True
Section 14.2 1
8
x = 4 2y
8
(b)
1
(8, 2)
2
3
−a
a
x
−a
0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A107
Answers to Odd-Numbered Exercises
冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 3
11.
5
xy dy dx
0 0 5 3
1 x 2 y
0 4
y dx dy x2 y2
4
3x兾4
0
0
0
x
21. 4
23. 1 xy dy dx 8
0
3 8
0
共2x x2 y2兲 dy dx
33. 2
0 0 2 冪4x 2
共x2 y2兲 dy dx
35. 4
0 0 2 冪22共y1兲 2
0
冪22共y1兲 2
39. 81兾2 y 45.
arccos y
0
sin x冪1 sin2 x dx dy
0
1 共2冪2 1兲 3
8 51. 2 53. 3 55. 共e 1兲2 57. 25,645.24 59. See “Definition of Double Integral” on page 976. The double integral of a function f 共x, y兲 0 over the region of integration yields the volume of that region. 61. No; 6 is the greatest possible value. 63. Proof; 51 7 65. Proof; 27 67. 400; 272
冕冕
冪1y 2
1
69. False. V 8
冪1 x2 y2 dx dy
0
71. R: x 2 y 2 9 73. 12共1 e兲 75. Putnam Problem A2, 1989
(page 991)
Section 14.3
2 3
16 共x y兲 dy dx 3
0 2 冪1 共x1兲 2
冕冕 1
x
π
0
25. 1
冪1 x 2 dy dx
0 冪4x 2
2
x dy dx 25
0
0 1 x
29. 2
37.
冪25x 2
4
冕冕 冕冕 冕冕 冕冕 冕冕 冕冕 1
冕冕 5
x dy dx
19. 4
6 5
x dx dy 25
4y兾3
π 2
2
2y dx dy
冪25y2
0
31.
冕冕 4
4y
3
27.
y 1 5 dy dx ln x2 y2 2 2
冪4y
3
17.
y = cos x 1
y 1 5 ln 2 y 2 dx dy 2 x 2 1 2 y兾2 2 4x 6 2y dy dx 5 4x
1 1
15.
2
225 xy dx dy 4
0 0 2 2x
13.
y
49.
225 4
1. Rectangular 3. Polar 5. The region R is a half-circle of radius 8. It can be described in polar coordinates as R 再共r, 兲: 0 r 8, 0 冎. 7. R 再共r, 兲: 4 r 8, 0 兾2冎 9. 兾4 11. 0 π 2
π 2
共4y x2 2y2兲 dx dy
41. 1.2315
43. Proof
0
0 1
4
2
y = 2x 1
15. 共9兾8兲 共3 2兾32兲
13. 5冪5兾6
1 2
π 2
π 2
x 1 2
冕冕 1
0
1
1兾2
ex dx dy 1 e1兾4 ⬇ 0.221 2
y兾2 y
47.
0 1
x2 + y2 = 4
3
−1
1
25. 共兾2兲 sin 1
3
冕 冕 冕 冕 兾2
−3
冕冕 2
19. 4
17. a3兾3 x
−1
冪4x2
2 冪4x2
3
0 1
1 −3
2
29. 冪4 y2 dy dx
64 3
0
冕 冕 兾4
27.
0
r 2 dr d
0
1
r 共cos sin 兲 dr d 2
r dr d
3 2 64
23.
2冪2
2
0 0 兾4 2
31.
21. 243兾10
33.
1 8
3 37. 64 39. 2冪4 2冪 2 9 共3 4兲 43. 3兾2 45.
2
2 3
4冪2 3
16 3 35.
250 3
41. 9
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A108
Answers to Odd-Numbered Exercises π 2
47.
π 2
49.
冢
r = 3 cos θ
r = 2 cos θ
0
0
3
1
r=1
r = 1 + cos θ
冪3 3 2
π 2
51.
37. 2k
b 0
3
39.
53. Let R be a region bounded by the graphs of r g1共 兲 and r g2共 兲 and the lines a and b. When using polar coordinates to evaluate a double integral over R, R can be partitioned into small polar sectors. 55. r-simple regions have fixed bounds for and variable bounds for r.
-simple regions have variable bounds for and fixed bounds for r. 57. 486,788 59. 1.2858 61. 56.051 63. False. Let f 共r, 兲 r 1 and let R be a sector where 0 r 6 and 0 . 65. (a) 2 (b) 冪2
冕冕 冕 冕 冕 冕 4
2
冪3x
f dy dx
(b)
2兾冪3 2
兾3
(c)
兾4
69.
冕 冕 4兾冪3
2 csc
冕 冕 4
f dy dx
x
2
4 csc
冪3x
4兾冪3
fr dr d
7.
1. m 4 3. m 18 5. (a) m ka2, 共a兾2, a兾2兲 (b) m ka3兾2, 共a兾2, 2a兾3兲 (c) m ka3兾2, 共2a兾3, a兾2兲 7. (a) m ka2兾2, 共a兾3, 2a兾3兲 (b) m ka3兾3, 共3a兾8, 3a兾4兲 (c) m ka3兾6, 共a兾2, 3a兾4兲 a a a 2a 9. (a) (b) 5, 5, 2 2 2 3 2共a2 15a 75兲 a (c) , 3共a 10兲 2 11. m k兾4, 共2兾3, 8兾15兲 13. m 30k, 共14兾5, 4兾5兲 1 e1 15. m k共e 1兲, , e1 4 256k 16 2kL L 17. m 19. m , 0, , , 15 7 2 8 k a 2 4冪2a 4a共2 冪2 兲 21. m , , 8 3 3
冢 冢
冣
冣
冢
冣
冢
冢 冢
冣
冣
冢 冣
冣
1 2
5.
共31冪31 8兲
关4冪17 ln共4 冪17兲兴 11. 冪2
9. 冪2 1
冕冕 冕冕 0 3
21.
15. 48冪14 17. 20 x 冪 27 5 5 冪5 4x 2 dy dx ⬇ 1.3183 12 0 冪9x 2
冪1 4x 2 4y 2 dy dx
3 冪9x 2
冕冕 冕冕 冕冕
共37冪37 1兲 ⬇ 117.3187 6
1
冪1 4x 2 4y 2 dy dx ⬇ 1.8616
0 0 4 10
25.
冪1 e2xy共x2 y2兲 dy dx
0 0 2 冪4x 2
冪1 e2x dy dx
2 冪4x 2
29. If f and its first partial derivatives are continuous on the closed region R in the xy-plane, then the area of the surface S given by z f 共x, y兲 over R is
(page 1000)
Section 14.4
4 27
1
19.
27.
4
冢56 6015冣
13. 2a共a 冪a2 b2 兲
4
f dy dx
k b 2 2 共b 4a 2兲 4
(page 1007)
3. 12
1. 24
23.
x
2
31. x a兾2 y a兾2 Ix 16k Iy 512k兾5 I0 592k兾5 x 4冪15兾5 y 冪6兾2
共x a兲2 dy dx
Section 14.5
1
f dx dy
6
6
L L 41. 43. 45. See definitions on page 996. 3 2 47. Answers will vary. 49. Proof
y
y兾冪3
2
冤 ee 5e7 冥冣
ky共y a兲2 dy dx ka5
a 0
4
r=2
67. (a)
冪a2 x 2
a
0 1
冕冕 冕冕
冪b2 x 2
b
4 2冪3 3
r = 4 sin 3θ
k e4 13 8 23. m 共1 5e4兲, 4 , 8 e 5 27 25. m k兾3, 共81冪3兾共40兲, 0兲 27. x 冪3b兾3 29. x a兾2 y 冪3h兾3 y a兾2 33. Ix 32k兾3 35. Iy 16k兾3 I0 16k x 2冪3兾3 y 2冪6兾3
冕冕 R
冪1 关 fx共x, y兲兴2 关 fy共x, y兲兴2 dA.
31. No. The size and shape of the graph stay the same; just the position is changed. So, the surface area does not increase. 33. (a) 812冪609 cm3 (b) 100冪609 cm2 35. 16
Section 14.6 1. 18
3.
冕冕 冕 冕 冕 冕 冕冕 冕 5
11. V 13. V 15. V 17.
256 15
(page 1017) 5. 共15兾2兲共1 1兾e兲
1 10
5x
7. 40 3
9.
324 5
5xy
dz dy dx
0 0 冪6
0 冪6y 2
6x 2 y 2
dz dx dy
冪6 冪6y 2 0 4
冪16x2
冪80x2 y2
4 冪16x2 共x2 y2兲兾2
19. 4a3兾3
21.
dz dy dx 256 15
23. 10
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A109
Answers to Odd-Numbered Exercises z
25.
冕冕冕 1
1
0
−1
dy dz dx
1
0
y
1 x z
27.
冕冕 0
冕
共124z兲兾3
3
3
0
x will be greater than 2, and y and z will be unchanged. x and z will be unchanged, and y will be greater than 0. 3 5 共0, 0, 3h兾4兲 49. 共0, 0, 2 兲 51. 共5, 6, 4 兲 5 (a) Ix 2ka 兾3 55. (a) Ix 256k Iy 2ka 5兾3 Iy 512k兾3 Iz 2ka 5兾3 Iz 256k (b) Ix ka 8兾8 (b) Ix 2048k兾3 Iy ka 8兾8 Iy 1024k兾3 Iz ka 8兾8 Iz 2048k兾3 57. Proof 43. 45. 47. 53.
冪z
1
共124z3x兲兾6
dy dx dz
0
59. 2
y
3
1
1
1x
1
1
共x 2 y2兲冪x 2 y2 z 2 dz dy dx
0
x
冕冕冕 1
1
0
z
冪1y 2
x
dz dy dx
0 0
1 m
(c) Iz 1
31.
冕冕冕 冕冕冕 冕冕冕 冕冕 冕 冕 冕冕 冕冕 冕 冕冕 冕 冕冕 冕 冕冕 冕 冕冕 冕
y
3
0 0 0 1 3 x
xyz dy dz dx,
0
y
冪9x2
33.
3
4
3
冪9y 2
3
1
2
1
0
2zz
0
1
0
1
0
1z
1
冪1x
2 1
xyz dy dx dz,
冪9y 2
1
1
2zz2
1
0
3
1
x
冕 冕冕 冕冕 冕 冕冕 冕 冕 冕冕 冕 冕冕 2
0 2
1 dy dx dz,
0
0
3 37. m 8k, x 2
冕冕冕 冕冕冕 冕冕冕 冕冕冕 b
b
0
1 dy dz dx,
2
0
0
b
x 2 y dz dy dx
0
b
2
xy dz dy dx
0 0 0 b b b
Mxy k
0
0
0 cot csc
arctan共1兾2兲 0
2
a
0
0
兾4
2
0
xyz dz dy dx
3 sin2 cos d d d 0
a 冪a 2 r 2
r 2 cos dz dr d 0
a
2a cos
a sec
0
3 sin2 cos d d d
3 sin2 cos d d d 0
17. 共2a 3兾9兲(3 4兲 19. 兾16 21. 共2a 3兾9兲共3 4兲 2 23. 48k 25. r0 h 兾3 27. 共0, 0, h兾5兲
冕 冕冕
29. Iz 4k
b
Mxz k
0
Spherical:
0 0 b b b
Myz k
4 sec
15. Cylindrical:
xy dz dy dx
0
r 2 cos dz dr d 0
0 r2 arctan共1兾2兲
兾2
0
39. m 128k兾3, z 1
b
41. m k
4
Spherical:
1y
dz dy dx
2
13. Cylindrical:
冪1x
0
y
共1 e9兲兾4
0
0
4
4
64冪3兾3
3
xyz dx dz dy
冪1x
1 冪1x
x
y
2
冪9y 2
1 dy dz dx
0
xyz dz dx dy,
0
0
1z
1 冪1x
0
1 dy dx dz
0
0
7. 共e 4 3兲 z 11.
5. 兾8
4
1y 2
1y
kz共x 2 y 2兲 dz dy dx
3
dx dz dy,
0
0
冪9x 2
4
xyz dx dy dz,
1y 2
0
3
0
kz2 dz dy dx
(page 1025)
52 45
3.
冪9x 2
3
2 冪4x 2 0
z
4
冪9y 2 0
xyz dy dz dx,
dx dy dz,
0
冕冕 冕 冕冕 冕 冕 冕冕 冕冕 冕 冕冕 冕 冕冕 冕 冪9y 2
3
冪9y 2
1z
1. 27 9.
y
4
冪9x 2
3
0
4x 2 y 2
2 冪4x 2 0 冪4x 2 2 4x 2 y 2
Section 14.7
xyz dx dz dy
3
0
1
xyz dy dx dz,
3
冪9x 2
4
xyz dz dx dy,
xyz dz dy dx,
3
0
35.
0
4
冪9x2 0
3
3
0 0 0 1 3 1
xyz dx dy dz,
3
1
0 y 0 3 1 x
0 0 0 3 1 1 0
冕冕冕 冕冕冕 冕冕冕 1
xyz dz dy dx,
冪4x 2
2
63. 13 65. 32 3 67. See “Definition of Triple Integral” on page 1009 and Theorem 14.4, “Evaluation by Iterated Integrals,” on page 1010. 69. (a) 71. Q: 3z 2 y 2 2x 2 1; 4冪6 兾45 ⬇ 0.684 73. Putnam Problem B1, 1965
1
x
x
kz dz dy dx
2 冪4x 2 0
(b) x y 0, by symmetry.
z
1
4x 2 y 2
冪4x 2
2
61. (a) m
4
29.
冕冕冕 冕冕 冕 冕冕 冕 冕冕 冕
兾2
0
r0
0
h 共r0 r兲兾r0
0
r 3 dz dr d 3mr02兾10
31. Proof 33. 9冪2 35. 16 2 39. 共0, 0, 3r兾8兲 41. k兾192
37. ka 4
0
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A110
Answers to Odd-Numbered Exercises
43. Rectangular to cylindrical: r2 x2 y2 tan y兾x zz 45.
冕冕 冕 g2共 兲
2
1
h 2 共r cos , r sin 兲
h1共r cos , r sin 兲
g1共 兲
y
11.
Cylindrical to rectangular: x r cos
y r sin
zz
5 4 3 2
f 共r cos , r sin , z兲r dz dr d
1 x −1 −1
47. (a) r constant: right circular cylinder about z-axis
constant: plane parallel to z-axis z constant: plane parallel to xy-plane (b) constant: sphere
constant: plane parallel to z-axis
constant: cone 49. Putnam Problem A1, 2006
1
冕冕 4
2
2
3
5
4
5
冕冕 5
dx dy
1
4
dy dx 8
1
2
6
8
y
13. 10 8 6 4
(page 1032)
Section 14.8 1. 9.
12
3. 1 2v
2
7. e
5. 1
2u
−4 −2
v
冕冕 冕冕 4
15.
0
(1, 0)
19.
u
1
4
8
冕冕 冕冕 8
dy dx
dx dy 16
0
0
4
2
4xy dy dx
0
40 3
y兾2
0
40 3
21.
4xy dx dy 64
17. 21
0
23. 13.67C
25. 共h3兾6兲关ln共冪2 1兲 冪2兴
v
11.
2
0 2x 2 4
(0, 1)
1
x −2
27.
81 5
29. 9兾2
π 2
31. 1
r=3 (1, 0)
(3, 0) u
2 −1
(1, −1)
0
(3, −1)
1
2
−2
13.
冕冕
冕 冕 2兾3
3xy dA
R
冕冕 4兾3
15. 83 17. 36 23. 12共e 4 1兲 31. (a)
3xy dy dx
2兾3 1x
共1兾2兲x2
共1兾2兲x
2兾3
r = 2 + 2 cos θ
共1兾2兲x2
冕冕 8兾3
3xy dy dx
4兾3
9冪3 2
4x
3xy dy dx
共1兾2兲x
19. 共e1兾2 e2兲 ln 8 ⬇ 0.9798 25. 100 27. 25 a 5兾2 29. One 9 y
164 9
33. (a) r 3冪cos 2
4
21. 96 −6
v
1
−4
b S
R x
u
1
a
(b) ab (c) ab 33. See “Definition of the Jacobian” on page 1027. 35. u2 v 37. uv 39. 2 sin 41. Putnam Problem A2, 1994
Review Exercises for Chapter 14 1.
4x5
3.
29 6
5. 36
7.
3 2
9. 16
6
(page 1034)
(c) 3共3 16冪2 20兲 ⬇ 20.392 k 32 64 32k 5 5 35. m 37. m , , , , 5 3 2 4 45 55 39. Ix ka2b3兾6 Iy ka4b兾4 I0 共2ka2b3 3ka4b兲兾12 x a兾冪2 y b兾冪3 (b) 9
冢 冣
41.
共101冪101 1兲
6 45. (a) 30,415.74 ft3
冢
冣
1 共37冪37 1兲 6 (b) 2081.53 ft2 47. 56 43.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A111
Answers to Odd-Numbered Exercises 49.
abc 2 共a b 2 c 2兲 3
51.
8 5
9. 冪3
53. 36
y
11. z 2
z
55.
y
4
1
4
1
x −2
−4
−1
1
2
−1 1
4
y
−4
1
冕冕冕 0
冪1x
1
x
2
2
dz dy dx
0
1
500k 5 57. m ,x 3 2
2 2 61. 3
59. 12冪3
3 冪13 63. 3冪13 4 ln ⬇ 48.995 2 2 2 67. 9 69. sin cos 71. 5 ln 5 3 ln 3 2 ⬇ 2.751 73. 81
冣冥
冢
冤
P.S. Problem Solving
65. 16
(page 1037)
15. 19. 23. 25. 31. 35. 37. 41. 43. 49. 51. 53. 57. 65.
5 4
67. 69. 77. 81.
(0, 0, 0) 2
x
(0, 6, 0)
(3, 3, 0)
6
冕冕 冕 3
0
2x
0
y
6x
83.
dy dz dx 18
x
Chapter 15 Section 15.1 1. d 5. 冪2
2. c
(page 1049) 3. a
4. b
ⱍⱍ
1. r共t兲
z 4
1 x
−4
x
−4
2
4
y
2xi 4yj 17. 共10x 3y兲 i 共3x 2y兲 j 2 2 6yz i 6xz j 6xyk 21. 2xye x i e x j k 关xy兾共 x y兲 y ln共x y兲兴 i 关xy兾共 x y兲 x ln共x y兲兴 j Conservative 27. Conservative 29. Conservative Not conservative 33. Conservative: f 共x, y兲 xy K Conservative: f 共x, y兲 x 2 y K Not conservative 39. Not conservative Conservative: f 共x, y兲 ex cos y K 45. 2k 47. 2x兾共x 2 y 2兲 k 4i j 3k cos共 y z兲 i cos共z x兲 j cos共x y兲k Conservative: f 共x, y, z兲 12共x2y2z2兲 K Not conservative 55. Conservative: f 共x, y, z兲 xz兾y K 59. cos x sin y 2z 61. 4 63. 0 2x 4y See “Definition of Vector Field” on page 1040. Some physical examples of vector fields include velocity fields, gravitational fields, and electric force fields. See “Definition of Curl of a Vector Field” on page 1046. 71. z j yk 73. 3z 2x 75. 0 9x j 2yk (a)–(h) Proofs 79. True False. Curl f is meaningful only for vector fields, when direction is involved. M 3mxy共x2 y2兲5兾2 M兾y 3mx共x2 4y2兲兾共x2 y2兲7兾2 N m共2y2 x2兲共x2 y2兲5兾2 N兾x 3mx共x2 4y2兲兾共x2 y2兲7兾2 Therefore, N兾x M兾y and F is conservative.
Section 15.2 7. 3 y
y
2
x
6
(3, 3, 6)
1
1 2
1 1. 8共2 冪2兲 3. 3 5. (a)–(g) Proofs 7. The results are not the same. Fubini’s Theorem is not valid because f is not continuous on the region 0 x 1, 0 y 1. 9. 冪兾4 11. If a, k > 0, then 1 ka2 or a 1兾冪k. 13. Answers will vary. 15. The greater the angle between the given plane and the xy-plane, the greater the surface area. So z2 < z1 < z4 < z3. z 17.
3
z
13.
x
1
−2
x
y
(page 1061)
冦ti共2 ttj,兲i
冦
0 t 1 1 t 2
冪2 tj,
t i, 3i 共t 3兲j, 3. r共t兲 共9 t兲i 3j, 共12 t兲j,
0 3 6 9
t t t t
3 6 9 12
5. r共t兲 3 cos t i 3 sin tj, 0 t 2 7. 20 9. 5兾2 11. (a) C: r共t兲 t i t j, 0 t 1 (b) 2冪2兾3
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A112
Answers to Odd-Numbered Exercises
13. (a) C: r共t兲 cos t i sin t j, 0 t 兾2 15. (a) C: r共t兲 t i, 0 t 1 (b) 1兾2
(b) 兾2
冦
t i, 0 t 1 17. (a) C: r共t兲 共2 t兲i 共t 1兲j, 1 t 2 2 t 3 共3 t兲j, (b)
19 6
共1 冪2 兲
冦
(b)
23 6
9 1 共k兾12兲共41冪41 27兲 27. 1 29. 2 31. 4 About 249.49 35. 66 37. 0 39. 10 2 Positive 43. Zero (a) 236 3 ; Orientation is from left to right, so the value is positive. (b) 236 3 ; Orientation is from right to left, so the value is negative. 47. F共t兲 2t i tj r 共t兲 i 2 j F共t兲 r 共t兲 2t 2t 0
C
F dr 0
49. F共t兲 共t 3 2t2兲 i 共t t 2兾2兲 j r 共t兲 i 2tj F共t兲 r 共t兲 t 3 2t 2 2t 2 t 3 0
冕
C
F dr 0
61. 69.
1
t 2 dt
0
C
87. 12
85. False. The orientations are different.
(page 1072)
冕 冕 冕 冕冤 1
1. (a)
共t 2 2t 4兲 dt 11 15
0 兾2
(b)
共sin2 cos 2 sin4 cos 兲 d 11 15
0 兾3
3. (a)
共sec tan2 sec3 兲 d ⬇ 1.317
0
冥
3
冪t 冪t 1 dt ⬇ 1.317 2冪t 1 2冪t Conservative 7. Not conservative Conservative 11. (a) 1 (b) 1 (a) 0 (b) 13 (c) 12 (a) 64 (b) 0 (c) 0 (d) 0 17. (a) 64 (b) 64 3 3 (a) 32 (b) 32 21. (a) 32 (b) 17 23. (a) 0 (b) 0 6 72 27. 1 29. 0 31. (a) 2 (b) 2 (c) 2 11 35. 30,366 37. 0
(b)
0
5. 9. 13. 15. 19. 25. 33.
冕
50
51. 1010 316 3 1 120
冕
xy ds 冪2
Section 15.3
23. 2 2
25. 33. 41. 45.
冕
冕
83. False.
t i, 0 t 1 19. (a) C: r共t兲 i t k, 0 t 1 i tj k, 0 t 1 21. 8冪5 共1 4 2兾3兲 ⬇ 795.7
77. 1750 ft-lb 79. See “Definition of Line Integral” on page 1052 and Theorem 15.4, “Evaluation of a Line Integral as a Definite Integral,” on page 1053. 81. z 3, z 1, z 2, z 4; The greater the height of the surface over the curve y 冪x, the greater the lateral surface area.
53.
190 3
55. 25
63. 5h
65.
共25冪5 11兲
1 2
57.
59. 11 6
63 2
67. 共h兾4兲 关2冪5 ln共2 冪5 兲兴
71. (a) 12 ⬇ 37.70 cm2 (b) 12兾5 ⬇ 7.54 cm3 z (c) 4
−3 3
y
x
73. Ix Iy a 3 75. (a)
z
Section 15.4
2 1
3
4
4
y
x
(b) 9 cm2 ⬇ 28.274 cm2 (c) Volume 2
冕
3
0
冕
1 (b) dr 共i 25 共50 t兲j兲 dt ⇒ 7
冢
3
3
175 dt 8750 ft-lb
0
50
共50 t兲 dt
0
8750 ft-lb 41. See Theorem 15.5, “Fundamental Theorem of Line Integrals,” on page 1066. 43. (a) 2 (b) 2 (c) 2 (d) 0 45. Yes, because the work required to get from point to point is independent of the path taken. 47. False. It would be true if F were conservative. 49. True 51. Proof 53. (a) Proof (b) (c) (d) 2 ; does not contradict Theorem 15.7 because F is not continuous at 共0, 0兲 in R enclosed by C. x 1兾y x兾y2 (e) arctan i j 2 y 1 共x兾y兲 1 共x兾y兲2
5
3
39. (a) dr 共i j兲 dt ⇒
冤
2冪9 y 2 1 4
27兾2 ⬇ 42.412 cm3
冢
y2 y2 1 9 9
冣冥 dy
1. 11. 21. 29. 33. 39. 43.
1 30 4 3
冣
(page 1081) 9
3. 0 5. About 19.99 7. 2 9. 56 1 13. 0 15. 0 17. 12 19. 32 23. 225 25. a2 27. 92 2 See Theorem 15.8 on page 1075. 31. Proof 共0, 85 兲 35. 共158 , 218 兲 37. 3a 2兾2 41. (a) 51兾2 (b) 243兾2 3冪3兾2 46
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Answers to Odd-Numbered Exercises
冕
F dr
冕
M dx N dy
冕冕冢
z
(c)
N M dA 0; x y C C R I 2 when C is a circle that contains the origin. 47–51. Proofs 45.
冣
A113
9
3 3
(page 1091)
Section 15.5
y
x
1. e 2. f 7. y 2z 0 Plane
3. b
4. a
5. d 6. c 9. x 2 z2 4 Cylinder
−9 z
(d) 12
z
z 3 2
3
−4 3 4 5
y
5
z
11.
5
5
x
x
12
y x
12
−3 z
13.
3
The radius of the generating circle that is revolved about the z-axis is b, and its center is a units from the axis of revolution. 55. 400 m2 z 57.
9 6
2 1 6
9
y −12
3
6
4π
y
9
x 2 2
x
y 2π
z
15. 5
−4
4
−2
3
−2 2
−1
1
2
−3
3
y
x
The paraboloid is reflected (inverted) through the xy-plane. The height of the paraboloid is increased from 4 to 9. r共u, v兲 ui vj vk r共u, v兲 12u cos vi uj 13u sin vk, u 0, 0 v 2 or r共x, y兲 xi 冪4x2 9y2 j zk 25. r共u, v兲 5 cos ui 5 sin uj vk 27. r共u, v兲 ui vj u2 k 29. r共u, v兲 v cos ui v sin uj 4k, 0 v 3 u u 31. x u, y cos v, z sin v, 0 u 6, 0 v 2 2 2 33. x sin u cos v, y sin u sin v, z u 0 u , 0 v 2 35. x y 2z 0 37. 4y 3z 12 39. 8冪2 41. 2ab 43. ab 2冪a 2 1 45. 共兾6兲 共17冪17 1兲 ⬇ 36.177 47. See “Definition of Parametric Surface” on page 1084. 49–51. Proofs z z 53. (a) (b) 17. 19. 21. 23.
4
4
y
−4
(page 1104)
Section 15.6
1. 12冪2 3. 2 5. 27冪3兾8 冪 7. 共391 17 1兲兾240 9. About 11.47 364 3
13. 12冪5 15. 8 17. 冪3 21. 486 23. 43 25. 3兾2 32兾3 29. 384 31. 0 33. Proof 20 37. 64 2a3h See Theorem 15.10, “Evaluating a Surface Integral,” on page 1094. 41. See “Definition of Flux Integral” on page 1100; see Theorem 15.11, “Evaluating a Flux Integral,” on page 1100. z 43. (a) 11. 19. 27. 35. 39.
4 −6
−6 6 −4
6 6
2 关32冪13 2 ln共3 冪13 兲 2 ln 2兴 59. Answers will vary. Sample answer: Let x 共2 u兲共5 cos v兲 cos 3 u y 共2 u兲共5 cos v兲 sin 3 u z 5u 共2 u兲 sin v where u and v .
x
4
−6
−6
x
2
4 x
3
−3
−2
y
x
6
6
y
6
y
(b) No. If a normal vector at a point P on the surface is moved around the Möbius strip once, it will point in the opposite direction.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A114
Answers to Odd-Numbered Exercises z
(c)
z
51.
4
6
−2 2 2
x
−4 y
2
1. 3. 9. 21. 23.
z
53. (a) 3
−4 −4 4
4
(page 1112)
−3 z
(b) 3
−4 −4 −3 4
(page 1119)
3
2
−1
2
4
3
−3
2
4
y
−4
3
−2
(page 1120)
−3 z
(d) 3
−4
4
−2
1
−4 −3
2
3
2 −2
x
3
4
y
−3
2 3
21. 27. 31. 39. 43. 45.
−2
−4 −3
z
19.
y
3
3
17.
4
z
(c)
1. 冪x2 5
3. 5. 7. 9. 11. 13. 15.
3
−3
x
x
2
−2
x
共xz ez兲 i 共 yz 1兲 j 2k 2 2 2 2 5. 18 7. 0 z共x 2e y z 兲i yzj 2ye x y k 11. 2 13. 0 15. 83 17. a 5兾4 19. 0 12 See Theorem 15.13, “Stokes’s Theorem,” on page 1114. Proof 25. Putnam Problem A5, 1987
Review Exercises for Chapter 15
y
−2
x
1. a 4 3. 18 5. 7. 3a 4 9. 0 11. 108 13. 0 15. 18共e4 5兲 17. 0 19. See Theorem 15.12, “The Divergence Theorem,” on page 1106. 21–27. Proofs
Section 15.8
y
4
−2
x
Circle (d) Construction (e) A strip with a double twist that is twice as long as the Möbius strip.
Section 15.7
2
4
−4
4
Circle (e) About 14.436 (f) About 4.269
y
共4x y兲 i xj 2zk Conservative: f 共x, y兲 y兾x K Conservative: f 共x, y兲 12 x2y2 13 x3 13 y3 K Not conservative Conservative: f 共x, y, z兲 x兾共 yz兲 K (a) div F 2x 2xy x2 (b) curl F 2xz j y2k (a) div F y sin x x cos y xy (b) curl F xz i yz j 1 (a) div F 2xy 2yz 冪1 x2 2 2 (b) curl F z i y k 2x 2y 2x 2y (a) div F 2 1 (b) curl F 2 k x y2 x y2 125 (a) 3 (b) 2 23. 6 25. (a) 18 (b) 18 29. 共冪5兾3兲共19 cos 6兲 ⬇ 13.446 9a 2兾5 1 33. 2 2 35. 36 37. 43 8 41. 6 3 共3 4冪2 兲 ⬇ 7.085 (a) 15 (b) 15 (c) 15 1 47. 0 49. 0
z
55. 2
−3
−3 3
3
x
y
−2
0 57. 66
59. 2a 6兾5
61. Proof
P.S. Problem Solving 1. (a) 共25冪2兾6兲 k
共
兲
(page 1123)
(b) 共25冪2兾6兲 k
3. Ix 冪13兾3 共27 32 2兲; Iy 共冪13兾3兲 共27 32 2兲; Iz 18冪13 5. (a)–(d) Proofs 7. 3a2 5 13 9. (a) 1 (b) 15 (c) 2 11. Proof 13. (a)–(b) Proofs
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A115
Answers to Odd-Numbered Exercises
Section 16.2 (page 1139)
Chapter 16 Section 16.1 (page 1131)
1.
Exact; My ⫽ 2xy ⫽ Nx Not exact; My ⫽ x cos y, Nx ⫽ cos y x2 ⫺ 3xy ⫹ y2 ⫽ C 7. 3xy2 ⫹ 5x2y2 ⫺ 2y ⫽ C Not exact 11. arctan共x兾y兲 ⫽ C 13. Not exact 11 15. (a) Answers will vary. (b) x2 tan y ⫹ 5x ⫽ 4 y 4 (c) 1. 3. 5. 9.
−5
5
y2 −5
y approaches zero as x → ⬁. 4 3. y1: C1 ⫽ 1, C2 ⫽ ⫺1 y3 y y2 1 y2: C1 ⫽ ⫺1, C2 ⫽ 1 y − 3: C1 ⫽ 2, C2 ⫽ 3 6
−4
x −4
−4
−4
17. y ln共x ⫺ 1兲 ⫹ ⫽ 16 19. e sin 3y ⫽ 0 21. x 2 y ⫺ 3x3 ⫹ y2 ⫹ y ⫽ 6 23. Integrating factor: 1兾y2 25. Integrating factor: 1兾x2 y x ⫺ 6y ⫽ C ⫹ 5x ⫽ C y x 27. Integrating factor: cos x y sin x ⫹ x sin x ⫹ cos x ⫽ C 29. Integrating factor: 1/y 31. Integrating factor: 1兾冪y xy ⫺ ln y ⫽ C x冪y ⫹ cos 冪y ⫽ C 2 y x 33. x 4 y3 ⫹ x2y 4 ⫽ C 35. 37. Proof ⫹ 2⫽C x y 39. x2 ⫹ y2 ⫽ C 41. 2x2y4 ⫹ x2 ⫽ C y2
3x
ⱍⱍ
4
c=4
2
c=4
c=9
−6
c=2
43. x ⫺ 2xy ⫹
3y2
冢
⫽3
53.
2
−1
5
−1
5
−1
(b) y2共2x2 ⫹ y2兲 ⫽ 9 2 49. (a)
(c)
59. 61. 63.
35. y ⫽ 12 sin 4x e ⫺ 3 3x 3 ⫺ e3 x 39. y ⫽ y ⫽ 2e x兾3 ⫹ 13 xe x兾3 e ⫹ e e ⫺ e3 e ⫺ e3 1 No solution 43. y ⫽ 2e7x兾2 ⫹ ⫺ 7/2 ⫺ 2 xe7x兾2 e Answers will vary. See Theorem 16.4 on page 1135. Damped 49. b 50. d 51. c 52. a 冪3 2 1 55. y ⫽ ⫺ cos 4冪3t ⫹ y ⫽ 2 cos 4冪3t sin 4冪3t 3 24 冪12,287 t 冪12,287 冪12,287 t e⫺t兾16 y⫽ cos ⫹ sin 2 16 12,287 16 Proof False. The general solution is y ⫽ C1e3x ⫹ C2xe3x. True 65–67. Proofs
冢
5
−1
冢
冣
冣
冢
冣
Section 16.3 (page 1147)
2
−1
冣
冢
1–3. Proofs −1
(b) y ⫽ 15 sin 10x
3 (c) y ⫽ ⫺cos 10x ⫹ 10 sin 10x
57. −1
冪7x
31. (a) y ⫽ 2 cos 10x
45. 47.
5共x2 ⫹ 冪x 4 ⫺ 1,000,000x 兲 45. C ⫽ x 2 47. (a) (c)
冣
冪7x ⫹ C2 cos 3 3 25. y ⫽ C1e x ⫹ C2e⫺x ⫹ C3 sin x ⫹ C4 cos x 27. y ⫽ C1e x ⫹ C2e2x ⫹ C3e3x 29. y ⫽ C1e x ⫹ e x 共C2 sin 2x ⫹ C3 cos 2x兲
23. y ⫽ e2x兾3 C1 sin
41.
−2
−4
2
21. y ⫽ C1e共3⫹ 冪5兲x兾2 ⫹ C2e共3⫺ 冪5兲x兾2
37.
3
c=1
5. 9. 13. 17. 19.
The graphs are basically the same shape, with left and right shifts and varying ranges. 7. y ⫽ C1e3x ⫹ C2e⫺2x y ⫽ C1 ⫹ C2ex x兾2 ⫺2x 11. y ⫽ C1e⫺3x ⫹ C2xe⫺3x y ⫽ C1e ⫹ C2e x兾4 x兾4 15. y ⫽ C1 sin x ⫹ C2 cos x y ⫽ C1e ⫹ C2xe y ⫽ C1e3x ⫹ C2e⫺3x y ⫽ ex 共C1 sin 冪3 x ⫹ C2 cos 冪3 x兲
1 6x 33. y ⫽ 11 共e ⫹ 10e⫺5x兲
c=6
−3
6
y1: C1 ⫽ 0, C2 ⫽ 1 y2: C1 ⫽ 1, C2 ⫽ 1 y3: C1 ⫽ ⫺1, C2 ⫽ ⫺2
y1
4
−6
5
y3
5
−1
(b) y2共2x2 ⫹ y2兲 ⫽ 9 Less accurate 51. See Theorem 16.1 on page 1126. ⭸M ⭸N 53. False. 55. True 57. k ⫽ 3 ⫽ 2x, ⫽ ⫺2x ⭸y ⭸x 59. f 共x兲 ⫽ ⫺cos x ⫹ C1, g共 y兲 ⫽ 13 y3 ⫹ C2
1 1 5. yp ⫽ 4 x ⫺ 16
7. yp ⫽ e3x
4 1 9. yp ⫽ ⫺ 65 sin x ⫹ 130 cos x 3 11. y ⫽ C1ex ⫹ C2e2x ⫹ x ⫹ 2
2 13. y ⫽ C1 ⫹ C2e⫺2x ⫹ 3e x
3 1 15. y ⫽ 共C1 ⫹ C2 x兲e5x ⫹ 8e x ⫹ 5 x 17. y ⫽ C1 ⫺ cos 3x ⫹ C2 sin 3x 6 19. y ⫽ 6 sin x ⫹ cos x ⫹ x3 ⫺ 6x 21. y ⫽ ⫺1 ⫹ 2e⫺x ⫺ sin x ⫺ cos x
冢
冣
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A116
27. 29. 31.
冢49 ⫺ 21 x 冣e
1 共1 ⫹ 3x兲e x 9 y ⫽ 共C1 ⫹ ln cos x 兲 cos x ⫹ 共C2 ⫹ x兲sin x x 1 y ⫽ C1 ⫺ cos 2x ⫹ C2 ⫹ ln sin 2x sin 2x 2 4 x2e x 共ln x2 ⫺ 3兲 y ⫽ 共C1 ⫹ C2x兲e x ⫹ 4 (a) yp ⫽ Ax2 ⫹ Bx ⫹ C; This is a generalized form of F共x兲 ⫽ x2. (b) Because yh ⫽ C1e4x ⫹ C2e⫺3x, let yp ⫽ Axe4x.
23. y ⫽ 25.
Answers to Odd-Numbered Exercises 2
ⱍ
冢
33. q ⫽ 35. y ⫽
4x
冣
⫺
ⱍ
冢
3 ⫺5t ⫹ 5te⫺5t ⫺ cos 5t兲 25 共e 1 1 4 cos 8t ⫺ 2 sin 8t ⫹ sin 4t
2
2
0
(b)
ⱍ冣
ⱍ
x3 x5 ⫹ ⫹. . . 3 20
17. (a) y ⫽ 2x ⫹
(c) The solution is symmetric about the origin.
12
−4
4
P3(x)
P5(x) −12
冢冣
3x 2x3 12x 4 16x6 120x7 1 ⫹ ⫺ ⫹ ⫺ ;y ⬇ 0.253 1! 3! 4! 6! 7! 4 2x 3x2 2x3 1 21. y ⬇ 3 ⫹ ⫹ ⫹ ;y ⬇ 3.846 1! 2! 3! 3 23–25. Proofs a a a a 27. y ⫽ a0 ⫹ a1x ⫹ 0 x3 ⫹ 1 x 4 ⫹ 0 x6 ⫹ 1 x7 6 12 180 504 19. y ⬇ 1 ⫺
冢冣
(page 1153)
Review Exercises for Chapter 16
⭸M ⭸N ⫽ ⭸y ⭸x 3. Exact; 16xy ⫹ 10x2 ⫹ 4x ⫹ 5y2 ⫹ 4y ⫽ C 5. Exact; ⫺2xy ⫺ 3y2 ⫹ 4y ⫹ x2 ⫺ 10x ⫽ C 7. Not exact 9. (a) Answers will vary. (b) x2 ⫹ y2 ⫺ xy ⫽ 4 y 4 (c)
−2
9 3 1 37. y ⫽ 共32 ⫺ 4t兲e⫺8t ⫺ 32 cos 8t
1. Not exact;
0.3
0
4
4 −0.05
冪5
冢
sin 8t ⫹ ⫺ arctan
Section 16.4 (page 1152) 1–5. Proofs
共⫺3兲k 2k x k k⫽0 2 k! Interval of convergence: 共⫺ ⬁, ⬁兲 ⬁ x2x⫹1 y ⫽ a0 ⫹ a1 k k⫽0 2 共k!兲共2k ⫹ 1兲 Interval of convergence: 共⫺ ⬁, ⬁兲 x2 x4 y ⫽ a0 1 ⫺ ⫹ ⫺. . . 8 128 x3 7x5 ⫹ ⫺. . . ⫹ a1 x ⫺ 24 1920 Taylor’s Theorem: 2x 2x2 10x3 2x4 y⫽2⫹ ⫺ ⫺ ⫹ ⫹. . . 1! 2! 3! 4! 1 ⬇ 2.547 y 2 1 Euler’s Method: y ⬇ 2.672 2 Given a differential equation, assume that the solution is of the form y ⫽ an x n. Then substitute y and its derivatives into the differential equation. You should then be able to determine the coefficients 共a0, a1, . . .兲 for the solution series.
7. y ⫽ a0
9.
11.
⬁
兺
兺
冢
冣
冢
13.
−6
冣
冪5 1 ⬇ sin共8t ⫹ 2.6779兲 4 2 4 41. (a) Undamped (b) Damped (c) b > 52; No oscillations in this case. 43. True 45. (a) No (b) Yes Putnam Problem A3, 1987
39. y ⫽
冣
x
−4
15.
兺
2
−4
4
−4
11. 2xy ⫹ 2x ⫺ 6x ⫺ 3y2 ⫹ 2y ⫽ ⫺4 y2 13. 3x ⫹ ⫽ C 15. xe3y ⫺ ey ⫽ C x 7 17. y1: C1 ⫽ C2 ⫽ 1 y1 y2: C1 ⫽ 1, C2 ⫽ 0 y3: C1 ⫽ 0, C2 ⫽ 1 y2 y3 Answers will vary. 2
−5
5
−1
21. y ⫽ 12e⫺3x ⫹ 32ex
19. y ⫽ e2x ⫺ e⫺x 8
−12
8
12
−6
−8
23. y ⫽
冢冣
冢冣
6
6 0
4e 共e 冢sin 2 ⫺ tan 4 cos 2 冣
⫺x
sin 2x ⫺ tan 4e⫺x cos 2x兲
4
−1
5
−2
25. No; f 共x兲 ⫽ 0.
27. y ⫽ C1 sin x ⫹ C2 cos x ⫺ 5x ⫹ x3
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Answers to Odd-Numbered Exercises 29. y ⫽ 共C1 ⫹ x兲sin x ⫹ C2 cos x 1 31. y ⫽ 共C1 ⫹ C2 x ⫹ 3 x3兲ex
35. y ⫽ 37. y ⫽ 39. y ⫽
11 33. y ⫽ 5 共2e3x ⫹ 3e⫺2x兲 ⫺ 9
17 1 3 cos 2x ⫺ 3 sin 2x ⫹ 3 cos x 1 ⫺x ⫺ 12 ⫺ 27 e ⫺ 19 xe⫺x ⫺ 16 x2e⫺x 1 2 cos 2冪6 t
共
兲
41. (a) (i) y ⫽
⫹
9. (a) 共t兲 ⫽ C1 cos (b) (c) (d) (f) 11. (a) (c)
83 2x 54 e
1 12 24 cos 2t ⫹ 2 sin 2t ⫹ sin t 2 ⫺4 4 ⫺ 2
12
A117
冢冪Lg t冣 ⫹ C sin冢冪Lg t冣; Proof 2
共t兲 ⫽ 0.128 cos关冪39.2共t ⫺ 0.108兲兴 Period ⬇ 1 sec 0.128 (e) 0.358 sec; 0.860 sec ⬘共0.358兲 ⫽ ⫺0.8012; ⬘共0.860兲 ⫽ 0.8012 Critically damped (b) y ⫽ e⫺4t ⫹ 5te⫺4t 2
Answers will vary. 0
10
0
2 0
−12
1 (ii) y ⫽ 2 关共1 ⫺ 6冪2t兲cos共2冪2t兲 ⫹ 3 sin共2冪2t兲兴
13. (a) Overdamped (b) y ⫽ e⫺16t ⫹ e⫺4t (c) 2 Answers will vary.
60
0
14
0
2 0
−60
(iii) y ⫽
冪199 t 冪199 t e⫺t兾5 ⫹ 冪199 sin 199 cos 398 5 5
冤
冥
1
0
8
−1
1 (iv) y ⫽ 2e⫺2t共cos 2t ⫹ sin 2t) 0.6
0
3
−0.2
a0 共a ⫹ a1兲 共x ⫺ 1兲2 ⫹ 0 共x ⫺ 1兲3 2 6 共a ⫹ 2a1兲 共4a0 ⫹ a1兲 ⫹ 0 共x ⫺ 1兲4 ⫹ 共x ⫺ 1兲5 24 120 共5a0 ⫹ 6a1兲 共9a0 ⫹ 11a1兲 ⫹ 共x ⫺ 1兲6 ⫹ 共x ⫺ 1兲7 720 5040 Answers will vary. ⬁ 共⫺1兲k x2k 17. (a) y ⫽ a0 (b) y ⫽ a0 J0共x兲 2k 2 k⫽0 2 共k!兲 19. (a) y ⫽ 16x4 ⫺ 48x2 ⫹ 12 (b) H0共x兲 ⫽ 1 H1共x兲 ⫽ 2x H2共x兲 ⫽ 4x2 ⫺ 2 H3共x兲 ⫽ 8x3 ⫺ 12x H4共x兲 ⫽ 16x4 ⫺ 48x2 ⫹ 12 15. y ⫽ a0 ⫹ a1共x ⫺ 1兲 ⫹
兺
(b) The object comes to rest more quickly. It might not oscillate at all, as in part (iv). (c) The object oscillates more rapidly. (d) Part (ii). The amplitude becomes increasingly large. 43. (a) Only a second derivative is used, so a cosine is unnecessary. (b) yp ⫽ 52 cos x (c) If yp ⫽ A cos x ⫹ B sin x, then yp⬙ ⫽ ⫺A cos x ⫺ B sin x. So it would be more difficult to solve for A and B. ⬁ xn 45. y ⫽ a0 n n⫽0 4 2 2x 4x 4 4x5 1 47. y ⬇ 2 ⫹ ⫹ ⫹ ;y ⬇ 2.063 2! 4! 5! 4
兺
冢冣
P.S. Problem Solving
(page 1155)
1. k ⫽ ⫺5; 6x ⫹ 10y ⫺ 15x y ⫽ C 3. y ⫽ B1eax ⫹ B2e⫺ax; Proof 5. Proof n 2 7. (a) and (b) Proofs (c) a ⫽ , n is an integer. L 3
3
2 2
冢 冣
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Index
A119
Index A Abel, Niels Henrik (1802–1829), 228 Absolute convergence, 622 Absolute maximum of a function, 162 of two variables, 936 Absolute minimum of a function, 162 of two variables, 936 Absolute value, 50 derivative involving, 324 function, 22 Absolute Value Theorem, 588 Absolute zero, 74 Absolutely convergent series, 622 Acceleration, 124, 833, 857 centripetal component of, 846 tangential and normal components of, 845, 846, 859 vector, 845, 859 Accumulation function, 283 Addition of vectors, 750, 760 Additive Identity Property of Vectors, 751 Additive Interval Property, 271 Additive Inverse Property of Vectors, 751 Agnesi, Maria Gaetana (1718–1799), 198 Airy’s equation, 1152, 1156 d’Alembert, Jean Le Rond (1717–1783), 890 Algebraic function(s), 24, 25, 371 derivatives of, 135 Algebraic properties of the cross product, 776 Alternating series, 619 geometric, 619 harmonic, 620, 622, 624 Alternating Series Remainder, 621 Alternating Series Test, 619 Alternative form of the derivative, 101 of the directional derivative, 918 of Green’s Theorem, 1080 of Log Rule for Integration, 328 of Mean Value Theorem, 173 Angle between two nonzero vectors, 767 between two planes, 785 of incidence, 684 of inclination of a plane, 931 of reflection, 684 Angular speed, 999 Antiderivative, 244 of f with respect to x, 245 finding by integration by parts, 515 general, 245 notation for, 245 representation of, 244 of a vector-valued function, 828 Antidifferentiation, 245 of a composite function, 292 Aphelion, 694, 741 Apogee, 694
Approximating zeros bisection method, 78 Intermediate Value Theorem, 77 Newton’s Method, 225 Approximation linear, 231, 902 Padé, 395 polynomial, 636 Stirling’s, 517 tangent line, 231 Two-point Gaussian Quadrature, 315 Arc length, 466, 467, 852 derivative of, 852 parameter, 852, 853 in parametric form, 709 of a polar curve, 729 of a space curve, 851 in the xy-plane, 1003 Arccosecant function, 366 Arccosine function, 366 Arccotangent function, 366 Archimedes (287–212 B.C.), 256 Principle, 506 spiral of, 717, 733 Arcsecant function, 366 Arcsine function, 366 series for, 670 Arctangent function, 366 series for, 670 Area found by exhaustion method, 256 line integral for, 1078 of a parametric surface, 1088 in polar coordinates, 725 problem, 45, 46 of a rectangle, 256 of a region between two curves, 437 of a region in the plane, 260 of a surface of revolution, 471 in parametric form, 710 in polar coordinates, 730 of the surface 1003 in the xy-plane, 1003 Associative Property of Vector Addition, 751 Astroid, 145 Astroidal sphere, 1093 Asymptote(s) horizontal, 196 of a hyperbola, 689 slant, 208 vertical, 85 Average rate of change, 12 Average value of a function on an interval, 281 over a region R, 982 over a solid region Q, 1019 Average velocity, 112 Axis conjugate, of a hyperbola, 689
major, of an ellipse, 685 minor, of an ellipse, 685 of a parabola, 683 polar, 715 of revolution, 446 transverse, of a hyperbola, 689 B Barrow, Isaac (1630–1677), 144 Base(s), 321, 356 of the natural exponential function, 356 of a natural logarithm, 321 other than e derivatives for, 358 exponential function, 356 logarithmic function, 357 Basic differentiation rules for elementary functions, 371 Basic equation obtained in a partial fraction decomposition, 544 guidelines for solving, 548 Basic integration rules, 246, 378, 508 procedures for fitting integrands to, 511 Basic limits, 59 Basic types of transformations, 23 Bearing, 754 Bernoulli equation, 430 general solution of, 430 Bernoulli, James (1654–1705), 702 Bernoulli, John (1667–1748), 542 Bessel function, 655 Bessel’s equation, 1156 Bifolium, 145 Binomial series, 669 Binormal vector, 849, 866 Bisection method, 78 Bose-Einstein condensate, 74 Boundary point of a region, 880 Bounded above, 591 below, 591 monotonic sequence, 591 region 936 sequence, 591 Brachistochrone problem, 702 Breteuil, Emilie de (1706–1749), 478 Bullet-nose curve, 137 C Cantor set, 679 Capillary action, 1008 Cardioid, 720, 721 Carrying capacity, 417, 419 Catenary, 386 Cauchy, Augustin-Louis (1789–1857), 75 Cauchy-Riemann differential equations, 914 Cauchy-Schwarz Inequality, 774 Cavalieri’s Theorem, 456
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A120
Index
Center of curvature, 856 of an ellipse, 685 of gravity, 488, 489 of a one-dimensional system, 488 of a two-dimensional system, 489 of a hyperbola, 689 of mass, 487, 488, 489 of a one-dimensional system, 487, 488 of a planar lamina, 490 of variable density, 996 of a solid region 1014 of a two-dimensional system, 489 of a power series, 647 Centered at c, 636 Central force field, 1041 Centripetal component of acceleration, 846 Centripetal force, 850 Centroid, 491 of a simple region, 996 Chain Rule, 129, 130, 135 implicit differentiation, 912 one independent variable, 907 three or more independent variables, 910 and trigonometric functions, 134 two independent variables, 909 Change in x, 97 Change in y, 97 Change of variables, 295 for definite integrals, 298 for double integrals, 1029 guidelines for making, 296 for homogeneous equations, 423 to polar form, 988 using a Jacobian, 1027 Characteristic equation of a differential equation, 1134 Charles, Jacques (1746–1823), 74 Charles’s Law, 74 Chebyshev’s equation, 1156 Circle, 145, 682, 721 Circle of curvature, 159, 856 Circulation of F around C␣ , 1117 Circumscribed rectangle, 258 Cissoid, 145 of Diocles, 746 Classification of conics by eccentricity, 734 Closed curve, 1070 disk, 880 region R, 880 surface, 1106 Cobb-Douglas production function, 873 Coefficient, 24 correlation, 31 leading, 24 Collinear, 17 Combinations of functions, 25 Common logarithmic function, 357 Common types of behavior associated with nonexistence of a limit, 51
Commutative Property of the dot product, 766 of vector addition, 751 Comparison Test Direct, 612 for improper integrals, 576 Limit, 614 Completeness, 77, 591 Completing the square, 377 Component of acceleration centripetal, 846 normal, 845, 846, 859 tangential, 845, 846, 859 Component form of a vector in the plane, 749 Component functions, 816 Components of a vector, 770 along v, 770 in the direction of v, 771 orthogonal to v, 770 in the plane, 749 Composite function, 25 antidifferentiation of, 292 continuity of, 75 derivative of, 129 limit of, 61 of two variables, 869 continuity of, 885 Composition of functions, 25, 869 Compound interest formulas, 360 Compounding, continuous, 360 Computer graphics, 874 Concave downward, 187 Concave upward, 187 Concavity, 187 test for, 188 Conditional convergence, 622 Conditionally convergent series, 622 Conic(s), 682 circle, 682 classification by eccentricity, 734 degenerate, 682 directrix of, 734 eccentricity, 734 ellipse, 682, 685 focus of, 734 hyperbola, 682, 689 parabola, 682, 683 polar equations of, 735 Conic section, 682 Conjugate axis of a hyperbola, 689 Connected region, 1068 Conservative vector field, 1043, 1065 independence of path, 1068 test for, 1044, 1047 Constant Euler’s, 611 force, 477 function, 24 gravitational, 479 of integration, 245 Multiple Rule, 109, 135 differential form, 234
Rule, 106, 135 spring, 34 term of a polynomial function, 24 Constraint, 952 Continued fraction expansion, 679 Continuity on a closed interval, 73 of a composite function, 75 of two variables, 885 differentiability implies, 102 and differentiability of inverse functions, 341 implies integrability, 268 properties of, 75 of a vector-valued function, 820 Continuous, 70 at c, 59, 70 on the closed interval 关a, b兴, 73 compounding, 360 everywhere, 70 function of two variables, 884 on an interval, 820 from the left and from the right, 73 on an open interval 共a, b兲, 70 in the open region R, 884, 886 at a point, 820, 884, 886 vector field, 1040 Continuously differentiable, 466 Contour lines, 871 Converge, 227, 585, 595 Convergence absolute, 622 conditional, 622 endpoint, 650 of a geometric series, 597 of improper integral with infinite discontinuities, 571 integration limits, 568 interval of, 648, 652 of Newton’s Method, 227, 228 of a power series, 648 of p-series, 607 radius of, 648, 652 of a sequence, 585 of a series, 595 of Taylor series, 666 tests for series Alternating Series Test, 619 Direct Comparison Test, 612 geometric series, 597 guidelines, 631 Integral Test, 605 Limit Comparison Test, 614 p-series, 607 Ratio Test, 627 Root Test, 630 summary, 632 Convergent power series, form of, 664 Convergent series, limit of nth term of, 599 Convex limaçon, 721 Coordinate conversion cylindrical to rectangular, 804 cylindrical to spherical, 807
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Index polar to rectangular, 716 rectangular to cylindrical, 804 rectangular to polar, 716 rectangular to spherical, 807 spherical to cylindrical, 807 spherical to rectangular, 807 Coordinate planes, 758 xy-plane, 758 xz-plane, 758 yz-plane, 758 Coordinate system cylindrical, 804 polar, 715 spherical, 807 three-dimensional, 758 Coordinates, polar, 715 area in, 725 area of a surface of revolution in, 730 converting to rectangular, 716 Distance Formula in, 722 Coordinates, rectangular, converting to polar, 716 Copernicus, Nicolaus (1473–1543), 685 Cornu spiral, 745, 865 Correlation coefficient, 31 Cosecant function derivative of, 122, 135 integral of, 333 inverse of, 366 derivative of, 369 Cosine function, 22 derivative of, 111, 135 integral of, 333 inverse of, 366 derivative of, 369 series for, 670 Cotangent function derivative of, 122, 135 integral of, 333 inverse of, 366 derivative of, 369 Coulomb’s Law, 479, 1041 Critical number(s) of a function, 164 relative extrema occur only at, 164 Critical point(s) of a function of two variables, 937 relative extrema occur only at, 937 Cross product of two vectors in space, 775 algebraic properties of, 776 determinant form, 775 geometric properties of, 777 torque, 779 Cruciform, 145 Cubic function, 24 Cubing function, 22 Curl of a vector field, 1046 and divergence, 1048 Curtate cycloid, 704 Curvature, 854 center of, 856 circle of, 159, 856 formulas for, 855, 859
radius of, 856 in rectangular coordinates, 856, 859 related to acceleration and speed, 857 Curve astroid, 145 bifolium, 145 bullet-nose, 137 cissoid, 145 closed, 1070 cruciform, 145 equipotential, 418 folium of Descartes, 145, 733 isothermal, 418 kappa, 144, 146 lateral surface area over, 1063 lemniscate, 40, 143, 146, 721 level, 871 logistic, 419, 550 natural equation for, 865 orientation of, 1051 piecewise smooth, 701, 1051 plane, 696, 816 pursuit, 388 rectifiable, 466 rose, 718, 721 simple, 1075 smooth, 466, 701, 826, 841, 1051 piecewise, 701, 1051 space, 816 tangent line to, 842 Curve sketching, summary of, 206 Cusps, 826 Cycloid, 701, 705 curtate, 704 prolate, 708 Cylinder, 794 directrix of, 794 equations of, 794 generating curve of, 794 right, 794 rulings of, 794 Cylindrical coordinate system, 804 pole of, 804 Cylindrical coordinates converting to rectangular, 804 converting to spherical, 807 Cylindrical surface, 794 D Damped motion of a spring, 1139 Darboux’s Theorem, 242 Decay model, exponential, 408 Decomposition of N共x兲兾D共x兲 into partial fractions, 543 Decreasing function, 177 test for, 177 Definite integral(s), 268 approximating Midpoint Rule, 262, 307 Simpson’s Rule, 308 Trapezoidal Rule, 306 as the area of a region, 269
A121
change of variables, 298 evaluation of a line integral as a, 1053 properties of, 272 two special, 271 of a vector-valued function, 828 Degenerate conic, 682 line, 682 point, 682 two intersecting lines, 682 Degree of a polynomial function, 24 Delta, ␦ , ␦ -neighborhood, 880 Demand, 18 Density, 490 Density function , 994, 1014 Dependent variable, 19 of a function of two variables, 868 Derivative(s) of algebraic functions, 135 alternative form, 101 of arc length function, 852 Chain Rule, 129, 130, 135 implicit differentiation, 912 one independent variable, 907 three or more independent variables, 910 two independent variables, 909 of a composite function, 129 Constant Multiple Rule, 109, 135 Constant Rule, 106, 135 of cosecant function, 122, 135 of cosine function, 111, 135 of cotangent function, 122, 135 Difference Rule, 110, 135 directional, 915, 916, 923 of an exponential function, base a, 358 of a function, 99 General Power Rule, 131, 135 higher-order, 124 of hyperbolic functions, 385 implicit, 141 of an inverse function, 341 of inverse trigonometric functions, 369 involving absolute value, 324 from the left and from the right, 101 of a logarithmic function, base a, 358 of the natural exponential function, 348 of the natural logarithmic function, 322 notation, 99 parametric form, 706 partial, 890 Power Rule, 107, 135 Product Rule, 118, 135 Quotient Rule, 120, 135 of secant function, 122, 135 second, 124 Simple Power Rule, 107, 135 simplifying, 133 of sine function, 111, 135 Sum Rule, 110, 135 of tangent function, 122, 135 third, 124 of trigonometric functions, 122, 135 of a vector-valued function, 824
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A122
Index
higher-order, 825 properties of, 826 Descartes, René (1596–1650), 2 Determinant form of cross product, 775 Difference quotient, 20, 97 Difference Rule, 110, 135 differential form, 234 Difference of two functions, 25 Difference of two vectors, 750 Differentiability implies continuity, 102, 903 and continuity of inverse functions, 341 sufficient condition for, 901 Differentiable at x, 99 Differentiable, continuously, 466 Differentiable function on the closed interval 关a, b兴, 101 on an open interval 共a, b兲, 99 in a region R, 901 of three variables, 902 of two variables, 901 vector-valued, 824 Differential, 232 as an approximation, 902 function of three or more variables, 900 function of three variables, 902 function of two variables, 900 of x, 232 of y, 232 Differential equation, 245, 398 Airy’s, 1152, 1156 Bernoulli equation, 430 Bessel’s of order one, 1156 of order zero, 1156 Cauchy-Riemann, 914 Chebyshev’s, 1156 doomsday, 433 Euler’s equation, 1140 Euler’s Method, 402 exact, 1126 first-order linear, 424 general solution of, 245, 398 Gompertz, 433 Hermite’s, 1156 higher-order linear homogeneous, 1137 homogeneous, 423, 1133 change of variables, 423 initial condition, 249, 399 integrating factor, 424, 1129 Laguerre’s, 1156 logistic, 241, 419 nonhomogeneous, 1133, 1141 solution of, 1141 order of, 398 particular solution of, 249, 399 power series solution of, 1149 second-order, 1133 separable, 415 separation of variables, 407, 415 singular solution of, 398 solution of, 398 Taylor series solution of, 1149
test for exactness, 1126 Differential form, 234 of a line integral, 1059 Differential formulas, 234 constant multiple, 234 product, 234 quotient, 234 sum or difference, 234 Differential operator, 1046, 1048 Laplacian, 1123 Differentiation, 99 Applied minimum and maximum problems, guidelines for solving, 216 basic rules for elementary functions, 371 implicit, 140 Chain Rule, 912 guidelines for, 141 involving inverse hyperbolic functions, 389 logarithmic, 323 numerical, 102 partial, 890 of power series, 652 of a vector-valued function, 824 Differentiation rules basic, 371 Chain, 129, 130, 135 Constant, 106, 135 Constant Multiple, 109, 135 cosecant function, 122, 135 cosine function, 111, 135 cotangent function, 122, 135 Difference, 110, 135 general, 135 General Power, 131, 135 Power, 107, 135 for Real Exponents, 359 Product, 118, 135 Quotient, 120, 135 secant function, 122, 135 Simple Power, 107, 135 sine function, 111, 135 Sum, 110, 135 summary of, 135 tangent function, 122, 135 Diminishing returns, point of, 223 Dimpled limaçon, 721 Direct Comparison Test, 612 Direct substitution, 59, 60 Directed distance, 489 Directed line segment, 748 equivalent, 748 initial point of, 748 length of, 748 magnitude of, 748 terminal point of, 748 Direction angles of a vector, 769 Direction cosines of a vector, 769 Direction field, 251, 319, 400 Direction of motion, 832 Direction numbers, 783
Direction vector, 783 Directional derivative, 915, 916 alternative form of, 918 of f in the direction of u, 916, 923 of a function in three variables, 923 Directrix of a conic, 734 of a cylinder, 794 of a parabola, 683 Dirichlet, Peter Gustav (1805–1859), 51 Dirichlet function, 51 Discontinuity, 71 infinite, 568 nonremovable, 71 removable, 71 Disk, 446, 880 closed, 880 method, 447 compared to shell, 459 open, 880 Displacement of a particle, 286, 287 Distance between a point and a line in space, 789 between a point and a plane, 788 directed, 489 total, traveled on 关a, b兴, 287 Distance Formula in polar coordinates, 722 in space, 759 Distributive Property for the dot product, 766 for vectors, 751 Diverge, 585, 595 Divergence of improper integral with infinite discontinuities, 571 integration limits, 568 of a sequence, 585 of a series, 595 tests for series Direct Comparison Test, 612 geometric series, 597 guidelines, 631 Integral Test, 605 Limit Comparison Test, 614 nth-Term Test, 599 p-series, 607 Ratio Test, 627 Root Test, 630 summary, 632 of a vector field, 1048 and curl, 1048 Divergence Theorem, 1080, 1106 Divergence-free vector field, 1048 Divide out like factors, 63 Domain feasible, 215 of a function, 19 explicitly defined, 21 of two variables, 868 implied, 21 of a power series, 648
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Index of a vector-valued function, 817 Doomsday equation, 433 Dot product Commutative Property of, 766 Distributive Property for, 766 form of work, 772 projection using the, 771 properties of, 766 of two vectors, 766 Double integral, 974, 975, 976 change of variables for, 1029 of over R, 976 properties of, 976 Doyle Log Rule, 878 Dummy variable, 270 Dyne, 477 E e, the number, 321 limit involving, 360 Eccentricity, 734 classification of conics by, 734 of an ellipse, 687 of a hyperbola, 690 Eight curve, 159 Elasticity of cost, 1132 Electric force field, 1041 Elementary function(s), 24, 371 basic differentiation rules for, 371 polynomial approximation of, 636 power series for, 670 Eliminating the parameter, 698 Ellipse, 682, 685 center of, 685 eccentricity of, 687 foci of, 685 major axis of, 685 minor axis of, 685 reflective property of, 687 rotated, 145 standard equation of, 685 vertices of, 685 Ellipsoid, 795, 796 Elliptic cone, 795, 797 Elliptic integral, 311 Elliptic paraboloid, 795, 797 Endpoint convergence, 650 Endpoint extrema, 162 Energy kinetic, 1071 potential, 1071 Epicycloid, 704, 705, 709 Epsilon-delta, - ␦, definition of limit, 52 Equal vectors, 749, 760 Equality of mixed partial derivatives, 895 Equation(s) Airy’s, 1152, 1156 basic, 544 guidelines for solving, 548 Bernoulli, 430 Bessel’s, 1156 characteristic, 1134
Chebyshev’s, 1156 of conics, polar, 735 of a cylinder, 794 doomsday, 433 of an ellipse, 685 general second-degree, 682 Gompertz, 433 graph of, 2 harmonic, 1123 Hermite’s, 1156 homogeneous, 1133 of a hyperbola, 689 Laguerre’s, 1156 Laplace’s, 1123 of a line general form, 14 horizontal, 14 point-slope form, 11, 14 slope-intercept form, 13, 14 in space, parametric, 783 in space, symmetric, 783 summary, 14 vertical, 14 nonhomogeneous, 1133 of a parabola, 683 parametric, 696, 1084 finding, 700 graph of, 696 of a plane in space general form, 784 standard form, 784 primary, 215, 216 related-rate, 148 secondary, 216 separable, 415 solution point of, 2 of tangent plane, 928 Equilibrium, 487 Equipotential curves, 418 lines, 871 Equivalent conditions, 1070 directed line segments, 748 Error in approximating a Taylor polynomial, 642 in measurement, 233 percent error, 233 propagated error, 233 relative error, 233 in Simpson’s Rule, 309 in Trapezoidal Rule, 309 Escape velocity, 94 Euler, Leonhard (1707–1783), 24 Euler’s constant, 611 differential equation, 1140 Method, 402, 1132 Evaluate a function, 19 Evaluating a flux integral, 1100 a surface integral, 1094
A123
Evaluation by iterated integrals, 1010 of a line integral as a definite integral, 1053 Even function, 26 integration of, 300 test for, 26 Everywhere continuous, 70 Exact differential equation, 1126 Exactness, test for, 1126 Existence of an inverse function, 339 of a limit, 73 theorem, 77, 162 Expanded about c, approximating polynomial, 636 Explicit form of a function, 19, 140 Explicitly defined domain, 21 Exponential decay, 408 Exponential function, 24 to base a, 356 derivative of, 358 integration rules, 350 natural, 346 derivative of, 348 properties of, 347 operations with, 347 series for, 670 Exponential growth and decay model, 408 initial value, 408 proportionality constant, 408 Exponentiate, 347 Extended Mean Value Theorem, 241, 558 Extrema endpoint, 162 of a function, 162, 936 guidelines for finding, 165 relative, 163 Extreme Value Theorem, 162, 936 Extreme values of a function, 162 F Factorial, 587 Family of functions, 268 Famous curves astroid, 145 bifolium, 145 bullet-nose curve, 137 circle, 145, 682, 721 cissoid, 145 cruciform, 145 eight curve, 159 folium of Descartes, 145, 7331 kappa curve, 144, 146 lemniscate, 40, 143, 146, 721 parabola, 2, 145, 682, 683 pear-shaped quartic, 159 rotated ellipse, 145 rotated hyperbola, 145 serpentine, 126 top half of circle, 137 witch of Agnesi, 126, 145, 198, 823
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A124
Index
Faraday, Michael (1791–1867), 1071 Feasible domain, 215 Fermat, Pierre de (1601–1665), 164 Fibonacci sequence, 594, 604 Field central force, 1041 direction, 251, 319, 400 electric force, 1041 force, 1040 gravitational, 1041 inverse square, 1041 slope, 251, 301, 319, 400 vector, 1040 over a plane region R, 1040 over a solid region Q, 1040 velocity, 1040, 1041 Finite Fourier series, 532 First Derivative Test, 179 First moments, 998, 1014 First partial derivatives, 890 notation for, 891 First-order differential equations linear, 424 solution of, 425 Fitting integrands to basic rules, 511 Fixed plane, 862 Fixed point, 229 Fluid(s) force, 498 pressure, 497 weight-densities of, 497 Flux integral, 1100 evaluating, 1100 Focal chord of a parabola, 683 Focus of a conic, 734 of an ellipse, 685 of a hyperbola, 689 of a parabola, 683 Folium of Descartes, 145, 733 Force, 477 constant, 477 exerted by a fluid, 498 of friction, 858 resultant, 754 variable, 478 Force field, 1040 central, 1041 electric, 1041 work, 1056 Forced motion of a spring, 1141 Form of a convergent power series, 664 Fourier, Joseph (1768–1830), 657 Fourier series, finite, 532 Fourier Sine Series, 523 Fraction expansion, continued, 679 Fractions, partial, 542 decomposition of N共x兲兾D共x兲, into, 543 method of, 542 Free motion of a spring, 1141 Frenet-Serret formulas, 866 Fresnel function, 315 Friction, 858
Fubini’s Theorem, 978 for a triple integral, 1010 Function(s), 6, 19 absolute maximum of, 162 absolute minimum of, 162 absolute value, 22 acceleration, 124 accumulation, 283 addition of, 25 algebraic, 24, 25, 371 antiderivative of, 244 arc length, 466, 467, 852 arccosecant, 366 arccosine, 366 arccotangent, 366 arcsecant, 366 arcsine, 366 arctangent, 366 average value of, 281, 982 Bessel, 655 Cobb-Douglas production, 873 combinations of, 25 common logarithmic, 357 component, 816 composite, 25, 869 composition of, 25, 869 concave downward, 187 concave upward, 187 constant, 24 continuous, 70 continuously differentiable, 466 cosine, 22 critical number of, 164 cubic, 24 cubing, 22 decreasing, 177 test for, 177 defined by power series, properties of, 652 density, 994, 1014 derivative of, 99 difference of, 25 differentiable, 99, 101 Dirichlet, 51 domain of, 19 elementary, 24, 371 algebraic, 24, 25 exponential, 24 logarithmic, 24 trigonometric, 24 evaluate, 19 even, 26 explicit form, 19, 140 exponential to base a, 356 extrema of, 162 extreme values of, 162 family of, 268 feasible domain of, 215 Fresnel, 315 Gamma, 566, 578 global maximum of, 162 global minimum of, 162 graph of, guidelines for analyzing, 206
greatest integer, 72 Gudermannian, 396 Heaviside, 39 homogeneous, 423, 913 hyperbolic, 383 identity, 22 implicit form, 19 implicitly defined, 140 increasing, 177 test for, 177 inner product of two, 532 integrable, 268 inverse, 337 inverse hyperbolic, 387 inverse trigonometric, 366 involving a radical, limit of, 60 jerk, 160 limit of, 48 linear, 24 linearly dependent, 1133 linearly independent, 1133 local extrema of, 163 local maximum of, 163 local minimum of, 163 logarithmic, 318 to base a, 357 logistic growth, 361 natural exponential, 346 natural logarithmic, 318 notation, 19 odd, 26 one-to-one, 21 onto, 21 orthogonal, 532 point of inflection, 189, 190 polynomial, 24, 60, 869 position, 32, 112, 837 potential, 1043 product of, 25 pulse, 94 quadratic, 24 quotient of, 25 radius, 800 range of, 19 rational, 22, 25, 869 real-valued, 19 relative extrema of, 163, 936 relative maximum of, 163, 936 relative minimum of, 163, 936 representation by power series, 657 Riemann zeta, 611 signum, 82 sine, 22 sine integral, 316 square root, 22 squaring, 22 standard normal probability density, 349 step, 72 strictly monotonic, 178, 339 sum of, 25 that agree at all but one point, 62 of three variables continuity of, 886
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Index directional derivative of, 923 gradient of, 923 transcendental, 25, 371 transformation of a graph of, 23 horizontal shift, 23 reflection about origin, 23 reflection about x-axis, 23 reflection about y-axis, 23 reflection in the line y ⫽ x, 338 vertical shift, 23 trigonometric, 24 of two variables, 868 absolute maximum of, 936 absolute minimum of, 936 continuity of, 884 critical point of, 937 dependent variable, 868 differentiability implies continuity, 903 differentiable, 901 differential of, 900 domain of, 868 gradient of, 918 graph of, 870 independent variables, 868 limit of, 881 maximum of, 936 minimum of, 936 nonremovable discontinuity of, 884 partial derivative of, 890 range of, 868 relative extrema of, 936 relative maximum of, 936, 939 relative minimum of, 936, 939 removable discontinuity of, 884 total differential of, 900 unit pulse, 94 vector-valued, 816 Vertical Line Test, 22 Wronskian of two, 1140 of x and y, 868 zero of, 26 approximating with Newton’s Method, 225 Fundamental Theorem of Algebra, 1106 of Calculus, 277, 278 guidelines for using, 278 Second, 284 of Line Integrals, 1065, 1066 G Gabriel’s Horn, 574, 1086 Galilei, Galileo (1564–1642), 371 Galois, Evariste (1811–1832), 228 Gamma Function, 566, 578 Gauss, Carl Friedrich (1777–1855), 255, 1106 Gaussian Quadrature Approximation, two-point, 315 Gauss’s Law, 1103 Gauss’s Theorem, 1106
General antiderivative, 245 General differentiation rules, 135 General form of the equation of a line, 14 of the equation of a plane in space, 784 of the equation of a quadric surface, 795 of a second-degree equation, 682 General harmonic series, 607 General partition, 267 General Power Rule for differentiation, 131, 135 for Integration, 297 General second-degree equation, 682 General solution of the Bernoulli equation, 430 of a differential equation, 245, 398 of a second-order nonhomogeneous linear differential equation, 1141 Generating curve of a cylinder, 794 Geometric power series, 657 Geometric properties of the cross product, 777 Geometric property of triple scalar product, 780 Geometric series, 597 alternating, 619 convergence of, 597 divergence of, 597 Germain, Sophie (1776–1831), 1141 Gibbs, Josiah Willard (1839–1903), 1051 Global maximum of a function, 162 Global minimum of a function, 162 Golden ratio, 594 Gompertz equation, 433 Grad, 918 Gradient, 1040, 1043 of a function of three variables, 923 of a function of two variables, 918 normal to level curves, 921 normal to level surfaces, 832 properties of, 919 recovering a function from, 1047 Graph(s) of absolute value function, 22 of cosine function, 22 of cubing function, 22 of an equation, 2 of a function guidelines for analyzing, 206 transformation of, 23 of two variables, 870 of hyperbolic functions, 384 of identity function, 22 intercept of, 4 of inverse hyperbolic functions, 388 of inverse trigonometric functions, 367 orthogonal, 146 of parametric equations, 696 polar, 717 points of intersection, 727 special polar graphs, 721 of rational function, 22 of sine function, 22
A125
of square root function, 22 of squaring function, 22 symmetry of, 5 Gravitational constant, 479 field, 1041 Greatest integer function, 72 Green, George (1793–1841), 1076 Green’s Theorem, 1075 alternative forms of, 1080 Gregory, James (1638–1675), 652 Gudermannian function, 396 Guidelines for analyzing the graph of a function, 206 for evaluating integrals involving secant and tangent, 527 for evaluating integrals involving sine and cosine, 524 for finding extrema on a closed interval, 165 for finding intervals on which a function is increasing or decreasing, 178 for finding an inverse function, 339 for finding limits at infinity of rational functions, 198 for finding a Taylor series, 668 for implicit differentiation, 141 for integration, 331 for integration by parts, 515 for making a change of variables, 296 for solving applied minimum and maximum problems, 216 for solving the basic equation, 548 for solving related-rate problems, 149 for testing a series for convergence or divergence, 631 for using the Fundamental Theorem of Calculus, 278 Gyration, radius of, 999 H Half-life, 356, 409 Hamilton,William Rowan (1805–1865), 750 Harmonic equation, 1123 Harmonic series, 607 alternating, 620, 622, 624 Heat flow, 1103 Heat flux, 1103 Heaviside, Oliver (1850–1925), 39 Heaviside function, 39 Helix, 817 Hermite’s equation, 1156 Heron’s Formula, 963 Herschel, Caroline (1750–1848), 691 Higher-order derivative, 124 linear differential equations, 1137 partial, 894 of a vector-valued function, 825 Homogeneous of degree n, 423, 913 Homogeneous differential equation, 423
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A126
Index
change of variables for, 423 Homogeneous equation, 1137 Homogeneous function, 423, 913 Hooke’s Law, 479, 1138 Horizontal asymptote, 196 Horizontal component of a vector, 753 Horizontal line, 14 Horizontal Line Test, 339 Horizontal shift of a graph of a function, 23 Horizontally simple region of integration, 968 Huygens, Christian (1629–1795), 466 Hypatia (370–415 A.D.), 682 Hyperbola, 682, 689 asymptotes of, 689 center of, 689 conjugate axis of, 689 eccentricity of, 690 foci of, 689 rotated, 145 standard equation of, 689 transverse axis of, 689 vertices of, 689 Hyperbolic functions, 383 derivatives of, 385 graphs of, 384 identities, 384 integrals of, 385 inverse, 387 differentiation involving, 389 graphs of, 388 integration involving, 389 Hyperbolic identities, 384 Hyperbolic paraboloid, 795, 797 Hyperboloid of one sheet, 795, 796 of two sheets, 795, 796 Hypocycloid, 705 I Identities, hyperbolic, 384 Identity function, 22 If and only if, 14 Image of x under f, 19 Implicit derivative, 141 Implicit differentiation, 140, 912 Chain Rule, 912 guidelines for, 141 Implicit form of a function, 19 Implicitly defined function, 140 Implied domain, 21 Improper integral, 568 comparison test for, 576 with infinite discontinuities, 571 convergence of, 571 divergence of, 571 with infinite integration limits, 568 convergence of, 568 divergence of, 568 special type, 574 Incidence, angle of, 684 Inclination of a plane, angle of, 931
Incompressible, 1048, 1111 Increasing function, 177 test for, 177 Increment of z, 900 Increments of x and y, 900 Indefinite integral, 245 pattern recognition, 282 of a vector-valued function, 828 Indefinite integration, 245 Independence of path and conservative vector fields, 1068 Independent of path, 1068 Independent variable, 19 of a function of two variables, 868 Indeterminate form, 63, 86, 197, 211, 557, 560 Index of summation, 254 Inductive reasoning, 589 Inequality Cauchy-Schwarz, 774 Napier’s, 336 preservation of, 272 triangle, 753 Inertia, moment of, 998, 1014 polar, 998 Infinite discontinuities, 568 improper integrals with, 571 convergence of, 571 divergence of, 571 Infinite integration limits, 568 improper integrals with, 568 convergence of, 568 divergence of, 568 Infinite interval, 195 Infinite limit(s), 83 at infinity, 201 from the left and from the right, 83 properties of, 87 Infinite series (or series), 595 absolutely convergent, 622 alternating, 619 geometric, 619 harmonic, 620, 622 remainder, 621 conditionally convergent, 622 convergence of, 595 convergent, limit of nth term, 599 divergence of, 595 nth term test for, 599 geometric, 597 guidelines for testing for convergence or divergence of, 631 harmonic, 607 alternating, 620, 622, 624 nth partial sum, 595 properties of, 599 p-series, 607 rearrangement of, 624 sum of, 595 telescoping, 596 terms of, 595 Infinity infinite limit at, 201
limit at, 195, 196 Inflection point, 189, 190 Initial condition(s), 249, 399 Initial point, directed line segment, 748 Initial value, 408 Inner partition, 974, 1009 polar, 987 Inner product of two functions, 532 of two vectors, 766 Inner radius of a solid of revolution, 449 Inscribed rectangle, 258 Inside limits of integration, 967 Instantaneous rate of change, 112 Instantaneous velocity, 113 Integrability and continuity, 268 Integrable function, 268, 976 Integral(s) definite, 268 properties of, 272 two special, 271 double, 974, 975, 976 elliptic, 311 flux, 1100 of hyperbolic functions, 385 improper, 568 indefinite, 245 involving inverse trigonometric functions, 375 involving secant and tangent, guidelines for evaluating, 527 involving sine and cosine, guidelines for evaluating, 524 iterated, 967 line, 1052 Mean Value Theorem, 280 of p共x兲 ⫽ Ax 2 ⫹ Bx ⫹ C, 307 single, 976 of the six basic trigonometric functions, 333 surface, 1094 trigonometric, 524 triple, 1009 Integral Test, 605 Integrand(s), procedures for fitting to basic rules, 511 Integrating factor, 424, 1129 Integration as an accumulation process, 441 Additive Interval Property, 271 basic rules of, 246, 378, 508 change of variables, 295 guidelines for, 296 constant of, 245 of even and odd functions, 300 guidelines for, 331 indefinite, 245 pattern recognition, 292 involving inverse hyperbolic functions, 389 Log Rule, 328 lower limit of, 268 of power series, 652
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Index preservation of inequality, 272 region R of, 967 rules for exponential functions, 350 upper limit of, 268 of a vector-valued function, 828 Integration by parts, 515 guidelines for, 515 summary of common integrals using, 520 tabular method, 520 Integration by tables, 551 Integration formulas reduction formulas, 553 special, 537 summary of, 1118 Integration rules basic, 246, 378, 508 General Power Rule, 297 Power Rule, 246 Integration techniques basic integration rules, 246, 378, 508 integration by parts, 515 method of partial fractions, 542 substitution for rational functions of sine and cosine, 554 tables, 551 trigonometric substitution, 533 Intercept(s), 4 x-intercept, 4 y-intercept, 4 Interest formulas, summary of, 360 Interior point of a region R, 880, 886 Intermediate Value Theorem, 77 Interpretation of concavity, 187 Interval of convergence, 648 Interval, infinite, 195 Inverse function, 337 continuity and differentiability of, 341 derivative of, 341 existence of, 339 guidelines for finding, 339 Horizontal Line Test, 339 properties of, 357 reflective property of, 338 Inverse hyperbolic functions, 387 differentiation involving, 389 graphs of, 388 integration involving, 389 Inverse square field, 1041 Inverse trigonometric functions, 366 derivatives of, 369 graphs of, 367 integrals involving, 375 properties of, 368 Irrotational vector field, 1046 Isobars, 871 Isothermal curves, 418 Isothermal surface, 874 Isotherms, 871 Iterated integral, 967 evaluation by, 1010 inside limits of integration, 967 outside limits of integration, 967
Iteration, 225 ith term of a sum, 254 J Jacobi, Carl Gustav (1804–1851), 1027 Jacobian, 1027 Jerk function, 160 K Kappa curve, 144, 146 Kepler, Johannes, (1571–1630), 737 Kepler’s Laws, 737 Kinetic energy, 1071 Kirchhoff’s Second Law, 426 Kovalevsky, Sonya (1850–1891), 880 L Lagrange, Joseph-Louis (1736–1813), 172, 952 Lagrange form of the remainder, 642 Lagrange multiplier, 952, 953 Lagrange’s Theorem, 953 Laguerre’s equation, 1156 Lambert, Johann Heinrich (1728–1777), 383 Lamina, planar, 490 Laplace, Pierre Simon de (1749–1827), 1020 Laplace Transform, 578 Laplace’s equation, 1123 Laplacian, 1123 Lateral surface area over a curve, 1063 Latus rectum, of a parabola, 683 Law of Conservation of Energy, 1071 Leading coefficient of a polynomial function, 24 test, 24 Least squares method of, 946 regression, 7 line, 946, 947 Least upper bound, 591 Left-hand limit, 72 Left-handed orientation, 758 Legendre, Adrien-Marie (1752–1833), 947 Leibniz, Gottfried Wilhelm (1646–1716), 234 Leibniz notation, 234 Lemniscate, 40, 143, 146, 721 Length of an arc, 466, 467 parametric form, 709 polar form, 729 of a directed line segment, 748 of the moment arm, 487 of a scalar multiple, 752 of a vector in the plane, 749 of a vector in space, 760 on x-axis, 1003
A127
Level curve, 871 gradient is normal to, 921 Level surface, 873 gradient is normal to, 932 L’Hôpital, Guillaume (1661–1704), 558 L’Hôpital’s Rule, 558 Limaçon, 721 convex, 721 dimpled, 721 with inner loop, 721 Limit(s), 45, 48 basic, 59 of a composite function, 61 definition of, 52 -␦ definition of, 52 evaluating direct substitution, 59, 60 divide out like factors, 63 rationalize the numerator, 63, 64 existence of, 73 of a function involving a radical, 60 of a function of two variables, 881 indeterminate form, 63 infinite, 83 from the left and from the right, 83 properties of, 87 at infinity, 195, 196 infinite, 201 of a rational function, guidelines for finding, 198 of integration inside, 967 lower, 268 outside, 967 upper, 268 involving e, 360 from the left and from the right, 72 of the lower and upper sums, 260 nonexistence of, common types of behavior, 51 of nth term of a convergent series, 599 one-sided, 72 of polynomial and rational functions, 60 properties of, 59 of a sequence, 585 properties of, 586 strategy for finding, 62 of trigonometric functions, 61 two special trigonometric, 65 of a vector-valued function, 819 Limit Comparison Test, 614 Line(s) contour, 871 as a degenerate conic, 682 equation of general form, 14 horizontal, 14 point-slope form, 11, 14 slope-intercept form, 13, 14 summary, 14 vertical, 14 equipotential, 871 least squares regression, 946, 947
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A128
Index
moment about, 487 normal, 927, 928 at a point, 146 parallel, 14 perpendicular, 14 radial, 715 secant, 45, 97 slope of, 10 in space direction number of, 783 direction vector of, 783 parametric equations of, 783 symmetric equations of, 783 tangent, 45, 97 approximation, 231 at the pole, 720 with slope 97 vertical, 98 Line of impact, 927 Line integral, 1052 for area, 1078 differential form of, 1059 evaluation of as a definite integral, 1053 of f along C, 1052 independent of path, 1068 summary of, 1103 of a vector field, 1056 Line segment, directed, 748 Linear approximation, 231, 902 Linear combination of i and j, 753 Linear combination of solutions, 1134 Linear function, 24 Linearly dependent functions, 1133 Linearly independent functions, 1133 Local maximum, 163 Local minimum, 163 Locus, 682 Log Rule for Integration, 328 Logarithmic differentiation, 323 Logarithmic function, 24, 318 to base a, 357 derivative of, 358 common, 357 natural, 318 derivative of, 322 properties of, 319 Logarithmic properties, 319 Logarithmic spiral, 733 Logistic curve, 419, 550 Logistic differential equation, 241, 419 carrying capacity, 419 Logistic growth function, 361 Lorenz curves, 444 Lower bound of a sequence, 591 Lower bound of summation, 254 Lower limit of integration, 268 Lower sum, 258 limit of, 260 Lune, 541 M Macintyre, Sheila Scott (1910–1960), 524 Maclaurin, Colin, (1698–1746), 664
Maclaurin polynomial, 638 Maclaurin series, 665 Magnitude of a directed line segment, 748 of a vector in the plane, 749 Major axis of an ellipse, 685 Marginal productivity of money, 955 Mass, 486, 1100 center of, 487, 488, 489 of a one-dimensional system, 487, 488 of a planar lamina, 490 of variable density, 996, 1014 of a solid region Q, 1014 of a two-dimensional system, 489 moments of, 996 of a planar lamina of variable density, 994 pound mass, 486 total, 488, 489 Mathematical model, 7, 946 Mathematical modeling, 33 Maximum absolute, 162 of f on I, 162 of a function of two variables, 936 global, 162 local, 163 relative, 163 Mean Value Theorem, 172 alternative form of, 173 Extended, 241, 558 for Integrals, 280 Measurement, error in, 233 Mechanic’s Rule, 229 Method of Lagrange Multipliers, 952, 953 least squares, 946 partial fractions, 542 undetermined coefficients, 1142 Midpoint Formula, 759 Midpoint Rule, 262, 307 Minimum absolute, 162 of f on I, 162 of a function of two variables, 936 global, 162 local, 163 relative, 163 Minor axis of an ellipse, 685 Mixed partial derivatives, 894 equality of, 895 Möbius Strip, 1093 Model exponential growth and decay, 408 mathematical, 7, 946 Modeling, mathematical, 33 Moment(s) about a line, 487 about the origin, 487, 488 about a point, 487 about the x-axis of a planar lamina, 490 of a two-dimensional system, 489
about the y-axis of a planar lamina, 490 of a two-dimensional system, 489 arm, length of, 487 first, 1014 of a force about a point, 779 of inertia, 998, 1014, 1123 polar, 998 for a space curve, 1064 of mass, 996 of a one-dimensional system, 488 of a planar lamina, 490 second, 998, 1014 Monotonic sequence, 590 bounded, 591 Monotonic, strictly, 178, 339 Motion of a spring damped, 1138 forced, 1141 free, 1141 undamped, 1138 Mutually orthogonal, 418 N n factorial, 587 Napier, John (1550–1617), 318 Napier’s Inequality, 336 Natural equation for a curve, 865 Natural exponential function, 346 derivative of, 348 integration rules, 350 operations with, 347 properties of, 347 series for, 670 Natural logarithmic base, 321 Natural logarithmic function, 318 base of, 321 derivative of, 322 properties of, 319 series for, 670 Negative of a vector, 750 Net change, 286 Net Change Theorem, 286 Newton (unit of force), 477 Newton, Isaac (1642–1727), 96, 225 Newton’s Law of Cooling, 411 Newton’s Law of Gravitation, 1041 Newton’s Law of Universal Gravitation, 479 Newton’s Method for approximating the zeros of a function, 225 convergence of, 227, 228 iteration, 225 Newton’s Second Law of Motion, 425, 836, 1138 Nodes, 826 Noether, Emmy (1882–1935), 751 Nonexistence of a limit, common types of behavior, 51 Nonhomogeneous equation, 1133 Nonhomogeneous linear equations, 1141 Nonremovable discontinuity, 71, 804
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Index Norm of a partition, 267, 974, 987, 1009 polar, 987 of a vector in the plane, 749 Normal component of acceleration, 845, 846, 859 of a vector field, 1100 Normal line, 927, 928 at a point, 146 Normal probability density function, 349 Normal vector(s), 768 principal unit, 842, 859 to a smooth parametric surface, 1087 Normalization of v, 752 Notation antiderivative, 245 derivative, 99 for first partial derivatives, 891 function, 19 Leibniz, 234 sigma, 254 nth Maclaurin polynomial for f at c, 638 nth partial sum, 595 nth Taylor polynomial for f at c, 638 nth term of a convergent series, 599 of a sequence, 584 nth-Term Test for Divergence, 599 Number, critical, 164 Number e, 321 limit involving, 360 Numerical differentiation, 103 O Octants, 758 Odd function, 26 integration of, 300 test for, 26 Ohm’s Law, 237 One-dimensional system center of gravity of, 488 center of mass of, 487, 488 moment of, 487, 488 total mass of, 488 One-sided limit, 72 One-to-one function, 21 Onto function, 21 Open disk, 880 Open interval continuous on, 70 differentiable on, 99 Open region R, 880, 886 continuous in, 884, 886 Open sphere, 886 Operations with exponential functions, 347 with power series, 659 Order of a differential equation, 398 Orientable surface, 1099 Orientation of a curve, 1051 of a plane curve, 697 of a space curve, 816
Oriented surface, 1099 Origin moment about, 487, 488 of a polar coordinate system, 715 reflection about, 23 symmetry, 5 Orthogonal functions, 532 graphs, 146 trajectory, 146, 418 vectors, 768 Ostrogradsky, Michel (1801–1861), 1106 Ostrogradsky’s Theorem, 1106 Outer radius of a solid of revolution, 449 Outside limits of integration, 967 P Padé approximation, 395 Pappus Second Theorem of, 496 Theorem of, 493 Parabola, 2, 145, 682, 683 axis of, 683 directrix of, 683 focal chord of, 683 focus of, 683 latus rectum of, 683 reflective property of, 684 standard equation of, 683 vertex of, 683 Parabolic spandrel, 495 Parallel lines, 14 planes, 785 vectors, 761 Parameter, 696 arc length, 852, 853 eliminating, 698 Parametric equations, 696 finding, 700 graph of, 696 of a line in space, 783 for a surface, 1084 Parametric form of arc length, 709 of the area of a surface of revolution, 710 of the derivative, 706 Parametric surface, 1084 area of, 1088 equations for, 1084 partial derivatives of, 1087 smooth, 1087 normal vector to, 1087 surface area of, 1088 Partial derivatives, 890 first, 890 of a function of three or more variables, 893 of a function of two variables, 890 higher-order, 894 mixed, 894 equality of, 895
A129
notation for, 891 of a parametric surface, 1087 Partial differentiation, 890 Partial fractions, 542 decomposition of N共x兲兾D共x兲 into, 543 method of, 542 Partial sums, sequence of, 595 Particular solution of a differential equation, 249, 399 of a nonhomogeneous linear equation, 1141 Partition general, 267 inner, 974, 1009 polar, 987 norm of, 267, 974, 1009 polar, 987 regular, 267 Pascal, Blaise (1623–1662), 497 Pascal’s Principle, 497 Path, 881, 1051 Pear-shaped quartic, 159 Percent error, 233 Perigee, 694 Perihelion, 694, 741 Perpendicular lines, 14 planes, 785 vectors, 768 Piecewise smooth curve, 701, 1051 Planar lamina, 490 center of mass of, 490 moment of, 490 Plane angle of inclination of, 931 distance between a point and, 788 region area of, 260 simply connected, 1044, 1075 tangent, 928 equation of, 928 vector in, 748 Plane curve, 696, 816 orientation of, 697 smooth, 1051 Plane in space angle between two, 785 equation of general form, 784 standard form, 784 parallel, 785 to the axis, 787 to the coordinate plane, 787 perpendicular, 785 trace of, 787 Planimeter, 1122 Point as a degenerate conic, 682 of diminishing returns, 223 fixed, 229 of inflection, 189, 190 of intersection, 6 of polar graphs, 727 moment about, 487
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A130
Index
in a vector field incompressible, 1111 sink, 1111 source, 1111 Point-slope equation of a line, 11, 14 Polar axis, 715 Polar coordinate system, 715 polar axis of, 715 pole (or origin), 715 Polar coordinates, 715 area in, 725 area of a surface of revolution in, 730 converting to rectangular, 716 Distance Formula in, 722 Polar curve, arc length of, 729 Polar equations of conics, 735 Polar form of slope, 719 Polar graphs, 717 cardioid, 720, 721 circle, 721 convex limaçon, 721 dimpled limaçon, 721 lemniscate, 721 limaçon with inner loop, 721 points of intersection, 727 rose curve, 718, 721 Polar moment of inertia, 998 Polar sectors, 986 Pole, 715 of cylindrical coordinate system, 804 tangent lines at, 720 Polynomial Maclaurin, 638 Taylor, 159, 638 Polynomial approximation, 636 centered at c, 636 expanded about c, 636 Polynomial function, 24, 60 constant term of, 24 degree of, 24 leading coefficient of, 24 limit of, 60 of two variables, 869 zero, 24 Position function, 32, 112, 124 for a projectile, 837 Potential energy, 1071 Potential function for a vector field, 1043 Pound mass, 486 Power Rule for differentiation, 107, 135 for integration, 246, 297 for Real Exponents, 359 Power series, 647 centered at c, 647 convergence of, 648 convergent, form of, 664 differentiation of, 652 domain of, 648 for elementary functions, 670 endpoint convergence, 650 geometric, 657 integration of, 652
interval of convergence, 648 operations with, 659 properties of functions defined by, 652 interval of convergence of, 652 radius of convergence of, 652 radius of convergence, 648 representation of functions by, 657 solution of a differential equation, 1149 Preservation of inequality, 272 Pressure, fluid, 497 Primary equation, 215, 216 Prime Number Theorem, 327 Principal unit normal vector, 842, 859 Probability density function, 349 Procedures for fitting integrands to basic rules, 511 Product of two functions, 25 inner, 532 of two vectors in space, 775 Product Rule, 118, 135 differential form, 234 Projectile, position function for, 837 Projection form of work, 772 Projection of u onto v, 770 using the dot product, 771 Prolate cycloid, 708 Propagated error, 233 Properties of continuity, 75 of the cross product algebraic, 776 geometric, 777 of definite integrals, 272 of the derivative of a vector-valued function, 826 of the dot product, 766 of double integrals, 976 of functions defined by power series, 652 of the gradient, 919 of infinite limits, 87 of infinite series, 599 of inverse functions, 357 of inverse trigonometric functions, 368 of limits, 59 of limits of sequences, 586 logarithmic, 319 of the natural exponential function, 319, 347 of the natural logarithmic function, 319 of vector operations, 751 Proportionality constant, 408 p-series, 607 convergence of, 607 divergence of, 607 harmonic, 607 Pulse function, 94 unit, 94 Pursuit curve, 388
Q Quadratic function, 24 Quadric surface, 795 ellipsoid, 795, 796 elliptic cone, 795, 797 elliptic paraboloid, 795, 797 general form of the equation of, 795 hyperbolic paraboloid, 795, 797 hyperboloid of one sheet, 795, 796 hyperboloid of two sheets, 795, 796 standard form of the equations of, 795, 796, 797 Quaternions, 750 Quotient, difference, 20, 97 Quotient Rule, 120, 135 differential form, 234 Quotient of two functions, 25 R Radial lines, 715 Radian measure, 367 Radical, limit of a function involving a, 60 Radicals, solution by, 228 Radioactive isotopes, half-lives of, 409 Radius of convergence, 648 of curvature, 856 function, 800 of gyration, 999 inner, 449 outer, 449 Ramanujan, Srinivasa (1887–1920), 661 Range of a function, 19 of two variables, 868 Raphson, Joseph (1648–1715), 225 Rate of change, 12, 893 average, 12 instantaneous, 12, 112 Ratio, 12 golden, 594 Ratio Test, 627 Rational function, 22, 25 guidelines for finding limits at infinity of, 198 limit of, 60 of two variables, 869 Rationalize the numerator, 63, 64 Rationalizing technique, 64 Real Exponents, Power Rule, 359 Real numbers, completeness of, 77, 591 Real-valued function f of a real variable x, 19 Reasoning, inductive, 589 Recovering a function from its gradient, 1047 Rectangle area of, 256 circumscribed, 258 inscribed, 258 representative, 436 Rectangular coordinates converting to cylindrical, 804
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Index converting to polar, 716 converting to spherical, 807 curvature in, 856, 859 Rectifiable curve, 466 Recursion formula, 1149 Recursively defined sequence, 584 Reduction formulas, 553 Reflection about the origin, 23 about the x-axis, 23 about the y-axis, 23 angle of, 684 in the line y ⫽ x, 338 Reflective property of an ellipse, 687 of inverse functions, 338 of a parabola, 684 Reflective surface, 684 Refraction, 223, 959 Region of integration R, 967 horizontally simple, 968 r-simple, 988 -simple, 988 vertically simple, 968 Region in the plane area of, 260, 968 between two curves, 437 centroid of, 491 connected, 1068 Region R boundary point of, 880 bounded, 936 closed, 880 differentiable function in, 901 interior point of, 880, 886 open, 880, 886 continuous in, 884, 886 simply connected, 1044, 1075 Regression, line, least squares, 7, 946, 947 Regular partition, 267 Related-rate equation, 148 Related-rate problems, guidelines for solving, 149 Relation, 19 Relative error, 233 Relative extrema First Derivative Test for, 179 of a function, 163, 936 occur only at critical numbers, 164 occur only at critical points, 937 Second Derivative Test for, 191 Second Partials Test for, 939 Relative maximum at 共c, f 共c兲兲, 163 First Derivative Test for, 179 of a function, 163, 936, 939 Second Derivative Test for, 191 Second Partials Test for, 939 Relative minimum at 共c, f 共c兲兲, 163 First Derivative Test for, 179 of a function, 163, 936, 939 Second Derivative Test for, 191
Second Partials Test for, 939 Remainder alternating series, 621 of a Taylor polynomial, 642 Removable discontinuity, 71 of a function of two variables, 884 Representation of antiderivatives, 244 Representative element, 441 disk, 446 rectangle, 436 shell, 457 washer, 449 Resultant force, 754 Resultant vector, 750 Return wave method, 532 Review of basic differentiation rules, 371 of basic integration rules, 378, 508 Revolution axis of, 446 solid of, 446 surface of, 470 area of, 471, 710, 730 volume of solid of disk method, 446 shell method, 457, 458 washer method, 449 Riemann, Georg Friedrich Bernhard (1826–1866), 267, 624 Riemann sum, 267 Riemann zeta function, 611 Right cylinder, 794 Right-hand limit, 72 Right-handed orientation, 758 Rolle, Michel (1652–1719), 170 Rolle’s Theorem, 170 Root Test, 630 Rose curve, 718, 721 Rotated ellipse, 145 Rotated hyperbola, 145 Rotation of F about N, 1117 r-simple region of integration, 988 Rulings of a cylinder, 794 S Saddle point, 939 Scalar, 748 field, 871 multiple, 750 multiplication, 750, 760 product of two vectors, 766 quantity, 748 Secant function derivative of, 122, 135 integral of, 333 inverse of, 366 derivative of, 369 Secant line, 45, 97 Second derivative, 124 Second Derivative Test, 191 Second Fundamental Theorem of Calculus, 284
A131
Second moment, 998, 1014 Second Partials Test, 939 Second Theorem of Pappus, 496 Secondary equation, 216 Second-degree equation, general, 682 Second-order homogeneous linear differential equation, 1133 linear differential equation, 1133 nonhomogeneous linear differential equation, 1133, 1141 solution of, 1141 Separable differential equation, 415 Separation of variables, 407, 415 Sequence, 584 Absolute Value Theorem, 588 bounded, 591 bounded above, 591 bounded below, 591 bounded monotonic, 591 convergence of, 585 divergence of, 585 Fibonacci, 594, 604 least upper bound of, 591 limit of, 585 properties of, 586 lower bound of, 591 monotonic, 590 nth term of, 584 of partial sums, 595 pattern recognition for, 588 recursively defined, 584 Squeeze Theorem, 587 terms of, 584 upper bound of, 591 Series, 595 absolutely convergent, 622 alternating, 619 geometric, 619 harmonic, 620, 622, 624 Alternating Series Test, 619 binomial, 669 conditionally convergent, 622 convergence of, 595 convergent, limit of nth term, 599 Direct Comparison Test, 612 divergence of, 595 nth term test for, 599 finite Fourier, 532 Fourier Sine, 523 geometric, 597 alternating, 619 convergence of, 597 divergence of, 597 guidelines for testing for convergence or divergence, 631 harmonic, 607 alternating, 620, 622, 624 infinite, 595 properties of, 599 Integral Test, 605 Limit Comparison Test, 614 Maclaurin, 665
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A132
Index
nth partial sum, 595 nth term of convergent, 599 power, 647 p-series, 607 Ratio Test, 627 rearrangement of, 624 Root Test, 630 sum of, 595 summary of tests for, 632 Taylor, 664, 665 telescoping, 596 terms of, 595 Serpentine, 126 Shell method, 457, 458 and disk method, comparison of, 459 Shift of a graph horizontal, 23 vertical, 23 Sigma notation, 254 index of summation, 254 ith term, 254 lower bound of summation, 254 upper bound of summation, 254 Signum function, 82 Simple curve, 1075 Simple Power Rule, 107, 135 Simple solid region, 1107 Simply connected plane region, 1075 Simpson’s Rule, 308 error in, 309 Sine function, 22 derivative of, 111, 135 integral of, 333 inverse of, 366 derivative of, 369 series for, 670 Sine integral function, 316 Sine Series, Fourier, 523 Single integral, 976 Singular solution, differential equation, 398 Sink, 1111 Slant asymptote, 208 Slope(s) field, 251, 301, 319, 400 of the graph of f at x ⫽ c, 97 of a line, 10 of a surface in x- and y-directions, 891 of a tangent line, 97 parametric form, 706 polar form, 719 Slope-intercept equation of a line, 13, 14 Smooth curve, 466, 701, 826, 841 on an open interval, 826 piecewise, 701 parametric surface, 1087 plane curve, 1051 space curve, 1051 Snell’s Law of Refraction, 223, 959 Solenoidal, 1048 Solid region, simple, 1107 Solid of revolution, 446
volume of disk method, 446 shell method, 457, 458 washer method, 449 Solution curves, 399 of a differential equation, 398 Bernoulli, 430 Euler’s Method, 402 first-order linear, 425 general, 245, 398, 1141 linear combinations of, 1134 particular, 249, 399, 1141 second-order linear nonhomogeneous, 1141 singular, 398 of y⬙ ⫹ ay⬘ ⫹ by ⫽ 0, 1135 point of an equation, 2 by radicals, 228 Some basic limits, 59 Somerville, Mary Fairfax (1780–1872), 868 Source, 1111 Space curve, 816 arc length of, 851 moments of inertia for, 1064 smooth, 1051 Spandrel, parabolic, 495 Special integration formulas, 537 Special polar graphs, 721 Special type of improper integral, 574 Speed, 113, 832, 833, 857, 859 angular, 999 Sphere, 759 astroidal, 1093 open, 886 standard equation of, 759 Spherical coordinate system, 807 converting to cylindrical coordinates, 807 converting to rectangular coordinates, 807 Spiral of Archimedes, 717, 733 cornu, 745, 865 logarithmic, 733 Spring constant, 34, 1138 Square root function, 22 Squared errors, sum of, 946 Squaring function, 22 Squeeze Theorem, 65 for Sequences, 587 Standard equation of an ellipse, 685 a hyperbola, 689 a parabola, 683 a sphere, 759 Standard form of the equation of an ellipse, 685 a hyperbola, 689 a parabola, 683 a plane in space, 784 a quadric surface, 795, 796, 797
Standard form of a first-order linear differential equation, 424 Standard normal probability density function, 349 Standard position of a vector, 749 Standard unit vector, 753 notation, 760 Step function, 72 Stirling’s approximation, 517 Stirling’s Formula, 354 Stokes, George Gabriel (1819–1903), 1114 Stokes’s Theorem, 1080, 1114 Strategy for finding limits, 62 Strictly monotonic function, 178, 339 Strophoid, 745 Substitution for rational functions of sine and cosine, 554 Sufficient condition for differentiability, 901 Sum(s) ith term of, 254 lower, 258 limit of, 260 nth partial, 595 Riemann, 267 Rule, 110, 135 differential form, 234 of a series, 595 sequence of partial, 595 of the squared errors, 946 of two functions, 25 of two vectors, 750 upper, 258 limit of, 260 Summary of common integrals using integration by parts, 520 of compound interest formulas, 360 of curve sketching, 206 of differentiation rules, 135 of equations of lines, 14 of integration formulas, 1118 of line and surface integrals, 1103 of tests for series, 632 of velocity, acceleration, and curvature, 859 Summation formulas, 255 index of, 254 lower bound of, 254 upper bound of, 254 Surface closed, 1106 cylindrical, 794 isothermal, 874 level, 873 orientable, 1099 oriented, 1099 parametric, 1084 parametric equations for, 1084 quadric, 795 reflective, 684 trace of, 795
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Index Surface area of a parametric surface, 1088 of a solid, 1002, 1003 Surface integral, 1094 evaluating, 1094 summary of, 1103 Surface of revolution, 470, 800 area of, 471 parametric form, 710 polar form, 730 Symmetric equations, line in space, 783 Symmetry tests for, 5 with respect to the origin, 5 with respect to the point 共a, b兲, 395 with respect to the x-axis, 5 with respect to the y-axis, 5 T Table of values, 2 Tables, integration by, 551 Tabular method for integration by parts, 520 Tangent function derivative of, 122, 135 integral of, 333 inverse of, 366 derivative of, 369 Tangent line(s), 45, 97 approximation of f at c, 231 to a curve, 842 at the pole, 720 problem, 45 slope of, 97 parametric form, 706 polar form, 719 with slope m, 97 vertical, 98 Tangent plane, 928 equation of, 928 Tangent vector, 832 Tangential component of acceleration, 845, 845, 846, 859 Tautochrone problem, 702 Taylor, Brook (1685–1731), 638 Taylor polynomial, 159, 638 error in approximating, 642 remainder, Lagrange form of, 642 Taylor series, 664, 665 convergence of, 666 guidelines for finding, 668 solution of a differential equation, 1149 Taylor’s Theorem, 642 Telescoping series, 596 Terminal point, directed line segment, 748 Terms for exactness, 1126 of a sequence, 584 of a series, 595 Test(s) comparison, for improper integrals, 576 for concavity, 188
conservative vector field in the plane, 1044 conservative vector field in space, 1047 for convergence Alternating Series, 619 Direct Comparison, 612 geometric series, 597 guidelines, 631 Integral, 605 Limit Comparison, 614 p-series, 607 Ratio, 627 Root, 630 summary, 632 for even and odd functions, 26 First Derivative, 179 Horizontal Line, 339 for increasing and decreasing functions, 177 Leading Coefficient, 24 Second Derivative, 191 for symmetry, 5 Vertical Line, 22 Theorem Absolute Value, 588 of Calculus, Fundamental, 277, 278 guidelines for using, 278 of Calculus, Second Fundamental, 284 Cavalieri’s, 456 Darboux’s, 242 existence, 77, 162 Extended Mean Value, 241, 558 Extreme Value, 162, 936 Fubini’s, 978 for a triple integral, 1010 Intermediate Value, 77 Mean Value, 172 alternative form, 173 Extended, 241, 558 for Integrals, 280 Net Change, 286 of Pappus, 493 Second, 496 Prime Number, 327 Rolle’s, 170 Squeeze, 65 for sequences, 587 Taylor’s, 642 Theta, simple region of integration, 988 Third derivative, 124 Three-dimensional coordinate system, 758 left-handed orientation, 758 right-handed orientation, 758 Top half of circle, 137 Topographic map, 871 Torque, 488, 779 Torricelli’s Law, 433 Torsion, 866 Total differential, 900 Total distance traveled on 关a, b兴, 287 Total mass, 488, 489 of a one-dimensional system, 488
A133
of a two-dimensional system, 489 Trace of a plane in space, 787 of a surface, 795 Tractrix, 327, 388 Trajectories, orthogonal, 146, 418 Transcendental function, 25, 371 Transformation, 23, 1028 Transformation of a graph of a function, 23 basic types, 23 horizontal shift, 23 reflection about origin, 23 reflection about x-axis, 23 reflection about y-axis, 23 reflection in the line y ⫽ x, 338 vertical shift, 23 Transverse axis of a hyperbola, 689 Trapezoidal Rule, 306 error in, 309 Triangle inequality, 753 Trigonometric function(s), 24 and the Chain Rule, 134 cosine, 22 derivative of, 122, 135 integrals of the six basic, 333 inverse, 366 derivatives of, 369 graphs of, 367 integrals involving, 375 properties of, 368 limit of, 61 sine, 22 Trigonometric integrals, 524 Trigonometric substitution, 533 Triple integral, 1009 in cylindrical coordinates, 1020 in spherical coordinates, 1023 Triple scalar product, 779 geometric property of, 780 Two-dimensional system center of gravity of, 489 center of mass of, 489 moment of, 489 total mass of, 489 Two-Point Gaussian Quadrature Approximation, 315 Two special definite integrals, 271 Two special trigonometric limits, 65 U Undamped motion of a spring, 1138 Undetermined coefficients, 1142 Unit pulse function, 94 Unit tangent vector, 841, 859 Unit vector, 749 in the direction of 752, 760 standard, 753 Universal Gravitation, Newton’s Law, 479 Upper bound least, 591 of a sequence, 591 of summation, 254
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A134
Index
Upper limit of integration, 268 Upper sum, 258 limit of, 260 u-substitution, 292 V Value of f at x, 19 Variable dependent, 19 dummy, 270 force, 478 independent, 19 Variation of parameters, 1145 Vector(s) acceleration, 845, 859 addition, 750, 751 associative property of, 751 commutative property of, 751 Additive Identity Property, 751 Additive Inverse Property, 751 angle between two, 767 binormal, 849, 866 component of u along v, 770 of u orthogonal to v, 770 component form of, 749 components, 749, 770 cross product of, 775 difference of two, 750 direction, 783 direction angles of, 769 direction cosines of, 769 Distributive Property, 751 dot product of, 766 equal, 749, 760 horizontal component of, 753 initial point, 748 inner product of, 766 length of, 749, 760 linear combination of, 753 magnitude of, 749 negative of, 750 norm of, 749 normal, 768 normalization of, 752 operations, properties of, 751 orthogonal, 768 parallel, 761 perpendicular, 768 in the plane, 748 principal unit normal, 842, 859 product of two vectors in space, 775 projection of, 770 resultant, 750 scalar multiplication, 750, 760 scalar product of, 766 in space, 760 standard position, 749 standard unit notation, 760 sum, 750 tangent, 832 terminal point, 748
triple scalar product, 779 unit, 749 in the direction of v, 752, 760 standard, 753 unit tangent, 841, 859 velocity, 832, 859 vertical component of, 753 zero, 749, 760 Vector field, 1040 circulation of, 1117 conservative, 1043, 1065 test for, 1044, 1047 continuous, 1040 curl of, 1046 divergence of, 1048 divergence-free, 1048 incompressible, 1111 irrotational, 1046 line integral of, 1056 normal component of, 1100 over a plane region R, 1040 over a solid region Q, 1040 potential function for, 1043 rotation of, 1117 sink, 1111 solenoidal, 1048 source, 1111 Vector space, 752 axioms, 752 Vector-valued function(s), 816 antiderivative of, 828 continuity of, 820 continuous on an interval, 820 continuous at a point, 820 definite integral of, 828 derivative of, 824 higher-order, 825 properties of, 826 differentiation of, 824 domain of, 817 indefinite integral of, 828 integration of, 828 limit of, 819 Velocity, 113, 833 average, 112 escape, 94 function, 124 instantaneous, 113 potential curves, 418 Velocity field, 1040, 1041 incompressible, 1048 Velocity vector, 832, 859 Vertéré, 198 Vertex of an ellipse, 685 of a hyperbola, 689 of a parabola, 683 Vertical asymptote, 85 Vertical component of a vector, 753 Vertical line, 14 Vertical Line Test, 22 Vertical shift of a graph of a function, 23 Vertical tangent line, 98
Vertically simple region of integration, 968 Volume of a solid disk method, 447 with known cross sections, 451 shell method, 457, 458 washer method, 449 Volume of a solid region, 976, 1009 W Wallis, John (1616–1703), 526 Wallis’s Formulas, 526, 532 Washer, 449 Washer method, 449 Weierstrass, Karl (1815–1897), 937 Weight-densities of fluids, 497 Wheeler, Anna Johnson Pell (1883–1966), 424 Witch of Agnesi, 126, 145, 198, 823 Work, 477, 772 done by a constant force, 477 done by a variable force, 478 dot product form, 772 force field, 1056 projection form, 772 Wronskian of two functions, 1140 X x-axis moment about, of a planar lamina, 490 moment about, of a two-dimensional system, 489 reflection about, 23 symmetry, 5 x-intercept, 4 xy-plane, 758 xz-plane, 758 Y y -axis moment about, of a planar lamina, 490 moment about, of a two-dimensional system, 489 reflection about, 23 symmetry, 5 y -intercept, 4 Young, Grace Chisholm (1868–1944), 45 yz-plane, 758 Z Zero factorial, 587 Zero of a function, 26 approximating bisection method, 78 Intermediate Value Theorem, 77 with Newton’s Method, 225 Zero polynomial, 24 Zero vector, 749, 760
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ALGEBRA Factors and Zeros of Polynomials Let p共x兲 an x n an1x n1 . . . a1x a0 be a polynomial. If p共a兲 0, then a is a zero of the polynomial and a solution of the equation p共x兲 0. Furthermore, 共x a兲 is a factor of the polynomial.
Fundamental Theorem of Algebra An nth degree polynomial has n (not necessarily distinct) zeros. Although all of these zeros may be imaginary, a real polynomial of odd degree must have at least one real zero.
Quadratic Formula
If p共x兲 ax 2 bx c, and 0 b2 4ac, then the real zeros of p are x 共b ± 冪b2 4ac兲兾2a.
Special Factors x 2 a 2 共x a兲共x a兲
x 3 a 3 共x a兲共x 2 ax a 2兲
x 3 a3 共x a兲共x 2 ax a 2兲
x 4 a 4 共x 2 a 2兲共x 2 a 2兲
Binomial Theorem 共x y兲2 x 2 2xy y 2
共x y兲2 x 2 2xy y 2
共x y兲3 x 3 3x 2y 3xy 2 y 3
共x y兲3 x 3 3x 2y 3xy 2 y 3
共x y兲4 x 4 4x 3y 6x 2y 2 4xy3 y 4
共x y兲4 x 4 4x 3y 6x 2y 2 4xy 3 y 4
n共n 1兲 n2 2 . . . x y nxy n1 y n 2! n共n 1兲 n2 2 . . . 共x y兲n x n nx n1y x y ± nxy n1 y n 2!
共x y兲n x n nx n1y
Rational Zero Theorem If p共x兲 an x n a n1x n1 . . . a1x a0 has integer coefficients, then every rational zero of p is of the form x r兾s, where r is a factor of a0 and s is a factor of an.
Factoring by Grouping acx 3 adx 2 bcx bd ax 2共cx d兲 b共cx d兲 共ax 2 b兲共cx d兲
Arithmetic Operations
© Brooks/Cole, Cengage Learning
ab ac a共b c兲
冢ab冣 a d ad 冢 冣冢 冣 c 冢d冣 b c bc b ab a冢 冣 c c
a c ad bc b d bd a b a c bc
ab a b c c c
ab ba cd dc
ab ac bc a
冢冣
a ac b b c
冢冣
Exponents and Radicals a0 1, a 0
共ab兲 x a xb x
a xa y a xy
冢ab冣
n am am兾n 冪
ax
x
ax bx
1 ax
冪a a1兾2
ax a xy ay
n a a1兾n 冪
n ab 冪 n a冪 n b 冪
共ax兲y a xy
冪ab n
n a 冪 n b 冪
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FORMULAS FROM GEOMETRY Sector of Circular Ring c
共 p average radius, w width of ring, in radians兲
a θ
h b
Area ab
a
Circumference ⬇ 2
b
Equilateral Triangle
b
冪a
2
s
b 2
共A area of base兲 Ah Volume 3
s
冪3s2
h
4
A
Right Circular Cone
Trapezoid
a
h Area 共a b兲 2
h
h
r 2h 3 Lateral Surface Area r冪r2 h2
Volume
h b
r
Frustum of Right Circular Cone
r
共r 2 rR R 2兲h 3 Lateral Surface Area s共R r兲
s
Volume
b a h
b
Circle
h
Right Circular Cylinder
Area r 2
Volume r 2h
r
Circumference 2 r
Lateral Surface Area 2 rh
Sector of Circle
Sphere
R
r h
4 Volume r 3 3 Surface Area 4 r 2
s θ
r
Circular Ring 共 p average radius, w width of ring兲 Area 共R 2 r 2兲 2 pw
h
s
Parallelogram
共 in radians兲 r2 Area 2 s r
a
2
Cone
冪3 s
Area bh
w
Ellipse c
(Pythagorean Theorem) c2 a 2 b2
Area
θ
Area pw
Right Triangle
2
p
r
Wedge r p R
w
共A area of upper face, B area of base兲 A B sec
A
θ
B
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© Brooks/Cole, Cengage Learning
h a sin 1 Area bh 2 (Law of Cosines) c2 a 2 b2 2ab cos
h
Tear out Formula Cards for Homework Success.
Triangle
Index of Applications Speed, 29, 175, 861, 951 of sound, 282 Statics problems, 494 Stopping distance, 117, 128, 237 Surface area, 153, 158 of a dome, 1093 of an oil spill, 443 of a pond, 503 of a satellite-signal receiving dish, 694 Surveying, 311 Suspension bridge, 476 Temperature, 18, 176, 204, 344, 405, 959 at which water boils, 327 Temperature distribution, 878, 898, 920, 926, 963 Topography, 871, 926 Torque, 779, 781, 812 Tossing bales, 839 Velocity, 117, 176, 289, 312 of a diver, 113 of a piston, 152 of a rocket, 582 Velocity and acceleration, 312, 316, 423 on the moon, 160 Velocity in a resisting medium, 566 Vertical motion, 116, 157, 174, 175, 250, 252, 382, 392 Vibrating spring, 157, 1140, 1147, 1148, 1154 Vibrating string, 523 Volume, 82, 116, 126, 153 of a box, 30, 903 of fluid in a storage tank, 540 of a glacier, 993 of the Great Salt Lake, 1038 of a pond, 465 of a pontoon, 461 of a pyramid, 452 of a shampoo bottle, 222 of a spherical ring, 505 of a trough, 906 of water in a conical tank, 148 Water flow, 291 Water running into a vase, 193 Water tower, 455 Wave equation, 897, 964 Wheelchair ramp, 12 Wind chill, 906 Work, 311, 504 done by aircraft engines, 1121 done in closing a door, 772 done by an expanding gas, 482 done by a force field, 1062, 1064, 1073, 1121, 1124 done by a hydraulic cylinder, 556 done in lifting a chain, 482, 484, 504 done in moving a particle, 1081, 1124 done by a person walking up a staircase, 1064
(continued from front inside cover) done in pulling a sled, 774 done in pulling a toy wagon, 774 done in splitting a piece of wood, 485 done in towing a car, 774 Wrinkled and bumpy spheres, 1026
Business and Economics Annuities, 603 Average production, 984 Average profit, 1034 Average sales, 289 Break-even analysis, 37 Break-even point, 9 Capitalized cost, 577 Cobb-Douglas production function, 873, 878, 955, 963 Compound interest, 361, 363, 364, 394, 413, 432, 566, 593, 676, 677 Consumer and producer surpluses, 506 Cost, 138, 344 Declining sales, 410 Demand function, 240 Depreciation, 18, 303, 353, 363, 394, 602, 676 Elasticity of cost, 1132 Eliminating budget deficits, 444 Energy consumption, 34 Federal debt, 594 Gross Domestic Product (GDP), 9 Gross income tax collections, 950 Health Care Expenditures, 127 Home mortgage, 327, 396 Inflation, 363, 593 Inventory cost, 239 Inventory management, 81, 117 Inventory replenishment, 126 Investment, 878, 898 Investment growth, 428 Manufacturing, 451, 455 Marginal costs, 898 Marginal productivity, 898 Marginal productivity of money, 955 Marginal revenue, 897 Marginal utility, 898 Marketing, 602 Maximum profit, 223, 945, 949, 962 Maximum revenue, 949, 962 Median income, 38 Minimum cost, 222, 949, 959 Present value, 523, 603 Profit, 444 Revenue, 444, 774 Salary, 603 Sales, 175, 303, 336, 432, 433 Apple, Inc., 879 Sales growth, 194, 239 Value of a mid-sized sedan, 354 Wages, 34
Social and Behavioral Sciences Cellular phone subscribers, 9 Crime, 230 Health maintenance organizations, 36 Learning curve, 413, 414, 428 Memory model, 523 Outlays for national defense, 239 Population, 413, 992 of California, 349 of Colorado, 12 of United States, 16, 414 Population growth, 428, 431 Psychology, intelligence test, 898 World population, 951
Life Sciences Agronomy, 962 Bacterial culture growth, 139, 361, 413, 422 Blood flow, 289 Carbon dioxide concentration, 7 Concentration of a tracer drug in a fluid, 434 DNA molecule, 817 Endangered species, 422 Epidemic model, 550 Forestry, 414, 878 Intravenous feeding, 429 Models for tumors, 1026 Pancreas transplants, 364 Population, 556 Population growth, 680 of bacteria, 126, 252, 336 of brook trout, 432 of coyotes, 417 of fish, 364 of fruit flies, 410 Respiratory cycle, 289, 314 Trachea contraction, 185
General Applicants to a university, 898 Average typing speed, 194, 204 Dental inlays, 814 Folding paper, 242 Möbius Strip, 1093 Probability, 303, 355, 577, 602, 603, 663, 674, 985, 993 Queuing model, 878 School commute, 27 Sphereflake, 603 Throwing a dart, 265
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.