Manuale Soluzioni Thomas' Calculus Multivariabile

INSTRUCTOR’S SOLUTIONS MANUAL MULTIVARIABLE WILLIAM ARDIS Collin County Community College THOMAS’ CALCULUS TWELFTH EDITION BASED ON THE ORIGINAL WORK BY George B. Thomas, Jr. Massachusetts Institute of Technology AS REVISED BY Maurice D. Weir Naval Postgraduate School Joel Hass University of California, Davis The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Addison-Wesley from electronic files supplied by the author. Copyright © 2010, 2005, 2001 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-321-60072-1 ISBN-10: 0-321-60072-X 1 2 3 4 5 6 BB 14 13 12 11 10 PREFACE TO THE INSTRUCTOR This Instructor's Solutions Manual contains the solutions to every exercise in the 12th Edition of THOMAS' CALCULUS by Maurice Weir and Joel Hass, including the Computer Algebra System (CAS) exercises. The corresponding Student's Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the CAS exercises (because the CAS command templates would give them all away). In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or rewritten every solution which appeared in previous solutions manuals to ensure that each solution ì conforms exactly to the methods, procedures and steps presented in the text ì is mathematically correct ì includes all of the steps necessary so a typical calculus student can follow the logical argument and algebra ì includes a graph or figure whenever called for by the exercise, or if needed to help with the explanation ì is formatted in an appropriate style to aid in its understanding Every CAS exercise is solved in both the MAPLE and MATHEMATICA computer algebra systems. A template showing an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within the text grouping require a change only in the input function or other numerical input parameters associated with the problem (such as the interval endpoints or the number of iterations). For more information about other resources available with Thomas' Calculus, visit http://pearsonhighered.com. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. TABLE OF CONTENTS 10 Infinite Sequences and Series 569 10.1 Sequences 569 10.2 Infinite Series 577 10.3 The Integral Test 583 10.4 Comparison Tests 590 10.5 The Ratio and Root Tests 597 10.6 Alternating Series, Absolute and Conditional Convergence 602 10.7 Power Series 608 10.8 Taylor and Maclaurin Series 617 10.9 Convergence of Taylor Series 621 10.10 The Binomial Series and Applications of Taylor Series 627 Practice Exercises 634 Additional and Advanced Exercises 642 11 Parametric Equations and Polar Coordinates 647 11.1 Parametrizations of Plane Curves 647 11.2 Calculus with Parametric Curves 654 11.3 Polar Coordinates 662 11.4 Graphing in Polar Coordinates 667 11.5 Areas and Lengths in Polar Coordinates 674 11.6 Conic Sections 679 11.7 Conics in Polar Coordinates 689 Practice Exercises 699 Additional and Advanced Exercises 709 12 Vectors and the Geometry of Space 715 12.1 12.2 12.3 12.4 12.5 12.6 Three-Dimensional Coordinate Systems 715 Vectors 718 The Dot Product 723 The Cross Product 728 Lines and Planes in Space 734 Cylinders and Quadric Surfaces 741 Practice Exercises 746 Additional Exercises 754 13 Vector-Valued Functions and Motion in Space 759 13.1 13.2 13.3 13.4 13.5 13.6 Curves in Space and Their Tangents 759 Integrals of Vector Functions; Projectile Motion 764 Arc Length in Space 770 Curvature and Normal Vectors of a Curve 773 Tangential and Normal Components of Acceleration 778 Velocity and Acceleration in Polar Coordinates 784 Practice Exercises 785 Additional Exercises 791 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 14 Partial Derivatives 795 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10 Functions of Several Variables 795 Limits and Continuity in Higher Dimensions 804 Partial Derivatives 810 The Chain Rule 816 Directional Derivatives and Gradient Vectors 824 Tangent Planes and Differentials 829 Extreme Values and Saddle Points 836 Lagrange Multipliers 849 Taylor's Formula for Two Variables 857 Partial Derivatives with Constrained Variables 859 Practice Exercises 862 Additional Exercises 876 15 Multiple Integrals 881 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 Double and Iterated Integrals over Rectangles 881 Double Integrals over General Regions 882 Area by Double Integration 896 Double Integrals in Polar Form 900 Triple Integrals in Rectangular Coordinates 904 Moments and Centers of Mass 909 Triple Integrals in Cylindrical and Spherical Coordinates 914 Substitutions in Multiple Integrals 922 Practice Exercises 927 Additional Exercises 933 16 Integration in Vector Fields 939 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8 Line Integrals 939 Vector Fields and Line Integrals; Work, Circulation, and Flux 944 Path Independence, Potential Functions, and Conservative Fields 952 Green's Theorem in the Plane 957 Surfaces and Area 963 Surface Integrals 972 Stokes's Theorem 980 The Divergence Theorem and a Unified Theory 984 Practice Exercises 989 Additional Exercises 997 Copyright © 2010 Pearson Education Inc. 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CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 SEQUENCES 2 13 2 14 3 " 1. a" œ 11#1 œ 0, a# œ " ## œ 4 , a$ œ 3# œ 9 , a% œ 4# œ 16 1 2. a" œ 1!1 œ 1, a# œ #"! œ 2" , a$ œ 3!1 œ 61 , a% œ 4!1 œ 24 3. # $ % & ) (1) (1) " " " a" œ (#1)1 œ 1, a# œ (4" 1 œ 3 , a$ œ 6 1 œ 5 , a% œ 8 1 œ 7 4. a" œ 2 (1)" œ 1, a# œ 2 (1)# œ 3, a$ œ 2 (1)$ œ 1, a% œ 2 (1)% œ 3 # $ % 5. a" œ #2# œ #" , a# œ 22$ œ #" , a$ œ 2#% œ #" , a% œ 22& œ #" # $ % 15 6. a" œ 2 # " œ "# , a# œ 2 2# 1 œ 43 , a$ œ 2 2$ 1 œ 87 , a% œ 2 2% " œ 16 15 31 63 " 7. a" œ 1, a# œ 1 "# œ #3 , a$ œ #3 #"# œ 74 , a% œ 74 #"$ œ 15 8 , a& œ 8 #% œ 16 , a' œ 32 , 255 511 1023 a( œ 127 64 , a) œ 128 , a* œ 256 , a"! œ 512 ˆ #" ‰ ˆ "6 ‰ " 3 œ 6 , a% œ 4 " " a* œ 362,880 , a"! œ 3,628,800 8. a" œ 1, a# œ "# , a$ œ œ #"4 , a& œ ˆ #"4 ‰ 5 " " œ 1#"0 , a' œ 7#"0 , a( œ 5040 , a) œ 40,320 , 9. a" œ 2, a# œ (1)# (2) œ 1, a$ œ (1)2 (1) œ "# , a% œ (1)% ˆ "# ‰ (1)& ˆ "4 ‰ œ 4" , a& œ œ 8" , # # 10. a" œ 2, a# œ 1†(#2) œ 1, a$ œ 2†(31) œ 32 , a% œ 3†ˆ 23 ‰ 4†ˆ " ‰ œ "# , a& œ 5 # œ 52 , a' œ 3" , 4 # $ " " " a' œ 16 , a( œ 3"# , a) œ 64 , a* œ 1#"8 , a"! œ 256 a( œ 27 , a) œ "4 , a* œ 29 , a"! œ "5 11. a" œ 1, a# œ 1, a$ œ 1 1 œ 2, a% œ 2 1 œ 3, a& œ 3 2 œ 5, a' œ 8, a( œ 13, a) œ 21, a* œ 34, a"! œ 55 12. a" œ 2, a# œ 1, a$ œ "# , a% œ ˆ "# ‰ 1 ˆ"‰ œ "# , a& œ ˆ# " ‰ œ 1, a' œ 2, a( œ 2, a) œ 1, a* œ "# , a"! œ "# # 13. an œ (1)n1 , n œ 1, 2, á 14. an œ (1)n , n œ 1, 2, á 15. an œ (1)n1 n# , n œ 1, 2, á 16. an œ ("n)# n1 n 1 , n œ 1, 2, á 17. an œ 3a2n 2b , n œ 1, 2, á 5 18. an œ n2n an 1b , n œ 1, 2, á 19. an œ n# 1, n œ 1, 2, á 20. an œ n 4 , n œ 1, 2, á 21. an œ 4n 3, n œ 1, 2, á 22. an œ 4n 2 , n œ 1, 2, á 23. an œ 3nn! 2 , n œ 1, 2, á 24. an œ 5nn 1 , n œ 1, 2, á 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 570 Chapter 10 Infinite Sequences and Series 25. an œ 1 (#1) n 1 , n œ 1, 2, á 26. an œ 27. n lim 2 (0.1)n œ 2 Ê converges Ä_ n "# (1)n ˆ "# ‰ œ Ú n# Û, n œ 1, 2, á # (Theorem 5, #4) n (") 28. n lim œ n lim 1 (n1) œ 1 Ê converges n Ä_ Ä_ n n ˆ"‰ 2 " 2n 2 n 29. n lim œ n lim œ n lim œ 1 Ê converges Ä _ 1 #n Ä _ ˆ"‰ 2 Ä_ # n 2n " 30. n lim œ n lim Ä _ 1 3È n Ä_ % " 5n 31. n lim œ n lim Ä _ n% 8n$ Ä_ 2Èn Š È"n ‹ Š È"n 3‹ Š n"% ‹ 5 1 ˆ 8n ‰ œ _ Ê diverges œ 5 Ê converges n3 n3 " 32. n lim œ n lim œ n lim œ 0 Ê converges Ä _ n# 5n 6 Ä _ (n 3)(n 2) Ä _ n# (n 1)(n 1) n 2n 1 33. n lim œ n lim œ n lim (n 1) œ _ Ê diverges n1 n1 Ä_ Ä_ Ä_ # Š "# ‹ n $ "n n 34 n lim œ n lim œ _ Ê diverges Ä _ 70 4n# Ä _ Š 70# ‹ 4 n 36. n lim (1)n ˆ1 "n ‰ does not exist Ê diverges Ä_ 35. n lim a1 (1)n b does not exist Ê diverges Ä_ ˆ n #n " ‰ ˆ1 "n ‰ œ lim ˆ "# #"n ‰ ˆ1 n" ‰ œ #" Ê converges 37. n lim Ä_ nÄ_ nb1 (") œ 0 Ê converges 39. n lim Ä _ #n 1 ˆ2 #"n ‰ ˆ3 #"n ‰ œ 6 Ê converges 38. n lim Ä_ ) ˆ "# ‰ œ lim (" œ 0 Ê converges 40. n lim Ä_ n Ä _ #n n n 2n É n 2n 41. n lim œ Ên lim Š 2 ‹ œ È2 Ê converges 1 œ É n lim Ä_ Ä _ n1 Ä _ 1 " n " ˆ "0 ‰ œ _ Ê diverges 42. n lim œ n lim Ä _ (0.9)n Ä_ 9 n ˆ 1 n" ‰‹ œ sin 1# œ 1 Ê converges 43. n lim sin ˆ 1# n" ‰ œ sin Šn lim Ä_ Ä_ # 44. n lim n1 cos (n1) œ n lim (n1)(1)n does not exist Ê diverges Ä_ Ä_ sin n 45. n lim œ 0 because n" Ÿ sinn n Ÿ n" Ê converges by the Sandwich Theorem for sequences Ä_ n # # sin n 46. n lim œ 0 because 0 Ÿ sin#n n Ÿ #"n Ê converges by the Sandwich Theorem for sequences Ä _ #n n " ^ 47. n lim œ n lim œ 0 Ê converges (using l'Hopital's rule) Ä _ #n Ä _ #n ln 2 Copyright © 2010 Pearson Education Inc. 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Section 10.1 Sequences n n n # $ n 3 (ln 3) 3 (ln 3) 3 3 ln 3 ^ 48. n lim œ n lim œ n lim œ n lim œ _ Ê diverges (using l'Hopital's rule) 6n 6 Ä _ n$ Ä _ 3n# Ä_ Ä_ ln (n ") 49. n lim œ n lim Èn Ä_ Ä_ ˆn " 1‰ " ‹ Š #È n 2È n œ n lim œ n lim Ä _ n1 Ä_ Š È2n ‹ 1 Š n" ‹ œ 0 Ê converges ˆ"‰ ln n n 50. n lim œ n lim œ 1 Ê converges Ä _ ln 2n Ä_ ˆ2‰ 2n 51. n lim 81În œ 1 Ê converges Ä_ (Theorem 5, #3) 52. n lim (0.03)1În œ 1 Ê converges Ä_ (Theorem 5, #3) ˆ1 7n ‰ œ e( Ê converges 53. n lim Ä_ (Theorem 5, #5) n n ) " ˆ1 "n ‰ œ lim ’1 (" Ê converges 54. n lim n “ œe Ä_ nÄ_ n (Theorem 5, #5) n È 55. n lim 10n œ n lim 101În † n1În œ 1 † 1 œ 1 Ê converges Ä_ Ä_ # n n È ˆÈ 56. n lim n# œ n lim n‰ œ 1# œ 1 Ê converges Ä_ Ä_ ˆ3‰ 57. n lim Ä_ n 1 În Ä_ 1În œ " œ 1 Ê converges œ nlim 1 n lim 31În n (Theorem 5, #3 and #2) (Theorem 5, #2) (Theorem 5, #3 and #2) Ä_ 58. n lim (n 4)1ÎÐn4Ñ œ x lim x1Îx œ 1 Ê converges; (let x œ n 4, then use Theorem 5, #2) Ä_ Ä_ ln n Ä_ 1În œ _ œ _ Ê diverges 59. n lim œ nlim 1 n Ä _ n1În lim ln n n Ä_ (Theorem 5, #2) n 60. n lim cln n ln (n 1)d œ n lim ln ˆ n n 1 ‰ œ ln Šn lim ‹ œ ln 1 œ 0 Ê converges Ä_ Ä_ Ä _ n1 n n È 61. n lim 4n n œ n lim 4È n œ 4 † 1 œ 4 Ê converges Ä_ Ä_ (Theorem 5, #2) n È 62. n lim 32n1 œ n lim 32 a1Înb œ n lim 3# † 31În œ 9 † 1 œ 9 Ê converges Ä_ Ä_ Ä_ "†2†3â(n 1)(n) n! ˆ " ‰ œ 0 and nn!n 63. n lim œ n lim Ÿ n lim n†n†nân†n Ä _ nn Ä_ Ä_ n (4) 64. n lim œ 0 Ê converges Ä _ n! n " œ_ 'n Š (10n! ) ‹ n! 66. n lim œ n lim Ä _ 2n 3n Ä_ " œ_ ˆ 6n!n ‰ 1ÎÐln nÑ n! 0 Ê n lim œ 0 Ê converges Ä _ nn (Theorem 5, #6) n! 65. n lim œ n lim Ä _ 106n Ä_ ˆ"‰ 67. n lim Ä_ n (Theorem 5, #3) Ê diverges Ê diverges (Theorem 5, #6) (Theorem 5, #6) œ n lim exp ˆ ln"n ln ˆ n" ‰‰ œ n lim exp ˆ ln 1lnnln n ‰ œ e" Ê converges Ä_ Ä_ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 571 572 Chapter 10 Infinite Sequences and Series n ˆ1 n" ‰n ‹ œ ln e œ 1 Ê converges 68. n lim ln ˆ1 "n ‰ œ ln Šn lim Ä_ Ä_ (Theorem 5, #5) " ‰‰ ˆ 3n " ‰ œ lim exp ˆn ln ˆ 3n 69. n lim œ n lim exp Š ln (3n 1) " ln (3n 1) ‹ 3n 1 Ä _ 3n 1 nÄ_ Ä_ n n 3 3 6n #Î$ ˆ6‰ œ n lim exp 3n 1 "3n 1 œ n lim exp Š (3n 1)(3n Ê converges 1) ‹ œ exp 9 œ e Ä_ Ä_ Š ‹ # n# " ˆ n ‰ œ lim exp ˆn ln ˆ n n 1 ‰‰ œ lim exp Š ln n ln" (n 1) ‹ œ lim exp n 70. n lim ˆ ‰ Ä _ n1 nÄ_ nÄ_ nÄ_ n n"1 Š n"# ‹ n # œ n lim exp Š n(nn 1) ‹ œ e" Ê converges Ä_ n ˆ x ‰ 71. n lim Ä _ 2n 1 1 În œ n lim x ˆ #n " 1 ‰ Ä_ 1În 1) œ x n lim exp ˆ "n ln ˆ #n " 1 ‰‰ œ x n lim exp Š ln (2n ‹ n Ä_ Ä_ œ x n lim exp ˆ 2n21 ‰ œ xe! œ x, x 0 Ê converges Ä_ n ˆ1 n"# ‰ œ lim exp ˆn ln ˆ1 n"# ‰‰ œ lim exp 72. n lim Ä_ nÄ_ nÄ_ ln Š1 n"# ‹ ˆ n" ‰ Š 2$ ‹‚Š1 n"# ‹ exp – n œ n lim Ä_ Š n"# ‹ — œ n lim exp ˆ n# 2n1 ‰ œ e! œ 1 Ê converges Ä_ 3 †6 36 73. n lim œ n lim œ 0 Ê converges Ä _ 2cn †n! Ä _ n! n n n ˆ 10 ‰n (Theorem 5, #6) ˆ 12 ‰n ˆ 10 ‰n ˆ 120 ‰n 11 11 11 121 74. n lim lim lim œ 0 Ê converges n n œ n n n n œ n Ä _ ˆ 9 ‰ ˆ 11 ‰ n Ä _ ˆ 12 ‰ ˆ 9 ‰ ˆ 12 ‰ ˆ 11 ‰ n Ä _ ˆ 108 ‰ 1 10 12 11 10 11 12 110 (Theorem 5, #4) e e e " 2e 75. n lim tanh n œ n lim œ n lim œ n lim œ n lim " œ 1 Ê converges Ä_ Ä _ en e n Ä _ e2n 1 Ä _ 2e2n Ä_ n n e 76. n lim sinh (ln n) œ n lim Ä_ Ä_ 77. n lim Ä_ 2n ln n 2n n ˆ n" ‰ e ln n œ n lim œ_ 2 # Ä_ Ê diverges ˆcos ˆ "n ‰‰ Š n"# ‹ cos ˆ n" ‰ sin ˆ "n ‰ n# sin ˆ n" ‰ œ lim œ lim œ "# 2n 1 œ n lim Ä _ Š2 " ‹ nÄ_ n Ä _ # ˆ 2n ‰ Š 2# 2$ ‹ # n n 78. n lim n ˆ1 cos "n ‰ œ n lim Ä_ Ä_ Èn sinŠ È1 ‹ œ lim 79. n lim n Ä_ nÄ_ n Ê converges n sin ˆ "n ‰‘ Š "# ‹ ˆ" cos "n ‰ n œ œ n lim lim sin ˆ "n ‰ œ 0 Ê converges ˆ "n ‰ nÄ_ Ä_ Š"‹ n# sinŠ È1n ‹ Èn 1 œ n lim Ä_ cos Š È1n ‹Š 1 2n3Î2 1 ‹ 2n3Î2 œ n lim cos Š È1n ‹ œ cos 0 œ 1 Ê converges Ä_ 80. n lim a3n 5n b1În œ n lim exp’lna3n 5n b1În “ œ n lim exp’ lna3 n 5 b “ œ n lim exp– Ä_ Ä_ Ä_ Ä_ n n œ n lim exp’ Ä_ Š 35n ‹ln 3 ln 5 ˆ 35nn ‰ 1 n 3n ln 3 b 5n ln 5 3n b 5n 1 — ˆ 3 ‰n ln 3 ln 5 exp’ 5 ˆ 3 ‰n 1 “ œ expaln 5b œ 5 “ œ n lim Ä_ 81. n lim tan" n œ 1# Ê converges Ä_ 5 " 82. n lim tan" n œ 0 † 1# œ 0 Ê converges Ä _ Èn Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.1 Sequences 573 n ˆ " ‰n È" n œ lim Šˆ 3" ‰n Š È" ‹ ‹ œ 0 Ê converges 83. n lim Ä_ 3 nÄ_ 2 (Theorem 5, #4) 2 # n 1‰ ! È 84. n lim n# n œ n lim exp ’ ln ann nb “ œ n lim exp ˆ 2n n# n œ e œ 1 Ê converges Ä_ Ä_ Ä_ #!! "** (ln n) 85. n lim n Ä_ 200 (ln n) œ n lim n Ä_ & Š 5(lnnn) ‹ "*) 200†199 (ln n) œ n lim n Ä_ 200! œ á œ n lim œ 0 Ê converges Ä_ n % (ln n) 86. n lim œ n lim Ä _ Èn Ä_ – " Š #Èn % $ 10(ln n) 80(ln n) 3840 œ n lim œ n lim œ á œ n lim œ 0 Ê converges Èn Èn Ä_ Ä_ Ä _ Èn ‹ — È # 87. n lim Šn Èn# n‹ œ n lim Šn Èn# n‹ Š n Èn# n ‹ œ n lim Ä_ Ä_ Ä_ n n n n " œ n lim Ä _ 1 É1 " n È n# n n œ "# Ê converges È # È n# 1 È n# n È # " " œ n lim Š ‹ Š Èn# 1 Èn# n ‹ œ n lim 1 n Ä _ È n# 1 È n# n Ä_ n 1 È n# n n 1 n n 88. n lim Ä_ È # œ n lim Ä_ É1 n"# É1 "n ˆ n" 1‰ œ 2 Ê converges " ' " ln n " 89. n lim dx œ n lim œ n lim œ 0 Ê converges Ä_ n 1 x Ä_ n Ä_ n n (Theorem 5, #1) " ˆ " ' " dx œ n lim 90. n lim 1‰ œ p" 1 if p 1 Ê converges ’ " " “ œ n lim Ä _ 1 xp Ä _ 1p xpc1 1 Ä _ 1p npc1 n n 72 2 91. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ê L œ 1 72 L Ê La1 Lb œ 72 Ê L L 72 œ 0 Ä_ n Ä _ n1 Ä _ 1 an Ê L œ 9 or L œ 8; since an 0 for n 1ÊLœ8 an 6 6 2 92. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ê L œ LL 2 Ê LaL 2b œ L 6 Ê L L 6 œ 0 Ä_ n Ä _ n1 Ä _ an 2 Ê L œ 3 or L œ 2; since an 0 for n 2ÊLœ2 È8 2an Ê L œ È8 2L Ê L2 2L 8 œ 0 Ê L œ 2 93. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ä_ n Ä _ n1 Ä_ or L œ 4; since an 0 for n 3ÊLœ4 È8 2an Ê L œ È8 2L Ê L2 2L 8 œ 0 Ê L œ 2 94. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ä_ n Ä _ n1 Ä_ or L œ 4; since an 0 for n 2ÊLœ4 È5an Ê L œ È5L Ê L2 5L œ 0 Ê L œ 0 or L œ 5; since 95. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ä_ n Ä _ n1 Ä_ an 0 for n 1ÊLœ5 ˆ12 Èan ‰ Ê L œ Š12 ÈL‹ Ê L2 25L 144 œ 0 96. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ä_ n Ä _ n1 Ä_ Ê L œ 9 or L œ 16; since 12 Èan 12 for n 97. an 1 œ 2 a1n , n 1ÊLœ9 1, a1 œ 2. Since an converges Ê n lim a œ L Ê n lim a œ n lim Š2 a1n ‹ Ê L œ 2 L1 Ä_ n Ä _ n1 Ä_ Ê L2 2L 1 œ 0 Ê L œ 1 „ È2; since an 0 for n 1 Ê L œ 1 È2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 574 Chapter 10 Infinite Sequences and Series È1 an Ê L œ È1 L 1, a1 œ È1. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ä_ n Ä _ n1 Ä_ 98. an 1 œ È1 an , n È È Ê L2 L 1 œ 0 Ê L œ 1 „2 5 ; since an 0 for n 1 Ê L œ 1 2 5 99. 1, 1, 2, 4, 8, 16, 32, á œ 1, 2! , 2" , 2# , 2$ , 2% , 2& , á Ê x" œ 1 and xn œ 2nc2 for n 2 100. (a) 1# 2(1)# œ 1, 3# 2(2)# œ 1; let f(aß b) œ (a 2b)# 2(a b)# œ a# 4ab 4b# 2a# 4ab 2b# œ 2b# a# ; a# 2b# œ 1 Ê f(aß b) œ 2b# a# œ 1; a# 2b# œ 1 Ê f(aß b) œ 2b# a# œ 1 # # # # " 2a 4ab 2b ‰ 2 œ a 4ab 4b(a (b) r#n 2 œ ˆ aa2b œ a(aa b)2b# b œ „" b)# b y# Ê rn œ Ê 2 „ Š yn ‹ # # # # n n. Let ba represent the (n 1)th fraction where ba In the first and second fractions, yn for n a positive integer lim rn œ È2. 3. Now the nth fraction is aa2b b and a b 2b 2n 2 1 and b n Ê yn n1 n. Thus, nÄ_ 101. (a) f(x) œ x# 2; the sequence converges to 1.414213562 ¸ È2 (b) f(x) œ tan (x) 1; the sequence converges to 0.7853981635 ¸ 14 (c) f(x) œ ex ; the sequence 1, 0, 1, 2, 3, 4, 5, á diverges 102. (a) n lim nf ˆ n" ‰ œ Ä_ f(?x) ?x œ lim ?x Ä !b lim ?x Ä !b f(0?x) f(0) œ f w (0), where ?x œ n" ?x (b) n lim n tan" ˆ "n ‰ œ f w (0) œ 1 " 0# œ 1, f(x) œ tan" x Ä_ (c) n lim n ae1În 1b œ f w (0) œ e! œ 1, f(x) œ ex 1 Ä_ (d) n lim n ln ˆ1 2n ‰ œ f w (0) œ 1 22(0) œ 2, f(x) œ ln (1 2x) Ä_ # # # 103. (a) If a œ 2n 1, then b œ Ú a# Û œ Ú 4n #4n 1 Û œ Ú2n# 2n "# Û œ 2n# 2n, c œ Ü a# Ý œ Ü2n# 2n "# Ý # œ 2n# 2n 1 and a# b# œ (2n 1)# a2n# 2nb œ 4n# 4n 1 4n% 8n$ 4n# # œ 4n% 8n$ 8n# 4n 1 œ a2n# 2n 1b œ c# . (b) a lim Ä_ # Ú a# Û # Ü a# Ý # 2n 2n œ a lim œ 1 or a lim Ä _ 2n# 2n 1 Ä_ # Ú a# Û # Ü a# Ý œ a lim sin ) œ Ä_ lim ) Ä 1 Î2 sin ) œ 1 Š 21 ‹ 2n1 ‰ 104. (a) n lim (2n1)1Î a2nb œ n lim exp ˆ ln2n œ n lim exp 2n#1 œ n lim exp ˆ #"n ‰ œ e! œ 1; Ä_ Ä_ Ä_ Ä_ n n n! ¸ ˆ ne ‰ È 2n1 , Stirlings approximation Ê È n! ¸ ˆ ne ‰ (2n1)1Î a2nb ¸ ne for large values of n (b) n 40 50 60 n È n! 15.76852702 19.48325423 23.19189561 n e 14.71517765 18.39397206 22.07276647 ˆ"‰ ln n " n 105. (a) n lim œ n lim œ n lim œ0 Ä _ nc Ä _ cncc1 Ä _ cnc (b) For all % 0, there exists an N œ eÐln %ÑÎc such that n eÐln %ÑÎc Ê ln n lnc % Ê ln nc ln ˆ "% ‰ Ê nc "% Ê n"c % Ê ¸ n"c 0¸ % Ê lim n"c œ 0 nÄ_ 106. Let {an } and {bn } be sequences both converging to L. Define {cn } by c2n œ bn and c2nc1 œ an , where n œ 1, 2, 3, á . For all % 0 there exists N" such that when n N" then kan Lk % and there exists N# such that when n N# then kbn Lk %. If n 1 2max{N" ß N# }, then kcn Lk %, so {cn } converges to L. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.1 Sequences 575 107. n lim n1În œ n lim exp ˆ "n ln n‰ œ n lim exp ˆ n" ‰ œ e! œ 1 Ä_ Ä_ Ä_ 108. n lim x1În œ n lim exp ˆ "n ln x‰ œ e! œ 1, because x remains fixed while n gets large Ä_ Ä_ 109. Assume the hypotheses of the theorem and let % be a positive number. For all % there exists a N" such that when n N" then kan Lk % Ê % an L % Ê L % an , and there exists a N# such that when n N# then kcn Lk % Ê % cn L % Ê cn L %. If n max{N" ß N# }, then L % an Ÿ bn Ÿ cn L % Ê kbn Lk % Ê n lim b œ L. Ä_ n 110. Let % !. We have f continuous at L Ê there exists $ so that kx Lk $ Ê kf(x) f(L)k %. Also, an Ä L Ê there exists N so that for n N kan Lk $ . Thus for n N, kf(an ) f(L)k % Ê f(an ) Ä f(L). 1) 1 3n 1 3n 4 3n 1 # # an Ê 3(n (n 1) 1 n 1 Ê n # n 1 Ê 3n 3n 4n 4 3n 6n n 2 111. an1 " Ê 4 2; the steps are reversible so the sequence is nondecreasing; 3n n 1 3 Ê 3n 1 3n 3 Ê 1 3; the steps are reversible so the sequence is bounded above by 3 1) 3)! (2n 3)! (2n 5)! (2n 3)! (2n 5)! (n 2)! an Ê (2(n ((n 1) 1)! (n 1)! Ê (n 2)! (n 1)! Ê (2n 3)! (n 1)! 112. an1 Ê (2n 5)(2n 4) n 2; the steps are reversible so the sequence is nondecreasing; the sequence is not 3)! bounded since (2n (n 1)! œ (2n 3)(2n 2)â(n 2) can become as large as we please nb1 nb1 n n nb1 nb1 113. an1 Ÿ an Ê 2(n 31)! Ÿ 2n!3 Ê 2 2n 33n Ÿ (n n!1)! Ê 2 † 3 Ÿ n 1 which is true for n 5; the steps are reversible so the sequence is decreasing after a& , but it is not nondecreasing for all its terms; a" œ 6, a# œ 18, a$ œ 36, a% œ 54, a& œ 324 5 œ 64.8 Ê the sequence is bounded from above by 64.8 an Ê 2 n2 1 #n"b1 114. an1 2 n2 #"n Ê 2n n2 1 " " #nb1 #n Ê n(n 2 1) #n"b1 ; the steps are reversible so the sequence is nondecreasing; 2 2n #"n Ÿ 2 Ê the sequence is bounded from above 115. an œ 1 "n converges because n" Ä 0 by Example 1; also it is a nondecreasing sequence bounded above by 1 116. an œ n "n diverges because n Ä _ and n" Ä 0 by Example 1, so the sequence is unbounded 117. an œ 2 2n 1 œ 1 #"n and 0 #"n "n ; since "n Ä 0 (by Example 1) Ê #"n Ä 0, the sequence converges; also it is n a nondecreasing sequence bounded above by 1 n 118. an œ 2 3n 1 œ ˆ 23 ‰ 3"n ; the sequence converges to ! by Theorem 5, #4 n 119. an œ a(1)n 1b ˆ nn 1 ‰ diverges because an œ 0 for n odd, while for n even an œ 2 ˆ1 n" ‰ converges to 2; it diverges by definition of divergence 120. xn œ max {cos 1ß cos 2ß cos 3ß á ß cos n} and xn1 œ max {cos 1ß cos 2ß cos 3ß á ß cos (n 1)} so the sequence is nondecreasing and bounded above by 1 Ê the sequence converges. 121. an È an1 Í 1 Èn 2n È and 1 Èn 2n " È2(n 1) Èn 1 Í Èn 1 È2n# 2n xn with xn Ÿ 1 Èn È2n# 2n Í Èn 1 È2 ; thus the sequence is nonincreasing and bounded below by È2 Ê it converges Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Èn 576 Chapter 10 Infinite Sequences and Series 122. an an1 Í n n 1 (n 1) " n1 Í n# 2n 1 n# 2n Í 1 0 and n n 1 1; thus the sequence is nonincreasing and bounded below by 1 Ê it converges 123. n n n" n 4nb1 3n ˆ 34 ‰n1 Í 1 43 and œ 4 ˆ 34 ‰ so an an1 Í 4 ˆ 34 ‰ 4 ˆ 34 ‰ Í ˆ 34 ‰ 4n n 4 ˆ 34 ‰ 4; thus the sequence is nonincreasing and bounded below by 4 Ê it converges 124. a" œ 1, a# œ 2 3, a$ œ 2(2 3) 3 œ 2# a22 "b † 3, a% œ 2 a2# a22 "b † 3b 3 œ 2$ a2$ 1b 3, a& œ 2 c2$ a2$ 1b 3d 3 œ 2% a2% 1b 3, á , an œ 2n" a2n" 1b 3 œ 2n" 3 † 2n1 3 œ 2n1 (1 3) 3 œ 2n 3; an an1 Í 2n 3 2n1 3 Í 2n 2n1 Í 1 Ÿ 2 so the sequence is nonincreasing but not bounded below and therefore diverges 125. Let 0 M 1 and let N be an integer greater than 1 MM . Then n N Ê n 1 MM Ê n nM M Ê n M nM Ê n M(n 1) Ê n n 1 M. 126. Since M" is a least upper bound and M# is an upper bound, M" Ÿ M# . Since M# is a least upper bound and M" is an upper bound, M# Ÿ M" . We conclude that M" œ M# so the least upper bound is unique. 127. The sequence an œ 1 ("#) is the sequence #" , #3 , #" , #3 , á . This sequence is bounded above by #3 , n but it clearly does not converge, by definition of convergence. 128. Let L be the limit of the convergent sequence {an }. Then by definition of convergence, for #% there corresponds an N such that for all m and n, m N Ê kam Lk #% and n N Ê kan Lk #% . Now kam an k œ kam L L an k Ÿ kam Lk kL an k #% #% œ % whenever m N and n N. 129. Given an % 0, by definition of convergence there corresponds an N such that for all n N, kL" an k % and kL# an k %. Now kL# L" k œ kL# an an L" k Ÿ kL# an k kan L" k % % œ 2%. kL# L" k 2% says that the difference between two fixed values is smaller than any positive number 2%. The only nonnegative number smaller than every positive number is 0, so kL" L# k œ 0 or L" œ L# . 130. Let k(n) and i(n) be two order-preserving functions whose domains are the set of positive integers and whose ranges are a subset of the positive integers. Consider the two subsequences akÐnÑ and aiÐnÑ , where akÐnÑ Ä L" , aiÐnÑ Ä L# and L" Á L# . Thus ¸akÐnÑ aiÐnÑ ¸ Ä kL" L# k 0. So there does not exist N such that for all m, n N Ê kam an k %. So by Exercise 128, the sequence Öan × is not convergent and hence diverges. 131. a2k Ä L Í given an % 0 there corresponds an N" such that c2k N" Ê ka2k Lk %d . Similarly, a2k1 Ä L Í c2k 1 N# Ê ka2k1 Lk %d . Let N œ max{N" ß N# }. Then n N Ê kan Lk % whether n is even or odd, and hence an Ä L. 132. Assume an Ä 0. This implies that given an % 0 there corresponds an N such that n N Ê kan 0k % Ê kan k % Ê kkan kk % Ê kkan k 0k % Ê kan k Ä 0. On the other hand, assume kan k Ä 0. This implies that given an % 0 there corresponds an N such that for n N, kkan k 0k % Ê kkan kk % Ê kan k % Ê kan 0k % Ê an Ä 0. # # # # 133. (a) f(x) œ x# a Ê f w (x) œ 2x Ê xn1 œ xn xn#xn a Ê xn1 œ 2xn 2xaxnn ab œ xn2xn a œ ˆxn xa ‰ n # (b) x" œ 2, x# œ 1.75, x$ œ 1.732142857, x% œ 1.73205081, x& œ 1.732050808; we are finding the positive number where x# 3 œ 0; that is, where x# œ 3, x 0, or where x œ È3 . Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.2 Infinite Series 577 134. x" œ 1, x# œ 1 cos (1) œ 1.540302306, x$ œ 1.540302306 cos (1 cos (1)) œ 1.570791601, x% œ 1.570791601 cos (1.570791601) œ 1.570796327 œ 1# to 9 decimal places. After a few steps, the arc axnc1 b and line segment cos axnc1 b are nearly the same as the quarter circle. 135-146. Example CAS Commands: Mathematica: (sequence functions may vary): Clear[a, n] a[n_]; = n1 / n first25= Table[N[a[n]],{n, 1, 25}] Limit[a[n], n Ä 8] Mathematica: (sequence functions may vary): Clear[a, n] a[n_]; = n1 / n first25= Table[N[a[n]],{n, 1, 25}] Limit[a[n], n Ä 8] The last command (Limit) will not always work in Mathematica. You could also explore the limit by enlarging your table to more than the first 25 values. If you know the limit (1 in the above example), to determine how far to go to have all further terms within 0.01 of the limit, do the following. Clear[minN, lim] lim= 1 Do[{diff=Abs[a[n] lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}] minN For sequences that are given recursively, the following code is suggested. The portion of the command a[n_]:=a[n] stores the elements of the sequence and helps to streamline computation. Clear[a, n] a[1]= 1; a[n_]; = a[n]= a[n 1] (1/5)(n1) first25= Table[N[a[n]], {n, 1, 25}] The limit command does not work in this case, but the limit can be observed as 1.25. Clear[minN, lim] lim= 1.25 Do[{diff=Abs[a[n] lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}] minN 10.2 INFINITE SERIES n a1 r b 1. sn œ a (1 r) œ 2 ˆ1 ˆ "3 ‰ ‰ 1 ˆ "3 ‰ a1 r b 2. sn œ a (1 r) œ " ‰n ‰ 9 ‰ˆ ˆ 100 1 ˆ 100 " 1 ˆ 100 ‰ n n 1 ˆ " ‰ Ê n lim s œ 1 2ˆ " ‰ œ 3 Ä_ n 3 ˆ 9 ‰ " Ê n lim s œ 1 100 ˆ " ‰ œ 11 Ä_ n 100 n a1 r b # 3. sn œ a (1 s œ ˆ "3 ‰ œ 32 r) œ 1 ˆ "# ‰ Ê n lim Ä_ n # n 4. sn œ 11 ((2)2) , a geometric series where krk 1 Ê divergence n 5. " " " (n 1)(n #) œ n 1 n # Ê sn œ ˆ #" 3" ‰ ˆ 3" 4" ‰ á ˆ n " 1 n " # ‰ œ #" n " # Ê n lim s œ #" Ä_ n Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 578 6. Chapter 10 Infinite Sequences and Series 5 5 5 n(n 1) œ n n 1 Ê sn œ ˆ5 25 ‰ ˆ 25 35 ‰ ˆ 35 45 ‰ á ˆ n 5 1 n5 ‰ ˆ n5 n 5 1 ‰ œ 5 n 5 1 Ê n lim s œ5 Ä_ n " " 7. 1 "4 16 64 á , the sum of this geometric series is 1 ˆ" " ‰ œ 1 "ˆ " ‰ œ 45 4 4 8. " ‰ ˆ 16 " " " " 16 64 256 á , the sum of this geometric series is 1 ˆ "4 ‰ œ 1# 9. ˆ 74 ‰ 7 7 7 7 4 16 64 á , the sum of this geometric series is 1 ˆ "4 ‰ œ 3 5 5 10. 5 54 16 64 á , the sum of this geometric series is 1 ˆ5 " ‰ œ 4 4 11. (5 1) ˆ 5# "3 ‰ ˆ 54 9" ‰ ˆ 58 #"7 ‰ á , is the sum of two geometric series; the sum is 5 1 "ˆ " ‰ œ 10 #3 œ 23 # 1 ˆ "# ‰ 3 12. (5 1) ˆ 5# "3 ‰ ˆ 54 9" ‰ ˆ 58 #"7 ‰ á , is the difference of two geometric series; the sum is 5 1 "ˆ " ‰ œ 10 #3 œ 17 # 1 ˆ "# ‰ 3 " ‰ 13. (1 1) ˆ 1# "5 ‰ ˆ 14 25 ˆ 18 1#"5 ‰ á , is the sum of two geometric series; the sum is 1 1 "ˆ " ‰ œ 2 56 œ 17 6 1 ˆ "# ‰ 5 8 16 4 8 14. 2 45 25 125 á œ 2 ˆ1 25 25 125 á ‰ ; the sum of this geometric series is 2 Š 1 "ˆ 2 ‰ ‹ œ 10 3 5 15. Series is geometric with r œ 25 Ê ¹ 25 ¹ 1 Ê Converges to 1 1 2 œ 53 5 16. Series is geometric with r œ 3 Ê ¹3¹ 1 Ê Diverges 1 17. Series is geometric with r œ 18 Ê ¹ 18 ¹ 1 Ê Converges to 1 8 1 œ 17 8 2 18. Series is geometric with r œ 23 Ê ¹ 23 ¹ 1 Ê Converges to 1 ˆ3 2 ‰ œ 25 3 _ n _ Š 23 ‹ 23 ˆ " ‰ 19. 0.23 œ ! 100 œ 1 100 œ 23 10# 99 ˆ " ‰ n œ0 n œ0 100 _ n 7 ˆ " ‰ 21. 0.7 œ ! 10 œ 10 nœ0 _ 7 Š 10 ‹ " 1 Š 10 ‹ n 1 ‰ˆ 6 ‰ˆ " ‰ 23. 0.06 œ ! ˆ 10 œ 10 10 n œ0 _ n nœ0 6 Š 100 ‹ " 1 Š 10 ‹ 414 ˆ " ‰ 24. 1.414 œ 1 ! 1000 œ1 10$ n œ0 _ n d ˆ " ‰ 22. 0.d œ ! 10 œ 10 œ 79 234 Š 1000 ‹ n 234 ˆ " ‰ 20. 0.234 œ ! 1000 œ 10$ " 1 Š 1000 ‹ d Š 10 ‹ " 1 Š 10 ‹ œ d9 6 " œ 90 œ 15 414 Š 1000 ‹ " 1 Š 1000 ‹ "413 œ 1 414 999 œ 999 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. œ 234 999 Section 10.2 Infinite Series _ Š 123& ‹ n ! 123& ˆ " $ ‰ œ 124 25. 1.24123 œ 124 100 10 10 100 10 1 Š "$ ‹ n œ0 _ Š 142,857 ' ‹ n 10 1 Š "' ‹ n œ0 28. lim n œ lim 11 œ 1 Á 0 Ê diverges nÄ_ n 10 nÄ_ lim n an 1 b n 2n 1 2 œ lim n2 n 5n 6 œ lim 2n 5 œ lim 2 œ 1 Á 0 Ê diverges 2 nÄ_ an 2ban 3b nÄ_ nÄ_ lim 1 œ 0 Ê test inconclusive nÄ_ n 4 30. 1 lim 2 n œ lim 2n œ 0 Ê test inconclusive nÄ_ n 3 nÄ_ 32. 33. 34. 3,142,854 116,402 œ 3 142,857 10' 1 œ 999,999 œ 37,037 10 29. 31. 123,999 41,333 124 124 123 œ 100 10&123 10# œ 100 99,900 œ 99,900 œ 33,300 10 ˆ 10" ' ‰ œ 3 26. 3.142857 œ 3 ! 142,857 10' 27. 579 nÄ_ lim cos 1n œ cos 0 œ 1 Á 0 Ê diverges nÄ_ n lim ne œ nÄ_ e n n lim n e nÄ_ e 1 n œ lim een œ lim 11 œ 1 Á 0 Ê diverges nÄ_ nÄ_ lim ln 1n œ _ Á 0 Ê diverges nÄ_ lim cos n 1 œ does not exist Ê diverges nÄ_ 35. sk œ ˆ1 "2 ‰ ˆ "2 "3 ‰ ˆ "3 "4 ‰ á ˆ k " 1 k" ‰ ˆ k" k " 1 ‰ œ 1 k " 1 Ê œ lim ˆ1 k " 1 ‰ œ 1, series converges to 1 lim sk kÄ_ kÄ_ 3 ‰ 36. sk œ ˆ 31 34 ‰ ˆ 34 39 ‰ ˆ 39 16 á Š ak 3 1b2 k32 ‹ Š k32 ak 3 1b2 ‹ œ 3 ak 3 1b2 Ê lim sk kÄ_ œ lim Š3 ak 3 1b2 ‹ œ 3, series converges to 3 kÄ_ 37. sk œ ŠlnÈ2 lnÈ1‹ ŠlnÈ3 lnÈ2‹ ŠlnÈ4 lnÈ3‹ á ŠlnÈk lnÈk 1‹ ŠlnÈk 1 lnÈk‹ œ lnÈk 1 lnÈ1 œ lnÈk 1 Ê lim sk œ lim lnÈk 1 œ _; series diverges kÄ_ kÄ_ 38. sk œ atan 1 tan 0b atan 2 tan 1b atan 3 tan 2b á atan k tan ak 1bb atan ak 1b tan kb œ tan ak 1b tan 0 œ tan ak 1b Ê lim sk œ lim tan ak 1b œ does not exist; series diverges kÄ_ kÄ_ 39. sk œ ˆcos1 ˆ 12 ‰ cos1 ˆ 13 ‰‰ ˆcos1 ˆ 13 ‰ cos1 ˆ 14 ‰‰ ˆcos1 ˆ 14 ‰ cos1 ˆ 15 ‰‰ á ˆcos1 ˆ 1k ‰ cos1 ˆ k 1 1 ‰‰ ˆcos1 ˆ k 1 1 ‰ cos1 ˆ k 1 # ‰‰ œ 13 cos1 ˆ k 1 # ‰ Ê lim sk œ lim ’ 13 cos1 ˆ k 1 # ‰“ œ 13 12 œ 16 , series converges to 16 kÄ_ kÄ_ 40. sk œ ŠÈ5 È4‹ ŠÈ6 È5‹ ŠÈ7 È6‹ á ŠÈk 3 Èk 2‹ ŠÈk 4 Èk 3‹ œ Èk 4 2 Ê lim sk œ lim ’Èk 4 2“ œ _; series diverges kÄ_ kÄ_ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 580 Chapter 10 Infinite Sequences and Series 4 " " "‰ " ‰ " ‰ ˆ ˆ" "‰ ˆ" ˆ " 41. (4n 3)(4n 1) œ 4n 3 4n 1 Ê sk œ 1 5 5 9 9 13 á 4k 7 4k 3 ˆ 4k " 3 4k " 1 ‰ œ 1 4k " 1 Ê lim sk œ lim ˆ1 4k " 1 ‰ œ 1 kÄ_ kÄ_ A(2n 1) B(2n 1) 6 A B 42. (2n 1)(2n Ê A(2n 1) B(2n 1) œ 6 Ê (2A 2B)n (A B) œ 6 1) œ 2n 1 2n 1 œ (2n 1)(2n 1) Ê œ k k 2A 2B œ 0 ABœ0 6 Êœ Ê 2A œ 6 Ê A œ 3 and B œ 3. Hence, ! (2n 1)(2n œ 3 ! ˆ #n " 1 #n " 1 ‰ 1) A Bœ6 ABœ6 n œ1 nœ1 œ 3 Š "1 "3 "3 "5 "5 "7 á #(k "1) 1 2k " 1 #k " 1 ‹ œ 3 ˆ1 #k " 1 ‰ Ê the sum is lim 3 ˆ1 #k " 1 ‰ œ 3 kÄ_ # # # # A(2n1)(2n1) B(2n1) C(2n1)(2n1) D(2n1) A B C D 43. (2n1)40n # (2n1)# œ (2n1) (2n1)# (2n1) (2n1)# œ (2n1)# (2n1)# Ê A(2n 1)(2n 1)# B(2n 1)# C(2n 1)(2n 1)# D(2n 1)# œ 40n Ê A a8n$ 4n# 2n 1b B a4n# 4n 1b C a8n$ 4n# 2n 1b œ D a4n# 4n 1b œ 40n Ê (8A 8C)n$ (4A 4B 4C 4D)n# (2A 4B 2C 4D)n (A B C D) œ 40n Ú Ú 8A 8C œ 0 8A 8C œ 0 Ý Ý Ý Ý B Dœ 0 4A 4B 4C 4D œ 0 A BC Dœ 0 Ê œ Ê 4B œ 20 Ê B œ 5 Ê Û Ê Û œ œ 2D œ 20 2A 4B 2C 4D 40 A 2 B C 2D 20 2B Ý Ý Ý Ý Ü A B C D œ 0 Ü A B C D œ 0 and D œ 5 Ê œ k ACœ0 Ê C œ 0 and A œ 0. Hence, ! ’ (#n1)40n # (2n1)# “ A 5 C 5 œ 0 n œ1 k œ 5 ! ’ (#n" 1)# (#n" 1)# “ œ 5 Š 1" 9" 9" #"5 #"5 á (2(k1)" 1)# (#k" 1)# (#k" 1)# ‹ n œ1 œ 5 Š1 (2k" 1)# ‹ Ê the sum is n lim 5 Š1 (2k" 1)# ‹ œ 5 Ä_ 1 " " "‰ " ‰ " " " " ˆ ˆ" "‰ ˆ" 44. n#2n (n 1)# œ n# (n 1)# Ê sk œ 1 4 4 9 9 16 á ’ (k 1)# k# “ ’ k# (k 1)# “ Ê lim sk œ lim ’1 (k " 1)# “ œ 1 kÄ_ kÄ_ 45. sk œ Š1 È"2 ‹ Š È"2 È"3 ‹ Š È"3 È"4 ‹ á Š È " k1 Ê È" ‹ Š È" Èk" 1 ‹ œ 1 Èk" 1 k k lim sk œ lim Š1 Èk" 1 ‹ œ 1 kÄ_ kÄ_ " " ‰ " " ‰ 46. sk œ ˆ "# #""Î# ‰ ˆ #"Î# #"Î$ ˆ #"Î$ #"Î% á ˆ #1ÎÐ"k 1Ñ #1"Îk ‰ ˆ #1"Îk #1ÎÐ"k1Ñ ‰ œ #" #1ÎÐ"k1Ñ Ê lim sk œ "# "1 œ "# kÄ_ 47. sk œ ˆ ln"3 ln"# ‰ ˆ ln"4 ln"3 ‰ ˆ ln"5 ln"4 ‰ á Š ln (k" 1) ln"k ‹ Š ln (k" 2) ln (k" 1) ‹ œ ln"# ln (k" 2) Ê lim sk œ ln"# kÄ_ 48. sk œ ctan" (1) tan" (2)d ctan" (2) tan" (3)d á ctan" (k 1) tan" (k)d ctan" (k) tan" (k 1)d œ tan" (1) tan" (k 1) Ê lim sk œ tan" (1) 1# œ 14 1# œ 14 kÄ_ 49. convergent geometric series with sum È " œ È22 1 œ 2 È2 " 1 ŠÈ ‹ 2 50. divergent geometric series with krk œ È2 1 51. convergent geometric series with sum Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Š 3# ‹ 1 Š "# ‹ œ1 Section 10.2 Infinite Series 52. n lim (1)n1 n Á 0 Ê diverges Ä_ 581 53. n lim cos (n1) œ n lim (1)n Á 0 Ê diverges Ä_ Ä_ 54. cos (n1) œ (1)n Ê convergent geometric series with sum 55. convergent geometric series with sum " œ 56 1 Š "5 ‹ # " œ e#e1 1 Š "# ‹ e 56. n lim ln 3"n œ _ Á 0 Ê diverges Ä_ 57. convergent geometric series with sum 2 18 2 2 œ 20 " 9 9 œ 9 1 Š 10 ‹ 58. convergent geometric series with sum " œ x x 1 1 Š "x ‹ 59. difference of two geometric series with sum n " " " œ 3 3# œ 3# 1 Š 23 ‹ 1 Š3‹ n ˆ1 "n ‰ œ lim ˆ1 " ‰ œ e" Á 0 Ê diverges 60. n lim n Ä_ nÄ_ n n †n â n 62. n lim œ n lim n lim n œ _ Ê diverges Ä _ n! Ä _ 1†#ân Ä_ n n! 61. n lim œ _ Á 0 Ê diverges Ä _ 1000n _ _ _ _ _ n n _ n _ n 63. ! 2 4n 3 œ ! 42n ! 43n œ ! ˆ 21 ‰ ! ˆ 43 ‰ ; both œ ! ˆ 21 ‰ and ! ˆ 43 ‰ are geometric series, and both converge n n n œ1 n n œ1 n nœ1 nœ1 nœ1 n œ1 n œ1 _ 1 _ 3 n n since r œ 12 Ê ¹ 12 ¹ 1 and r œ 34 Ê ¹ 34 ¹ 1, respectivley Ê ! ˆ 12 ‰ œ 1 2 1 œ 1 and ! ˆ 34 ‰ œ 1 4 3 œ 3 Ê 2 4 n œ1 _ nœ1 ! 2 n 3 œ 1 3 œ 4 by Theorem 8, part (1) n œ1 64. n n 4 4 lim 23n 4n œ n 2n n ˆ 1 ‰n " " lim 43nn " œ lim ˆ 23 ‰n " œ 11 œ 1 Á 0 Ê diverges by nth term test for divergence nÄ_ nÄ_ 4n _ _ n œ1 n œ1 nÄ_ 4 65. ! ln ˆ n n 1 ‰ œ ! cln (n) ln (n 1)d Ê sk œ cln (1) ln (2)d cln (2) ln (3)d cln (3) ln (4)d á cln (k 1) ln (k)d cln (k) ln (k 1)d œ ln (k 1) Ê lim sk œ _, Ê diverges kÄ_ 66. n lim a œ n lim ln ˆ 2n n 1 ‰ œ ln ˆ "# ‰ Á 0 Ê diverges Ä_ n Ä_ 1 67. convergent geometric series with sum 1 "ˆ e ‰ œ 1 e 1 1 68. divergent geometric series with krk œ 1e e ¸ 23.141 22.459 1 _ _ n œ0 n œ0 _ _ n œ0 n œ0 69. ! (1)n xn œ ! (x)n ; a œ 1, r œ x; converges to 1 "(x) œ 1 " x for kxk 1 70. ! (1)n x2n œ ! ax# b ; a œ 1, r œ x# ; converges to 1 " x# for kxk 1 n Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 582 Chapter 10 Infinite Sequences and Series 71. a œ 3, r œ x # 1 ; converges to _ _ 72. ! (#1) ˆ 3 "sin x ‰ œ ! n n n œ0 n œ0 3 œ 3 6 x for 1 x" # 1 or 1 x 3 1 Šx # "‹ ˆ "# ‰ " ˆ " ‰n " " " # 3 sin x ; a œ # , r œ 3 sin x ; converges to 1 Š 3 sin x ‹ 3 sin x 3 sin x " " " ˆ ‰ œ 2(4 sin x) œ 8 2 sin x for all x since 4 Ÿ 3 sin x Ÿ # for all x 73. a œ 1, r œ 2x; converges to 1 " 2x for k2xk 1 or kxk #" 74. a œ 1, r œ x"# ; converges to # " œ x#x 1 for ¸ x1# ¸ 1 or kxk 1. 1 Š #" ‹ x 75. a œ 1, r œ (x 1)n ; converges to 1 (x" 1) œ # " x for kx 1k 1 or 2 x 0 76. a œ 1, r œ 3 # x ; converges to " œ x 2 1 for ¸ 3 # x ¸ 1 or 1 x 5 1 Š3 # x‹ 77. a œ 1, r œ sin x; converges to 1 "sin x for x Á (2k 1) 1# , k an integer 78. a œ 1, r œ ln x; converges to 1 "ln x for kln xk 1 or e" x e _ 79. (a) ! nœ2 _ 80. (a) ! nœ1 _ " (n 4)(n 5) " (b) ! (n 2)(n 3) 5 (n 2)(n 3) 5 (b) ! (n 2)(n 1) n œ0 _ (c) ! n œ3 Š "# ‹ 1 Š "# ‹ 3 (b) one example is 3# 34 38 16 á œ n 1 82. The series ! kˆ 12 ‰ Š 3# ‹ 1 Š "# ‹ n œ0 _ _ nœ1 nœ1 nœ1 _ _ _ nœ1 nœ1 nœ1 5 (n 19)(n 18) œ 3 Š "# ‹ 1 Š "# ‹ is a geometric series whose sum is _ nœ20 œ1 " (c) one example is 1 "# 4" 8" 16 á œ 1 n n œ5 _ " 81. (a) one example is "# 4" 8" 16 á œ _ _ " (c) ! (n 3)(n #) œ 0. Š k# ‹ 1 Š "# ‹ n œ k where k can be any positive or negative number. _ _ nœ1 nœ1 _ _ nœ1 nœ1 83. Let an œ bn œ ˆ "# ‰ . Then ! an œ ! bn œ ! ˆ "# ‰ œ 1, while ! Š bann ‹ œ ! (1) diverges. n n n 84. Let an œ bn œ ˆ "# ‰ . Then ! an œ ! bn œ ! ˆ "# ‰ œ 1, while ! aan bn b œ ! ˆ 4" ‰ œ 3" Á AB. n n _ _ _ _ n œ1 nœ1 nœ1 nœ1 n 85. Let an œ ˆ 4" ‰ and bn œ ˆ "# ‰ . Then A œ ! an œ 3" , B œ ! bn œ 1 and ! Š bann ‹ œ ! ˆ "# ‰ œ 1 Á AB . 86. Yes: ! Š a"n ‹ diverges. The reasoning: ! an converges Ê an Ä 0 Ê a"n Ä _ Ê ! Š a"n ‹ diverges by the nth-Term Test. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.3 The Integral Test 87. Since the sum of a finite number of terms is finite, adding or subtracting a finite number of terms from a series that diverges does not change the divergence of the series. 88. Let An œ a" a# á an and n lim A œ A. Assume ! aan bn b converges to S. Let Ä_ n Sn œ (a" b" ) (a# b# ) á (an bn ) Ê Sn œ (a" a# á an ) (b" b# á bn ) Ê b" b# á bn œ Sn An Ê n lim ab" b# á bn b œ S A Ê ! bn converges. This Ä_ contradicts the assumption that ! bn diverges; therefore, ! aan bn b diverges. 89. (a) (b) 2 1r œ 5 Š 13 2 ‹ 1r # Ê 52 œ 1 r Ê r œ 53 ; 2 2 ˆ 35 ‰ 2 ˆ 35 ‰ á # $ 3 13 13 ˆ 3 ‰ 13 ˆ 3 ‰ ˆ 3 ‰ á œ 5 Ê 13 13 10 œ 1 r Ê r œ 10 ; 2 # 10 # 10 # 10 90. 1 eb e2b á œ 1"eb œ 9 Ê "9 œ 1 eb Ê eb œ 98 Ê b œ ln ˆ 89 ‰ 91. sn œ 1 2r r# 2r$ r% 2r& á r2n 2r2n1 , n œ 0, 1, á Ê sn œ a1 r# r% á r2n b a2r 2r$ 2r& á 2r2n1 b Ê n lim s œ 1 " r# 1 2rr# Ä_ n 2r # œ 11 r# , if kr k 1 or krk 1 92. L sn œ 1 a r a a11 rr b œ 1arr n n # # 93. area œ 2# ŠÈ2‹ (1)# Š È"2 ‹ á œ 4 2 1 #" á œ 1 4 " œ 8 m# # # 94. (a) L" œ 3, L# œ 3 ˆ 43 ‰ , L$ œ 3 ˆ 43 ‰ , á , Ln œ 3 ˆ 43 ‰ (b) nc1 Ê n lim L œ n lim 3 ˆ 43 ‰ Ä_ n Ä_ nc1 œ_ È È Using the fact that the area of an equilateral triangle of side length s is 43 s2 , we see that A" œ 43 , È È È È È È È 2 2 A# œ A" 3Š 43 ‹ˆ "3 ‰ œ 43 1#3 , A$ œ A# 3a4bŠ 43 ‹ˆ 3"2 ‰ œ 43 123 #73 , È È 2 2 A% œ A$ 3a4b2 Š 43 ‹ˆ 3"3 ‰ , A5 œ A4 3a4b3 Š 43 ‹ˆ 3"4 ‰ , . . . , An œ n n n È3 ! 3a4bk2 Š È3 ‹ˆ "2 ‰k1 œ È3 ! 3È3a4bk$ ˆ " ‰k1 œ È3 3È3Œ! 4kkc$1 . 4 4 3 4 9 4 9 kœ2 kœ2 kœ2 n 1 È3 kc$ È È È 1 ‰ lim An œ n lim 3È3Œ! 49k 1 œ 43 3È3Œ 1 36 4 œ 43 3È3ˆ 20 œ 43 ˆ1 53 ‰ nÄ_ Ä_ Œ 4 9 kœ2 È œ 43 ˆ 85 ‰ œ 85 A" 10.3 THE INTEGRAL TEST _1 1; '1 1. faxb œ x12 is positive, continuous, and decreasing for x œ lim bÄ_ ˆ 1b 1‰ œ 1 Ê ' _1 1 x _ bÄ_ ˆ 54 b0.8 54 ‰ œ _ Ê ' '1b x1 dx œ lim 2 bÄ_ b ’ 1x “ 1 ! 12 converges 2 dx converges Ê n œ1 n _ 1 1; '1 2. faxb œ x10.2 is positive, continuous, and decreasing for x œ lim x2 dx œ b lim Ä_ _ 1 1 x0.2 dx diverges Ê _ x0.2 dx œ b lim Ä_ '1b x1 dx œ lim 0.2 bÄ_ ! 10.2 diverges n œ1 n Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. b ’ 54 x0.8 “ 1 583 584 Chapter 10 Infinite Sequences and Series _ 1; '1 3. faxb œ x2 1 4 is positive, continuous, and decreasing for x œ lim ˆ 12 tan1 b2 12 tan1 12 ‰ œ 14 12 tan1 12 Ê bÄ_ '1b x 1 4 dx œ lim ’lnlx 4l“ '1b e2x dx œ lim ’ 12 e2x “ _ 2 _ 1 1; '1 x 4 dx œ b lim Ä_ _ b bÄ_ 1 2 n œ1 _ alnlb 4l ln 5b œ _ Ê '1 x 1 4 dx diverges Ê ! n 1 4 diverges bÄ_ ’ 21 tan1 x2 “ 2 '1_ x 1 4 dx converges Ê ! n 1 4 converges 4. faxb œ x 1 4 is positive, continuous, and decreasing for x œ lim '1b x 1 4 dx œ lim 1 x2 4 dx œ b lim Ä_ b bÄ_ 1 n œ1 _ 1; '1 e2x dx œ lim 5. faxb œ e2x is positive, continuous, and decreasing for x _ 2x ˆ 2e12b 2e12 ‰ œ 2e12 Ê ' e 1 œ lim bÄ_ _ ˆ ln1b ln12 ‰ œ ln12 Ê ' œ lim bÄ_ _ 2; '2 '3 x x 4 dx œ lim '3 x x 4 dx œ lim 2 _ _ n œ3 n œ1 '2b xaln1xb dx œ lim 2 b ’ ln1x “ bÄ_ 1, f w axb œ ax42x4b2 0 for x 2, thus f is decreasing for x 2 b b bÄ_ 1 dx œ lim xaln xb2 bÄ_ 2 1 dx converges Ê ! naln1nb2 converges xaln xb2 n œ2 7. faxb œ x2 x 4 is positive and continuous for x 2 1 n œ1 _ 2 _ b bÄ_ dx converges Ê ! e2n converges 6. faxb œ xaln1xb2 is positive, continuous, and decreasing for x _ bÄ_ bÄ_ ’ 21 lnax2 4b“ œ lim bÄ_ 3 _ 3; ˆ 12 lnab2 4b 12 lna13b‰ œ _ Ê ' _ 3 x x2 4 dx diverges Ê ! n2 n 4 diverges Ê ! n2 n 4 œ 15 28 ! n2 n 4 diverges n œ3 2, f w axb œ 2 xln2 x 0 for x e, thus f is decreasing for x 2 2 8. faxb œ lnxx is positive and continuous for x _ ln x2 '3 dx œ lim x bÄ_ _ '3 lnxx dx œ lim b _ 2 _ ln x2 b 2 bÄ_ ’2aln xb“ œ lim _ 2 bÄ_ 3 a2aln bb 2aln 3bb œ _ Ê '3 x 3; dx 2 diverges Ê ! lnnn diverges Ê ! lnnn œ ln24 ! lnnn diverges n œ3 n œ2 n œ3 2 9. faxb œ exxÎ3 is positive and continuous for x _ x2 '7 e xÎ3 dx œ lim bÄ_ 2 xÎ3 bÄ_ 18b lim Š 3a6b ‹ e327 7Î3 œ ebÎ3 œ lim bÄ_ _ '7 ex dx œ lim b bÄ_ ax 6 b 1, f w axb œ x3e 0 for x 6, thus f is decreasing for x xÎ3 b 2 54 lim ’ e3xxÎ3 e18x xÎ3 exÎ3 “ œ 7 327 ' ˆ b54 ‰ 327 Î3 7Î3 œ 7Î3 Ê e e e _ 2 bÄ_ _ x2 7 exÎ3 7; 54 e327 Š 3b eb18b 7Î3 ‹ œ Î3 2 _ 2 dx converges Ê ! ennÎ3 converges n œ7 2 25 36 ! nnÎ3 converges Ê ! ennÎ3 œ e11Î3 e24Î3 e91 e16 4Î3 e5Î3 e2 e n œ7 n œ1 10. faxb œ x2 x 2x4 1 œ axx14b2 is continuous for x decreasing for x œ lim bÄ_ _ _ x4 8; '8 x1 3 1 3 ”'8 ax 1b2 dx '8 ax 1b2 dx• œ b lim ”'8 x 1 dx '8 ax 1b2 dx• bÄ_ Ä_ dx œ lim ax 1b 2 b ’lnlx 1l x 3 1 “ œ lim 8 _ 2, f is positive for x 4, and f w axb œ ax71xb3 0 for x 7, thus f is bÄ_ b b b _ x4 ˆlnlb 1l b 3 1 ln 7 37 ‰ œ _ Ê ' 8 ax 1b2 b dx diverges _ 1 2 3 Ê ! n2 n 2n4 1 diverges Ê ! n2 n 2n4 1 œ 2 14 0 16 25 36 ! n2 n 2n4 1 diverges nœ8 nœ2 " 11. converges; a geometric series with r œ 10 1 n œ8 12. converges; a geometric series with r œ "e 1 n 13. diverges; by the nth-Term Test for Divergence, n lim œ1Á0 Ä _ n1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.3 The Integral Test _ 5 14. diverges by the Integral Test; '1 x 5 1 dx œ 5 ln (n 1) 5 ln 2 Ê '1 n _ _ n œ1 nœ1 x 1 dx Ä _ 15. diverges; ! È3n œ 3 ! È"n , which is a divergent p-series (p œ #" ) _ _ n œ1 n œ1 " 16. converges; ! nÈ2n œ 2 ! n$Î# , which is a convergent p-series (p œ 3# ) 17. converges; a geometric series with r œ "8 1 _ _ _ _ n œ1 nœ1 nœ1 n œ1 18. diverges; ! n8 œ 8 ! 1n and since ! "n diverges, 8 ! 1n diverges _ ln x 19. diverges by the Integral Test: '2 lnxx dx œ "# aln# n ln 2b Ê '2 n x dx Ä _ Ô t œ lndxx × _ b dt œ x Ä 'ln 2 tetÎ2 dt œ lim 2tetÎ2 4etÎ2 ‘ ln 2 bÄ_ Õ dx œ et dt Ø œ lim 2ebÎ2 (b 2) 2eÐln 2ÑÎ2 (ln 2 2)‘ œ _ _ ln x 20. diverges by the Integral Test: '2 Èx dx; bÄ_ 21. converges; a geometric series with r œ 23 1 n n n 5 5 ln 5 ˆ ln 5 ‰ ˆ 54 ‰ œ _ Á 0 22. diverges; n lim œ n lim œ n lim Ä _ 4n 3 Ä _ 4n ln 4 Ä _ ln 4 _ _ n œ0 n œ0 23. diverges; ! n21 œ 2 ! n " 1 , which diverges by the Integral Test 24. diverges by the Integral Test: '1 2xdx 1 œ "# ln (2n 1) Ä _ as n Ä _ n n n 2 2 ln 2 25. diverges; n lim a œ n lim œ n lim œ_Á0 1 Ä_ n Ä _ n1 Ä_ dx 26. diverges by the Integral Test: '1 Èx ˆÈ ; x 1‰ – n Èn 27. diverges; n lim œ n lim Ä _ ln n Ä_ " Š 2È ‹ n Š "n ‹ œ n lim Ä_ Ènb1 du u œ Èx " ' ˆÈn 1‰ ln 2 Ä _ as n Ä _ Ä dx u œ ln 2 du œ Èx — Èn # œ_Á0 n ˆ1 n" ‰ œ e Á 0 28. diverges; n lim a œ n lim Ä_ n Ä_ 29. diverges; a geometric series with r œ ln"# ¸ 1.44 1 30. converges; a geometric series with r œ ln"3 ¸ 0.91 1 _ 31. converges by the Integral Test: '3 Š "x ‹ u œ ln x dx; ” (ln x) È(ln x)# 1 du œ "x dx • _ Ä 'ln 3 È "# u u 1 du Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 585 586 Chapter 10 Infinite Sequences and Series b œ lim csec" kukd ln 3 œ lim csec" b sec" (ln 3)d œ lim cos" ˆ "b ‰ sec" (ln 3)‘ bÄ_ bÄ_ bÄ_ œ cos" (0) sec" (ln 3) œ 1# sec" (ln 3) ¸ 1.1439 _ 32. converges by the Integral Test: '1 " x a1 ln# xb dx œ '1_ 1 Š(ln‹x) dx; ” u œ ln" x • Ä '0_ 1"u du " x du œ x dx # # œ lim ctan" ud 0 œ lim atan" b tan" 0b œ 1# 0 œ 1# b bÄ_ bÄ_ 33. diverges by the nth-Term Test for divergence; n lim n sin ˆ "n ‰ œ n lim Ä_ Ä_ sin ˆ "n ‰ sin x œ1Á0 ˆ "n ‰ œ xlim Ä0 x 34. diverges by the nth-Term Test for divergence; n lim n tan ˆ "n ‰ œ n lim Ä_ Ä_ Š n"# ‹ sec# ˆ n" ‰ tan ˆ "n ‰ œ lim ˆ "n ‰ nÄ_ Š " ‹ n# œ n lim sec# ˆ "n ‰ œ sec# 0 œ 1 Á 0 Ä_ _ 35. converges by the Integral Test: '1 uœe ex 1 e2x dx; ” du œ ex dx • x Ä 'e _ b " ctan" ud e 1 u# du œ n lim Ä_ œ lim atan" b tan" eb œ 1# tan" e ¸ 0.35 bÄ_ _ 36. converges by the Integral Test: '1 u œ ex × _ _ 2 du œ ex dx Ä ' u(1 2 u) du œ ' ˆ 2u u 2 1 ‰ du 1 ex dx; e e Õ dx œ " du Ø u Ô b œ lim 2 ln u u 1 ‘ e œ lim 2 ln ˆ b b 1 ‰ 2 ln ˆ e e 1 ‰ œ 2 ln 1 2 ln ˆ e e 1 ‰ œ 2 ln ˆ e e 1 ‰ ¸ 0.63 bÄ_ bÄ_ _ 8 tanc" x 37. converges by the Integral Test: '1 1 x# _ x 38. diverges by the Integral Test: '1 dx; ” u œ tan" x '11ÎÎ42 8u du œ c4u# d 11ÎÎ24 œ 4 Š 14# 161# ‹ œ 341# dx • Ä du œ 1 x# u œ x# 1 x# 1 dx; ” du œ 2x dx • _ 39. converges by the Integral Test: '1 sech x dx œ 2 lim _ du Ä #" '2 4 b œ lim #" ln u‘ 2 œ lim " (ln b ln 2) œ _ bÄ_ # bÄ_ '1b 1 eae b dx œ 2 lim ctan" ex d b1 x x # bÄ_ œ 2 lim atan" eb tan" eb œ 1 2 tan" e ¸ 0.71 bÄ_ bÄ_ _ 40. converges by the Integral Test: '1 sech# x dx œ lim bÄ_ œ 1 tanh 1 ¸ 0.76 _ '1b sech# x dx œ lim ctanh xd b1 œ lim (tanh b tanh 1) bÄ_ bÄ_ 41. '1 ˆ x a 2 x " 4 ‰ dx œ lim ca ln kx 2k ln kx 4kd 1 œ lim ln (bb2)4 ln ˆ 35 ‰ ; a b bÄ_ a _, a 1 lim (bb2)4 œ a lim (b 2)a1 œ œ 1, a œ 1 bÄ_ bÄ_ a bÄ_ Ê the series converges to ln ˆ 53 ‰ if a œ 1 and diverges to _ if a 1. If a 1, the terms of the series eventually become negative and the Integral Test does not apply. From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges. _ x1 b1 ‰ ˆ2‰ 42. '3 ˆ x " 1 x 2a 1 dx œ lim ’ln ¹ (x 1)2a ¹“ œ lim ln (b 1)2a ln 42a ; lim b bÄ_ 1, a œ "# " œ 2a c 1 _, a " b Ä _ #a(b 1) # œ lim 3 bÄ_ b" 2a b Ä _ (b 1) Ê the series converges to ln ˆ #4 ‰ œ ln 2 if a œ #" and diverges to _ if Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.3 The Integral Test 587 if a "# . If a "# , the terms of the series eventually become negative and the Integral Test does not apply. From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges. 43. (a) (b) There are (13)(365)(24)(60)(60) a10* b seconds in 13 billion years; by part (a) sn Ÿ 1 ln n where n œ (13)(365)(24)(60)(60) a10* b Ê sn Ÿ 1 ln a(13)(365)(24)(60)(60) a10* bb œ 1 ln (13) ln (365) ln (24) 2 ln (60) 9 ln (10) ¸ 41.55 _ _ _ n œ1 n œ1 n œ1 " 44. No, because ! nx œ x" ! n" and ! n" diverges _ _ _ nœ1 nœ1 nœ1 45. Yes. If ! an is a divergent series of positive numbers, then ˆ "# ‰ ! an œ ! ˆ a#n ‰ also diverges and a#n an . _ There is no “smallest" divergent series of positive numbers: for any divergent series ! an of positive numbers n œ1 _ ! ˆ an ‰ has smaller terms and still diverges. n œ1 # _ _ _ nœ1 nœ1 nœ1 46. No, if ! an is a convergent series of positive numbers, then 2 ! an œ ! 2an also converges, and 2an an . There is no “largest" convergent series of positive numbers. 47. (a) Both integrals can represent the area under the curve faxb œ Èx1 1 , and the sum s50 can be considered an 50 approximation of either integral using rectangles with ?x œ 1. The sum s50 œ ! Èn1 1 is an overestimate of the n œ1 51 integral '1 Èx1 1 dx. The sum s50 represents a left-hand sum (that is, the we are choosing the left-hand endpoint of each subinterval for ci ) and because f is a decreasing function, the value of f is a maximum at the left-hand endpoint of each sub interval. The area of each rectangle overestimates the true area, thus '1 51 50 1 ! È 1 . In a similar Èx 1 dx n1 n œ1 50 manner, s50 underestimates the integral '0 Èx1 1 dx. In this case, the sum s50 represents a right-hand sum and because f is a decreasing function, the value of f is aminimum at the right-hand endpoint of each subinterval. The area of each rectangle underestimates the true area, thus ! Èn1 1 '0 50 50 n œ1 œ ’2Èx 1“ œ 2È52 2È2 ¸ 11.6 and '0 51 50 1 1 Èx 1 dx. Evaluating the integrals we find 50 1 È Èx 1 dx œ ’2 x 1“ 0 '151 Èx1 1 dx œ 2È51 2È1 ¸ 12.3. Thus, 50 11.6 ! Èn1 1 12.3. n œ1 nb1 (b) sn 1000 Ê '1 Ên 2 nb1 1 È œ 2Èn 1 2È2 1000 Ê n Š500 2È2‹ Èx 1 dx œ ’2 x 1“ 1 251415. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. ¸ 251414.2 588 Chapter 10 Infinite Sequences and Series 30 _ n œ1 n œ1 _ 48. (a) Since we are using s30 œ ! n14 to estimate ! n14 , the error is given by ! n14 . We can consider this sum as an estimate nœ31 of the area under the curve faxb œ x14 when x 30 using rectangles with ?x œ 1 and ci is the right-hand endpoint of each subinterval. Since f is a decreasing function, the value of f is a minimum at the right-hand endpoint of each 1 subinterval, thus ! n14 '30 x14 dx œ lim '30 x14 dx œ lim ’ 3x1 3 “ œ lim Š 3b1 3 3a30 ‹ ¸ 1.23 ‚ 105 . b3 _ _ b b bÄ_ nœ31 bÄ_ bÄ_ 30 5 Thus the error 1.23 ‚ 10 Þ (b) We want S sn 0.000001 Ê 'n _ œ 1 x4 dx 0.000001 Ê 4 _ 1 x3 dx 0.01 Ê œ 2n1 2 0.01 Ê n È50 ¸ 7.071 Ê n 4 bÄ_ 3 1000000 lim ˆ 3b1 3 3n1 3 ‰ œ 3n1 3 0.000001 Ê n É ¸ 69.336 Ê n 3 bÄ_ 49. We want S sn 0.01 Ê 'n 3 bÄ_ n 70. 'n_ x1 dx œ lim 'nb x1 dx œ lim ’ 2x1 “b œ lim ˆ 2b1 2n1 ‰ 3 bÄ_ 3 bÄ_ 2 2 bÄ_ n 2 8 8 Ê S ¸ s8 œ ! n13 ¸ 1.195 n œ1 50. We want S sn 0.1 Ê 'n ' _ œ 'n_ x1 dx œ lim 'nb x1 dx œ lim ’ 3x1 “b b b 1 1 1 1 ˆ x ‰ lim tan ’ “ 2 4 dx œ x2 4 dx 0.1 Ê blim x 2 2 Ä_ n bÄ_ n 1 1 1 1 1 1 ˆ b ‰ 1 ˆ n ‰‰ 1 ˆ n ‰ ˆ ˆ ‰ lim tan œ 4 2 tan 2 2 tan 2 2 0.1 Ê n 2tan 2 0.2 ¸ 9.867 Ê n bÄ_ 2 10 Ê S ¸ s10 10 œ ! n2 1 4 ¸ 0.57 n œ1 51. S sn 0.00001 Ê 'n _ 1 x1.1 dx 0.00001 Ê 'n_ x1 dx œ lim 'nb x1 dx œ lim ’ x10 “b œ lim ˆ b10 n10 ‰ 1.1 bÄ_ 1.1 bÄ_ 0.1 bÄ_ n 0.1 0.1 10 œ n10 Ê n 1060 0.1 0.00001 Ê n 1000000 52. S sn 0.01 Ê 'n _ œ 1 dx 0.01 Ê xaln xb3 'n_ xaln1xb dx œ lim 'nb xaln1xb dx œ lim ’ 2aln1xb “b 3 bÄ_ 3 È lim Š 2aln1bb2 2aln1nb2 ‹ œ 2aln1nb2 0.01 Ê n e 50 ¸ 1177.405 Ê n bÄ_ n n k œ1 k œ1 bÄ_ 2 n 1178 53. Let An œ ! ak and Bn œ ! 2k aa2k b , where {ak } is a nonincreasing sequence of positive terms converging to 0. Note that {An } and {Bn } are nondecreasing sequences of positive terms. Now, Bn œ 2a# 4a% 8a) á 2n aa2n b œ 2a# a2a% 2a% b a2a) 2a) 2a) 2a) b á ˆ2aa2n b 2aa2n b á 2aa2n b ‰ Ÿ 2a" 2a# a2a$ 2a% b a2a& 2a' 2a( 2a) b á ðóóóóóóóóóóóóóóñóóóóóóóóóóóóóóò 2n1 terms _ ˆ2aa2nc1 b 2aa2nc1 1b á 2aa2n b ‰ œ 2Aa2n b Ÿ 2 ! ak . Therefore if ! ak converges, k œ1 then {Bn } is bounded above Ê ! 2k aa2k b converges. Conversely, _ An œ a" aa# a$ b aa% a& a' a( b á an a" 2a# 4a% á 2n aa2n b œ a" Bn a" ! 2k aa2k b . k œ1 _ Therefore, if ! 2 aa2k b converges, then {An } is bounded above and hence converges. k k œ1 _ _ _ n œ2 n œ2 n œ2 " " ! " ! 2 n a a2 n b œ ! 2 n n " 54. (a) aa2n b œ 2n ln"a2n b œ 2n †n(ln # †n(ln 2) œ ln # 2) Ê n , which diverges _ Ê ! n ln" n diverges. n œ2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.3 The Integral Test _ _ _ _ n œ1 nœ1 nœ1 nœ1 589 n (b) aa2n b œ #"np Ê ! 2n aa2n b œ ! 2n † #"np œ ! a2n"bpc1 œ ! ˆ #p"c1 ‰ , a geometric series that converges if #p"c1 1 or p 1, but diverges if p Ÿ 1. _ 55. (a) '2 '_ u œ ln x cpb1 dx ucp du œ lim ’ up 1 “ œ lim Š 1 " p ‹ cbp1 (ln 2)p1 d x(ln x)p ; ” du œ dx • Ä ln 2 b Ä _ bÄ_ ln 2 x " c pb1 ,p1 p 1 (ln 2) Ê the improper integral converges if p 1 and diverges if p 1. œœ b _, p " _ dx For p œ 1: '2 p œ 1. cln (ln x)d 2 œ lim cln (ln b) ln (ln 2)d œ _, so the improper integral diverges if x ln x œ b lim Ä_ bÄ_ b _ (b) Since the series and the integral converge or diverge together, ! n(ln" n)p converges if and only if p 1. n œ2 56. (a) p œ 1 Ê the series diverges (b) p œ 1.01 Ê the series converges _ _ n œ2 n œ2 (c) ! n aln" n$ b œ 3" ! n(ln" n) ; p œ 1 Ê the series diverges (d) p œ 3 Ê the series converges nb1 57. (a) From Fig. 10.11(a) in the text with f(x) œ "x and ak œ k" , we have '1 " " " " x dx Ÿ 1 # 3 á n Ÿ 1 '1 f(x) dx Ê ln (n 1) Ÿ 1 "# "3 á "n Ÿ 1 ln n Ê 0 Ÿ ln (n 1) ln n n Ÿ ˆ1 "# 3" á n" ‰ ln n Ÿ 1. Therefore the sequence ˜ˆ1 #" 3" á n" ‰ ln n™ is bounded above by 1 and below by 0. nb1 (b) From the graph in Fig. 10.11(b) with f(x) œ "x , n " 1 'n " x dx œ ln (n 1) ln n Ê 0 n " 1 cln (n 1) ln nd œ ˆ1 #" "3 á n" 1 ln (n 1)‰ ˆ1 #" 3" á n" ln n‰ . If we define an œ 1 "# œ 3" "n ln n, then 0 an1 an Ê an1 an Ê {an } is a decreasing sequence of nonnegative terms. # 58. ex Ÿ ex for x _ _ # b 1, and '1 ecx dx œ lim cex d " œ lim êb e1 ‰ œ ec1 Ê '1 ecx dx converges by bÄ_ bÄ_ _ n# the Comparison Test for improper integrals Ê ! e n œ0 _ 59. (a) s10 œ ! n"3 œ 1.97531986; '11 x13 dx œ lim 10 bÄ_ n œ1 _ # œ 1 ! en converges by the Integral Test. nœ1 1 ‰ 1 '11b x3 dx œ lim ’ x2c “ b œ lim ˆ 2b1 242 œ 242 and 2 bÄ_ 11 1 ‰ 1 '10_ x1 dx œ lim '10b x3 dx œ lim ’ x2c “ b œ lim ˆ 2b1 200 œ 200 bÄ_ 2 2 3 (b) bÄ_ bÄ_ 10 bÄ_ 2 1 1 Ê 1.97531986 242 s 1.97531986 200 Ê 1.20166 s 1.20253 _ s œ ! n"3 ¸ 1.20166 2 1.20253 œ 1.202095; error Ÿ 1.20253 2 1.20166 œ 0.000435 n œ1 _ 60. (a) s10 œ ! n"4 œ 1.082036583; '11 x14 dx œ lim 10 bÄ_ nœ1 1 ‰ 1 '11b x4 dx œ lim ’ x3c “ b œ lim ˆ 3b1 3993 œ 3993 and 3 bÄ_ 11 1 ‰ 1 '10_ x1 dx œ lim '10b x4 dx œ lim ’ x3c “ b œ lim ˆ 3b1 3000 œ 3000 bÄ_ 3 4 bÄ_ bÄ_ 10 bÄ_ 3 1 1 Ê 1.082036583 3993 s 1.082036583 3000 Ê 1.08229 s 1.08237 _ (b) s œ ! n"4 ¸ 1.08229 2 1.08237 œ 1.08233; error Ÿ 1.08237 2 1.08229 œ 0.00004 n œ1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 3 590 Chapter 10 Infinite Sequences and Series 10.4 COMPARISON TESTS _ 1. Compare with ! n"2 , which is a convergent p-series, since p œ 2 1. Both series have nonnegative terms for n 1. For n œ1 n 1, we have n2 Ÿ n2 30 Ê n12 1 n2 30 . _ Then by Comparison Test, ! n2 1 30 converges. n œ1 _ 2. Compare with ! n"3 , which is a convergent p-series, since p œ 3 1. Both series have nonnegative terms for n 1. For n œ1 n 1, we have n4 Ÿ n4 2 Ê n14 1 n n 4 2 Ê n 4 n 1 n 4 2 Ê n 3 n n 4 2 _ n1 n 4 2 . Then by Comparison Test, ! nn412 n œ1 converges. _ 3. Compare with ! È"n , which is a divergent p-series, since p œ #" Ÿ 1. Both series have nonnegative terms for n 2. For n œ2 n 2, we have Èn 1 Ÿ Èn Ê Èn1 1 1 Èn . _ Then by Comparison Test, ! Èn1 1 diverges. n œ2 _ 4. Compare with ! "n , which is a divergent p-series, since p œ 1 Ÿ 1. Both series have nonnegative terms for n 2. For n œ2 n 2, we have n2 n Ÿ n2 Ê n2 1 n n 1 n2 n2 œ n Ê n2 n 1 n n2 Ê n2 n n 1 n. n2 n _ Thus ! nn22n diverges. n œ2 _ 5. Compare with ! n3"Î2 , which is a convergent p-series, since p œ 32 1. Both series have nonnegative terms for n 1. n œ1 _ 2 2 n n 1, we have 0 Ÿ cos2 n Ÿ 1 Ê cos Ÿ n31Î2 . Then by Comparison Test, ! cos converges. n3Î2 n3Î2 For n n œ1 _ 6. Compare with ! 3"n , which is a convergent geometric series, since lrl œ ¹ 13 ¹ 1. Both series have nonnegative terms for n œ1 n 1. For n _ 1, we have n † 3n 3n Ê n†13n Ÿ 31n . Then by Comparison Test, ! n†13n converges. _ _ _ nœ1 nœ1 È n œ1 È 7. Compare with ! n3Î52 . The series ! n31Î2 is a convergent p-series, since p œ 32 1, and the series ! n3Î52 nœ1 _ œ È5 ! n31Î2 converges by Theorem 8 part 3. Both series have nonnegative terms for n 1. For n 1, we have n œ1 n3 Ÿ n4 Ê 4n3 Ÿ 4n4 Ê n4 4n3 Ÿ n4 4n4 œ 5n4 Ê n4 4n3 Ÿ 5n4 20 œ 5an4 4b Ê nn44n4 Ÿ 5. 4 3 _ È Ê n na4n44b Ÿ 5 Ê nn444 Ÿ n53 Ê É nn444 Ÿ É n53 œ n3Î52 Then by Comparison Test, ! É nn444 converges. 3 n œ1 _ 8. Compare with ! È"n , which is a divergent p-series, since p œ #" Ÿ 1. Both series have nonnegative terms for n n œ1 n 1, we have Èn Ê n2 2 nÈn n Èn 1 ÊÈ2 n 3 1 Èn . 1 Ê 2È n n2 3 Ê 2 Ê 2È n 1 n ˆn 2 È n 1 ‰ n2 3 3 Ê nˆ2Èn 1‰ 1Ê _ Èn 1 Then by Comparison Test, ! È 2 n œ1 n 2È n 1 n2 3 n 3 3 Ê 2 nÈ n n 3n ˆÈ n 1 ‰ 1 n Ê n2 3 2 ˆÈ ‰ 1 Ê nn2 31 n Ê diverges. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 3 2 É 1n 1. For Section 10.4 Comparison Tests _ 9. Compare with ! n"2 , which is a convergent p-series, since p œ 2 1. Both series have positive terms for n nc2 n3 c n2 b 3 1 În 2 œ lim _ ! n œ1 nÄ_ n œ1 591 1. lim bann nÄ_ 3n 4n 6n 4 6 œ lim n3nn22n 3 œ lim 3n lim 6n 2 2n œ 2 œ lim 6 œ 1 0. Then by Limit Comparison Test, 3 2 2 nÄ_ nÄ_ nÄ_ nÄ_ n2 n3 n2 3 converges. _ 10. Compare with ! È"n , which is a divergent p-series, since p œ #" Ÿ 1. Both series have positive terms for n É nn2bb12 n œ1 1. lim bann nÄ_ n n n 2n 1 2 È1 œ 1 0. Then by Limit Comparison œ lim 1ÎÈn œ lim É nn2 2 œ É lim n2 2 œ É lim 2n œ É lim 2 œ 2 nÄ_ 2 nÄ_ nÄ_ nÄ_ nÄ_ _ Test, ! É nn212 diverges. n œ1 _ 11. Compare with ! "n , which is a divergent p-series, since p œ 1 Ÿ 1. Both series have positive terms for n nan b 1b Šn2 œ lim n œ2 b 1‹an c 1b 3 2 2 nÄ_ _ Test, nÄ_ 2n 2 6 œ lim n3 nn2+nn 1 œ lim 3n3n lim 6n 2 2n 1 œ 6n 2 œ lim 6 œ 1 0. Then by Limit Comparison 1 În nÄ_ 2. lim bann nÄ_ nÄ_ nÄ_ ! 2 nan 1b diverges. an 1ban 1b n œ2 _ 12. Compare with ! 2"n , which is a convergent geometric series, since lrl œ ¹ 12 ¹ 1. Both series have positive terms for n œ1 2n n lim an œ lim 13 bÎ24nn nÄ_ bn nÄ_ 1. n _ n n 4 ! 2 n converges. œ lim 3 4 4n œ lim 44n ln ln 4 œ 1 0. Then by Limit Comparison Test, 34 nÄ_ nÄ_ n œ1 _ 13. Compare with ! È"n , which is a divergent p-series, since p œ 12 Ÿ 1. Both series have positive terms for n n œ1 5n n†4n È œ lim 1ÎÈn œ lim 54n œ n nÄ_ nÄ_ _ n lim ˆ 5 ‰ œ nÄ_ 4 _ 1. lim bann nÄ_ n _. Then by Limit Comparison Test, ! È5n†4n diverges. n œ1 n 14. Compare with ! ˆ 25 ‰ , which is a convergent geometric series, since lrl œ ¹ 25 ¹ 1. Both series have positive terms for n œ1 ˆ 2n b 3 ‰ 15 ‰n 15 ‰n 15 ‰ n 1. lim bann œ lim a5n2Îb54bn œ lim ˆ 10n œ exp lim lnˆ 10n œ exp lim n lnˆ 10n 10n 8 10n 8 nÄ_ nÄ_ nÄ_ 10n 8 nÄ_ nÄ_ b 15 ‰ 10 lnˆ 10n 10b 8 70n2 70n2 10n b 8 œ exp lim œ exp lim 10n b151În10n œ exp lim a10n 15 2 2 1 În ba10n 8b œ exp nlim nÄ_ nÄ_ nÄ_ Ä_ 100n 230n 120 n _ n 140 3‰ 7Î10 œ exp lim 200n140n 0. Then by Limit Comparison Test, ! ˆ 2n converges. 230 œ exp lim 200 œ e 5n 4 nÄ_ nÄ_ n œ1 _ 15. Compare with ! "n , which is a divergent p-series, since p œ 1 Ÿ 1. Both series have positive terms for n n œ2 2. lim bann nÄ_ _ " ln n œ lim 1În œ lim lnnn œ lim 11În œ lim n œ _. Then by Limit Comparison Test, ! ln"n diverges. nÄ_ nÄ_ nÄ_ nÄ_ n œ2 _ 16. Compare with ! n"2 , which is a convergent p-series, since p œ 2 1. Both series have positive terms for n n œ1 œ lim nÄ_ lnŠ1 n"2 ‹ 1În2 2 " Š n3 ‹ 1 n2 Š n23 ‹ 1 œ lim nÄ_ _ 1. lim bann nÄ_ œ lim 1 1 " œ 1 0. Then by Limit Comparison Test, ! lnˆ1 n"2 ‰ converges. nÄ_ n2 n œ1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 592 Chapter 10 Infinite Sequences and Series _ 17. diverges by the Limit Comparison Test (part 1) when compared with ! È"n , a divergent p-series: n œ1 " lim nÄ_ Œ #Èn È $ n Š " ‹ Èn Èn œ n lim lim ˆ " ‰ œ #" $ n œ Ä _ 2È n È n Ä _ # n 1Î6 18. diverges by the Direct Comparison Test since n n n n Èn 0 Ê n 3Èn "n , which is the nth _ term of the divergent series ! "n or use Limit Comparison Test with bn œ n" nœ1 # 19. converges by the Direct Comparison Test; sin2n n Ÿ #"n , which is the nth term of a convergent geometric series n 20. converges by the Direct Comparison Test; 1 ncos Ÿ n2# and the p-series ! n"# converges # 2n 21. diverges since n lim œ 23 Á 0 Ä _ 3n 1 " 22. converges by the Limit Comparison Test (part 1) with n$Î# , the nth term of a convergent p-series: lim nÄ_ Š nn# È"n ‹ ˆ n n " ‰ œ 1 œ n lim Ä_ " ‹ Š $Î# n 23. converges by the Limit Comparison Test (part 1) with n"# , the nth term of a convergent p-series: lim nÄ_ Š n(n 10n1)(n" 2) ‹ Š n"# ‹ # 10n n 20n 1 20 œ n lim œ n lim œ n lim œ 10 Ä _ n# 3n 2 Ä _ 2n 3 Ä_ 2 24. converges by the Limit Comparison Test (part 1) with n"# , the nth term of a convergent p-series: lim n# (n 5n$ 3n 2) Šn# 5‹ Š n"# ‹ nÄ_ $ # 5n 3n 15n 3 30n œ n lim œ n lim œ n lim œ5 Ä _ n$ 2n# 5n 10 Ä _ 3n# 4n 5 Ä _ 6n 4 n n n n ‰ 25. converges by the Direct Comparison Test; ˆ 3n n 1 ‰ ˆ 3n œ ˆ "3 ‰ , the nth term of a convergent geometric series " 26. converges by the Limit Comparison Test (part 1) with n$Î# , the nth term of a convergent p-series: lim nÄ_ " Š $Î# ‹ $ n Š È $" n 2 ‹ É n n$ 2 œ lim É1 n2$ œ 1 œ n lim Ä_ nÄ_ _ " ! " diverges 27. diverges by the Direct Comparison Test; n ln n Ê ln n ln ln n Ê "n ln"n ln (ln n) and n n œ3 _ 28. converges by the Limit Comparison Test (part 2) when compared with ! n"# , a convergent p-series: n œ1 # lim nÄ_ ’ (lnn$n) “ Š n"# ‹ # (ln n) œ n lim œ n lim n Ä_ Ä_ 2(ln n) Š n" ‹ 1 ln n œ 2 n lim œ0 Ä_ n 29. diverges by the Limit Comparison Test (part 3) with "n , the nth term of the divergent harmonic series: lim nÄ_ ’ Èn1ln n “ ˆ n" ‰ Èn œ n lim œ n lim Ä _ ln n Ä_ " Š 2È ‹ n ˆ n" ‰ œ n lim Ä_ Èn 2 œ_ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.4 Comparison Tests 593 " 30. converges by the Limit Comparison Test (part 2) with n&Î% , the nth term of a convergent p-series: # lim nÄ_ n) ’ (ln$Î# “ n " Š &Î% ‹ n ˆ 2 lnn n ‰ # (ln n) œ n lim œ n lim Ä _ n"Î% Ä_ " Š $Î% ‹ 4n ln n œ 8 n lim œ 8 n lim Ä _ n"Î% Ä_ ˆ n" ‰ Š " ‹ 4n$Î% " œ 32 n lÄ im_ n"Î% œ 32 † 0 œ 0 31. diverges by the Limit Comparison Test (part 3) with "n , the nth term of the divergent harmonic series: lim nÄ_ ˆ 1 "ln n ‰ n " œ n lim œ n lim nœ_ ˆ "n ‰ œ n lim Ä _ 1 ln n Ä _ Š"‹ Ä_ n _ ln (x 1) 32. diverges by the Integral Test: '2 x1 _ b dx œ 'ln 3 u du œ lim "2 u# ‘ ln 3 œ lim " ab bÄ_ # bÄ_ # ln# 3b œ _ " 33. converges by the Direct Comparison Test with n$Î# , the nth term of a convergent p-series: n# 1 n for " " n 2 Ê n# an# 1b n$ Ê nÈn# 1 n$Î# Ê $Î# or use Limit Comparison Test with 1# . nÈ n# 1 n n " 34. converges by the Direct Comparison Test with n$Î# , the nth term of a convergent p-series: n# 1 n# Èn # " " Ê n# 1 Ènn$Î# Ê nÈn1 n$Î# Ê n# 1 n$Î# or use Limit Comparison Test with n$Î# . _ _ _ n œ1 n œ1 n œ1 35. converges because ! "n2nn œ ! n2" n ! " #n which is the sum of two convergent series: _ ! nœ1 _ " " " ! "n is a convergent geometric series n2n converges by the Direct Comparison Test since n#n #n , and 2 nœ1 _ _ 36. converges by the Direct Comparison Test: ! nn# 22n œ ! ˆ n2" n n"# ‰ and n2" n n"# Ÿ #"n n"# , the sum of n n œ1 nœ1 the nth terms of a convergent geometric series and a convergent p-series 37. converges by the Direct Comparison Test: 3nc1" 1 3n"c1 , which is the nth term of a convergent geometric series nc1 ˆ " 3"n ‰ œ 3" Á 0 38. diverges; n lim Š 3 3n" ‹ œ n lim Ä_ Ä_ 3 _ n 39. converges by Limit Comparison Test: compare with ! ˆ 15 ‰ , which is a convergent geometric series with lrl œ 15 1, n œ1 lim nÄ_ 1 1 Š n2n b b 3n † 5n ‹ a 1 Î5 b n n1 1 œ n lim œ n lim œ 0. Ä _ n2 3n Ä _ 2n 3 _ n 40. converges by Limit Comparison Test: compare with ! ˆ 34 ‰ , which is a convergent geometric series with lrl œ 15 1, n œ1 3 Š 23n b b 4n ‹ n n lim n n Ä _ a 3 Î4 b 8 ‰n ˆ 12 1 8 12 œ n lim lim œ 11 œ 1 0. n n 12n œ 9 ˆ Ä_ n Ä _ 129 ‰ 1 n n Š 2n 2cnn ‹ n _ 41. diverges by Limit Comparison Test: compare with ! 1n , which is a divergent p-series, n lim Ä_ n œ1 † 1 În 2 n œ n lim Ä _ 2n n 2n aln 2b2 2n ln 2 1 œ n lim œ n lim œ 1 0. Ä _ 2n ln 2 Ä _ 2n aln 2b2 _ _ n œ1 n œ1 42. diverges by the definition of an infinite series: ! lnˆ n n 1 ‰ œ ! ln n ln an 1b‘, sk œ aln 1 ln 2b aln 2 ln 3b Þ Þ Þ alnak 1b ln kb aln k ln ak 1bb œ ln ak 1b Ê lim sk œ _ kÄ_ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 594 Chapter 10 Infinite Sequences and Series _ _ _ n œ2 n œ2 nœ2 43. converges by Comparison Test with ! nan 1 1b which converges since ! nan 1 1b œ ! ’ n 1 1 n1 “, and sk œ ˆ1 12 ‰ ˆ 12 13 ‰ Þ Þ Þ ˆ k 1 2 k 1 1 ‰ ˆ k 1 1 k1 ‰ œ 1 k1 Ê lim sk œ 1; for n Ê nan 1ban 2b! nan 1b Ê n! 2, an 2b! kÄ_ nan 1b Ê n!1 Ÿ nan11b _ an c 1bx n 2x 44. converges by Limit Comparison Test: compare with ! n13 , which is a convergent p-series, n lim Ä _ 1 În 3 a b n œ1 n a n 1 bx n 2n 2 œ n lim œ n lim œ n lim œ n lim œ10 Ä _ an 2ban 1bnan 1bx Ä _ n2 3n 2 Ä _ 2n 3 Ä_ 2 3 2 45. diverges by the Limit Comparison Test (part 1) with "n , the nth term of the divergent harmonic series: lim nÄ_ ˆsin "n ‰ sin x œ1 ˆ "n ‰ œ xlim Ä0 x 46. diverges by the Limit Comparison Test (part 1) with "n , the nth term of the divergent harmonic series: lim nÄ_ ˆtan "n ‰ ˆsin " ‰ Š " ‹ ˆ " ‰n œ lim ˆ cos" x ‰ ˆ sinx x ‰ œ 1 † 1 œ 1 ˆ "n ‰ œ n lim Ä _ cos " xÄ0 n n c" 1 _ _ 1 47. converges by the Direct Comparison Test: tann1.1 n n#1.1 and ! n#1.1 œ 1# ! n"1.1 is the product of a n œ1 nœ1 convergent p-series and a nonzero constant c" ˆ1‰ Þ Þ _ 48. converges by the Direct Comparison Test: sec" n 1# Ê secn1 3 n n#1 3 and ! n œ1 _ ˆ 1# ‰ 1 ! " n1 3 œ # n1 3 is the Þ Þ n œ1 product of a convergent p-series and a nonzero constant 49. converges by the Limit Comparison Test (part 1) with n"# : n lim Ä_ n Š coth ‹ n# Š n"# ‹ c2n cn e e œ n lim coth n œ n lim Ä_ Ä _ en ecn n "e œ n lim œ1 Ä _ 1 ec2n 50. converges by the Limit Comparison Test (part 1) with n"# : n lim Ä_ n Š tanh ‹ n# Š n"# ‹ c2n e e œ n lim tanh n œ n lim Ä_ Ä _ en e n n n "e œ n lim œ1 Ä _ 1 ec2n 51. diverges by the Limit Comparison Test (part 1) with 1n : n lim Ä_ 52. 1 Š nÈ n n‹ ˆ 1n ‰ converges by the Limit Comparison Test (part 1) with n"# : n lim Ä_ 1 œ n lim n n œ 1. Ä_ È Èn n Š n# ‹ Š n"# ‹ n È œ n lim nœ1 Ä_ 53. 1 2 3" á n œ ˆ n(n" 1) ‰ œ n(n 2 1) . The series converges by the Limit Comparison Test (part 1) with n"# : # lim nÄ_ Š nan 2b 1b ‹ Š n"# ‹ # 2n 4n 4 œ n lim œ n lim œ n lim œ 2. Ä _ n# n Ä _ 2n 1 Ä_ 2 " 6 6 54. 1 2# 3"# á n# œ n(n b 1)(2n b 1) œ n(n 1)(2n 1) Ÿ n$ Ê the series converges by the Direct Comparison Test 6 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 1 Section 10.4 Comparison Tests 595 an 55. (a) If n lim œ 0, then there exists an integer N such that for all n N, ¹ bann 0¹ 1 Ê 1 bann 1 Ä _ bn Ê an bn . Thus, if ! bn converges, then ! an converges by the Direct Comparison Test. an (b) If n lim œ _, then there exists an integer N such that for all n N, bann 1 Ê an bn . Thus, if Ä _ bn ! bn diverges, then ! an diverges by the Direct Comparison Test. _ 56. Yes, ! ann converges by the Direct Comparison Test because ann an n œ1 an 57. n lim œ _ Ê there exists an integer N such that for all n N, bann 1 Ê an bn . If ! an converges, Ä _ bn then ! bn converges by the Direct Comparison Test 58. ! an converges Ê n lim a œ 0 Ê there exists an integer N such that for all n N, 0 Ÿ an 1 Ê an# an Ä_ n Ê ! a#n converges by the Direct Comparison Test 59. Since an 0 and n lim a œ _ Á 0, by nth term test for divergence, ! an diverges. Ä_ n an 60. Since an 0 and n lim a n2 † an b œ 0, compare !an with ! n"# , which is a convergent p-series; n lim Ä_ Ä _ 1 În 2 œ n lim a n2 † an b œ 0 Ê !an converges by Limit Comparison Test Ä_ _ q _ 61. Let _ q _ and p 1. If q œ 0, then ! alnnnp b œ ! n1p , which is a convergent p-series. If q Á 0, compare with n œ2 _ ! n œ2 1 nr where 1 r p, then n lim Ä_ aln nbq np 1 În r nœ2 aln nbq aln nbq œ n lim , and p r 0. If q 0 Ê q 0 and n lim Ä _ npcr Ä _ npcr qaln nbqc1 ˆ 1 ‰ q qc1 qaln nb aln nb 1 n œ n lim œ 0. If q 0, n lim œ n lim œ n lim . If q 1 Ÿ 0 Ê 1 q cq Ä _ aln nb npcr Ä _ npcr Ä _ ap rbnpcrc1 Ä _ ap rbnpcr lim qaln nbqc1 n Ä _ ap rbnpcr q œ n lim œ 0, otherwise, we apply L'Hopital's Rule again. n lim Ä _ ap rbnpcr aln nb1cq Ä_ qc2 qaq 1baln nb œ n lim . If q 2 Ÿ 0 Ê 2 q Ä _ ap rb2 npcr qc2 qaq 1baln nb 0 and n lim Ä _ ap rb2 npcr 0 and qaq 1baln nbqc2 ˆ 1n ‰ ap rb2 npcrc1 q aq 1 b œ n lim œ 0; otherwise, we Ä _ ap rb2 npcr aln nb2cq apply L'Hopital's Rule again. Since q is finite, there is a positive integer k such that q k Ÿ 0 Ê k q 0. Thus, after k qaq 1bâaq k 1baln nbqck q a q 1 bâa q k 1 b applications of L'Hopital's Rule we obtain n lim œ n lim œ 0. Since the limit is Ä_ Ä _ ap rbk npcr aln nbkcq ap rbk npcr _ q 0 in every case, by Limit Comparison Test, the series ! alnnnpb converges. n œ1 _ q _ 62. Let _ q _ and p Ÿ 1. If q œ 0, then ! alnnnp b œ ! n1p , which is a divergent p-series. If q 0, compare with n œ2 _ ! nœ2 1 np , which is a divergent p-series. Then n lim Ä_ where 0 p r Ÿ 1. n lim Ä_ lim ar pbn rcpc1 n Ä _ aqbaln nbcqc1 ˆ 1n ‰ aln nbq np 1 În r nœ2 aln nbq np 1 În p _ œ n lim aln nbq œ _. If q 0 Ê q 0, compare with ! n1r , Ä_ nœ2 aln nbq nrcp œ n lim œ n lim cq since r p 0. Apply L'Hopital's to obtain Ä _ npcr Ä _ aln nb rcp ar pbn œ n lim . If q 1 Ÿ 0 Ê q 1 Ä _ aqbaln nbcqc1 rcp ar pbn aln nb 0 and n lim aqb Ä_ 2 rcpc1 qb1 œ _, 2 rcp a r pb n a r pb n otherwise, we apply L'Hopital's Rule again to obtain n lim œ n lim . If Ä _ aqbaq 1baln nbcqc2 ˆ 1 ‰ Ä _ aqbaq 1baln nbcqc2 n q 2 Ÿ 0 Ê q 2 a r pb2 nrcp a r pb2 nrcp aln nbqb2 0 and n lim œ n lim œ _, otherwise, we aqbaq 1b Ä _ aqbaq 1baln nbcqc2 Ä_ apply L'Hopital's Rule again. Since q is finite, there is a positive integer k such that q k Ÿ 0 Ê q k k rcp k rcp 0. Thus, after qbk a r pb n a r pb n aln nb k applications of L'Hopital's Rule we obtain n lim œ n lim œ _. Ä _ aqbaq 1bâaq k 1baln nbcqck Ä _ aqbaq 1bâaq k 1b Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 596 Chapter 10 Infinite Sequences and Series _ q Since the limit is _ if q 0 or if q 0 and p 1, by Limit comparison test, the series ! alnnpncbr diverges. Finally if q 0 _ q _ _ q nœ1 and p œ 1 then ! alnnnp b œ ! alnnnb . Compare with ! 1n , which is a divergent p-series. For n nœ2 Ê aln nbq _ nœ2 q 1 Ê alnnnb 1 n . Thus 3, ln n 1 nœ2 _ q ! aln nb diverges by Comparison Test. Thus, if _ q _ and p Ÿ 1, n œ2 n q the series ! alnnpncbr diverges. nœ1 63. Converges by Exercise 61 with q œ 3 and p œ 4. 64. Diverges by Exercise 62 with q œ 12 and p œ 12 . 65. Converges by Exercise 61 with q œ 1000 and p œ 1.001. 66. Diverges by Exercise 62 with q œ 15 and p œ 0.99. 67. Converges by Exercise 61 with q œ 3 and p œ 1.1. 68. Diverges by Exercise 62 with q œ 12 and p œ 12 . 69. Example CAS commands: Maple: a := n -> 1./n^3/sin(n)^2; s := k -> sum( a(n), n=1..k ); # (a)] limit( s(k), k=infinity ); pts := [seq( [k,s(k)], k=1..100 )]: # (b) plot( pts, style=point, title="#69(b) (Section 10.4)" ); pts := [seq( [k,s(k)], k=1..200 )]: # (c) plot( pts, style=point, title="#69(c) (Section 10.4)" ); pts := [seq( [k,s(k)], k=1..400 )]: # (d) plot( pts, style=point, title="#69(d) (Section 10.4)" ); evalf( 355/113 ); Mathematica: Clear[a, n, s, k, p] a[n_]:= 1 / ( n3 Sin[n]2 ) s[k_]= Sum[ a[n], {n, 1, k}] points[p_]:= Table[{k, N[s[k]]}, {k, 1, p}] points[100] ListPlot[points[100]] points[200] ListPlot[points[200] points[400] ListPlot[points[400], PlotRange Ä All] To investigate what is happening around k = 355, you could do the following. N[355/113] N[1 355/113] Sin[355]//N a[355]//N N[s[354]] Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.5 The Ratio and Root Tests 597 N[s[355]] N[s[356]] _ _ 70. (a) Let S œ ! n12 , which is a convergent p-series. By Example 5 in Section 10.2, ! nan 1 1b converges to 1. By Theorem 8, n œ1 n œ1 _ _ _ _ n œ1 _ n œ1 n œ1 nœ1 S œ ! n12 œ ! nan 1 1b ! n12 ! nan 1 1b _ _ nœ1 nœ1 œ ! nan 1 1b ! Š n12 nan 1 1b ‹ also converges. _ _ n œ1 n œ1 (b) Since ! nan 1 1b converges to 1 (from Example 5 in Section 10.2), S œ 1 ! Š n12 nan 1 1b ‹ œ 1 ! n2 an1 1b n œ1 _ (c) _ Ä 0 faster than the terms of ! n12 . The new series is comparible to ! n13 , so it will converge faster because its terms n œ1 1000 n œ1 1000 (d) The series 1 ! n2 an1 1b gives a better approximation. Using Mathematica, 1 ! n2 an1 1b œ 1.644933568, while nœ1 1000000 ! n œ1 nœ1 1 1 n2 œ 1.644933067. Note that 6 œ 1.644934067. The error is 4.99 ‚ 107 compared with 1 ‚ 106 . 2 10.5 THE RATIO AND ROOT TESTS 1. 2. 3. 2n n! 0 for all n 2nb" 1; lim Œ n2n" ! œ nÄ_ a n nÄ_ n! an b1b b 2 1; lim Œ 3nnbb21 œ n2 3n 0 for all n nÄ_ b1bc1b! b2 b b anb1b2 _ nœ1 _ nÄ_ aan a n n 3n aa nÄ_ 3 3 ‰ n3 ‰ ! n n 2 converges ˆ1‰ 1 lim ˆ n3 œ lim ˆ 3n n †3 † n 2 6 œ lim 3 œ 3 1 Ê 3 1; lim Œ nbnc1 1b!1 œ nÄ_ an 1 b! 0 for all n an 1b2 _ 2 †2 n! ! 2 converges ˆ 2 ‰ lim Š an" b†n! † 2n ‹ œ lim n " œ 0 1 Ê n! b nÄ_ nÄ_ 2 nœ1 "b n 2n n 3n 4n 1 lim Š na†nan21b2b! † aann 1b! ‹ œ lim Š n2 4n 4 ‹ œ lim Š 2n 4 ‹ 3 nÄ_ 2 2 nÄ_ nÄ_ 1b! œ lim ˆ 6n 2 4 ‰ œ _ 1 Ê ! aann diverges 1 b2 nÄ_ 4. n œ1 2nb1 n†3n 1 0 for all n _ 2an1b1 b an1b 1 1; lim n1 2†3n1 nÄ_ n 1 a n†3 nb1 lim Š an21b†3†n2 1 †3 † n2†3n1 ‹ œ lim ˆ 3n2n 3 ‰ œ lim ˆ 23 ‰ œ 23 1 œ n 1 nÄ_ nÄ_ nÄ_ nb1 Ê ! n2†3nc1 converges nœ1 5. n4 4n 0 for all n b1b4 an 1; lim Œ 4nnb4 1 œ nÄ_ n 4 4 lim Š an4n †14b † 4n4 ‹ œ lim Š n 4n 4n6n4 4n 1 ‹ n nÄ_ 4 3 2 nÄ_ _ 4 œ lim ˆ 14 1n 2n3 2 n13 4n1 4 ‰ œ 14 1 Ê ! n4n converges nÄ_ 6. n œ1 3nb2 ln n 0 for all n 3anb1bb2 2; lim Œ ln3nnbb21 œ nÄ_ a ln n _ nb2 3 lim Š ln3an †31b † 3lnnbn2 ‹ œ lim Š ln 3anlnn1b ‹ œ lim Š n1 ‹ œ lim ˆ 3n n 3 ‰ b nÄ_ nÄ_ nÄ_ nb1 nÄ_ nb2 œ lim ˆ 31 ‰ œ 3 1 Ê ! 3ln n diverges nÄ_ 7. n œ2 n 2 a n 2 b! 0 for all n nx32n an 1; lim nÄ_ b 1b2aan b 1b b 2b! b 1bx32an 1b an n2 an 2b! nx32n œ 2 3ban 2b! lim Š an an1ba1nb † n2 annx3 2b! ‹ œ lim Š n 9n5n3 9n7n2 3 ‹ †nx32n †32 2n nÄ_ 3 nÄ_ _ 7 ! n an 2n2b! converges ˆ 6n 15 ‰ ˆ6‰ 1 œ lim Š 3n27n2 15n 18n ‹ œ lim 54n 18 œ lim 54 œ 9 1 Ê n x3 2 nÄ_ nÄ_ nÄ_ 2 n œ1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 2 598 8. Chapter 10 Infinite Sequences and Series an n †5 n a2n 3b lnan 1b 0 for all n a a nÄ_ 1b†a2n 3b lim Š 5an † lnlnaann 21bb ‹ œ na2n 5b œ b lim Œ 2 n 1 1; b 1b†5n 1 3b lnaan 1b n†5n 3b lnan 1b a2n a2n 3b lnan 1b n 1b†5 †5 lim Š a2na ‹ n †5 n 5b lnan 2b † n œ 1b nÄ_ 1 lnan 1b 15 25 ‰ lim Š 10n 2n2 25n † lim Š n b1 1 ‹ ‹ † lim Š lnan 2b ‹ œ lim ˆ 20n 5n 4n 5 2 nÄ_ nÄ_ nÄ_ _ nÄ_ nÄ_ nb2 n †5 ! ‰ ˆ n2‰ ˆ 1‰ œ lim ˆ 20 4 † lim n 1 œ 5 † lim 1 œ 5 † 1 œ 5 1 Ê a2n 3b lnan 1b diverges nÄ_ 9. 7 a2n 5bn nÄ_ nÄ_ n 10. a3n4 bn 0 for all n 3‰ 11. ˆ 4n 3n 5 n n 1 _ nÄ_ n œ1 _ 4 n 1; lim É œ a3n bn nÄ_ nÄ_ n œ1 n n 3‰ ˆ 4n 2; lim É œ 3n 5 n 1 n nÄ_ diverges 8 2n 3 1n ‰ n 1; lim É ˆ n 3‰ lim ˆ 43 ‰ œ 43 1 Ê ! ˆ 4n diverges 3n 5 nÄ_ 1; lim Ê’lnê2 1n ‰“ Ê ! ’lnê2 1n ‰“ _ 3‰ lim ˆ 4n 3n 5 œ nÄ_ 0 for all n n 4 ‰ lim ˆ 3n œ 0 1 Ê ! a3n4 bn converges n n 1 _ n È lim Š 2n 7 5 ‹ œ 0 1 Ê ! a2n 7 5bn converges nÄ_ 0 for all n 12. ’lnê2 1n ‰“ n œ2 7 n 1; lim É œ a2n 5bn 0 for all n n nÄ_ n œ1 1 1 În œ lim ’lnê2 1n ‰“ nÄ_ œ lnae2 b œ 2 1 n œ1 13. ˆ 0 for all n nÄ_ n 14. ’sinŠ È1n ‹“ n2 15. ˆ1 n1 ‰ 16. n1"bn 8 2n œ 3 1n ‰ lim Œ ˆ nÄ_ n È 8 0 for all n 17. converges by the Ratio Test: n nÄ_ n œ1 _ n n2 lim ˆ1 n1 ‰ œ e1 1 Ê ! ˆ1 n1 ‰ converges nÄ_ " n 2; lim É n1bn œ nÄ_ n2 lim Éˆ1 n1 ‰ œ n n lim sinŠ È1n ‹ œ sina0b œ 0 1 Ê ! ’sinŠ È1n ‹“ converges nÄ_ 1; 8 n œ1 _ n 0 for all n 1 n n 1; lim Ê ’sinŠ È1n ‹“ œ 0 for all n _ œ 9 1 Ê ! ˆ3 1 ‰2n converges 3 1 ‰ 2 nÄ_ nœ1 n È nÄ_ ” lim anb1 œ n lim n Ä _ an Ä_ _ n È 1 ! 1"bn converges lim Š n È n n‹ œ 0 1 Ê n lim Š n1În 1 1 ‹ œ nÄ_ È (n b 1) 2 2 n b1 • È n 2 n œ2 È È n (n 1) 2 ˆ1 n" ‰ 2 ˆ "# ‰ œ "# 1 œ n lim † 2È2 œ n lim Ä _ #nb1 Ä_ n ” #n • Š (nenbb1)1 ‹ 2 18. converges by the Ratio Test: lim anb1 œ n lim n Ä _ an Ä_ anb1 19. diverges by the Ratio Test: n lim œ n lim Ä _ an Ä_ anb1 20. diverges by the Ratio Test: n lim œ n lim Ä _ an Ä_ # Š nen ‹ Š (nenbb1)! 1 ‹ ˆ en!n ‰ b 1)! ‹ Š (n 10nb1 ˆ 10n!n ‰ "! 21. converges by the Ratio Test: lim anb1 œ n lim n Ä _ an Ä_ "! 2 n (n ")! e n" œ n lim † n! œ n lim e œ_ Ä _ enb1 Ä_ n (n ")! 10 n œ n lim † n! œ n lim œ_ Ä _ 10nb1 Ä _ 10 Š (n10bn1)1 ‹ Š n10n ‹ # (n 1) œ n lim † ne2 œ n Ä lim ˆ1 n" ‰ ˆ "e ‰ œ "e 1 _ Ä _ enb1 n "! (n ") " ˆ1 n" ‰"! ˆ 1"0 ‰ œ 10 œ n lim † 10 1 n"! œ n lim Ä _ 10n 1 Ä_ n Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.5 The Ratio and Root Tests n 599 n ˆ nn 2 ‰ œ lim ˆ1 n2 ‰ œ e# Á 0 22. diverges; n lim a œ n lim Ä_ n Ä_ nÄ_ (1) ˆ 45 ‰ c2 (1)n d Ÿ ˆ 45 ‰ (3) which is the nth term of a convergent 23. converges by the Direct Comparison Test: 2(1.25) n œ n n n geometric series 24. converges; a geometric series with krk œ ¸ 23 ¸ 1 n n ˆ1 3n ‰ œ lim ˆ1 n3 ‰ œ e$ ¸ 0.05 Á 0 25. diverges; n lim a œ n lim Ä_ n Ä_ nÄ_ " ‰n ˆ1 3n 26. diverges; n lim a œ n lim œ n lim 1 Ä_ n Ä_ Ä_ Š "3 ‹ n n "Î$ œe 27. converges by the Direct Comparison Test: lnn$n nn$ œ n"# for n ¸ 0.72 Á 0 2, the nth term of a convergent p-series. n 1În n a(ln n) b n (ln n) n È É 28. converges by the nth-Root Test: n lim an œ n lim n 1În nn œ n lim Ä_ Ä_ Ä_ an b ln n œ n lim œ n lim Ä_ n Ä_ Š "n ‹ 1 œ01 29. diverges by the Direct Comparison Test: "n n"# œ n n# 1 "# ˆ "n ‰ for n 2 or by the Limit Comparison Test (part 1) with "n . n 1În n n n ˆ n" n"# ‰ œ lim ˆˆ n" n"# ‰ ‰ È É 30. converges by the nth-Root Test: n lim an œ n lim Ä_ Ä_ nÄ_ 31. diverges by the Direct Comparison Test: lnnn n" for n ˆ " n"# ‰ œ 0 1 œ n lim Ä_ n 3 (n 1) ln (n 1) anb1 32. converges by the Ratio Test: n lim œ n lim † n ln2 (n) œ "# 1 #nb1 Ä _ an Ä_ n (n 2)(n 3) anb1 n! 33. converges by the Ratio Test: n lim œ n lim † (n 1)(n (n 1)! 2) œ 0 1 Ä _ an Ä_ $ (n 1) anb1 34. converges by the Ratio Test: n lim œ n lim † ne$ œ "e 1 Ä _ an Ä _ en 1 n (n 4)! anb1 n4 35. converges by the Ratio Test: n lim œ n lim † 3! n! 3 œ n lim œ 3" 1 Ä _ an Ä _ 3! (n 1)! 3nb1 (n 3)! Ä _ 3(n 1) n nb1 (n 1)2 (n 2)! anb1 2 2‰ ˆ n n 1 ‰ ˆ 32 ‰ ˆ nn 36. converges by the Ratio Test: n lim œ n lim † n2n3(n n! 3nb1 (n 1)! 1 œ 3 1 1)! œ n lim Ä _ an Ä_ Ä_ n (n 1)! 1)! anb1 n" 37. converges by the Ratio Test: n lim œ n lim † (2n n! œ n lim œ01 Ä _ an Ä _ (2n 3)! Ä _ (2n 3)(2n 2) (n 1)! anb1 " ˆ n ‰ œ lim 38. converges by the Ratio Test: n lim œ n lim † n œ n lim Ä _ an Ä _ (n 1)nb1 n! Ä _ n1 n Ä _ ˆ n b " ‰n n n n œ n lim Ä_ ˆ " n 1 "n ‰ œ "e 1 n n È " n n n È 39. converges by the Root Test: n lim an œ n lim œ n lim œ n lim œ01 Ä_ Ä _ É (ln n)n Ä _ ln n Ä _ ln n Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 600 Chapter 10 Infinite Sequences and Series n n È n n n È 40. converges by the Root Test: n lim an œ n lim œ n lim Ä_ Ä _ É (ln n)nÎ2 Ä_ È ln n Ä_ Ä_ n n lim È n Èln n œ 0 1 lim n œ n È n œ 1‹ Šn lim Ä_ n! ln n ln n n " " 41. converges by the Direct Comparison Test: n(n 2)! œ n(n 1)(n 2) n(n 1)(n 2) œ (n 1)(n #) n# which is the nth-term of a convergent p-series $ n n 1 3 an 1 3 n ˆ 3 ‰ œ 3# 1 42. diverges by the Ratio Test: n lim œ n lim † n32n œ n lim Ä _ an Ä _ (n 1)$ 2n 1 Ä _ (n 1)3 # an1bx‘2 2 an1b anb1 n 2n 1 43. converges by the Ratio Test: n lim œ n lim † a2nbx œ n lim œ n lim œ 41 1 Ä _ an Ä _ 2(n 1)‘x ‘2 Ä _ (2n 2)(2n 1) Ä _ 4n2 6n 2 2 nx anb1 44. converges by the Ratio Test: n lim œ n lim Ä _ an Ä_ n n n n a2n 5bˆ2nb1 3‰ † a2n 33ba22n 3b œ n lim ’ 2n 5 † 2†6 4†2 3†3 6 “ 3nb1 2 Ä _ 2n 3 3†6n 9†3n 2†2n 6 œ n lim ’ 2n 5 “ † n lim ’ 2†6 4†2 3†3 6 “ œ 1 † 23 œ 23 1 Ä _ 2n 3 Ä _ 3 † 6 n 9 † 3 n 2† 2 n 6 n n n anb1 45. converges by the Ratio Test: n lim œ n lim Ä _ an Ä_ anb1 46. converges by the Ratio Test: n lim œ n lim Ä _ an Ä_ ˆ 1 b nsin n ‰ an œ01 an "n Š 1 b tan n ‹ an an approaches 1 1# while the denominator tends to _ anb1 47. diverges by the Ratio Test: n lim œ n lim Ä _ an Ä_ " tan œ n lim n Ä_ "n œ 0 since the numerator c1‰ ˆ 3n 3n 1 2n b 5 an œ n lim œ 3# 1 an Ä _ 2n 5 2 ‰ 48. diverges; an1 œ n n 1 an Ê an1 œ ˆ n n 1 ‰ ˆ n n 1 an1 ‰ Ê an1 œ ˆ n n 1 ‰ ˆ n n 1 ‰ ˆ nn 1 an2 a " n n 1 n 2 3 " Ê an1 œ ˆ n 1 ‰ ˆ n ‰ ˆ n 1 ‰ â ˆ # ‰ a" Ê an1 œ n 1 Ê an1 œ n 1 , which is a constant times the general term of the diverging harmonic series anb1 49. converges by the Ratio Test: n lim œ n lim Ä _ an Ä_ 50. converges by the Ratio Test: lim anb1 œ n lim n Ä _ an Ä_ anb1 51. converges by the Ratio Test: n lim œ n lim Ä _ an Ä_ Š 2n ‹ an an 2 œ n lim œ01 Ä_ n Èn n Œ # an an œ n lim Ä_ Š 1 bnln n ‹ an an n n È n œ "# 1 "ln n " œ n lim œ n lim œ01 n Ä_ Ä_ n 52. nnln10n 0 and a" œ #" Ê an 0; ln n 10 for n e"! Ê n ln n n 10 Ê nnln10n 1 Ê an1 œ nnln10n an an ; thus an1 an " # Ê n lim a Á 0, so the series diverges by the nth-Term Test Ä_ n 3 3 6 " %! " 2 " 2 " 2 " É É É É 53. diverges by the nth-Term Test: a" œ "3 , a# œ É 3 , a$ œ Ê 3 œ 3 , a% œ ËÊ 3 œ 3 ,á , % n! " n! " n " É an œ É a œ 1 because šÉ 3 Ê n lim 3 › is a subsequence of š 3 › whose limit is 1 by Table 8.1 Ä_ n Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.5 The Ratio and Root Tests # $ # ' % ' #% 54. converges by the Direct Comparison Test: a" œ "# , a# œ ˆ "# ‰ , a$ œ Šˆ "# ‰ ‹ œ ˆ "# ‰ , a% œ Šˆ "# ‰ ‹ œ ˆ "# ‰ , á n! n Ê an œ ˆ "# ‰ ˆ "# ‰ which is the nth-term of a convergent geometric series 2 anb1 55. converges by the Ratio Test: n lim œ n lim Ä _ an Ä_ nb1 n" œ n lim œ #" 1 Ä _ 2n 1 (n 1)! (n 1)! † 2(2n)! lim 2(n 1)(n 1) n n! n! œ (2n 2)! n Ä _ (2n #)(2n 1) (3n 3)! 1)! (n 2)! anb1 56. diverges by the Ratio Test: n lim œ n lim † n! (n (3n)! Ä _ an Ä _ (n 1)! (n 2)! (n 3)! (3n 3)(3 2)(3n 1) 2 ‰ ˆ 3n 1 ‰ œ n lim œ n lim 3 ˆ 3n n# n 3 œ 3 † 3 † 3 œ 27 1 Ä _ (n 1)(n 2)(n 3) Ä_ n n! n (n!) n È 57. diverges by the Root Test: n lim an ´ n lim œ n lim œ_1 Ä_ Ä _ É an n b# Ä _ n# n! n (n!) n (n!) ˆ " ‰ ˆ 2n ‰ ˆ 3n ‰ â ˆ n n 1 ‰ ˆ nn ‰ É 58. converges by the Root Test: n lim œ n lim œ n lim œ n lim É ann bn Ä_ Ä_ Ä _ nn Ä_ n nn# n n " Ÿ n lim œ01 Ä_ n " n n n n È 59. converges by the Root Test: n lim an œ n lim œ n lim œ n lim œ01 Ä_ Ä _ É 2 n# Ä _ #n Ä _ #n ln 2 n n n n n È 60. diverges by the Root Test: n lim an œ n lim œ n lim œ_1 Ä_ Ä _ É a#n b# Ä_ 4 n 1†3† â †(2n 1)(2n 1) anb1 4 2 n! 2n " 61. converges by the Ratio Test: n lim œ n lim † 1 †3 † â œ 4" 1 4nb1 2nb1 (n 1)! †(2n 1) œ n lim Ä _ an Ä_ Ä _ (4†#)(n 1) n n (2n 1) 1†2†3†4 â (2n 1)(2n) (2n)! 62. converges by the Ratio Test: an œ (2†14†3ââ#n) a3n 1b œ (2†4 â 2n)# a3n 1b œ a2n n!b# a3n 1b # a3 1 b (2n ")(2n 2) a3 1b (2n 2)! Ê n lim † a2 n!b(2n)! œ n lim 2# (n 1)# a3n 1 1b Ä _ c2nb1 (n 1)!d# a3nb1 1b Ä_ n n n cn œ n lim Š 4n 6n 2 ‹ a1 3 b œ 1 † 3" œ 3" 1 Ä _ 4n# 8n 4 a3 3cn b # anb1 " ˆ n ‰ œ 1p œ 1 Ê no conclusion 63. Ratio: n lim œ n lim † n œ n lim Ä _ an Ä _ (n 1)p 1 Ä_ n1 p p " " n " n È É Root: n lim an œ n lim n n‰p œ (1)p œ 1 Ê no conclusion np œ n lim Ä_ Ä_ Ä _ ˆÈ p ˆ"‰ p anb1 ln n n" " n 64. Ratio: n lim œ n lim † (ln1n) œ ’n lim œ Šn lim “ œ ”n lim ‹ Ä _ an Ä _ (ln (n 1))p Ä _ ln (n 1) Ä _ ˆ n b" 1 ‰ • Ä_ n p œ (1)p œ 1 Ê no conclusion " n n È Root: n lim an œ n lim É (ln n)p œ Ä_ Ä_ " p lim (ln n)1În ‹ ŠnÄ_ ln (ln n) Ê n lim ln f(n) œ n lim œ n lim n Ä_ Ä_ Ä_ ˆ n ln" n ‰ 1 n È œ n lim eln fÐnÑ œ e! œ 1; therefore n lim an œ Ä_ Ä_ n) ; let f(n) œ (ln n)1În , then ln f(n) œ ln (ln n " œ n lim œ 0 Ê n lim (ln n)1În Ä _ n ln n Ä_ " p lim (ln n)1În ‹ ŠnÄ_ _ œ (1)" p œ 1 Ê no conclusion (n ") 2 65. an Ÿ 2nn for every n and the series ! #nn converges by the Ratio Test since n lim † n œ "# 1 Ä _ 2nb1 _ n nœ1 Ê ! an converges by the Direct Comparison Test nœ1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. p 601 602 Chapter 10 Infinite Sequences and Series 2 2anb1b n2 n2 b2nb1 1; lim n2bn21 ! œ nÄ_ 66. 2n! 0 for all n a nÄ_ n! _ 2nb1 †4 ‰ ˆ 2†4 1ln 4 ‰ lim Š a2n1b†n! † 2n!n2 ‹ œ lim Š 2n1 ‹ œ lim ˆ n2 1 œ lim b n nÄ_ n nÄ_ nÄ_ n2 œ _ 1 Ê ! 2n! diverges n œ1 10.6 ALTERNATING SERIES, ABSOLUTE AND CONDITIONAL CONVERGENCE 1. converges by the Alternating Convergence Test since: un œ È1n 0 for all n Ê Èn11 Ÿ È1n Ê un1 Ÿ un ; lim un œ nÄ_ 1; n 1Ê n1 _ _ n œ1 n œ1 n Ê Èn 1 Èn lim 1 œ 0. nÄ_ Èn " 2. converges absolutely Ê converges by the Alternating Convergence Test since ! kan k œ ! n$Î# which is a convergent p-series 3. converges Ê converges by Alternating Series Test since: un œ n31 n 0 for all n Ê an 1b3n1 n 3n Ê an11b3nb1 Ÿ n13n Ê un1 Ÿ un ; lim un œ lim un œ nÄ_ ln n Ê aln an 1bb2 2; n 3n 2Ên1 n aln nb2 Ê aln an11bb2 Ÿ aln1nb2 Ê aln an41bb2 Ÿ aln4nb2 Ê un1 Ÿ un ; lim 4 2 œ 0. nÄ_ aln nb 5. converges Ê converges by Alternating Series Test since: un œ n2 n 1 0 for all n Ê n3 2n2 2n Ê n2n1 n Ê 3n 1 lim 1 n œ 0. nÄ_ n 3 nÄ_ 4. converges Ê converges by Alternating Series Test since: un œ aln4nb2 0 for all n Ê ln an 1b 1Ên1 1; n n3 n2 n 1 Ê nan2 2n 2b n 1 Ê un1 Ÿ un ; an1b2 1 lim un œ nÄ_ 1 Ê 2n2 2n 1; n n3 n2 n 1 Ê nŠan 1b2 1‹ n2 n 1 an2 1ban 1b lim 2 n œ 0. nÄ_ n 1 6. diverges Ê diverges by nth Term Test for Divergence since: 2 lim n2 5 œ 1 Ê nÄ_ n 4 7. diverges Ê diverges by nth Term Test for Divergence since: lim 2n2 œ _ Ê n nÄ_ 5 lim a1bn1 nn2 4 œ does not exist 2 nÄ_ lim a1bn1 2n2 œ does not exist n nÄ_ _ _ n œ1 n œ1 n 8. converges absolutely Ê converges by the Absolute Convergence Test since ! kan k œ ! an10 1bx , which converges by the Ratio Test, since lim anabn 1 œ nÄ_ lim n 10 2 œ 0 1 nÄ_ _ n n n n ‰ ˆ n ‰ Á 0 Ê ! (1)n1 ˆ 10 9. diverges by the nth-Term Test since for n 10 Ê 10 1 Ê n lim diverges Ä _ 10 n œ1 10. converges by the Alternating Series Test because f(x) œ ln x is an increasing function of x Ê ln"x is decreasing Ê un un1 for n 1; also un 0 for n " 1 and n lim œ0 Ä _ ln n 11. converges by the Alternating Series Test since f(x) œ lnxx Ê f w (x) œ 1 x#ln x 0 when x e Ê f(x) is decreasing Ê un un1 ; also un 0 for n ln n 1 and n lim u œ n lim œ n lim Ä_ n Ä_ n Ä_ Š "n ‹ 1 œ0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.6 Alternating Series, Absolute and Conditional Convergence 603 12. converges by the Alternating Series Test since f(x) œ ln a1 x" b Ê f w (x) œ x(x" 1) 0 for x 0 Ê f(x) is decreasing Ê un un1 ; also un 0 for n ˆ1 n" ‰‹ œ ln 1 œ 0 1 and n lim u œ n lim ln ˆ1 "n ‰ œ ln Šn lim Ä_ n Ä_ Ä_ 13. converges by the Alternating Series Test since f(x) œ Ê un unb1 ; also un 0 for n Èx " x1 1 and n lim u œ n lim Ä_ n Ä_ È 3 n1 14. diverges by the nth-Term Test since n lim œ n lim Ä _ Èn 1 Ä_ _ _ nœ1 nœ1 1 x 2È x Ê f w (x) œ 2Èx (x 1)# 0 Ê f(x) is decreasing Èn " n1 œ 0 3É1 "n 1 Š È"n ‹ œ3Á0 n " ‰ 15. converges absolutely since ! kan k œ ! ˆ 10 a convergent geometric series nb1 " ‰ 16. converges absolutely by the Direct Comparison Test since ¹ (1) n (0.1) ¹ œ (10)" n n ˆ 10 which is the nth term n n of a convergent geometric series _ _ n œ1 n œ1 " " 17. converges conditionally since È"n Èn" 1 0 and n lim œ 0 Ê convergence; but ! kan k œ ! n"Î# Ä _ Èn is a divergent p-series " 18. converges conditionally since 1 "Èn 1 È"n 1 0 and n lim œ 0 Ê convergence; but Ä _ 1 Èn _ _ nœ1 nœ1 ! kan k œ ! " is a divergent series since 1"Èn 1 Èn _ _ n œ1 nœ1 _ " " and ! n"Î# is a divergent p-series #È n nœ1 19. converges absolutely since ! kan k œ ! n$n1 and n$n1 n"# which is the nth-term of a converging p-series n! 20. diverges by the nth-Term Test since n lim œ_ Ä _ #n _ " " 21. converges conditionally since n " 3 (n 1) œ 0 Ê convergence; but ! kan k 3 0 and n lim Ä _ n 3 n œ1 _ œ! n œ1 " " n 3 diverges because n 3 _ " ! " is a divergent series 4n and n n œ1 _ 22. converges absolutely because the series ! ¸ sinn# n ¸ converges by the Direct Comparison Test since ¸ sinn# n ¸ Ÿ n"# n œ1 3n 23. diverges by the nth-Term Test since n lim œ1Á0 Ä _ 5n nb1 nb1 24. converges absolutely by the Direct Comparison Test since ¹ (n2)5n ¹ œ n25n 2 ˆ 25 ‰ which is the nth term n of a convergent geometric series 25. converges conditionally since f(x) œ x"# "x Ê f w (x) œ ˆ x2$ x"# ‰ 0 Ê f(x) is decreasing and hence un unb1 0 for n _ _ n œ1 nœ1 _ _ n œ1 n œ1 ˆ " "n ‰ œ 0 Ê convergence; but ! kan k œ ! 1n#n 1 and n lim Ä _ n# œ ! n"# ! "n is the sum of a convergent and divergent series, and hence diverges Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 604 Chapter 10 Infinite Sequences and Series 26. diverges by the nth-Term Test since n lim a œ n lim 101În œ 1 Á 0 Ä_ n Ä_ 27. converges absolutely by the Ratio Test: n lim Š uunbn 1 ‹ œ n lim Ä_ Ä_ ” (n")# ˆ 23 ‰ n n# ˆ 23 ‰ n 1 •œ 3 1 2 1d 28. converges conditionally since f(x) œ x ln" x Ê f w (x) œ cln(x(x) ln x)# 0 Ê f(x) is decreasing " 2 and n lim œ 0 Ê convergence; but by the Integral Test, Ä _ n ln n Ê un unb1 0 for n '2_ x dxln x œ lim '2b Šln x‹ dx œ lim cln (ln x)d b2 œ lim cln (ln b) ln (ln 2)d œ _ " x bÄ_ _ _ n œ1 n œ1 bÄ_ bÄ_ Ê ! kan k œ ! n ln" n diverges _ # 29. converges absolutely by the Integral Test since '1 atan" xb ˆ 1 " x# ‰ dx œ lim ’ atan # xb “ " bÄ_ œ lim ’atan bÄ_ " # " bb atan œ (x ln x)# œ n lim Ä_ Š "n ‹ 1 Š n" ‹ _ _ n œ1 nœ1 ! kan k œ ! 1 # # 1# 1b “ œ "# ’ˆ 1# ‰ ˆ 14 ‰ “ œ 332 # 30. converges conditionally since f(x) œ x lnlnx x Ê f w (x) œ 1 Š lnxx ‹ ln x Š lnxx ‹ b œ (x1lnln x)x # 0 Ê un Š "x ‹ (x ln x) (ln x) Š1 x" ‹ (x ln x)# ln n un1 0 when n e and n lim Ä _ n ln n œ 0 Ê convergence; but n ln n n Ê n"ln n "n Ê nlnlnn n "n so that ln n n ln n diverges by the Direct Comparison Test n 31. diverges by the nth-Term Test since n lim œ1Á0 Ä _ n1 _ _ n œ1 nœ1 n 32. converges absolutely since ! kan k œ ! ˆ "5 ‰ is a convergent geometric series nb1 ("00) n! 33. converges absolutely by the Ratio Test: n lim † (100) lim "00 œ 0 1 Š uunbn 1 ‹ œ n lim n œ Ä_ Ä _ (n1)! n Ä _ n1 _ _ n œ1 n œ1 " " " 34. converges absolutely by the Direct Comparison Test since ! kan k œ ! n# 2n 1 and n# 2n 1 n# which is the nth-term of a convergent p-series _ _ n œ1 n œ1 _ " 35. converges absolutely since ! kan k œ ! ¹ (nÈ1)n ¹ œ ! n$Î# is a convergent p-series n n œ1 _ _ n œ1 nœ1 36. converges conditionally since ! cosnn1 œ ! (n1) is the convergent alternating harmonic series, but _ _ n œ1 n œ1 ! kan k œ ! n " n diverges 1) n È kan k œ n lim 37. converges absolutely by the Root Test: n lim Š (n(2n) n ‹ Ä_ Ä_ n 1 În n" œ n lim œ "# 1 Ä _ #n Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.6 Alternating Series, Absolute and Conditional Convergence # 605 # a(n 1)!b (n 1) 38. converges absolutely by the Ratio Test: n lim † (2n)! œ n lim œ "4 1 ¹ anb1 ¹ œ n lim Ä _ an Ä _ ((2n 2)!) (n!)# Ä _ (2n 2)(2n 1) (2n)! (n ")(n 2)â(2n) 39. diverges by the nth-Term Test since n lim kan k œ n lim œ n lim 2n n Ä_ Ä _ 2n n! n Ä_ (n 1)(n 2)â(n (n 1)) ˆ n # 1 ‰ œ n lim n lim #nc1 Ä_ Ä_ n 1 œ_Á0 (n 1)! (n 1)! 3 40. converges absolutely by the Ratio Test: n lim ¹ anabn 1 ¹ œ n lim (2n 3)! Ä_ Ä_ nb1 1)! † (2n n! n! 3n # (n 1) 3 œ n lim œ 43 1 Ä _ (2n 2)(2n 3) Èn 1 Èn Èn 1 Èn † Èn 1 Èn œ Èn "1 Èn and š Èn 1" Èn › is a 1 41. converges conditionally since _ (") decreasing sequence of positive terms which converges to 0 Ê ! Èn converges; but 1 Èn n n œ1 _ _ n œ1 nœ1 " Èn 1 Èn ! kan k œ ! diverges by the Limit Comparison Test (part 1) with È"n ; a divergent p-series: " lim nÄ_ Èn 1 Èn œ lim " Èn n Ä _ Èn Èn 1 Èn œ n lim Ä_ 1 œ "# É1 1n 1 È # n n 42. diverges by the nth-Term Test since n lim ŠÈn# n n‹ œ n lim ŠÈn# n n‹ † Š Ènn# ‹ Ä_ Ä_ n n n œ n lim œ n lim Ä _ È n# n n Ä_ " œ "# Á 0 É1 "n 1 43. diverges by the nth-Term Test since n lim ŠÉn Èn Èn‹ œ n lim ŠÉ n È n È n ‹ Ä_ Ä_ – œ n lim Ä_ Èn É n Èn Èn œ n lim Ä_ É n Èn Èn Én Èn Èn — " œ #" Á 0 " É 1 Èn 1 44. converges conditionally since š Èn "Èn 1 › is a decreasing sequence of positive terms converging to 0 _ (") Ê ! Èn Èn 1 converges; but n lim Ä_ n n œ1 _ " Èn 1 ‹ Èn " œ n lim œ n lim œ "# Ä _ È n È n 1 Ä _ 1 É 1 " Š È"n ‹ n Š Èn _ so that ! Èn "Èn 1 diverges by the Limit Comparison Test with ! È"n which is a divergent p-series nœ1 nœ1 n n 2 45. converges absolutely by the Direct Comparison Test since sech (n) œ en 2ecn œ e2n2e 1 2e e2n œ en which is the nth term of a convergent geometric series _ _ n œ1 nœ1 46. converges absolutely by the Limit Comparison Test (part 1): ! kan k œ ! en 2ecn Apply the Limit Comparison Test with e1n , the n-th term of a convergent geometric series: 2 cn 2e 2 e ce lim œ n lim œ n lim œ2 n Ä _ Œ 1n Ä _ en ecn Ä _ 1 ec2n n n e _ nb1 ) 1 1 1 1 47. 14 16 18 10 12 14 Þ Þ Þ œ ! 2(" an 1b ; converges by Alternating Series Test since: un œ 2an 1b 0 for all n n œ1 n2 n 1 Ê 2 an 2 b 2an 1b Ê 2aan 11b 1b Ÿ 2an11b Ê un1 Ÿ un ; lim un œ nÄ_ lim 1 œ 0. nÄ_ 2an1b Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 1; 606 Chapter 10 Infinite Sequences and Series _ _ _ n œ1 n œ1 n œ1 1 1 1 1 1 48. 1 14 91 16 25 36 49 64 Þ Þ Þ œ ! an ; converges by the Absolute Convergence Test since ! kan k œ ! n"# which is a convergent p-series 49. kerrork ¸(1)' ˆ "5 ‰¸ œ 0.2 50. kerrork ¸(1)' ˆ 10" & ‰¸ œ 0.00001 & "" 51. kerrork ¹(1)' (0.01) 5 ¹ œ 2 ‚ 10 52. kerrork k(1)% t% k œ t% 1 53. kerrork 0.001 Ê un1 0.001 Ê an 11b2 3 0.001 Ê an 1b2 3 1000 Ê n 1 È997 ¸ 30.5753 Ê n 54. kerrork 0.001 Ê un1 0.001 Ê an n 1b21 1 0.001 Ê an 1b2 1 1000an 1b Ê n ¸ 998.9999 Ê n 31 998È9982 4a998b 2 999 55. kerrork 0.001 Ê un1 0.001 Ê 1 È n 1‹ 3 0.001 Ê Šan 1b 3 ân 1b 3Èn 1‰ 3 1000 2 È Ê ŠÈn 1‹ 3Èn 1 10 0 Ê Èn 1 œ 3 29 40 œ 2 Ê n œ 3 Ê n 4 56. kerrork 0.001 Ê un1 0.001 Ê lnalnan1 3bb 0.001 Ê lnalnan 3bb 1000 Ê n 3 ee 1000 ¸ 5.297 ‚ 10323228467 which is the maximum arbitrary-precision number represented by Mathematica on the particular computer solving this problem.. ' " 57. (2n)! 105 ' Ê (2n)! 105 œ 200,000 Ê n ' 58. n!" 105 ' Ê 105 n! Ê n 59. (a) an 9 Ê 1 1 #"! 3!" 4!" 5!" 6!" 7!" 8!" ¸ 0.367881944 an1 fails since 3" #" _ _ nœ1 nœ1 5 Ê 1 #"! 4!" 6!" 8!" ¸ 0.54030 n _ n _ n n (b) Since ! kan k œ ! ˆ 3" ‰ ˆ "# ‰ ‘ œ ! ˆ "3 ‰ ! ˆ "# ‰ is the sum of two absolutely convergent nœ1 nœ1 series, we can rearrange the terms of the original series to find its sum: " ˆ "3 "9 27 á ‰ ˆ "# "4 "8 á ‰ œ ˆ "3 ‰ 1 ˆ 3" ‰ ˆ"‰ 1 #ˆ " ‰ œ #" 1 œ #" # " " 60. s#! œ 1 "# 3" 4" á 19 20 ¸ 0.6687714032 Ê s#! #" † #"1 ¸ 0.692580927 _ 61. The unused terms are ! (1)j 1 aj œ (1)n 1 aan 1 an 2 b (1)n 3 aan 3 an 4 b á jœn 1 œ (1)n 1 caan 1 an 2 b aan 3 an 4 b á d . Each grouped term is positive, so the remainder has the same sign as (1)n 1 , which is the sign of the first unused term. n n k œ1 k œ1 62. sn œ 1"†2 #"†3 3"†4 á n(n " 1) œ ! k(k " 1) œ ! ˆ k" k " 1 ‰ œ ˆ1 "# ‰ ˆ "# 3" ‰ ˆ 3" 4" ‰ ˆ 4" 5" ‰ á ˆ n" n " 1 ‰ which are the first 2n terms of the first series, hence the two series are the same. Yes, for n sn œ ! ˆ k" k" 1 ‰ œ ˆ1 "# ‰ ˆ "# 3" ‰ ˆ 3" 4" ‰ ˆ 4" 5" ‰ á ˆ n " 1 n" ‰ ˆ n" n " 1 ‰ œ 1 n " 1 k œ1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.6 Alternating Series, Absolute and Conditional Convergence 607 ˆ1 n " 1 ‰ œ 1 Ê both series converge to 1. The sum of the first 2n 1 terms of the first Ê n lim s œ n lim Ä_ n Ä_ ˆ1 n " 1 ‰ œ 1. series is ˆ1 n " 1 ‰ n " 1 œ 1. Their sum is n lim s œ n lim Ä_ n Ä_ _ _ _ _ n œ1 n œ1 n œ1 n œ1 63. Theorem 16 states that ! kan k converges Ê ! an converges. But this is equivalent to ! an diverges Ê ! kan k diverges _ _ n œ1 n œ1 64. ka" a# á an k Ÿ ka" k ka# k á kan k for all n; then ! kan k converges Ê ! an converges and these imply that _ _ n œ1 n œ1 º ! an º Ÿ ! kan k _ 65. (a) ! kan bn k converges by the Direct Comparison Test since kan bn k Ÿ kan k kbn k and hence n œ1 _ ! aan bn b converges absolutely n œ1 _ _ _ (b) ! kbn k converges Ê ! bn converges absolutely; since ! an converges absolutely and nœ1 _ nœ1 nœ1 _ _ ! bn converges absolutely, we have ! can (bn )d œ ! aan bn b converges absolutely by part (a) nœ1 _ _ _ nœ1 nœ1 _ n œ1 nœ1 nœ1 nœ1 (c) ! kan k converges Ê kkk ! kan k œ ! kkan k converges Ê ! kan converges absolutely _ _ _ nœ1 nœ1 nœ1 66. If an œ bn œ (1)n È"n , then ! (1)n È"n converges, but ! an bn œ ! n" diverges 67. s" œ "# , s# œ "# 1 œ "# , " " " " s$ œ "# 1 "4 "6 8" 10 1"# 14 16 18 #"0 2"# ¸ 0.5099, s% œ s$ 3" ¸ 0.1766, " " " " " " " s& œ s% #"4 #"6 #"8 30 3"# 34 36 38 40 42 44 ¸ 0.512, s' œ s& 5" ¸ 0.312, " " " " " " " " " " " s( œ s' 46 48 50 52 54 56 58 60 62 64 66 ¸ 0.51106 N" 1 68. (a) Since ! kan k converges, say to M, for % 0 there is an integer N" such that º ! kan k Mº #% nœ1 N" 1 N" 1 _ _ _ nœ1 nœ1 nœN" nœN" nœN" Í » ! kan k ! kan k ! kan k » #% Í » ! kan k» #% Í ! kan k #% . Also, ! an converges to L Í for % 0 there is an integer N# (which we can choose greater than or equal to N" ) such _ that ksN# Lk #% . Therefore, ! kan k #% and ksN# Lk #% . nœN" Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 608 Chapter 10 Infinite Sequences and Series _ k nœ1 nœ1 (b) The series ! kan k converges absolutely, say to M. Thus, there exists N" such that º ! kan k Mº % whenever k N" . Now all of the terms in the sequence ekbn kf appear in ekan kf. Sum together all of the N terms in ekbn kf, in order, until you include all of the terms ekan kf nœ" 1 , and let N# be the largest index in the N# N# _ n œ1 nœ1 n œ1 sum ! kbn k so obtained. Then º ! kbn k Mº % as well Ê ! kbn k converges to M. 10.7 POWER SERIES _ nb1 1. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ xxn ¹ 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have ! (1)n , a divergent Ä_ Ä_ n œ1 _ series; when x œ 1 we have ! 1, a divergent series n œ1 (a) the radius is 1; the interval of convergence is 1 x 1 (b) the interval of absolute convergence is 1 x 1 (c) there are no values for which the series converges conditionally nb1 2. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x(x5)5)n ¹ 1 Ê kx 5k 1 Ê 6 x 4; when x œ 6 we have Ä_ Ä_ _ _ n œ1 nœ1 ! (1)n , a divergent series; when x œ 4 we have ! 1, a divergent series (a) the radius is 1; the interval of convergence is 6 x 4 (b) the interval of absolute convergence is 6 x 4 (c) there are no values for which the series converges conditionally nb1 1) " " 3. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (4x (4x 1)n ¹ 1 Ê k4x 1k 1 Ê 1 4x 1 1 Ê # x 0; when x œ # we Ä_ Ä_ _ _ _ _ _ n œ1 n œ1 n œ1 n œ1 n œ1 have ! (1)n (1)n œ ! (1)2n œ ! 1n , a divergent series; when x œ 0 we have ! (1)n (1)n œ ! (1)n , a divergent series (a) the radius is "4 ; the interval of convergence is "# x 0 (b) the interval of absolute convergence is "# x 0 (c) there are no values for which the series converges conditionally nb1 4. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (3xn2)1 Ä_ Ä_ ˆ n ‰ 1 Ê k3x 2k 1 † (3x n 2)n ¹ 1 Ê k3x 2k n lim Ä _ n1 _ Ê 1 3x 2 1 Ê "3 x 1; when x œ 3" we have ! ("n ) which is the alternating harmonic series and is n n œ1 _ conditionally convergent; when x œ 1 we have ! "n , the divergent harmonic series n œ1 (a) the radius is "3 ; the interval of convergence is 3" Ÿ x 1 (b) the interval of absolute convergence is "3 x 1 (c) the series converges conditionally at x œ "3 nb1 kx 2 k 2) 5. n lim † (x 10 ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 10 nb1 10 1 Ê kx 2k 10 Ê 10 x 2 10 2)n ¹ 1 Ê Ä_ Ä_ n _ _ nœ1 nœ1 Ê 8 x 12; when x œ 8 we have ! (")n , a divergent series; when x œ 12 we have ! 1, a divergent series (a) the radius is "0; the interval of convergence is 8 x 12 (b) the interval of absolute convergence is 8 x 12 (c) there are no values for which the series converges conditionally Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.7 Power Series nb1 609 6. n lim k2xk 1 Ê k2xk 1 Ê "# x "# ; when x œ "# we have ¹ uunbn 1 ¹ 1 Ê n lim ¹ (2x) (2x)n ¹ 1 Ê n lim Ä_ Ä_ Ä_ _ _ ! (")n , a divergent series; when x œ " we have ! 1, a divergent series # n œ1 n œ1 (a) the radius is "# ; the interval of convergence is "# x "# (b) the interval of absolute convergence is "# x "# (c) there are no values for which the series converges conditionally nb1 (n 2) (n 1)(n 2) 1)x 7. n lim 1 Ê kxk 1 ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n (n 3) † nxn ¹ 1 Ê kxk n lim Ä_ Ä_ Ä _ (n 3)(n) _ Ê 1 x 1; when x œ 1 we have ! (")n n n # , a divergent series by the nth-term Test; when x œ " we n œ1 _ have ! n n # , a divergent series n œ1 (a) the radius is "; the interval of convergence is " x " (b) the interval of absolute convergence is " x " (c) there are no values for which the series converges conditionally nb1 8. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x n2)1 Ä_ Ä_ ˆ n ‰ 1 Ê kx 2k 1 † (x n 2)n ¹ 1 Ê kx 2k n lim Ä _ n1 _ Ê 1 x 2 1 Ê 3 x 1; when x œ 3 we have ! "n , a divergent series; when x œ " we have n œ1 _ ! n œ1 (1)n n , a convergent series (a) the radius is "; the interval of convergence is 3 x Ÿ " (b) the interval of absolute convergence is 3 x " (c) the series converges conditionally at x œ 1 nb1 x 9. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ (n 1)Èn 1 3nb1 nÈ n 3n xn ¹ 1 n n Ê kx3k Šn lim ‹ ŠÉn lim ‹1 Ä _ n1 Ä _ n1 _ ) Ê kx3k (1)(1) 1 Ê kxk 3 Ê 3 x 3; when x œ 3 we have ! (" , an absolutely convergent series; n$Î# n n œ1 _ 1 when x œ 3 we have ! n$Î# , a convergent p-series n œ1 (a) the radius is 3; the interval of convergence is 3 Ÿ x Ÿ 3 (b) the interval of absolute convergence is 3 Ÿ x Ÿ 3 (c) there are no values for which the series converges conditionally nb1 Èn n 10. n lim 1 Ê kx 1k 1 ¹ uunbn 1 ¹ 1 Ê n lim ¹ (xÈn1) 1 † (x 1)n ¹ 1 Ê kx 1k Én lim Ä_ Ä_ Ä _ n1 _ ) Ê 1 x 1 1 Ê 0 x 2; when x œ 0 we have ! (" , a conditionally convergent series; when x œ 2 n"Î# n n œ1 _ we have ! n"Î# , a divergent series 1 n œ1 (a) the radius is 1; the interval of convergence is 0 Ÿ x 2 (b) the interval of absolute convergence is 0 x 2 (c) the series converges conditionally at x œ 0 nb1 ˆ " ‰ 1 for all x 11. n lim † n! ¹ 1 Ê kxk n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ x Ä_ Ä _ (n 1)! xn Ä _ n1 (a) the radius is _; the series converges for all x Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 610 Chapter 10 Infinite Sequences and Series (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally nb1 nb1 ˆ " ‰ 1 for all x 12. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ 3 x † 3nn!xn ¹ 1 Ê 3 kxk n lim Ä_ Ä _ (n 1)! Ä _ n1 (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally nb1 2nb2 ˆ 4n ‰ œ 4x# 1 Ê x# 41 13. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ 4 n x 1 † 4n nx2n ¹ 1 Ê x# n lim Ä_ Ä_ Ä _ n1 _ n Ê 12 x 12 ; when x œ 12 we have ! 4n ˆ 12 ‰ 2n n œ1 _ ! n œ1 _ œ ! n1 , a divergent p-series; when x œ 12 we have nœ1 _ 4 ˆ 1 ‰2n œ ! n1 , a divergent p-series n 2 n n œ1 (a) the radius is 12 ; the interval of convergence is 12 x 12 (b) the interval of absolute convergence is 12 x 12 (c) there are no values for which the series converges conditionally nb1 14. n lim † n 3 ¹ 1 Ê lx 1l n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 1) Š n ‹ œ 13 lx 1l 1 Ä_ Ä _ an 1b2 3nb1 (x 1)n Ä _ 3an 1b2 2 _ n 2 _ Ê 2 x 4; when x œ 2 we have ! (n2 3)3n œ ! (n1) , an absolutely convergent series; when x œ 4 we have 2 n n œ1 _ n nœ1 _ n ! (3) ! 12 , an absolutely convergent series. 2 n œ n œ1 n 3 n œ1 n (a) the radius is 3; the interval of convergence is 2 Ÿ x Ÿ 4 (b) the interval of absolute convergence is 2 Ÿ x Ÿ 4 (c) there are no values for which the series converges conditionally nb1 x 15. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ È(n 1)# 3 È n# 3 ¹1 xn _ # n 3 Ê kxk Én lim " Ê kxk 1 Ä _ n# 2n 4 Ê 1 x 1; when x œ 1 we have ! È("# ) , a conditionally convergent series; when x œ 1 we have n œ1 _ " ! È# n œ1 n 3 n n 3 , a divergent series (a) the radius is 1; the interval of convergence is 1 Ÿ x 1 (b) the interval of absolute convergence is 1 x 1 (c) the series converges conditionally at x œ 1 n 1 x 16. n lim † ¹ uun n 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ È(n 1)# 3 È n# 3 ¹1 xn # n 3 Ê kxk Én lim " Ê kxk 1 Ä _ n# 2n 4 _ _ nœ1 nœ1 ) Ê 1 x 1; when x œ 1 we have ! Èn"# 3 , a divergent series; when x œ 1 we have ! È(" , n# 3 a conditionally convergent series (a) the radius is 1; the interval of convergence is 1 x Ÿ 1 (b) the interval of absolute convergence is 1 x 1 (c) the series converges conditionally at x œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. n Section 10.7 Power Series 3) 17. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n 1)(x 5nb1 Ä_ Ä_ nb1 611 ˆ n n " ‰ 1 Ê kx 5 3k 1 † n(x 5 3)n ¹ 1 Ê kx 5 3k n lim Ä_ n _ _ Ê kx 3k 5 Ê 5 x 3 5 Ê 8 x 2; when x œ 8 we have ! n(5n5) œ ! (1)n n, a divergent n n œ1 _ n œ1 _ n series; when x œ 2 we have ! n5n œ ! n, a divergent series n œ1 5 n œ1 (a) the radius is 5; the interval of convergence is 8 x 2 (b) the interval of absolute convergence is 8 x 2 (c) there are no values for which the series converges conditionally nb1 # # (n 1) n 1 18. n lim † 4 annxn 1b ¹ 1 Ê kx4k n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n 1)x ¹ a b ¹ 1 Ê kxk 4 Ä_ Ä _ 4nb1 an# 2n 2b Ä _ n an# 2n 2b n _ _ Ê 4 x 4; when x œ 4 we have ! n(n#1)1 , a conditionally convergent series; when x œ 4 we have ! n# n 1 , n n œ1 n œ1 a divergent series (a) the radius is 4; the interval of convergence is 4 Ÿ x 4 (b) the interval of absolute convergence is 4 x 4 (c) the series converges conditionally at x œ 4 19. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä_ Èn 1 xnb1 n † È3n xn ¹ 1 3nb1 ˆ n n 1 ‰ 1 Ê k3xk 1 Ê kxk 3 Ê k3xk Én lim Ä_ _ _ n œ1 n œ1 Ê 3 x 3; when x œ 3 we have ! (1)n Èn , a divergent series; when x œ 3 we have ! Èn, a divergent series (a) the radius is 3; the interval of convergence is 3 x 3 (b) the interval of absolute convergence is 3 x 3 (c) there are no values for which the series converges conditionally 20. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä_ nbÈ 1 n 1 (2x5)nb1 ¹1 n n (2x5)n È Ê k2x 5k n lim Š Ä_ nbÈ 1 n1 ‹1 n n È t lim È t Ä_ n 1 Ê k2x 5k 1 Ê 1 2x 5 1 Ê 3 x 2; when x œ 3 we have Ê k2x 5k Œ tlim Èn n _ Ä_ _ n n n ! (1) È È n, a divergent series since n lim n œ 1; when x œ 2 we have ! È n, a divergent series Ä_ n œ1 n œ1 (a) the radius is "# ; the interval of convergence is 3 x 2 (b) the interval of absolute convergence is 3 x 2 (c) there are no values for which the series converges conditionally _ _ _ n œ1 n œ1 21. First, rewrite the series as ! a2 (1)n bax 1bn1 œ ! 2ax 1bn1 ! (1)n ax 1bn1 . For the series n œ1 _ n ! 2ax 1bn1 : lim ¹ unb1 ¹ 1 Ê lim ¹ 2ax1nbc1 ¹ 1 Ê lx 1l lim 1 œ lx 1l 1 Ê 2 x 0; For the un nÄ_ n Ä _ 2 ax 1 b nÄ_ n œ1 _ nb1 n (1) ax1b series ! (1)n ax 1bn1 : n lim 1 œ lx 1l 1 ¹ uunbn 1 ¹ 1 Ê n lim ¹ ¹ 1 Ê lx 1ln lim Ä_ Ä _ (1)n ax1bnc1 Ä_ n œ1 _ Ê 2 x 0; when x œ 2 we have ! a2 (1)n ba1bn1 , a divergent series; when x œ 0 we have n œ1 _ ! a2 (1)n b, a divergent series n œ1 (a) the radius is 1; the interval of convergence is 2 x 0 (b) the interval of absolute convergence is 2 x 0 (c) there are no values for which the series converges conditionally Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 612 Chapter 10 Infinite Sequences and Series ( 1) 22. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä_ 3 ax 2bnb1 9n † (1)n 32n3nax 2bn ¹ 1 Ê lx 2ln lim œ 9lx 2l 1 3an 1b Ä _ n1 nb1 2nb2 _ _ 19 17 ! (1) 3 ˆ 1 ‰ œ ! 1 , a divergent series; when x œ 19 we have Ê 17 9 x 9 ; when x œ 9 we have 3n 9 3n 9 n 2n n œ1 _ _ n nœ1 ! (1) 3 ˆ 1 ‰n œ ! (1) , a conditionally convergent series. n œ1 (a) (b) (c) n 2n 3n 9 n 3n n œ1 19 the radius is 19 ; the interval of convergence is 17 9 xŸ 9 19 the interval of absolute convergence is 17 9 x 9 the series converges conditionally at x œ 19 9 nb1 23. lim ¹ uunbn 1 ¹ 1 nÄ_ Ê n lim Ä_ » Š1 n " 1 ‹ xnb1 Š1 "n ‹ xn n " t lim Š1 t ‹ e Ä_ » 1 Ê kxk lim Š1 " ‹n 1 Ê kxk ˆ e ‰ 1 Ê kxk 1 n nÄ_ t _ n Ê 1 x 1; when x œ 1 we have ! (1)n ˆ1 "n ‰ , a divergent series by the nth-Term Test since n œ1 _ n n lim ˆ1 "n ‰ œ e Á 0; when x œ 1 we have ! ˆ1 n" ‰ , a divergent series nÄ_ n œ1 (a) the radius is "; the interval of convergence is 1 x 1 (b) the interval of absolute convergence is 1 x 1 (c) there are no values for which the series converges conditionally 24. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ ln (nxnln1)xn Ä_ Ä_ nb1 ˆ " ‰ n 1 ˆ n ‰ 1 Ê kxk 1 1 Ê kxk n lim ¹ 1 Ê kxk n lim Ä _ º ˆ"‰ º Ä _ n1 n _ Ê 1 x 1; when x œ 1 we have ! (1)n ln n, a divergent series by the nth-Term Test since n lim ln n Á 0; Ä_ n œ1 _ when x œ 1 we have ! ln n, a divergent series n œ1 (a) the radius is 1; the interval of convergence is 1 x 1 (b) the interval of absolute convergence is 1 x 1 (c) there are no values for which the series converges conditionally nb1 nb1 x 25. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n 1) nn xn Ä_ Ä_ n ˆ1 n" ‰ ‹ Š lim (n 1)‹ 1 ¹ 1 Ê kxk Šn lim Ä_ nÄ_ Ê e kxk n lim (n 1) 1 Ê only x œ 0 satisfies this inequality Ä_ (a) the radius is 0; the series converges only for x œ 0 (b) the series converges absolutely only for x œ 0 (c) there are no values for which the series converges conditionally 26. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n n!1)!(x(x4)4)n Ä_ Ä_ nb1 (n 1) 1 Ê only x œ 4 satisfies this inequality ¹ 1 Ê kx 4k n lim Ä_ (a) the radius is 0; the series converges only for x œ 4 (b) the series converges absolutely only for x œ 4 (c) there are no values for which the series converges conditionally nb1 ˆ n ‰ 1 Ê kx # 2k 1 Ê kx 2k 2 27. n lim † n2 ¹ 1 Ê kx # 2k n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 2) Ä_ Ä _ (n 1) 2nb1 (x 2)n Ä _ n1 n _ _ n œ1 n œ1 ! (1) Ê 2 x 2 2 Ê 4 x 0; when x œ 4 we have ! " n , a divergent series; when x œ 0 we have n the alternating harmonic series which converges conditionally (a) the radius is 2; the interval of convergence is 4 x Ÿ 0 (b) the interval of absolute convergence is 4 x 0 (c) the series converges conditionally at x œ 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. nb1 , Section 10.7 Power Series nb1 613 nb1 (n 2)(x 1) ˆ n 2 ‰ 1 Ê 2 kx 1k 1 28. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ ((2)2)n (n 1)(x 1)n ¹ 1 Ê 2 kx 1k n lim Ä_ Ä_ Ä _ n1 _ Ê kx 1k "# Ê "# x 1 "# Ê "# x 3# ; when x œ #" we have ! (n 1) , a divergent series; when x œ 3# _ n œ1 we have ! (1) (n 1), a divergent series n n œ1 (a) the radius is "# ; the interval of convergence is "# x #3 (b) the interval of absolute convergence is "# x 3# (c) there are no values for which the series converges conditionally nb1 # # x n ln n 29. n lim † n(lnxnn) ¹ 1 Ê kxk Šn lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ ‹ Šn lim ‹ 1 Ä_ Ä _ (n 1) aln (n 1)b# Ä _ n1 Ä _ ln (n 1) ˆ"‰ # # n1 n Ê kxk (1) Œn lim 1 Ê kxk Šn lim ‹ 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have Ä _ ˆ " ‰ Ä_ n nb1 _ _ nœ1 nœ1 n " ! (1) # which converges absolutely; when x œ 1 we have ! n(ln n) n(ln n)# which converges (a) the radius is "; the interval of convergence is 1 Ÿ x Ÿ 1 (b) the interval of absolute convergence is 1 Ÿ x Ÿ 1 (c) there are no values for which the series converges conditionally nb1 ln (n) x n 30. n lim † n lnxn(n) ¹ 1 Ê kxk Šn lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ ‹ Šn lim ‹1 Ä_ Ä _ (n 1) ln (n 1) Ä _ n1 Ä _ ln (n 1) _ Ê kxk (1)(1) 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have ! (nln1)n , a convergent alternating series; n n œ2 _ when x œ 1 we have ! n ln" n which diverges by Exercise 38, Section 9.3 n œ2 (a) the radius is "; the interval of convergence is 1 Ÿ x 1 (b) the interval of absolute convergence is 1 x 1 (c) the series converges conditionally at x œ 1 2nb3 $Î# 5) n 31. n lim † (4x n 5)2n 1 ¹ 1 Ê (4x 5)# Šn lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (4x ‹ (n 1)$Î# Ä_ Ä_ Ä _ n1 _ $Î# 2nb1 Ê k4x 5k 1 Ê 1 4x 5 1 Ê 1 x 3# ; when x œ 1 we have ! (n1)$Î# n œ1 _ 1 Ê (4x 5)# 1 _ œ ! n" $Î# which is n œ1 2nb1 absolutely convergent; when x œ 3# we have ! ("n)$Î# , a convergent p-series n œ1 (a) the radius is "4 ; the interval of convergence is 1 Ÿ x Ÿ 3# (b) the interval of absolute convergence is 1 Ÿ x Ÿ 3# (c) there are no values for which the series converges conditionally nb2 32. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (3x2n1)4 Ä_ Ä_ ˆ 2n 2 ‰ 1 Ê k3x 1k 1 † (3x2n1)2nb1 ¹ 1 Ê k3x 1k n lim Ä _ 2n 4 _ nb1 1) Ê 1 3x 1 1 Ê 23 x 0; when x œ 23 we have ! (2n 1 , a conditionally convergent series; n œ1 _ when x œ 0 we have ! n œ1 _ (")nb1 ! " , a divergent series 2n 1 œ #n 1 nœ1 (a) the radius is "3 ; the interval of convergence is 32 Ÿ x 0 (b) the interval of absolute convergence is 23 x 0 (c) the series converges conditionally at x œ 23 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 614 Chapter 10 Infinite Sequences and Series nb1 x ˆ 1 ‰ 1 for all x 33. n lim † 2†4†6xân a2nb ¹ 1 Ê kxk n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ 2†4†6âa2nba2an 1bb Ä _ 2n 2 (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally 3 5 7 2n 1 2 n 1 1x 34. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ † † âa an ba1ba2 2nb1 b b Ä_ Ä_ nb2 2n 3 n † 3†5†7âan2n2 1bxnb1 ¹ 1 Ê kxk n lim Š a2an 1bb2 ‹ 1 Ê only Ä_ 2 2 n x œ 0 satisfies this inequality (a) the radius is 0; the series converges only for x œ 0 (b) the series converges absolutely only for x œ 0 (c) there are no values for which the series converges conditionally _ nan 1b n n 35. For the series ! 121222â and 12 22 â n2 œ nan 1ba6 2n 1b so that we can â n2 x , recall 1 2 â n œ 2 n œ1 nan b 1b _ _ nb1 rewrite the series as ! Œ n n b 1 2 2n b 1 xn œ ! ˆ 2n 3 1 ‰xn ; then n lim † a2n3xn 1b ¹ 1 ¹ uunbn 1 ¹ 1 Ê n lim ¹ 3x Ä_ Ä _ a2 an 1 b 1 b a n œ1 Ê kx k ba b nœ1 6 _ lim ¹ a2n 1b ¹ 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have ! ˆ 2n 3 1 ‰a1bn , a conditionally n Ä _ a2n 3b n œ1 _ convergent series; when x œ 1 we have ! ˆ n œ1 3 ‰ 2n 1 , a divergent series. (a) the radius is 1; the interval of convergence is 1 Ÿ x 1 (b) the interval of absolute convergence is 1 x 1 (c) the series converges conditionally at x œ 1 _ 36. For the series ! ŠÈn 1 Èn‹ax 3bn , note that Èn 1 Èn œ n œ1 _ Èn 1 Èn Èn 1 Èn † Èn 1 Èn œ Èn 11 Èn so that we 1 nb1 n can rewrite the series as ! Ènax13b Èn ; then n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ ax 3 b Ä_ Ä _ Èn 2 Èn 1 Èn 1 Èn n œ1 ax 3bn _ Èn 1 Èn ¹1 n b Ê lx 3ln lim 1 Ê lx 3l 1 Ê 2 x 4; when x œ 2 we have ! Èn a11 , a conditionally Èn Ä _ Èn 2 Èn 1 n œ1 _ convergent series; when x œ 4 we have ! Èn 11 Èn , a divergent series; n œ1 (a) the radius is 1; the interval of convergence is 2 Ÿ x 4 (b) the interval of absolute convergence is 2 x 4 (c) the series converges conditionally at x œ 2 nb1 lx l an 1bxx an 1 b 37. n lim † 3†6†n9xâxna3nb ¹ 1 Ê lxln lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ ¹ ¹ 1 Ê 3 1 Ê lxl 3 Ê R œ 3 Ä_ Ä _ 3†6†9âa3nba3an 1bb Ä _ 3 an 1 b 2 nb1 2 2 4 lx l 2 4 6 2n 2 n 1 x 38. n lim † aa22†5†4†8†6ââaa3n2nbb12 xbbn ¹ 1 Ê lxln lim ¹ a2n 2b ¹ 1 Ê 9 1 ¹ uunbn 1 ¹ 1 Ê n lim ¹ a † † âa ba a bbb Ä_ Ä _ a2†5†8âa3n 1ba3an 1b 1bb2 Ä _ a3n 2b2 Ê lxl 94 Ê R œ 94 2 nb1 2 lx l n 1 x an 1 b 39. n lim † 2 a2nbx ¹ 1 Ê lxln lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ aa bxb ¹ ¹ 1 Ê 8 1 Ê lxl 8 Ê R œ 8 Ä_ Ä _ 2nb1 a2an 1bbx anxb2 xn Ä _ 2a2n 2ba2n 1b n2 n n n n ‰ n Éˆ ˆ n ‰ 1 Ê lxle1 1 Ê lxl e Ê R œ e È 40. n lim un 1 Ê n lim xn 1 Ê lxl n lim n1 Ä_ Ä_ Ä _ n1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.7 Power Series nb1 615 nb1 41. n lim 3 1 Ê lxl 31 Ê 31 x 31 ; at x œ 31 we have ¹ uunbn 1 ¹ 1 Ê n lim ¹ 3 3n xxn ¹ 1 Ê lxl n lim Ä_ Ä_ Ä_ _ _ _ _ _ ! 3n ˆ 1 ‰n œ ! a1bn , which diverges; at x œ 1 we have ! 3n ˆ 1 ‰n œ ! 1 , which diverges. The series ! 3n xn 3 n œ0 3 n œ0 3 n œ0 nœ0 _ œ ! a3xbn is a convergent geometric series when 1 x 1 and the sum is 3 n œ0 3 nœ0 1 1 3x . nb1 e 4 42. n lim 1 1 Ê lex 4l 1 Ê 3 ex 5 Ê ln 3 x ln 5; ¹ uunbn 1 ¹ 1 Ê n lim ¹ a aex 4b bn ¹ 1 Ê lex 4l n lim Ä_ Ä_ Ä_ x _ n _ _ nœ0 nœ0 _ n at x œ ln 3 we have ! êln 3 4‰ œ ! a1bn , which diverges; at x œ ln 5 we have ! êln 5 4‰ œ ! 1, which nœ0 _ nœ0 diverges. The series ! aex 4bn is a convergent geometric series when ln 3 x ln 5 and the sum is 1 a1ex 4b œ 5 1 ex . n œ0 2nb2 43. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 4n1)1 Ä_ Ä_ n † (x 4 1)2n ¹ 1 Ê (x 1)# lim 4 nÄ_ _ k1k 1 Ê (x 1)# 4 Ê kx 1k 2 _ _ Ê 2 x 1 2 Ê 1 x 3; at x œ 1 we have ! (42)n œ ! 44n œ ! 1, which diverges; at x œ 3 2n n œ0 _ _ 2n n nœ0 n œ0 _ n we have ! 24n œ ! 44n œ ! 1, a divergent series; the interval of convergence is 1 x 3; the series n œ0 _ ! n œ0 nœ0 nœ0 _ # n (x ")2n œ ! Šˆ x # 1 ‰ ‹ is a convergent geometric series when 1 x 3 and the sum is 4n nœ0 " œ " 1 Šxc # ‹ # " ’ 4 (x 4 ")# “ œ 4 x# 2x 1 œ 3 2x x# 4 4 2nb2 44. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 9n1)1 Ä_ Ä_ n † (x 9 1)2n ¹ 1 Ê (x 1)# lim k1k 1 9 nÄ_ _ Ê (x 1)# 9 Ê kx 1k 3 _ Ê 3 x 1 3 Ê 4 x 2; when x œ 4 we have ! (93)n œ ! 1 which diverges; at x œ 2 we have n œ0 _ 2n n œ0 _ ! 32nn œ ! " which also diverges; the interval of convergence is 4 x 2; the series 9 n œ0 _ nœ0 _ n ! (x n1) œ ! Šˆ x1 ‰# ‹ is a convergent geometric series when 4 x 2 and the sum is 9 3 2n n œ0 n œ0 " 1Š # œ xb1 ‹ 3 " ’ 9 (x 1)# “ 9 œ 9 x# 9 2x 1 œ 8 2x9 x# 45. n lim ¹ uunbn 1 ¹ 1 Ê n lim Ä_ Ä_ º n ˆÈx 2‰nb1 † ˆÈx2 2‰n º 1 2nb1 Ê ¸Èx 2¸ 2 Ê 2 Èx 2 2 Ê 0 Èx 4 _ _ Ê 0 x 16; when x œ 0 we have ! (1)n , a divergent series; when x œ 16 we have ! (1)n , a divergent nœ0 nœ0 _ Èx 2 n series; the interval of convergence is 0 x 16; the series ! Š # ‹ is a convergent geometric series when n œ0 0 x 16 and its sum is " 1Œ Èx c 2 œ # Œ 2c " Èx 2 # œ 4 2Èx nb1 46. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (ln(lnx)x)n ¹ 1 Ê kln xk 1 Ê 1 ln x 1 Ê e" x e; when x œ e" or e we Ä_ Ä_ _ _ _ nœ0 nœ0 nœ0 obtain the series ! 1n and ! (1)n which both diverge; the interval of convergence is e" x e; ! (ln x)n œ 1 "ln x " when e xe Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 616 Chapter 10 Infinite Sequences and Series # 47. n lim ¹ uunbn 1 ¹ 1 Ê n lim Š x 3 1 ‹ Ä_ Ä_ º n1 # † ˆ x# 3 1 ‰ º 1 Ê ax 3 1b n lim k1k 1 Ê x 3 " 1 Ê x# 2 Ä_ n # _ Ê kxk È2 Ê È2 x È2 ; at x œ „ È2 we have ! (1)n which diverges; the interval of convergence is n œ0 _ # È2 x È2 ; the series ! Š x 3 1 ‹ n is a convergent geometric series when È2 x È2 and its sum is n œ0 " œ 3 c x"# c 1 œ # 3 x# # 1 Š x 3b 1 ‹ Š ‹ 3 # ax 1 b 48. n lim ¹ uun n 1 ¹ 1 Ê n lim ¹ 2n 1 Ä_ Ä_ n 1 † ax# 2 1bn ¹ 1 Ê kx# 1k 2 Ê È3 x È3 ; when x œ „ È3 we n _ _ n # have ! 1n , a divergent series; the interval of convergence is È3 x È3 ; the series ! Š x 2 1 ‹ is a n œ0 n œ0 convergent geometric series when È3 x È3 and its sum is nb1 49. n lim ¹ (x #n3) b1 Ä_ " " œ # 1 Šx 2 1‹ 2 œ 3 2 x# Šx# 1 ‹ # _ n † (x 2 3)n ¹ 1 Ê kx 3k 2 Ê 1 x 5; when x œ 1 we have ! (1)n which diverges; nœ1 _ when x œ 5 we have ! (1) which also diverges; the interval of convergence is 1 x 5; the sum of this n n œ1 " œ x2 1 . 3 1 Šxc # ‹ convergent geometric series is n If f(x) œ 1 #" (x 3) 4" (x 3)# á ˆ #" ‰ (x 3)n á n œ x 2 1 then f w (x) œ #" "# (x 3) á ˆ #" ‰ n(x 3)n1 á is convergent when 1 x 5, and diverges 2 2 when x œ 1 or 5. The sum for f w (x) is (x 1)# , the derivative of x 1 . 50. If f(x) œ 1 "# (x 3) 4" (x 3)# á ˆ "# ‰ (x 3)n á œ x 2 1 then ' f(x) dx n $ # œ x (x 4 3) (x 123) á ˆ "# ‰ _ n (x 3)n 1 á . n 1 _ At x œ 1 the series ! n21 diverges; at x œ 5 n œ1 2 the series ! (n1) 1 converges. Therefore the interval of convergence is 1 x Ÿ 5 and the sum is n n œ1 2 ln kx 1k (3 ln 4), since ' x 2 1 dx œ 2 ln kx 1k C, where C œ 3 ln 4 when x œ 3. # % ' ) "! 5x 7x 9x 11x 51. (a) Differentiate the series for sin x to get cos x œ 1 3x 3! 5! 7! 9! 11! á # % ' ) "! x œ 1 x#! x4! x6! x8! 10! á . The series converges for all values of x since lim nÄ_ 2nb2 " ¹ (2nx 2)! † a#xn#8b! ¹ œ x2 n lim Š ‹ œ 0 1 for all x. Ä _ a2n 1ba2n 2b $ $ & & ( ( * * "" "" $ & ( * "" 32x 128x 512x 2048x (b) sin 2x œ 2x 23!x 25!x 27!x 29!x 2 11!x á œ 2x 8x 3! 5! 7! 9! 11! á " "‰ $ ‰ # ˆ (c) 2 sin x cos x œ 2 (0 † 1) (0 † 0 1 † 1)x ˆ0 † " # 1 † 0 0 † 1 x 0 † 0 1 † # 0 † 0 1 † 3! x ˆ0 † 4!" 1 † 0 0 † #" 0 † 3!" 0 † 1‰ x% ˆ0 † 0 1 † 4!" 0 † 0 #" † 3!" 0 † 0 1 † 5!" ‰ x& $ & 16x ˆ0 † 6!" 1 † 0 0 † 4!" 0 † 3!" 0 † #" 0 † 5!" 0 † 1‰ x' á ‘ œ 2 ’x 4x 3! 5! á “ $ $ & & ( ( * * "" "" œ 2x 23!x 25!x 27!x 29!x 2 11!x á 52. (a) d 2x 3x# 4x$ 5x% x# x$ x% x x x x x ae b œ 1 2! 3! 4! 5! á œ 1 x #! 3! 4! á œ e ; thus the derivative of e is e itself (b) ' ex dx œ ex C œ x x# x3! x4! x5! á C, which is the general antiderivative of ex # $ % & (c) ex œ 1 x x#! x3! x4! x5! á ; ecx † ex œ 1 † 1 (1 † 1 1 † 1)x ˆ1 † #"! 1 † 1 #"! † 1‰ x# ˆ1 † 3!" 1 † #"! #"! † 1 3!" † 1‰ x$ ˆ1 † 4!" 1 † 3!" #"! † #"! 3!" † 1 4!" † 1‰ x% # $ % & ˆ1 † 5!" 1 † 4!" #"! † 3!" 3!" † #"! 4!" † 1 5!" † 1‰ x& á œ 1 0 0 0 0 0 á Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.8 Taylor and Maclaurin Series 617 17x 62x 53. (a) ln ksec xk C œ ' tan x dx œ ' Šx x3 2x 15 315 2835 á ‹ dx $ # % ' ) & ( * "! # % ' ) "! x 17x 31x x x 17x 31x œ x# 1x# 45 2520 14,175 á C; x œ 0 Ê C œ 0 Ê ln ksec xk œ x# 12 45 2520 14,175 á , converges when 1# x 1# $ & ( * % ' ) x) d 17x 62x 2x 17x 62x # (b) sec# x œ d(tan œ dx Šx x3 2x dx 15 315 2835 á ‹ œ 1 x 3 45 315 á , converges when 1# x 1# # % ' # % ' 61x x 5x 61x (c) sec# x œ (sec x)(sec x) œ Š1 x# 5x 24 720 á ‹ Š1 # 24 720 á ‹ 5 5 ‰ % 61 5 5 61 ‰ ' œ 1 ˆ "# "# ‰ x# ˆ 24 "4 24 x ˆ 720 48 48 720 x á % ' ) 62x 1 1 œ 1 x# 2x3 17x 45 315 á , # x # 61x 54. (a) ln ksec x tan xk C œ ' sec x dx œ ' Š1 x2 5x 24 720 á ‹ dx # $ & ( * $ & ( * % ' x 61x 277x œ x x6 24 5040 72,576 á C; x œ 0 Ê C œ 0 Ê ln ksec x tan xk x 61x 277x œ x x6 24 5040 72,576 á , converges when 1# x 1# # % ' $ & ( x) d 61x 5x 61x 277x (b) sec x tan x œ d(sec œ dx Š1 x# 5x dx 24 720 á ‹ œ x 6 120 1008 á , converges when 1# x 1# # % ' $ & ( 61x x 2x 17x (c) (sec x)(tan x) œ Š1 x# 5x 24 720 á ‹ Šx 3 15 315 á ‹ $ & ( 2 5 ‰ & 17 " 5 61 ‰ ( 277x œ x ˆ "3 #" ‰ x$ ˆ 15 "6 24 x ˆ 315 15 72 720 x á œ x 5x6 61x 120 1008 á , 1# x 1# _ _ n œ0 n œk _ 55. (a) If f(x) œ ! an xn , then f ÐkÑ (x) œ ! n(n 1)(n 2)â(n (k 1)) an xnk and f ÐkÑ (0) œ k!ak Ê ÐkÑ ÐkÑ ak œ f k!(0) ; likewise if f(x) œ ! bn xn , then bk œ f k!(0) n œ0 Ê ak œ bk for every nonnegative integer k _ (b) If f(x) œ ! an xn œ 0 for all x, then f ÐkÑ (x) œ 0 for all x Ê from part (a) that ak œ 0 for every nonnegative integer k n œ0 10.8 TAYLOR AND MACLAURIN SERIES 1. f(x) œ e2x , f w (x) œ 2e2x , f ww (x) œ 4e2x , f www (x) œ 8e2x ; f(0) œ e2a0b œ ", f w (0) œ 2, f ww (0) œ 4, f www (0) œ 8 Ê P! (x) œ 1, P" (x) œ 1 2x, P# (x) œ 1 x 2x# , P$ (x) œ 1 x 2x# 43 x3 2. f(x) œ sin x, f w (x) œ cos x , f ww (x) œ sin x , f www (x) œ cos x; f(0) œ sin 0 œ 0, f w (0) œ 1, f ww (0) œ 0, f www (0) œ 1 Ê P! (x) œ 0, P" (x) œ x, P# (x) œ x, P$ (x) œ x 16 x3 3. f(x) œ ln x, f w (x) œ "x , f ww (x) œ x"# , f www (x) œ x2$ ; f(1) œ ln 1 œ 0, f w (1) œ 1, f ww (1) œ 1, f www (1) œ 2 Ê P! (x) œ 0, P" (x) œ (x 1), P# (x) œ (x 1) "# (x 1)# , P$ (x) œ (x 1) "# (x 1)# 3" (x 1)$ 4. f(x) œ ln (1 x), f w (x) œ 1 " x œ (1 x)" , f ww (x) œ (1 x)# , f www (x) œ 2(1 x)$ ; f(0) œ ln 1 œ 0, # # $ f w (0) œ 11 œ 1, f ww (0) œ (1)# œ 1, f www (0) œ 2(1)$ œ 2 Ê P! (x) œ 0, P" (x) œ x, P# (x) œ x x# , P$ (x) œ x x# x3 5. f(x) œ "x œ x" , f w (x) œ x# , f ww (x) œ 2x$ , f www (x) œ 6x% ; f(2) œ "# , f w (2) œ 4" , f ww (2) œ 4" , f www (x) œ 83 Ê P! (x) œ "# , P" (x) œ "# "4 (x 2), P# (x) œ "# "4 (x 2) "8 (x 2)# , " P$ (x) œ "# "4 (x 2) "8 (x 2)# 16 (x 2)$ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 618 Chapter 10 Infinite Sequences and Series 6. f(x) œ (x 2)" , f w (x) œ (x 2)# , f ww (x) œ 2(x 2)$ , f www (x) œ 6(x 2)% ; f(0) œ (2)" œ "# , f w (0) œ (2)# # œ 4" , f ww (0) œ 2(2)$ œ 4" , f www (0) œ 6(2)% œ 38 Ê P! (x) œ #" , P" (x) œ #" x4 , P# (x) œ #" x4 x8 , # $ x P$ (x) œ "# x4 x8 16 È2 È2 1 w ˆ1‰ # ,f 4 œ cos 4 œ # , È È È È È f ww ˆ 14 ‰ œ sin 14 œ #2 , f www ˆ 14 ‰ œ cos 14 œ #2 Ê P! œ #2 , P" (x) œ #2 #2 ˆx 14 ‰ , È È È È È È È # # $ P# (x) œ #2 #2 ˆx 14 ‰ 42 ˆx 14 ‰ , P$ (x) œ #2 #2 ˆx 14 ‰ 42 ˆx 14 ‰ 1#2 ˆx 14 ‰ 7. f(x) œ sin x, f w (x) œ cos x, f ww (x) œ sin x, f www (x) œ cos x; f ˆ 14 ‰ œ sin 14 œ 8. f(x) œ tan x, f w (x) œ sec2 x, f ww (x) œ 2sec2 x tan x, f www (x) œ 2sec4 x 4sec2 x tan2 x; f ˆ 14 ‰ œ tan 14 œ 1 , f w ˆ 14 ‰ œ sec2 ˆ 14 ‰ œ 2 , f ww ˆ 14 ‰ œ 2sec2 ˆ 14 ‰ tan ˆ 14 ‰ œ 4 , f www ˆ 14 ‰ œ 2sec4 ˆ 14 ‰ 4sec2 ˆ 14 ‰ tan2 ˆ 14 ‰ œ 16 Ê P! (x) œ 1 , 2 2 P" (x) œ 1 2 ˆx 14 ‰ , P# (x) œ 1 2 ˆx 14 ‰ 2 ˆx 14 ‰ , P$ (x) œ 1 2 ˆx 14 ‰ 2 ˆx 14 ‰ 83 ˆx 14 ‰ 3 9. f(x) œ Èx œ x"Î# , f w (x) œ ˆ "# ‰ x"Î# , f ww (x) œ ˆ 4" ‰ x$Î# , f www (x) œ ˆ 83 ‰ x&Î# ; f(4) œ È4 œ 2, " 3 f w (4) œ ˆ "# ‰ 4"Î# œ "4 , f ww (4) œ ˆ "4 ‰ 4$Î# œ 32 ,f www (4) œ ˆ 38 ‰ 4&Î# œ 256 Ê P! (x) œ 2, P" (x) œ 2 "4 (x 4), " " P# (x) œ 2 4" (x 4) 64 (x 4)# , P$ (x) œ 2 "4 (x 4) 64 (x 4)# 51"# (x 4)$ 10. f(x) œ (1 x)"Î# , f w (x) œ "# (1 x)"Î# , f ww (x) œ "4 (1 x)$Î# , f www (x) œ 38 (1 x)&Î# ; f(0) œ (1)"Î# œ 1, f w (0) œ "# (1)"Î# œ "# , f ww (0) œ "4 (1)$Î# œ "4 , f www (0) œ 83 (1)&Î# œ 83 Ê P! (x) œ 1, 1 $ P" (x) œ 1 2" x, P# (x) œ 1 2" x 8" x# , P$ (x) œ 1 2" x 8" x# 16 x 11. f(x) œ ex , f w (x) œ ex , f ww (x) œ ex , f www (x) œ ex Ê á f ÐkÑ (x) œ a1bk ex ; f(0) œ ea0b œ ", f w (0) œ 1, _ f ww (0) œ 1, f www (0) œ 1, á ß f ÐkÑ (0) œ (1)k Ê ex œ 1 x 12 x# 16 x3 á œ ! (n!1) xn n n œ0 12. f(x) œ x ex , f w (x) œ x ex ex , f ww (x) œ x ex 2ex , f www (x) œ x ex 3ex Ê á f ÐkÑ (x) œ x ex k ex ; f(0) œ a0bea0b œ 0, _ f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3, á ß f ÐkÑ (0) œ k Ê x x# 12 x3 á œ ! an 1 1b! xn n œ0 13. f(x) œ (1 x)" Ê f w (x) œ (1 x)# , f ww (x) œ 2(1 x)$ , f www (x) œ 3!(1 x)% Ê á f ÐkÑ (x) œ (1)k k!(1 x)k1 ; f(0) œ 1, f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3!, á ß f ÐkÑ (0) œ (1)k k! _ _ n œ0 nœ0 Ê 1 x x# x$ á œ ! (x)n œ ! (1)n xn x 3 w ww $ www % 14. f(x) œ 21 Ê á f ÐkÑ (x) œ 3ak!b(1 x) k 1 ; f(0) œ 2, x Ê f (x) œ (1 x)# , f (x) œ 6(1 x) , f (x) œ 18(1 x) _ f w (0) œ 3, f ww (0) œ 6, f www (0) œ 18, á ß f ÐkÑ (0) œ 3ak!b Ê 2 3x 3x# 3x$ á œ 2 ! 3xn n œ1 _ n 2nb1 ) x 15. sin x œ ! (" (#n1)! n œ0 _ n 2nb1 ) x 16. sin x œ ! (" (#n1)! nœ0 _ 2nb1 Ê sin 3x œ ! ("(#) n(3x) 1)! n n œ0 _ Ê sin x# œ ! nœ0 _ 2n 1 (")n ˆ x# ‰ (#n1)! _ n 2nb1 2nb1 œ ! (")(#3n1)!x n œ0 _ n 2nb1 $ $ & & œ 3x 33!x 35!x á $ & œ ! #(2n"1)(2nx 1)! œ #x 2$x†3! 2&x†5! á nœ0 )x 7x 7x 7x 17. 7 cos (x) œ 7 cos x œ 7 ! (" (2n)! œ 7 #! 4! 6! á , since the cosine is an even function n 2n # % ' n œ0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.8 Taylor and Maclaurin Series _ 1) x 18. cos x œ ! ((2n)! n 2n n œ0 _ (1x) Ê 5 cos 1x œ 5 ! (1)(#n)! œ 5 512!x 514!x 516!x á n # # 2n % % ' ' nœ0 cx œ "# ’Š1 x# x#! x3! x4! á ‹ Š1 x x#! x3! x4! á ‹“ œ 1 x#! x4! x6! á cx œ "# ’Š1 x x#! x3! x4! á ‹ Š1 x x#! x3! x4! á ‹“ œ x x3! x5! x6! á 19. cosh x œ e #e x _ 619 # $ % # $ % # % ' 2n x œ ! (2n)! n œ0 20. sinh x œ e #e x _ # $ % # $ % $ & ' 2n 1 œ ! (2nx 1)! n œ0 21. f(x) œ x% 2x$ 5x 4 Ê f w (x) œ 4x$ 6x# 5, f ww (x) œ 12x# 12x, f www (x) œ 24x 12, f Ð4Ñ (x) œ 24 Ê f ÐnÑ (x) œ 0 if n 5; f(0) œ 4, f w (0) œ 5, f ww (0) œ 0, f www (0) œ 12, f Ð4Ñ (0) œ 24, f ÐnÑ (0) œ 0 if n 5 24 % $ % $ Ê x% 2x$ 5x 4 œ 4 5x 12 3! x 4! x œ x 2x 5x 4 n x 6 22. f(x) œ x x 1 Ê f w (x) œ a2x ; f ww (x) œ ax 2 1b3 ; f www (x) œ ax Ê f ÐnÑ (x) œ axa11bbnnbx 1 ; f(0) œ 0, f w (0) œ 0, f ww (0) œ 2, 1 b4 x 1 b2 # # f www (0) œ 6, f ÐnÑ (0) œ a1bn nx if n _ 2 Ê x# x3 x4 x5 Þ Þ Þ œ ! a1bn xn n œ2 23. f(x) œ x$ 2x 4 Ê f w (x) œ 3x# 2, f ww (x) œ 6x, f www (x) œ 6 Ê f ÐnÑ (x) œ 0 if n 4; f(2) œ 8, f w (2) œ 10, 6 # $ f ww (2) œ 12, f www (2) œ 6, f ÐnÑ (2) œ 0 if n 4 Ê x$ 2x 4 œ 8 10(x 2) 12 2! (x 2) 3! (x 2) œ 8 10(x 2) 6(x 2)# (x 2)$ 24. f(x) œ 2x$ x# 3x 8 Ê f w (x) œ 6x# 2x 3, f ww (x) œ 12x 2, f www (x) œ 12 Ê f ÐnÑ (x) œ 0 if n f w (1) œ 11, f ww (1) œ 14, f www (1) œ 12, f ÐnÑ (1) œ 0 if n 4 Ê 2x$ x# 3x 8 12 # $ # $ œ 2 11(x 1) 14 2! (x 1) 3! (x 1) œ 2 11(x 1) 7(x 1) 2(x 1) 4; f(1) œ 2, 25. f(x) œ x% x# 1 Ê f w (x) œ 4x$ 2x, f ww (x) œ 12x# 2, f www (x) œ 24x, f Ð4Ñ (x) œ 24, f ÐnÑ (x) œ 0 if n 5; f(2) œ 21, f w (2) œ 36, f ww (2) œ 50, f www (2) œ 48, f Ð4Ñ (2) œ 24, f ÐnÑ (2) œ 0 if n 5 Ê x% x# 1 48 24 # $ % # $ % œ 21 36(x 2) 50 2! (x 2) 3! (x 2) 4! (x 2) œ 21 36(x 2) 25(x 2) 8(x 2) (x 2) 26. f(x) œ 3x& x% 2x$ x# 2 Ê f w (x) œ 15x% 4x$ 6x# 2x, f ww (x) œ 60x$ 12x# 12x 2, f www (x) œ 180x# 24x 12, f Ð4Ñ (x) œ 360x 24, f Ð5Ñ (x) œ 360, f ÐnÑ (x) œ 0 if n 6; f(1) œ 7, f w (1) œ 23, f ww (1) œ 82, f www (1) œ 216, f Ð4Ñ (1) œ 384, f Ð5Ñ (1) œ 360, f ÐnÑ (1) œ 0 if n 6 216 384 360 # $ % & Ê 3x& x% 2x$ x# 2 œ 7 23(x 1) 82 2! (x 1) 3! (x 1) 4! (x 1) 5! (x 1) œ 7 23(x 1) 41(x 1)# 36(x 1)$ 16(x 1)% 3(x 1)& 27. f(x) œ x# Ê f w (x) œ 2x$ , f ww (x) œ 3! x% , f www (x) œ 4! x& Ê f ÐnÑ (x) œ (1)n (n 1)! xn2 ; f(1) œ 1, f w (1) œ 2, f ww (1) œ 3!, f www (1) œ 4!, f ÐnÑ (1) œ (1)n (n 1)! Ê x"# _ œ 1 2(x 1) 3(x 1)# 4(x 1)$ á œ ! (1)n (n 1)(x 1)n n œ0 28. f(x) œ a1 1 xb3 Ê f w (x) œ 3(1 x)4 , f ww (x) œ 12(1 x)5 , f www (x) œ 60 (1 x)6 Ê f ÐnÑ (x) œ an 2 2b! (1 x)n3 ; fa0b œ 1, f w a0b œ 3, f ww a0b œ 12, f www a0b œ 60, á , f ÐnÑ a0b œ an 2 2b! Ê a1 1 xb3 œ 1 3x 6x# 10x3 á _ œ ! an 2ba2 n 1b xn n œ0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 620 Chapter 10 Infinite Sequences and Series 29. f(x) œ ex Ê f w (x) œ ex , f ww (x) œ ex Ê f ÐnÑ (x) œ ex ; f(2) œ e# , f w (2) œ e# , á f ÐnÑ (2) œ e# # _ $ # Ê ex œ e# e# (x 2) e# (x 2)# e3! (x 2)$ á œ ! en! (x 2)n n œ0 30. f(x) œ 2x Ê f w (x) œ 2x ln 2, f ww (x) œ 2x (ln 2)# , f www (x) œ 2x (ln 2)3 Ê f ÐnÑ (x) œ 2x (ln 2)n ; f(1) œ 2, f w (1) œ 2 ln 2, f ww (1) œ 2(ln 2)# , f www (1) œ 2(ln 2)$ , á , f ÐnÑ (1) œ 2(ln 2)n # _ Ê 2x œ 2 (2 ln 2)(x 1) 2(ln# 2) (x 1)# 2(ln3!2) (x 1)3 á œ ! 2(ln 2)n!(x1) 3 n n n œ0 31. f(x) œ cosˆ2x 12 ‰, f w (x) œ 2 sinˆ2x 12 ‰, f ww (x) œ 4 cosˆ2x 12 ‰, f www (x) œ 8 sinˆ2x 12 ‰, f a4b axb œ 24 cosˆ2x 12 ‰ß f a5b axb œ 25 sinˆ2x 12 ‰ß . . ; fˆ 14 ‰ œ 1, f w ˆ 14 ‰ œ 0, f ww ˆ 14 ‰ œ 4, f www ˆ 14 ‰ œ 0, f a4b ˆ 14 ‰ œ 24 , 2 4 f a5b ˆ 14 ‰ œ 0, . . ., f Ð2nÑ ˆ 14 ‰ œ a1bn 22n Ê cosˆ2x 12 ‰ œ 1 2ˆx 14 ‰ 32 ˆx 14 ‰ . . . _ n 2n œ ! aa12nb b2x ˆx 14 ‰ 2n n œ0 7 Î2 32. f(x) œ Èx 1, f w (x) œ 12 ax 1b1Î2 , f ww (x) œ 14 ax 1b3Î2 , f www (x) œ 38 ax 1b5Î2 , f a4b (x) œ 15 , . . .; 16 ax 1b 1 1 3 15 1 1 1 5 f(0) œ 1, f w (0) œ , f ww (0) œ , f www (0) œ , f a4b (0) œ , . . . Ê Èx 1 œ 1 x x2 x3 x4 Þ Þ Þ 2 4 8 16 _ 2 8 16 128 n 33. The Maclaurin series generated by cos x is ! aa2n1bbx x2n which converges on a_, _b and the Maclaurin series generated n œ0 _ by 1 2 x is 2 ! xn which converges on a1, 1b. Thus the Maclaurin series generated by faxb œ cos x 1 2 x is given by n œ0 _ _ n œ0 nœ0 n ! a1b x2n 2 ! xn œ 1 2x 5 x2 Þ Þ Þ Þ which converges on the intersection of a_, _b and a1, 1b, so the a2nbx 2 interval of convergence is a1, 1b. _ n 34. The Maclaurin series generated by ex is ! xnx which converges on a_, _b. The Maclaurin series generated by n œ0 _ n faxb œ a1 x x be is given by a1 x x2 b ! xnx œ 1 12 x2 23 x3 Þ Þ Þ Þ which converges on a_, _bÞ 2 x n œ0 _ n a1b 2n1 35. The Maclaurin series generated by sin x is ! a2n which converges on a_, _b and the Maclaurin series 1bx x n œ0 _ generated by lna1 xb is ! a1nb n œ1 nc1 xn which converges on a1, 1b. Thus the Maclaurin series genereated by _ _ n a1b a 1 b 2n1 faxb œ sin x † lna1 xb is given by Œ ! a2n Œ ! n 1bx x n œ0 n œ1 nc1 xn œ x2 12 x3 61 x4 Þ Þ Þ Þ which converges on the intersection of a_, _b and a1, 1b, so the interval of convergence is a1, 1b. _ n a1b 2n1 36. The Maclaurin series generated by sin x is ! a2n which converges on a_, _b. The Maclaurin series 1bx x n œ0 _ n 2 _ n _ n a 1 b a 1 b a 1 b 2n1 2n1 2n1 genereated by faxb œ x sin2 x is given by xŒ ! a2n œ xŒ ! a2n 1bx x Œ ! a2n 1bx x 1bx x n œ0 nœ0 n œ0 2 7 œ x3 13 x5 45 x . . . which converges on a_, _bÞ _ ÐnÑ 37. If ex œ ! f n!(a) (x a)n and f(x) œ ex , we have f ÐnÑ (a) œ ea f or all n œ 0, 1, 2, 3, á n œ0 ! " # # Ê e œ ea ’ (x 0!a) (x 1!a) (x 2!a) á “ œ ea ’1 (x a) (x 2!a) á “ at x œ a x Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.9 Convergence of Taylor Series 621 38. f(x) œ ex Ê f ÐnÑ (x) œ ex for all n Ê f ÐnÑ (1) œ e for all n œ 0, 1, 2, á # $ Ê ex œ e e(x 1) #e! (x 1)# 3!e (x 1)$ á œ e ’1 (x 1) (x 2!1) (x 3!1) á “ ww www 39. f(x) œ f(a) f w (a)(x a) f #(a) (x a)# f 3!(a) (x a)$ á Ê f w (x) www Ð4Ñ œ f w (a) f ww (a)(x a) f 3!(a) 3(x a)# á Ê f ww (x) œ f ww (a) f www (a)(x a) f 4!(a) 4 † 3(x a)# á Ê f ÐnÑ (x) œ f ÐnÑ (a) f Ðn1Ñ (a)(x a) f Ðn 2Ñ (a) Ê f(a) œ f(a) 0, f w (a) œ f w (a) 0, á , f # Ðn Ñ (x a)# á (a) œ f ÐnÑ (a) 0 40. E(x) œ f(x) b! b" (x a) b# (x a)# b$ (x a)$ á bn (x a)n Ê 0 œ E(a) œ f(a) b! Ê b! œ f(a); from condition (b), lim xÄa f(x) f(a) b" (x a) b# (x a)# b$ (x a)$ á bn (x a)n œ0 (x a)n w # f (x) b" 2b# (x a) 3b$ (x a) á nbn (x a) Ê xlim n(x a)n 1 Äa n 1 œ0 f ww (x) 2b# 3! b$ (x a) á n(n ")bn (x a)n 2 Ê b" œ f (a) Ê xlim œ0 n(n 1)(x a)n 2 Äa n 3 www f (x) 3! b$ á n(n 1)(n 2)bn (x a) Ê b# œ "# f ww (a) Ê xlim œ0 n(n 1)(n #)(x a)n 3 Äa w f œ b$ œ 3!" f www (a) Ê xlim Äa ÐnÑ (x) n! bn œ0 n! Ê bn œ n!" f ÐnÑ (a); therefore, ww ÐnÑ g(x) œ f(a) f w (a)(x a) f 2!(a) (x a)# á f n!(a) (x a)n œ Pn (x) # 41. f(x) œ ln (cos x) Ê f w (x) œ tan x and f ww (x) œ sec# x; f(0) œ 0, f w (0) œ 0, f ww (0) œ 1 Ê L(x) œ 0 and Q(x) œ x2 42. f(x) œ esin x Ê f w (x) œ (cos x)esin x and f ww (x) œ ( sin x)esin x (cos x)# esin x ; f(0) œ 1, f w (0) œ 1, f ww (0) œ 1 # Ê L(x) œ 1 x and Q(x) œ 1 x x# "Î# 43. f(x) œ a1 x# b Ê f w (x) œ x a1 x# b $Î# and f ww (x) œ a1 x# b $Î# 3x# a1 x# b &Î# ; f(0) œ 1, f w (0) œ 0, # f ww (0) œ 1 Ê L(x) œ 1 and Q(x) œ 1 x# # 44. f(x) œ cosh x Ê f w (x) œ sinh x and f ww (x) œ cosh x; f(0) œ 1, f w (0) œ 0, f ww (0) œ 1 Ê L(x) œ 1 and Q(x) œ 1 x# 45. f(x) œ sin x Ê f w (x) œ cos x and f ww (x) œ sin x; f(0) œ 0, f w (0) œ 1, f ww (0) œ 0 Ê L(x) œ x and Q(x) œ x 46. f(x) œ tan x Ê f w (x) œ sec# x and f ww (x) œ 2 sec# x tan x; f(0) œ 0, f w (0) œ 1, f ww œ 0 Ê L(x) œ x and Q(x) œ x 10.9 CONVERGENCE OF TAYLOR SERIES _ # _ 1. ex œ 1 x x#! á œ ! xn! Ê e5x œ 1 (5x) (#5x) á œ 1 5x 5#x! 53!x á œ ! (1)n!5 x ! # n # # $ $ nœ0 _ # nœ0 2. ex œ 1 x x#! á œ ! xn! Ê exÎ2 œ 1 ˆ #x ‰ n nœ0 & _ 3. sin x œ x x3! x5! á œ ! ((#1)nx1)! $ n 2n 1 n œ0 & _ 4. sin x œ x x3! x5! á œ ! ((#1)nx1)! $ n n n nœ0 n 2n 1 _ n n ˆ x# ‰# # $ á œ 1 x# 2x# #! 2x$ 3! á œ ! (21)n n!x #! nœ0 $ _ & x Ê 5 sin (x) œ 5 ’(x) (3!x) (5!x) á “ œ ! 5((1) #n1)! n 1 2n 1 n œ0 Ê sin 1#x œ 1#x _ n 2n 1 2n 1 ˆ 1#x ‰$ ˆ 1#x ‰& ˆ 1#x ‰( 1 x ! (1) 3! 5! 7! á œ 22n 1 (#n1)! nœ0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 622 Chapter 10 Infinite Sequences and Series _ 1) x 5. cos x œ ! ((2n)! n 2n n œ0 _ Ê cos 5x2 œ ! n œ0 5 x 625x 15625x œ ! (1)(2n)! œ 1 25x á 6! #! 4! n 2n 4n 4 $ 1bn x2n cos x œ ! a(2n)! Ê n œ0 8 12 n œ0 2n "Î# n _ a1b ŒŠ x# ‹ "Î# $Î# $ cos Š xÈ ‹ œ cos ŒŠ x# ‹ œ ! (#n)! 2 _ 6. _ 2n (1)n 5x2 ‘ (2n)! _ (1)n x3n 2n (2n)! œ! nœ0 nœ0 $ ' * œ 1 2x†2! 2#x†4! 2$x†6! á _ a1bnc1 ˆx2 ‰ n _ nc1 n 7. lna1 xb œ ! a1bn x Ê lna1 x2 b œ ! n œ1 nœ1 _ n 2nb1 1b x 8. tan1 x œ ! a2n 1 nœ0 9. _ a1bn ˆ3x4 ‰ 2n 1 Ê tan1 a3x4 b œ ! nœ0 _ n nc1 œ ! a1b n x œ x2 x2 x3 x4 . . . 2n 4 6 8 nœ1 2nb1 _ 2nb1 8nb4 n œ ! a1b 3 n x nœ0 _ _ _ n œ0 nœ0 nœ0 2187 28 20 œ 3x4 9x12 243 5 x 7 x ... 1 ! a1bn xn Ê 13 3 œ ! a1bn ˆ 3 x3 ‰n œ ! a1bn ˆ 3 ‰n x3n œ 1 3 x3 9 x6 27 x9 . . . 1x œ 4 4 4 16 64 1 4x _ _ _ n 10. 1 1 x œ ! xn Ê 2 1 x œ #" 1 1 " x œ #" ! ˆ #" x‰ œ ! ˆ #" ‰ # n œ0 _ nœ0 _ n _ n œ0 _ n 2nb1 1) x 12. sin x œ ! ((2n 1)! nœ0 _ 1) x 13. cos x œ ! ((2n)! n 2n n œ0 1 3 x œ #" 14 x 18 x2 16 x ... nœ0 xn ! xnb1 œ x x# x$ x% x& á #! n! œ n! 3! 4! 11. ex œ ! xn! Ê xex œ x Œ ! n œ0 n 1 n n œ0 _ _ n 2nb1 n 2nb3 1) x x x x $ Ê x# sin x œ x# Œ ! ((#1)nx1)! œ ! ((2n 1)! œ x 3! 5! 7! á nœ0 & ( * nœ0 _ x x x x x x x Ê x# 1 cos x œ x# 1 ! ((1) #n)! œ # 1 1 2 4! 6! 8! 10! á # # # n 2n # % ' ) "! n œ0 _ x x œ x4! x6! x8! 10! á œ ! ((1) #n)! % ' ) "! n 2n n œ2 _ n 2nb1 1) x 14. sin x œ ! ((2n 1)! n œ0 $ & _ $ n 2nb1 $ Ê sin x x x3! œ Œ ! ((#1)nx1)! x x3! n œ0 ( * "" $ & ( * _ "" 1) x x x œ Šx x3! x5! x7! x9! 11! á ‹ x x3! œ x5! x7! x9! 11! á œ ! ((2n 1)! n 2n 1 n œ2 _ 1) x 15. cos x œ ! ((2n)! n 2n n œ0 _ 1) x 16. cos x œ ! ((2n)! n 2n n œ0 _ _ n 2n 2nb1 (1x) Ê x cos 1x œ x ! (1)(#n)! œ ! (")(#1n)!x n 2n nœ0 nœ0 _ _ # 2n ax b x Ê x# cos ax# b œ x# ! (1)(#n)! œ ! ("(#) n)! n n œ0 _ n 4n 2 n œ0 % & ' ( ' "! "% œ x# x2! x4! x6! á % # # $ œ x 12!x 14!x 16!x á ' ) (2x) (2x) (2x) (2x) 17. cos# x œ "# cos#2x œ "# "# ! (1)(2n)! œ #" "# ’1 (2x) 2! 4! 6! 8! á “ n 2n n œ0 _ _ n œ1 n œ1 # n 2n n 2n 1 2n (2x)% (2x)' (2x)) ! (1) (2x) œ 1 ! (1) 2 x œ 1 (2x) 2†2! 2†4! 2†6! 2†8! á œ 1 2†(2n)! (2n)! # % ' # % ' (2x) (2x) (2x) (2x) (2x) 2x ‰ 18. sin# x œ ˆ 1cos œ "# "# cos 2x œ "# "# Š1 (2x) # #! 4! 6! á ‹ œ 2†2! 2†4! 2†6! á _ nb1 _ (2x) 2 x œ ! (1)#†(2n)! œ ! (1) (2n)! n œ1 2n n 2n 1 2n nœ1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.9 Convergence of Taylor Series _ _ n œ0 n œ0 # 19. 1x 2x œ x# ˆ 1"2x ‰ œ x# ! (2x)n œ ! 2n xn2 œ x# 2x$ 2# x% 2$ x& á _ nc1 _ nc1 n n 1 20. x ln (1 2x) œ x ! (1) n (2x) œ ! (1) n2 x n œ1 n n œ1 # $ $ % % & œ 2x# 2 #x 2 4x 2 5x á _ _ _ nœ0 nœ1 nœ0 " d ˆ " ‰ # ! nxn1 œ ! (n 1)xn 21. 1 " x œ ! xn œ 1 x x# x$ á Ê dx 1x œ (1x)# œ 1 2x 3x á œ _ # d ˆ " ‰ d " d # # ! n(n 1)xn2 22. a12xb$ œ dx # 1x œ dx Š (1x)# ‹ œ dx a1 2x 3x á b œ 2 6x 12x á œ n œ2 _ œ ! (n 2)(n 1)xn n œ0 3 5 7 23. tan1 x œ x 13 x3 15 x5 17 x7 Þ Þ Þ Ê x tan1 x2 œ xŠx2 13 ax2 b 15 ax2 b 17 ax2 b Þ Þ Þ ‹ _ n 4nc1 1b x œ x3 13 x7 15 x11 17 x15 Þ Þ Þ œ ! a2n 1 n œ1 3 5 7 b a2xb a2xb 24. sin x œ x x3! x5! x7! á Ê sin x † cos x œ "# sin 2x œ "# Š2x a2x 3! 5! 7! á ‹ 3 5 7 _ 2nb1 4x ! (1) 2 x œ x 43!x 165!x 647!x á œ x 23x 2x 15 315 á œ (#n1)! 3 5 7 3 5 7 n 2n n œ0 2 3 2 3 25. ex œ 1 x x2! x3! á and 1 1 x œ 1 x x2 x3 á Ê ex 1 1 x _ 4 ! ˆ 1 a1bn ‰xn œ Š1 x x2! x3! á ‹ a1 x x2 x3 á b œ 2 32 x2 56 x3 25 24 x á œ n! n œ0 3 5 7 2 4 6 26. sin x œ x x3! x5! x7! á and cos x œ 1 x2! x4! x6! á Ê cos x sin x 2 4 6 3 5 7 2 3 4 5 6 7 œ Š1 x2! x4! x6! á ‹ Šx x3! x5! x7! á ‹ œ 1 x x2! x3! x4! x5! x6! x7! á _ n 2nb1 1) x œ ! Š ((2n)! ((#1)nx1)! ‹ n 2n n œ0 2 3 4 27. lna1 xb œ x 12 x2 13 x3 14 x4 á Ê x3 lna1 x2 b œ x3 Šx2 12 ax2 b 13 ax2 b 14 ax2 b á ‹ _ nc1 1 9 œ 13 x3 16 x5 19 x7 12 x á œ ! a13nb x2n1 n œ1 28. lna1 xb œ x 12 x2 13 x3 14 x4 á and lna1 xb œ x 12 x2 13 x3 14 x4 á Ê lna1 xb lna1 xb _ œ ˆx 12 x2 13 x3 14 x4 á ‰ ˆx 12 x2 13 x3 14 x4 á ‰ œ 2x 23 x3 25 x5 á œ ! 2n 2 1 x2n1 n œ0 2 3 2 3 3 5 7 29. ex œ 1 x x2! x3! á and sin x œ x x3! x5! x7! á Ê ex † sin x 3 5 7 1 5 œ Š1 x x2! x3! á ‹Šx x3! x5! x7! á ‹ œ x x2 13 x3 30 x ÞÞÞÞ 30. lna1 xb œ x 12 x2 31 x3 41 x4 á and 1 " x œ 1 x x# x$ á Ê ln1a1xxb œ lna1 xb † 1 " x 7 4 œ ˆx 12 x2 13 x3 14 x4 á ‰a1 x x# x$ á b œ x 12 x2 56 x3 12 x ÞÞÞÞ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 623 624 Chapter 10 Infinite Sequences and Series 2 31. tan1 x œ x 13 x3 15 x5 17 x7 Þ Þ Þ Ê atan1 xb œ atan1 xbatan1 xb 44 8 6 œ ˆx 13 x3 15 x5 17 x7 Þ Þ Þ‰ˆx 13 x3 15 x5 17 x7 Þ Þ Þ‰ œ x2 23 x4 23 45 x 105 x Þ Þ Þ Þ 3 5 7 2 4 6 32. sin x œ x x3! x5! x7! á and cos x œ 1 x2! x4! x6! á Ê cos2 x † sin x œ cos x † cos x † sin x 3 5 7 b a2xb a2xb 7 3 61 5 1247 7 œ cos x † "# sin 2x œ "# Š1 x2! x4! x6! á ‹Š2x a2x 3! 5! 7! á ‹ œ x 6 x 120 x 5040 x Þ Þ Þ 2 3 5 4 6 7 2 3 33. sin x œ x x3! x5! x7! á and ex œ 1 x x2! x3! á 3 5 7 3 5 2 7 3 5 3 7 Ê esin x œ 1 Šx x3! x5! x7! á ‹ 12 Šx x3! x5! x7! á ‹ 16 Šx x3! x5! x7! á ‹ á œ 1 x 12 x2 18 x4 Þ Þ Þ Þ 34. sin x œ x x3! x5! x7! á and tan1 x œ x 13 x3 51 x5 71 x7 Þ Þ Þ Ê sinatan1 xb œ ˆx 13 x3 51 x5 71 x7 Þ Þ Þ‰ 3 5 7 3 5 7 1 ˆ 1 ˆ 16 ˆx 31 x3 51 x5 71 x7 Þ Þ Þ‰ 120 x 13 x3 15 x5 17 x7 Þ Þ Þ‰ 5040 x 13 x3 15 x5 17 x7 Þ Þ Þ‰ á 5 7 œ x 12 x3 38 x5 16 x ÞÞÞ 35. Since n œ 3, then f a4b axb œ sin x, lf a4b axbl Ÿ M on Ò0, 0.1Ó Ê lsin xl Ÿ 1 on Ò0, 0.1Ó Ê M œ 1. Then lR3 a0.1bl Ÿ 1 l0.14x 0l 4 œ 4.2 ‚ 106 Ê error Ÿ 4.2 ‚ 106 36. Since n œ 4, then f a5b axb œ ex , lf a5b axbl Ÿ M on Ò0, 0.5Ó Ê lex l Ÿ Èe on Ò0, 0.5Ó Ê M œ 2.7. Then lR4 a0.5bl Ÿ 2.7 l0.55x 0l œ 7.03 ‚ 104 Ê error Ÿ 7.03 ‚ 104 5 & 37. By the Alternating Series Estimation Theorem, the error is less than kx5!k Ê kxk& a5!b a5 ‚ 10% b Ê kxk& 600 ‚ 10% 5 Ê kxk È 6 ‚ 10# ¸ 0.56968 % # 38. If cos x œ 1 x# and kxk 0.5, then the error is less than ¹ (.5) 24 ¹ œ 0.0026, by Alternating Series Estimation Theorem; # since the next term in the series is positive, the approximation 1 x# is too small, by the Alternating Series Estimation Theorem c$ $ 39. If sin x œ x and kxk 10$ , then the error is less than a103! b ¸ 1.67 ‚ 1010 , by Alternating Series Estimation Theorem; $ The Alternating Series Estimation Theorem says R# (x) has the same sign as x3! . Moreover, x sin x Ê 0 sin x x œ R# (x) Ê x 0 Ê 10$ x 0. # $ # # x 40. È1 x œ 1 x# x8 16 á . By the Alternating Series Estimation Theorem the kerrork ¹ 8x ¹ (0.01) 8 œ 1.25 ‚ 10& c $ Ð0Þ1Ñ c $ $ 41. kR# (x)k œ ¹ e3!x ¹ 3 (0.1)$ 1.87 ‚ 104 , where c is between 0 and x 3! % 42. kR# (x)k œ ¹ e3!x ¹ (0.1) 3! œ 1.67 ‚ 10 , where c is between 0 and x # % ' # $ % & ' (2x) (2x) 2x ‰ 2x 2 x 2 x 43. sin# x œ ˆ 1 cos œ "# "# cos 2x œ "# "# Š1 (2x) # 2! 4! 6! á ‹ œ #! 4! 6! á # $ % & ' $ & ( (2x) (2x) (2x) d d 2 x 2 x Ê dx asin# xb œ dx Š 2x 2! 4! 6! á ‹ œ 2x 3! 5! 7! á Ê 2 sin x cos x $ & ( (2x) (2x) œ 2x (2x) 3! 5! 7! á œ sin 2x, which checks Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.9 Convergence of Taylor Series # % ' ) # $ % & ' 625 ( ) (2x) (2x) (2x) 2x 2 x 2 x 2 x 44. cos# x œ cos 2x sin# x œ Š1 (2x) #! 4! 6! 8! á ‹ Š #! 4! 6! 8! á ‹ # $ % & ' " % " 2 x 2 x 2 ' # ) œ 1 2x #! 4! 6! á œ 1 x 3 x 45 x 315 x á 45. A special case of Taylor's Theorem is f(b) œ f(a) f w (c)(b a), where c is between a and b Ê f(b) f(a) œ f w (c)(b a), the Mean Value Theorem. 46. If f(x) is twice differentiable and at x œ a there is a point of inflection, then f ww (a) œ 0. Therefore, L(x) œ Q(x) œ f(a) f w (a)(x a). ww 47. (a) f ww Ÿ 0, f w (a) œ 0 and x œ a interior to the interval I Ê f(x) f(a) œ f (c# # ) (x a)# Ÿ 0 throughout I Ê f(x) Ÿ f(a) throughout I Ê f has a local maximum at x œ a ww (b) similar reasoning gives f(x) f(a) œ f (c# # ) (x a)# 0 throughout I Ê f(x) f(a) throughout I Ê f has a local minimum at x œ a 48. f(x) œ (1 x)" Ê f w (x) œ (1 x)# Ê f ww (x) œ 2(1 x)$ Ê f Ð3Ñ (x) œ 6(1 x)% 10 " " ˆ 10 ‰ Ê f Ð4Ñ (x) œ 24(1 x)& ; therefore 1" x ¸ 1 x x# x$ . kxk 0.1 Ê 10 11 1x 9 Ê ¹ (1x)& ¹ 9 & % Ð4Ñ & % & Ð4Ñ ‰ Ê the error e$ Ÿ ¹ max f 4! (x) x ¹ (0.1)% ˆ 10 ‰ œ 0.00016935 0.00017, since ¹ f 4!(x) ¹ œ ¹ (1"x)& ¹ . Ê ¹ (1x x)& ¹ x% ˆ 10 9 9 49. (a) f(x) œ (1 x)k Ê f w (x) œ k(1 x)k1 Ê f ww (x) œ k(k 1)(1 x)k2 ; f(0) œ 1, f w (0) œ k, and f ww (0) œ k(k 1) Ê Q(x) œ 1 kx k(k # ") x# " " (b) kR# (x)k œ ¸ 3†3!2†" x$ ¸ 100 Ê kx$ k 100 Ê 0x " or 0 x .21544 100"Î$ 50. (a) Let P œ x 1 Ê kxk œ kP 1k .5 ‚ 10n since P approximates 1 accurate to n decimals. Then, P sin P œ (1 x) sin (1 x) œ (1 x) sin x œ 1 (x sin x) Ê k(P sin P) 1k $ 3n œ ksin x xk Ÿ kx3!k 0.125 .5 ‚ 103n Ê P sin P gives an approximation to 1 correct to 3n decimals. 3! ‚ 10 _ _ n œ0 n œk 51. If f(x) œ ! an xn , then f ÐkÑ (x) œ ! n(n 1)(n 2)â(n k 1)an xnk and f ÐkÑ (0) œ k! ak Ê ÐkÑ ak œ f k!(0) for k a nonnegative integer. Therefore, the coefficients of f(x) are identical with the corresponding coefficients in the Maclaurin series of f(x) and the statement follows. 52. Note: f even Ê f(x) œ f(x) Ê f w (x) œ f w (x) Ê f w (x) œ f w (x) Ê f w odd; f odd Ê f(x) œ f(x) Ê f w (x) œ f w (x) Ê f w (x) œ f w (x) Ê f w even; also, f odd Ê f(0) œ f(0) Ê 2f(0) œ 0 Ê f(0) œ 0 (a) If f(x) is even, then any odd-order derivative is odd and equal to 0 at x œ 0. Therefore, a" œ a$ œ a& œ á œ 0; that is, the Maclaurin series for f contains only even powers. (b) If f(x) is odd, then any even-order derivative is odd and equal to 0 at x œ 0. Therefore, a! œ a# œ a% œ á œ 0; that is, the Maclaurin series for f contains only odd powers. 53-58. Example CAS commands: Maple: f := x -> 1/sqrt(1+x); x0 := -3/4; x1 := 3/4; # Step 1: plot( f(x), x=x0..x1, title="Step 1: #53 (Section 10.9)" ); Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 626 Chapter 10 Infinite Sequences and Series # Step 2: P1 := unapply( TaylorApproximation(f(x), x = 0, order=1), x ); P2 := unapply( TaylorApproximation(f(x), x = 0, order=2), x ); P3 := unapply( TaylorApproximation(f(x), x = 0, order=3), x ); # Step 3: D2f := D(D(f)); D3f := D(D(D(f))); D4f := D(D(D(D(f)))); plot( [D2f(x),D3f(x),D4f(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 3: #57 (Section 9.9)" ); c1 := x0; M1 := abs( D2f(c1) ); c2 := x0; M2 := abs( D3f(c2) ); c3 := x0; M3 := abs( D4f(c3) ); # Step 4: R1 := unapply( abs(M1/2!*(x-0)^2), x ); R2 := unapply( abs(M2/3!*(x-0)^3), x ); R3 := unapply( abs(M3/4!*(x-0)^4), x ); plot( [R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 4: #53 (Section 10.9)" ); # Step 5: E1 := unapply( abs(f(x)-P1(x)), x ); E2 := unapply( abs(f(x)-P2(x)), x ); E3 := unapply( abs(f(x)-P3(x)), x ); plot( [E1(x),E2(x),E3(x),R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], linestyle=[1,1,1,3,3,3], title="Step 5: #53 (Section 10.9)" ); # Step 6: TaylorApproximation( f(x), view=[x0..x1,DEFAULT], x=0, output=animation, order=1..3 ); L1 := fsolve( abs(f(x)-P1(x))=0.01, x=x0/2 ); # (a) R1 := fsolve( abs(f(x)-P1(x))=0.01, x=x1/2 ); L2 := fsolve( abs(f(x)-P2(x))=0.01, x=x0/2 ); R2 := fsolve( abs(f(x)-P2(x))=0.01, x=x1/2 ); L3 := fsolve( abs(f(x)-P3(x))=0.01, x=x0/2 ); R3 := fsolve( abs(f(x)-P3(x))=0.01, x=x1/2 ); plot( [E1(x),E2(x),E3(x),0.01], x=min(L1,L2,L3)..max(R1,R2,R3), thickness=[0,2,4,0], linestyle=[0,0,0,2], color=[red,blue,green,black], view=[DEFAULT,0..0.01], title="#53(a) (Section 10.9)" ); abs(`f(x)`-`P`[1](x) ) <= evalf( E1(x0) ); # (b) abs(`f(x)`-`P`[2](x) ) <= evalf( E2(x0) ); abs(`f(x)`-`P`[3](x) ) <= evalf( E3(x0) ); Mathematica: (assigned function and values for a, b, c, and n may vary) Clear[x, f, c] f[x_]= (1 x)3/2 {a, b}= {1/2, 2}; pf=Plot[ f[x], {x, a, b}]; poly1[x_]=Series[f[x], {x,0,1}]//Normal poly2[x_]=Series[f[x], {x,0,2}]//Normal poly3[x_]=Series[f[x], {x,0,3}]//Normal Plot[{f[x], poly1[x], poly2[x], poly3[x]}, {x, a, b}, PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,1,0], RGBColor[0,0,1], RGBColor[0,.5,.5]}]; Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.10 The Binomial Series 627 The above defines the approximations. The following analyzes the derivatives to determine their maximum values. f''[c] Plot[f''[x], {x, a, b}]; f'''[c] Plot[f'''[x], {x, a, b}]; f''''[c] Plot[f''''[x], {x, a, b}]; Noting the upper bound for each of the above derivatives occurs at x = a, the upper bounds m1, m2, and m3 can be defined and bounds for remainders viewed as functions of x. m1=f''[a] m2=-f'''[a] m3=f''''[a] r1[x_]=m1 x2 /2! Plot[r1[x], {x, a, b}]; r2[x_]=m2 x3 /3! Plot[r2[x], {x, a, b}]; r3[x_]=m3 x4 /4! Plot[r3[x], {x, a, b}]; A three dimensional look at the error functions, allowing both c and x to vary can also be viewed. Recall that c must be a value between 0 and x, so some points on the surfaces where c is not in that interval are meaningless. Plot3D[f''[c] x2 /2!, {x, a, b}, {c, a, b}, PlotRange Ä All] Plot3D[f'''[c] x3 /3!, {x, a, b}, {c, a, b}, PlotRange Ä All] Plot3D[f''''[c] x4 /4!, {x, a, b}, {c, a, b}, PlotRange Ä All] 10.10 THE BINOMIAL SERIES 1. (1 x)"Î# œ 1 "# x ˆ "# ‰ ˆ "# ‰ x# 2. (1 x)"Î$ œ 1 "3 x ˆ "3 ‰ ˆ 23 ‰ x# #! #! 3. (1 x)"Î# œ 1 "# (x) ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ x$ 3! ˆ 3" ‰ ˆ 32 ‰ ˆ 53 ‰ x$ #! #! ˆ x ‰# x ‰# ˆ (4)(3) 3 4 6. ˆ1 x3 ‰ œ 1 4 ˆ x3 ‰ #! "Î# 8. a1 x# b "Î$ œ 1 "# x$ œ 1 "3 x# "Î# 9. ˆ1 1x ‰ œ 1 "# ˆ x1 ‰ ˆ "# ‰ ˆ "# ‰ (2x)# (2)(3) # # 5. ˆ1 x# ‰ œ 1 # ˆ x# ‰ #! 7. a1 x$ b 3! ˆ "# ‰ ˆ 3# ‰ (x)# 4. (1 2x)"Î# œ 1 "# (2x) ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ (x)$ 3! Š "# ‹ Š "# ‹ Š #3 ‹ (2x)$ 3! $ 3! (4)(3)(2) ˆ x3 ‰ #! ˆ "3 ‰ ˆ 43 ‰ ax# b# #! ˆ "# ‰ ˆ "# ‰ ˆ 1x ‰# 5 $ á œ 1 3" x 9" x# 81 x á (2)(3)(4) ˆ x# ‰ ˆ "# ‰ ˆ 3# ‰ ax$ b# #! " $ á œ 1 "# x "8 x# 16 x á 3! $ á œ 1 x 12 x# 12 x$ á á œ 1 x 34 x# "# x$ (4)(3)(2)(1) ˆ x3 ‰ 4! ˆ #" ‰ ˆ #3 ‰ ˆ #5 ‰ ax$ b$ 3! ˆ 3" ‰ ˆ 43 ‰ ˆ 73 ‰ ax# b$ 3! ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ ˆ 1x ‰$ 3! 5 $ á œ 1 "# x 38 x# 16 x á 4 4 3 1 4 0 á œ 1 34 x 32 x2 27 x 81 x 5 * á œ 1 "# x$ 38 x' 16 x á ' á œ 1 "3 x# 29 x% 14 81 x á " á œ 1 #"x 8x1 # 16x $ á Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 628 Chapter 10 Infinite Sequences and Series ˆ "3 ‰ ˆ 43 ‰ x# x 10. È œ xa1 xb"Î3 œ xŒ1 ˆc "3 ‰x 3 1x # $ ˆ "3 ‰ ˆ 43 ‰ ˆ 73 ‰ x$ #! 14 4 á œ x 31 x# 92 x3 81 x á 3! % 11. (1 x)% œ 1 4x (4)(3)x (4)(3)(2)x (4)(3)(2)x œ 1 4x 6x# 4x$ x% #! 3! 4! # # $ # $ ax b 12. a1 x# b œ 1 3x# (3)(2)#!ax b (3)(2)(1) œ 1 3x# 3x% x' 3! # $ 2x) 2x) 13. (1 2x)$ œ 1 3(2x) (3)(2)(# (3)(2)(1)( œ 1 6x 12x# 8x$ ! 3! % 14. ˆ1 #x ‰ œ 1 4 ˆ #x ‰ (4)(3) ˆ x# ‰ # #! (4)(3)(2) ˆ x# ‰ 3! $ (4)(3)(2)(1) ˆ x# ‰ 15. '0 sin x# dx œ '0 Šx# x3! x5! á ‹ dx œ ’ x3 7x†3! á “ 0Þ2 0 Þ2 ' "! % 4! $ !Þ# ( ! ( " % œ 1 2x 23 x# "# x$ 16 x $ ¸ ’ x3 “ !Þ# ! ¸ 0.00267 with error kEk Ÿ (.2) 7†3! ¸ 0.0000003 16. '0 0Þ2 ecx " dx œ x '00 2 x" Š1 x x#! x3! x4! á 1‹ dx œ '00 2 Š1 x# x6 24x á ‹ dx Þ # # $ x œ ’x x4 18 á“ 17. '0 0Þ1 !Þ# ! $ Þ % % '00 1 Š1 x2 3x8 á ‹ dx œ ’x 10x á “ !Þ" ¸ [x]!Þ" ! ¸ 0.1 with error Þ " È1 x% dx œ !Þ#& $È $ ¸ 0.19044 with error kEk Ÿ (0.2) 96 ¸ 0.00002 % ) & ! & kEk Ÿ (0.1) 10 œ 0.000001 18. '! # 1 x# dx œ '0 0Þ25 # % $ & x á“ Š1 x3 x9 á ‹ dx œ ’x x9 45 & !Þ#& ! $ ¸ ’x x9 “ !Þ#& ! ¸ 0.25174 with error kEk Ÿ (0.25) 45 ¸ 0.0000217 19. '0 0Þ1 sin x x dx œ '00 1 Š1 x3! x5! x7! á ‹ dx œ ’x 3x†3! 5x†5! 7x†7! á “ !Þ" ¸ ’x 3x†3! 5x†5! “ !Þ" Þ # % ' $ & ( $ & ! ! 7 12 ¸ 0.0999444611, kEk Ÿ (0.1) 7†7! ¸ 2.8 ‚ 10 x x 20. '0 exp ax# b dx œ '0 Š1 x# x2! x3! x4! á ‹ dx œ ’x x3 10 42 á“ 0Þ1 0 Þ1 % ' ) $ & ( !Þ" ! $ & 9 12 ¸ 0.0996676643, kEk Ÿ (0.1) 216 ¸ 4.6 ‚ 10 21. a1 x% b "Î# œ (1)"Î# ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ 4! Š "# ‹ 1 (1)"Î# ax% b ˆ "# ‰ ˆ "# ‰ #! % % # (1)$Î# ax% b ) % 3! (1)&Î# ax% b ) "' "# "' & !Þ" ! 9 11 ¸ 0.100001, kEk Ÿ (0.1) 72 ¸ 1.39 ‚ 10 x‰ x 22. '0 ˆ 1 xcos dx œ '0 Š "# x4! x6! x8! 10! á ‹ dx ¸ ’ x# 3x†4! 5x†6! 7x†8! 9†x10! “ # 1 1 ¸ 0.4863853764, # % ' ) $ & 1 ( kEk Ÿ 11†112! ¸ 1.9 ‚ 1010 t t 23. '0 cos t# dt œ '0 Š1 t# 4! t6! á ‹ dt œ ’t 10 9t†4! 13t †6! á “ 1 $ (1)(Î# ax% b á œ 1 x# x8 x16 5x 128 á x Ê '0 Š1 x# x8 x16 5x 128 á ‹ dx ¸ ’x 10 “ 0Þ1 "# ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ % ) "# & * "$ " ! * ( x x ¸ ’x x3 10 42 “ " ! Ê kerrork 13"†6! ¸ .00011 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. !Þ" ! Section 10.10 The Binomial Series t t t 24. '0 cos Èt dt œ '0 Š1 #t 4! 6! 8! á ‹ dt œ ’t t4 3t†4! 4t†6! 5t†8! á “ 1 1 # $ % # $ % & Ê kerrork 5†"8! ¸ 0.000004960 t 25. F(x) œ '0 Št# 3! t5! t7! á ‹ dt œ ’ t3 7t†3! 11t †5! 15t †7! á “ x ' "! "% $ ( "" "& Ê kerrork 15"†7! ¸ 0.000013 x $ " ! ( "" ¸ x3 7x†3! 11x †5! ! t t 26. F(x) œ '0 Št# t% 2! 3! t4! t5! á ‹ dt œ ’ t3 t5 7t†2! 9t†3! 11t †4! 13t †5! á “ x $ ' & ( ) * "! "# $ & ( 629 * "" "$ "" x ! ¸ x3 x5 7x†2! 9x†3! 11x †4! Ê kerrork 13"†5! ¸ 0.00064 t 27. (a) F(x) œ '0 Št t3 t5 t7 á ‹ dt œ ’ t2 1t # 30 á “ ¸ x# 1x# Ê kerrork (0.5) 30 ¸ .00052 x $ & ( # # % x ' % ' # ' % ! ) $# (b) kerrork 33"†34 ¸ .00089 when F(x) ¸ x# 3x†4 5x†6 7x†8 á (1)"& 31x†32 28. (a) F(x) œ '0 Š1 2t t3 t4 á ‹ dt œ ’t 2t†2 3t†3 4t†4 5t†5 á “ ¸ x x## 3x# 4x# 5x# x # $ # $ % x & # $ % & ! ' Ê kerrork (0.5) 6# ¸ .00043 # $ % $" x (b) kerrork 32" # ¸ .00097 when F(x) ¸ x x## x3# x4# á (1)$" 31 # # $ # 29. x"# aex (1 x)b œ x"# ŠŠ1 x x# x3! á ‹ 1 x‹ œ #" 3!x x4! á Ê lim xÄ0 ex (1 x) x# # œ lim Š "# 3!x x4! á ‹ œ "# xÄ0 $ # % # $ % $ & ( 2x 2x 30. x" aex ex b œ x" ’Š1 x x#! x3! x4! á ‹ Š1 x x#! x3! x4! á ‹“ œ x" Š2x 2x 3! 5! 7! á ‹ # % ' # ex e x 2x% 2x' œ x lim Š2 2x x 3! 5! 7! á ‹ œ 2 Ä_ 2x 2x œ 2 2x 3! 5! 7! á Ê lim xÄ0 # # # % ' # # % t t t t 31. t"% Š1 cos t t# ‹ œ t"% ’1 t# Š1 t# 4! 6! á ‹“ œ 4!" 6! 8! á Ê lim " cos t Š t# ‹ t% tÄ0 # % t t " œ lim Š 4!" 6! 8! á ‹ œ 24 tÄ0 $ $ $ & # $ % 32. )"& Š) )6 sin )‹ œ )"& Š) )6 ) )3! )5! á ‹ œ 5!" )7! )9! á Ê lim )Ä0 )% # sin ) ) Š )6 ‹ )& œ lim Š 5!" )7! 9! á ‹ œ 1"#0 )Ä0 $ & # % 33. y"$ ay tan" yb œ y"$ ’y Šy y3 y5 á ‹“ œ "3 y5 y7 á Ê lim yÄ0 # % y tan " y œ lim Š 3" y5 y7 á ‹ y$ yÄ0 œ "3 34. tanc" y sin y œ y$ cos y Ê lim yÄ0 y$ y& y$ y& Œy 3 5 á Œy 3! 5! á y$ cos y tanc" y sin y œ lim y$ cos y yÄ0 23y# " Œ 6 5! á cos y y$ œ Œ 6 23y& 5! á y$ cos y œ " Œ 6 23y# 5! á cos y œ 6" # # 35. x# Š1 e1Îx ‹ œ x# ˆ1 1 x"# #"x% 6x" ' á ‰ œ 1 #"x# 6x" % á Ê x lim x# Še1Îx 1‹ Ä_ ˆ1 #x" # 6x" % á ‰ œ 1 œ x lim Ä_ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 630 Chapter 10 Infinite Sequences and Series 36. (x 1) sin ˆ x " 1 ‰ œ (x 1) Š x " 1 3!(x " 1)$ 5!(x " 1)& á ‹ œ 1 3!(x " 1)# 5!(x " 1)% á Ê x lim (x 1) sin ˆ x " 1 ‰ œ x lim Š1 3!(x " 1)# 5!(x " 1)% á ‹ œ 1 Ä_ Ä_ % 37. ' # % # % x x x x x x # Œx # 3 á Œ1 # 3 á Œ1 # 3 á ln a1 x# b ln a1 x# b œ œ Ê lim œ lim œ 2! œ 2 # # # % 1 cos x x Ä 0 1 cos x xÄ0 1 Š1 x#! x4! á‹ Š #"! x4! á‹ Š #"! x4! á‹ (x 2)(x 2) # 38. lnx(x41) œ ’(x 2) (x c 2)# # x 2 œ lim (x c 2)$ á“ 3 # x Ä 2 ’1 x c# 2 (x 32) á“ œ x2 ’1 x # 2 Ê lim (x c 2)# á“ 3 x# 4 x Ä 2 ln (x 1) œ4 81 10 4 6 39. sin 3x2 œ 3x2 92 x6 40 x . . . and 1 cos 2x œ 2x2 23 x4 45 x . . . Ê lim sin 3x2 x Ä 0 1 cos 2x 10 8 3x2 92 x6 81 3 92 x4 81 40 x . . . 40 x . . . œ 32 2 2 4 4 2 2 x4 4 x6 . . . œ lim 2x 2 x x . . . xÄ0 xÄ0 3 45 3 45 œ lim 6 9 12 1 11 1 40. lna1 x3 b œ x3 x2 x3 x4 . . . and x sin x2 œ x3 16 x7 120 x 5040 x15 Þ Þ Þ Ê lim xÄ0 œ lim xÄ0 6 9 12 x3 x2 x3 x4 . . . 1 7 1 11 1 3 x 6 x 120 x 5040 x15 Þ Þ Þ 3 6 9 1 x2 x3 x4 . . . 1 4 1 8 1 1 6 x 120 x 5040 x12 Þ Þ Þ œ lim xÄ0 œ1 41. 1 1 21x 31x 41x Þ Þ Þ œ e1 œ e 3 4 5 3 2 1 1 1 4 1 42. ˆ 14 ‰ ˆ 14 ‰ ˆ 14 ‰ Þ Þ Þ œ ˆ 14 ‰ ”1 ˆ 14 ‰ ˆ 14 ‰ Þ Þ Þ • œ 64 1 1Î4 œ 64 3 œ 48 2 4 2 6 4 6 43. 1 432 2x 434 4x 436 6x Þ Þ Þ œ 1 21x ˆ 34 ‰ 41x ˆ 34 ‰ 61x ˆ 34 ‰ Þ Þ Þ œ cosˆ 34 ‰ 2 3 4 5 7 5 7 44. 21 2†122 3†123 4†124 Þ Þ Þ œ ˆ 21 ‰ 21 ˆ #1 ‰ 31 ˆ #1 ‰ 41 ˆ #1 ‰ Þ Þ Þ œ lnˆ1 21 ‰ œ lnˆ 23 ‰ 3 45. 13 313 3x 315 5x 317 7x Þ Þ Þ œ 13 31x ˆ 13 ‰ 51x ˆ 13 ‰ 71x ˆ 13 ‰ Þ Þ Þ œ sinˆ 13 ‰ œ 3 5 7 3 È3 2 46. 23 323 †3 325 †5 327 †7 Þ Þ Þ œ ˆ 32 ‰ 31 ˆ 32 ‰ 51 ˆ 32 ‰ 71 ˆ 32 ‰ Þ Þ Þ œ tan1 ˆ 23 ‰ 3 5 7 3 47. x3 x4 x5 x6 Þ Þ Þ œ x3 a1 x x2 x3 Þ Þ Þ b œ x3 ˆ 1 1 x ‰ œ 1 x x 2 2 4 4 6 6 48. 1 32xx 34xx 36xx Þ Þ Þ œ 1 21x a3xb2 41x a3xb4 61x a3xb6 Þ Þ Þ œ cosa3xb 2 3 3 49. x3 x5 x7 x9 Þ Þ Þ œ x3 Š1 x2 ax2 b ax2 b Þ Þ Þ ‹ œ x3 ˆ 1 +1x2 ‰ œ 1 +x x2 2 3 4 50. x2 2x3 22xx 23xx 24xx Þ Þ Þ œ x2 Š1 2x a2x2xb a2x3xb a2x4xb Þ Þ Þ ‹ œ x2 e2x 2 4 3 5 4 6 d d ˆ 1 ‰ 1 51. 1 2x 3x2 4x3 5x4 Þ Þ Þ œ dx a1 x x2 x3 x4 x5 Þ Þ Þ b œ dx 1 x œ a1 x b 2 52. 1 x2 x3 x4 x5 Þ Þ œ 1x Šx x2 x3 x4 x5 Þ Þ ‹ œ 1x lna1 xb œ lna1xxb 2 3 4 2 3 4 5 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. lnˆ1 x3 ‰ x sin x2 Section 10.10 The Binomial Series $ # % # $ % $ 631 & x‰ x x x x x x x x 53. ln ˆ 11 x œ ln (1 x) ln (1 x) œ Šx # 3 4 á ‹ Šx # 3 4 á ‹ œ 2 Šx 3 5 á ‹ # $ % " 54. ln (1 x) œ x x# x3 x4 á (1)n x á Ê kerrork œ ¹ (")n x ¹ œ n10 n when x œ 0.1; " " n10n 10) n 1 n Ê n10n 10) when n $ & ( n 1 n 8 Ê 7 terms * ) x 55. tan" x œ x x3 x5 x7 x9 á ("2n 1 " " #n1 10$ n 1 2n 1 á Ê kerrork œ ¹ (1)2nx1 n 1 2n 1 ¹ œ #n"1 when x œ 1; st Ê n 1001 # œ 500.5 Ê the first term not used is the 501 Ê we must use 500 terms & ( n1 2n1 * 56. tan" x œ x x3 x5 x7 x9 á (1)2nx1 $ ¸ 2n 1 ¸ œ x# á and n lim † 2n 1 ¹ œ x# n lim ¹x Ä _ 2n 1 x2n1 Ä _ #n 1 2n 1 _ Ê tan" x converges for kxk 1; when x œ 1 we have ! (2n1)1 which is a convergent series; when x œ 1 n n œ1 _ nc1 we have ! (2n1)1 which is a convergent series Ê the series representing tan" x diverges for kxk 1 n œ1 $ & ( * 57. tan" x œ x x3 x5 x7 x9 á (1)2n x1 n 1 2n 1 " ‰ á and when the series representing 48 tan" ˆ 18 has an error less than "3 † 10' , then the series representing the sum " ‰ " ‰ 48 tan" ˆ 18 32 tan" ˆ 57 20 tan" ˆ #"39 ‰ also has an error of magnitude less than 10' ; thus 2nc1 Š"‹ " kerrork œ 48 #18n 1 3†10 ' Ê n 4 using a calculator Ê 4 terms x x 58. ln (sec x) œ '0 tan t dt œ '0 Št t3 2t15 á ‹ dt ¸ x# 12 45 á x 59. (a) a1 x# b lim nÄ_ "Î# x # $ % & # ' % $ ' & ( 5x " ¸ 1 x# 3x8 5x x ¸ x x6 3x 16 Ê sin 40 112 ; Using the Ratio Test: 2nb3 (2n 1)(2n1) (2n 1)(2n 1)x 2†4†6â(2n)(2n ") # ¹ 12††34††56â ¹ ¹1 â(2n)(2n 2)(2n 3) † 1†3†5â(2n 1)x2nb1 ¹ 1 Ê x n lim Ä _ (2n 2)(2n 3) Ê kxk 1 Ê the radius of convergence is 1. See Exercise 69. (b) "Î# d " xb œ a1 x# b dx acos 60. (a) a1 t# b "Î# $ & ( $ & "‰ ˆ ¸ (1)"Î# ˆ "# ‰ (1)$Î# at# b # ˆ #3 ‰ (1) &Î# at# b #! # ˆ #" ‰ ˆ #3 ‰ ˆ #5 ‰ (1) (Î# at# b$ 3! 3†5t t 3t 5t x 3x 5x " œ 1 t# 23t # †2! 2$ †3! Ê sinh x ¸ '0 Š1 # 8 16 ‹ dt œ x 6 40 112 # % ' x # % ' $ & ( " 3 (b) sinh" ˆ 4" ‰ ¸ 4" 384 40,960 œ 0.24746908; the error is less than the absolute value of the first unused ( ( 5x 1 x 3x 5x Ê cos" x œ 1# sin" x ¸ 1# Šx x6 3x 40 112 ‹ ¸ # x 6 40 112 5x term, 112 , evaluated at t œ 4" since the series is alternating Ê kerrork 5 ˆ 4" ‰ 112 ( ¸ 2.725 ‚ 10' " d ˆ 1 ‰ " d # $ # $ 61. 1" x œ 1 (x) œ 1 x x x á Ê dx 1 x œ 1 x# œ dx a1 x x x á b œ 1 2x 3x# 4x$ á d ˆ " ‰ 2x d # % ' $ & 62. 1 " x# œ 1 x# x% x' á Ê dx 1 x# œ a1 x# b# œ dx a1 x x x á b œ 2x 4x 6x á Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 632 Chapter 10 Infinite Sequences and Series 8â(2n 2)†(2n) 63. Wallis' formula gives the approximation 1 ¸ 4 ’ 3†23††45††45††67††67†â (2n 1)†(2n 1) “ to produce the table n µ1 10 3.221088998 20 3.181104886 30 3.167880758 80 3.151425420 90 3.150331383 93 3.150049112 94 3.149959030 95 3.149870848 100 3.149456425 At n œ 1929 we obtain the first approximation accurate to 3 decimals: 3.141999845. At n œ 30,000 we still do not obtain accuracy to 4 decimals: 3.141617732, so the convergence to 1 is very slow. Here is a Maple CAS procedure to produce these approximations: pie := proc(n) local i,j; a(2) := evalf(8/9); for i from 3 to n do a(i) := evalf(2*(2*i2)*i/(2*i1)^2*a(i1)) od; [[j,4*a(j)] $ (j = n5 .. n)] end _ _ _ 64. (a) faxb œ 1 !ˆ mk ‰xk Ê f w axb œ !ˆ mk ‰k xk1 Ê a1 xb † f w axb œ a1 xb!ˆ mk ‰k xk1 kœ1 kœ1 _ _ kœ1 _ _ _ _ kœ2 kœ1 œ !ˆ mk ‰k xk 1 x † !ˆ mk ‰k xk1 œ !ˆ mk ‰k xk 1 !ˆ mk ‰k xk œ ˆ m1 ‰a1b x0 !ˆ mk ‰k xk1 !ˆ mk ‰k xk kœ1 kœ1 _ kœ1 _ œ m ! ˆ mk ‰k xk 1 ! ˆ mk ‰k xk kœ2 kœ1 w kœ1 _ _ kœ2 kœ1 _ _ kœ1 kœ1 k ‰ Note that: ! ˆ mk ‰k xk 1 œ ! ˆ k m 1 ak 1b x . _ _ kœ2 kœ1 k ! ˆ m ‰k xk ‰ Thus, a1 xb † f axb œ m ! ˆ mk ‰k xk 1 ! ˆ mk ‰k xk œ m ! ˆ k m 1 ak 1b x k _ _ kœ1 kœ1 k ! ’ˆˆ m ‰ak 1b ˆ m ‰k ‰xk “. ‰ ˆm‰ k œ m ! ’ˆ k m 1 ak 1b x k k x “ œ m k1 k m†am "bâam ak 1b 1b m ‰ ˆm‰ Note that: ˆ k ak 1b m†am "bâk!am k 1b k 1 ak 1b k k œ ak1b! œ m†am "k!bâam kb m†am "bâk!am k 1b k œ m†am "bâk!am k 1b aam kb kb œ m m†am "bâk!am k 1b œ mˆ mk ‰. _ _ _ kœ1 kœ1 kœ1 ! ’ˆmˆ m ‰ ‰xk “ œ m m!ˆ m ‰xk ‰ ˆm‰ ‰ k Thus, a1 xb † f w axb œ m ! ’ˆˆ k m 1 ak 1b k k x “ œ m k k _ †faxb œ mŒ1 !ˆ mk ‰xk œ m † faxb Ê f w axb œ m a1xb if " x 1. kœ1 (b) Let gaxb œ a1 xbm faxb Ê gw axb œ ma1 xbm1 faxb a1 xbm f w axb †faxb m1 œ ma1 xbm1 faxb a1 xbm † m faxb a1 xbm1 † m † faxb œ 0. a1xb œ ma1 xb _ (c) gw axb œ 0 Ê gaxb œ c Ê a1 xbm faxb œ c Ê faxb œ a1cxbcm œ ca1 xbm . Since faxb œ 1 !ˆ mk ‰xk kœ1 _ Ê fa0b œ 1 !ˆ mk ‰a0bk œ 1 0 œ 1 Ê ca1 0bm œ 1 Ê c œ 1 Ê faxb œ a1 xbm . kœ1 65. a1 x# b "Î# œ a1 ax# bb "Î# ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ (1) (Î# ax# b$ 3! œ (1)"Î# ˆ "# ‰ (1)$Î# ax# b ˆ "# ‰ ˆ 3# ‰ (1)c&Î# ax# b# _ #! 1†3†5x ! 1†3†5ân(2n1)x á œ 1 x# 12†#3x †#! 2$ †3! á œ 1 # †n! # % ' 2n n œ1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 10.10 The Binomial Series Ê sin" x œ '0 a1 t# b x 633 1†3†5â(2n 1)x 1)x dt œ '0 Œ1 ! 1†3†5â#(2n n †n! dt œ x ! #†4â(2n)(2n 1) , _ x "Î# _ 2n n œ1 2nb1 nœ1 where kxk 1 _ dt 66. ctan" td x œ 1# tan" x œ 'x _ _ 1 t# _ _ " Š 1# ‹ œ 'x – t " — dt œ 'x 1Š ‹ t# œ 'x ˆ t"# t"% t"' t") á ‰ dt œ lim bÄ_ t# ˆ1 t"# t"% t"' á ‰ dt "t 3t"$ 5t"& 7t"( á ‘ b œ "x 3x" $ 5x" & 7x" ( á x Ê tan" x œ 1# x" 3x" $ 5x" & á , x 1; ctan" td c_ œ tan" x 1# œ ' _ 1 dt t# x x œ lim b Ä _ "t 3t"$ 5t"& 7t"( á ‘ x œ "x 3x" $ 5x" & 7x" ( á Ê tan" x œ 1# x" 3x" $ 5x" & á , b x 1 67. (a) ei1 œ cos (1) i sin (1) œ 1 i(0) œ 1 (b) ei1Î4 œ cos ˆ 14 ‰ i sin ˆ 14 ‰ œ È" Èi œ Š È" ‹ (1 i) 2 2 2 (c) ei1Î2 œ cos ˆ 1# ‰ i sin ˆ 1# ‰ œ 0 i(1) œ i 68. ei) œ cos ) i sin ) Ê ei) œ ei()) œ cos ()) i sin ()) œ cos ) i sin ); i) ei) ei) œ cos ) i sin ) cos ) i sin ) œ 2 cos ) Ê cos ) œ e #e i) i) e e ci) ; i) ci) sin ) œ e #ie œ cos ) i sin ) (cos ) i sin )) œ 2i sin ) Ê # $ % )) )) )) 69. ex œ 1 x x#! x3! x4! á Ê ei) œ 1 i) (i2! (i3! (i4! á and # $ % # $ % # $ % )) )) ei) œ 1 i) (2!i)) (3!i)) (4!i)) á œ 1 i) (i#)!) (i3! (i4! á # $ % # $ % )) )) (i4! á‹ Š1 i) (i)) (i)) (i)) á‹ Š1 i) (i#)!) (i3! i) ci) #! 3! 4! Ê e #e œ # % ' œ 1 )#! )4! )6! á œ cos ); # $ # % # $ % )) )) (i4! á‹ Š1 i) (i)) (i)) (i)) á‹ Š1 i) (i#)!) (i3! #! 3! 4! ei) eci) œ #i $ & ( œ ) )3! )5! )7! á œ sin ) #i 70. ei) œ cos ) i sin ) Ê ei) œ eiÐ)Ñ œ cos ()) i sin ()) œ cos ) i sin ) i) (a) ei) ei) œ (cos ) i sin )) (cos ) i sin )) œ 2 cos ) Ê cos ) œ e #e (b) ei) ei) œ (cos ) i sin )) (cos ) i sin )) œ 2i sin ) Ê i sin ) œ # $ % $ & ci) œ cosh i) e eci) i) 2 œ sinh i) ( 71. ex sin x œ Š1 x x#! x3! x4! á ‹ Šx x3! x5! x7! á ‹ " & œ (1)x (1)x# ˆ "6 #" ‰ x$ ˆ 6" "6 ‰ x% ˆ 1#"0 1"# #"4 ‰ x& á œ x x# "3 x$ 30 x á ; ex † eix œ eÐ1iÑx œ ex (cos x i sin x) œ ex cos x i aex sin xb Ê ex sin x is the series of the imaginary part _ # $ % of eÐ1iÑx which we calculate next; eÐ1iÑx œ ! (xn!ix) œ 1 (x ix) (x #!ix) (x 3!ix) (x 4!ix) á n n œ0 œ 1 x ix #"! a2ix# b 3!" a2ix$ 2x$ b 4!" a4x% b 5!" a4x& 4ix& b 6!" a8ix' b á Ê the imaginary part " & " ' of eÐ1iÑx is x #2! x# 3!2 x$ 5!4 x& 6!8 x' á œ x x# "3 x$ 30 x 90 x á in agreement with our product calculation. The series for ex sin x converges for all values of x. d ˆ ÐaibÑ ‰ d ceax (cos bx i sin bx)d œ aeax (cos bx i sin bx) eax (b sin bx bi cos bx) 72. dx e œ dx œ aeax (cos bx i sin bx) bieax (cos bx i sin bx) œ aeÐaibÑx ibeÐaibÑx œ (a ib)eÐaibÑx Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 634 Chapter 10 Infinite Sequences and Series 73. (a) ei)" ei)# œ (cos )" i sin )" )(cos )# i sin )# ) œ (cos )" cos )# sin )" sin )# ) i(sin )" cos )# sin )# cos )" ) œ cos()" )# ) i sin()" )# ) œ eiÐ)" )# Ñ ) i sin ) ‰ " " (b) ei) œ cos()) i sin()) œ cos ) i sin ) œ (cos ) i sin )) ˆ cos cos ) i sin ) œ cos ) i sin ) œ ei) 74. aa# bib# eÐabiÑx C" iC# œ ˆ aa# bib# ‰ eax (cos bx i sin bx) C" iC# ax œ a# e b# (a cos bx ia sin bx ib cos bx b sin bx) C" iC# ax œ a# e b# [(a cos bx b sin bx) (a sin bx b cos bx)i] C" iC# œ e (a cosa#bxb#b sin bx) C" ie (a sina#bxb#b cos bx) iC# ; ax ax eÐabiÑx œ eax eibx œ eax (cos bx i sin bx) œ eax cos bx ieax sin bx, so that given ' eÐabiÑx dx œ aa bib eÐabiÑx C" iC# we conclude that ' eax cos bx dx œ e (a cosa bxb b sin bx) C" and ' eax sin bx dx œ e (a sinabxbb cos bx) C# ax # # # # ax # # CHAPTER 10 PRACTICE EXERCISES 1. converges to 1, since n lim a œ n lim Š1 (n1) ‹ œ 1 Ä_ n Ä_ n 2 2. converges to 0, since 0 Ÿ an Ÿ È2n , n lim 0 œ 0, n lim œ 0 using the Sandwich Theorem for Sequences Ä_ Ä _ Èn ˆ 1 2n2 ‰ œ lim ˆ #"n 1‰ œ 1 3. converges to 1, since n lim a œ n lim Ä_ n Ä_ nÄ_ n 4. converges to 1, since n lim a œ n lim c1 (0.9)n d œ 1 0 œ 1 Ä_ n Ä_ ™ œ e0ß 1ß 0ß 1ß 0ß 1ß á f 5. diverges, since ˜sin n1 # 6. converges to 0, since {sin n1} œ {0ß 0ß 0ß á } # ln n 7. converges to 0, since n lim a œ n lim œ 2 n lim Ä_ n Ä_ n Ä_ Š "n ‹ 1 Š 2n 2b 1 ‹ ln (2n") 8. converges to 0, since n lim a œ n lim œ n lim n Ä_ n Ä_ Ä_ 1 1Š "n ‹ ˆ n nln n ‰ œ lim 9. converges to 1, since n lim a œ n lim Ä_ n Ä_ nÄ_ $ œ0 1 ln a2n 1b 10. converges to 0, since n lim a œ n lim œ n lim n Ä_ n Ä_ Ä_ Š œ0 œ1 6n# ‹ 2n$ 1 1 12n 2 œ n lim œ n lim œ0 Ä _ 6n# Ä_ n n ˆ n n 5 ‰ œ lim Š1 (n5) ‹ œ ec5 by Theorem 5 11. converges to ec5 , since n lim a œ n lim Ä_ n Ä_ nÄ_ n ˆ1 "n ‰ 12. converges to "e , since n lim a œ n lim Ä_ n Ä_ n " " œ n lim n œ e by Theorem 5 Ä _ ˆ1 " ‰ n 1 În 3 œ n lim œ 13 œ 3 by Theorem 5 Ä _ n1În 1 În 3 œ n lim œ 11 œ 1 by Theorem 5 Ä _ n1În ˆ3 ‰ 13. converges to 3, since n lim a œ n lim Ä_ n Ä_ n ˆ3‰ 14. converges to 1, since n lim a œ n lim Ä_ n Ä_ n cn 1În Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 10 Practice Exercises 1În 2 1 15. converges to ln 2, since n lim a œ n lim n a21În 1b œ n lim œ n lim " Ä_ n Ä_ Ä_ Ä_ – Š 21În ln 2‹ n# Š n#" ‹ Šn‹ — 635 œ n lim 21În ln 2 Ä_ œ 2! † ln 2 œ ln 2 2 n È 16. converges to 1, since n lim a œ n lim 2n 1 œ n lim exp Š ln (2nn 1) ‹ œ n lim exp Œ 2n1b 1 œ e! œ 1 Ä_ n Ä_ Ä_ Ä_ (n 1)! 17. diverges, since n lim a œ n lim œ n lim (n 1) œ _ n! Ä_ n Ä_ Ä_ (4) 18. converges to 0, since n lim a œ n lim œ 0 by Theorem 5 Ä_ n Ä _ n! n Š"‹ Š"‹ # # " 19. (2n 3)(2n 1) œ #n 3 2n 1 Ê sn œ – Š "# ‹ 3 Š "# ‹ 5 —– Š "# ‹ 5 Š "# ‹ 7 Š "# ‹ Š "# ‹ — á – #n 3 2n 1 — œ Š "# ‹ 3 Š "‹ # 2n 1 Š "‹ # " " Ê n lim s œ n lim 2n 1— œ 6 Ä_ n Ä _ –6 20. n(n2 1) œ n2 n 2 1 Ê sn œ ˆ #2 32 ‰ ˆ 32 42 ‰ á ˆ n2 n 2 1 ‰ œ #2 n 2 1 Ê n lim s Ä_ n ˆ1 n 2 1 ‰ œ 1 œ n lim Ä_ 9 3 3 3 ‰ 3 ‰ ˆ3 3‰ ˆ3 3‰ ˆ3 ˆ 3 21. (3n 1)(3n 2) œ 3n 1 3n 2 Ê sn œ # 5 5 8 8 11 á 3n 1 3n 2 ˆ 3 3n 3 2 ‰ œ 3# œ 3# 3n3# Ê n lim s œ n lim Ä_ n Ä_ # 8 2 2 2 ‰ 2 ‰ 2 ‰ ˆ 92 13 22. (4n 3)(4n ˆ 132 17 ˆ 172 21 á ˆ 4n2 3 4n 2 1 ‰ 1) œ 4n 3 4n 1 Ê sn œ ˆ 29 4n21 ‰ œ 29 œ 29 4n21 Ê n lim s œ n lim Ä_ n Ä_ _ _ n œ0 nœ0 23. ! en œ ! e"n , a convergent geometric series with r œ "e and a œ 1 Ê the sum is _ _ n œ1 n œ0 " œ e e 1 1 Š "e ‹ ‰ a convergent geometric series with r œ 4" and a œ 43 Ê the sum is 24. ! (1)n 43n œ ! ˆ 43 ‰ ˆ " 4 n ˆ 34 ‰ œ 35 1ˆ c4" ‰ 25. diverges, a p-series with p œ "# _ _ n œ1 nœ1 26. ! n5 œ 5 ! "n , diverges since it is a nonzero multiple of the divergent harmonic series " 27. Since f(x) œ x"Î# Ê f w (x) œ #x"$Î# 0 Ê f(x) is decreasing Ê an1 an , and _ _ 1 lim a œ lim œ 0, the n Ä _ n n Ä _ Èn ) ! È" diverges, the given series converges conditionally. series ! (" Èn converges by the Alternating Series Test. Since n nœ1 n nœ1 28. converges absolutely by the Direct Comparison Test since #n" $ n"$ for n 1, which is the nth term of a convergent p-series Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 636 Chapter 10 Infinite Sequences and Series 29. The given series does not converge absolutely by the Direct Comparison Test since ln (n" 1) n " 1 , which is " the nth term of a divergent series. Since f(x) œ ln (x" 1) Ê f w (x) œ (ln (x 1)) # (x 1) 0 Ê f(x) is decreasing " Ê an1 an , and n lim a œ n lim œ 0, the given series converges conditionally by the Alternating Ä_ n Ä _ ln (n 1) Series Test. _ 30. '2 " x(ln x)# dx œ b lim Ä_ '2b x(ln" x) dx œ lim c(ln x)" d b2 œ lim ˆ ln"b ln"2 ‰ œ ln"# Ê the series # bÄ_ bÄ_ converges absolutely by the Integral Test 31. converges absolutely by the Direct Comparison Test since lnn$n nn$ œ n"# , the nth term of a convergent p-series n n 32. diverges by the Direct Comparison Test for en n Ê ln ên ‰ ln n Ê nn ln n Ê ln nn ln (ln n) Ê n ln n ln (ln n) Ê lnln(lnnn) "n , the nth term of the divergent harmonic series 33. n lim Ä_ Š È "# n n Š n"# ‹ 1 ‹ # n œ Én lim œ È1 œ 1 Ê converges absolutely by the Limit Comparison Test Ä _ n# 1 $ 34. Since f(x) œ x$3x 1 Ê f w (x) œ 3xaxa$21xb# b 0 when x # # 2 Ê an1 an for n 3n 2 and n lim œ 0, the Ä _ n$ 1 series converges by the Alternating Series Test. The series does not converge absolutely: By the Limit # Comparison Test, n lim Ä_ Š n$3n 1 ‹ ˆ n" ‰ $ 3n œ n lim œ 3. Therefore the convergence is conditional. Ä _ n$ 1 n2 35. converges absolutely by the Ratio Test since n lim œ01 ’ n2 † n! “ œ n lim Ä _ (n 1)! n 1 Ä _ (n 1)# # (") an 1b 36. diverges since n lim a œ n lim does not exist Ä_ n Ä _ 2n# n 1 n nb1 3 37. converges absolutely by the Ratio Test since n lim † n! “ œ n lim œ01 ’ 3 Ä _ (n 1)! 3n Ä _ n1 n 2 3 6 n È É 38. converges absolutely by the Root Test since n lim an œ n lim œ01 nn œ n lim Ä_ Ä_ Ä_ n n n 39. converges absolutely by the Limit Comparison Test since n lim Ä_ 40. converges absolutely by the Limit Comparison Test since n lim Ä_ nb1 " Š $Î# ‹ n Š Èn(n "1)(n Š n"# ‹ Š È "# n n ‹ 2) n(n 1)(n 2) œ Én lim œ1 n$ Ä_ # 1 ‹ # n an 1 b œ Én lim œ1 n% Ä_ kx 4 k ˆ n ‰ 1 Ê kx 3 4k 1 41. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 4) † n3 ¹ 1 Ê 3 n lim Ä_ Ä _ (n 1)3nb1 (x 4)n Ä _ n1 n _ _ 1) 3 Ê kx 4k 3 Ê 3 x 4 3 Ê 7 x 1; at x œ 7 we have ! (n3 œ ! ("n ) , the alternating n _ n œ1 _ n n n n œ1 3 ! " , the divergent harmonic series harmonic series, which converges conditionally; at x œ 1 we have! n3 n œ n n œ1 n n œ1 (a) the radius is 3; the interval of convergence is 7 Ÿ x 1 (b) the interval of absolute convergence is 7 x 1 (c) the series converges conditionally at x œ 7 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 10 Practice Exercises " 42. n lim œ 0 1, which holds for all x ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x1) † (2n1)! ¹ 1 Ê (x 1)# n lim Ä_ Ä _ (2n1)! (x1)2nc2 Ä _ (#n)(2n1) 2n (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally nb1 n 43. n lim 1 Ê k3x 1k 1 ¹ uunbn 1 ¹ 1 Ê n lim ¹ (3x(n1)1)# † (3x n 1)n ¹ 1 Ê k3x 1k n lim Ä_ Ä_ Ä _ (n 1)# # # _ nc1 _ 2nc1 Ê 1 3x 1 1 Ê 0 3x 2 Ê 0 x 23 ; at x œ 0 we have ! (1) n# (1) œ ! ("n)# n n œ1 n œ1 _ œ ! n"# , a nonzero constant multiple of a convergent p-series, which is absolutely convergent; at x œ 23 we n œ1 _ _ have ! (1)n# (1) œ ! ("n)# n 1 n n œ1 n 1 , which converges absolutely nœ1 (a) the radius is "3 ; the interval of convergence is 0 Ÿ x Ÿ 32 (b) the interval of absolute convergence is 0 Ÿ x Ÿ 23 (c) there are no values for which the series converges conditionally 44. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ n 2 † (2x 2nb1)1 Ä_ Ä _ 2n 3 nb1 k2x 1k 2 1 † 2n lim ¸ n 2 † 2n " ¸ 1 n 1 † (2x 1)n ¹ 1 Ê 2 n Ä _ 2n 3 n 1 n Ê k2x # 1k (1) 1 Ê k2x 1k 2 Ê 2 2x 1 2 Ê 3 2x 1 Ê #3 x "# ; at x œ #3 we have _ _ ! n 1 † (2) œ ! (") (n1) which diverges by the nth-Term Test for Divergence since n n œ1 n 2n 1 # n œ1 n 2n 1 _ _ n1 2 ! n " , which diverges by the nth-Term Test lim ˆ n 1 ‰ œ "# Á 0; at x œ #" we have ! 2n 1 † #n œ 2n 1 n Ä _ 2n 1 n n œ1 n œ1 (a) the radius is 1; the interval of convergence is 3# x "# (b) the interval of absolute convergence is 3# x "# (c) there are no values for which the series converges conditionally nb1 ¸ˆ n ‰ ˆ n " 1 ‰¸ 1 Ê kxek lim ˆ n " 1 ‰ 1 45. n lim † n ¹ 1 Ê kxk n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ x Ä_ Ä _ (n 1)nb1 xn Ä _ n1 nÄ_ n n Ê kxek † 0 1, which holds for all x (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally nb1 Èn n 46. n lim † xn ¹ 1 Ê kxk n lim 1 Ê kxk 1; when x œ 1 we have ¹ uunbn 1 ¹ 1 Ê n lim ¹ x Ä_ Ä _ Èn 1 Ä _ Én1 _ _ ! (È1) , which converges by the Alternating Series Test; when x œ 1 we have ! È" , a divergent p-series n œ1 n n nœ1 n (a) the radius is 1; the interval of convergence is 1 Ÿ x 1 (b) the interval of absolute convergence is 1 x 1 (c) the series converges conditionally at x œ 1 2nb1 47. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n 32)x nb1 Ä_ Ä_ _ _ n œ1 n œ1 # 3 x ˆ n 2 ‰ 1 Ê È3 x È3; † (n 1)x 2n 1 ¹ 1 Ê 3 n lim Ä _ n1 n the series ! nÈ31 and ! nÈ31 , obtained with x œ „ È3, both diverge (a) the radius is È3; the interval of convergence is È3 x È3 (b) the interval of absolute convergence is È3 x È3 (c) there are no values for which the series converges conditionally Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 637 638 Chapter 10 Infinite Sequences and Series 2nb3 48. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 2n1)x 3 Ä_ Ä_ 1 # ˆ 2n 1 ‰ 1 Ê (x 1)# (1) 1 † (x 2n 1)2nb1 ¹ 1 Ê (x 1) n lim Ä _ 2n 3 _ Ê (x 1)# 1 Ê kx 1k 1 Ê 1 x 1 1 Ê 0 x 2; at x œ 0 we have ! (1)#n(1)1 n 2nb1 n œ1 _ œ ! (1) n œ1 _ 3nb1 2n 1 _ œ ! (1) n œ1 nc1 2n 1 which converges conditionally by the Alternating Series Test and the fact _ 2nb1 that ! 2n " 1 diverges; at x œ 2 we have ! (1)2n (1) 1 n œ1 n nœ1 _ (1) œ ! 2n 1 , which also converges conditionally n nœ1 (a) the radius is 1; the interval of convergence is 0 Ÿ x Ÿ 2 (b) the interval of absolute convergence is 0 x 2 (c) the series converges conditionally at x œ 0 and x œ 2 (n 1)x 49. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ cschcsch (n)xn Ä_ Ä_ c" nb1 Š enb1 c ecnc1 ‹ 1 ¹ 1 Ê kxk n lim Ä_ » ˆ 2 ‰ » 2 en c ecn _ kx k Ê kxk n lim ¹ e1 ee 2n 2 ¹ 1 Ê e 1 Ê e x e; the series ! a „ ebn csch n, obtained with x œ „ e, Ä_ 2n 1 n œ1 both diverge since n lim a „ e) csch n Á 0 Ä_ n (a) the radius is e; the interval of convergence is e x e (b) the interval of absolute convergence is e x e (c) there are no values for which the series converges conditionally nb1 c2nc2 c2n (n 1) 50. n lim † 1 e ¹ 1 Ê kxk 1 ¹ uunbn 1 ¹ 1 Ê n lim ¹ x xncoth ¹1e coth (n) ¹ 1 Ê kxk n lim Ä_ Ä_ Ä _ 1 ec2nc2 1 ec2n _ Ê 1 x 1; the series ! a „ 1bn coth n, obtained with x œ „ 1, both diverge since n lim a „ 1bn coth n Á 0 Ä_ n œ1 (a) the radius is 1; the interval of convergence is 1 x 1 (b) the interval of absolute convergence is 1 x 1 (c) there are no values for which the series converges conditionally 51. The given series has the form 1 x x# x$ á (x)n á œ 1 " x , where x œ 4" ; the sum is 1 "ˆ " ‰ œ 54 4 # $ n $ & 2n 1 # % 2n 52. The given series has the form x x# x3 á (1)n1 xn á œ ln (1 x), where x œ 23 ; the sum is ln ˆ 53 ‰ ¸ 0.510825624 53. The given series has the form x x3! x5! á (1)n (2nx 1)! á œ sin x, where x œ 1; the sum is sin 1 œ 0 x 54. The given series has the form 1 x2! x4! á (1)n (2n)! á œ cos x, where x œ 13 ; the sum is cos 13 œ "# # # 55. The given series has the form 1 x x2! x3! á xn! á œ ex , where x œ ln 2; the sum is eln Ð2Ñ œ 2 $ n & x " 56. The given series has the form x x3 x5 á (1)n (2n x, where x œ È"3 ; the sum is 1) á œ tan 2n 1 tan" Š È"3 ‹ œ 16 57. Consider 1 " 2x as the sum of a convergent geometric series with a œ 1 and r œ 2x Ê 1 " 2x _ _ n œ0 nœ0 œ 1 (2x) (2x)# (2x)$ á œ ! (2x)n œ ! 2n xn where k2xk 1 Ê kxk "# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 10 Practice Exercises 639 58. Consider 1 " x$ as the sum of a convergent geometric series with a œ 1 and r œ x$ Ê 1 " x$ œ 1 a"x$ b # _ $ œ 1 ax$ b ax$ b ax$ b á œ ! (1)n x3n where kx$ k 1 Ê kx$ k 1 Ê kxk 1 n œ0 _ _ n 2nb1 2nb1 1) (1x) 59. sin x œ ! ((2n1)x 1)! Ê sin 1x œ ! ((2n 1)! nœ0 n nœ0 _ nœ0 2nb1 _ n 2n n œ0 _ Ê cos ˆx5Î3 ‰ œ ! nœ0 (1)n ˆx5Î3 ‰ (2n)! 2n _ (1)n Š x3 ‹ È Ê cosŠ Èx 5 ‹ œ ! _ _ ˆ 1 x ‰n n 2n n œ0 3 5 (2n)! nœ0 63. ex œ ! xn! Ê eÐ1xÎ2Ñ œ ! n n œ0 n œ0 _ _ n! œ! # n _ # n œ0 n 2nb1 2nb1 nœ0 _ n 10nÎ3 x œ ! (1)(#n)! nœ0 2n _ 1) x 62. cos x œ ! ((2n)! _ _ œ ! (32n1)b12(#n x 1)! (2n 1)! nœ0 n 2nb1 2nb1 nœ0 _ (1)n Š 2x ‹ 3 n 2nb1 ! 60. sin x œ ! ((2n1)x 1)! Ê sin 2x 3 œ 1) x 61. cos x œ ! ((2n)! _ œ ! (1)(#1n 1)!x _ x œ ! (5n1)(#n)! n 6n nœ0 1 n xn #n n! 64. ex œ ! xn! Ê ex œ ! an!x b œ ! (1)n! x # n n œ0 nœ0 n œ0 "Î# 65. f(x) œ È3 x# œ a3 x# b Ê f www (x) œ 3x$ a3 x# b 3 9 f www (1) œ 32 83 œ 32 n 2n Ê f w (x) œ x a3 x# b &Î# "Î# Ê f ww (x) œ x# a3 x# b $Î# a3 x# b "Î# $Î# 3x a3 x# b ; f(1) œ 2, f w (1) œ "# , f ww (1) œ 8" #" œ 83 , # $ Ê È3 x# œ 2 (x 1) 3(x$ 1) 9(x& 1) á 2†1! 2 †2! 2 †3! 66. f(x) œ 1 " x œ (1 x)" Ê f w (x) œ (1 x)# Ê f ww (x) œ 2(1 x)$ Ê f www (x) œ 6(1 x)% ; f(2) œ 1, f w (2) œ 1, f ww (2) œ 2, f www (2) œ 6 Ê 1" x œ 1 (x 2) (x 2)# (x 2)$ á 67. f(x) œ x " 1 œ (x 1)" Ê f w (x) œ (x 1)# Ê f ww (x) œ 2(x 1)$ Ê f www (x) œ 6(x 1)% ; f(3) œ 4" , f w (3) œ 4"# , f ww (3) œ 42$ , f www (2) œ 4%6 Ê x" 1 œ "4 4"# (x 3) 4"$ (x 3)# 4"% (x 3)$ á 68. f(x) œ x" œ x" Ê f w (x) œ x# Ê f ww (x) œ 2x$ Ê f www (x) œ 6x% ; f(a) œ "a , f w (a) œ a"# , f ww (a) œ a2$ , f www (a) œ a%6 Ê "x œ "a a"# (x a) a"$ (x a)# a"% (x a)$ á 69. '0 exp ax$ b dx œ '0 Š1 x$ x2! x3! x4! á ‹ dx œ ’x x4 7x†2! 10x †3! 13x †4! á “ 1Î2 1Î2 ' * "# % ( "! "$ " " " ¸ "# #%"†4 #( †"7†2! 2"! †10 †3! 2"$ †13†4! 2"' †16†5! ¸ 0.484917143 "Î# ! 70. '0 x sin ax$ b dx œ '0 x Šx$ x3! x5! x7! x9! á ‹ dx œ '0 Šx% x3! x5! x7! x9! á ‹ dx 1 1 & "" "( * #$ "& #" #( 1 "! "' ## #) " #* œ ’ x5 11x †3! 17x †5! 23x †7! 29x †9! á “ ¸ 0.185330149 ! 71. '1 1Î2 tanc" x dx œ x "Î# x '11Î2 Š1 x3 x5 x7 x9 x11 á ‹ dx œ ’x x9 25x 49x 81x 121 á“ # % ' ) "! $ & ( * "" ! ¸ "# 9†"2$ 5#"†#& 7#"†#( 9#"†2* 11#"†2"" 13#"†2"$ 15#"†2"& 17#"†#"( 19#"†#"* 21#"†##" ¸ 0.4872223583 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 640 Chapter 10 Infinite Sequences and Series 72. '0 1Î64 tan " x Èx dx œ '01Î64 È"x Šx x3$ x5& x7( á ‹ dx œ '01Î64 ˆx"Î# "3 x&Î# "5 x*Î# 7" x"$Î# á ‰ dx "Î'% 2 ""Î# 2 œ 23 x$Î# #21 x(Î# 55 x 105 x"&Î# á ‘ ! $ & œ ˆ 3†28$ 212†8( 55†28"" 1052†8"& á ‰ ¸ 0.0013020379 # 7 Šx x x á ‹ % 7 Š1 x x á ‹ 3! 5! 3! 5! 7 sin x œ lim œ lim œ #7 2x # # $ $ # $ # x Ä 0 e 1 x Ä 0 Š2x 2 2!x 2 3!x á‹ x Ä 0 Š2 2#!x 2 3!x á‹ 73. lim 74. # $ # $ Š1 ) )#! )3! á‹ Š1 ) )#! )3! á‹ 2) ) c) lim e )e sin)2) œ lim )Ä0 )Ä0 $ & ) Š) )3! )5! á‹ $ & $ & 2 Š )3! )5! á‹ œ lim )Ä0 Š )3! )5! á‹ # œ lim )Ä0 2 Š 3!" )5! á‹ œ2 # Š 3!" )5! á‹ # 75. # 2 cos t lim ˆ " t"# ‰ œ lim t 2t# (12 cos t) œ t lim t Ä 0 # 2 cos t tÄ0 Ä0 % t á t# 2 2 Œ1 t# 4! # % 2t# Š1 1 t# t4! á‹ % œ lim tÄ0 ' 2 Œ t4! t6! á ' Št% 2t4! á‹ # œ lim tÄ0 76. lim " 2 Š 4! t6! á‹ # Š1 2t4! á‹ Š sinh h ‹ cos h hÄ0 œ lim h# œ 1"# h# œ lim h% h# h# hÄ0 h# h# h% h% h' h' Œ #! 3! 5! 4! 6! 7! á h# hÄ0 h% Œ1 3! 5! á Œ1 #! 4! á # # % % Š #"! 3!" h5! h4! h6! h7! á ‹ œ 3" œ lim hÄ0 % % 1 Š1 z# z3 á‹ Šz# z3 á‹ " cos# z œ lim œ lim # $ $ & # $ % z Ä 0 ln (1 z) sin z z Ä 0 Šz z# z3 á‹ Šz z3! z5! á‹ z Ä 0 Š z# 2z3 z4 á‹ 77. lim # œ lim Š1 z3 á‹ # z z Ä 0 Š "# 2z 3 4 á‹ œ 2 y# y# y# œ lim œ lim # % ' # % ' # cos y cosh y y y y y y y 2y 2y' yÄ0 y Ä 0 Œ1 á Œ1 á y Ä 0 Œ 4! 6! 4! 6! # #! # 6! á 78. lim œ lim " y Ä 0 Œ1 2y% á 6! œ 1 $ 79. lim ˆ sinx$3x xr# s‰ œ lim – xÄ0 xÄ0 & (3x) Š3x (3x) 6 120 á‹ x$ # r xr# s— œ lim Š x3# #9 81x 40 á x# s‹ œ 0 xÄ0 Ê xr# x3# œ 0 and s #9 œ 0 Ê r œ 3 and s œ #9 80. The approximation sin x ¸ 6 6xx# is better than sin x ¸ x. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 10 Practice Exercises (3n 1)(3n 2)x 81. n lim ¹ 2†5†8#â †4†6â(2n)(2n 2) Ä_ nb1 â(2n) ¸ 3n 2 ¸ 1 Ê kxk 23 † #†52†8†4â†6(3n 1)xn ¹ 1 Ê kxk n lim Ä _ 2n 2 Ê the radius of convergence is 23 nb1 (2n1)(2n3)(x1) 82. n lim ¹ 3†5†47†â 9†14â(5n1)(5n4) Ä_ ¸ 2n 3 ¸ 1 Ê kxk 52 † 34††59††714ââ(2n(5n1)x1)n ¹ 1 Ê kxk n lim Ä _ 5n 4 Ê the radius of convergence is 52 n n n kœ2 kœ2 kœ2 83. ! ln ˆ1 k"# ‰ œ ! ln ˆ1 "k ‰ ln ˆ1 "k ‰‘ œ ! cln (k 1) ln k ln (k 1) ln kd œ cln 3 ln 2 ln 1 ln 2d cln 4 ln 3 ln 2 ln 3d cln 5 ln 4 ln 3 ln 4d cln 6 ln 5 ln 4 ln 5d á cln (n 1) ln n ln (n 1) ln nd œ cln 1 ln 2d cln (n 1) ln nd after cancellation n _ k œ2 k œ2 1 ‰ 1‰ Ê ! ln ˆ1 k"# ‰ œ ln ˆ n2n Ê ! ln ˆ1 k"# ‰ œ n lim ln ˆ n 2n œ ln "# is the sum Ä_ n n k œ2 k œ2 84. ! k# " 1 - œ "# ! ˆ k " 1 k " 1 ‰ œ #" ˆ 11 3" ‰ ˆ #" 4" ‰ ˆ "3 5" ‰ ˆ 4" 6" ‰ á ˆ n " # n" ‰ # 2(n 1) 2n ˆ n " 1 n " 1 ‰‘ œ #" ˆ 1" #" n" n " 1 ‰ œ #" ˆ #3 n" n " 1 ‰ œ #" ’ 3n(n 1)2n(n “ œ 3n4n(n n 1)2 1) _ "ˆ3 Ê ! k#"1 œ n lim 1n n 1 1 ‰ œ 43 Ä_ # 2 k œ2 1)x 85. (a) n lim ¹ 1†4†7â(3n(3n2)(3n 3)! Ä_ 3nb3 (3n ") $ † 1†4†7â(3n)! (3n 2)x3n ¹ 1 Ê kx k n lim Ä _ (3n 1)(3n 2)(3n 3) œ kx$ k † 0 1 Ê the radius of convergence is _ _ _ n œ1 _ n œ1 _ n œ1 nœ2 (3n 2) 3n ! 1†4†7â(3n 2) x3nc1 (b) y œ 1 ! 1†4†7â x Ê dy (3n)! dx œ (3n 1)! Ê d# y ! 1†4†7â(3n 2) x3nc2 œ x ! 1†4†7â(3n5) x3nc2 dx# œ (3n 2)! (3n3)! _ (3n 2) 3n œ x Œ1 ! 1†4†7â x œ xy 0 (3n)! n œ1 Ê a œ 1 and b œ 0 _ 86. (a) x# x# # # # # # $ # $ % & ! (1)n xn which 1 x œ 1 (x) œ x x (x) x (x) x (x) á œ x x x x á œ n œ2 converges absolutely for kxk 1 _ _ n œ2 nœ2 (b) x œ 1 Ê ! (1)n xn œ ! (1)n which diverges _ _ 87. Yes, the series ! an bn converges as we now show. Since ! an converges it follows that an Ä 0 Ê an 1 n œ1 nœ1 _ _ n œ1 n œ1 for n some index N Ê an bn bn for n N Ê ! an bn converges by the Direct Comparison Test with ! bn _ 88. No, the series ! an bn might diverge (as it would if an and bn both equaled n) or it might converge (as it would if n œ1 an and bn both equaled "n ). _ _ n œ1 k œ1 !(xk1 xk ) œ lim (xn1 x" ) œ lim (xn1 ) x" Ê both the series and 89. ! (xn1 xn ) œ n lim Ä_ nÄ_ nÄ_ sequence must either converge or diverge. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 641 642 Chapter 10 Infinite Sequences and Series 90. It converges by the Limit Comparison Test since n lim Ä_ Š 1 banan ‹ an _ " œ n lim œ 1 because ! an converges Ä _ 1 a n n œ1 and so an Ä 0. _ a" ˆ #" ‰ a# ˆ "3 "4 ‰ a% ˆ "5 "6 "7 "8 ‰ a) 91. ! ann œ a" a## a3$ a4% á n œ1 " " " ‰ ˆ 9" 10 11 á 16 a"' á 92. an œ ln"n for n 2 Ê a# a$ " # (a# a% a) a"' á ) which is a divergent series á , and ln"# ln"4 ln"8 á œ ln"# # ln" 2 3 ln" 2 á a% _ œ ln"# ˆ1 "# "3 á ‰ which diverges so that 1 ! n ln" n diverges by the Integral Test. n œ2 CHAPTER 10 ADDITIONAL AND ADVANCED EXERCISES _ 1. converges since (3n #")Ð2n 1ÑÎ2 (3n "2)$Î# and ! (3n "2)$Î# converges by the Limit Comparison Test: n œ1 lim nÄ_ Š " Š $Î# ‹ n (3n " ‹ 2)$Î# ˆ 3n n 2 ‰ œ n lim Ä_ $Î# œ 3$Î# _ 2. converges by the Integral Test: '1 atan" xb $ $ # dx x # 1 " $ " b $ $ 1 œ lim ’ atan 3 xb “ œ lim ’ atan 3 bb 192 “ bÄ_ " bÄ_ $ 71 1 1 œ Š 24 192 ‹ œ 192 c2n e 3. diverges by the nth-Term Test since n lim a œ n lim (1)n tanh n œ lim (1)n Š 11 (1)n ec2n ‹ œ n lim Ä_ n Ä_ Ä_ bÄ_ does not exist 4. converges by the Direct Comparison Test: n! nn Ê ln (n!) n ln (n) Ê lnln(n!) (n) n Ê logn (n!) n Ê lognn $(n!) n"# , which is the nth-term of a convergent p-series 12 1†2 12 ˆ 42††53 ‰ ˆ 31††42 ‰ 5. converges by the Direct Comparison Test: a" œ 1 œ (1)(3)(2) # , a# œ 3†4 œ (2)(4)(3)# , a$ œ _ 12 12 "2 ! ˆ 35††64 ‰ ˆ 24††53 ‰ ˆ 13††24 ‰ œ (4)(6)(5) œ (3)(5)(4) # , a% œ # , á Ê 1 (n 1)(n 3)(n 2)# represents the n œ1 given series and (n 1)(n 123)(n 2)# 12 , which is the nth-term of a convergent p-series n% anb1 n 6. converges by the Ratio Test: n lim œ n lim œ01 Ä _ an Ä _ (n 1)(n 1) 7. diverges by the nth-Term Test since if an Ä L as n Ä _, then L œ 1 " L Ê L# L 1 œ 0 Ê L œ 1 „# _ _ n œ1 n œ1 8. Split the given series into ! 32n"b1 and ! 32n2n ; the first subseries is a convergent geometric series and the n 2n É second converges by the Root Test: n lim 32n œ n lim Ä_ Ä_ n n n È 2È È 9 œ "9†1 œ "9 1 9. f(x) œ cos x with a œ 13 Ê f ˆ 13 ‰ œ 0.5, f w ˆ 13 ‰ œ #3 , f ww ˆ 13 ‰ œ 0.5, f www ˆ 13 ‰ œ È # È È3 Ð4Ñ ˆ 13 ‰ œ 0.5; # ,f $ cos x œ "# #3 ˆx 13 ‰ 4" ˆx 13 ‰ 1#3 ˆx 13 ‰ á Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. È5 Á0 Chapter 10 Additional and Advanced Exercises 643 10. f(x) œ sin x with a œ 21 Ê f(21) œ 0, f w (21) œ 1, f ww (21) œ 0, f www (21) œ 1, f Ð4Ñ (21) œ 0, f Ð5Ñ (21) œ 1, $ & ( f Ð6Ñ (21) œ 0, f Ð7Ñ (21) œ 1; sin x œ (x 21) (x 3!21) (x 5!21) (x 7!21) á # $ 11. ex œ 1 x x#! x3! á with a œ 0 12. f(x) œ ln x with a œ 1 Ê f(1) œ 0, f w (1) œ 1, f ww (1) œ 1,f www (1) œ 2, f Ð4Ñ (1) œ 6; # $ % ln x œ (x 1) (x # 1) (x 3 1) (x 4 1) á 13. f(x) œ cos x with a œ 221 Ê f(221) œ 1, f w (221) œ 0, f ww (221) œ 1, f www (221) œ 0, f Ð4Ñ (221) œ ", f Ð5Ñ (221) œ 0, f Ð6Ñ (221) œ 1; cos x œ 1 "# (x 221)# 4!" (x 221)% 6!" (x 221)' á 14. f(x) œ tan" x with a œ 1 Ê f(1) œ 14 , f w (1) œ "# , f ww (1) œ "# , f www (1) œ "# ; # $ tan" x œ 14 (x 2 1) (x 4 1) (x 121) á n 15. Yes, the sequence converges: cn œ aan bn b1În Ê cn œ b ˆˆ ba ‰ 1‰ ˆ a ‰n ln ˆ a ‰ œ ln b lim ˆb a ‰n 1b œ ln b nÄ_ b 1 În n lnˆˆ ba ‰ 1‰ n nÄ_ Ê n lim c œ ln b lim Ä_ n 0†ln ˆ ba ‰ c œ eln b œ b. 0 1 œ ln b since 0 a b. Thus, n lim Ä_ n _ _ _ 2 16. 1 10 103 # 107 $ 102 % 103 & 107 ' á œ 1 ! 103n2 2 ! 103n3 1 ! 1073n _ _ _ n œ0 n œ0 n œ0 œ 1 ! 103n2 1 ! 103n3 2 ! 103n7 3 œ 1 n œ1 n œ1 2 ‰ ˆ 10 Š 103# ‹ "‰ 1 ˆ 10 $ "‰ 1 ˆ 10 n œ1 Š 107$ ‹ $ "‰ 1 ˆ 10 $ 30 7 999237 œ 1 200 œ 412 999 999 999 œ 999 333 kb1 17. sn œ ! 'k n 1 k œ0 dx 1 x# Ê sn œ '0 1 dxx# '1 1 dxx# á 'nc1 1 dxx# Ê sn œ '0 1 dxx# 1 2 n n Ê n lim s œ n lim atan" n tan" 0b œ 1# Ä_ n Ä_ nb1 # 1) 18. n lim † (n 1)(2x ¹ uunbn 1 ¹ œ n lim ¹ (n 1)x ¹ œ n lim ¹ x † (n 1) ¹ œ ¸ 2x x 1 ¸ 1 nxn Ä_ Ä _ (n 2)(2x 1)n 1 Ä _ 2x 1 n(n 2) n Ê kxk k2x 1k ; if x 0, kxk k2x 1k Ê x 2x 1 Ê x 1; if "# x 0, kxk k2x 1k Ê x 2x 1 Ê 3x 1 Ê x "3 ; if x #" , kxk k2x 1k Ê x 2x 1 Ê x 1. Therefore, the series converges absolutely for x 1 and x "3 . 19. (a) No, the limit does not appear to depend on the value of the constant a (b) Yes, the limit depends on the value of b (c) cos ˆ a ‰ n s œ Š1 n n ‹ œ n lim Ä_ a â‰ â‰ 01 n sin n cos n cos ˆ na ‰ 10 1 n a‰ n ˆ cos n 1Îb lim Š1 nÄ_ Ê ln s œ ln Œ1 œ bn cos ˆ na ‰ n ˆ n" ‰ Ê n lim ln s œ Ä_ œ 1 Ê n lim sœe Ä_ " 1 " cos ˆ na ‰ Œ n a a a n sin ˆ n ‰ cos ˆ n ‰ n# Š "# ‹ n ¸ 0.3678794412; similarly, ‹ œe n 1În _ an ‰ 20. ! an converges Ê n lim a œ 0; n lim ’ˆ 1 sin “ # Ä_ n Ä_ n œ1 an ‰ ˆ 1sin œ n lim œ # Ä_ Ä_ 1sin Šnlim an ‹ # œ "# Ê the series converges by the nth-Root Test Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 0 œ 1sin # 644 Chapter 10 Infinite Sequences and Series nb1 nb1 21. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ b x † ln n ¹ 1 Ê kbxk 1 Ê b" x b" œ 5 Ê b œ „ "5 Ä_ Ä _ ln (n 1) bn xn 22. A polynomial has only a finite number of nonzero terms in its Taylor series, but the functions sin x, ln x and ex have infinitely many nonzero terms in their Taylor expansions. $ $ $ Šax a 3!x á‹ Šx x3! á‹ x $ & sin (ax) sin xx œ lim œ lim ’ a x# 2 a3! 3!" Š a5! 5!" ‹ x# á “ $ $ x x xÄ0 xÄ0 xÄ0 $ xx is finite if a 2 œ 0 Ê a œ 2; lim sin 2x xsin œ 23! 3!" œ 76 $ xÄ0 23. lim # # 24. lim cos#axx# b œ 1 xÄ0 Ê lim Ê b œ 1 and a œ „ 2 25. (a) (b) #x# xÄ0 (n 1)# un œ 1 2n n"# unb1 œ n# un n1 1 0 unb1 œ n œ 1 n n# % % a x a x Œ1 # 4! á b # xÄ0 _ Ê C œ 2 1 and ! n"# converges n œ1 _ Ê C œ 1 Ÿ 1 and ! "n diverges n œ1 6 26. # # œ 1 Ê lim Š "2x#b a4 a48x á ‹ œ 1 5n# 3 Š4‹ Š#‹ 2n(2n 1) un 4n# 2n 5 unb1 œ (2n 1)# œ 4n# 4n 1 œ 1 n 4n# 4n 1 œ 1 n # Ê C œ 3# 1 and kf(n)k œ 4n# 5n4n 1 œ 5 Ÿ5 Š4 4n "# ‹ _ _ n œ1 n œ1 nœ1 n# after long division _ Ê ! un converges by Raabe's Test n _ – Š4n# c 4n b 1‹ — n œ1 27. (a) ! an œ L Ê an# Ÿ an ! an œ an L Ê ! an# converges by the Direct Comparison Test (b) converges by the Limit Comparison Test: n lim Ä_ an Š1c an ‹ _ an " œ n lim œ 1 since ! an converges and Ä _ 1 an # $ n œ1 therefore x lim a œ0 Ä_ n 28. If 0 an 1 then kln (1 an )k œ ln (1 an ) œ an a#n a3n á an an# an$ á œ 1 anan , a positive term of a convergent series, by the Limit Comparison Test and Exercise 27b _ _ n œ1 nœ1 d 29. (1 x)" œ 1 ! xn where kxk 1 Ê (1"x)# œ dx (1 x)" œ ! nxn1 and when x œ #" we have # $ n 1 4 œ 1 2 ˆ "# ‰ 3 ˆ "# ‰ 4 ˆ "# ‰ á n ˆ "# ‰ á _ _ _ _ nœ1 nœ1 nœ1 # 2xx# ! n(n 1)xn1 œ 2 $ Ê ! n(n 1)xn œ 2x $ 30. (a) ! xn1 œ 1xx Ê ! (n 1)xn œ (1 x)# Ê (1x) (1x) nœ1 Ê _ ! n(n n 1) œ x n œ1 2 x 2x# $ œ (x 1)$ , kxk 1 Š1 x" ‹ _ $ # (b) x œ ! n(nxn ") Ê x œ (x 2x 1)$ Ê x 3x x 1 œ 0 Ê x œ 1 Š1 # n œ1 "Î$ È57 "Î$ È Š1 957 ‹ 9 ‹ ¸ 2.769292, using a CAS or calculator _ 31. (a) " d ˆ " ‰ d # $ # $ ! nxn1 (1x)# œ dx 1x œ dx a1 x x x á b œ 1 2x 3x 4x á œ n œ1 _ (b) 2 n 1 from part (a) we have ! n ˆ 56 ‰ ˆ "6 ‰ œ ˆ "6 ‰ ’ 1 "ˆ 5 ‰ “ œ 6 6 n œ1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 10 Additional and Advanced Exercises _ (c) from part (a) we have ! npn1 q œ (1 q p)# œ qq# œ q" n œ1 _ _ k œ1 k œ1 " _ _ _ k œ1 kœ1 kœ1 ˆ ‰ 32. (a) ! pk œ ! 2k œ 1#ˆ " ‰ œ 1 and E(x) œ ! kpk œ ! k2k œ "# ! k21k œ ˆ "# ‰ # " # œ 2 1ˆ "# ‰‘ by Exercise 31(a) _ _ k œ1 k œ1 _ kc1 k ˆ5‰ _ _ k œ1 kœ1 _ kc1 (b) ! pk œ ! 56k œ "5 ! ˆ 65 ‰ œ ˆ 5" ‰ ’ 1 6ˆ 5 ‰ “ œ 1 and E(x) œ ! kpk œ ! k 56k œ 6" ! k ˆ 65 ‰ k œ1 6 k1 k œ1 œ ˆ "6 ‰ " # œ 6 1ˆ 56 ‰‘ _ _ _ kœ1 kœ1 (c) ! pk œ ! k(k"1) œ ! ˆ k" k " 1 ‰ œ kœ1 _ œ! k œ1 _ _ lim ˆ1 k " 1 ‰ œ 1 and E(x) œ ! kpk œ ! k Š k(k " 1) ‹ kÄ_ kœ1 kœ1 " k 1 , a divergent series so that E(x) does not exist C! e kt! ˆ1 e nkt! ‰ 1e kt! kt! ! lim R œ 1C! ee kt! œ ektC! 1 nÄ_ n e 1 a1 e n b e " a1 e "! b " (b) Rn œ 1 e " Ê R" œ e ¸ 0.36787944 and R"! œ 1 e " ¸ 0.58195028; 33. (a) Rn œ C! ekt! C! e2kt! á C! enkt! œ Ê Rœ R œ e" 1 ¸ 0.58197671; R R"! ¸ 0.00002643 Ê R RR"! 0.0001 Þ1n e Þ1 ˆ1 e Þ1n ‰ R , # œ #" ˆ eÞ1 " 1 ‰ ¸ 4.7541659; Rn R# Ê 1eÞ1 e 1 ˆ #" ‰ ˆ eÞ1 " 1 ‰ 1 e Þ1 n n 1 enÎ10 "# Ê enÎ10 "# Ê 10 ln ˆ "# ‰ Ê 10 ln ˆ "# ‰ Ê n 6.93 (c) Rn œ Ê Ê nœ7 ! 34. (a) R œ ektC! Ê Rekt! œ R C! œ CH Ê ekt! œ CCHL Ê t! œ k" ln Š CCHL ‹ 1 " (b) t! œ 0.05 ln e œ 20 hrs 2 ‰ (c) Give an initial dose that produces a concentration of 2 mg/ml followed every t! œ 0.0" # ln ˆ 0.5 ¸ 69.31 hrs by a dose that raises the concentration by 1.5 mg/ml 0.1 ‰ " ‰ (d) t! œ 0.2 ln ˆ 0.03 œ 5 ln ˆ 10 3 ¸ 6 hrs Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 645 646 Chapter 10 Infinite Sequences and Series NOTES: Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 11.1 PARAMETRIZATIONS OF PLANE CURVES 1. x œ 3t, y œ 9t# , _ t _ Ê y œ x# 2. x œ Èt , y œ t, t 0 Ê x œ È y # or y œ x , x Ÿ 0 3. x œ 2t 5, y œ 4t 7, _ t _ Ê x 5 œ 2t Ê 2(x 5) œ 4t Ê y œ 2(x 5) 7 Ê y œ 2x 3 5. x œ cos 2t, y œ sin 2t, 0 Ÿ t Ÿ 1 Ê cos# 2t sin# 2t œ 1 Ê x# y# œ 1 4. x œ 3 3t, y œ 2t, 0 Ÿ t Ÿ 1 Ê y# œ t Ê x œ 3 3 ˆ y# ‰ Ê 2x œ 6 3y Ê y œ 2 23 x, ! Ÿ x Ÿ $ 6. x œ cos (1 t), y œ sin (1 t), 0 Ÿ t Ÿ 1 Ê cos# (1 t) sin# (1 t) œ 1 Ê x# y# œ 1, y ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 648 Chapter 11 Parametric Equations and Polar Coordinates 7. x œ 4 cos t, y œ 2 sin t, 0 Ÿ t Ÿ 21 Ê # 16 cos# t t 4 sin œ1 16 4 Ê y# x# 16 4 œ 1 9. x œ sin t, y œ cos 2t, 12 Ÿ t Ÿ 12 Ê y œ cos 2t œ 1 2sin# t Ê y œ 1 2x2 11. x œ t2 , y œ t6 2t4 , _ t _ 3 2 Ê y œ at2 b 2at2 b Ê y œ x3 2x2 13. x œ t, y œ È1 t# , 1 Ÿ t Ÿ 0 Ê y œ È1 x# 8. x œ 4 sin t, y œ 5 cos t, 0 Ÿ t Ÿ 21 # # # # sin t cos t x Ê 16 16 25 25 œ 1 Ê 16 #y5 œ 1 10. x œ 1 sin t, y œ cos t 2, 0 Ÿ t Ÿ 1 Ê sin# t cos# t œ 1 Ê ax 1b# ay 2b# œ 1 2 12. x œ t t 1 , y œ tt 1 , 1 t 1 2x Ê t œ x x 1 Ê y œ 2x 1 14. x œ Èt 1, y œ Èt, t 0 Ê y# œ t Ê x œ Èy# 1, y Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 0 Section 11.1 Parametrizations of Plane Curves 15. x œ sec# t 1, y œ tan t, 1# t 1# 16. x œ sec t, y œ tan t, 1# t 1# 17. x œ cosh t, y œ sinh t, _ 1 _ Ê cosh# t sinh# t œ 1 Ê x# y# œ 1 18. x œ 2 sinh t, y œ 2 cosh t, _ t _ Ê 4 cosh# t 4 sinh# t œ 4 Ê y# x# œ 4 19. (a) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21 20. (a) x œ a sin t, y œ b cos t, 1# Ÿ t Ÿ 5#1 Ê sec# t 1 œ tan# t Ê x œ y# (b) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21 (c) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41 649 Ê sec# t tan# t œ 1 Ê x# y# œ 1 (d) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41 (b) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 21 (c) x œ a sin t, y œ b cos t, 1# Ÿ t Ÿ 9#1 (d) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 41 21. Using a"ß $b we create the parametric equations x œ " at and y œ $ bt, representing a line which goes through a"ß $b at t œ !. We determine a and b so that the line goes through a%ß "b when t œ ". Since % œ " a Ê a œ &. Since " œ $ b Ê b œ %. Therefore, one possible parameterization is x œ " &t, y œ $ %t, 0 Ÿ t Ÿ ". 22. Using a"ß $b we create the parametric equations x œ " at and y œ $ bt, representing a line which goes through a"ß $b at t œ !. We determine a and b so that the line goes through a$ß #b when t œ ". Since $ œ " a Ê a œ %. Since # œ $ b Ê b œ &. Therefore, one possible parameterization is x œ " %t, y œ $ &t, 0 Ÿ t Ÿ ". 23. The lower half of the parabola is given by x œ y# " for y Ÿ !. Substituting t for y, we obtain one possible parameterization x œ t# ", y œ t, t Ÿ 0Þ 24. The vertex of the parabola is at a"ß "b, so the left half of the parabola is given by y œ x# #x for x Ÿ ". Substituting t for x, we obtain one possible parametrization: x œ t, y œ t# #t, t Ÿ ". 25. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a#ß $b for t œ ! and passes through a"ß "b at t œ ". Then x œ fatb, where fa!b œ # and fa"b œ ". Since slope œ ??xt œ "# "! œ $, x œ fatb œ $t # œ # $t. Also, y œ gatb, where ga!b œ $ and ga"b œ ". 3 Since slope œ ??yt œ " "! œ 4. y œ gatb œ %t $ œ $ %t. One possible parameterization is: x œ # $t, y œ $ %t, t !. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 650 Chapter 11 Parametric Equations and Polar Coordinates 26. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a"ß #b for t œ ! and passes through a!ß !b at t œ ". Then x œ fatb, where fa!b œ " and fa"b œ !. a"b Since slope œ ??xt œ ! "! œ ", x œ fatb œ "t a"b œ " t. Also, y œ gatb, where ga!b œ # and ga"b œ !. # Since slope œ ??yt œ !"! œ #. y œ gatb œ #t # œ # #t. One possible parameterization is: x œ " t, y œ # #t, t 27. Since we only want the top half of a circle, y !. 0, so let x œ 2cos t, y œ 2lsin tl, 0 Ÿ t Ÿ 41 28. Since we want x to stay between 3 and 3, let x œ 3 sin t, then y œ a3 sin tb2 œ 9sin# t, thus x œ 3 sin t, y œ 9sin# t, 0Ÿt_ dy dy x x 29. x# y# œ a# Ê 2x 2y dy dx œ 0 Ê dx œ y ; let t œ dx Ê y œ t Ê x œ yt. Substitution yields y# t# y# œ a# Ê y œ È1a t# and x œ È1att , _ t _ 30. In terms of ), parametric equations for the circle are x œ a cos ), y œ a sin ), 0 Ÿ ) 21. Since ) œ as , the arc length parametrizations are: x œ a cos as , y œ a sin as , and 0 Ÿ as 21 Ê 0 Ÿ s Ÿ 21a is the interval for s. 31. Drop a vertical line from the point ax, yb to the x-axis, then ) is an angle in a right triangle, and from trigonometry we know that tan ) œ yx Ê y œ x tan ). The equation of the line through a0, 2b and a4, 0b is given by y œ 12 x 2. Thus 4 tan ) 1 x tan ) œ 12 x 2 Ê x œ 2 tan4) 1 and y œ 2 tan ) 1 where 0 Ÿ ) 2 . 32. Drop a vertical line from the point ax, yb to the x-axis, then ) is an angle in a right triangle, and from trigonometry we know that tan ) œ yx Ê y œ x tan ). Since y œ Èx Ê y2 œ x Ê ax tan )b2 œ x Ê x œ cot2 ) Ê y œ cot ) where 0 ) Ÿ 12 . 33. The equation of the circle is given by ax 2b2 y2 œ 1. Drop a vertical line from the point ax, yb on the circle to the x-axis, then ) is an angle in a right triangle. So that we can start at a1, 0b and rotate in a clockwise direction, let x œ 2 cos ), y œ sin ), 0 Ÿ ) Ÿ 21. 34. Drop a vertical line from the point ax, yb to the x-axis, then ) is an angle in a right triangle, whose height is y and whose base is x 2. By trigonometry we have tan ) œ x y 2 Ê y œ ax 2b tan ). The equation of the circle is given by x2 y2 œ 1 Ê x2 aax 2btan )b2 œ 1 Ê x2 sec2 ) 4x tan2 ) 4tan2 ) 1 œ 0. Solving for x we obtain xœ 2 4tan2 ) „ Éa4tan2 )b2 4 sec2 ) a4tan2 ) 1b 2È1 3tan2 ) œ 4tan ) „2 sec œ 2sin2 ) „ cos )Ècos2 ) 3sin2 ) 2) 2 sec2 ) œ 2 2cos2 ) „ cos )È4cos2 ) 3 and y œ Š2 2cos2 ) „ cos )È4cos2 ) 3 2‹ tan ) œ 2sin ) cos ) „ sin )È4cos2 ) 3. Since we only need to go from a1, 0b to a0, 1b, let x œ 2 2cos2 ) cos )È4cos2 ) 3, y œ 2sin ) cos ) sin )È4cos2 ) 3, 0 Ÿ ) Ÿ tan1 ˆ 1 ‰. 2 To obtain the upper limit for ), note that x œ 0 and y œ 1, using y œ ax 2b tan ) Ê 1 œ 2 tan ) Ê ) œ tan1 ˆ 12 ‰. 35. Extend the vertical line through A to the x-axis and let C be the point of intersection. Then OC œ AQ œ x 2 2 and tan t œ OC œ x2 Ê x œ tan2 t œ 2 cot t; sin t œ OA Ê OA œ sin2 t ; and (AB)(OA) œ (AQ)# Ê AB ˆ sin2 t ‰ œ x# # 2 2 2 sin t Ê AB ˆ sin t ‰ œ ˆ tan t ‰ Ê AB œ tan# t . Next y œ 2 AB sin t Ê y œ 2 ˆ 2tansin# tt ‰ sin t œ # sin t # # # 2 2tan # t œ 2 2 cos t œ 2 sin t. Therefore let x œ 2 cot t and y œ 2 sin t, 0 t 1 . Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.1 Parametrizations of Plane Curves 36. Arc PF œ Arc AF since each is the distance rolled and Arc PF œ nFCP Ê Arc PF œ b(nFCP); ArcaAF œ ) b Ê Arc AF œ a) Ê a) œ b(nFCP) Ê nFCP œ ba ); nOCG œ 1# ); nOCG œ nOCP nPCE œ nOCP ˆ 1# !‰ . Now nOCP œ 1 nFCP œ 1 ba ). Thus nOCG œ 1 ba ) 1# ! Ê 1# ) œ 1 ba ) 1# ! Ê ! œ 1 ba ) ) œ 1 ˆ ab b )‰ . Then x œ OG BG œ OG PE œ (a b) cos ) b cos ! œ (a b) cos ) b cos ˆ1 a b b )‰ œ (a b) cos ) b cos ˆ a b b )‰ . Also y œ EG œ CG CE œ (a b) sin ) b sin ! œ (a b) sin ) b sin ˆ1 a b b )‰ œ (a b) sin ) b sin ˆ a b b )‰ . Therefore x œ (a b) cos ) b cos ˆ a b b )‰ and y œ (a b) sin ) b sin ˆ a b b )‰ . aâ‰ If b œ 4a , then x œ â 4a ‰ cos ) 4a cos Š ˆ a ‰4 )‹ 4 a 3a a œ 3a 4 cos ) 4 cos 3) œ 4 cos ) 4 (cos ) cos 2) sin ) sin 2)) a # # œ 3a 4 cos ) 4 a(cos )) acos ) sin )b (sin ))(2 sin ) cos ))b a a 2a $ # # œ 3a 4 cos ) 4 cos ) 4 cos ) sin ) 4 sin ) cos ) a 3a $ # $ œ 3a 4 cos ) 4 cos ) 4 (cos )) a1 cos )b œ a cos ); aâ‰ a 3a a y œ â 4a ‰ sin ) 4a sin Š ˆ a ‰4 )‹ œ 3a 4 sin ) 4 sin 3) œ 4 sin ) 4 (sin ) cos 2) cos ) sin 2)) 4 a # # œ 3a 4 sin ) 4 a(sin )) acos ) sin )b (cos ))(2 sin ) cos ))b a a 2a # $ # œ 3a 4 sin ) 4 sin ) cos ) 4 sin ) 4 cos ) sin ) 3a a # $ œ 3a 4 sin ) 4 sin ) cos ) 4 sin ) 3a a # $ $ œ 3a 4 sin ) 4 (sin )) a1 sin )b 4 sin ) œ a sin ). 37. Draw line AM in the figure and note that nAMO is a right angle since it is an inscribed angle which spans the diameter of a circle. Then AN# œ MN# AM# . Now, OA œ a, AN AM a œ tan t, and a œ sin t. Next MN œ OP Ê OP# œ AN# AM# œ a# tan# t a# sin# t Ê OP œ Èa# tan# t a# sin# t # sin t œ (a sin t)Èsec# t 1 œ acos t . In triangle BPO, $ sin t # x œ OP sin t œ acos t œ a sin t tan t and y œ OP cos t œ a sin# t Ê x œ a sin# t tan t and y œ a sin# t. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 651 652 Chapter 11 Parametric Equations and Polar Coordinates 38. Let the x-axis be the line the wheel rolls along with the y-axis through a low point of the trochoid (see the accompanying figure). Let ) denote the angle through which the wheel turns. Then h œ a) and k œ a. Next introduce xw yw -axes parallel to the xy-axes and having their origin at the center C of the wheel. Then xw œ b cos ! and yw œ b sin !, where ! œ 3#1 ). It follows that xw œ b cos ˆ 3#1 )‰ œ b sin ) and yw œ b sin ˆ 3#1 )‰ œ b cos ) Ê x œ h xw œ a) b sin ) and y œ k yw œ a b cos ) are parametric equations of the trochoid. # # # 39. D œ É(x 2)# ˆy "# ‰ Ê D# œ (x 2)# ˆy "# ‰ œ (t 2)# ˆt# "# ‰ Ê D# œ t% 4t 17 4 # Ê d adtD b œ 4t$ 4 œ 0 Ê t œ 1. The second derivative is always positive for t Á 0 Ê t œ 1 gives a local minimum for D# (and hence D) which is an absolute minimum since it is the only extremum Ê the closest point on the parabola is (1ß 1). # # # 40. D œ Éˆ2 cos t 34 ‰ (sin t 0)# Ê D# œ ˆ2 cos t 34 ‰ sin# t Ê d adtD b œ 2 ˆ2 cos t 34 ‰ (2 sin t) 2 sin t cos t œ (2 sin t) ˆ3 cos t 3# ‰ œ 0 Ê 2 sin t œ 0 or 3 cos t 3# œ 0 # # # # Ê t œ 0, 1 or t œ 13 , 531 . Now d dtaD# b œ 6 cos# t 3 cos t 6 sin# t so that d dtaD# b (0) œ 3 Ê relative # # # # maximum, d dtaD# b (1) œ 9 Ê relative maximum, d dtaD# b ˆ 13 ‰ œ 92 Ê relative minimum, and d # aD # b ˆ 5 1 ‰ œ 9# dt# 3 Ê relative minimum. Therefore both t œ 13 and t œ 531 give points on the ellipse closest to È È the point ˆ 34 ß !‰ Ê Š1ß #3 ‹ and Š1ß #3 ‹ are the desired points. 41. (a) (b) (c) 42. (a) (b) (c) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.1 Parametrizations of Plane Curves 43. 44. (a) (b) (c) 45. (a) (b) 46. (a) (b) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 653 654 Chapter 11 Parametric Equations and Polar Coordinates 47. (a) (b) (c) 48. (a) (b) (c) (d) 11.2 CALCULUS WITH PARAMETRIC CURVES dy dy dy/dt 2 cos t 1. t œ 14 Ê x œ 2 cos 14 œ È2, y œ 2 sin 14 œ È2; dx dt œ 2 sin t, dt œ 2 cos t Ê dx œ dx/dt œ 2 sin t œ cot t Ê dy dx ¹ # w tœ 14 # œ cot 14 œ 1; tangent line is y È2 œ 1 Šx È2‹ or y œ x 2È2 ; dy dt œ csc t w # # /dt d y " csc t Ê ddxy# œ dy dx/dt œ 2 sin t œ 2 sin$ t Ê dx# ¹ tœ 14 œ È 2 È 2. t œ 6" Ê x œ sin ˆ21 ˆ 6" ‰‰ œ sin ˆ 13 ‰ œ #3 , y œ cos ˆ21 ˆ 6" ‰‰ œ cos ˆ 13 ‰ œ #" ; dx dt œ 21 cos 21t, dy dy sin 21t Ê dx œ 2211cos 21t œ tan 21t Ê dx ¹ dy dt œ 21 sin 21t È tœc 16 œ tan ˆ21 ˆ 6" ‰‰ œ tan ˆ 13 ‰ œ È3; w # # d y 21 sec 21t # tangent line is y "# œ È3 ’x Š #3 ‹“ or y œ È3x 2; dy dt œ 21 sec 21t Ê dx# œ 21 cos 21t # œ cos$"21t Ê ddxy# ¹ tœc 16 œ 8 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.2 Calculus With Parametric Curves dy dy dy/dt 2 sin t 3. t œ 14 Ê x œ 4 sin 14 œ 2È2, y œ 2 cos 14 œ È2; dx dt œ 4 cos t, dt œ 2 sin t Ê dx œ dx/dt œ 4 cos t œ "# tan t Ê dy dx ¹ dyw " # dt œ # sec t tœ 14 œ "# tan 14 œ "# ; tangent line is y È2 œ "# Šx 2È2‹ or y œ "# x 2È2 ; # w d y dy /dt Ê dx # œ dx/dt œ "# sec# t " 4 cos t œ 8 cos$ t # d y Ê dx #¹ È tœ 14 œ 42 È È 3 sin t dy dy È È3 4. t œ 231 Ê x œ cos 231 œ "# , y œ È3 cos 231 œ #3 ; dx dt œ sin t, dt œ 3 sin t Ê dx œ sin t œ Ê dy dx ¹ w È tœ 231 # Ê ddxy# ¹ # d y 0 œ È3 ; tangent line is y Š #3 ‹ œ È3 x ˆ #" ‰‘ or y œ È3 x; dy dt œ 0 Ê dx# œ sin t œ 0 œ0 tœ 231 dy dy dy/dt dy " 1 5. t œ 14 Ê x œ 14 , y œ "# ; dx dt œ 1, dt œ #Èt Ê dx œ dx/dt œ 2Èt Ê dx ¹ w # tœ 14 œ " #É "4 œ 1; tangent line is w # /dt " $Î# " $Î# y "# œ 1 † ˆx 4" ‰ or y œ x 4" ; dy Ê ddxy# œ dy Ê ddxy# ¹ dt œ 4 t dx/dt œ 4 t tœ 14 œ 2 dy # # 6. t œ 14 Ê x œ sec# ˆ 14 ‰ 1 œ 1, y œ tan ˆ 14 ‰ œ 1; dx dt œ 2 sec t tan t, dt œ sec t # dy " " sec t Ê dy dx œ 2 sec# t tan t œ 2 tan t œ # cot t Ê dx ¹ tœc 14 œ "# cot ˆ 14 ‰ œ "# ; tangent line is w " csc# t # d y " " # $ # y (1) œ "# (x 1) or y œ "# x "# ; dy dt œ # csc t Ê dx# œ 2 sec# t tan t œ 4 cot t # Ê ddxy# ¹ tœc 14 œ "4 dy dy dy/dt # 7. t œ 16 Ê x œ sec 16 œ È23 , y œ tan 16 œ È"3 ; dx dt œ sec t tan t, dt œ sec t Ê dx œ dx/dt # dy œ secsect tant t œ csc t Ê dx ¹ dyw dt œ csc t cot t tœ 16 œ csc 16 œ 2; tangent line is y È"3 œ 2 Šx È23 ‹ or y œ 2x È3 ; # w # /dt d y csc t cot t $ Ê ddxy# œ dy dx/dt œ sec t tan t œ cot t Ê dx# ¹ tœ 16 œ 3È3 ˆ 3 ‰ (3t) "Î# " "Î# dy # 8. t œ 3 Ê x œ È3 1 œ 2, y œ È3(3) œ 3; dx , dt œ 3# (3t)"Î# Ê dy dt œ # (t 1) dx œ ˆ " ‰ (t1) "Î# # È œ 3 Èt3t 1 œ dy dx ¹ È tœ3 31 œ 3È3(3) œ 2; tangent line is y 3 œ 2[x (2)] or y œ 2x 1; È3t 3 (t 1) "Î# ‘3Èt 1 3 (3t) "Î# ‘ # # œ 2tÈ3t3Èt1 dt œ 3t dyw # Ê ddxy# ¹ tœ3 # Ê ddxy# œ Š 2tÈ3t3Èt 1 ‹ Š 2Èt1 1 ‹ œ tÈ33t œ "3 $ dy dy dy/dt dy 4t $ # 9. t œ 1 Ê x œ 5, y œ 1; dx dt œ 4t, dt œ 4t Ê dx œ dx/dt œ 4t œ t Ê dx ¹ w # w # d y dy /dt d y " 2t y 1 œ 1 † (x 5) or y œ x 4; dy dt œ 2t Ê dx# œ dx/dt œ 4t œ # Ê dx# ¹ dy " dy " 10. t œ 1 Ê x œ 1, y œ 2; dx dt œ t# , dt œ t Ê dx œ w # tœc1 ˆ "t ‰ Š t"# ‹ d y y (2) œ 1(x 1) or y œ x 1; dy dt œ 1 Ê dx# œ œ t Ê dy dx ¹ 1 œ t# Š t"# ‹ tœc1 tœ1 œ (1)# œ 1; tangent line is œ "# œ 1; tangent line is # Ê ddxy# ¹ tœ1 œ1 È dy dy dy/dt 11. t œ 13 Ê x œ 13 sin 13 œ 13 #3 , y œ 1 cos 13 œ 1 #" œ #" ; dx dt œ 1 cos t, dt œ sin t Ê dx œ dx/dt È Š #3 ‹ È sin ˆ 1 ‰ dy œ 1 sincost t Ê dx ¹ 1 œ 1cos 3ˆ 1 ‰ œ ˆ " ‰ œ È3 ; tangent line is y "# œ È3 Šx 13 #3 ‹ tœ 3 3 # Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 655 656 Chapter 11 Parametric Equations and Polar Coordinates w È # w 1 ˆ ‰ (1 cos t)(cos t) (sin t)(sin t) d y dy /dt 1 1 cos t Ê y œ È3x 1 3 3 2; dy œ 1 cos dt œ (1cos t)# t Ê dx# œ dx/dt œ 1 cos t # d y 1 œ (1 cos t)# Ê dx# ¹ tœ 13 œ 4 dy dy cos t 12. t œ 12 Ê x œ cos 12 œ 0, y œ 1 sin 12 œ 2; dx dt œ sin t, dt œ cos t Ê dx œ sin t œ cot t dy Ê dx ¹ w tœ 2 1 # # # d y d y csc t # $ œ cot 1# œ 0; tangent line is y œ 2; dy dt œ csc t Ê dx# œ sin t œ csc t Ê dx# ¹ 2 at 1b dy dy dy 1 1 13. t œ 2 Ê x œ 2 1 1 œ 31 , y œ 2 2 1 œ 2; dx dt œ at 1b2 , dt œ at 1b2 Ê dx œ at 1b2 Ê dx ¹ tœ 12 œ 1 2 tœ2 1b œ aa22 œ 9; 1 b2 w 4 at 1 b 4 at 1 b3 4a 2 1b3 d# y d# y tangent line is y œ 9x 1; dy dt œ at 1b3 Ê dx# œ at 1b3 Ê dx# ¹ tœ2 œ a2 1b3 œ 108 dy dy e t dy t 14. t œ 0 Ê x œ 0 e0 œ 1, y œ 1 e0 œ 0; dx dt œ 1 e , dt œ e Ê dx œ 1 et Ê dx ¹ t w # # d y d y e e tangent line is y œ 12 x 12 ; dy dt œ a1 et b2 Ê dx# œ a1 et b3 Ê dx# ¹ t t œ 1ee0 œ 12 ; 0 tœ0 e œ a1 œ 18 e 0 b3 0 tœ0 dx 4t 2 dx 15. x3 2t# œ 9 Ê 3x2 dx dt 4t œ 0 Ê 3x dt œ 4t Ê dt œ 3x2 ; dy dy dy/dt 6t t 2y$ 3t# œ 4 Ê 6y# dy dt 6t œ 0 Ê dt œ 6y# œ y# ; thus dx œ dx/dt œ # Š yt# ‹ t(3x2 ) 3x2 œ y# (4t) œ 4y# ; t œ 2 Š c4t ‹ 3x2 Ê x 2(2) œ 9 Ê x 8 œ 9 Ê x œ 1 Ê x œ 1; t œ 2 Ê 2y 3(2)# œ 4 3 3 $ 3 Ê 2y$ œ 16 Ê y$ œ 8 Ê y œ 2; therefore dy dx ¹ " ˆ Èt‰ 16. x œ É5 Èt Ê dx dt œ # 5 "Î# 2 tœ2 3 œ 34aa"#bb# œ 16 ˆ "# t"Î# ‰ œ " 4È t É 5 È t " "Î# ; y(t 1) œ Èt Ê y (t 1) dy dt œ # t " #yÈt " dy " #yÈt 4Èt É5 Èt " #yÈt dy dy #Èt y #tÈt 2Èt " dt Ê at 1b dy œ È † " dt œ #Èt y Ê dt œ at 1b œ #tÈt 2Èt ; thus dx œ dx œ " dt œ 4 Èt É5 Èt # tat" b #ˆ" #yÈt‰É& Èt ; t œ 4 Ê x œ É5 È4 œ È3; t œ 4 Ê y † 3 œ È4 Ê y œ 23 "t therefore, dy dx ¹ tœ4 œ 2Š" 2a 23 bÈ4‹É& È4 "4 œ 10È3 9 dx 2t1 "Î# dx "Î# ‰ dx È ˆ È 17. x 2x$Î# œ t# t Ê dx dt 3x dt œ 2t 1 Ê 1 3x dt œ 2t 1 Ê dt œ 13x"Î# ; y t 1 2t y œ 4 Èt 1 y ˆ " ‰ (t 1)"Î# 2Èy 2t ˆ " y"Î# ‰ dy œ 0 Ê dy Èt 1 y 2Èy Š t ‹ dy œ 0 Ê dy Èy # # dt dt dt dt 2È t 1 y dy È Ê ŠÈt 1 Èt y ‹ dy dt œ 2Èt1 2 y Ê dt œ cyÈy c 4yÈt b 1 dy dy/dt dx œ dx/dt œ Œ 2Èy (t b 1) b 2tÈt b 1 Š 2t b 1 ‹ 1 b 3x"Î# Š 2Èct yb 1 2Èy‹ ŠÈt 1 Èy ‹ t œ yÈ y 4yÈt 1 ; thus 2È y (t 1) 2tÈt 1 ; t œ 0 Ê x 2x$Î# œ 0 Ê x ˆ1 2x"Î# ‰ œ 0 Ê x œ 0; t œ 0 tœ0 œ È4 4(4)È0 1 Œ 2È4(0 1) 2(0)È0 1 4 Ê yÈ0 1 2(0)Èy œ 4 Ê y œ 4; therefore dy dx ¹ 2(0) 1 Œ 1 3(0)"Î# œ 6 dx dx dx 1 x cos t 18. x sin t 2x œ t Ê dx dt sin t x cos t 2 dt œ 1 Ê (sin t 2) dt œ 1 x cos t Ê dt œ sin t2 ; dy sin t t cos t 2 t sin t 2t œ y Ê sin t t cos t 2 œ dy ; t œ 1 Ê x sin 1 2x œ 1 dt ; thus dx œ ˆ 1 c x cos t ‰ sin t b 2 Ê sin 1 1 cos 1 2 x œ 1# ; therefore dy œ 2411 8 œ 4 dx ¹ tœ1 œ 1 Š 1 ‹ cos 1 – # sin 1 2 — Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.2 Calculus With Parametric Curves 657 dy dy dx 2 2 2 2 19. x œ t3 t, y 2t3 œ 2x t2 Ê dx dt œ 3t 1, dt 6t œ 2 dt 2t Ê dt œ 2a3t 1b 2t 6t œ 2t 2 dy 2t 2 Ê dy dx œ 3t2 1 Ê dx ¹ tœ1 œ 32aa11bb221 œ 1 dy 1 ˆ dx dx dx t t ‰ 20. t œ lnax tb, y œ t et Ê 1 œ x t dt 1 Ê x t œ dt 1 Ê dt œ x t 1, dt œ t e e ; dy te e Ê dy dx œ x t 1 ; t œ 0 Ê 0 œ lnax 0b Ê x œ 1 Ê dx ¹ t 21. A œ '0 21 œ a10be0e1 œ 12 0 t tœ0 0 y dx œ '0 aa1 cos tbaa1 cos tbdt œ a2 '0 a1 cos tb2 dt œ a2 '0 a1 2cos t cos2 tbdt 21 21 21 2t ‰ œ a2 '0 ˆ1 2cos t 1 cos dt œ a2 '0 ˆ 23 2cos t 12 cos 2t‰dt œ a2 ” 23 t 2sin t 14 sin 2t• 2 21 21 21 0 œ a2 a31 0 0b 0 œ 31 a2 22. A œ '0 x dy œ '0 at t2 baet bdt ”u œ t t2 Ê du œ a1 2tbdt; dv œ aet bdt Ê v œ et • 1 1 1 œ et at t2 bº '0 et a1 2tbdt 1 t t ”u œ 1 2t Ê du œ 2dt; dv œ e dt Ê v œ e • 0 1 1 0 0 œ et at t2 bº ”et a1 2tbº '0 2et dt• œ ”et at t2 b et a1 2tb 2et •º 1 1 0 œ ae1 a0b e1 a1b 2e1 b ae0 a0b e0 a1b 2e0 b œ 1 3e1 œ 1 3e 23. A œ 2'1 y dx œ 2'1 ab sin tbaa sin tbdt œ 2ab'0 sin2 t dt œ 2ab'0 0 1 0 1 1 ' 1 1 cos 2t dt œ ab 0 a1 cos 2tb dt 2 œ ab’t 12 sin 2t“ œ abaa1 0b !b œ 1 ab 0 24. (a) x œ t2 , y œ t6 , 0 Ÿ t Ÿ 1 Ê A œ '0 y dx œ '0 at6 b2t dt œ '0 2t7 dt œ ’ 14 t8 “ œ 14 0 œ 14 1 1 1 1 (b) x œ t , y œ t , 0 Ÿ t Ÿ 1 Ê A œ '0 y dx œ '0 at b3t dt œ '0 3t 3 1 9 1 9 2 1 11 0 1 1 12 dt œ ’ 4 t “ 0 œ 14 0 œ 14 # # dy ˆ dx ‰ Š dy Éasin tb# a1 cos tb# œ È2 2 cos t 25. dx dt œ sin t and dt œ 1 cos t Ê Ê dt dt ‹ œ cos t ‰ È2 ' É sin t dt Ê Length œ '0 È2 2 cos t dt œ È2 '0 Éˆ 11 cos t (1 cos t) dt œ 1 cos t 0 1 1 t œ È2 '0 È1sin dt (since sin t cos t 1 1 # 0 on [0ß 1]); [u œ 1 cos t Ê du œ sin t dt; t œ 0 Ê u œ 0, t œ 1 Ê u œ 2] Ä È2 '0 u"Î# du œ È2 2u"Î# ‘ ! œ 4 2 # # # dy # ˆ dx ‰ Š dy Éa3t# b# (3t)# œ È9t% 9t# œ 3tÈt# 1 Šsince t 26. dx dt œ 3t and dt œ 3t Ê Ê dt dt ‹ œ È3 Ê Length œ '0 Ä '1 4 0 on ’0ß È3“‹ 3tÈt# 1 dt; ’u œ t# 1 Ê 3# du œ 3t dt; t œ 0 Ê u œ 1, t œ È3 Ê u œ 4“ % 3 "Î# du œ u$Î# ‘ " œ (8 1) œ 7 # u # dy "Î# Èt# a2t 1b œ Éat 1b# œ kt 1k œ t 1 since 0 Ÿ t Ÿ 4 ‰# Š dy 27. dx Ê Êˆ dx dt œ t and dt œ (2t 1) dt dt ‹ œ Ê Length œ '0 at 1b dt œ ’ t2 t“ œ a8 4b œ 12 4 # % ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 658 Chapter 11 Parametric Equations and Polar Coordinates # "Î# ˆ dx ‰# Š dy Éa2t 3b a1 tb# œ Èt# 4t 4 œ kt 2k œ t 2 28. dx and dy dt œ a2t 3b dt œ 1 t Ê Ê dt dt ‹ œ since 0 Ÿ t Ÿ 3 Ê Length œ '0 (t 2) dt œ ’ t2 2t“ œ 21 # 3 3 # ! # # dy ˆ dx ‰ Š dy Éa8t cos tb# a8t sin tb# œ È64t# cos# t 64t# sin# t 29. dx dt œ 8t cos t and dt œ 8t sin t Ê Ê dt dt ‹ œ œ k8tk œ 8t since 0 Ÿ t Ÿ 1# Ê Length œ '0 1Î2 1Î# 8t dt œ c4t# d ! œ 1# # # dy # ˆ " ‰ ˆ dx ‰ Š dy 30. dx dt œ sec ttan t asec t tan t sec tb cos t œ sec t cos t and dt œ sin t Ê Ê dt dt ‹ œ Éasec t cos tb# asin tb# œ Èsec# t 1 œ Ètan# t œ ktan tk œ tan t since 0 Ÿ t Ÿ 13 Ê Length œ '0 1Î3 tan t dt œ '0 1Î3 1Î$ sin t œ ln "# ln 1 œ ln 2 cos t dt œ c ln kcos tkd ! dy ˆ dx ‰ Š dy Éasin tb# acos tb# œ 1 Ê Area œ ' 21y ds 31. dx dt œ sin t and dt œ cos t Ê Ê dt dt ‹ œ # # œ '0 21a2 sin tba1bdt œ 21 c2t cos td #!1 œ 21[a41 1b a0 1b] œ 81# 21 "Î# "Î# Èt t" œ É t# 1 Ê Area œ ' 21x ds ‰# Š dy 32. dx and dy Ê Êˆ dx dt œ t dt œ t dt dt ‹ œ t # È3 œ '0 È3 21 ˆ 23 t$Î# ‰ É t t " dt œ 431 '0 # tÈt# 1 dt; cu œ t# 1 Ê du œ 2t dt; t œ 0 Ê u œ 1, ’t œ È3 Ê u œ 4“ Ä '1 231 Èu du œ 491 u$Î# ‘ " œ 2891 4 È3 % Note: '0 21 ˆ 23 t$Î# ‰ É t t 1 dt is an improper integral but limb fatb exists and is equal to 0, where tÄ! # # fatb œ 21 ˆ 23 t$Î# ‰ É t t " . Thus the discontinuity is removable: define Fatb œ fatb for t 0 and Fa0b œ 0 È3 Ê '0 Fatb dt œ 2891 . # # dy È2 Ê Êˆ dx ‰# Š dy ‹ œ Ê1# Št È2‹ œ Ét# 2È2 t 3 Ê Area œ ' 21x ds 33. dx dt œ 1 and dt œ t dt dt È2 œ 'cÈ2 21 Št È2‹ Ét# 2È2 t 3 dt; ’u œ t# 2È2 t 3 Ê du œ Š2t 2È2‹ dt; t œ È2 Ê u œ 1, ’t œ È2 Ê u œ 9“ Ä '1 1Èu du œ 23 1u$Î# ‘ " œ 231 a27 1b œ 5231 9 * ' 21y ds œ '0 ‰ Š dy 34. From Exercise 30, Êˆ dx dt dt ‹ œ tan t Ê Area œ # 1Î$ œ 21 c cos td ! # 1 Î3 21 cos t tan t dt œ 21 '0 1 Î3 sin t dt œ 21 "# (1)‘ œ 1 dy È2# 1# œ È5 Ê Area œ ' 21y ds œ ' 21at 1bÈ5 dt ˆ dx ‰# Š dy 35. dx dt œ 2 and dt œ 1 Ê Ê dt dt ‹ œ 0 # # 1 " œ 21È5 ’ t2 t“ œ 31È5. Check: slant height is È5 Ê Area is 1a1 2bÈ5 œ 31È5 . ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.2 Calculus With Parametric Curves 659 dy Èh# r# Ê Area œ ' 21y ds œ ' 21rtÈh# r# dt ˆ dx ‰# Š dy 36. dx dt œ h and dt œ r Ê Ê dt dt ‹ œ 0 # 1 œ 21rÈh# r# '0 t dt œ 21rÈh# r# ’ t2 “ œ 1rÈh# r# . Check: slant height is Èh# r# Ê Area is 1 # " ! 1rÈh# r# . dy 37. Let the density be $ œ 1. Then x œ cos t t sin t Ê dx dt œ t cos t, and y œ sin t t cos t Ê dt œ t sin t # # È(t cos t)# (t sin t)# œ ktk dt œ t dt since 0 Ÿ t Ÿ 1 . The curve's mass is ‰ Š dy Ê dm œ 1 † ds œ Êˆ dx dt dt ‹ dt œ # M œ ' dm œ '0 1Î2 1Î2 1Î2 # t dt œ 18 . Also Mx œ ' µ y dm œ '0 asin t t cos tb t dt œ '0 t sin t dt '0 t# cos t dt 1Î2 1Î# œ csin t t cos td ! # x yœ M M œ Š3 14 ‹ 1 Î2 1Î# œ ccos t t sin td ! y xœ M M œ # œ 3 14 , where we integrated by parts. Therefore, ' µx dm œ '0 œ 24 1# 2. Next, My œ # Š 18 ‹ ˆ 3#1 3‰ # Š 18 ‹ 1Î# ct# sin t 2 sin t 2t cos td ! 1Î# ct# cos t 2 cos t 2t sin td ! acos t t sin tb t dt œ '0 1 Î2 t cos t dt '0 1 Î2 t# sin t dt œ 3#1 3, again integrating by parts. Hence 24 ˆ 12 24 24 ‰ œ 12 1 1# . Therefore axß yb œ 1 1# ß 1# 2 . dy t t t t t 38. Let the density be $ œ 1. Then x œ et cos t Ê dx dt œ e cos t e sin t, and y œ e sin t Ê dt œ e sin t e cos t # ‰# Š dy Éaet cos t et sin tb# aet sin t et cos tb# dt œ È2e2t dt œ È2 et dt. Ê dm œ 1 † ds œ Êˆ dx dt dt ‹ dt œ 1 1 The curve's mass is M œ ' dm œ '0 È2 et dt œ È2 e1 È2 . Also Mx œ ' µ y dm œ '0 aet sin tb ŠÈ2 et ‹ dt È2 Š e21 " ‹ x œ '0 È2 e2t sin t dt œ È2 ’ e5 (2 sin t cos t)“ œ È2 Š e5 5" ‹ Ê y œ M M œ È 1 1 2t 21 ! 5 5 2 e1 È 2 21 œ 5 eae1 "1b . 1 1 1 2t 21 Next My œ ' µ x dm œ '0 aet cos tb ŠÈ2 et ‹ dt œ '0 È2 e2t cos t dt œ È2 ’ e5 a2 cos t sin tb“ œ È2 Š 2e5 52 ‹ ! 21 y Ê xœ M M œ È2 Š 2e5 25 ‹ È 2 e1 È 2 21 21 21 œ 52eae1 12b . Therefore axß yb œ Š 52eae1 12b ß 5 eae1 11b ‹. dy 39. Let the density be $ œ 1. Then x œ cos t Ê dx dt œ sin t, and y œ t sin t Ê dt œ 1 cos t # ‰# Š dy Éasin tb# a1 cos tb# dt œ È2 2 cos t dt. The curve's mass Ê dm œ 1 † ds œ Êˆ dx dt dt ‹ dt œ is M œ ' dm œ '0 È2 2 cos t dt œ È2'0 È1 cos t dt œ È2 '0 É2 cos# ˆ #t ‰ dt œ 2 '0 ¸cos ˆ #t ‰¸ dt 1 1 1 1 œ 2 '0 cos ˆ #t ‰ dt ˆsince 0 Ÿ t Ÿ 1 Ê 0 Ÿ #t Ÿ 1# ‰ œ 2 2 sin ˆ 2t ‰‘ ! œ 4. Also Mx œ ' µ y dm 1 1 œ '0 at sin tb ˆ2 cos #t ‰ dt œ '0 2t cos ˆ #t ‰ dt '0 2 sin t cos ˆ #t ‰ dt 1 1 1 1 1 Mx œ 2 4 cos ˆ 2t ‰ 2t sin ˆ #t ‰‘ ! 2 "3 cos ˆ #3 t‰ cos ˆ "# t‰‘ ! œ 41 16 3 Ê yœ M œ ˆ41 16 ‰ 3 œ 1 43 . 4 1 1 sin ˆ 3 t‰ cos t cos ˆ #t ‰ dt œ 2 ’sin ˆ 2t ‰ 3# “ œ 2 23 0 ! Next My œ ' µ x dm œ '0 acos tbˆ2 cos #t ‰ dt œ ' 1 ˆ 43 ‰ " 4 œ 3 . y œ 43 Ê x œ M M œ Therefore axß yb œ ˆ 3" ß 1 43 ‰. # dy 3t # 40. Let the density be $ œ 1. Then x œ t$ Ê dx dt œ 3t , and y œ # Ê dt œ 3t Ê dm œ 1 † ds # ‰# Š dy Éa3t# b# (3t)# dt œ 3 ktk Èt# 1 dt œ 3tÈt# 1 dt since 0 Ÿ t Ÿ È3. The curve's mass œ Êˆ dx dt dt ‹ dt œ È3 is M œ ' dm œ '0 È3 œ #9 '0 $Î# 3tÈt# 1 dt œ ’at# 1b “ È3 ! È 3 # œ 7. Also Mx œ ' µ y dm œ '0 3t# Š3tÈt# 1‹ dt Mx 17.4 ' µx dm t$ Èt# 1 dt œ 87 5 œ 17.4 (by computer) Ê y œ M œ 7 ¸ 2.49. Next My œ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 660 Chapter 11 Parametric Equations and Polar Coordinates È3 È3 œ '0 t$ † 3t Èat# 1b dt œ 3 '0 t% Èt# 1 dt ¸ 16.4849 (by computer) Ê x œ My œ 16.4849 ¸ 2.35. 7 M Therefore, axß yb ¸ a2.35ß 2.49b. 41. (a) ‰ Š dy Éa2 sin 2tb# a2 cos 2tb# œ 2 Ê Êˆ dx dt dt ‹ œ Ê Length œ '0 œ1 1Î2 (b) # # dy dx dt œ 2 sin 2t and dt œ 2 cos 2t 1Î# 2 dt œ c2td ! dy dx ˆ dx ‰# Š dy dt œ 1 cos 1t and dt œ 1 sin 1t Ê Ê dt dt ‹ # œ Éa1 cos 1tb# a1 sin 1tb# œ 1 Ê Length œ '1Î2 1 dt œ c1td "Î# œ 1 1 Î2 "Î# dy w 42. (a) x œ gayb has the parametrization x œ gayb and y œ y for c Ÿ y Ÿ d Ê dx dy œ g ayb and dy œ 1; then dx dx ' ' È1 [gw ayb]# dy Length œ 'c ÊŠ dy dy ‹ Š dy ‹ dy œ c Ê1 Š dy ‹ dy œ c d # # # d 3 1Î2 (b) x œ y3Î2 , 0 Ÿ y Ÿ 43 Ê dx Ê L œ '0 dy œ 2 y 4 Î3 d É1 ˆ 32 y1Î2 ‰# dy œ ' 4Î3 0 É1 94 y dy œ ” 49 † 23 ˆ1 94 y‰ 3Î2 4Î3 • 0 8 8 œ 27 a4b3Î2 27 a1b3Î2 œ 56 27 dx (c) x œ 23 y2Î3 , 0 Ÿ y Ÿ 1 Ê dy œ y1Î3 Ê L œ '0 É1 ay1Î3 b dy œ '0 É1 y21Î3 dy œ lim 'a É y y2Î3 1 dy 1 1 # 1 2Î3 a Ä0 œ lim 32 'a ˆy2Î3 1‰ 1 1 Î2 a Ä0 1 ˆ 23 y1Î3 ‰ dy œ lim ” 32 † 23 ˆy2Î3 1‰3Î2 • œ lim Ša2b3Î2 â2Î3 1‰3Î2 ‹ œ 2È2 1 a Ä0 a Ä0 a dy 2 43. x œ a1 2 sin )bcos ), y œ a1 2 sin )bsin ) Ê dx d) œ 2cos ) sin )a1 2 sin )b, d) œ 2cos ) sin ) cos )a1 2 sin )b 2cos ) sin ) cos )a1 2 sin )b 4cos ) sin ) cos ) 2 sin 2) cos ) Ê dy œ 2cos 2 ) 2sin2 ) sin ) œ 2 cos 2) sin ) dx œ 2cos2 ) sin )a1 2 sin )b (a) x œ a1 2 sina0bbcosa0b œ 1, y œ a1 2 sina0bbsina0b œ 0; dy dx º )œ0 (b) sina2a0bb cosa0b 01 1 œ 22 cos a2a0bb sina0b œ 2 0 œ 2 2 sinˆ2ˆ 1 ‰‰ cosˆ 1 ‰ x œ ˆ1 2 sinˆ 1# ‰‰cosˆ 1# ‰ œ 0, y œ ˆ1 2 sinˆ 1# ‰‰sinˆ 1# ‰ œ 3; dy œ 2 cosˆ2ˆ#1 ‰‰ sinˆ 1# ‰ œ 0201 œ 0 dx º # # )œ1/2 (c) x œ ˆ1 2 sinˆ 431 ‰‰cosˆ 431 ‰ œ È3 1 È dy , y œ ˆ1 2 sinˆ 431 ‰‰sinˆ 431 ‰ œ 3 2 3 ; dx º 2 2 sinˆ2ˆ 41 ‰‰ cosˆ 41 ‰ )œ41/3 œ 2 cosˆ2ˆ 431 ‰‰ sinˆ 431 ‰ 3 3 È3 1 2È 3 1 œ È2 œ È3 2 œ Š4 3È3‹ 1 3 2 2 dy dy d y sin t d dy cos t 44. x œ t, y œ 1 cos t, 0 Ÿ t Ÿ 21 Ê dx dt œ 1, dt œ sin t Ê dx œ 1 œ sin t Ê dt Š dx ‹ œ cos t Ê dx2 œ 1 œ cos t. The 2 d y maximum and minimum slope will occur at points that maximize/minimize dy dx , in other words, points where dx2 œ 0 2 d y Ê cos t œ 0 Ê t œ 12 or t œ 321 Ê dx ± ± 2 œ 31Î2 1 Î2 (a) the maximum slope is dy dx º tœ1Î2 (a) œ sinˆ 12 ‰ œ 1, which occurs at x œ 12 , y œ 1 cosˆ 12 ‰ œ 1 the minimum slope is dy œ sinˆ 321 ‰ œ 1, which occurs at x œ 321 , y œ 1 cosˆ 321 ‰ œ 1 dx º tœ31Î2 # # 2 a2 cos t 1b t 1b dy dy dy/dt dy 2 cos 2t 45. dx ; then dx œ 0 Ê 2 a2 cos œ0 dt œ cos t and dt œ 2 cos 2t Ê dx œ dx/dt œ cos t œ cos t cos t Ê 2 cos# t 1 œ 0 Ê cos t œ „ È"2 Ê t œ 14 , 341 , 541 , 741 . In the 1st quadrant: t œ 14 Ê x œ sin 14 œ È È2 # and y œ sin 2 ˆ 14 ‰ œ 1 Ê Š #2 ß 1‹ is the point where the tangent line is horizontal. At the origin: x œ 0 and y œ 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.2 Calculus With Parametric Curves 661 Ê sin t œ 0 Ê t œ 0 or t œ 1 and sin 2t œ 0 Ê t œ 0, 1# , 1, 3#1 ; thus t œ 0 and t œ 1 give the tangent lines at the origin. Tangents at origin: dy dx ¹ œ 2 Ê y œ 2x and dy dx ¹ tœ0 tœ1 œ 2 Ê y œ 2x dy dy dy/dt 3(cos 2t cos t sin 2t sin t) 3 cos 3t 46. dx dt œ 2 cos 2t and dt œ 3 cos 3t Ê dx œ dx/dt œ 2 cos 2t œ 2 a2 cos# t1b # # # # t) 2 sin t cos t sin td t 1 2 sin tb t) a4 cos t 3b œ 3 ca2 cos t 12ba(cos œ (3 cos t) 2a2a2cos œ (3 cos ; then 2 cos# t 1b cos# t 1b 2 a2 cos# t 1b dy dx œ 0 # t) a4 cos t3b Ê (3 cos œ 0 Ê 3 cos t œ 0 or 4 cos# t 3 œ 0: 3 cos t œ 0 Ê t œ 1# , 3#1 and 2 a2 cos# t 1b È 4 cos# t 3 œ 0 Ê cos t œ „ #3 Ê t œ 16 , 561 , 761 , 1161 . In the 1st quadrant: t œ 16 Ê x œ sin 2 ˆ 16 ‰ œ È3 # È and y œ sin 3 ˆ 16 ‰ œ 1 Ê Š #3 ß 1‹ is the point where the graph has a horizontal tangent. At the origin: x œ 0 and y œ 0 Ê sin 2t œ 0 and sin 3t œ 0 Ê t œ 0, 1# , 1, 3#1 and t œ 0, 13 , 231 , 1, 431 , 531 Ê t œ 0 and t œ 1 give the tangent lines at the origin. Tangents at the origin: dy dx ¹ tœ0 (31) 3 3 œ 32 cos cos (21) œ # Ê y œ # x dy cos 0 3 3 œ 23 cos 0 œ # Ê y œ # x, and dx ¹ tœ1 dy 47. (a) x œ aat sin tb, y œ aa1 cos tb, 0 Ÿ t Ÿ 21 Ê dx dt œ aa1 cos tb, dt œ a sin t Ê Length œ '0 Éaaa1 cos tbb# aa sin tb# dt œ '0 Èa# 2a# cos t a# cos# t a# sin# t dt 21 21 œ aÈ2'0 È1 cos t dt œ aÈ2'0 É2 sin2 ˆ 2t ‰ dt œ 2a'0 sinˆ 2t ‰ dt œ ’4a cosˆ 2t ‰“ 21 21 21 21 0 œ 4a cos 1 4a cosa0b œ 8a dy (b) a œ 1 Ê x œ t sin t, y œ 1 cos t, 0 Ÿ t Ÿ 21 Ê dx dt œ 1 cos t, dt œ sin t Ê Surface area œ œ '0 21a1 cos tbÉa1 cos tb# asin tb# dt œ '0 21a1 cos tbÈ1 2 cos t cos# t sin# t dt 21 21 3Î2 œ 21'0 a1 cos tbÈ2 2 cos t dt œ 2È21'0 a1 cos tb3Î2 dt œ 2È21'0 ˆ1 cos ˆ2 † 2t ‰‰ dt 21 21 21 3 Î2 œ 2È21'0 ˆ2 sin2 ˆ 2t ‰‰ dt œ 81'0 sin3 ˆ 2t ‰ dt 21 21 ’u œ 2t Ê du œ 21 dt Ê dt œ 2 du; t œ 0 Ê u œ 0, t œ 21 Ê u œ 1“ œ 161'0 sin3 u du œ 161'0 sin2 u sin u du œ 161'0 a1 cos2 u bsin u du œ 161'0 sin u du 161'0 cos2 u sin u du 1 1 1 1 1 œ ’161cos u 1631 cos3 u“ œ ˆ161 1631 ‰ ˆ161 1631 ‰ œ 6431 0 48. x œ t sin t, y œ 1 cos t, 0 Ÿ t Ÿ 21; Volume œ '0 1 y2 dx œ '0 1a1 cos tb2 a1 cos tbdt 21 21 2t ‰ œ 1'0 a1 3cos t 3cos2 t cos3 tbdt œ 1'0 ˆ1 3cos t 3ˆ 1 cos cos2 t cos t‰dt 2 21 21 œ 1'0 ˆ 52 3cos t 32 cos 2t a1 sin2 tb cos t‰dt œ 1'0 ˆ 52 4cos t 32 cos 2t sin2 t cos t‰dt 21 21 21 œ 1’ 52 t 4sin t 34 sin 2t 31 sin3 t “ 0 œ 1a51 0 0 0b 0 œ 512 47-50. Example CAS commands: Maple: with( plots ); with( student ); x := t -> t^3/3; y := t -> t^2/2; a := 0; b := 1; N := [2, 4, 8 ]; for n in N do Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 1 662 Chapter 11 Parametric Equations and Polar Coordinates tt := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x(t),y(t)],t=tt)]; L := simplify(add( student[distance](pts[i+1],pts[i]), i=1..n )); # (b) T := sprintf("#47(a) (Section 11.2)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [[x(t),y(t),t=a..b],pts], title=T ): # (a) end do: display( [seq(P[n],n=N)], insequence=true ); ds := t ->sqrt( simplify(D(x)(t)^2 + D(y)(t)^2) ): # (c) L := Int( ds(t), t=a..b ): L = evalf(L); 11.3 POLAR COORDINATES 1. a, e; b, g; c, h; d, f 2. a, f; b, h; c, g; d, e 3. (a) ˆ2ß 1# 2n1‰ and ˆ2ß 1# (2n 1)1‰ , n an integer (b) (#ß 2n1) and (#ß (2n 1)1), n an integer (c) ˆ2ß 3#1 2n1‰ and ˆ2ß 3#1 (2n 1)1‰ , n an integer (d) (#ß (2n 1)1) and (#ß 2n1), n an integer 4. (a) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer (b) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer (c) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer (d) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer 5. (a) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0) (b) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0) (c) x œ r cos ) œ 2 cos 21 œ 1, y œ r sin ) œ 2 sin 21 œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3 (d) 3 x œ r cos ) œ 2 cos 731 œ 1, y œ r sin ) œ 2 sin 731 œ È3 Ê Cartesian coordinates are Š1ß È3‹ (e) x œ r cos ) œ 3 cos 1 œ 3, y œ r sin ) œ 3 sin 1 œ 0 Ê Cartesian coordinates are (3ß 0) (f) x œ r cos ) œ 2 cos 1 œ 1, y œ r sin ) œ 2 sin 1 œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3 3 (g) x œ r cos ) œ 3 cos 21 œ 3, y œ r sin ) œ 3 sin 21 œ 0 Ê Cartesian coordinates are (3ß 0) (h) x œ r cos ) œ 2 cos ˆ 1 ‰ œ 1, y œ r sin ) œ 2 sin ˆ 1 ‰ œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3 3 6. (a) x œ È2 cos 14 œ 1, y œ È2 sin 14 œ 1 Ê Cartesian coordinates are (1ß 1) (b) x œ 1 cos 0 œ 1, y œ 1 sin 0 œ 0 Ê Cartesian coordinates are (1ß 0) (c) x œ 0 cos 1# œ 0, y œ 0 sin 1# œ 0 Ê Cartesian coordinates are (!ß 0) (d) x œ È2 cos ˆ 1 ‰ œ 1, y œ È2 sin ˆ 1 ‰ œ 1 Ê Cartesian coordinates are (1ß 1) (e) 4 4 3È 3 51 51 x œ 3 cos 6 œ 2 , y œ 3 sin 6 œ 3# È Ê Cartesian coordinates are Š 3 # 3 ß 3# ‹ (f) x œ 5 cos ˆtan" 43 ‰ œ 3, y œ 5 sin ˆtan" 43 ‰ œ 4 Ê Cartesian coordinates are ($ß 4) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.3 Polar Coordinates 663 (g) x œ 1 cos 71 œ 1, y œ 1 sin 71 œ 0 Ê Cartesian coordinates are (1ß 0) (h) x œ 2È3 cos 21 œ È3, y œ 2È3 sin 21 œ 3 Ê Cartesian coordinates are ŠÈ3ß 3‹ 3 3 7. (a) a1, 1b Ê r œ È12 12 œ È2, sin ) œ È12 and cos ) œ È12 Ê ) œ 14 Ê Polar coordinates are ŠÈ2, 14 ‹ (b) a3, 0b Ê r œ Éa3b2 02 œ 3, sin ) œ 0 and cos ) œ 1 Ê ) œ 1 Ê Polar coordinates are a3, 1b 2 (c) ŠÈ3, 1‹ Ê r œ ÊŠÈ3‹ a1b2 œ 2, sin ) œ 12 and cos ) œ È3 111 ˆ 111 ‰ 2 Ê ) œ 6 Ê Polar coordinates are 2, 6 (d) a3, 4b Ê r œ Éa3b2 42 œ 5, sin ) œ 45 and cos ) œ 35 Ê ) œ 1 arctanˆ 43 ‰ Ê Polar coordinates are ˆ5, 1 arctanˆ 43 ‰‰ 8. (a) a2, 2b Ê r œ Éa2b2 a2b2 œ 2È2, sin ) œ È12 and cos ) œ È12 Ê ) œ 341 Ê Polar coordinates are Š2È2, 341 ‹ (b) a0, 3b Ê r œ È02 32 œ 3, sin ) œ 1 and cos ) œ 0 Ê ) œ 12 Ê Polar coordinates are ˆ3, 12 ‰ 2 È (c) ŠÈ3, 1‹ Ê r œ ÊŠÈ3‹ 12 œ 2, sin ) œ 12 and cos ) œ 23 Ê ) œ 561 Ê Polar coordinates are ˆ2, 561 ‰ 5 ˆ 12 ‰ (d) a5, 12b Ê r œ É52 a12b2 œ 13, sin ) œ 12 13 and cos ) œ 12 Ê ) œ arctan 5 Ê Polar coordinates are ˆ13, arctanˆ 12 ‰‰ 5 9. (a) a3, 3b Ê r œ È32 32 œ 3È2, sin ) œ È12 and cos ) œ È12 Ê ) œ 541 Ê Polar coordinates are Š3È2, 541 ‹ (b) a1, 0b Ê r œ Éa1b2 02 œ 1, sin ) œ 0 and cos ) œ 1 Ê ) œ 0 Ê Polar coordinates are a1, 0b 2 È (c) Š1, È3‹ Ê r œ Êa1b2 ŠÈ3‹ œ 2, sin ) œ 23 and cos ) œ 12 Ê ) œ 531 Ê Polar coordinates are ˆ2, 531 ‰ (d) a4, 3b Ê r œ É42 a3b2 œ 5, sin ) œ 35 and cos ) œ 45 Ê ) œ 1 arctanˆ 34 ‰ Ê Polar coordinates are ˆ5, 1 arctanˆ 43 ‰‰ 10. (a) a2, 0b Ê r œ Éa2b2 02 œ 2, sin ) œ 0 and cos ) œ 1 Ê ) œ 0 Ê Polar coordinates are a2, 0b (b) a1, 0b Ê r œ È12 02 œ 1, sin ) œ 0 and cos ) œ 1 Ê ) œ 1 or ) œ 1 Ê Polar coordinates are a1, 1b or a1, 1b (c) a0, 3b Ê r œ É02 a3b2 œ 3, sin ) œ 1 and cos ) œ 0 Ê ) œ 12 Ê Polar coordinates are ˆ3, 12 ‰ È È 2 2 È (d) Š 23 , 12 ‹ Ê r œ ÊŠ 23 ‹ ˆ 21 ‰ œ 1, sin ) œ 12 and cos ) œ 23 Ê ) œ 761 or ) œ 561 Ê Polar coordinates are ˆ1, 761 ‰ or ˆ1, 561 ‰ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 664 Chapter 11 Parametric Equations and Polar Coordinates 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.3 Polar Coordinates 665 26. 27. r cos ) œ 2 Ê x œ 2, vertical line through (#ß 0) 28. r sin ) œ 1 Ê y œ 1, horizontal line through (0ß 1) 29. r sin ) œ 0 Ê y œ 0, the x-axis 30. r cos ) œ 0 Ê x œ 0, the y-axis 31. r œ 4 csc ) Ê r œ sin4 ) Ê r sin ) œ 4 Ê y œ 4, a horizontal line through (0ß 4) 3 32. r œ 3 sec ) Ê r œ cos ) Ê r cos ) œ 3 Ê x œ 3, a vertical line through (3ß 0) 33. r cos ) r sin ) œ 1 Ê x y œ 1, line with slope m œ 1 and intercept b œ 1 34. r sin ) œ r cos ) Ê y œ x, line with slope m œ 1 and intercept b œ 0 35. r# œ 1 Ê x# y# œ 1, circle with center C œ (!ß 0) and radius 1 36. r# œ 4r sin ) Ê x# y# œ 4y Ê x# y# 4y 4 œ 4 Ê x# (y 2)# œ 4, circle with center C œ (0ß 2) and radius 2 37. r œ sin )52 cos ) Ê r sin ) 2r cos ) œ 5 Ê y 2x œ 5, line with slope m œ 2 and intercept b œ 5 38. r# sin 2) œ 2 Ê 2r# sin ) cos ) œ 2 Ê (r sin ))(r cos )) œ 1 Ê xy œ 1, hyperbola with focal axis y œ x )‰ˆ " ‰ 39. r œ cot ) csc ) œ ˆ cos Ê r sin# ) œ cos ) Ê r# sin# ) œ r cos ) Ê y# œ x, parabola with vertex (0ß 0) sin ) sin ) which opens to the right sin ) ‰ 40. r œ 4 tan ) sec ) Ê r œ 4 ˆ cos Ê r cos# ) œ 4 sin ) Ê r# cos# ) œ 4r sin ) Ê x# œ 4y, parabola with #) vertex œ (!ß 0) which opens upward 41. r œ (csc )) er cos ) Ê r sin ) œ er cos ) Ê y œ ex , graph of the natural exponential function 42. r sin ) œ ln r ln cos ) œ ln (r cos )) Ê y œ ln x, graph of the natural logarithm function 43. r# 2r# cos ) sin ) œ 1 Ê x# y# 2xy œ 1 Ê x# 2xy y# œ 1 Ê (x y)# œ 1 Ê x y œ „ 1, two parallel straight lines of slope 1 and y-intercepts b œ „ 1 44. cos# ) œ sin# ) Ê r# cos# ) œ r# sin# ) Ê x# œ y# Ê kxk œ kyk Ê „ x œ y, two perpendicular lines through the origin with slopes 1 and 1, respectively. 45. r# œ 4r cos ) Ê x# y# œ 4x Ê x# 4x y# œ 0 Ê x# 4x 4 y# œ 4 Ê (x 2)# y# œ 4, a circle with center C(2ß 0) and radius 2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 666 Chapter 11 Parametric Equations and Polar Coordinates 46. r# œ 6r sin ) Ê x# y# œ 6y Ê x# y# 6y œ 0 Ê x# y# 6y 9 œ 9 Ê x# (y 3)# œ 9, a circle with center C(0ß 3) and radius 3 47. r œ 8 sin ) Ê r# œ 8r sin ) Ê x# y# œ 8y Ê x# y# 8y œ 0 Ê x# y# 8y 16 œ 16 Ê x# (y 4)# œ 16, a circle with center C(0ß 4) and radius 4 48. r œ 3 cos ) Ê r# œ 3r cos ) Ê x# y# œ 3x Ê x# y# 3x œ 0 Ê x# 3x 94 y# œ 94 # Ê ˆx 3# ‰ y# œ 49 , a circle with center C ˆ 3# ß !‰ and radius 3# 49. r œ 2 cos ) 2 sin ) Ê r# œ 2r cos ) 2r sin ) Ê x# y# œ 2x 2y Ê x# 2x y# 2y œ 0 Ê (x 1)# (y 1)# œ 2, a circle with center C(1ß 1) and radius È2 50. r œ 2 cos ) sin ) Ê r# œ 2r cos ) r sin ) Ê x# y# œ 2x y Ê x# 2x y# y œ 0 # Ê (x 1)# ˆy "# ‰ œ 54 , a circle with center C ˆ1ß "# ‰ and radius È5 # È 51. r sin ˆ) 16 ‰ œ 2 Ê r ˆsin ) cos 16 cos ) sin 16 ‰ œ 2 Ê #3 r sin ) "# r cos ) œ 2 Ê Ê È3 y x œ 4, line with slope m œ " and intercept b œ 4 È3 È3 " # y # xœ 2 È3 È 52. r sin ˆ 231 )‰ œ 5 Ê r ˆsin 231 cos ) cos 231 sin )‰ œ 5 Ê #3 r cos ) "# r sin ) œ 5 Ê Ê È3 x y œ 10, line with slope m œ È3 and intercept b œ 10 È3 " # x # yœ 5 53. x œ 7 Ê r cos ) œ 7 54. y œ 1 Ê r sin ) œ 1 55. x œ y Ê r cos ) œ r sin ) Ê ) œ 14 56. x y œ 3 Ê r cos ) r sin ) œ 3 57. x# y# œ 4 Ê r# œ 4 Ê r œ 2 or r œ 2 58. x# y# œ 1 Ê r# cos# ) r# sin# ) œ 1 Ê r# acos# ) sin# )b œ 1 Ê r# cos 2) œ 1 # # 59. x9 y4 œ 1 Ê 4x# 9y# œ 36 Ê 4r# cos# ) 9r# sin# ) œ 36 60. xy œ 2 Ê (r cos ))(r sin )) œ 2 Ê r# cos ) sin ) œ 2 Ê 2r# cos ) sin ) œ 4 Ê r# sin 2) œ 4 61. y# œ 4x Ê r# sin# ) œ 4r cos ) Ê r sin# ) œ 4 cos ) 62. x# xy y# œ 1 Ê x# y# xy œ 1 Ê r# r# sin ) cos ) œ 1 Ê r# (1 sin ) cos )) œ 1 63. x# (y 2)# œ 4 Ê x# y# 4y 4 œ 4 Ê x# y# œ 4y Ê r# œ 4r sin ) Ê r œ 4 sin ) 64. (x 5)# y# œ 25 Ê x# 10x 25 y# œ 25 Ê x# y# œ 10x Ê r# œ 10r cos ) Ê r œ 10 cos ) 65. (x 3)# (y 1)# œ 4 Ê x# 6x 9 y# 2y 1 œ 4 Ê x# y# œ 6x 2y 6 Ê r# œ 6r cos ) 2r sin ) 6 66. (x 2)# (y 5)# œ 16 Ê x# 4x 4 y# 10y 25 œ 16 Ê x# y# œ 4x 10y 13 Ê r# œ 4r cos ) 10r sin ) 13 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.4 Graphing in Polar Coordinates 67. (!ß )) where ) is any angle 68. (a) x œ a Ê r cos ) œ a Ê r œ cosa ) Ê r œ a sec ) (b) y œ b Ê r sin ) œ b Ê r œ sinb ) Ê r œ b csc ) 11.4 GRAPHING IN POLAR COORDINATES 1. 1 cos ()) œ 1 cos ) œ r Ê symmetric about the x-axis; 1 cos ()) Á r and 1 cos (1 )) œ 1 cos ) Á r Ê not symmetric about the y-axis; therefore not symmetric about the origin 2. 2 2 cos ()) œ 2 2 cos ) œ r Ê symmetric about the x-axis; 2 # cos ()) Á r and 2 2 cos (1 )) œ 2 2 cos ) Á r Ê not symmetric about the y-axis; therefore not symmetric about the origin 3. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 )) œ 1 sin ) Á r Ê not symmetric about the x-axis; 1 sin (1 )) œ 1 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin 4. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 )) œ 1 sin ) Á r Ê not symmetric about the x-axis; 1 sin (1 )) œ 1 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 667 668 Chapter 11 Parametric Equatins and Polar Coordinates 5. 2 sin ()) œ 2 sin ) Á r and 2 sin (1 )) œ 2 sin ) Á r Ê not symmetric about the x-axis; 2 sin (1 )) œ 2 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin 6. 1 2 sin ()) œ 1 2 sin ) Á r and 1 2 sin (1 )) œ 1 2 sin ) Á r Ê not symmetric about the x-axis; 1 2 sin (1 )) œ 1 2 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin 7. sin ˆ #) ‰ œ sin ˆ #) ‰ œ r Ê symmetric about the y-axis; sin ˆ 21#) ‰ œ sin ˆ 2) ‰ , so the graph is symmetric about the x-axis, and hence the origin. 8. cos ˆ #) ‰ œ cos ˆ #) ‰ œ r Ê symmetric about the x-axis; cos ˆ 21#) ‰ œ cos ˆ 2) ‰ , so the graph is symmetric about the y-axis, and hence the origin. 9. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the graph when (rß )) is on the graph Ê symmetric about the x-axis and the y-axis; therefore symmetric about the origin Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.4 Graphing in Polar Coordinates 10. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 )) are on the graph when (rß )) is on the graph Ê symmetric about the y-axis and the x-axis; therefore symmetric about the origin 11. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 )) are on the graph when (rß )) is on the graph Ê symmetric about the y-axis and the x-axis; therefore symmetric about the origin 12. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the graph when (rß )) is on the graph Ê symmetric about the x-axis and the y-axis; therefore symmetric about the origin 13. Since a „ rß )b are on the graph when (rß )) is on the graph â „ rb# œ 4 cos 2( )) Ê r# œ 4 cos 2)‰ , the graph is symmetric about the x-axis and the y-axis Ê the graph is symmetric about the origin 14. Since (rß )) on the graph Ê (rß )) is on the graph â „ rb# œ 4 sin 2) Ê r# œ 4 sin 2)‰ , the graph is symmetric about the origin. But 4 sin 2()) œ 4 sin 2) Á r# and 4 sin 2(1 )) œ 4 sin (21 2)) œ 4 sin (2)) œ 4 sin 2) Á r# Ê the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis 15. Since (rß )) on the graph Ê (rß )) is on the graph â „ rb# œ sin 2) Ê r# œ sin 2)‰ , the graph is symmetric about the origin. But sin 2()) œ ( sin 2)) sin 2) Á r# and sin 2(1 )) œ sin (21 2)) œ sin (2)) œ ( sin 2)) œ sin 2) Á r# Ê the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 669 670 Chapter 11 Parametric Equatins and Polar Coordinates 16. Sincea „ rß )b are on the graph when (rß )) is on the graph â „ rb# œ cos 2()) Ê r# œ cos 2)‰, the graph is symmetric about the x-axis and the y-axis Ê the graph is symmetric about the origin. 17. ) œ 1# Ê r œ 1 Ê ˆ1ß 1# ‰ , and ) œ 1# Ê r œ 1 w )r cos ) Ê ˆ1ß 1# ‰ ; rw œ ddr) œ sin ); Slope œ rrw sin cos )r sin ) # sin )r cos ) 1‰ ˆ œ sin ) cos )r sin ) Ê Slope at 1ß # is sin# ˆ 1# ‰(1) cos 1# sin 1# cos 1# (1) sin 1# œ 1; Slope at ˆ1ß 1# ‰ is sin# ˆ 1# ‰(1) cos ˆ 1# ‰ œ1 sin ˆ 1# ‰ cos ˆ 1# ‰(1) sin ˆ 1# ‰ 18. ) œ 0 Ê r œ 1 Ê ("ß 0), and ) œ 1 Ê r œ 1 dr Ê ("ß 1); rw œ d) œ cos ); w )r cos ) cos ) sin )r cos ) Slope œ rrw sin cos )r sin ) œ cos ) cos )r sin ) 0 sin 0(1) cos 0 ) sin )r cos ) œ coscos Ê Slope at ("ß 0) is coscos # )r sin ) # 0(1) sin 0 1 sin 1(1) cos 1 œ 1; Slope at ("ß 1) is coscos œ1 # 1(1) sin 1 19. ) œ 14 Ê r œ 1 Ê ˆ"ß 14 ‰ ; ) œ 14 Ê r œ 1 Ê ˆ1ß 14 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ"ß 341 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ1ß 341 ‰ ; dr rw œ d) œ 2 cos 2); )r cos ) 2 cos 2) sin )r cos ) Slope œ rr sin cos )r sin ) œ 2 cos 2) cos )r sin ) w w 2 cos ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰ Ê Slope at ˆ1ß 14 ‰ is 2 cos ˆ 1# ‰ cos ˆ41 ‰(1) sin ˆ 14 ‰ œ 1; # 4 4 2 cos ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰ Slope at ˆ1ß 14 ‰ is 2 cos ˆ 1# ‰ cos ˆ 41 ‰(1) sin ˆ 41 ‰ œ 1; # Slope at ˆ1ß 341 ‰ is Slope at ˆ1ß 341 ‰ is 4 2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹ 2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹ 4 œ 1; 2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹ 2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹ œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.4 Graphing in Polar Coordinates 20. ) œ 0 Ê r œ 1 Ê (1ß 0); ) œ 12 Ê r œ 1 Ê ˆ1ß 12 ‰ ; ) œ 1# Ê r œ 1 Ê ˆ"ß 12 ‰ ; ) œ 1 Ê r œ 1 dr Ê (1ß 1); rw œ d) œ 2 sin 2); )r cos ) 2 sin 2) sin )r cos ) Slope œ rr sin cos )r sin ) œ 2 sin 2) cos )r sin ) w w 2 sin 0 sin 0cos 0 Ê Slope at (1ß 0) is 2 sin 0 cos 0sin 0 , which is undefined; 2 sin 2 ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰ Slope at ˆ1ß 12 ‰ is 2 sin 2 ˆ 12 ‰ cos ˆ21 ‰(1) sin ˆ 21 ‰ œ 0; 2 2 2 2 sin 2 ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰ Slope at ˆ1ß 12 ‰ is 2 sin 2 ˆ 1# ‰ cos ˆ #1 ‰(1) sin ˆ #1 ‰ œ 0; # # # 2 sin 21 sin 1cos 1 Slope at ("ß 1) is 2 sin 21 cos 1sin 1 , which is undefined 21. (a) (b) 22. (a) (b) 23. (a) (b) 24. (a) (b) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 671 672 Chapter 11 Parametric Equatins and Polar Coordinates 25. 26. r œ 2 sec ) Ê r œ cos2 ) Ê r cos ) œ 2 Ê x œ 2 27. 28. 29. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ 1 cos ) Í 1 cos () 1) œ 1 (cos ) cos 1 sin ) sin 1) œ 1 cos ) œ (1 cos )) œ r; therefore (rß )) is on the graph of r œ 1 cos ) Í (rß ) 1) is on the graph of r œ 1 cos ) Ê the answer is (a). Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.4 Graphing in Polar Coordinates 30. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ cos 2) Í sin ˆ2() 1)) 1# ‰ œ sin ˆ2) 5#1 ‰ œ sin (2)) cos ˆ 5#1 ‰ cos (2)) sin ˆ 5#1 ‰ œ cos 2) œ r; therefore (rß )) is on the graph of r œ sin ˆ2) 1# ‰ Ê the answer is (a). 31. 33. (a) 34. (a) 32. (b) (c) (b) (d) (c) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 673 674 Chapter 11 Parametric Equatins and Polar Coordinates (d) (e) 11.5 AREA AND LENGTHS IN POLAR COORDINATES 1. A œ '0 "# )# d) œ 16 )3 ‘ ! œ 16 1 1 3 2) 2. A œ '1Î4 "# a2 sin )b# d) œ 2'1Î4 sin2 ) d) œ 2'1Î4 1 cos d) œ '1Î4 a1 cos 2)bd) œ ) 12 sin 2)‘1Î4 2 1Î2 1Î2 1Î2 1Î2 1Î2 œ ˆ 12 0‰ ˆ 14 12 ‰ œ 14 12 3. A œ '0 21 " # # (4 2 cos )) d) œ 2 ) ‰‘ '021 "# a16 16 cos ) 4 cos# )b d) œ '021 8 8 cos ) 2 ˆ 1 cos d) # œ '0 (9 8 cos ) cos 2)) d) œ 9) 8 sin ) "2 sin 2)‘ ! œ 181 21 #1 4. A œ '0 21 ' ' 21 21 " " # " # 2) ‰ # # ˆ1 2 cos ) 1 cos d) # [a(1 cos ))] d) œ 0 # a a1 2 cos ) cos )b d) œ # a # 0 21 ˆ #3 2 cos ) #" cos 2)‰ d) œ #" a# #3 ) 2 sin ) "4 sin 2)‘ #1 œ 3# 1a# ! 0 œ "# a# ' 5. A œ 2 '0 1Î4 " # # cos 2) d) œ 1Î% 4) '01Î4 1 cos d) œ "# ) sin44) ‘ ! œ 18 # 6) 6. A œ '1Î6 "# acos 3)b2 d) œ "# ' 1Î6 cos2 3) d) œ "# '1Î6 1 cos d) œ "4 '1Î6 a1 cos 6)b d) 2 1 Î6 1 Î6 1Î6 1 Î6 1 Î6 1 œ "4 ) 16 sin 6)‘ 1Î6 œ "4 ˆ 16 0‰ "4 ˆ 16 0‰ œ 12 7. A œ '0 1Î2 " # (4 sin 2)) d) œ 8. A œ (6)(2)'0 1Î6 '01Î2 2 sin 2) d) œ c cos 2)d 1! Î# œ 2 " # (2 sin 3)) d) œ 12 '01Î6 sin 3) d) œ 12 cos33) ‘ 1! Î' œ 4 9. r œ 2 cos ) and r œ 2 sin ) Ê 2 cos ) œ 2 sin ) Ê cos ) œ sin ) Ê ) œ 14 ; therefore A œ 2 '0 1Î4 œ '0 1Î4 " # # (2 sin )) d) œ 2) ‰ 4 ˆ 1 cos d) œ '0 # '01Î4 4 sin# ) d) 1Î4 (2 2 cos 2)) d) 1Î% œ c2) sin 2)d ! œ 1# 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.5 Area and Lengths in Polar Coordinates 10. r œ 1 and r œ 2 sin ) Ê 2 sin ) œ 1 Ê sin ) œ "# Ê ) œ 16 or 561 ; therefore A œ 1(1)# '1Î6 51Î6 " # # # c(2 sin )) 1 d d) œ 1 '1Î6 ˆ2 sin# ) "# ‰ d) 51Î6 œ 1 '1Î6 ˆ1 cos 2) "# ‰ d) 51Î6 &1Î' œ 1 '1Î6 ˆ "# cos 2)‰ d) œ 1 "2 ) sin#2) ‘ 1Î' 51Î6 1 œ 1 ˆ 511# "# sin 531 ‰ ˆ 12 "# sin 13 ‰ œ 41 63 È3 11. r œ 2 and r œ 2(1 cos )) Ê 2 œ 2(1 cos )) Ê cos ) œ 0 Ê ) œ „ 1# ; therefore A œ 2 '0 1Î2 œ '0 1Î2 œ '0 1Î2 œ '0 1Î2 " " # # [2(1 cos ))] d) # area of the circle 4 a1 2 cos ) cos# )b d) ˆ "# 1‰ (2)# 2) ‰ 4 ˆ1 2 cos ) 1 cos d) 21 # (4 8 cos ) 2 2 cos 2)) d) 21 1Î# œ c6) 8 sin ) sin 2)d ! 2 1 œ 51 8 12. r œ 2(1 cos )) and r œ 2(1 cos )) Ê 1 cos ) œ 1 cos ) Ê cos ) œ 0 Ê ) œ 1# or 3#1 ; the graph also gives the point of intersection (0ß 0); therefore A œ 2 '0 1Î2 " # # [2(1 cos ))] d) 2 œ '0 4a1 2cos ) cos# )bd) '11Î2 "# [2(1 cos ))]# d) 1Î2 '1Î2 4 a1 2 cos ) cos# )bd) 1 œ '0 1Î2 œ '0 1Î2 2) ‰ 2) ‰ 4 ˆ1 2 cos ) 1 cos d) '1Î2 4 ˆ1 2 cos ) 1 cos d) # # 1 (6 8 cos ) 2 cos 2)) d) '1Î2 (6 8 cos ) 2 cos 2)) d) 1 1Î# œ c6) 8 sin ) sin 2)d ! c6) 8 sin ) sin 2)d 11Î# œ 61 16 13. r œ È3 and r# œ 6 cos 2) Ê 3 œ 6 cos 2) Ê cos 2) œ "# Ê ) œ 16 (in the 1st quadrant); we use symmetry of the graph to find the area, so A œ 4 '0 ” "# (6 cos 2)) "# ŠÈ3‹ • d) 1Î6 # œ 2 '0 (6 cos 2) 3) d) œ 2 c3 sin 2) 3)d ! 1Î6 1Î' œ 3È 3 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 675 676 Chapter 11 Parametric Equatins and Polar Coordinates 14. r œ 3a cos ) and r œ a(1 cos )) Ê 3a cos ) œ a(1 cos )) Ê 3 cos ) œ 1 cos ) Ê cos ) œ "# Ê ) œ 13 or 13 ; the graph also gives the point of intersection (0ß 0); therefore A œ 2 '0 1Î3 " # # # # c(3a cos )) a (1 cos )) d d) œ '0 a9a# cos# ) a# 2a# cos ) a# cos# )b d) 1Î3 œ '0 1Î3 a8a# cos# ) 2a# cos ) a# b d) œ '0 c4a# (1 cos 2)) 2a# cos ) a# d d) 1Î3 œ '0 a3a# 4a# cos 2) 2a# cos )b d) 1Î3 1Î$ œ c3a# ) 2a# sin 2) 2a# sin )d ! È œ 1a# 2a# ˆ "# ‰ 2a# Š #3 ‹ œ a# Š1 1 È3‹ 15. r œ 1 and r œ 2 cos ) Ê 1 œ 2 cos ) Ê cos ) œ "# Ê ) œ 231 in quadrant II; therefore A œ 2'21Î3 "# c(2 cos ))# 1# d d) œ '21Î3 a4 cos# ) 1b d) 1 1 œ '21Î3 [2(1 cos 2)) 1] d) œ '21Î3 (1 2 cos 2)) d) 1 1 È œ c) sin 2)d 1#1Î$ œ 13 #3 16. r œ 6 and r œ 3 csc ) Ê 6 sin ) œ 3 Ê sin ) œ "# Ê ) œ 16 or 561 ; therefore A œ '1Î6 51Î6 " # # # a6 9 csc )b d) &1Î' œ '1Î6 ˆ18 9# csc# )‰ d) œ 18) 9# cot )‘ 1Î' 51Î6 œ Š151 9# È3‹ Š31 9# È3‹ œ 121 9È3 17. r œ sec ) and r œ 4 cos ) Ê 4 cos ) œ sec ) Ê cos2 ) œ 14 Ê ) œ 13 , 231 , 431 , or 531 ; therefore A œ 2'0 1Î3 œ '0 1Î3 " # # # a16 cos ) sec )b d) a8 8 cos 2) sec# )b d) 1Î3 œ c8) 4 sin 2) tan )d0 œ Š 831 2È3 È3‹ a0 0 0b œ 831 È3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.5 Area and Lengths in Polar Coordinates 677 18. r œ 3 csc ) and r œ 4 sin ) Ê 4 sin ) œ 3 csc ) Ê sin2 ) œ 34 Ê ) œ 13 , 231 , 431 , or 531 ; therefore A œ 41 2'1Î3 "# a16 sin# ) 9 csc# )b d) 1 Î2 œ 41 '1Î3 a8 8 cos 2) 9 csc# )b d) 1Î2 1Î2 œ 41 c8) 4 sin 2) 9 cot )d1Î3 œ 41 ’a41 0 0b Š 831 2È3 3È3‹“ œ 831 È3 È È 19. (a) r œ tan ) and r œ Š #2 ‹ csc ) Ê tan ) œ Š #2 ‹ csc ) È È Ê sin# ) œ Š #2 ‹ cos ) Ê 1 cos# ) œ Š #2 ‹ cos ) È Ê cos# ) Š #2 ‹cos ) 1 œ 0 Ê cos ) œ È2 or È2 # (use the quadratic formula) Ê ) œ 14 (the solution in the first quadrant); therefore the area of R" is A" œ '0 1Î4 " " # # tan ) d) œ # '01Î4 asec# ) 1b d) œ "# ctan ) )d 1! Î% œ "# ˆtan 14 14 ‰ œ "# 18 ; AO œ Š È#2 ‹ csc 1# È2 È2 È2 # È2 È2 1 " È2 " # # and OB œ Š # ‹ csc 4 œ 1 Ê AB œ Ê1 Š # ‹ œ # Ê the area of R# is A# œ # Š # ‹ Š # ‹ œ 4 ; œ therefore the area of the region shaded in the text is 2 ˆ "# 18 "4 ‰ œ #3 14 . Note: The area must be found this way since no common interval generates the region. For example, the interval 0 Ÿ ) Ÿ 14 generates the arc OB of r œ tan ) È2 # csc ). but does not generate the segment AB of the liner œ È _ on the line r œ #2 csc ). (b) lim ) Ä 1Î2 œ lim Instead the interval generates the half-line from B to tan ) œ _ and the line x œ 1 is r œ sec ) in polar coordinates; then ) Ä 1Î2c sin ) " ‰ ˆ cos ) cos ) œ lim ) Ä 1 Î2 c ˆ sincos) ) 1 ‰ œ lim ) Ä 1Î2c (tan ) sec )) ) ‰ ˆ cos sin ) œ 0 Ê r œ tan ) approaches lim ) Ä 1 Î2 c c r œ sec ) as ) Ä 1# Ê r œ sec ) (or x œ 1) is a vertical asymptote of r œ tan ). Similarly, r œ sec ) (or x œ 1) is a vertical asymptote of r œ tan ). 20. It is not because the circle is generated twice from ) œ 0 to 21. The area of the cardioid is 2) A œ 2 '0 "# (cos ) 1)# d) œ '0 acos# ) 2 cos ) 1b d) œ '0 ˆ 1 cos 2 cos ) 1‰ d) # 1 1 1 1 # œ 32) sin42) 2 sin )‘ ! œ 3#1 . The area of the circle is A œ 1 ˆ "# ‰ œ 14 Ê the area requested is actually 3#1 14 œ 541 È5 dr 21. r œ )# , 0 Ÿ ) Ÿ È5 Ê d) œ 2); therefore Length œ '0 È5 œ '0 k)k È)# 4 d) œ (since ) È5 0) '0 Éa)# b# (2))# d) œ ' È5 0 È ) % 4) # d) )È)# 4 d); u œ )# 4 Ê "# du œ ) d); ) œ 0 Ê u œ 4, ) œ È5 Ê u œ 9“ Ä '4 "# Èu du œ "# 23 u$Î# ‘ % œ 19 3 * 9 22. r œ Èe 2 , 0 Ÿ ) Ÿ 1 Ê ddr) œ Èe 2 ; therefore Length œ '0 ÊŠ Èe 2 ‹ Š Èe 2 ‹ d) œ '0 Ê2 Š e# ‹ d) ) ) 1 ) # ) # 1 œ '0 e) d) œ e) ‘ ! œ e1 1 1 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 2) 678 Chapter 11 Parametric Equatins and Polar Coordinates dr 23. r œ 1 cos ) Ê d) œ sin ); therefore Length œ '0 È(1 cos ))# ( sin ))# d) 21 1 œ 2 '0 È2 2 cos ) d) œ 2'0 É 4(1 #cos )) d) œ 4 '0 É 1 #cos ) d) œ 4 '0 cos ˆ #) ‰ d) œ 4 2 sin 2) ‘ ! œ 8 1 1 1 1 24. r œ a sin# #) , 0 Ÿ ) Ÿ 1, a 0 Ê ddr) œ a sin #) cos #) ; therefore Length œ '0 Éâ sin# #) ‰ â sin #) cos #) ‰ d) 1 # # 1 œ '0 Éa# sin% #) a# sin# #) cos# #) d) œ '0 a ¸sin #) ¸ Ésin# #) cos# #) d) œ (since 0 Ÿ ) Ÿ 1) a ' sin ˆ #) ‰ d) 1 1 0 1 œ 2a cos 2) ‘ ! œ 2a 6 sin ) ' 25. r œ 1 6cos ) , 0 Ÿ ) Ÿ 1# Ê ddr) œ (1 cos ))# ; therefore Length œ 0 1Î2 œ '0 1Î2 36 sin ) ' É (1 36 cos ))# a1 cos )b% d) œ 6 0 1Î2 # # # 6 sin ) Êˆ 1 6cos ) ‰ Š (1 cos ))# ‹ d) " sin# ) ¸ 1cos ¸ ) É 1 (1 cos ))# d) cos# ) sin# ) œ ˆsince 1 "cos ) 0 on 0 Ÿ ) Ÿ 1# ‰ 6 '0 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos d) ) )# 1Î2 cos ) È ' œ 6 '0 ˆ 1 "cos ) ‰ É (12 2cos ) ) # d) œ 6 2 0 1Î2 1Î2 œ 3'0 sec$ #) d) œ 6'0 1Î2 1Î4 d) œ 6È 2 (1 cos ))$Î# '01Î2 1Î% sec$ u du œ (use tables) 6 Œ sec u2tan u ‘ ! ' 1Î2 ¸sec$ #) ¸ d) d) œ3 0 ˆ2 cos# #) ‰$Î# "# '0 1Î4 sec u du 1Î% œ 6 Š È"2 2" ln ksec u tan uk‘ ! ‹ œ 3 ’È2 ln Š1 È2‹“ sin ) sin ) ' ˆ 2 ‰# Š (12cos 26. r œ 1 2cos ) , 1# Ÿ ) Ÿ 1 Ê ddr) œ (12cos ))# ; therefore Length œ 1Î2 Ê 1 cos ) ) )# ‹ d ) # 1 ) 4 sin ) ' ¸ 2 ¸ É (1 (1cos )cos) ))sin œ '1Î2 Ê (1 cos d) # ))# Š1 a1 cos )b# ‹ d) œ 1Î2 1 cos ) 1 1 # œ ˆsince 1 cos ) # # cos ) sin ) 0 on 1# Ÿ ) Ÿ 1‰ 2 '1Î2 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos d) ))# 1 # cos ) d) È ' È ' œ 2 '1Î2 ˆ 1 "cos ) ‰ É (12 2cos ))# d) œ 2 2 1Î2 (1 cos ))$Î# œ 2 2 1Î2 1 1 œ '1Î2 csc$ ˆ #) ‰ d) œ ˆsince csc #) 1 1 # d) œ ˆ2 sin# )# ‰$Î# '11Î2 ¸csc$ #) ¸ d) 0 on 1# Ÿ ) Ÿ 1‰ 2 '1Î4 csc$ u du œ (use tables) 1Î2 2Œ csc u2cot u ‘ 1Î% #" '1Î4 csc u du œ 2 Š È" 2" ln kcsc u cot uk‘ 1Î% ‹ œ 2 ’ È" "# ln ŠÈ2 1‹“ 2 2 1Î2 1Î# 1Î# œ È2 ln Š1 È2‹ 27. r œ cos$ 3) Ê ddr) œ sin 3) cos# 3) ; therefore Length œ '0 1Î4 œ '0 1Î4 œ '0 Écos' ˆ 3) ‰ sin# ˆ 3) ‰ cos% ˆ 3) ‰ d) œ '0 1Î4 1Î4 1cos ˆ 2) ‰ 3 # 1Î% d) œ "# ) 32 sin 23) ‘ ! Éˆcos$ 3) ‰# ˆ sin 3) cos# 3) ‰# d) ˆcos# 3) ‰ Écos# ˆ 3) ‰ sin# ˆ 3) ‰ d) œ ' 1Î4 0 cos# ˆ 3) ‰ d) œ 18 38 dr 28. r œ È1 sin 2) , 0 Ÿ ) Ÿ 1È2 Ê d) œ "# (1 sin 2))"Î# (2 cos 2)) œ (cos 2))(1 sin 2))"Î# ; therefore Length œ '0 È 1 2 œ '0 È 1 2 2) ' É(1 sin 2)) (1 cos sin 2)) d) œ 0 # sin 2) ' É 212sin 2 ) d) œ 0 È 1 2 È2 d) œ ’È2 )“ È 1 2 1È# ! # # sin 2) cos 2) É 1 2 sin 2)1 d) sin 2) œ 21 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.6 Conic Sections 679 # w ˆ dx ‰ œ cf w ()) cos ) f()) sin )d# 29. Let r œ f()). Then x œ f()) cos ) Ê dx d) œ f ()) cos ) f()) sin ) Ê d) w œ cf w ())d# cos# ) 2f w ()) f()) sin ) cos ) [f())]# sin# ); y œ f()) sin ) Ê dy d) œ f ()) sin ) f()) cos ) # # # w w # w # # Ê Š dy d) ‹ œ cf ()) sin ) f()) cos )d œ cf ())d sin ) 2f ())f()) sin ) cos ) [f())] cos ). Therefore # # # w # # # # # w # # ˆ dx ‰# Š dy ˆ dr ‰# d) d) ‹ œ cf ())d acos ) sin )b [f())] acos ) sin )b œ cf ())d [f())] œ r d) . ' Ér# ˆ ddr) ‰ d). ‰ Š dy Thus, L œ '! Êˆ dx d) d) ‹ d) œ ! " # " # # dr 30. (a) r œ a Ê d) œ 0; Length œ '0 Èa# 0# d) œ '0 kak d) œ ca)d #!1 œ 21a 21 21 dr (b) r œ a cos ) Ê d) œ a sin ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d) 1 œ '0 kak d) œ ca)d 1! œ 1a 1 1 dr (c) r œ a sin ) Ê d) œ a cos ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d) 1 1 œ '0 kak d) œ ca)d 1! œ 1a 1 31. (a) rav œ 21"0 '0 a(1 cos )) d) œ 2a1 c) sin )d #!1 œ a 21 (b) rav œ 21"0 '0 a d) œ #"1 ca)d #!1 œ a 21 (c) rav œ ˆ 1 ‰"ˆ 1 ‰ 'c1Î2 a cos ) d) œ 1" ca sin )d 1Î# œ 2a 1 1Î2 # 1Î# # 32. r œ 2f()), ! Ÿ ) Ÿ " Ê ddr) œ 2f w ()) Ê r# ˆ ddr) ‰ œ [2f())]# c2f w ())d# Ê Length œ '! É4[f())]# 4 cf w ())d# d) # " œ 2 '! É[f())]# cf w ())d# d) which is twice the length of the curve r œ f()) for ! Ÿ ) Ÿ " . " 11.6 CONIC SECTIONS # 1. x œ y8 Ê 4p œ 8 Ê p œ 2; focus is (2ß 0), directrix is x œ 2 # 2. x œ y4 Ê 4p œ 4 Ê p œ 1; focus is (1ß 0), directrix is x œ 1 # 3. y œ x6 Ê 4p œ 6 Ê p œ 3# ; focus is ˆ!ß 3# ‰ , directrix is y œ 3# # 4. y œ x2 Ê 4p œ 2 Ê p œ 1# ; focus is ˆ!ß #1 ‰ , directrix is y œ 1# 5. y# x# 4 9 œ1 Ê c œ È4 9 œ È13 Ê foci are Š „ È13ß !‹ ; vertices are a „ 2ß 0b ; asymptotes are y œ „ 3# x 6. y# x# 4 9 œ1 Ê c œ È9 4 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a0ß „ 3b 7. x# # 2 y œ1 Ê c œ È2 1 œ 1 Ê foci are a „ 1ß 0b ; vertices are Š „ È2ß !‹ 8. y# # 4 x œ1 Ê c œ È4 1 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a!ß „ 2b ; asymptotes are y œ „ 2x Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 680 Chapter 11 Parametric Equatins and Polar Coordinates # 9. y# œ 12x Ê x œ 1y# Ê 4p œ 12 Ê p œ 3; focus is ($ß !), directrix is x œ 3 # 11. x# œ 8y Ê y œ x 8 Ê 4p œ 8 Ê p œ 2; focus is (!ß 2), directrix is y œ 2 # " 13. y œ 4x# Ê y œ ˆx" ‰ Ê 4p œ "4 Ê p œ 16 ; 4 " ‰ " focus is ˆ!ß 16 , directrix is y œ 16 # 15. x œ 3y# Ê x œ ˆy" ‰ Ê 4p œ "3 Ê p œ 1"# ; 3 focus is ˆ 1"# ß !‰ , directrix is x œ 1"# # 10. x# œ 6y Ê y œ x6 Ê 4p œ 6 Ê p œ 3# ; focus is ˆ!ß 3# ‰ , directrix is y œ 3# # y 12. y# œ 2x Ê x œ # Ê 4p œ 2 Ê p œ #" ; focus is ˆ "# ß !‰ , directrix is x œ "# # " 14. y œ 8x# Ê y œ ˆx" ‰ Ê 4p œ 8" Ê p œ 32 ; 8 " ‰ focus is ˆ!ß 32 , directrix is y œ 3"# # 16. x œ 2y# Ê x œ ˆy" ‰ Ê 4p œ #" Ê p œ "8 ; # focus is ˆ "8 ß !‰ , directrix is x œ 8" Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.6 Conic Sections # # y 17. 16x# 25y# œ 400 Ê #x5 16 œ1 Ê c œ Èa# b# œ È25 16 œ 3 # 19. 2x# y# œ 2 Ê x# y# œ 1 Ê c œ Èa# b# œ È2 1 œ 1 # # 21. 3x# 2y# œ 6 Ê x# y3 œ 1 Ê c œ Èa# b# œ È3 2 œ 1 # # x 18. 7x# 16y# œ 112 Ê 16 y7 œ 1 Ê c œ Èa# b# œ È16 7 œ 3 # # 20. 2x# y# œ 4 Ê x# y4 œ 1 Ê c œ Èa# b# œ È4 2 œ È2 # # x 22. 9x# 10y# œ 90 Ê 10 y9 œ 1 Ê c œ Èa# b# œ È10 9 œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 681 682 Chapter 11 Parametric Equations and Polar Coordinates # # 23. 6x# 9y# œ 54 Ê x9 y6 œ 1 Ê c œ Èa# b# œ È9 6 œ È3 # # y x 24. 169x# 25y# œ 4225 Ê 25 169 œ1 Ê c œ Èa# b# œ È169 25 œ 12 # 25. Foci: Š „ È2ß !‹ , Vertices: a „ 2ß 0b Ê a œ 2, c œ È2 Ê b# œ a# c# œ 4 ŠÈ2‹ œ 2 Ê x4 y# œ 1 # # # # 26. Foci: a!ß „ 4b , Vertices: a0ß „ 5b Ê a œ 5, c œ 4 Ê b# œ 25 16 œ 9 Ê x9 #y5 œ 1 27. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 ; asymptotes are y œ „ x # # x 28. 9x# 16y# œ 144 Ê 16 y9 œ 1 Ê c œ Èa# b# œ È16 9 œ 5; asymptotes are y œ „ 34 x # # 29. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b# œ È8 8 œ 4; asymptotes are y œ „ x # # 30. y# x# œ 4 Ê y4 x4 œ 1 Ê c œ Èa# b# œ È4 4 œ 2È2; asymptotes are y œ „ x Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.6 Conic Sections 683 # # 31. 8x# 2y# œ 16 Ê x# y8 œ 1 Ê c œ Èa# b# œ È2 8 œ È10 ; asymptotes are y œ „ 2x 32. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b# œ È3 1 œ 2; asymptotes are y œ „ È3x # # 33. 8y# 2x# œ 16 Ê y# x8 œ 1 Ê c œ Èa# b# œ È2 8 œ È10 ; asymptotes are y œ „ x y x 34. 64x# 36y# œ 2304 Ê 36 64 œ 1 Ê c œ È a# b # œ È36 64 œ 10; asymptotes are y œ „ 4 # # # # 3 35. Foci: Š!ß „ È2‹ , Asymptotes: y œ „ x Ê c œ È2 and ba œ 1 Ê a œ b Ê c# œ a# b# œ 2a# Ê 2 œ 2a# Ê a œ 1 Ê b œ 1 Ê y# x# œ 1 # 36. Foci: a „ 2ß !b , Asymptotes: y œ „ È"3 x Ê c œ 2 and ba œ È"3 Ê b œ Èa3 Ê c# œ a# b# œ a# a3 œ 4a3 Ê 4 œ 4a3 Ê a# œ 3 Ê a œ È3 Ê b œ 1 Ê x3 y# œ 1 # # # # y 37. Vertices: a „ 3ß 0b , Asymptotes: y œ „ 43 x Ê a œ 3 and ba œ 43 Ê b œ 43 (3) œ 4 Ê x9 16 œ1 # # x 38. Vertices: a!ß „ 2b , Asymptotes: y œ „ 12 x Ê a œ 2 and ba œ 12 Ê b œ 2(2) œ 4 Ê y4 16 œ1 39. (a) y# œ 8x Ê 4p œ 8 Ê p œ 2 Ê directrix is x œ 2, focus is (#ß !), and vertex is (!ß 0); therefore the new directrix is x œ 1, the new focus is (3ß 2), and the new vertex is (1ß 2) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. # 684 Chapter 11 Parametric Equations and Polar Coordinates 40. (a) x# œ 4y Ê 4p œ 4 Ê p œ 1 Ê directrix is y œ 1, focus is (!ß 1), and vertex is (!ß 0); therefore the new directrix is y œ 4, the new focus is (1ß 2), and the new vertex is (1ß 3) 41. (a) y# x# 16 9 œ 1 Ê center is (!ß 0), vertices are (4ß 0) (b) (b) and (%ß !); c œ Èa# b# œ È7 Ê foci are ŠÈ7ß 0‹ and ŠÈ7ß !‹ ; therefore the new center is (%ß $), the new vertices are (!ß 3) and (8ß 3), and the new foci are Š4 „ È7ß $‹ 42. (a) y# x# 9 25 œ 1 Ê center is (!ß 0), vertices are (0ß 5) and (0ß 5); c œ Èa# b# œ È16 œ 4 Ê foci are (b) (!ß 4) and (!ß 4) ; therefore the new center is (3ß 2), the new vertices are (3ß 3) and (3ß 7), and the new foci are (3ß 2) and (3ß 6) 43. (a) y# x# 16 9 œ 1 Ê center is (!ß 0), vertices are (4ß 0) (b) and (4ß 0), and the asymptotes are x4 œ „ y3 or Èa# b# œ È25 œ 5 Ê foci are y œ „ 3x 4 ;cœ (5ß 0) and (5ß 0) ; therefore the new center is (2ß 0), the new vertices are (2ß 0) and (6ß 0), the new foci are (3ß 0) and (7ß 0), and the new asymptotes are y œ „ 3(x 4 2) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.6 Conic Sections 44. (a) y# x# 4 5 œ1 Ê center is (!ß 0), vertices are (0ß 2) and (0ß 2), and the asymptotes are y2 œ 685 (b) „ Èx5 or 2x yœ „È ; c œ Èa# b# œ È9 œ 3 Ê foci are 5 (0ß 3) and (0ß 3) ; therefore the new center is (0ß 2), the new vertices are (0ß 4) and (0ß 0), the new foci are (0ß 1) and (0ß 5), and the new asymptotes are 2x y2œ „ È 5 45. y# œ 4x Ê 4p œ 4 Ê p œ 1 Ê focus is ("ß 0), directrix is x œ 1, and vertex is (0ß 0); therefore the new vertex is (2ß 3), the new focus is (1ß 3), and the new directrix is x œ 3; the new equation is (y 3)# œ 4(x 2) 46. y# œ 12x Ê 4p œ 12 Ê p œ 3 Ê focus is (3ß 0), directrix is x œ 3, and vertex is (0ß 0); therefore the new vertex is (4ß 3), the new focus is (1ß 3), and the new directrix is x œ 7; the new equation is (y 3)# œ 12(x 4) 47. x# œ 8y Ê 4p œ 8 Ê p œ 2 Ê focus is (0ß 2), directrix is y œ 2, and vertex is (0ß 0); therefore the new vertex is (1ß 7), the new focus is (1ß 5), and the new directrix is y œ 9; the new equation is (x 1)# œ 8(y 7) 48. x# œ 6y Ê 4p œ 6 Ê p œ #3 Ê focus is ˆ!ß #3 ‰ , directrix is y œ #3 , and vertex is (0ß 0); therefore the new vertex is (3ß 2), the new focus is ˆ3ß "# ‰ , and the new directrix is y œ 7# ; the new equation is (x 3)# œ 6(y 2) 49. x6 y9 œ 1 Ê center is (!ß 0), vertices are (0ß 3) and (!ß 3); c œ Èa# b# œ È9 6 œ È3 Ê foci are Š!ß È3‹ # # and Š!ß È3‹ ; therefore the new center is (#ß 1), the new vertices are (2ß 2) and (#ß 4), and the new foci are Š#ß 1 „ È3‹ ; the new equation is (x 6 2) (y 9 1) œ 1 # # 50. x2 y# œ 1 Ê center is (!ß 0), vertices are ŠÈ2ß !‹ and ŠÈ2ß !‹ ; c œ Èa# b# œ È2 1 œ 1 Ê foci are # (1ß 0) and ("ß !); therefore the new center is (3ß 4), the new vertices are Š3 „ È2ß 4‹ , and the new foci are (2ß 4) # and (4ß 4); the new equation is (x # 3) (y 4)# œ 1 51. x3 y# œ 1 Ê center is (!ß 0), vertices are ŠÈ3ß !‹ and ŠÈ3ß !‹ ; c œ Èa# b# œ È3 2 œ 1 Ê foci are # # (1ß 0) and ("ß !); therefore the new center is (2ß 3), the new vertices are Š2 „ È3ß 3‹ , and the new foci are (1ß 3) # # and (3ß 3); the new equation is (x 3 2) (y # 3) œ 1 x 52. 16 #y5 œ 1 Ê center is (!ß 0), vertices are (!ß &) and (!ß 5); c œ Èa# b# œ È25 16 œ 3 Ê foci are # # (0ß 3) and (0ß 3); therefore the new center is (4ß 5), the new vertices are (4ß 0) and (4ß 10), and the new # # foci are (4ß 2) and (4ß 8); the new equation is (x 164) (y #55) œ 1 53. x4 y5 œ 1 Ê center is (!ß 0), vertices are (2ß 0) and (2ß 0); c œ Èa# b# œ È4 5 œ 3 Ê foci are ($ß !) and # # (3ß 0); the asymptotes are „ x# œ Èy5 Ê y œ „ È5x # ; therefore the new center is (2ß 2), the new vertices are Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 686 Chapter 11 Parametric Equations and Polar Coordinates (4ß 2) and (0ß 2), and the new foci are (5ß 2) and (1ß 2); the new asymptotes are y 2 œ „ # # È5 (x 2) ; the new # equation is (x 4 2) (y 5 2) œ 1 x 54. 16 y9 œ 1 Ê center is (!ß 0), vertices are (4ß 0) and (4ß 0); c œ Èa# b# œ È16 9 œ 5 Ê foci are (5ß !) # # and (5ß 0); the asymptotes are „ x4 œ y3 Ê y œ „ 3x 4 ; therefore the new center is (5ß 1), the new vertices are (1ß 1) and (9ß 1), and the new foci are (10ß 1) and (0ß 1); the new asymptotes are y 1 œ „ 3(x 4 5) ; # # the new equation is (x 165) (y 9 1) œ 1 55. y# x# œ 1 Ê center is (!ß 0), vertices are (0ß 1) and (0ß 1); c œ Èa# b# œ È1 1 œ È2 Ê foci are Š!ß „ È2‹ ; the asymptotes are y œ „ x; therefore the new center is (1ß 1), the new vertices are (1ß 0) and (1ß 2), and the new foci are Š1ß 1 „ È2‹ ; the new asymptotes are y 1 œ „ (x 1); the new equation is (y 1)# (x 1)# œ 1 56. y3 x# œ 1 Ê center is (!ß 0), vertices are Š0ß È3‹ and Š!ß È3‹ ; c œ Èa# b# œ È3 1 œ 2 Ê foci are (!ß #) # and (!ß 2); the asymptotes are „ x œ Èy3 Ê y œ „ È3x; therefore the new center is (1ß 3), the new vertices are Š"ß $ „ È3‹ , and the new foci are ("ß &) and (1ß 1); the new asymptotes are y 3 œ „ È3 (x 1); the new equation is (y 3)# (x 1)# œ 1 3 57. x# 4x y# œ 12 Ê x# 4x 4 y# œ 12 4 Ê (x 2)# y# œ 16; this is a circle: center at C(2ß 0), a œ 4 58. 2x# 2y# 28x 12y 114 œ 0 Ê x# 14x 49 y# 6y 9 œ 57 49 9 Ê (x 7)# (y 3)# œ 1; this is a circle: center at C(7ß 3), a œ 1 59. x# 2x 4y 3 œ 0 Ê x# 2x 1 œ 4y 3 1 Ê (x 1)# œ 4(y 1); this is a parabola: V(1ß 1), F(1ß 0) 60. y# 4y 8x 12 œ 0 Ê y# 4y 4 œ 8x 12 4 Ê (y 2)# œ 8(x 2); this is a parabola: V(#ß 2), F(!ß #) # 61. x# 5y# 4x œ 1 Ê x# 4x 4 5y# œ 5 Ê (x 2)# 5y# œ 5 Ê (x 5 2) y# œ 1; this is an ellipse: the center is (2ß 0), the vertices are Š2 „ È5ß 0‹ ; c œ Èa# b# œ È5 1 œ 2 Ê the foci are (4ß 0) and (!ß 0) # 62. 9x# 6y# 36y œ 0 Ê 9x# 6 ay# 6y 9b œ 54 Ê 9x# 6(y 3)# œ 54 Ê x6 (y 9 3) œ 1; this is an ellipse: # the center is (0ß 3), the vertices are (!ß 0) and (!ß 6); c œ Èa# b# œ È9 6 œ È3 Ê the foci are Š0ß 3 „ È3‹ 63. x# 2y# 2x 4y œ 1 Ê x# 2x 1 2 ay# 2y 1b œ 2 Ê (x 1)# 2(y 1)# œ 2 # Ê (x1) (y 1)# œ 1; this is an ellipse: the center is (1ß 1), the vertices are Š" „ È2ß "‹ ; 2 c œ Èa# b# œ È2 1 œ 1 Ê the foci are (2ß 1) and (0ß 1) 64. 4x# y# 8x 2y œ 1 Ê 4 ax# 2x 1b y# 2y 1 œ 4 Ê 4(x 1)# (y 1)# œ 4 # Ê (x 1)# (y41) œ 1; this is an ellipse: the center is (1ß 1), the vertices are (1ß 3) and (1ß 1); c œ Èa# b# œ È4 1 œ È3 Ê the foci are Š1ß " „ È3‹ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.6 Conic Sections 65. x# y# 2x 4y œ 4 Ê x# 2x 1 ay# 4y 4b œ 1 Ê (x 1)# (y 2)# œ 1; this is a hyperbola: the center is (1ß 2), the vertices are (2ß 2) and (!ß 2); c œ Èa# b# œ È1 1 œ È2 Ê the foci are Š1 „ È2ß #‹ ; the asymptotes are y 2 œ „ (x 1) 66. x# y# 4x 6y œ 6 Ê x# 4x 4 ay# 6y 9b œ 1 Ê (x 2)# (y 3)# œ 1; this is a hyperbola: the center is (2ß 3), the vertices are (1ß 3) and (3ß 3); c œ Èa# b# œ È1 1 œ È2 Ê the foci are Š2 „ È2ß 3‹ ; the asymptotes are y 3 œ „ (x 2) # 67. 2x# y# 6y œ 3 Ê 2x# ay# 6y 9b œ 6 Ê (y 6 3) x3 œ 1; this is a hyperbola: the center is (!ß $), # the vertices are Š!ß 3 „ È6‹ ; c œ Èa# b# œ È6 3 œ 3 Ê the foci are (0ß 6) and (!ß 0); the asymptotes are y 3 È6 œ „ Èx3 Ê y œ „ È2x 3 # # 68. y# 4x# 16x œ 24 Ê y# 4 ax# 4x 4b œ 8 Ê y8 (x 2 2) œ 1; this is a hyperbola: the center is (2ß 0), the vertices are Š2ß „ È8‹ ; c œ Èa# b# œ È8 2 œ È10 Ê the foci are Š2ß „ È10‹ ; the asymptotes are y È8 œ „ xÈ2 Ê y œ „ 2(x 2) 2 # 69. (a) y# œ kx Ê x œ yk ; the volume of the solid formed by Èkx revolving R" about the y-axis is V" œ '0 Èkx œ k1# '0 # # 1 Š yk ‹ dy #È y% dy œ 1x 5 kx ; the volume of the right circular cylinder formed by revolving PQ about the y-axis is V# œ 1x# Èkx Ê the volume of the solid formed by revolving R# about the y-axis is #È V$ œ V# V" œ 41x 5 kx . Therefore we can see the ratio of V$ to V" is 4:1. (b) The volume of the solid formed by revolving R# about the x-axis is V" œ '0 1 ŠÈkt‹ dt œ 1k'0 t dt x # x # œ 1kx # . The volume of the right circular cylinder formed by revolving PS about the x-axis is # V# œ 1 ŠÈkx‹ x œ 1kx# Ê the volume of the solid formed by revolving R" about the x-axis is # # 1kx V$ œ V# V" œ 1kx# 1kx # œ # . Therefore the ratio of V$ to V" is 1:1. w(0) w x wx wx 70. y œ ' w H x dx œ H Š # ‹ C œ 2H C; y œ 0 when x œ 0 Ê 0 œ 2H C Ê C œ 0; therefore y œ 2H is the # # # # equation of the cable's curve 71. x# œ 4py and y œ p Ê x# œ 4p# Ê x œ „ 2p. Therefore the line y œ p cuts the parabola at points (2pß p) and (2pß p), and these points are È[2p (2p)]# (p p)# œ 4p units apart. 72. x lim Š b x ba Èx# a# ‹ œ ba x lim Šx Èx# a# ‹ œ ba x lim Ä_ a Ä_ Ä_ – # # # x x a œ ba x lim ’ x Èax # a #b “ œ ba x lim ’ Ä_ Ä_ Šx Èx# a# ‹ Šx Èx# a# ‹ x È x # a# a# “œ0 x È x# a# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. — 687 688 Chapter 11 Parametric Equations and Polar Coordinates # 73. Let y œ É1 x4 on the interval 0 Ÿ x Ÿ 2. The area of the inscribed rectangle is given by # # A(x) œ 2x Š2É1 x4 ‹ œ 4xÉ1 x4 (since the length is 2x and the height is 2y) # Ê Aw (x) œ 4É1 x4 x# . É1 x4# # Thus Aw (x) œ 0 Ê 4É1 x4 x# É1 x4# # œ 0 Ê 4 Š1 x4 ‹ x# œ 0 Ê x# œ 2 Ê x œ È2 (only the positive square root lies in the interval). Since A(0) œ A(2) œ 0 we have that A ŠÈ2‹ œ 4 is the maximum area when the length is 2È2 and the height is È2. 74. (a) Around the x-axis: 9x# 4y# œ 36 Ê y# œ 9 94 x# Ê y œ „ É9 94 x# and we use the positive root # # Ê V œ 2 '0 1 ŠÉ9 94 x# ‹ dx œ 2 '0 1 ˆ9 94 x# ‰ dx œ 21 9x 34 x$ ‘ ! œ 241 2 2 (b) Around the y-axis: 9x# 4y# œ 36 Ê x# œ 4 49 y# Ê x œ „ É4 49 y# and we use the positive root # 4 $‘ Ê V œ 2'0 1 ŠÉ4 49 y# ‹ dy œ 2 '0 1 ˆ4 49 y# ‰ dy œ 21 4y 27 y ! œ 161 3 3 $ 75. 9x# 4y# œ 36 Ê y# œ 9x 4 36 Ê y œ „ 3# Èx# 4 on the interval 2 Ÿ x Ÿ 4 Ê V œ '2 1 Š #3 Èx# 4‹ dx # 4 # ‰ ˆ8 ‰‘ œ 941 ˆ 56 ‰ 31 œ 941 '2 ax# 4b dx œ 941 ’ x3 4x“ œ 941 ˆ 64 3 16 3 8 3 8 œ 4 (56 24) œ 241 4 % $ # 76. Let P" (pß y" ) be any point on x œ p, and let P(xß y) be a point where a tangent intersects y# œ 4px. Now dy 2p y y" dy 2p y# œ 4px Ê 2y dy dx œ 4p Ê dx œ y ; then the slope of a tangent line from P" is x (p) œ dx œ y # # y y Ê y# yy" œ 2px 2p# . Since x œ 4p , we have y# yy" œ 2p Š 4p ‹ 2p# Ê y# yy" œ "# y# 2p# È4y#" 16p# œ y" „ Èy"# 4p# . Therefore the slopes of the two # # and m# œ y È2p Ê m" m# œ y# ay4p# 4p# b œ 1 y#" 4p# " " " Ê "# y# yy" 2p# œ 0 Ê y œ 2y" „ tangents from P" are m" œ y È2p y# 4p# " " Ê the lines are perpendicular dy x2 # # 77. (x 2)# (y 1)# œ 5 Ê 2(x 2) 2(y 1) dy dx œ 0 Ê dx œ y1 ; y œ 0 Ê (x 2) (0 1) œ 5 Ê (x 2)# œ 4 Ê x œ 4 or x œ 0 Ê the circle crosses the x-axis at (4ß 0) and (!ß 0); x œ 0 Ê (0 2)# (y 1)# œ 5 Ê (y 1)# œ 1 Ê y œ 2 or y œ 0 Ê the circle crosses the y-axis at (!ß 2) and (!ß !). 4 2 At (4ß 0): dy dx œ 01 œ 2 Ê the tangent line is y œ 2(x 4) or y œ 2x 8 0 2 At (!ß !): dy dx œ 01 œ 2 Ê the tangent line is y œ 2x 0 2 At (!ß #): dy dx œ 21 œ 2 Ê the tangent line is y 2 œ 2x or y œ 2x 2 78. x# y# œ 1 Ê x œ „ È1 y# on the interval 3 Ÿ y Ÿ 3 Ê V œ 'c3 1 ˆÈ1 y# ‰ dy œ 2'0 1 ˆÈ1 y# ‰ dy 3 # # 3 œ 21'0 a1 y# b dy œ 21 ’y y3 “ œ 241 3 $ $ ! # 79. Let y œ É16 16 9 x on the interval 3 Ÿ x Ÿ 3. Since the plate is symmetric about the y-axis, x œ 0. For a vertical strip: aµ x ßµ y b œ xß # É16 16 9 x # , length œ É16 9 x , width œ dx Ê area œ dA œ É16 9 x dx 16 # # Ê mass œ dm œ $ dA œ $É16 16 9 x dx. Moment of the strip about the x-axis: É16 9 x µ 8 #‰ # ˆ y dm œ dx so the moment of the plate about the x-axis is Š$ É16 16 # 9 x ‹ dx œ $ 8 9 x 16 # Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 16 # Section 11.7 Conics in Polar Coordinates 689 8 $‘ Mx œ ' µ y dm œ 'c3 $ ˆ8 89 x# ‰ dx œ $ 8x 27 x $ œ 32$ ; also the mass of the plate is 3 $ # ' É1 ˆ "3 x‰ dx œ 4$ 'c1 3È1 u# du where u œ x3 Ê 3 du œ dx; x œ 3 M œ 'c3 $ É16 16 9 x dx œ c3 4$ 3 3 1 # Ê u œ 1 and x œ 3 Ê u œ 1. Hence, 4$ 'c1 3È1 u# du œ 12$ 'c1 È1 u# du 1 œ 12$ ’ "2 ŠuÈ1 u# sin" u‹“ " # 80. y œ Èx# 1 Ê dy dx œ # ax 1b È2 1 ' œ É 2x x# 1 Ê S œ 0 # – " " 1 œ 61$ Ê y œ MMx œ 6321$$ œ 3161 . Therefore the center of mass is ˆ!ß 3161 ‰ . "Î# # (2x) œ È #x x 1 # # # dy x É1 x#x 1 Ê Š dy dx ‹ œ x# 1 Ê Ê1 Š dx ‹ œ È2 È2 1 È # È # ' ' É 2x 21yÊ1 Š dy dx ‹ dx œ 0 21 x 1 x# 1 dx œ 0 21 2x 1 dx ; # # # 2 u œ È2x 21 ' È # 21 " Ä È u 1 du œ È ’ 2 ŠuÈu# 1 ln Šu Èu# 1‹‹“ œ È12 ’2È5 ln Š2 È5‹“ — 0 2 2 ! du œ È2 dx 81. (a) tan " œ mL Ê tan " œ f w (x! ) where f(x) œ È4px ; f w (x) œ "# (4px)"Î# (4p) œ È2p Ê f w (x! ) œ È2p 4px 4px ! 2p œ 2p y! Ê tan " œ y! . 0 y! (b) tan 9 œ mFP œ xy!! p œ x! p Š x ! p c y2p ‹ y ! 9 tan " (c) tan ! œ 1tan tan 9 tan " œ ! 1 b Š x ! p ‹ Š y2p ‹ y ! ! # y# 2p(x p) 2p ! 2p ! p) œ y!! (x! p ! 2p) œ 4px!y! (x2px œ 2p(x y! (x! p) œ y! ! p) 11.7 CONICS IN POLAR COORDINATES # y# 1. 16x# 25y# œ 400 Ê #x5 16 œ 1 Ê c œ È a# b # œ È25 16 œ 3 Ê e œ c œ 3 ; F a „ 3ß 0b ; a 5 directrices are x œ 0 „ ae œ „ ˆ 53 ‰ œ „ 25 3 5 # x# 2. 7x# 16y# œ 112 Ê 16 y7 œ 1 Ê c œ Èa# b# œ È16 7 œ 3 Ê e œ c œ 3 ; F a „ 3ß 0b ; a 4 directrices are x œ 0 „ ae œ „ ˆ 43 ‰ œ „ 16 3 4 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 690 Chapter 11 Parametric Equations and Polar Coordinates 3. 2x# y# œ 2 Ê x# y2 œ 1 Ê c œ Èa# b# œ È2 1 œ 1 Ê e œ c œ 1 ; F a0ß „ 1b ; # È2 a directrices are y œ 0 „ ae œ „ È2 Š È12 ‹ œ „2 4. 2x# y# œ 4 Ê x# y4 œ 1 Ê c œ Èa# b# # # œ È4 2 œ È2 Ê e œ ca œ È2 È 2‹ ; 2 ; F Š0ß „ directrices are y œ 0 „ ae œ „ È22 œ „ 2È2 Š ‹ 2 # # 5. 3x# 2y# œ 6 Ê x# y3 œ 1 Ê c œ Èa# b# œ È3 2 œ 1 Ê e œ c œ 1 ; F a0ß „ 1b ; È3 a directrices are y œ 0 „ ae œ „ È3 Š È13 ‹ œ „3 x 6. 9x# 10y# œ 90 Ê 10 y9 œ 1 Ê c œ Èa# b# œ È10 9 œ 1 Ê e œ c œ 1 ; F a „ 1ß 0b ; # # a directrices are x œ 0 „ ae œ „ È10 È10 Š È110 ‹ œ „ 10 7. 6x# 9y# œ 54 Ê x9 y6 œ 1 Ê c œ Èa# b# # # œ È9 6 œ È3 Ê e œ ca œ È3 È3ß 0‹ ; 3 ; FŠ „ directrices are x œ 0 „ ae œ „ È33 œ „ 3È3 Š ‹ 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.7 Conics in Polar Coordinates 691 y# x# 8. 169x# 25y# œ 4225 Ê 25 169 œ 1 Ê c œ Èa# b# œ È169 25 œ 12 Ê e œ c œ 12 ; F a0ß „ 12b ; a 13 169 directrices are y œ 0 „ ae œ „ ˆ13 12 ‰ œ „ 12 13 # # y 3 9. Foci: a0ß „ 3b , e œ 0.5 Ê c œ 3 and a œ ce œ 0.5 œ 6 Ê b# œ 36 9 œ 27 Ê #x7 36 œ1 # # y x 10. Foci: a „ 8ß 0b , e œ 0.2 Ê c œ 8 and a œ ce œ 0.8# œ 40 Ê b# œ 1600 64 œ 1536 Ê 1600 1536 œ1 # # y x 11. Vertices: a0ß „ 70b , e œ 0.1 Ê a œ 70 and c œ ae œ 70(0.1) œ 7 Ê b# œ 4900 49 œ 4851 Ê 4851 4900 œ1 # # y x 12. Vertices: a „ 10ß 0b , e œ 0.24 Ê a œ 10 and c œ ae œ 10(0.24) œ 2.4 Ê b# œ 100 5.76 œ 94.24 Ê 100 94.24 œ1 9 13. Focus: ŠÈ5ß !‹ , Directrix: x œ È95 Ê c œ ae œ È5 and ea œ È95 Ê ae e# œ È 5 Ê Ê eœ È5 3 . Then PF œ È5 3 PD # Ê ÊŠx È5‹ (y 0)# œ È5 9 3 ¹x È5 ¹ È5 9 e# œ È 5 Ê e# œ 95 # Ê Šx È5‹ y# œ 59 Šx È95 ‹ # y 18 4 # x # Ê x# 2È5 x 5 y# œ 59 Šx# È x 81 5 ‹ Ê 9 x y œ 4 Ê 9 4 œ 1 5 # # a 16 ae 16 4 16 3 # 14. Focus: (%ß 0), Directrix: x œ 16 3 Ê c œ ae œ 4 and e œ 3 Ê e# œ 3 Ê e# œ 3 Ê e œ 4 Ê e œ È3 È(x 4)# (y 0)# œ È3 ¸x 16 ¸ Ê (x 4)# y# œ 3 ˆx 16 ‰# # PD Ê # 3 4 3 y# " # 3 ˆ # 32 256 ‰ 16 x# # œ 4 x 3 x 9 Ê 4 x y œ 3 Ê ˆ 64 ‰ ˆ 16 ‰ œ 1 3 3 PF œ È3 # . Then Ê x# 8x 16 y# " 4 1 # 15. Focus: (%ß 0), Directrix: x œ 16 Ê c œ ae œ 4 and ae œ 16 Ê ae e# œ 16 Ê e# œ 16 Ê e œ 4 Ê e œ # . Then PF œ 1 PD Ê È(x 4)# (y 0)# œ 1 kx 16k Ê (x 4)# y# œ 1 (x 16)# Ê x# 8x 16 y# # # 4 # # y x œ 14 ax# 32x 256b Ê 43 x# y# œ 48 Ê 64 48 œ1 È2 " # È È 16. Focus: ŠÈ2ß !‹ , Directrix: x œ 2È2 Ê c œ ae œ È2 and ae œ 2È2 Ê ae e# œ 2 2 Ê e# œ 2 2 Ê e œ # # # Ê e œ È12 . Then PF œ È12 PD Ê ÊŠx È2‹ (y 0)# œ È12 ¹x 2È2¹ Ê Šx È2‹ y# # œ "# Šx 2È2‹ Ê x# 2È2 x 2 y# œ "# Šx# 4È2 x 8‹ Ê "# x# y# œ 2 Ê x4 y# œ 1 # Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. # 692 Chapter 11 Parametric Equations and Polar Coordinates 17. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 Ê e œ ca œ È2 È2 ; asymptotes are y œ 1 œ directrices are x œ 0 „ ae œ „ È" „ x; F Š „ È2 ß !‹ ; 2 # x# 18. 9x# 16y# œ 144 Ê 16 y9 œ 1 Ê c œ Èa# b# œ È16 9 œ 5 Ê e œ c œ 5 ; asymptotes are a 4 y œ „ 34 x; F a „ 5ß !b ; directrices are x œ 0 „ ae œ „ "6 5 # # 19. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b# œ È8 8 œ 4 Ê e œ c œ 4 œ È2 ; asymptotes are È8 a y œ „ x; F a0ß „ 4b ; directrices are y œ 0 „ ae È œ „ È82 œ „ 2 20. y# x# œ 4 Ê y4 x4 œ 1 Ê c œ Èa# b# # # È œ È4 4 œ 2È2 Ê e œ ca œ 2 2 2 œ È2 ; asymptotes are y œ „ x; F Š0ß „ 2È2‹ ; directrices are y œ 0 „ ae œ „ È22 œ „ È2 21. 8x# 2y# œ 16 Ê x2 y8 œ 1 Ê c œ Èa# b# # # È œ È2 8 œ È10 Ê e œ ca œ È10 œ È5 ; asymptotes 2 are y œ „ 2x; F Š „ È10ß !‹ ; directrices are x œ 0 „ ae È œ „ È25 œ „ È210 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.7 Conics in Polar Coordinates 693 22. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b# œ È3 1 œ 2 Ê e œ c œ 2 ; asymptotes are # È3 a y œ „ È3 x; F a0ß „ 2b ; directrices are y œ 0 „ ae œ „ È3 Š È23 ‹ œ „ 3# 23. 8y# 2x# œ 16 Ê y2 x8 œ 1 Ê c œ Èa# b# # # È œ È2 8 œ È10 Ê e œ ca œ È10 œ È5 ; asymptotes 2 are y œ „ x# ; F Š0ß „ È10‹ ; directrices are y œ 0 „ ae È œ „ È25 œ „ È210 y x 24. 64x# 36y# œ 2304 Ê 36 64 œ 1 Ê c œ È a# b # œ È36 64 œ 10 Ê e œ c œ 10 œ 5 ; asymptotes are # # a 6 3 y œ „ 43 x; F a „ 10ß !b ; directrices are x œ 0 „ ae œ „ ˆ 65 ‰ œ „ 18 5 3 # 25. Vertices a!ß „ 1b and e œ 3 Ê a œ 1 and e œ ca œ 3 Ê c œ 3a œ 3 Ê b# œ c# a# œ 9 1 œ 8 Ê y# x8 œ 1 # # 26. Vertices a „ 2ß !b and e œ 2 Ê a œ 2 and e œ ca œ 2 Ê c œ 2a œ 4 Ê b# œ c# a# œ 16 4 œ 12 Ê x4 1y# œ 1 # 27. Foci a „ 3ß !b and e œ 3 Ê c œ 3 and e œ ca œ 3 Ê c œ 3a Ê a œ 1 Ê b# œ c# a# œ 9 1 œ 8 Ê x# y8 œ 1 28. Foci a!ß „ 5b and e œ 1.25 Ê c œ 5 and e œ ca œ 1.25 œ 54 Ê c œ 54 a Ê 5 œ 54 a Ê a œ 4 Ê b# œ c# a# # # y œ 25 16 œ 9 Ê 16 x9 œ 1 2(1) 2 29. e œ 1, x œ 2 Ê k œ 2 Ê r œ 1 (1) cos ) œ 1cos ) 2(1) 2 30. e œ 1, y œ 2 Ê k œ 2 Ê r œ 1 (1) sin ) œ 1sin ) 30 31. e œ 5, y œ 6 Ê k œ 6 Ê r œ 1 6(5) 5 sin ) œ 15 sin ) 8 32. e œ 2, x œ 4 Ê k œ 4 Ê r œ 1 4(2) 2 cos ) œ 12 cos ) ˆ " ‰ (1) 1 33. e œ "# , x œ 1 Ê k œ 1 Ê r œ 1 ˆ#" ‰ cos ) œ 2cos ) # Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 694 Chapter 11 Parametric Equations and Polar Coordinates ˆ " ‰ (2) 2 34. e œ "4 , x œ 2 Ê k œ 2 Ê r œ 1 ˆ4" ‰ cos ) œ 4cos ) 4 ˆ " ‰ (10) 35. e œ 5" , y œ 10 Ê k œ 10 Ê r œ 1 5ˆ " ‰ sin ) œ 510 sin ) 5 ˆ " ‰ (6) 6 36. e œ "3 , y œ 6 Ê k œ 6 Ê r œ 1 ˆ3 " ‰ sin ) œ 3sin ) 3 37. r œ 1 "cos ) Ê e œ 1, k œ 1 Ê x œ 1 38. r œ 2 6cos ) œ 1 ˆ "3‰ cos ) Ê e œ "# , k œ 6 Ê x œ 6; # # a a1 e b œ ke Ê a ’1 ˆ "# ‰ “ œ 3 Ê 34 a œ 3 # Ê a œ 4 Ê ea œ 2 ˆ 25 ‰ ˆ5‰ 10 # 10 # 39. r œ 10 25 5 cos ) Ê r œ 1 ˆ 5 ‰ cos ) œ 1 ˆ " ‰ cos ) Ê e œ "# , k œ 5 Ê x œ 5; a a1 e# b œ ke # 5 Ê a ’1 ˆ "# ‰ “ œ 5# Ê 34 a œ 5# Ê a œ 10 3 Ê ea œ 3 2 40. r œ 224cos ) Ê r œ 1cos ) Ê e œ 1, k œ 2 Ê x œ 2 ˆ 400 ‰ 25 16 41. r œ 16 400 8 sin ) Ê r œ 1 ˆ 8 ‰ sin ) Ê r œ 1 ˆ " ‰ sin ) 16 e œ "# , k œ 50 Ê y œ 50; a a1 e# b œ ke # # Ê a ’1 ˆ "# ‰ “ œ 25 Ê 34 a œ 25 Ê a œ 100 3 Ê ea œ 50 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.7 Conics in Polar Coordinates 42. r œ 3 312sin ) Ê r œ 1 4sin ) Ê e œ 1, 43. r œ 2 28sin ) Ê r œ 1 4sin ) Ê e œ 1, kœ4 Ê yœ4 k œ 4 Ê y œ 4 44. r œ 2 4sin ) Ê r œ 1 ˆ "2‰ sin ) Ê e œ #" , k œ 4 # # Ê y œ 4; a a1 e# b œ ke Ê a ’1 ˆ "# ‰ “ œ 2 Ê 34 a œ 2 Ê a œ 83 Ê ea œ 43 45. r cos ˆ) 14 ‰ œ È2 Ê r ˆcos ) cos 14 sin ) sin 14 ‰ œ È2 Ê " r cos ) " r sin ) œ È2 Ê " x " y È2 È2 È2 È2 œ È2 Ê x y œ 2 Ê y œ 2 x 46. r cos ˆ) 341 ‰ œ 1 Ê r ˆcos ) cos 341 sin ) sin 341 ‰ œ 1 È È Ê 22 r cos ) 22 r sin ) œ 1 Ê x y œ È2 Ê y œ x È 2 47. r cos ˆ) 231 ‰ œ 3 Ê r ˆcos ) cos 231 sin ) sin 231 ‰ œ 3 È È Ê 12 r cos ) 23 r sin ) œ 3 Ê "# x #3 y œ 3 Ê x È 3 y œ 6 Ê y œ È3 È 3 x2 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 695 696 Chapter 11 Parametric Equations and Polar Coordinates 48. r cos ˆ) 13 ‰ œ 2 Ê r ˆcos ) cos 13 sin ) sin 13 ‰ œ 2 È È Ê 12 r cos ) 23 r sin ) œ 2 Ê "# x #3 y œ 2 Ê x È3 y œ 4 Ê y œ È3 4È 3 3 x 3 È È 49. È2 x È2 y œ 6 Ê È2 r cos ) È2 r sin ) œ 6 Ê r Š #2 cos ) #2 sin )‹ œ 3 Ê r ˆcos 14 cos ) sin 14 sin )‰ œ 3 Ê r cos ˆ) 14 ‰ œ 3 È 50. È3 x y œ 1 Ê È3 r cos ) r sin ) œ 1 Ê r Š #3 cos ) 1# sin )‹ œ #" Ê r ˆcos 16 cos ) sin 16 sin )‰ œ "# Ê r cos ˆ) 16 ‰ œ "# 51. y œ 5 Ê r sin ) œ 5 Ê r sin ) œ 5 Ê r sin ()) œ 5 Ê r cos ˆ 1# ())‰ œ 5 Ê r cos ˆ) 1# ‰ œ 5 52. x œ 4 Ê r cos ) œ 4 Ê r cos ) œ 4 Ê r cos () 1) œ 4 53. 54. 55. 56. 57. (x 6)# y# œ 36 Ê C œ (6ß 0), a œ 6 Ê r œ 12 cos ) is the polar equation 58. (x 2)# y# œ 4 Ê C œ (2ß 0), a œ 2 Ê r œ 4 cos ) is the polar equation Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 11.7 Conics in Polar Coordinates 59. x# (y 5)# œ 25 Ê C œ (!ß 5), a œ 5 Ê r œ 10 sin ) is the polar equation 60. x# (y 7)# œ 49 Ê C œ (!ß 7), a œ 7 Ê r œ 14 sin ) is the polar equation 61. x# 2x y# œ 0 Ê (x 1)# y# œ 1 Ê C œ (1ß 0), a œ 1 Ê r œ 2 cos ) is the polar equation 62. x# 16x y# œ 0 Ê (x 8)# y# œ 64 Ê C œ (8ß 0), a œ 8 Ê r œ 16 cos ) is the polar equation # 63. x# y# y œ 0 Ê x# ˆy "# ‰ œ 4" Ê C œ ˆ!ß "# ‰ , a œ "# Ê r œ sin ) is the # 64. x# y# 34 y œ 0 Ê x# ˆy 32 ‰ œ 49 Ê C œ ˆ0ß 23 ‰ , a œ 23 Ê r œ 43 sin ) is the polar equation 65. polar equation 66. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 697 698 Chapter 11 Parametric Equations and Polar Coordinates 67. 68. 69. 70. 71. 72. 73. 74. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 11 Practice Exercises 75. (a) Perihelion œ a ae œ a(1 e), Aphelion œ ea a œ a(1 e) (b) Planet Perihelion Aphelion Mercury 0.3075 AU 0.4667 AU Venus 0.7184 AU 0.7282 AU Earth 0.9833 AU 1.0167 AU Mars 1.3817 AU 1.6663 AU Jupiter 4.9512 AU 5.4548 AU Saturn 9.0210 AU 10.0570 AU Uranus 18.2977 AU 20.0623 AU Neptune 29.8135 AU 30.3065 AU # a1 0.2056 b 0.3707 76. Mercury: r œ (0.3871) œ 1 0.2056 1 0.2056 cos ) cos ) # a1 0.0068 b 0.7233 Venus: r œ (0.7233) œ 1 0.0068 1 0.0068 cos ) cos ) # 0.0167 b 0.9997 Earth: r œ 11a10.0167 cos ) œ 1 0.0617 cos ) # a1 0.0934 b 1.511 Mars: r œ (1.524) œ 1 0.0934 1 0.0934 cos ) cos ) # a1 0.0484 b 5.191 Jupiter: r œ (5.203) œ 1 0.0484 1 0.0484 cos ) cos ) # a1 0.0543 b 9.511 Saturn: r œ (9.539) œ 1 0.0543 1 0.0543 cos ) cos ) # a1 0.0460 b 19.14 Uranus: r œ (19.18) œ 1 0.0460 1 0.0460 cos ) cos ) # a1 0.0082 b 30.06 Neptune: r œ (30.06) œ 1 0.0082 1 0.0082 cos ) cos ) CHAPTER 11 PRACTICE EXERCISES 1. x œ #t and y œ t 1 Ê 2x œ t Ê y œ 2x 1 2. x œ Èt and y œ 1 Èt Ê y œ 1 x 3. x œ "# tan t and y œ "# sec t Ê x# œ 4" tan# t 4. x œ 2 cos t and y œ 2 sin t Ê x# œ 4 cos# t and and y# œ "4 sec# t Ê 4x# œ tan# t and 4y# œ sec# t Ê 4x# 1 œ 4y# Ê 4y# 4x# œ 1 y# œ 4 sin# t Ê x# y# œ 4 699 700 Chapter 11 Parametric Equations and Polar Coordinates 5. x œ cos t and y œ cos# t Ê y œ (x)# œ x# 6. x œ 4 cos t and y œ 9 sin t Ê x# œ 6 cos# t and # # y x y# œ 81 sin# t Ê 16 81 œ1 # # y 7. 16x# 9y# œ 144 Ê x9 16 œ 1 Ê a œ 3 and b œ 4 Ê x œ 3 cos t and y œ 4 sin t, 0 Ÿ t Ÿ 21 8. x# y# œ 4 Ê x œ 2 cos t and y œ 2 sin t, 0 Ÿ t Ÿ 61 " dy/dt dy tan t # 9. x œ "# tan t, y œ "# sec t Ê dy œ sec " dx œ dx/dt œ t œ sin t Ê dx ¹ sec# t sec t tan t # Ê x œ "# tan 13 œ È3 " 1 # and y œ # sec 3 œ 1 tœ1Î3 œ sin 13 œ È3 1 # ;tœ 3 È3 dyw /dt " d# y cos t 3 # x 4 ; dx# œ dx/dt œ "# sec# t œ 2 cos t Ê yœ # Ê ddxy# ¹ tœ1Î3 œ 2 cos3 ˆ 13 ‰ œ "4 dy/dt 10. x œ " t"# , y œ " 3t Ê dy dx œ dx/dt œ Š t3# ‹ Š t2$ ‹ # œ 32 t Ê dy dx ¹ ˆ 3# ‰ w /dt y œ 1 #3 œ #" Ê y œ 3x "43 ; ddxy# œ dy dx/dt œ Èx Š t2$ ‹ Èx tœ2 œ 3# (2) œ 3; t œ 2 Ê x œ 1 #"# œ 54 and # œ 43 t$ Ê ddxy# ¹ 3 tœ2 œ 43 (2)$ œ 6 3Î2 11. (a) x œ 4t2 , y œ t3 1 Ê t œ „ 2 Ê y œ Š „ 2 ‹ 1 œ „ x 8 1 (b) x œ cos t, y œ tan t Ê sec t œ 1x Ê tan2 t 1 œ sec2 t Ê y2 œ x12 1 œ 1 x2x Ê y œ „ 2 È 1 x2 x 12. (a) The line through a1, 2b with slope 3 is y œ 3x 5 Ê x œ t, y œ 3t 5, _ t _ (b) ax 1b2 ay 2b2 œ 9 Ê x 1 œ 3 cos t, y 2 œ 3 sin t Ê x œ 1 3 cos t, y œ 2 3 sin t, 0 Ÿ t Ÿ 21 (c) y œ 4x2 x Ê x œ t, y œ 4t2 t, _ t _ 2 2 (d) 9x2 4y2 œ 36 Ê x4 y9 œ 1 Ê x œ 2 cos t, y œ 3 sin t, 0 Ÿ t Ÿ 21 $Î# 13. y œ x"Î# x 3 " "Î# " ˆ" ' É1 4" ˆ x" 2 x‰ dx ‰ Ê dy #" x"Î# Ê Š dy dx œ # x dx ‹ œ 4 x 2 x Ê L œ 1 # 4 Ê L œ '1 É 4" ˆ x" 2 x‰ dx œ '1 É 4" ax"Î# x"Î# b dx œ '1 #" ˆx"Î# x"Î# ‰ dx œ #" 2x"Î# 23 x$Î# ‘ " 4 4 % 4 # ‰ 10 œ "# ˆ4 23 † 8‰ ˆ2 32 ‰‘ œ "# ˆ2 14 3 œ 3 # #Î$ 2 "Î$ 4x 14. x œ y#Î$ Ê dx Ê Š dx dy œ 3 x dy ‹ œ 9 œ '1 ' É1 9x4#Î$ dy Ê L œ '1 Ê1 Š dx dy ‹ dy œ 1 8 # 8 '18 È9x#Î$ 4 ˆx"Î$ ‰ dx; u œ 9x#Î$ 4 Ê du œ 6y"Î$ dy; x œ 1 Ê u œ 13, 40 " ' " 2 $Î# ‘ %! x œ 8 Ê u œ 40d Ä L œ 18 u"Î# du œ 18 œ #"7 40$Î# 13$Î# ‘ ¸ 7.634 3 u "$ 13 8È 9x#Î$ 4 dx œ 3" 3x"Î$ # " "Î& " ˆ #Î& 5 'Î& 15. y œ 12 x 58 x%Î& Ê dy "# x"Î& Ê Š dy 2 x#Î& ‰ dx œ # x dx ‹ œ 4 x Ê L œ '1 É1 "4 ax#Î& 2 x#Î& b dx Ê L œ '1 É 4" ax#Î& 2 x#Î& b dx œ ' É "4 ax"Î& x"Î& b dx 32 32 32 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. # Chapter 11 Practice Exercises œ '1 $# " ˆ "Î& 75 ‰ x"Î& ‰ dx œ "# 56 x'Î& 54 x%Î& ‘ " œ "# ˆ 56 † 2' 54 † 2% ‰ ˆ 56 54 ‰‘ œ "# ˆ 315 # x 6 4 " 285 œ 48 (1260 450) œ 1710 48 œ 8 32 " # " " % " " " % " " dx ' 16. x œ 1"# y$ "y Ê dx dy œ 4 y y# Ê Š dy ‹ œ 16 y # y% Ê L œ 1 Ê1 Š 16 y # y% ‹ dy # 2 " % œ '1 É 16 y #" y"% dy œ '1 ÊŠ 4" y# y"# ‹ dy œ '1 Š 4" y# y"# ‹ dy œ ’ 1"# y$ y" “ 2 # 2 2 # " 8 œ ˆ 12 "# ‰ ˆ 1"# 1‰ œ 17# "# œ 13 12 # # dy ˆ dx ‰ Š dy 17. dx dt œ 5 sin t 5 sin 5t and dt œ 5 cos t 5 cos 5t Ê Ê dt dt ‹ œ Éa5 sin t 5 sin 5tb# a5 cos t 5 cos 5tb# œ 5Èsin# 5t #sin t sin 5t sin# t cos# t #cos t cos 5t cos# 5t œ &È# #asin t sin 5t cos t cos 5 tb œ 5È#a" cos %tb œ 5É%ˆ "# ‰a" cos %tb œ "!Èsin# #t œ "!lsin #tl œ "!sin #t (since ! Ÿ t Ÿ 1# ) Ê Length œ '! 1 Î2 1Î# "!sin #t dt œ c5 cos #td ! œ a&ba"b a&ba"b œ "! # # dy 2 2 ˆ dx ‰ Š dy Éa3t2 12tb# a3t2 12tb# œ È288t# "8t4 18. dx dt œ 3t 12t and dt œ 3t 12t Ê Ê dt dt ‹ œ œ 3È2 ktkÈ16 t2 Ê Length œ '! 3È2 ktkÈ16 t2 dt œ 3È2'! t È16 t2 dt; ’u œ 16 t2 Ê du œ 2t dt " " Ê "# du œ t dt; t œ 0 Ê u œ 16; t œ 1 Ê u œ 17“; 3 2 2 '16 Èu du œ 3 2 2 23 u3/2 ‘16 œ 3 2 2 Š 23 a17b3/2 23 a16b3/2 ‹ "7 È È È 17 È œ 3 2 2 † 23 Ša17b3/2 64‹ œ È2Ša17b3/2 64‹ ¸ 8.617. # # dy ˆ dx ‰ Š dy Éa$ sin )b# a$ cos )b# œ È$asin# ) cos# )b œ $ 19. dx d) œ $ sin ) and d) œ $ cos ) Ê Ê d) d) ‹ œ Ê Length œ '! $1Î2 20. x œ t œ # ' $ d) œ $'! $1Î2 d) œ $ˆ $#1 !‰ œ *#1 $ dy # and y œ t3 t, È3 Ÿ t Ÿ È3 Ê dx dt œ 2t and dt œ t " Ê Length œ È3 È3 Èt% #t# " dt œ ' È3 È 3 Èt% 2t# " dt œ ' È3 È 3 Éat# "b# dt œ ' È3 È 3 Éa2tb# at# "b# dt È 'È33 at# "b dt œ ’ t3 t“ È3 3 È 3 œ 4È 3 È5 dy ' 21. x œ t# and y œ 2t, 0 Ÿ t Ÿ È5 Ê dx dt œ t and dt œ 2 Ê Surface Area œ 0 # * 21(2t)Èt# 4 dt œ '4 21u"Î# du 9 œ 21 23 u$Î# ‘ % œ 7631 , where u œ t# 4 Ê du œ 2t dt; t œ 0 Ê u œ 4, t œ È5 Ê u œ 9 dy " 2 22. x œ t# 2t" and y œ 4Èt , È" Ÿ t Ÿ 1 Ê dx dt œ 2t 2t# and dt œ Èt 2 Ê Surface Area œ '1ÎÈ2 21 ˆt# #"t ‰ Êˆ2t 2t"# ‰ Š È2 t ‹ dt œ 21 '1ÎÈ2 ˆt# #"t ‰ Éˆ2t #"t# ‰ dt 1 # # 1 # œ 21 '1ÎÈ2 ˆt# 2t" ‰ ˆ2t 2t"# ‰ dt œ 21 '1ÎÈ2 ˆ2t$ 3# "4 t$ ‰ dt œ 21 "2 t% 3# t 8" t# ‘ "ÎÈ# 1 1 È œ 21 Š2 3 4 2 ‹ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. " 701 702 Chapter 11 Parametric Equations and Polar Coordinates 23. r cos ˆ) 13 ‰ œ 2È3 Ê r ˆcos ) cos 13 sin ) sin 13 ‰ È œ 2È3 Ê "# r cos ) #3 r sin ) œ 2È3 Ê r cos ) È3 r sin ) œ 4È3 Ê x È3 y œ 4È3 Ê yœ È3 3 x4 24. r cos ˆ) 341 ‰ œ œ È2 # È È2 # Ê r ˆcos ) cos 341 sin ) sin 341 ‰ È Ê #2 r cos ) #2 r sin ) œ Ê yœx1 È2 # Ê x y œ 1 25. r œ 2 sec ) Ê r œ cos2 ) Ê r cos ) œ 2 Ê x œ 2 26. r œ È2 sec ) Ê r cos ) œ È2 Ê x œ È2 27. r œ 3# csc ) Ê r sin ) œ 3# Ê y œ 3# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 11 Practice Exercises 28. r œ 3È3 csc ) Ê r sin ) œ 3È3 Ê y œ 3È3 29. r œ 4 sin ) Ê r# œ 4r sin ) Ê x# y# 4y œ 0 Ê x# (y 2)# œ 4; circle with center (!ß 2) and radius 2. 30. r œ 3È3 sin ) Ê r# œ 3È3 r sin ) # È Ê x# y# 3È3 y œ 0 Ê x# Šy 3 # 3 ‹ œ 27 4 ; È È circle with center Š!ß 3 # 3 ‹ and radius 3 # 3 31. r œ 2È2 cos ) Ê r# œ 2È2 r cos ) # Ê x# y# 2È2 x œ 0 Ê Šx È2‹ y# œ 2; circle with center ŠÈ2ß 0‹ and radius È2 32. r œ 6 cos ) Ê r# œ 6r cos ) Ê x# y# 6x œ 0 Ê (x 3)# y# œ 9; circle with center (3ß 0) and radius 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 703 704 Chapter 11 Parametric Equations and Polar Coordinates # 5‰ ˆ 33. x# y# 5y œ 0 Ê x# ˆy #5 ‰ œ 25 4 Ê C œ !ß # and a œ 5# ; r# 5r sin ) œ 0 Ê r œ 5 sin ) 34. x# y# 2y œ 0 Ê x# (y 1)# œ 1 Ê C œ (!ß 1) and a œ 1; r# 2r sin ) œ 0 Ê r œ 2 sin ) # 35. x# y# 3x œ 0 Ê ˆx 3# ‰ y# œ 49 Ê C œ ˆ 3# ß !‰ and a œ 3# ; r# 3r cos ) œ 0 Ê r œ 3 cos ) 36. x# y# 4x œ 0 Ê (x 2)# y# œ 4 Ê C œ (2ß 0) and a œ 2; r# 4r cos ) œ 0 Ê r œ 4 cos ) 37. 38. 39. d 40. e 41. l 42. f 43. k 44. h 45. i 46. j Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 11 Practice Exercises 2) ‰ 47. A œ 2'0 "# r# d) œ '0 (2 cos ))# d) œ '0 a4 4 cos ) cos# )b d) œ '0 ˆ4 4 cos ) 1 cos d) # 1 1 1 1 œ '0 ˆ 9# 4 cos ) cos# 2) ‰ d) œ 92 ) 4 sin ) sin42) ‘ ! œ 9# 1 1 1 48. A œ '0 1Î3 " # # asin 3)b d) œ 1Î$ 6) ‰ 1 '01Î3 ˆ 1 cos d) œ "4 ) "6 sin 6)‘ ! œ 12 # 49. r œ 1 cos 2) and r œ 1 Ê 1 œ 1 cos 2) Ê 0 œ cos 2) Ê 2) œ 1# Ê ) œ 14 ; therefore A œ 4'0 1Î4 " # # # c(1 cos 2)) 1 d d) œ 2 '01Î4 a1 2 cos 2) cos# 2) 1b d) 1Î% œ 2'0 ˆ2 cos 2) "# cos# 4) ‰ d) œ 2 sin 2) "2 ) sin84) ‘ ! œ 2 ˆ1 18 0‰ œ 2 14 1Î4 50. The circle lies interior to the cardioid. Thus, 1Î2 A œ 2 ' 1Î2 "# [2(1 sin ))]# d) 1 (the integral is the area of the cardioid minus the area of the circle) 1Î2 1Î2 œ ' 1Î2 4 a1 2 sin ) sin# )b d) 1 œ ' 1Î2 (6 8 sin ) 2 cos 2)) d) 1 œ c6) 8 cos ) sin 2)d 1Î# 1 1Î# œ c31 (31)d 1 œ 51 dr 51. r œ 1 cos ) Ê d) œ sin ); Length œ '0 È(1 cos ))# ( sin ))# d) œ '0 È2 2 cos ) d) 21 21 #1 œ '0 É 4(1 #cos )) d) œ '0 2 sin #) d) œ 4 cos 2) ‘ ! œ (4)(1) (4)(1) œ 8 21 21 # 52. r œ 2 sin ) 2 cos ), 0 Ÿ ) Ÿ 1# Ê ddr) œ 2 cos ) 2 sin ); r# ˆ ddr) ‰ œ (2 sin ) 2 cos ))# (2 cos ) 2 sin ))# œ 8 asin# ) cos# )b œ 8 Ê L œ '0 È8 d) œ ’2È2 )“ 1Î2 1Î# ! œ 2È2 ˆ 1# ‰ œ 1È2 # # # 53. r œ 8 sin$ ˆ 3) ‰ , 0 Ÿ ) Ÿ 14 Ê ddr) œ 8 sin# ˆ 3) ‰ cos ˆ 3) ‰ ; r# ˆ ddr) ‰ œ 8 sin$ ˆ 3) ‰‘ 8 sin# ˆ 3) ‰ cos ˆ 3) ‰‘ œ 64 sin% ˆ 3) ‰ Ê L œ '0 É64 sin% ˆ 3) ‰ d) œ '0 1Î4 1Î4 8 sin# ˆ 3) ‰ d) œ '0 8 ’ 1Î4 1cos ˆ 23) ‰ “ d) # 1Î% œ '0 4 4 cos ˆ 23) ‰‘ d) œ 4) 6 sin ˆ 23) ‰‘ ! œ 4 ˆ 14 ‰ 6 sin ˆ 16 ‰ 0 œ 1 3 1Î4 # # sin 2) 54. r œ È1 cos 2) Ê ddr) œ "# (1 cos 2))"Î# (2 sin 2)) œ È1 sincos2)2) Ê ˆ ddr) ‰ œ 1 cos 2) # # # (1 cos 2)) sin 2) sin 2) cos 2) sin 2) Ê r# ˆ ddr) ‰ œ 1 cos 2) 1 œ 1 2 cos 2)1cos cos 2) œ 1 cos 2) 2) # # # 1Î2 cos 2) ' 1Î2 È2 d) œ È2 1# ˆ 1# ‰‘ œ È2 1 œ 212cos 2) œ 2 Ê L œ # 55. x# œ 4y Ê y œ x4 Ê 4p œ 4 Ê p œ 1; therefore Focus is (0ß 1), Directrix is y œ 1 # 56. x# œ 2y Ê x# œ y Ê 4p œ 2 Ê p œ "# ; therefore Focus is ˆ!ß "# ‰; Directrix is y œ "# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 705 706 Chapter 11 Parametric Equations and Polar Coordinates # # 57. y# œ 3x Ê x œ y3 Ê 4p œ 3 Ê p œ 43 ; 58. y# œ 38 x Ê x œ ˆy8 ‰ Ê 4p œ 38 Ê p œ 32 ; 3 therefore Focus is ˆ 34 ß 0‰ , Directrix is x œ 34 # therefore Focus is ˆ 23 ß !‰ , Directrix is x œ 23 # # # y 59. 16x# 7y# œ 112 Ê x7 16 œ1 60. x# 2y# œ 4 Ê x4 y# œ 1 Ê c# œ 4 2 œ 2 Ê c# œ 16 7 œ 9 Ê c œ 3; e œ ca œ 34 Ê c œ È2 ; e œ ca œ # 61. 3x# y# œ 3 Ê x# y3 œ 1 Ê c# œ 1 3 œ 4 Ê c œ 2; e œ ca œ 21 œ 2; the asymptotes are # È2 # # 62. 5y# 4x# œ 20 Ê y4 x5 œ 1 Ê c# œ 4 5 œ 9 Ê c œ 3, e œ ca œ 3# ; the asymptotes are y œ „ È25 x y œ „ È3 x # 63. x# œ 12y Ê 1x# œ y Ê 4p œ 12 Ê p œ 3 Ê focus is (!ß 3), directrix is y œ 3, vertex is (0ß 0); therefore new vertex is (2ß 3), new focus is (2ß 0), new directrix is y œ 6, and the new equation is (x 2)# œ 12(y 3) # y 64. y# œ 10x Ê 10 œ x Ê 4p œ 10 Ê p œ 5# Ê focus is ˆ 5# ß 0‰ , directrix is x œ 5# , vertex is (0ß 0); therefore new vertex is ˆ "# ß 1‰ , new focus is (2ß 1), new directrix is x œ 3, and the new equation is (y 1)# œ 10 ˆx "# ‰ # # 65. x9 #y5 œ 1 Ê a œ 5 and b œ 3 Ê c œ È25 9 œ 4 Ê foci are a!ß „ 4b , vertices are a!ß „ 5b , center is (0ß 0); therefore the new center is ($ß 5), new foci are (3ß 1) and (3ß 9), new vertices are ($ß 10) and # # ($ß 0), and the new equation is (x 9 3) (y #55) œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 11 Practice Exercises 707 # # y x 66. 169 144 œ 1 Ê a œ 13 and b œ 12 Ê c œ È169 144 œ 5 Ê foci are a „ 5ß 0b , vertices are a „ 13ß 0b , center is (0ß 0); therefore the new center is (5ß 12), new foci are (10ß 12) and (0ß 12), new vertices are (18ß 12) and # # 5) (8ß 12), and the new equation is (x169 (y 14412) œ 1 # # 67. y8 x2 œ 1 Ê a œ 2È2 and b œ È2 Ê c œ È8 2 œ È10 Ê foci are Š0ß „ È10‹ , vertices are Š0ß „ 2È2‹ , center is (0ß 0), and the asymptotes are y œ „ 2x; therefore the new center is Š2ß 2È2‹, new foci are Š2ß 2È2 „ È10‹ , new vertices are Š2ß 4È2‹ and (#ß 0), the new asymptotes are y œ 2x 4 2È2 and # y 2È 2‹ Š y œ 2x 4 2È2; the new equation is # 8 # (x # 2) œ 1 # y x 68. 36 64 œ 1 Ê a œ 6 and b œ 8 Ê c œ È36 64 œ 10 Ê foci are a „ 10ß 0b , vertices are a „ 6ß 0b , the center is (0ß 0) and the asymptotes are y8 œ „ x6 or y œ „ 43 x; therefore the new center is (10ß 3), the new foci are (20ß 3) and (0ß 3), the new vertices are (16ß 3) and (4ß 3), the new asymptotes are y œ 43 x 31 3 and # # (x 10) y œ 43 x 49 (y 643) œ 1 3 ; the new equation is 36 # 69. x# 4x 4y# œ 0 Ê x# 4x 4 4y# œ 4 Ê (x 2)# 4y# œ 4 Ê (x 4 2) y# œ 1, a hyperbola; a œ 2 and b œ 1 Ê c œ È1 4 œ È5 ; the center is (2ß 0), the vertices are (!ß 0) and (4ß 0); the foci are Š2 „ È5 ß 0‹ and the asymptotes are y œ „ x #2 # 70. 4x# y# 4y œ 8 Ê 4x# y# 4y 4 œ 4 Ê 4x# (y 2)# œ 4 Ê x# (y 4 2) œ 1, a hyperbola; a œ 1 and b œ 2 Ê c œ È1 4 œ È5 ; the center is (!ß 2), the vertices are (1ß 2) and ("ß 2), the foci are Š „ È5ß 2‹ and the asymptotes are y œ „ 2x 2 71. y# 2y 16x œ 49 Ê y# 2y 1 œ 16x 48 Ê (y 1)# œ 16(x 3), a parabola; the vertex is ($ß 1); 4p œ 16 Ê p œ 4 Ê the focus is (7ß 1) and the directrix is x œ 1 72. x# 2x 8y œ 17 Ê x# 2x 1 œ 8y 16 Ê (x 1)# œ 8(y 2), a parabola; the vertex is (1ß 2); 4p œ 8 Ê p œ 2 Ê the focus is (1ß 4) and the directrix is y œ 0 73. 9x# 16y# 54x 64y œ 1 Ê 9 ax# 6xb 16 ay# 4yb œ 1 Ê 9 ax# 6x 9b 16 ay# 4y 4b œ 144 # # Ê 9(x 3)# 16(y 2)# œ 144 Ê (x 163) (y 9 2) œ 1, an ellipse; the center is (3ß 2); a œ 4 and b œ 3 Ê c œ È16 9 œ È7 ; the foci are Š$ „ È7ß 2‹ ; the vertices are (1ß 2) and (7ß 2) 74. 25x# 9y# 100x 54y œ 44 Ê 25 ax# 4xb 9 ay# 6yb œ 44 Ê 25 ax# 4x 4b 9 ay# 6y 9b œ 225 # # Ê (x 2) (y 3) œ 1, an ellipse; the center is (2ß 3); a œ 5 and b œ 3 Ê c œ È25 9 œ 4; the foci are 9 25 (2ß 1) and (2ß 7); the vertices are (2ß 2) and (2ß 8) 75. x# y# 2x 2y œ 0 Ê x# 2x 1 y# 2y 1 œ 2 Ê (x 1)# (y 1)# œ 2, a circle with center (1ß 1) and radius œ È2 76. x# y# 4x 2y œ 1 Ê x# 4x 4 y# 2y 1 œ 6 Ê (x 2)# (y 1)# œ 6, a circle with center (2ß 1) and radius œ È6 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 708 Chapter 11 Parametric Equations and Polar Coordinates 77. r œ 1 2cos ) Ê e œ 1 Ê parabola with vertex at (1ß 0) 78. r œ 2 8cos ) Ê r œ 1 ˆ "4‰ cos ) Ê e œ "# Ê ellipse; # ke œ 4 Ê "# k œ 4 Ê k œ 8; k œ ae ea Ê 8 œ ˆ a" ‰ "# a # ˆ " ‰ ˆ 16 ‰ 8 Ê a œ 16 3 Ê ea œ # 3 œ 3 ; therefore the center is ˆ 83 ß 1‰ ; vertices are ()ß 1) and ˆ 83 ß 0‰ 79. r œ 1 26cos ) Ê e œ 2 Ê hyperbola; ke œ 6 Ê 2k œ 6 Ê k œ 3 Ê vertices are (2ß 1) and (6ß 1) 80. r œ 3 12sin ) Ê r œ 1 ˆ "4‰ sin ) Ê e œ "3 ; ke œ 4 3 Ê " 3 kœ4 # Ê k œ 12; a a1 e# b œ 4 Ê a ’1 ˆ 3" ‰ “ œ 4 Ê a œ 9# Ê ea œ ˆ "3 ‰ ˆ 9# ‰ œ 3# ; therefore the center is ˆ 3# ß 3#1 ‰ ; vertices are ˆ3ß 1# ‰ and ˆ6ß 3#1 ‰ 81. e œ 2 and r cos ) œ 2 Ê x œ 2 is directrix Ê k œ 2; the conic is a hyperbola; r œ 1 ekecos ) Ê r œ 1 (2)(2) # cos ) Ê r œ 1 #4cos ) (4)(1) 82. e œ 1 and r cos ) œ 4 Ê x œ 4 is directrix Ê k œ 4; the conic is a parabola; r œ 1 ekecos ) Ê r œ 1 cos ) Ê r œ 1 4cos ) (2) ˆ " ‰ 83. e œ "# and r sin ) œ 2 Ê y œ 2 is directrix Ê k œ 2; the conic is an ellipse; r œ 1 ekesin ) Ê r œ 1 ˆ " ‰#sin ) # Ê r œ 2 2sin ) (6) ˆ " ‰ 84. e œ "3 and r sin ) œ 6 Ê y œ 6 is directrix Ê k œ 6; the conic is an ellipse; r œ 1 ekesin ) Ê r œ 1 ˆ " ‰3sin ) 3 Ê r œ 3 6sin ) 85. (a) Around the x-axis: 9x# 4y# œ 36 Ê y# œ 9 94 x# Ê y œ „ É9 94 x# and we use the positive root: # V œ 2 '0 1 ŠÉ9 94 x# ‹ dx œ 2 '0 1 ˆ9 94 x# ‰ dx œ 21 9x 34 x$ ‘ ! œ 241 2 2 # Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 11 Additional and Advanced Exercises 709 (b) Around the y-axis: 9x# 4y# œ 36 Ê x# œ 4 49 y# Ê x œ „ É4 49 y# and we use the positive root: # 4 $‘ $ V œ 2 '0 1 ŠÉ4 49 y# ‹ dy œ 2'0 1 ˆ4 49 y# ‰ dy œ 21 4y 27 y ! œ 161 3 86. 3 9x# 4y# œ 36, x œ 4 Ê y# œ 9x 4 36 Ê y œ #3 Èx# 4 ; V œ '2 1 Š #3 Èx# 4‹ dx œ 941 '2 ax# 4b dx 4 # # 4 % $ 24 ‰ 31 ‰ ˆ8 ‰‘ œ 941 ˆ 56 œ 941 ’ x3 4x“ œ 941 ˆ 64 3 16 3 8 3 3 œ 4 (32) œ 241 # 87. (a) r œ 1 ekcos ) Ê r er cos ) œ k Ê Èx# y# ex œ k Ê Èx# y# œ k ex Ê x# y# œ k# 2kex e# x# Ê x# e# x# y# 2kex k# œ 0 Ê a1 e# b x# y# 2kex k# œ 0 (b) e œ 0 Ê x# y# k# œ 0 Ê x# y# œ k# Ê circle; 0 e 1 Ê e# 1 Ê e# 1 0 Ê B# 4AC œ 0# 4 a1 e# b (1) œ 4 ae# 1b 0 Ê ellipse; e œ 1 Ê B# 4AC œ 0# 4(0)(1) œ 0 Ê parabola; e 1 Ê e# 1 Ê B# 4AC œ 0# 4 a1 e# b (1) œ 4e# 4 0 Ê hyperbola 88. Let (r" ß )" ) be a point on the graph where r" œ a)" . Let (r# ß )# ) be on the graph where r# œ a)# and )# œ )" 21. Then r" and r# lie on the same ray on consecutive turns of the spiral and the distance between the two points is r# r" œ a)# a)" œ a()# )" ) œ 21a, which is constant. CHAPTER 11 ADDITIONAL AND ADVANCED EXERCISES 1. Directrix x œ 3 and focus (4ß 0) Ê vertex is ˆ 7# ß !‰ # Ê p œ "# Ê the equation is x 7# œ y# # 2. x# 6x 12y 9 œ 0 Ê x# 6x 9 œ 12y Ê (x123) œ y Ê vertex is (3ß 0) and p œ 3 Ê focus is (3ß 3) and the directrix is y œ 3 3. x# œ 4y Ê vertex is (!ß 0) and p œ 1 Ê focus is (!ß 1); thus the distance from P(xß y) to the vertex is Èx# y# and the distance from P to the focus is Èx# (y 1)# Ê Èx# y# œ 2Èx# (y 1)# Ê x# y# œ 4 cx# (y 1)# d Ê x# y# œ 4x# 4y# 8y 4 Ê 3x# 3y# 8y 4 œ 0, which is a circle 4. Let the segment a b intersect the y-axis in point A and intersect the x-axis in point B so that PB œ b and PA œ a (see figure). Draw the horizontal line through P and let it intersect the y-axis in point C. Let nPBO œ ) Ê nAPC œ ). Then sin ) œ yb and cos ) œ xa # # Ê xa# by# œ cos# ) sin# ) œ 1. 5. Vertices are a!ß „ 2b Ê a œ 2; e œ ca Ê 0.5 œ #c Ê c œ 1 Ê foci are a0ß „ 1b Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 710 Chapter 11 Parametric Equations and Polar Coordinates 6. Let the center of the ellipse be (xß 0); directrix x œ 2, focus (4ß 0), and e œ 23 Ê ae c œ 2 Ê ae œ 2 c a Ê a œ 32 (2 c). Also c œ ae œ 32 a Ê a œ 32 ˆ2 32 a‰ Ê a œ 34 94 a Ê 95 a œ 34 Ê a œ 12 5 ;x2œ e 28 28 8 # # # ‰ ˆ 3# ‰ œ 18 ˆ 28 ‰ Ê x 2 œ ˆ 12 5 5 Ê x œ 5 Ê the center is 5 ß 0 ; x 4 œ c Ê c œ 5 4 œ 5 so that c œ a b # # 80 ‰ ˆ 58 ‰ œ 25 œ ˆ 12 ; therefore the equation is 5 # # ˆx 28 ‰# ‰# 25 ˆx 28 5 5 ˆy80 ‰ œ 1 or 5y ˆ 144 ‰ 144 16 œ 1 25 25 7. Let the center of the hyperbola be (0ß y). (a) Directrix y œ 1, focus (0ß 7) and e œ 2 Ê c ae œ 6 Ê ae œ c 6 Ê a œ 2c 12. Also c œ ae œ 2a Ê a œ 2(2a) 12 Ê a œ 4 Ê c œ 8; y (1) œ ae œ #4 œ 2 Ê y œ 1 Ê the center is (0ß 1); c# œ a# b# # x Ê b# œ c# a# œ 64 16 œ 48; therefore the equation is (y161) 48 œ1 # (b) e œ 5 Ê c ae œ 6 Ê ae œ c 6 Ê a œ 5c 30. Also, c œ ae œ 5a Ê a œ 5(5a) 30 Ê 24a œ 30 Ê a œ 54 a Ê c œ 25 4 ; y (1) œ e œ ˆ 54 ‰ " 5 œ 4 Ê y œ 43 Ê the center is ˆ!ß 43 ‰ ; c# œ a# b# Ê b# œ c# a# 25 75 œ 625 16 16 œ # ; therefore the equation is # # ˆy 34 ‰# 16 ˆy 34 ‰ x# œ 1 or 2x 75 25 ˆ#‰ ˆ 16 ‰ 25 75 œ 1 # # 144 8. The center is (0ß 0) and c œ 2 Ê 4 œ a# b# Ê b# œ 4 a# . The equation is ya# bx# œ 1 Ê 49 a# b# œ 1 144 # # # # # # # % % # Ê 49 a# a4a# b œ 1 Ê 49 a4 a b 144a œ a a4 a b Ê 196 49a 144a œ 4a a Ê a 197a 196 œ 0 Ê aa# 196b aa# 1b œ 0 Ê a œ 14 or a œ 1; a œ 14 Ê b# œ 4 (14)# 0 which is impossible; a œ 1 # Ê b# œ 4 1 œ 3; therefore the equation is y# x3 œ 1 # # b x" b x 9. b# x# a# y# œ a# b# Ê dy dx œ a# y ; at (x" ß y" ) the tangent line is y y" œ Š a# y" ‹ (x x" ) Ê a# yy" b# xx" œ b# x"# a# y"# œ a# b# Ê b# xx" a# yy" a# b# œ 0 # # b x" b x 10. b# x# a# y# œ a# b# Ê dy dx œ a# y ; at (x" ß y" ) the tangent line is y y" œ Š a# y" ‹ (x x" ) Ê b# xx" a# yy" œ b# x"# a# y"# œ a# b# Ê b# xx" a# yy" a# b# œ 0 11. 12. 13. 14. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 11 Additional and Advanced Exercises 711 15. a9x# 4y# 36b a4x# 9y# 16b Ÿ 0 Ê 9x# 4y# 36 Ÿ 0 and 4x# 9y# 16 0 or 9x# 4y# 36 0 and 4x# 9y# 16 Ÿ 0 16. a9x# 4y# 36b a4x# 9y# 16b 0, which is the complement of the set in Exercise 15 sin t 17. (a) x œ e2t cos t and y œ e2t sin t Ê x# y# œ e4t cos# t e4t sin# t œ e4t . Also yx œ ee2t cos t œ tan t " ˆ y ‰ # # % tan " ayÎxb # # Ê t œ tan Ê x y œe is the Cartesian equation. Since r œ x y# and x 2t ) œ tan" ˆ yx ‰ , the polar equation is r# œ e4) or r œ e2) for r 0 (b) ds# œ r# d)# dr# ; r œ e2) Ê dr œ 2e2) d) # # Ê ds# œ r# d)# ˆ2e2) d)‰ œ ê2) ‰ d)# 4e4) d)# œ 5e4) d)# Ê ds œ È5 e2) d) Ê L œ '0 È5 e2) d) 21 œ’ È5 e2) #1 È5 41 2 “ ! œ # ae 1b # 18. r œ 2 sin$ ˆ 3) ‰ Ê dr œ 2 sin# ˆ 3) ‰ cos ˆ 3) ‰ d) Ê ds# œ r# d)# dr# œ 2 sin$ ˆ 3) ‰‘ d)# 2 sin# ˆ 3) ‰ cos ˆ 3) ‰ d)‘ œ 4 sin' ˆ 3) ‰ d)# 4 sin% ˆ 3) ‰ cos# ˆ 3) ‰ d)# œ 4 sin% ˆ 3) ‰‘ sin# ˆ 3) ‰ cos# ˆ 3) ‰‘ d)# œ 4 sin% ˆ 3) ‰ d)# $1 Ê ds œ 2 sin# ˆ 3) ‰ d). Then L œ '0 2 sin# ˆ 3) ‰ d) œ '0 1 cos ˆ 23) ‰‘ d) œ ) 23 sin ˆ 23) ‰‘ ! œ 31 31 31 19. e œ 2 and r cos ) œ 2 Ê x œ 2 is the directrix Ê k œ 2; the conic is a hyperbola with r œ 1 ekecos ) 4 Ê r œ 1 (2)(2) 2 cos ) œ 1 2 cos ) 20. e œ 1 and r cos ) œ 4 Ê x œ 4 is the directrix Ê k œ 4; the conic is a parabola with r œ 1 ekecos ) (4)(1) 4 Ê r œ 1 cos ) œ 1 cos ) 21. e œ "# and r sin ) œ 2 Ê y œ 2 is the directrix Ê k œ 2; the conic is an ellipse with r œ 1 ekesin ) 2 ˆ"‰ Ê r œ 1 ˆ " ‰# sin ) œ 2 2sin ) # 22. e œ "3 and r sin ) œ 6 Ê y œ 6 is the directrix Ê k œ 6; the conic is an ellipse with r œ 1 ekesin ) 6 ˆ"‰ Ê r œ 1 ˆ " ‰3 sin ) œ 3 6sin ) 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. # 712 Chapter 11 Parametric Equations and Polar Coordinates 23. Arc PF œ Arc AF since each is the distance rolled; nPCF œ Arcb PF Ê Arc PF œ b(nPCF); ) œ ArcaAF Ê Arc AF œ a) Ê a) œ b(nPCF) Ê nPCF œ ˆ ba ‰ ); nOCB œ 1# ) and nOCB œ nPCF nPCE œ nPCF ˆ 1# !‰ œ ˆ ba ‰ ) ˆ 1# !‰ Ê 1# ) œ ˆ ba ‰ ) ˆ 1# !‰ Ê 1# ) œ ˆ ba ‰ ) 1# ! Ê ! œ 1 ) ˆ ba ‰ ) Ê ! œ 1 ˆ ab b ‰ ). Now x œ OB BD œ OB EP œ (a b) cos ) b cos ! œ (a b) cos ) b cos ˆ1 ˆ a b b ‰ )‰ œ (a b) cos ) b cos 1 cos ˆˆ a b b ‰ )‰ b sin 1 sin ˆˆ a b b ‰ )‰ œ (a b) cos ) b cos ˆˆ a b b ‰ )‰ and y œ PD œ CB CE œ (a b) sin ) b sin ! œ (a b) sin ) b sin ˆˆ a b b ‰ )‰ œ (a b) sin ) b sin 1 cos ˆˆ a b b ‰ )‰ b cos 1 sin ˆˆ a b b ‰ )‰ œ (a b) sin ) b sin ˆˆ a b b ‰ )‰ ; therefore x œ (a b) cos ) b cos ˆˆ a b b ‰ )‰ and y œ (a b) sin ) b sin ˆˆ a b b ‰ )‰ ˆ dx ‰ 24. x œ a(t sin t) Ê dx dt œ a(1 cos t) and let $ œ 1 Ê dm œ dA œ y dx œ y dt dt œ a(1 cos t) a (1 cos t) dt œ a# (1 cos t)# dt; then A œ '0 a# (1 cos t)# dt 21 œ a# '0 a1 2 cos t cos# tb dt œ a# '0 ˆ1 2 cos t "# "# cos 2t‰ dt œ a# 32 t 2 sin t sin4 2t ‘ ! 21 21 #1 œ 31a# ; µ x = x œ a(t sin t) and µ y = "# y œ "# a(1 cos t) Ê Mx œ ' µ y dm œ ' µ y $ dA œ '0 21 " " $ # # # a(1 cos t) a (1 cos t) dt œ # a '021 (1 cos t)$ dt œ a# '021 a1 3 cos t 3 cos# t cos$ tb dt $ 2t œ a# '0 1 3 cos t #3 3 cos a1 sin# tb (cos t)‘ dt œ a# ’ 25 t 3 sin t 3 sin4 2t sin t sin3 t “ # 21 $ $ œ 51#a . $ 51 a $ Š # ‹ Therefore y œ MMx œ 31a# œ 65 a. $ #1 ! Also, My œ ' µ x dm œ ' µ x $ dA œ '0 a(t sin t) a# (1 cos t)# dt œ a$ '0 at 2t cos t t cos# t sin t 2 sin t cos t sin t cos# tb dt 21 21 # $ œ a$ ’ t2 2 cos t 2t sin t 4" t# 8" cos 2t 4t sin 2t cos t sin# t cos3 t “ # $ #1 ! œ 31# a$ . Thus x œ My œ 3311aa# œ 1a Ê ˆ1aß 56 a‰ is the center of mass. M 25. <# tan <" " œ <# <" Ê tan " œ tan (<# <" ) œ 1tan tan <# tan <" ; the curves will be orthogonal when tan " is undefined, or when tan <# œ tan"<" Ê g (r)) œ " r w ’ f ()) “ w # w w Ê r œ f ( ) ) g ( ) ) 26. sin% ˆ ) ‰ 4 ˆ)‰ r œ sin% ˆ 4) ‰ Ê ddr) œ sin$ ˆ 4) ‰ cos ˆ 4) ‰ Ê tan < œ sin$ ˆ ) ‰ cos ˆ ) ‰ œ tan 4 4 27. 4 2a sin 3) " 1 " 1 1 r œ 2a sin 3) Ê ddr) œ 6a cos 3) Ê tan < œ ˆ drr ‰ œ 6a cos 3) œ 3 tan 3); when ) œ 6 , tan < œ 3 tan # Ê < œ # d) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 11 Additional and Advanced Exercises 28. 713 dr (b) r) œ 1 Ê r œ )" Ê d) œ )# Ê tan <k )œ1 (a) " œ )) # œ ) Ê lim tan < œ _ )Ä_ Ê < Ä 1# from the right as the spiral winds in around the origin. 29. È ) sin ) È3 at ) œ 1 ; since the product of tan <" œ È33cos œ cot ) is È"3 at ) œ 13 ; tan <# œ cos ) œ tan ) is 3 sin ) these slopes is 1, the tangents are perpendicular 30. tan < œ ˆ drr ‰ œ a(1asincos) )) is 1 at ) œ 1# Ê < œ 14 d) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 714 Chapter 11 Parametric Equations and Polar Coordinates NOTES: Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS 1. The line through the point a2ß 3ß 0b parallel to the z-axis 2. The line through the point a1ß 0ß 0b parallel to the y-axis 3. The x-axis 4. The line through the point a1ß 0ß 0b parallel to the z-axis 5. The circle x# y# œ 4 in the xy-plane 6. The circle x# y# œ 4 in the plane z = 2 7. The circle x# z# œ 4 in the xz-plane 8. The circle y# z# œ 1 in the yz-plane 9. The circle y# z# œ 1 in the yz-plane 10. The circle x# z# œ 9 in the plane y œ 4 11. The circle x# y# œ 16 in the xy-plane 12. The circle x# z# œ 3 in the xz-plane 13. The ellipse formed by the intersection of the cylinder x# y# œ 4 and the plane z œ y. 14. The circle formed by the intersection of the sphere x# y# z# œ 4 and the plane y œ x. 15. The parabola y œ x# in the the xy-plane. 16. The parabola z œ y# in the the plane x œ 1. 17. (a) The first quadrant of the xy-plane (b) The fourth quadrant of the xy-plane 18. (a) The slab bounded by the planes x œ 0 and x œ 1 (b) The square column bounded by the planes x œ 0, x œ 1, y œ 0, y œ 1 (c) The unit cube in the first octant having one vertex at the origin 19. (a) The solid ball of radius 1 centered at the origin (b) The exterior of the sphere of radius 1 centered at the origin 20. (a) The circumference and interior of the circle x# y# œ 1 in the xy-plane (b) The circumference and interior of the circle x# y# œ 1 in the plane z œ 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 716 Chapter 12 Vectors and the Geometry of Space (c) A solid cylindrical column of radius 1 whose axis is the z-axis 21. (a) The solid enclosed between the sphere of radius 1 and radius 2 centered at the origin (b) The solid upper hemisphere of radius 1 centered at the origin 22. (a) The line y œ x in the xy-plane (b) The plane y œ x consisting of all points of the form (xß xß z) 23. (a) The region on or inside the parabola y œ x# in the xy-plane and all points above this region. (b) The region on or to the left of the parabola x œ y# in the xy-plane and all points above it that are 2 units or less away from the xy-plane. 24. (a) All the points the lie on the plane z œ 1 y. (b) All points that lie on the curve z œ y3 in the plane x œ 2. 25. (a) x œ 3 (b) y œ 1 (c) z œ 2 26. (a) x œ 3 (b) y œ 1 (c) z œ 2 27. (a) z œ 1 (b) x œ 3 (c) y œ 1 28. (a) x# y# œ 4, z œ 0 (b) y# z# œ 4, x œ 0 (c) x# z# œ 4, y œ 0 29. (a) x# ay 2b# œ 4, z œ 0 (b) ay 2b# z# œ 4, x œ 0 (c) x# z# œ 4, y œ 2 30. (a) ax 3b# ay 4b# œ 1, z œ 1 (b) ay 4b# az 1b# œ 1, x œ 3 (c) ax 3b# az 1b# œ 1, y œ 4 31. (a) y œ 3, z œ 1 (b) x œ 1, z œ 1 (c) x œ 1, y œ 3 32. Èx# y# z# œ Éx# ay 2b# z# Ê x# y# z# œ x# ay 2b# z# Ê y# œ y# 4y 4 Ê y œ 1 33. x# y# z# œ 25, z œ 3 Ê x# y# œ 16 in the plane z œ 3 34. x# y# az 1b# œ 4 and x# y# az 1b# œ 4 Ê x# y# az 1b# œ x# y# az 1b# Ê z œ 0, x# y# œ 3 35. 0 Ÿ z Ÿ 1 36. 0 Ÿ x Ÿ 2, 0 Ÿ y Ÿ 2, 0 Ÿ z Ÿ 2 37. z Ÿ 0 38. z œ È1 x# y# 39. (a) ax 1b# ay 1b# az 1b# 1 (b) ax 1b# ay 1b# az 1b# 1 40. 1 Ÿ x# y# z# Ÿ 4 41. kP" P# k œ Éa3 1b# a3 1b# a0 1b# œ È9 œ 3 42. kP" P# k œ Éa2 1b# a5 1b# a0 5b# œ È50 œ 5È2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.1 Three-Dimensional Coordinate Systems 717 43. kP" P# k œ Éa4 1b# a2 4b# a7 5b# œ È49 œ 7 44. kP" P# k œ Éa2 3b# a3 4b# a4 5b# œ È3 45. kP" P# k œ Éa2 0b# a2 0b# a2 0b# œ È3 † 4 œ 2È3 46. kP" P# k œ Éa0 5b# a0 3b# a0 2b# œ È38 47. center (2ß 0ß 2), radius 2È2 48. center ˆ1ß "# ß 3‰ , radius 5 49. center ŠÈ2ß È2ß È2‹ , radius È2 50. center ˆ!ß "3 ß 3" ‰ , radius 34 51. ax 1b# ay 2b# az 3b# œ 14 52. x# ay 1b# az 5b# œ 4 # # 53. ax 1b# ˆy 12 ‰ ˆz 23 ‰ œ 16 81 54. x# ay 7b# z# œ 49 55. x# y# z# 4x 4z œ 0 Ê ax# 4x 4b y# az# 4z 4b œ 4 4 # Ê ax 2b# ay 0b# az 2b# œ ŠÈ8‹ Ê the center is at a2ß 0ß 2b and the radius is È8 56. x# y# z# 6y 8z œ 0 Ê x# ay# 6y 9b az# 8z 16b œ 9 16 Ê ax 0b# ay 3b# az 4b# œ 5# Ê the center is at a0ß 3ß 4b and the radius is 5 57. 2x# 2y# 2z# x y z œ 9 Ê x# "# x y# "# y z# "# z œ 9# # # # È # " ‰ " ‰ " ‰ 3 Ê ˆx# "# x 16 ˆy# "# y 16 ˆz# "# z 16 œ 9# 16 Ê ˆx 4" ‰ ˆy 4" ‰ ˆz 4" ‰ œ Š 5 4 3 ‹ È Ê the center is at ˆ "4 ß 4" ß 4" ‰ and the radius is 5 4 3 58. 3x# 3y# 3z# 2y 2z œ 9 Ê x# y# 23 y z# 23 z œ 3 Ê x# ˆy# 23 y "9 ‰ ˆz# 23 z "9 ‰ œ 3 29 # # È # Ê (x 0)# ˆy 3" ‰ ˆz 3" ‰ œ Š 329 ‹ Ê the center is at ˆ0ß 3" ß 3" ‰ and the radius is 59. (a) the distance between axß yß zb and axß 0ß 0b is Èy# z# (b) the distance between axß yß zb and a0ß yß 0b is Èx# z# (c) the distance between axß yß zb and a0ß 0ß zb is Èx# y# 60. (a) the distance between axß yß zb and axß yß 0b is z (b) the distance between axß yß zb and a0ß yß zb is x (c) the distance between axß yß zb and axß 0ß zb is y 61. kABk œ Éa1 a1bb# a1 2b# a3 1b# œ È4 9 4 œ È17 kBCk œ Éa3 1b# a4 a1bb# a5 3b# œ È4 25 4 œ È33 kCAk œ Éa1 3b# a2 4b# a1 5b# œ È16 4 16 œ È36 œ 6 Thus the perimeter of triangle ABC is È17 È33 6. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. È29 3 718 Chapter 12 Vectors and the Geometry of Space 62. kPAk œ Éa2 3b# a1 1b# a3 2b# œ È1 4 1 œ È6 kPBk œ Éa4 3b# a3 1b# a1 2b# œ È1 4 1 œ È6 Thus P is equidistant from A and B. 63. Éax xb# ay a1bb# az zb# œ Éax xb# ay 3b# az zb# Ê ay 1b# œ ay 3b# Ê 2y 1 œ 6y 9 Êyœ1 64. Éax 0b# ay 0b# az 2b# œ Éax xb# ay yb# az 0b# Ê x# y2 az 2b# œ z# # 2 Ê x# y2 4z 4 œ 0 Ê z œ x4 y4 1 65. (a) Since the entire sphere is below the xy-plane, the point on the sphere closest to the xy-plane is the point at the top of the sphere, which occurs when x œ 0 and y œ 3 Ê 02 a3 3b2 az 5b2 œ 4 Ê z œ 5 „ 2 Ê z œ 3 Ê a0, 3, 3b. (b) Both the center a0, 3, 5b and the the point a0, 7, 5b lie in the plane z œ 5, so the point on the sphere closest to a0, 7, 5b should also be in the same plane. In fact it should lie on the line segment between a0, 3, 5b and a0, 7, 5b, thus the point occurs when x œ 0 and z œ 5 Ê 02 ay 3b2 a5 5b2 œ 4 Ê y œ 3 „ 2 Ê y œ 5 Ê a0, 5, 5b. 66. Éax 0b# ay 0b# az 0b# œ Éax 0b# ay 4b# az 0b# œ Éax 3b# ay 0b# az 0b# œ Éax 2b# ay 2b# az 3b# Ê x# y2 z# œ x# y2 8y 16 z# œ x# 6x 9 y2 z# œ x# 4x y2 4y z# 6z 17 Solve: x# y2 z# œ x# y2 8y 16 z# Ê 0 œ 8y 16 Ê y œ 2 Solve: x# y2 z# œ x# 6x 9 y2 z# Ê 0 œ 6x 9 Ê x œ 32 Solve: x# y2 z# œ x# 4x y2 4y z# 6z 17 Ê 0 œ 4x 4y 6z 17 Ê 0 œ 4ˆ 32 ‰ 4a2b 6z 17 Ê z œ 12 Ê ˆ 32 , 2, 12 ‰ 12.2 VECTORS 1. (a) 3a3b, 3a2b¡ œ 9, 6¡ (b) É92 a6b2 œ È117 œ 3È13 3. (a) 3 a2b, 2 5¡ œ 1, 3¡ (b) È12 32 œ È10 5. (a) 2u œ 2a3b, 2a2b¡ œ 6, 4¡ 3v œ 3a2b, 3a5b¡ œ 6, 15¡ 2u 3v œ 6 a 6b, 4 15¡ œ 12, 19¡ (b) É122 a19b2 œ È505 2. (a) 2a2b, 2a5b¡ œ 4, 10¡ (b) É42 a10b2 œ È116 œ 2È29 4. (a) 3 a2b, 2 5¡ œ 5, 7¡ (b) É52 a7b2 œ È74 6. (a) 2u œ 2a3b, 2a2b¡ œ 6, 4¡ 5v œ 5a2b, 5a5b¡ œ 10, 25¡ 2u 5v œ 6 a10b, 4 25¡ œ 16, 29¡ (b) Éa16b2 292 œ È1097 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.2 Vectors 7. (a) 3 3 3 9 6 5 u œ ¢ 5 a3b, 5 a2b£ œ ¢ 5 , 5 £ 719 5 5 5 15 10 8. (a) 13 u œ ¢ 13 a3b, 13 a2b£ œ ¢ 13 , 13 £ 4 4 4 8 5 v œ ¢ 5 a2b, 5 a5b£ œ ¢ 5 , 4£ 12 12 12 24 60 13 v œ ¢ 13 a2b, 13 a5b£ œ ¢ 13 , 13 £ 3 4 9 1 14 ˆ 8‰ 6 5 u 5 v œ ¢ 5 5 , 5 4£ œ ¢ 5 , 5 £ 5 15 70 ˆ 24 ‰ 10 60 13 u 12 13 v œ ¢ 13 13 , 13 13 £ œ ¢3, 13 £ 2 2 ‰ œ (b) Éˆ 15 ‰ ˆ 14 5 È"97 5 9. 2 1, 1 3¡ œ 1, 4¡ 11. 0 2, 0 3¡ œ 2, 3¡ 2 70 ‰ (b) Éa3b2 ˆ 13 œ È6421 13 3 0£ œ 1, 1¡ 10. ¢ 2a#4b 0, " # Ä Ä Ä Ä 12. AB œ 2 1, 0 a1b¡ œ 1, 1¡, CD œ 2 a1b, 2 3¡ œ 1, 1¡, AB CD œ 0, 0¡ 13. ¢cos 231 , sin 231 £ œ ¢ "# , È3 2 £ 14. ¢cos ˆ 341 ‰, sin ˆ 341 ‰£ œ ¢ È"2 , È"2 £ 15. This is the unit vector which makes an angle of 120‰ 90‰ œ 210‰ with the positive x-axis; È cos 210‰ , sin 210‰ ¡ œ ¢ 23 , "# £ 16. cos 135‰ , sin 135‰ ¡ œ ¢ È"2 , È"2 £ Ä 17. P" P# œ a2 5bi a9 7bj a2 a1bbk œ 3i 2j k Ä 18. P" P# œ a3 1bi a0 2bj a5 0bk œ 4i 2j 5k Ä 19. AB œ a10 a7bbi a8 a8bbj a1 1bk œ 3i 16j Ä 20. AB œ a1 1bi a4 0bj a5 3bk œ 2i 4j 2k 21. 5u v œ 5 1, 1, 1¡ 2, 0, 3¡ œ 5, 5, 5¡ 2, 0, 3¡ œ 5 2, 5 0, 5 3¡ œ 3, 5, 8¡ œ 3i 5j 8k 22. 2u 3v œ 2 1, 0, 2¡ 3 1, 1, 1¡ œ 2, 0, 4¡ 3, 3, 3¡ œ 5, 3, 1¡ œ 5i 3j k 23. The vector v is horizontal and 1 in. long. The vectors u and w are "161 in. long. w is vertical and u makes a 45‰ angle with the horizontal. All vectors must be drawn to scale. (a) (b) (c) (d) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 720 Chapter 12 Vectors and the Geometry of Space 24. The angle between the vectors is 120‰ and vector u is horizontal. They are all 1 in. long. Draw to scale. (a) (b) (c) (d) 25. length œ k2i j 2kk œ È2# 1# (2)# œ 3, the direction is 23 i 3" j 32 k Ê 2i j 2k œ 3 ˆ 32 i 3" j 32 k‰ 9 2 6 26. length œ k9i 2j 6kk œ È81 4 36 œ 11, the direction is 11 i 11 j 11 k Ê 9i 2 j 6k 9 2 6 œ 11 ˆ 11 i 11 j 11 k‰ 27. length œ k5kk œ È25 œ 5, the direction is k Ê 5k œ 5(k) 9 16 28. length œ ¸ 35 i 45 k¸ œ É 25 25 œ 1, the direction is 35 i 45 k Ê 35 i 45 k œ 1 ˆ 35 i 45 k‰ # 29. length œ ¹ È16 i È16 j È"6 k¹ œ Ê3 Š È"6 ‹ œ É #" , the direction is È13 i È13 j È"3 k Ê È16 i È16 j È"6 k œ É "# Š È13 i È13 j È"3 k‹ # 30. length œ ¹ È13 i È13 j È"3 k¹ œ Ê3 Š È"3 ‹ œ 1, the direction is È13 i È13 j È"3 k Ê È13 i È13 j È"3 k œ 1 Š È13 i È13 j È"3 k‹ 31. (a) 2i (b) È3k 32. (a) 7j (b) 3 5 2 i 4 5 2 k È È (c) 3 2 10 j 5 k (d) 6i 2j 3k (c) 1 1 4 i 3 jk (d) a a a È2 i È3 j È6 k " " 7 33. kvk œ È12# 5# œ È169 œ 13; kvvk œ 13 (12i 5k) Ê the desired vector is 13 (12i 5k) v œ 13 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.2 Vectors 34. kvk œ É 4" 4" 4" œ È3 v 1 1 1 # ; kv k œ È 3 i È 3 j È 3 k 721 Ê the desired vector is 3 Š È"3 i È"3 j È"3 k‹ œ È 3i È 3 j È 3k 3 4 3 4 35. (a) 3i 4j 5k œ 5È2 Š 5È i 5È j È"2 k‹ Ê the direction is 5È i 5È j È"2 k 2 2 2 2 (b) the midpoint is ˆ "# ß 3ß 5# ‰ 36. (a) 3i 6j 2k œ 7 ˆ 37 i 67 j 27 k‰ Ê the direction is 37 i 67 j 27 k (b) the midpoint is ˆ 5# ß 1ß 6‰ 37. (a) i j k œ È3 Š È13 i È13 j È13 k‹ Ê the direction is È13 i È13 j È13 k (b) the midpoint is ˆ 5# ß 7# ß 9# ‰ 38. (a) 2i 2j 2k œ 2È3 Š È13 i È13 j È13 k‹ Ê the direction is È13 i È13 j È13 k (b) the midpoint is ("ß "ß 1) Ä 39. AB œ (5 a)i (1 b)j (3 c)k œ i 4j 2k Ê 5 a œ 1, 1 b œ 4, and 3 c œ 2 Ê a œ 4, b œ 3, and c œ 5 Ê A is the point (4ß 3ß 5) Ä 40. AB œ (a 2)i (b 3)j (c 6)k œ 7i 3j 8k Ê a 2 œ 7, b 3 œ 3, and c 6 œ 8 Ê a œ 9, b œ 0, and c œ 14 Ê B is the point (9ß 0ß 14) 41. 2i j œ a(i j) b(i j) œ (a b)i (a b)j Ê a b œ 2 and a b œ 1 Ê 2a œ 3 Ê a œ 3# and b œ a " œ "# 42. i 2j œ a(2i 3j) b(i j) œ (2a b)i (3a b)j Ê 2a b œ 1 and 3a b œ 2 Ê a œ 3 and b œ 1 #a œ 7 Ê u" œ a(2i 3j) œ 6i 9j and u# œ b(i j) œ 7i 7j 43. 25‰ west of north is 90‰ 25‰ œ 115‰ north of east. 800 cos 115‰ , sin 115‰ ¡ ¸ 338.095, 725.046¡ 44. Let u œ x, y¡ be represent the velocity of the plane alone, v œ 70 cos 60‰ , 70 sin 60‰ ¡ œ 35, 35È3¡, and let the resultant u v œ 500, 0¡. Then x, y¡ 35, 35È3¡ œ 500, 0¡ Ê x 35, y 35È3¡ œ 500, 0¡ Ê x 35 œ 500 and y 35È3 œ 0 Ê x œ 465 and y œ 35È3 Ê u œ 465, 35È3¡ 2 È 3 ‰ ‰ Ê lul œ Ê4652 Š35È3‹ ¸ 468.9 mph, and tan ) œ 35 465 Ê ) ¸ 7.4 Ê 7.4 south of east. È 45. F1 œ ¢lF1 lcos 30‰ , lF1 lsin 30‰ £ œ ¢ 23 lF1 l, 12 lF1 l£, F2 œ ¢lF2 lcos 45‰ , lF2 lsin 45‰ £ œ ¢ È12 lF2 l, È12 lF2 l£, and È w œ 0, 100¡. Since F1 F2 œ 0, 100¡ Ê ¢ 23 lF1 l È12 lF2 l, 21 lF1 l È12 lF2 l£ œ 0, 100¡ È Ê 23 lF1 l È12 lF2 l œ 0 and 21 lF1 l È12 lF2 l œ 100. Solving the first equation for lF2 l results in: lF2 l œ È È6 2 l F1 l . Substituting this result into the second equation gives us: 12 lF1 l È12 Š 26 lF1 l‹ œ 100 Ê lF1 l œ 1 200È3 ¸ 73.205 N È 6 Ê lF2 l œ 1100 ¸ 89.658 N Ê F1 ¸ 63.397, 36.603¡ and F2 œ ¸ 63.397, 63.397¡ È3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 722 Chapter 12 Vectors and the Geometry of Space 46. F1 œ 35 cos !, 35 sin !¡, F2 œ ¢lF2 lcos 60‰ , lF2 lsin 60‰ £ œ ¢ 12 lF2 l, È3 2 lF2 l£, and w œ 0, 50¡. Since È F1 F2 œ 0, 50¡ Ê 35 cos ! 12 lF2 l, 35 sin ! 23 lF2 l¡ œ 0, 50¡ Ê 35 cos ! 21 lF2 l œ 0 and È 35 sin ! 23 lF2 l œ 50. Solving the first equation for lF2 l results in: lF2 l œ 70 cos !. Substituting this result into the second equation gives us: 35 sin ! 35È3 cos ! œ 50 Ê È3 cos ! œ 10 sin ! Ê 3cos2 ! œ 100 20 sin ! sin2 ! 7 49 7 È 5„6 2 20 2 2 Ê 3a1 sin2 !b œ 100 . Since ! 0 Ê 49 7 sin ! sin ! Ê 196 sin ! 140 sin ! 47 œ 0 Ê sin ! œ 14 È sin ! 0 Ê sin ! œ 5 146 2 Ê ! ¸ 74.42‰ , and lF2 l œ 70 cos ! ¸ 18.81 N. 47. F1 œ ¢lF1 lcos 40‰ , lF1 lsin 40‰ £, F2 œ 100 cos 35‰ , 100 sin 35‰ ¡, and w œ 0, w¡. Since F1 F2 œ 0, w¡ Ê ¢lF1 lcos 40‰ 100 cos 35‰ , lF1 lsin 40‰ 100 sin 35‰ £ œ 0, w¡ Ê lF1 lcos 40‰ 100 cos 35‰ œ 0 and ‰ lF1 lsin 40‰ 100 sin 35‰ œ w. Solving the first equation for lF1 l results in: lF1 l œ 100coscos4035 ¸ 106.933 N. Substituting this ‰ result into the second equation gives us: w ¸ 126.093 N. 48. F1 œ lF1 l cos !, lF1 l sin !¡ œ 75 cos !, 75 sin !¡, F2 œ lF2 l cos " , lF2 l sin " ¡ œ 75 cos !, 75 sin !¡, and w œ 0, 25¡. Since F1 F2 œ 0, 25¡ Ê 75 cos ! 75 cos !, 75 sin ! 75 sin !¡ œ 0, 25¡ Ê 150 sin ! œ 25 Ê ! ¸ 9.59‰ . Ä È È 49. (a) The tree is located at the tip of the vector OP œ (5 cos 60°)i (5 sin 60°)j œ 5# i 5 # 3 j Ê P œ Š #5 ß 5 # 3 ‹ Ä Ä (b) The telephone pole is located at the point Q, which is the tip of the vector OP PQ È È È È œ Š 52 i 5 # 3 j‹ (10 cos 315°)i (10 sin 315°)j œ Š 5# 10# 2 ‹ i Š 5 # 3 10# 2 ‹ j Ê Q œ Š 5 10 # È2 ß 5 È3 10È2 ‹ # 50. Let t œ p q q and s œ p p q . Choose T on OP1 so that TQ is parallel to OP2 , so that ˜TP1 Q is similar to ÕP1 P2 . Then Ä Ä kOTk kOP1 k œ t Ê OT œ t OP1 so that T œ at x1 , t y1 , t z1 b. Ä Ä TQk Also, kkOP œ s Ê TQ œ s OP2 œ s x2 , y2 , z2 ¡. 2k Letting Q œ ax, y, zb, we have that Ä TQ œ x t x1 , y t y1 , z t z1 ¡ œ s x2 , y2 , z2 ¡ Thus x œ t x1 s x2 , y œ t y1 s y2 , z œ t z1 s z2 . (Note that if Q is the midpoint, then qp œ 1 and t œ s œ "# so that x œ "# x1 "# x2 œ x1 2 x2 , y œ y1 2 y2 , z œ z1 2 z2 so that this result agress with the midpoint formula.) Ä 51. (a) the midpoint of AB is M ˆ 5# ß 5# ß 0‰ and CM œ ˆ 5# 1‰ i ˆ 5# 1‰ j (! 3)k œ 3# i 3# j 3k Ä (b) the desired vector is ˆ 23 ‰ CM œ 23 ˆ #3 i #3 j 3k‰ œ i j 2k (c) the vector whose sum is the vector from the origin to C and the result of part (b) will terminate at the center of mass Ê the terminal point of (i j 3k) (i j 2k) œ 2i 2j k is the point (2ß 2ß 1), which is the location of the center of mass Ä 52. The midpoint of AB is M ˆ #3 ß 0ß 5# ‰ and ˆ 23 ‰ CM œ 23 ˆ #3 1‰ i (0 2)j ˆ #5 1‰ k‘ œ 32 ˆ 5# i 2j 7# k‰ Ä œ 53 i 43 j 73 k. The vector from the origin to the point of intersection of the medians is ˆ 53 i 43 j 73 k‰ OC œ ˆ 53 i 43 j 73 k‰ (i 2j k) œ 23 i 23 j 43 k . Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.3 The Dot Product 723 53. Without loss of generality we identify the vertices of the quadrilateral such that A(0ß 0ß 0), B(xb ß 0ß 0), C(xc ß yc ß 0) and D(xd ß yd ß zd ) Ê the midpoint of AB is MAB ˆ x#b ß 0ß 0‰ , the midpoint of BC is MBC ˆ xb # xc ß y#c ß 0‰ , the midpoint of CD is MCD ˆ xc # xd ß yc # yd ß z#d ‰ and the midpoint of AD is xb MAD ˆ x#d ß y#d ß z#d ‰ Ê the midpoint of MAB MCD is Œ # xb xc x x b c# d # ß yc 4 yd , z4d which is the same as the midpoint xd b of MAD MBC œ Œ # # # ß yc 4 yd , z4d . 54. Let V" , V# , V$ , á , Vn be the vertices of a regular n-sided polygon and vi denote the vector from the center to n Vi for i œ 1, 2, 3, á , n. If S œ ! vi and the polygon is rotated through an angle of i(2n1) where i œ 1, 2, 3, á , n, i œ1 then S would remain the same. Since the vector S does not change with these rotations we conclude that S œ 0. 55. Without loss of generality we can coordinatize the vertices of the triangle such that A(0ß 0), B(bß 0) and Ä C(xc ß yc ) Ê a is located at ˆ b # xc ß y#c ‰ , b is at ˆ x#c ß y#c ‰ and c is at ˆ b# ß 0‰ . Therefore, Aa œ ˆ b# x#c ‰ i ˆ y#c ‰ j , Ä Ä Ä Ä Ä Bb œ ˆ x#c b‰ i ˆ y#c ‰ j , and Cc œ ˆ b# xc ‰ i (yc ) j Ê Aa Bb Cc œ 0 . 56. Let u be any unit vector in the plane. If u is positioned so that its initial point is at the origin and terminal point is at ax, yb, then u makes an angle ) with i, measured in the counter-clockwise direction. Since kuk œ 1, we have that x œ cos ) and y œ sin ). Thus u œ cos ) i sin ) j. Since u was assumed to be any unit vector in the plane, this holds for every unit vector in the plane. 12.3 THE DOT PRODUCT ) NOTE: In Exercises 1-8 below we calculate projv u as the vector Š kuk kcos vk ‹ v , so the scalar multiplier of v is the number in column 5 divided by the number in column 2. v†u kvk kuk cos ) kuk cos ) projv u 1. 25 5 5 1 5 2i 4 j È 5 k 2. 3 1 13 3 13 3 3 ˆ 35 i 45 k‰ 3. 25 15 5 " 3 5 3 " 9 (10i 11j 2k) 4. 13 15 3 13 45 13 15 13 225 (2i 10j 11k) 5. 2 È34 È3 2 È3 È34 2 È34 " 17 (5j 3k) 6. È3 È2 È2 3 È3 È2 3È 2 È3 È2 È2 È3 È2 (i j) # 7. 10 È17 È26 È21 10 È17 È546 10 È17 È26 10 È17 (5i j) 26 È30 6 È30 6 " 5 " È30 " " " 5 ¢ È# , È3 £ 8. " 6 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 724 Chapter 12 Vectors and the Geometry of Space (1)(2) (0)(1) 9. ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È2#(2)(1) ‹ œ cos" Š È54È6 ‹ œ cos" Š È430 ‹ ¸ 0.75 rad 1# 0# È1# 2# ( 1)# (2)(0) (1)(4) 10. ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È2#(2)(3) ‹ œ cos" Š È910È25 ‹ œ cos" ˆ 23 ‰ ¸ 0.84 rad ( 2)# 1# È3# 0# 4# 11. ) œ cos" Š kuuk†kvvk ‹ œ cos" Î Ñ ŠÈ3‹ ŠÈ3‹ (7)(1) (0)(2) # # Ï ÊŠÈ3‹ (7)# 0# ÊŠÈ3‹ (1)# (2)# Ò 7 œ cos" Š È352È ‹ 8 1 œ cos" Š È26 ‹ ¸ 1.77 rad 12. ) œ cos" Š kuuk†kvvk ‹ œ cos" Î Ï Ñ (1)(1) ŠÈ2‹ (1) ŠÈ2‹ (1) # Ê(1)# ŠÈ2‹ # ŠÈ2‹ È(1)# (1)# (1)# Ò 1 œ cos" Š È ‹ 5 È3 1 œ cos" Š È15 ‹ ¸ 1.83 rad Ä Ä Ä Ä Ä Ä 13. AB œ 3, 1¡, BC œ 1, 3¡, and AC œ 2, 2¡. BA œ 3, 1¡, CB œ 1, 3¡, CA œ 2, 2¡. Ä Ä Ä Ä Ä Ä ¹ AB ¹ œ ¹ BA ¹ œ È10, ¹ BC ¹ œ ¹ CB ¹ œ È10, ¹ AC ¹ œ ¹ CA ¹ œ 2È2, Ä Ä 3a2b 1a2b AB†AC " Angle at A œ cos" Ä œ cos" Š È15 ‹ ¸ 63.435‰ Ä œ cos È È ¹AB¹ ¹AC¹ Š 10‹Š2 2‹ Ä Ä a1ba3b a3ba1b BC†BA " " 3 ‰ Angle at B œ cos" Ä Ä œ cos œ cos ˆ 5 ‰ ¸ 53.130 , and È È ¹BC¹ ¹BA¹ Š 10‹Š 10‹ Ä Ä 1a2b 3a2b CB†CA " Angle at C œ cos" Ä œ cos" Š È15 ‹ ¸ 63.435‰ Ä œ cos È È ¹CB¹ ¹CA¹ Š 10‹Š2 2‹ Ä Ä Ä Ä 14. AC œ 2, 4¡ and BD œ 4, 2¡. AC † BD œ 2a4b 4a2b œ 0, so the angle measures are all 90‰ . 15. (a) cos ! œ kiik†kvvk œ kvak , cos " œ kjjk†kvvk œ kbvk , cos # œ kkkk†kvvk œ kvck and # # # cos# ! cos# " cos# # œ Š kvak ‹ Š kvbk ‹ Š kvck ‹ œ a kvbk kvk c œ kkvvkk kkvvkk œ 1 # # # (b) kvk œ 1 Ê cos ! œ kvak œ a, cos " œ kvbk œ b and cos # œ kvck œ c are the direction cosines of v 16. u œ 10i 2k is parallel to the pipe in the north direction and v œ 10j k is parallel to the pipe in the east direction. The angle between the two pipes is ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È1042È101 ‹ ¸ 1.55 rad ¸ 88.88°. 17. The sum of two vectors of equal length is always orthogonal to their difference, as we can see from the equation (v" v# ) † (v" v# ) œ v" † v" v# † v" v" † v# v# † v# œ kv" k # kv# k # œ 0 Ä Ä 18. CA † CB œ (v (u)) † (v u) œ v † v v † u u † v u † u œ kvk # kuk # œ 0 because kuk œ kvk since both equal Ä Ä the radius of the circle. Therefore, CA and CB are orthogonal. 19. Let u and v be the sides of a rhombus Ê the diagonals are d" œ u v and d# œ u v Ê d" † d# œ (u v) † (u v) œ u † u u † v v † u v † v œ kvk # kuk # œ 0 because kuk œ kvk , since a rhombus has equal sides. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.3 The Dot Product 725 20. Suppose the diagonals of a rectangle are perpendicular, and let u and v be the sides of a rectangle Ê the diagonals are d" œ u v and d# œ u v. Since the diagonals are perpendicular we have d" † d# œ 0 Í (u v) † (u v) œ u † u u † v v † u v † v œ 0 Í kvk # kuk # œ 0 Í akvk kukb akvk kukb œ 0 Í akvk kukb œ 0 which is not possible, or akvk kukb œ 0 which is equivalent to kvk œ kuk Ê the rectangle is a square. 21. Clearly the diagonals of a rectangle are equal in length. What is not as obvious is the statement that equal diagonals happen only in a rectangle. We show this is true by letting the adjacent sides of a parallelogram be the vectors (v" i v# j) and (u" i u# j). The equal diagonals of the parallelogram are d" œ (v" i v# j) (u" i u# j) and d# œ (v" i v# j) (u" i u# j). Hence kd" k œ kd# k œ k(v" i v# j) (u" i u# j)k œ k(v" i v# j) (u" i u# j)k Ê k(v" u" )i (v# u# )jk œ k(v" u" )i (v# u# )jk Ê È(v" u" )# (v# u# )# œ È(v" u" )# (v# u# )# Ê v#1 2v" u" u1# v## 2v# u# u## œ v1# 2v" u" u1# v## 2v# u# u## Ê 2(v" u" v# u# ) œ 2(v" u" v# u# ) Ê v" u" v# u# œ 0 Ê (v" i v# j) † (u" i u# j) œ 0 Ê the vectors (v" i v# j) and (u" i u# j) are perpendicular and the parallelogram must be a rectangle. 22. If kuk œ kvk and u v is the indicated diagonal, then (u v) † u œ u † u v † u œ kuk # v † u œ u † v kvk # œ u † v v † v œ (u v) † v Ê the angle cos" Š k(uuvvk )k†uuk ‹ between the diagonal and u and the angle cos" Š k(uuvvk )k†vvk ‹ between the diagonal and v are equal because the inverse cosine function is one-to-one. Therefore, the diagonal bisects the angle between u and v. 23. horizontal component: 1200 cosa8‰ b ¸ 1188 ft/s; vertical component: 1200 sina8‰ b ¸ 167 ft/s 2.5 lb 2.5 lb ‰ ‰¡ 24. kwkcosa33‰ 15‰ b œ 2.5 lb, so kwk œ cos ¸ 2.205, 1.432¡ 18‰ . Then w œ cos 18‰ cos 33 , sin 33 25. (a) Since kcos )k Ÿ 1, we have ku † vk œ kuk kvk kcos )k Ÿ kuk kvk (1) œ kuk kvk . (b) We have equality precisely when kcos )k œ 1 or when one or both of u and v is 0. In the case of nonzero vectors, we have equality when ) œ 0 or 1, i.e., when the vectors are parallel. 26. (xi yj) † v œ kxi yjk kvk cos ) Ÿ 0 when 1# Ÿ ) Ÿ 1. This means (xß y) has to be a point whose position vector makes an angle with v that is a right angle or bigger. 27. v † u" œ (au" bu# ) † u" œ au" † u" bu# † u" œ a ku" k # b(u# † u" ) œ a(1)# b(0) œ a 28. No, v" need not equal v# . For example, i j Á i 2j but i † (i j) œ i † i i † j œ 1 0 œ 1 and i † (i 2j) œ i † i 2i † j œ 1 2 † 0 œ 1. 2 29. projv u œ luv†lv2 v Ê Šu luv†lv2 v‹ † Š luv†lv2 v‹ œ u † Š luv†lv2 v‹ Š luv†lv2 v‹ † Š luv†lv2 v‹ œ Š luv†lv2 ‹au † vb Š luv†lv2 ‹ av † vb 2 2 œ aulv†vl2b aulv†vl4b lvl2 œ 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 726 Chapter 12 Vectors and the Geometry of Space 30. F œ 2i j 3k and v œ 3i j Ê projv F œ lFv†lv2 v œ 5 2 ŠÈ"!‹ a3i jb œ 23 i 21 j, is the vector parallel to v. F projv F œ a2i j 3kb ˆ 32 i 12 j‰ œ 12 i 32 j 3k is the vector orthogonal to v. 31. P(x" ß y" ) œ P ˆx" , bc ba x" ‰ and Q(x# ß y# ) œ Q ˆx# ß bc ba x# ‰ are any two points P and Q on the line with b Á 0 Ä Ä Ê PQ œ (x# x" )i ba (x" x# )j Ê PQ † v œ (x# x" )i ba (x" x# )j‘ † (ai bj) œ a(x# x" ) b ˆ ba ‰ (x" x# ) Ä œ 0 Ê v is perpendicular to PQ for b Á 0. If b œ 0, then v œ ai is perpendicular to the vertical line ax œ c. Alternatively, the slope of v is ba and the slope of the line ax by œ c is ba , so the slopes are negative reciprocals Ê the vector v and the line are perpendicular. 32. The slope of v is ba and the slope of bx ay œ c is ba , provided that a Á 0. If a œ 0, then v œ bj is parallel to the vertical line bx œ c. In either case, the vector v is parallel to the line bx ay œ c. 33. v œ i 2j is perpendicular to the line x 2y œ c; P(2ß 1) on the line Ê 2 2 œ c Ê x 2y œ 4 34. v œ 2i j is perpendicular to the line 2x y œ c; P(1ß 2) on the line Ê (2)(1) 2 œ c Ê 2x y œ 0 35. v œ 2i j is perpendicular to the line 2x y œ c; P(2ß 7) on the line Ê (2)(2) 7 œ c Ê 2x y œ 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.3 The Dot Product 36. v œ 2i 3j is perpendicular to the line 2x 3y œ c; P(11ß 10) on the line Ê (2)(11) (3)(10) œ c Ê 2x 3y œ 8 37. v œ i j is parallel to the line x y œ c; P(2ß 1) on the line Ê a2b 1 œ c Ê x y œ 1 or x y œ 1. 38. v œ 2i 3j is parallel to the line 3x 2y œ c; P(0ß 2) on the line Ê 0 2(2) œ c Ê 3x 2y œ 4 39. v œ i 2j is parallel to the line 2x y œ c; P(1ß 2) on the line Ê 2(1) 2 œ c Ê 2x y œ 0 or 2x y œ 0. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 727 728 Chapter 12 Vectors and the Geometry of Space 40. v œ 3i 2j is parallel to the line 2x 3y œ c; P(1ß 3) on the line Ê (2)(1) (3)(3) œ c Ê 2x 3y œ 11 or 2x 3y œ 11 Ä Ä 41. P(0ß 0), Q(1ß 1) and F œ 5j Ê PQ œ i j and W œ F † PQ œ (5j) † (i j) œ 5 N † m œ 5 J 42. W œ kFk (distance) cos ) œ (602,148 N)(605 km)(cos 0) œ 364,299,540 N † km œ (364,299,540)(1000) N † m œ 3.6429954 ‚ 10"" J Ä 43. W œ kFk ¹ PQ ¹ cos ) œ (200)(20)(cos 30°) œ 2000È3 œ 3464.10 N † m œ 3464.10 J Ä 44. W œ kFk ¹ PQ ¹ cos ) œ (1000)(5280)(cos 60°) œ 2,640,000 ft † lb In Exercises 45-50 we use the fact that n œ ai bj is normal to the line ax by œ c. 1 45. n" œ 3i j and n# œ 2i j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È610È ‹ œ cos" Š È"2 ‹ œ 14 5 3 1 46. n" œ È3i j and n# œ È3i j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È ‹ œ cos" ˆ "2 ‰ œ 231 4 È4 È È È 47. n" œ È3i j and n# œ i È3j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È34È43 ‹ œ cos" Š 23 ‹ œ 16 48. n" œ i È3j and n# œ Š1 È3‹ i Š1 È3‹ j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" 1 È3 È3 3 4 œ cos" Š 2È ‹ œ cos" Š È"2 ‹ œ 14 8 È 1 3 É 1 2È 3 3 1 2È 3 3 4 7 49. n" œ 3i 4j and n# œ i j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È325È ‹ œ cos" Š 5È ‹ ¸ 0.14 rad 2 2 10 " 50. n" œ 12i 5j and n# œ 2i 2j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È24169 Š 2614 È8 ‹ œ cos È2 ‹ ¸ 1.18 rad 12.4 THE CROSS PRODUCT â â j k â âi â â 1. u ‚ v œ â 2 2 " â œ 3 ˆ 23 i 3" j 32 k‰ Ê length œ 3 and the direction is 23 i 3" j 32 k; â â â " 0 " â v ‚ u œ (u ‚ v) œ 3 ˆ 23 i "3 j 32 k‰ Ê length œ 3 and the direction is 32 i 3" j 32 k Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.4 The Cross Product â â i â 2. u ‚ v œ â 2 â â " â j kâ â 3 0 â œ 5(k) Ê length œ 5 and the direction is k â 1 0â v ‚ u œ (u ‚ v) œ 5(k) Ê length œ 5 and the direction is k â â j k â â i â â 3. u ‚ v œ â 2 2 4 â œ 0 Ê length œ 0 and has no direction â â â " 1 2 â v ‚ u œ (u ‚ v) œ 0 Ê length œ 0 and has no direction â âi â 4. u ‚ v œ â 1 â â0 â j k â â 1 1 â œ 0 Ê length œ 0 and has no direction â 0 0 â v ‚ u œ (u ‚ v) œ 0 Ê length œ 0 and has no direction â â j kâ âi â â 5. u ‚ v œ â 2 0 0 â œ 6(k) Ê length œ 6 and the direction is k â â â 0 3 0 â v ‚ u œ (u ‚ v) œ 6(k) Ê length œ 6 and the direction is k â âi j â 6. u ‚ v œ (i ‚ j) ‚ (j ‚ k) œ k ‚ i œ â 0 0 â â1 0 â kâ â 1 â œ j Ê length œ 1 and the direction is j â 0â v ‚ u œ (u ‚ v) œ j Ê length œ 1 and the direction is j â â j k â â i â â 7. u ‚ v œ â 8 2 4 â œ 6i 12k Ê length œ 6È5 and the direction is È"5 i È25 k â â 2 1 â â 2 v ‚ u œ (u ‚ v) œ (6i 12k) Ê length œ 6È5 and the direction is " i 2 k È5 âi â â 8. u ‚ v œ â 3# â â1 È5 k ââ â 1 â œ 2i 2j 2k Ê length œ 2È3 and the direction is È"3 i È13 j È"3 k â 2â v ‚ u œ (u ‚ v) œ (2i 2j 2k) Ê length œ 2È3 and the direction is " i 1 j " k j "# 1 È3 â âi â 9. u ‚ v œ â 1 â â0 â j kâ â 0 0â œ k â 1 0â â âi â 10. u ‚ v œ â 1 â â0 È3 È3 â j k â â 0 1 â œ i k â 1 0 â Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 729 730 Chapter 12 Vectors and the Geometry of Space â âi â 11. u ‚ v œ â 1 â â0 â j k â â 0 1 â œ i j k â 1 1 â â j âi â 12. u ‚ v œ â 2 1 â â1 2 â âi â 13. u ‚ v œ â 1 â â1 â j kâ â 1 0 â œ 2k â 1 0 â â âi â 14. u ‚ v œ â 0 â â1 â â i Ä Ä â 15. (a) PQ ‚ PR œ â 1 â â 1 Ä â kâ â 0 â œ 5k â 0â â j kâ â 1 2 â œ 2j k â 0 0â â j k â Ä Ä â 1 3 â œ 8i 4j 4k Ê Area œ "# ¹ PQ ‚ PR ¹ œ "# È64 16 16 œ 2È6 â 3 1 â Ä ‚PR " (b) u œ PQ Ä Ä œ È (2i j k) 6 ¹PQ‚PR¹ â âi Ä Ä â 16. (a) PQ ‚ PR œ â 1 â â2 Ä j 0 2 Ä â kâ Ä Ä â 2 â œ 4i 4j 2k Ê Area œ "# ¹ PQ ‚ PR ¹ œ "# È16 16 4 œ 3 â 0â ‚PR " (b) u œ PQ Ä Ä œ 3 (2i 2j k) ¹PQ‚PR¹ â âi Ä Ä â 17. (a) PQ ‚ PR œ â 1 â â1 Ä j 1 1 Ä â kâ Ä Ä È â 1 â œ i j Ê Area œ "# ¹ PQ ‚ PR ¹ œ "# È1 1 œ #2 â 0â ‚PR " " (b) u œ PQ Ä Ä œ È (i j ) œ È ( i j ) 2 2 ¹PQ‚PR¹ â âi Ä Ä â 18. (a) PQ ‚ PR œ â 2 â â1 Ä Ä j 1 0 â k â Ä Ä È â 1 â œ 2i 3j k Ê Area œ "# ¹ PQ ‚ PR ¹ œ "# È4 9 1 œ #14 â 2 â ‚PR " (b) u œ PQ (2i 3j k) Ä Ä œ È 14 ¹PQ‚PR¹ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.4 The Cross Product â â â a" a# a$ â â â 19. If u œ a" i a# j a$ k, v œ b" i b# j b$ k, and w œ c" i c# j c$ k, then (u ‚ v) † w œ â b" b# b$ â , â â â c" c# c$ â â â â â â b" b# b $ â â c" c # c $ â â â â â (v ‚ w) † u œ â c" c# c$ â and (w ‚ u) † v œ â a" a# a$ â which all have the same absolute value, since the â â â â â a" a# a$ â â b" b# b$ â interchanging of two rows in a determinant does not change its absolute value Ê the volume is â â â2 0 0â â â k(u ‚ v) † wk œ abs â 0 2 0 â œ 8 â â â0 0 2â â â â 1 1 1 â â â 1 2 â œ 4 (for details about verification, see Exercise 19) 20. k(u ‚ v) † wk œ abs â 2 â â â 1 2 1 â â â2 1 â 21. k(u ‚ v) † wk œ abs â 2 1 â â1 0 â 0â â 1 â œ k7k œ 7 (for details about verification, see Exercise 19) â 2â â â â 1 1 2 â â â 22. k(u ‚ v) † wk œ abs â 1 0 1 â œ 8 (for details about verification, see Exercise 19) â â â 2 4 2 â 23. (a) u † v œ 6, u † w œ 81, v † w œ 18 Ê none are perpendicular â â â â â â j k â j k â j k â âi â i â i â â â â â â 1 1 â œ 0, v ‚ w œ â 0 1 5 â Á 0 (b) u ‚ v œ â 5 1 1 â Á 0, u ‚ w œ â 5 â â â â â â â 0 1 5 â â 15 3 3 â â 15 3 3 â Ê u and w are parallel 24. (a) u † v œ 0, u ‚ w œ 0, u † r œ 31, v † w œ 0, v † r œ 0, w † r œ 0 Ê u ¼ v, u ¼ w, v ¼ w, v ¼ r and w ¼ r â â â â â â j k â j k â â i âi j k â â i â â â â â â 2 1 â œ 0 (b) u ‚ v œ â 1 2 1 â Á 0, u ‚ w œ â 1 2 1 â Á 0, u ‚ r œ â 1 â â â â â 1 1 1 â â 1 1 1 â â1 0 1 â â # # â â â â â â â j kâ j kâ j kâ â i â i â i â â â â â â 1 1 â Á 0, w ‚ r œ â 1 0 1âÁ0 v ‚ w œ â 1 1 1 â Á 0, v ‚ r œ â 1 â â â 1 1 1 â â 1 1 1 â â 1 0 1â â # â # # â # â Ê u and r are parallel Ä Ä È 25. ¹ PQ ‚ F¹ œ ¹ PQ ¹ kFk sin (60°) œ 23 † 30 † #3 ft † lb œ 10È3 ft † lb Ä Ä È 26. ¹ PQ ‚ F¹ œ ¹ PQ ¹ kFk sin (135°) œ 23 † 30 † #2 ft † lb œ 10È2 ft † lb 27. (a) true, kuk œ Èa#1 a## a3# œ Èu † u (b) not always true, u † u œ kuk # â â â j kâ â i â i â â â (c) true, u ‚ 0 œ â u" u# u$ â œ 0i 0j 0k œ 0 and 0 ‚ u œ â 0 â â â â0 0 0â â u" j 0 u# â kâ â 0 â œ 0i 0j 0k œ 0 â u$ â Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 731 732 Chapter 12 Vectors and the Geometry of Space â â i â (d) true, u ‚ (u) œ â u" â â u" j u# u# â k â â u$ â œ (u# u$ u# u$ )i (u" u$ u" u$ )j (u" u# u" u# )k œ 0 â u$ â (e) not always true, i ‚ j œ k Á k œ j ‚ i for example (f) true, distributive property of the cross product (g) true, (u ‚ v) † v œ u † (v ‚ v) œ u † 0 œ 0 (h) true, the volume of a parallelpiped with u, v, and w along the three edges is the same whether the plane containing u and v or the plane containing v and w is used as the base plane, and the dot product is commutative. 28. (a) true, u † v œ u" v" u# v# u$ v$ œ v" u" v# u# v$ u$ œ v † u â â â â j kâ j kâ â i â i â â â â (b) true, u ‚ v œ â u" u# u$ â œ â v" v# v$ â œ (v ‚ u) â â â â â v" v# v $ â â u" u # u $ â â â â â j k â j kâ â i â i â â â â (c) true, (u) ‚ v œ â u" u# u$ â œ â u" u# u$ â œ (u ‚ v) â â â â v# v$ â â v" â v" v# v$ â (d) true, (cu) † v œ (cu" )v" (cu# )v# (cu$ )v$ œ u" (cv" ) u# (cv# ) u$ (cv$ ) œ u † (cv) œ c(u" v" u# v# u$ v$ ) œ c(u † v) â â â â â â j kâ j k â j k â â i â i â i â â â â â â (e) true, c(u ‚ v) œ c â u" u# u$ â œ â cu" cu# cu$ â œ (cu) ‚ v œ â u" u# u$ â œ u ‚ (cv) â â â â â â â v" v# v $ â â v" v# v $ â â cv" cv# cv$ â # (f) true, u † u œ u1# u## u3# œ ˆÈu1# u## u3# ‰ œ kuk # (g) true, (u ‚ u) † u œ 0 † u œ 0 (h) true, u ‚ v ¼ u and u ‚ v ¼ v Ê (u ‚ v) † u œ v † (u ‚ v) œ 0 29. (a) projv u œ Š kvukk†vvk ‹ v (e) (u ‚ v) ‚ (u ‚ w) (b) (u ‚ v) (f) (c) a(u ‚ v) ‚ wb (d) k(u ‚ v) † wk kuk kvvk 30. ai ‚ jb ‚ j œ k ‚ j œ i; i ‚ aj ‚ jb œ i ‚ 0 œ 0. The cross product is not associative. 31. (a) yes, u ‚ v and w are both vectors (c) yes, u and u ‚ w are both vectors (b) no, u is a vector but v † w is a scalar (d) no, u is a vector but v † w is a scalar 32. (u ‚ v) ‚ w is perpendicular to u ‚ v, and u ‚ v is perpendicular to both u and v Ê (u ‚ v) ‚ w is parallel to a vector in the plane of u and v which means it lies in the plane determined by u and v. The situation is degenerate if u and v are parallel so u ‚ v œ 0 and the vectors do not determine a plane. Similar reasoning shows that u ‚ (v ‚ w) lies in the plane of v and w provided v and w are nonparallel. 33. No, v need not equal w. For example, i j Á i j, but i ‚ ai jb œ i ‚ i i ‚ j œ 0 k œ k and i ‚ ai jb œ i ‚ aib i ‚ j œ 0 k œ k. 34. Yes. If u ‚ v œ u ‚ w and u † v œ u † w, then u ‚ (v w) œ 0 and u † (v w) œ 0. Suppose now that v Á w. Then u ‚ (v w) œ 0 implies that v w œ ku for some real number k Á 0. This in turn implies that u † (v w) œ u † (ku) œ k kuk # œ 0, which implies that u œ 0. Since u Á 0, it cannot be true that v Á w, so v œ w. â â j kâ â i Ä Ä Ä Ä Ä Ä â â 35. AB œ i j and AD œ i j Ê AB ‚ AD œ â 1 1 0 â œ 2k Ê area œ ¹ AB ‚ AD ¹ œ 2 â â â 1 1 0 â Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.4 The Cross Product â âi Ä Ä Ä Ä â 36. AB œ 7i 3j and AD œ 2i 5j Ê AB ‚ AD œ â 7 â â2 j 3 5 â âi Ä Ä Ä Ä â 37. AB œ 3i 2j and AD œ 5i j Ê AB ‚ AD œ â 3 â â5 â j kâ Ä Ä â 2 0 â œ 13k Ê area œ ¹ AB ‚ AD ¹ œ 13 â 1 0â â âi Ä Ä Ä Ä â 38. AB œ 7i 4j and AD œ 2i 5j Ê AB ‚ AD œ â 7 â â2 â j kâ Ä Ä â 4 0 â œ 43k Ê area œ ¹ AB ‚ AD ¹ œ 43 â 5 0â 733 â kâ Ä Ä â 0 â œ 29k Ê area œ ¹ AB ‚ AD ¹ œ 29 â 0â Ä Ä Ä Ä Ä Ä Ä 39. AB œ 3i 2j 4k and DC œ 3i 2j 4k Ê AB is parallel to DC; BC œ 2i j and AD œ 2i j Ê BC is parallel to â â j kâ âi Ä Ä Ä Ä Ä â â AD. AB ‚ BC œ â 3 2 4 â œ 4i 8j 7k Ê area œ lAB ‚ BC l œ È129 â â â 2 1 0 â Ä Ä Ä Ä Ä Ä Ä 40. AC œ i 4j and DB œ i 4j Ê AC is parallel to DB; AD œ i 3j 3k and CB œ i 3j 3k Ê AD is parallel â â j kâ â i Ä Ä Ä Ä Ä â â to CB . AC ‚ AD œ â 1 4 0 â œ 12i 3j 7k Ê area œ lAC ‚ AD l œ È202 â â â 1 3 3 â â â i Ä Ä Ä Ä â 41. AB œ 2i 3j and AC œ 3i j Ê AB ‚ AC œ â 2 â â 3 â âi Ä Ä Ä Ä â 42. AB œ 4i 4j and AC œ 3i 2j Ê AB ‚ AC œ â 4 â â3 j 3 1 â kâ Ä Ä â 0 â œ 11k Ê area œ "# ¹ AB ‚ AC ¹ œ 11 # â 0â â j kâ Ä Ä â 4 0 â œ 4k Ê area œ "# ¹ AB ‚ AC ¹ œ 2 â 2 0â â â i Ä Ä Ä Ä â 43. AB œ 6i 5j and AC œ 11i 5j Ê AB ‚ AC œ â 6 â â 11 â j kâ Ä Ä â 5 0 â œ 25k Ê area œ "# ¹ AB ‚ AC ¹ œ 25 # â 5 0 â â â j kâ â i Ä Ä Ä Ä Ä Ä â â 44. AB œ 16i 5j and AC œ 4i 4j Ê AB ‚ AC œ â 16 5 0 â œ 84k Ê area œ "# ¹ AB ‚ AC ¹ œ 42 â â 4 0â â 4 â â i Ä Ä Ä Ä â 45. AB œ i 2j and AC œ i k Ê AB ‚ AC œ â 1 â â 1 â j k â Ä Ä â 2 0 â œ 2i j 2k Ê area œ "# ¹ AB ‚ AC ¹ œ #3 â 0 1 â â â i Ä Ä Ä Ä â 46. AB œ i j k and AC œ 3i 3k Ê AB ‚ AC œ â 1 â â 3 â j k â Ä Ä È â 1 1 â œ 3i 3k Ê area œ "# ¹ AB ‚ AC ¹ œ 3 # 2 â 0 3 â â â i Ä Ä Ä Ä â 47. AB œ i 2j and AC œ j 2k Ê AB ‚ AC œ â 1 â â 0 â k â Ä Ä È â 0 â œ 4i 2j k Ê area œ "# ¹ AB ‚ AC ¹ œ #21 â 2 â j 2 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 734 Chapter 12 Vectors and the Geometry of Space â â1 Ä Ä Ä Ä Ä Ä â 48. AB œ i 2j, AC œ 3j 2k and AD œ 3i 4j 5k Ê Š AB ‚ AC ‹ † AD œ â 0 â â3 Ä Ä Ä Ê volume œ ¹Š AB ‚ AC ‹ † AD ¹ œ 5 â â i â 49. If A œ a" i a# j and B œ b" i b# j, then A ‚ B œ â a" â â b" " # kA ‚ Bk œ a „ "# º " b" j a# b# â kâ a â 0â œ º " b â " 0â 2 3 4 â 0â â 2â œ 5 â 5â a# k and the triangle's area is b# º a# . The applicable sign is () if the acute angle from A to B runs counterclockwise b# º in the xy-plane, and () if it runs clockwise, because the area must be a nonnegative number. Ä Ä 50. If A œ a" i a# j, B œ b" i b# j, and C œ c" i c# j, then the area of the triangle is "# ¹ AB ‚ AC ¹ . Now, â â i j kâ â Ä Ä Ä Ä b a" b# a# â â k Ê "# ¹ AB ‚ AC ¹ AB ‚ AC œ â b" a" b# a# 0 â œ º " º c" a" c# a# â â â c" a" c# a# 0 â œ "# k(b" a" )(c# a# ) (c" a" )(b# a# )k œ "# ka" (b# c# ) a# (c" b" ) (b" c# c" b# )k â â â a" a# 1 â â " â œ „ # â b" b# 1 â . The applicable sign ensures the area formula gives a nonnegative number. â â â c" c# 1 â 12.5 LINES AND PLANES IN SPACE 1. The direction i j k and P(3ß 4ß 1) Ê x œ 3 t, y œ 4 t, z œ 1 t Ä 2. The direction PQ œ 2i 2j 2k and P(1ß 2ß 1) Ê x œ 1 2t, y œ 2 2t, z œ 1 2t Ä 3. The direction PQ œ 5i 5j 5k and P(2ß 0ß 3) Ê x œ 2 5t, y œ 5t, z œ 3 5t Ä 4. The direction PQ œ j k and P(1ß 2ß 0) Ê x œ 1, y œ 2 t, z œ t 5. The direction 2j k and P(!ß !ß !) Ê x œ 0, y œ 2t, z œ t 6. The direction 2i j 3k and P(3ß 2ß 1) Ê x œ 3 2t, y œ 2 t, z œ 1 3t 7. The direction k and P(1ß 1ß 1) Ê x œ 1, y œ 1, z œ 1 t 8. The direction 3i 7j 5k and P(2ß 4ß 5) Ê x œ 2 3t, y œ 4 7t, z œ 5 5t 9. The direction i 2j 2k and P(0ß 7ß 0) Ê x œ t, y œ 7 2t, z œ 2t â âi j â 10. The direction is u ‚ v œ â 1 2 â â3 4 â kâ â 3 â œ 2i 4j 2k and Pa2, 3, 0b Ê x œ 2 2t, y œ 3 4t, z œ 2t â 5â 11. The direction i and P(0ß 0ß 0) Ê x œ t, y œ 0, z œ 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.5 Lines and Planes in Space 12. The direction k and P(0ß 0ß 0) Ê x œ 0, y œ 0, z œ t Ä 13. The direction PQ œ i j 3# k and P(0ß 0ß 0) Ê x œ t, y œ t, z œ 3# t, where 0 Ÿ t Ÿ 1 Ä 14. The direction PQ œ i and P(0ß 0ß 0) Ê x œ t, y œ 0, z œ 0, where 0 Ÿ t Ÿ 1 Ä 15. The direction PQ œ j and P(1ß 1ß 0) Ê x œ 1, y œ 1 t, z œ 0, where 1 Ÿ t Ÿ 0 Ä 16. The direction PQ œ k and P(1ß 1ß 0) Ê x œ 1, y œ 1, z œ t, where 0 Ÿ t Ÿ 1 Ä 17. The direction PQ œ 2j and P(0ß 1ß 1) Ê x œ 0, y œ 1 2t, z œ 1, where 0 Ÿ t Ÿ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 735 736 Chapter 12 Vectors and the Geometry of Space Ä 18. The direction PQ œ $i 2j and P(0ß 2ß 0) Ê x œ 3t, y œ 2 2t, z œ 0, where 0 Ÿ t Ÿ 1 Ä 19. The direction PQ œ 2i 2j 2k and P(2ß 0ß 2) Ê x œ 2 2t, y œ 2t, z œ 2 2t, where 0 Ÿ t Ÿ 1 Ä 20. The direction PQ œ i 3j k and P(1ß 0ß 1) Ê x œ 1 t, y œ 3t, z œ 1 t, where 0 Ÿ t Ÿ 1 21. 3(x 0) (2)(y 2) (1)(z 1) œ 0 Ê 3x 2y z œ 3 22. 3(x 1) (1)(y 1) (1)(z 3) œ 0 Ê 3x y z œ 5 â â j kâ â i Ä Ä Ä Ä â â 23. PQ œ i j 3k, PS œ i 3j 2k Ê PQ ‚ PS œ â 1 1 3 â œ 7i 5j 4k is normal to the plane â â â 1 3 2 â Ê 7(x 2) (5)(y 0) (4)(z 2) œ 0 Ê 7x 5y 4z œ 6 â â j kâ â i Ä Ä Ä Ä â â 24. PQ œ i j 2k, PS œ 3i 2j 3k Ê PQ ‚ PS œ â 1 1 2 â œ i 3j k is normal to the plane â â â 3 2 3 â Ê (1)(x 1) (3)(y 5) (1)(z 7) œ 0 Ê x 3y z œ 9 25. n œ i 3j 4k, P(2ß 4ß 5) Ê (1)(x 2) (3)(y 4) (4)(z 5) œ 0 Ê x 3y 4z œ 34 26. n œ i 2j k, P(1ß 2ß 1) Ê (1)(x 1) (2)(y 2) (1)(z 1) œ 0 Ê x 2y z œ 6 27. œ x œ 2t 1 œ s 2 2t s œ 1 4t 2s œ 2 Ê œ Ê œ Ê t œ 0 and s œ 1; then z œ 4t 3 œ 4s 1 y œ 3t 2 œ 2s 4 3t 2s œ 2 3t 2s œ 2 Ê 4(0) 3 œ (4)(1) 1 is satisfied Ê the lines intersect when t œ 0 and s œ 1 Ê the point of intersection is x œ 1, y œ 2, and z œ 3 or P(1ß 2ß 3). A vector normal to the plane determined by these lines is Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.5 Lines and Planes in Space 737 â â âi j k â â â n" ‚ n# œ â 2 3 4 â œ 20i 12j k, where n" and n# are directions of the lines Ê the plane â â â 1 2 4 â containing the lines is represented by(20)(x 1) (12)(y 2) (1)(z 3) œ 0 Ê 20x 12y z œ 7. 28. œ œ 2s 2 xœ t t 2s œ 2 Ê œ Ê s œ 1 and t œ 0; then z œ t 1 œ 5s 6 Ê 0 1 œ 5(1) 6 t s œ 1 y œ t 2 œ s 3 is satisfied Ê the lines do intersect when s œ 1 and t œ 0 Ê the point of intersection is x œ 0, y œ 2 and z œ 1 â â j kâ âi â â or P(0ß 2ß 1). A vector normal to the plane determined by these lines is n" ‚ n# œ â 1 " 1 â œ 6i 3j 3k, â â â2 1 5â where n" and n# are directions of the lines Ê the plane containing the lines is represented by (6)(x 0) (3)(y 2) (3)(z 1) œ 0 Ê 6x 3y 3z œ 3. 29. The cross product of i j k and 4i 2j 2k has the same direction as the normal to the plane â â j k â â i â â Ê n œ â 1 " 1 â œ 6j 6k. Select a point on either line, such as P(1ß 2ß 1). Since the lines are given â â â 4 2 2 â to intersect, the desired plane is 0(x 1) 6(y 2) 6(z 1) œ 0 Ê 6y 6z œ 18 Ê y z œ 3. 30. The cross product of i 3j k and i j k has the same direction as the normal to the plane â â j k â âi â â n œ â 1 3 1 â œ 2i 2j 4k. Select a point on either line, such as P(0ß 3ß 2). Since the lines are â â 1 â â1 1 given to intersect, the desired plane is (2)(x 0) (2)(y 3) (4)(z 2) œ 0 Ê 2x 2y 4z œ 14 Ê x y 2z œ 7. â âi j â 31. n" ‚ n# œ â 2 1 â â1 2 â k â â 1 â œ 3i 3j 3k is a vector in the direction of the line of intersection of the planes â 1 â Ê 3(x 2) (3)(y 1) 3(z 1) œ 0 Ê 3x 3y 3z œ 0 Ê x y z œ 0 is the desired plane containing P! (2ß 1ß 1) â âi Ä â 32. A vector normal to the desired plane is P" P# ‚ n œ â 2 â â4 â j k â â 0 2 â œ 2i 12j 2k; choosing P" (1ß 2ß 3) as a point on â 1 2 â the plane Ê (2)(x 1) (12)(y 2) (2)(z 3) œ 0 Ê 2x 12y 2z œ 32 Ê x 6y z œ 16 is the desired plane â âi Ä â 33. S(0ß 0ß 12), P(0ß 0ß 0) and v œ 4i 2j 2k Ê PS ‚ v œ â 0 â â4 Ê dœ Ä ¹PS‚v¹ kvk È j 0 2 â kâ â 12 â œ 24i 48j œ 24(i 2j) â 2 â È 5 È5 † 24 œ 2È30 is the distance from S to the line œ È2416 1 44 4 œ 24 È24 œ â â j k â â i Ä â â 34. S(0ß 0ß 0), P(5ß 5ß 3) and v œ 3i 4j 5k Ê PS ‚ v œ â 5 5 3 â œ 13i 16j 5k â â 4 5 â â 3 Ê dœ Ä ¹PS‚v¹ kv k œ È169 256 25 È450 È9 œ 3 is the distance from S to the line È9 16 25 œ È50 œ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 738 Chapter 12 Vectors and the Geometry of Space Ä ¹PS‚v¹ Ä 35. S(2ß 1ß 3), P(2ß 1ß 3) and v œ 2i 6j Ê PS ‚ v œ 0 Ê d œ kvk œ È040 œ 0 is the distance from S to the line (i.e., the point S lies on the line) â âi Ä â 36. S(2ß 1ß 1), P(0ß 1ß 0) and v œ 2i 2j 2k Ê PS ‚ v œ â 2 â â2 Ê dœ Ä ¹PS‚v¹ kvk œ â k â â 1 â œ 2i 6j 4k â 2 â j 0 2 È4 36 16 È56 14 È4 4 4 œ È12 œ É 3 is the distance from S to the line â â j kâ â i Ä â â 37. S(3ß 1ß 4), P(4ß 3ß 5) and v œ i 2j 3k Ê PS ‚ v œ â 1 4 9 â œ 30i 6j 6k â â â 1 2 3 â Ê dœ Ä ¹PS‚v¹ kvk œ È900 36 36 È È È È œ È972 œ È486 œ È817†6 œ 9 742 is the distance from S to the line È1 4 9 7 14 â â i Ä â 38. S(1ß 4ß 3), P(10ß 3ß 0) and v œ 4i 4k Ê PS ‚ v œ â 11 â â 4 Ê dœ Ä ¹PS‚v¹ kv k j 7 0 â kâ â 3 â œ 28i 56j 28k œ 28(i 2j k) â 4â È œ 28 4È1 14 1 1 œ 7È3 is the distance from S to the line Ä 39. S(2ß 3ß 4), x 2y 2z œ 13 and P(13ß 0ß 0) is on the plane Ê PS œ 11i 3j 4k and n œ i 2j 2k Ä 9 Ê d œ ¹ PS † knnk ¹ œ ¹ È111 4648 ¹ œ ¹ È ¹œ3 9 Ä 40. S(0ß 0ß 0), 3x 2y 6z œ 6 and P(2ß 0ß 0) is on the plane Ê PS œ 2i and n œ 3i 2j 6k Ä Ê d œ ¹ PS † knnk ¹ œ ¹ È9 46 36 ¹ œ È649 œ 67 Ä 41. S(0ß 1ß 1), 4y 3z œ 12 and P(0ß 3ß 0) is on the plane Ê PS œ 4j k and n œ 4j 3k Ä Ê d œ ¹ PS † knnk ¹ œ ¹ È161639 ¹ œ 19 5 Ä 42. S(2ß 2ß 3), 2x y 2z œ 4 and P(2ß 0ß 0) is on the plane Ê PS œ 2j 3k and n œ 2i j 2k Ä Ê d œ ¹ PS † knnk ¹ œ ¹ È4216 4 ¹ œ 83 Ä 43. S(0ß 1ß 0), 2x y 2z œ 4 and P(2ß 0ß 0) is on the plane Ê PS œ 2i j and n œ 2i j 2k Ä 4 1 0 Ê d œ ¹ PS † knnk ¹ œ ¹ È ¹ œ 53 414 Ä 44. S(1ß 0ß 1), 4x y z œ 4 and P(1ß 0ß 0) is on the plane Ê PS œ 2i k and n œ 4i j k Ä È Ê d œ ¹ PS † knnk ¹ œ ¹ È16811 1 ¹ œ È918 œ 3 # 2 Ä 45. The point P(1ß 0ß 0) is on the first plane and S(10ß 0ß 0) is a point on the second plane Ê PS œ 9i, and Ä n œ i 2j 6k is normal to the first plane Ê the distance from S to the first plane is d œ ¹ PS † knnk ¹ œ ¹ È1 94 36 ¹ œ È9 , which is also the distance between the planes. 41 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.5 Lines and Planes in Space 739 46. The line is parallel to the plane since v † n œ î j "# k‰ † (i 2j 6k) œ 1 2 3 œ 0. Also the point Ä S(1ß 0ß 0) when t œ 1 lies on the line, and the point P(10ß 0ß 0) lies on the plane Ê PS œ 9i. The distance from Ä S to the plane is d œ ¹ PS † knnk ¹ œ ¹ È1 49 36 ¹ œ È9 , which is also the distance from the line to the plane. 41 47. n" œ i j and n# œ 2i j 2k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È22È1 9 ‹ œ cos" Š È"2 ‹ œ 14 523 48. n" œ 5i j k and n# œ i 2j 3k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È ‹ œ cos" (0) œ 1# 27 È14 442 49. n" œ 2i 2j 2k and n# œ 2i 2j k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È ‹ œ cos" Š 3" È3 ‹ ¸ 1.76 rad 12 È9 50. n" œ i j k and n# œ k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È31È1 ‹ ¸ 0.96 rad 5 51. n" œ 2i 2j k and n# œ i 2j k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š 2È94È61 ‹ œ cos" Š 3È ‹ ¸ 0.82 rad 6 18 26 ‰ 52. n" œ 4j 3k and n# œ 3i 2j 6k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È825È ¸ 0.73 rad ‹ œ cos" ˆ 35 49 53. 2x y 3z œ 6 Ê 2(1 t) (3t) 3(1 t) œ 6 Ê 2t 5 œ 6 Ê t œ "# Ê x œ #3 , y œ #3 and z œ "# Ê ˆ 3# ß 3# ß #" ‰ is the point 41 54. 6x 3y 4z œ 12 Ê 6(2) 3(3 2t) 4(2 2t) œ 12 Ê 14t 29 œ 12 Ê t œ 41 14 Ê x œ 2, y œ 3 7 , 20 27 ‰ ˆ and z œ 2 41 7 Ê 2ß 7 ß 7 is the point 55. x y z œ 2 Ê (1 2t) (1 5t) (3t) œ 2 Ê 10t 2 œ 2 Ê t œ 0 Ê x œ 1, y œ 1 and z œ 0 Ê (1ß 1ß 0) is the point 56. 2x 3z œ 7 Ê 2(1 3t) 3(5t) œ 7 Ê 9t 2 œ 7 Ê t œ 1 Ê x œ 1 3, y œ 2 and z œ 5 Ê (4ß 2ß 5) is the point â âi â 57. n" œ i j k and n# œ i j Ê n" ‚ n# œ â 1 â â1 â kâ â 1 â œ i j, the direction of the desired line; (1ß 1ß 1) â 0â is on both planes Ê the desired line is x œ 1 t, y œ 1 t, z œ 1 j 1 1 â â j k â âi â â 58. n" œ 3i 6j 2k and n# œ 2i j 2k Ê n" ‚ n# œ â 3 6 2 â œ 14i 2j 15k, the direction of the â â â 2 1 2 â desired line; (1ß 0ß 0) is on both planes Ê the desired line is x œ 1 14t, y œ 2t, z œ 15t â â j k â âi â â 59. n" œ i 2j 4k and n# œ i j 2k Ê n" ‚ n# œ â 1 2 4 â œ 6j 3k, the direction of the â â â 1 1 2 â desired line; (4ß 3ß 1) is on both planes Ê the desired line is x œ 4, y œ 3 6t, z œ 1 3t Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 740 Chapter 12 Vectors and the Geometry of Space â â j k â âi â â 60. n" œ 5i 2j and n# œ 4j 5k Ê n" ‚ n# œ â 5 2 0 â œ 10i 25j 20k, the direction of the â â â 0 4 5 â desired line; (1ß 3ß 1) is on both planes Ê the desired line is x œ 1 10t, y œ 3 25t, z œ 1 20t 61. L1 & L2: x œ 3 2t œ 1 4s and y œ 1 4t œ 1 2s Ê œ 2t 4s œ 2 2t 4s œ 2 Ê œ 4t 2s œ 2 2t s œ 1 Ê 3s œ 3 Ê s œ 1 and t œ 1 Ê on L1, z œ 1 and on L2, z œ 1 Ê L1 and L2 intersect at (5ß 3ß 1). L2 & L3: The direction of L2 is "6 (4i 2j 4k) œ 3" (2i j 2k) which is the same as the direction " 3 (2i j 2k) of L3; hence L2 and L3 are parallel. L1 & L3: x œ 3 2t œ 3 2r and y œ 1 4t œ 2 r Ê œ 2t 2r œ 0 t rœ0 Ê œ Ê 3t œ 3 4t r œ 3 4t r œ 3 Ê t œ 1 and r œ 1 Ê on L1, z œ 2 while on L3, z œ 0 Ê L1 and L2 do not intersect. The direction of L1 is È"21 (2i 4j k) while the direction of L3 is 3" (2i j 2k) and neither is a multiple of the other; hence L1 and L3 are skew. 62. L1 & L2: x œ 1 2t œ 2 s and y œ 1 t œ 3s Ê œ 2t s œ 1 Ê 5s œ 3 Ê s œ 35 and t œ 45 Ê on L1, t 3s œ 1 3 2 " z œ 12 5 while on L2, z œ 1 5 œ 5 Ê L1 and L2 do not intersect. The direction of L1 is È (2i j 3k) 14 while the direction of L2 is È"11 (i 3j k) and neither is a multiple of the other; hence, L1 and L2 are skew. L2 & L3: x œ 2 s œ 5 2r and y œ 3s œ 1 r Ê œ s 2r œ 3 Ê 5s œ 5 Ê s œ 1 and r œ 2 Ê on L2, 3s r œ 1 z œ 2 and on L3, z œ 2 Ê L2 and L3 intersect at (1ß 3ß 2). L1 & L3: L1 and L3 have the same direction È"14 (2i j 3k); hence L1 and L3 are parallel. 63. x œ 2 2t, y œ 4 t, z œ 7 3t; x œ 2 t, y œ 2 "# t, z œ 1 3# t 64. 1(x 4) 2(y 1) 1(z 5) œ 0 Ê x 4 2y 2 z 5 œ 0 Ê x 2y z œ 7; È2 (x 3) 2È2 (y 2) È2 (z 0) œ 0 Ê È2x 2È2y È2z œ 7È2 65. x œ 0 Ê t œ "# , y œ "# , z œ 3# Ê ˆ!ß "# ß 3# ‰ ; y œ 0 Ê t œ 1, x œ 1, z œ 3 Ê (1ß 0ß 3); z œ 0 Ê t œ 0, x œ 1, y œ 1 Ê (1ß 1ß 0) 66. The line contains (0ß 0ß 3) and ŠÈ3ß 1ß 3‹ because the projection of the line onto the xy-plane contains the origin and intersects the positive x-axis at a 30° angle. The direction of the line is È3i j 0k Ê the line in question is x œ È3t, y œ t, z œ 3. 67. With substitution of the line into the plane we have 2(1 2t) (2 5t) (3t) œ 8 Ê 2 4t 2 5t 3t œ 8 Ê 4t 4 œ 8 Ê t œ 1 Ê the point (1ß 7ß 3) is contained in both the line and plane, so they are not parallel. 68. The planes are parallel when either vector A" i B" j C" k or A# i B# j C# k is a multiple of the other or when (A" i B" j C" k) ‚ aA# i B# j C# kb œ 0. The planes are perpendicular when their normals are perpendicular, or(A" i B" j C" k) † (A# i B# j C# k) œ 0. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.6 Cylinders and Quadric Surfaces 69. There are many possible answers. One is found as follows: eliminate t to get t œ x 1 œ 2 y œ z # 3 Ê x 1 œ 2 y and 2 y œ z # 3 Ê x y œ 3 and 2y z œ 7 are two such planes. 70. Since the plane passes through the origin, its general equation is of the form Ax By Cz œ 0. Since it meets the plane M at a right angle, their normal vectors are perpendicular Ê 2A 3B C œ 0. One choice satisfying this equation is A œ 1, B œ 1 and C œ 1 Ê x y z œ 0. Any plane Ax By Cz œ 0 with 2A 3B C œ 0 will pass through the origin and be perpendicular to M. 71. The points (aß 0ß 0), (0ß bß 0) and (0ß 0ß c) are the x, y, and z intercepts of the plane. Since a, b, and c are all nonzero, the plane must intersect all three coordinate axes and cannot pass through the origin. Thus, y x z a b c œ 1 describes all planes except those through the origin or parallel to a coordinate axis. 72. Yes. If v" and v# are nonzero vectors parallel to the lines, then v" ‚ v# Á 0 is perpendicular to the lines. Ä Ä 73. (a) EP œ cEP" Ê x! i yj zk œ c c(x" x! )i y" j z" kd Ê x! œ c(x" x! ), y œ cy" and z œ cz" , where c is a positive real number x ! (b) At x" œ 0 Ê c œ 1 Ê y œ y" and z œ z" ; at x" œ x! Ê x! œ 0, y œ 0, z œ 0; x lim c œ x lim Ä_ Ä _ x" x! " œ x lim œ 1 Ê c Ä 1 so that y Ä y" and z Ä z" Ä _ 1 ! ! ! 74. The plane which contains the triangular plane is x y z œ 2. The line containing the endpoints of the line segment is x œ 1 t, y œ 2t, z œ 2t. The plane and the line intersect at ˆ 23 ß 23 ß 23 ‰ . The visible section of the line # # # segment is Éˆ "3 ‰ ˆ 32 ‰ ˆ 32 ‰ œ 1 unit in length. The length of the line segment is È1# 2# 2# œ 3 Ê 23 of the line segment is hidden from view. 12.6 CYLINDERS AND QUADRIC SURFACES 1. d, ellipsoid 2. i, hyperboloid 3. a, cylinder 4. g, cone 5. l, hyperbolic paraboloid 6. e, paraboloid 7. b, cylinder 8. j, hyperboloid 9. k, hyperbolic paraboloid 10. f, paraboloid 11. h, cone 12. c, ellipsoid 13. x# y# œ 4 14. z œ y# 1 15. x# 4z# œ 16 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 741 742 Chapter 12 Vectors and the Geometry of Space 16. 4x# y# œ 36 17. 9x# y# z# œ 9 18. 4x# 4y# z# œ 16 19. 4x# 9y# 4z# œ 36 20. 9x# 4y# 36z# œ 36 21. x# 4y# œ z 22. z œ 8 x# y# 23. x œ 4 4y# z# 24. y œ 1 x# z# 25. x# y# œ z# 26. 4x# 9z# œ 9y# 27. x# y# z# œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.6 Cylinders and Quadric Surfaces # # 28. y# z# x# œ 1 29. z# x# y# œ 1 30. y4 x4 z# œ 1 31. y# x# œ z 32. x# y# œ z 33. z œ 1 y# x# 34. 4x# 4y# œ z# 35. y œ ax# z# b 36. 16x# 4y# œ 1 37. x# y# z# œ 4 38. x# z# œ y 39. x# z# œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 743 744 Chapter 12 Vectors and the Geometry of Space 40. 16y# 9z# œ 4x# 41. z œ ax# y# b 43. 4y# z# 4x# œ 4 44. x# y# œ z # # # 42. y# x# z# œ 1 # 45. (a) If x# y4 z9 œ 1 and z œ c, then x# y4 œ 99c Ê œ 1Š x# Š c# 9 9 ‹ y# ’ 4 ˆ9 c# ‰ 9 “ œ 1 Ê A œ ab1 # È 9 c# È # ‹ Š 2 93 c ‹ œ 21 a99 c b 3 (b) From part (a), each slice has the area 21 a99 z b , where 3 Ÿ z Ÿ 3. Thus V œ 2 '0 291 a9 z# b dz 3 # œ 491 '0 a9 z# b dz œ 491 ’9z z3 “ œ 491 (27 9) œ 81 3 $ $ ! (c) y# x# z# a# b # c# œ 1 x# Ê a# Šc# c# – z# ‹ y# — – b# Šc# c# z# ‹ œ 1 Ê A œ 1 Ša È c# z# È # # ‹ Š b cc z ‹ c — 21ab z 21ab ˆ 2 $ ‰ # # # Ê V œ 2 '0 1cab œ 413abc . Note that if r œ a œ b œ c, # ac z b dz œ c# ’c z 3 “ œ c# 3 c c $ $ then V œ 413r , which is the volume of a sphere. # # c ! # 46. The ellipsoid has the form Rx # Ry # zc# œ 1. To determine c# we note that the point (0ß rß h) lies on the surface # # # # of the barrel. Thus, Rr # hc# œ 1 Ê c# œ Rh# Rr# . We calculate the volume by the disk method: V œ 1 'ch y# dz. Now, Ry # cz# œ 1 Ê y# œ R# Š1 cz# ‹ œ R# ’1 z ahR# R# r b “ œ R# Š R h# r ‹ z# h # # # # Ê V œ 1 'ch ’R# Š R h# r ‹ z# “ dz œ 1 ’R# z "3 Š R h# r ‹ z$ “ h # # # # h ch # # # # # # œ 21 R# h "3 aR# r# b h‘ œ 21 Š 2R3 h r3h ‹ œ 43 1R# h 23 1r# h, the volume of the barrel. If r œ R, then V œ 21R# h which is the volume of a cylinder of radius R and height 2h. If r œ 0 and h œ R, then V œ 43 1R$ which is the volume of a sphere. # # 47. We calculate the volume by the slicing method, taking slices parallel to the xy-plane. For fixed z, xa# by# œ cz gives the ellipse x# # Š zac ‹ y# # Š zbc ‹ œ 1. The area of this ellipse is 1 âÈ cz ‰ ˆbÈ cz ‰ œ 1abz c (see Exercise 45a). Hence 1abz 1abh the volume is given by V œ '0 1abz c dz œ ’ 2c “ œ c . Now the area of the elliptic base when z œ h is h A œ 1abh c , as determined previously. # h # ! 1abh# ‰ h œ "# (base)(altitude), as claimed. Thus, V œ c œ "# ˆ 1abh c Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 12.6 Cylinders and Quadric Surfaces # # # 48. (a) For each fixed value of z, the hyperboloid xa# by# cz# œ 1 results in a cross-sectional ellipse x# – a# Šc# c# z# ‹ y# — – b# Šc# z# ‹ c# œ 1. The area of the cross-sectional ellipse (see Exercise 45a) is — # # A(z) œ 1 Š ac Èc# z# ‹ Š bc Èc# z# ‹ œ 1cab # ac z b . The volume of the solid by the method of slices is V œ '0 A(z) dz œ '0 h (b) h " $‘ h " $‰ 1ab 1ab # 1ab ˆ # # # # # œ 13cabh # a3c h b c# ac z b dz œ c# c z 3 z ! œ c# c h 3 h 1abh # # # # A! œ A(0) œ 1ab and Ah œ A(h) œ 1cab # ac h b , from part (a) Ê V œ 3c# a3c h b # # # h 1abh c h h 1ab h # # ‘ œ 1abh 3 Š2 1 c # ‹ œ 3 Š 2 c# ‹ œ 3 21ab c# ac h b œ 3 (2A! Ah ) # h h 1ab # # # (c) Am œ A ˆ h# ‰ œ 1cab # Šc 4 ‹ œ 4c# a4c h b Ê 6 (A! 4Am Ah ) 1ab 1abh 1abh # # # # ‘ # # # # # # # œ h6 1ab 1cab # a4c h b c# ac h b œ 6c# ac 4c h c h b œ 6c# a6c 2h b # # œ 1abh 3c# a3c h b œ V from part (a) 49. z œ y# 50. z œ 1 y# 51. z œ x# y# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 745 746 Chapter 12 Vectors and the Geometry of Space 52. z œ x# 2y# (a) (b) (c) (d) 53-58. Example CAS commands: Maple: with( plots ); eq := x^2/9 + y^2/36 = 1 - z^2/25; implicitplot3d( eq, x=-3..3, y=-6..6, z=-5..5, scaling=constrained, shading=zhue, axes=boxed, title="#89 (Section 11.6)" ); Mathematica: (functions and domains may vary): In the following chapter, you will consider contours or level curves for surfaces in three dimensions. For the purposes of plotting the functions of two variables expressed implicitly in this section, we will call upon the function ContourPlot3D. To insert the stated function, write all terms on the same side of the equal sign and the default contour equating that expression to zero will be plotted. This built-in function requires the loading of a special graphics package. <<Graphics`ContourPlot3D` Clear[x, y, z] ContourPlot3D[x2 /9 y2 /16 z2 /2 1, {x, 9, 9}, {y, 12, 12}, {z, 5, 5}, Axes Ä True, AxesLabel Ä {x, y, z}, Boxed Ä False, PlotLabel Ä "Elliptic Hyperboloid of Two Sheets"] Your identification of the plot may or may not be able to be done without considering the graph. CHAPTER 12 PRACTICE EXERCISES 1. (a) 3 3, 4¡ 4 2, 5¡ œ 9 8, 12 20¡ œ 17, 32¡ (b) È172 322 œ È1313 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 12 Practice Exercises 2. (a) 3 2, 4 5¡ œ 1, 1¡ 3. (a) (b) Éa1b2 a1b2 œ È2 4. (a) 747 2a3b, 2a4b¡ œ 6, 8¡ (b) É62 a8b2 œ 10 5a2b, 5a5b¡ œ 10, 25¡ (b) É102 a25b2 œ È725 œ 5È29 5. 1 6 radians below the negative x-axis: È ¢ #3 , "# £ [assuming counterclockwise]. È 6. ¢ #3 , "# £ 7. 2Š È 21 4 12 ‹a4i jb œ Š È817 i È217 j‹ 8. 5 1 ˆ 3 i 54 j‰ œ 3i 4j Éˆ 35 ‰2 ˆ 45 ‰2 5 9. length œ ¹È2i È2j¹ œ È2 2 œ 2, È2i È2j œ 2 Š È"2 i È"2 j‹ Ê the direction is È"2 i È"2 j 10. length œ ki jk œ È1 1 œ È2, i j œ È2 Š È"2 i È"2 j‹ Ê the direction is È"2 i È"2 j 11. t œ 12 Ê v œ (2 sin 12 )i ˆ2 cos 12 ‰ j œ 2i; length œ k2ik œ È4 0 œ 2; 2i œ 2aib Ê the direction is i 12. t œ ln 2 Ê v œ êln 2 cosaln 2b eln 2 sinaln 2b‰ i êln 2 sinaln 2b eln 2 cosaln 2b‰ j œ a2 cosaln 2b 2 sinaln 2bb i a2 sinaln 2b 2 cosaln 2bb j œ 2c acosaln 2b sinaln 2bb i asinaln 2b cosaln 2bb j d length œ k2c acosaln 2b sinaln 2bb i asinaln 2b cosaln 2bb j dk œ 2Éacosaln 2b sinaln 2bb2 acosaln 2b sinaln 2bb2 œ 2È2cos2 aln 2b 2sin2 aln 2b œ 2È2; 2c acosaln 2b sinaln 2bb i asinaln 2b cosaln 2bb j d œ 2È2Š acosaln 2b sinaln 2bb iÈ asinaln 2b cosaln 2bb j ‹ 2 cosaln 2bb Ê direction œ acosaln 2bÈ sinaln 2bb i asinaln 2bÈ j 2 2 13. length œ k2i 3j 6kk œ È4 9 36 œ 7, 2i 3j 6k œ 7 ˆ 27 i 37 j 67 k‰ Ê the direction is 27 i 37 j 67 k 14. length œ ki 2j kk œ È1 4 1 œ È6, i 2j k œ È6 Š È16 i È26 j È16 k‹ Ê the direction is 1 2 1 È6 i È6 j È6 k j 4k 4k 15. 2 kvvk œ 2 † È44# i( œ 2 † 4i Èj33 œ È833 i È233 j È833 k 1)# 4# 16. 5 kvvk œ 5 † ˆ 35 ‰ i ˆ 45 ‰ k Éˆ 35 ‰# ˆ 45 ‰# œ 5 † ˆ 35 ‰ iˆ 45 ‰k 9 16 É 25 25 œ 3i 4k â âi â 17. kvk œ È1 1 œ È2, kuk œ È4 1 4 œ 3, v † u œ 3, u † v œ 3, v ‚ u œ â 1 â â2 j 1 1 â k â â 0 â œ 2i 2j k , â 2 â u ‚ v œ (v ‚ u) œ 2i 2j k, kv ‚ uk œ È4 4 1 œ 3, ) œ cos" Š kvvk†kuuk ‹ œ cos" Š È" ‹ œ 14 , 2 kuk cos ) œ È32 , projv u œ Š kvvkk†uvk ‹ v œ 32 (i j) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 748 Chapter 12 Vectors and the Geometry of Space 18. kvk œ È1# 1# 2# œ È6, kuk œ È(1)# (1)# œ È2, v † u œ (1)(1) (1)(0) (2)(1) œ 3, â â j k â â i â â u † v œ 3, v ‚ u œ â 1 1 2 â œ i j k , u ‚ v œ (v ‚ u) œ i j k, â â â 1 0 1 â 3 3 kv ‚ uk œ È(1)# (1)# 1# œ È3, ) œ cos" Š kvvk†kuuk ‹ œ cos" Š È ‹ œ cos" Š È12 ‹ 6 È2 È È È œ cos" Š #3 ‹ œ 561 , kuk cos ) œ È2 † Š 2 3 ‹ œ 26 , projv u œ Š kvvkk†uvk ‹ v œ 63 (i j 2k) œ "# (i j k) 19. projv u œ Š kvvkk†uvk ‹ v œ 43 a2i j kb where v † u œ 8 and v † v œ 6 20. projv u œ Š kvvkk†uvk ‹ v œ 13 ai 2jb where v † u œ 1 and v † v œ 3 â âi â 21. u ‚ v œ â 1 â â1 j 0 1 â kâ â 0â œ k â 0â â â j kâ âi â â 22. u ‚ v œ â 1 1 0 â œ 2k â â â1 1 0â 23. Let v œ v" i v# j v$ k and w œ w" i w# j w$ k. Then kv 2wk # œ k(v" i v# j v$ k) 2(w" i w# j w$ k)k # œ k(v" 2w" )i (v# 2w# )j (v$ 2w$ )kk # œ ˆÈ(v" 2w" )# (v# 2w# )# (v$ 2w$ )# ‰ # œ av#" v## v3# b 4(v" w" v# w# v$ w$ ) 4 aw#1 w## w#3 b œ kvk # 4v † w 4 kwk # œ kvk # 4 kvk kwk cos ) 4 kwk # œ 4 4(2)(3) ˆcos 13 ‰ 36 œ 40 24 ˆ "# ‰ œ 40 12 œ 28 Ê kv 2wk œ È28 œ 2È 7 â j â i â 2 4 24. u and v are parallel when u ‚ v œ 0 Ê â â â 4 8 Ê 4a 40 œ 0 and 20 2a œ 0 Ê a œ 10 â k â â 5 â œ 0 Ê (4a 40)i (20 2a)j (0)k œ 0 â a â â â âi j k â â â 25. (a) area œ ku ‚ vk œ abs â 1 1 1 â œ k2i 3j kk œ È4 9 1 œ È14 â â â2 1 1 â â â 1 1 â â 1 â â 1 1 â œ 1a3 2b 1a6 a1bb 1a4 1b œ 1 (b) volume œ au ‚ vb † w œ â 2 â â â 1 2 3 â Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 12 Practice Exercises â â â i j kâ â â 26. (a) area œ ku ‚ vk œ abs â 1 1 0 â œ kkk œ 1 â â â0 1 0â â â â1 1 0â â â (b) volume œ au ‚ vb † w œ â 0 1 0 â œ 1a1 0b 1a0 0b 0 œ 1 â â â1 1 1â 27. The desired vector is n ‚ v or v ‚ n since n ‚ v is perpendicular to both n and v and, therefore, also parallel to the plane. 28. If a œ 0 and b Á 0, then the line by œ c and i are parallel. If a Á 0 and b œ 0, then the line ax œ c and j are parallel. If a and b are both Á 0, then ax by œ c contains the points ˆ ca ß !‰ and ˆ0ß bc ‰ Ê the vector ab ˆ ca i bc j‰ œ c(bi aj) and the line are parallel. Therefore, the vector bi aj is parallel to the line ax by œ c in every case. Ä 29. The line L passes through the point Pa0ß 0ß 1b parallel to v œ i j k. With PS œ 2i 2j k and â â j kâ â i Ä â â PS ‚ v œ â 2 2 1 â œ a2 1bi (2 1)j a2 2bk œ i 3j 4k, we find the distance â â â 1 1 1 â dœ Ä ¹PS‚v¹ kv k È È œ È119116 œ È26 œ 1 3 È78 3 . Ä 30. The line L passes through the point Pa2ß 2ß 0b parallel to v œ i j k. With PS œ 2i 2j k and â â j kâ â i Ä â â PS ‚ v œ â 2 2 1 â œ a2 1bi a2 1bj a2 2bk œ i 3j 4k, we find the distance â â â 1 1 1â dœ Ä ¹PS‚v¹ kv k È È œ È119116 œ È26 œ 1 3 È78 3 . 31. Parametric equations for the line are x œ 1 3t, y œ 2, z œ 3 7t. Ä 32. The line is parallel to PQ œ 0i j k and contains the point P(1ß 2ß 0) Ê parametric equations are x œ 1, y œ 2 t, z œ t for 0 Ÿ t Ÿ 1. Ä 33. The point P(4ß 0ß 0) lies on the plane x y œ 4, and PS œ (6 4)i 0j (6 0)k œ 2i 6k with n œ i j Ê dœ Ä ¹n†PS¹ knk œ ¹ È210100 ¹ œ È22 œ È2. Ä 34. The point P(0ß 0ß 2) lies on the plane 2x 3y z œ 2, and PS œ (3 0)i (0 0)j (10 2)k œ 3i 8k with n œ 2i 3j k Ê d œ Ä ¹n†PS¹ knk œ ¹ È640981 ¹ œ È1414 œ È14. 35. P(3ß 2ß 1) and n œ 2i j k Ê (2)(x 3) (1)(y (2)) (1)(z 1) œ 0 Ê 2x y z œ 5 36. P(1ß 6ß 0) and n œ i 2j 3k Ê (1)(x (1)) (2)(y 6) (3)(z 0) œ 0 Ê x 2y 3z œ 13 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 749 750 Chapter 12 Vectors and the Geometry of Space Ä Ä Ä Ä 37. P(1ß 1ß 2), Q(2ß 1ß 3) and R(1ß 2ß 1) Ê PQ œ i 2j k, PR œ 2i 3j 3k and PQ ‚ PR â â j k â â i â â œ â 1 2 1 â œ 9i j 7k is normal to the plane Ê (9)(x 1) (1)(y 1) (7)(z 2) œ 0 â â â 2 3 3 â Ê 9x y 7z œ 4 Ä Ä Ä Ä 38. P(1ß 0ß 0), Q(0ß 1ß 0) and R(0ß 0ß 1) Ê PQ œ i j, PR œ i k and PQ ‚ PR â â j kâ â i â â œ â 1 1 0 â œ i j k is normal to the plane Ê (1)(x 1) (1)(y 0) (1)(z 0) œ 0 â â â 1 0 1 â Ê xyzœ1 39. ˆ0ß "# ß 3# ‰ , since t œ "# , y œ "# and z œ 3# when x œ 0; ("ß 0ß 3), since t œ 1, x œ 1 and z œ 3 when y œ 0; (1ß 1ß 0), since t œ 0, x œ 1 and y œ 1 when z œ 0 40. x œ 2t, y œ t, z œ t represents a line containing the origin and perpendicular to the plane 2x y z œ 4; this line intersects the plane 3x 5y 2z œ 6 when t is the solution of 3(2t) 5(t) 2(t) œ 6 Ê t œ 23 Ê ˆ 43 ß 23 ß 23 ‰ is the point of intersection 41. n" œ i and n# œ i j È2k Ê the desired angle is cos" Š knn""k†knn## k ‹ œ cos" ˆ "# ‰ œ 13 42. n" œ i j and n# œ j k Ê the desired angle is cos" Š knn""k†knn## k ‹ œ cos" ˆ "# ‰ œ 13 â âi â 43. The direction of the line is n" ‚ n# œ â 1 â â1 â j kâ â 2 1 â œ 5i j 3k. Since the point (5ß 3ß 0) is on â 1 2 â both planes, the desired line is x œ 5 5t, y œ 3 t, z œ 3t. â â j k â âi â â 44. The direction of the intersection is n" ‚ n# œ â 1 2 2 â œ 6i 9j 12k œ 3(2i 3j 4k) and is the â â â 5 2 1 â same as the direction of the given line. 45. (a) The corresponding normals are n" œ 3i 'k and n# œ 2i 2j k and since n" † n# œ (3)(2) (0)(2) (6)(1) œ 6 0 6 œ 0, we have that the planes are orthogonal â â âi j k â â â (b) The line of intersection is parallel to n" ‚ n# œ â 3 0 6 â œ 12i 15j 6k. Now to find a point in â â â 2 2 1 â the intersection, solve œ 3x 6z œ 1 3x 6z œ 1 Ê œ Ê 15x 12y œ 19 Ê x œ 0 and y œ 19 12 2x 2y z œ 3 12x 12y 6z œ 18 "‰ 19 " Ê ˆ!ß 19 1# ß 6 is a point on the line we seek. Therefore, the line is x œ 12t, y œ 12 15t and z œ 6 6t. â âi â 46. A vector in the direction of the plane's normal is n œ u ‚ v œ â 2 â â1 â j kâ â 3 1 â œ 7i 3j 5k and P("ß 2ß 3) on â 1 2 â the plane Ê 7(x 1) 3(y 2) 5(z 3) œ 0 Ê 7x 3y 5z œ 14. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 12 Practice Exercises 751 47. Yes; v † n œ (2i 4j k) † (2i j 0k) œ 2 † 2 4 † 1 1 † 0 œ 0 Ê the vector is orthogonal to the plane's normal Ê v is parallel to the plane Ä 48. n † PP! 0 represents the half-space of points lying on one side of the plane in the direction which the normal n points â âi Ä Ä â 49. A normal to the plane is n œ AB ‚ AC œ â 2 â â2 j 0 1 â k â Ä â 1 â œ i 2j 2k Ê the distance is d œ ¹ APn†n ¹ â 0 â j)†(i 2j #k) œ ¹ (i 4È ¹ œ ¸ 1 38 0 ¸ œ 3 144 Ä 50. P(0ß 0ß 0) lies on the plane 2x 3y 5z œ 0, and PS œ 2i 2j 3k with n œ 2i 3j 5k Ê Ä 4 6 15 25 d œ ¹ nk†PS nk ¹ œ ¹ È4 9 25 ¹ œ È38 â âi â 51. n œ 2i j k is normal to the plane Ê n ‚ v œ â 2 â â1 j 1 1 â k â â 1 â œ 0i 3j 3k œ 3j 3k is orthogonal â 1 â to v and parallel to the plane 52. The vector B ‚ C is normal to the plane of B and C Ê A ‚ (B ‚ C) is orthogonal to A and parallel to the plane of B and C: â â â â j k â âi j k â â i â â â â B ‚ C œ â 1 2 1 â œ 5i 3j k and A ‚ (B ‚ C) œ â 2 1 1 â œ 2i 3j k â â â â â 1 1 2 â â 5 3 1 â Ê kA ‚ (B ‚ C)k œ È4 9 1 œ È14 and u œ " (2i 3j k) is the desired unit vector. È14 â âi â 53. A vector parallel to the line of intersection is v œ n" ‚ n# œ â 1 â â1 j 2 1 â kâ â 1 â œ 5i j 3 k â 2â Ê kvk œ È25 1 9 œ È35 Ê 2 Š kvvk ‹ œ È235 (5i j 3k) is the desired vector. 54. The line containing (0ß 0ß 0) normal to the plane is represented by x œ 2t, y œ t, and z œ t. This line intersects the plane 3x 5y 2z œ 6 when 3(2t) 5(t) 2(t) œ 6 Ê t œ 23 Ê the point is ˆ 43 ß 23 ß 23 ‰ . 55. The line is represented by x œ 3 2t, y œ 2 t, and z œ 1 2t. It meets the plane 2x y 2z œ 2 when 26 7‰ 2(3 2t) (2 t) 2(" 2t) œ 2 Ê t œ 89 Ê the point is ˆ 11 9 ß 9 ß 9 . â âi â 56. The direction of the intersection is v œ n" ‚ n# œ â 2 â â1 j 1 1 â k â â 1 â œ 3i 5j k Ê ) œ cos" Š kvvk†kiik ‹ â 2 â œ cos" Š È335 ‹ ¸ 59.5° 57. The intersection occurs when (3 2t) 3(2t) t œ 4 Ê t œ 1 Ê the point is (1ß 2ß 1). The required line â â âi j k â â â must be perpendicular to both the given line and to the normal, and hence is parallel to â 2 2 1 â â â â 1 3 1 â œ 5i 3j 4k Ê the line is represented by x œ 1 5t, y œ 2 3t, and z œ 1 4t. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 752 Chapter 12 Vectors and the Geometry of Space 58. If P(aß bß c) is a point on the line of intersection, then P lies in both planes Ê a 2b c 3 œ 0 and 2a b c 1 œ 0 Ê (a 2b c 3) k(2a b c 1) œ 0 for all k. â i j k ââ â Ä Ä â 3 2 4 ââ œ 26 i j k 59. The vector AB ‚ CD œ â 5 (2 7 2 ) is normal to the plane and A(2ß 0ß 3) lies on the â 26 26 â 0 â 5 5 â plane Ê 2(x 2) 7(y 0) 2(z (3)) œ 0 Ê 2x 7y 2z 10 œ 0 is an equation of the plane. 60. Yes; the line's direction vector is 2i 3j 5k which is parallel to the line and also parallel to the normal 4i 6j 10k to the plane Ê the line is orthogonal to the plane. â j â i Ä Ä â 61. The vector PQ ‚ PR œ â 2 1 â â 3 0 â kâ â 3 â œ i 11j 3k is normal to the plane. â 1â Ä Ä (a) No, the plane is not orthogonal to PQ ‚ PR . (b) No, these equations represent a line, not a plane. Ä Ä (c) No, the plane (x 2) 11(y 1) 3z œ 0 has normal i 11j 3k which is not parallel to PQ ‚ PR . (d) No, this vector equation is equivalent to the equations 3y 3z œ 3, 3x 2z œ 6, and 3x 2y œ 4 Ê x œ 43 23 t, y œ t, z œ 1 t, which represents a line, not a plane. Ä Ä (e) Yes, this is a plane containing the point R(2ß 1ß 0) with normal PQ ‚ PR . 62. (a) The line through A and B is x œ 1 t, y œ t, z œ 1 5t; the line through C and D must be parallel and is L" : x œ 1 t, y œ 2 t, z œ 3 5t. The line through B and C is x œ 1, y œ 2 2s, z œ 3 4s; the line through A and D must be parallel and is L# : x œ 2, y œ 1 2s, z œ 4 4s. The lines L" and L# intersect at D(2ß 1ß 8) where t œ 1 and s œ 1. 4k)†(i j 5k) (b) cos ) œ (2j È œ È315 20 È27 Ä Ä Ä Ä Ä 18 Ä 9 †BC (c) Š BA Ä Ä ‹ BC œ 20 BC œ 5 (j 2k) where BA œ i j 5k and BC œ 2j 4k BC†BC (d) area œ k(2j 4k) ‚ (i j 5k)k œ k14i 4j 2kk œ 6È6 (e) From part (d), n œ 14i 4j 2k is normal to the plane Ê 14(x 1) 4(y 0) 2(z 1) œ 0 Ê 7x 2y z œ 8. (f) From part (d), n œ 14i 4j 2k Ê the area of the projection on the yz-plane is kn † ik œ 14; the area of the projection on the xy-plane is kn † jk œ 4; and the area of the projection on the xy-plane is kn † kk œ 2. â â i Ä Ä Ä â 63. AB œ 2i j k, CD œ i 4j k, and AC œ 2i j Ê n œ â 2 â â 1 â j k â â 1 1 â œ 5i j 9k Ê the distance is â 4 1 â †(5i j 9k) d œ ¹ (2i Èj)25 ¹ œ È11 1 81 107 â â i Ä Ä Ä â 64. AB œ 2i 4j k, CD œ i j 2k, and AC œ 3i 3j Ê n œ â 2 â â 1 j 4 1 â k â â 1 â œ 7i 3j 2k Ê the distance â 2 â j)†(7i 3j 2k) is d œ ¹ (3i È349 ¹ œ È1262 94 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 12 Practice Exercises 65. x# y# z# œ 4 66. x# (y 1)# z# œ 1 67. 4x# 4y# z# œ 4 68. 36x# 9y# 4z# œ 36 69. z œ ax# y# b 70. y œ ax# z# b 71. x# y# œ z# 72. x# z# œ y# 73. x# y# z# œ 4 74. 4y# z# 4x# œ 4 75. y# x# z# œ 1 76. z# x# y# œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 753 754 Chapter 12 Vectors and the Geometry of Space CHAPTER 12 ADDITIONAL AND ADVANCED EXERCISES 1. Information from ship A indicates the submarine is now on the line L" : x œ 4 2t, y œ 3t, z œ "3 t; information from ship B indicates the submarine is now on the line L# : x œ 18s, y œ 5 6s, z œ s. The current position of the sub is ˆ6ß 3ß "3 ‰ and occurs when the lines intersect at t œ 1 and s œ 3" . The straight line path of the submarine contains both points P ˆ2ß 1ß "3 ‰ and Q ˆ6ß 3ß 3" ‰ ; the line representing this path is L: x œ 2 4t, y œ 1 4t, z œ "3 . The submarine traveled the distance between P and Q in 4 minutes Ê a speed of minutes the submarine will move 20È2 thousand ft from Q along the line L Ä ¹PQ¹ 4 œ È32 È2 thousand ft/min. 4 œ In 20 Ê 20È2 œ È(2 4t 6)# (1 4t 3)# 0# Ê 800 œ 16(t 1)# 16(t 1)# œ 32(t 1)# Ê (t 1)# œ 800 32 œ 25 Ê t œ 6 Ê the submarine will be located at ˆ26ß 23ß "3 ‰ in 20 minutes. 2. H# stops its flight when 6 110t œ 446 Ê t œ 4 hours. After 6 hours, H" is at P(246ß 57ß 9) while H# is at (446ß 13ß 0). The distance between P and Q is È(246 446)# (57 13)# (9 0)# ¸ 204.98 miles. At 150 mph, it would take about 1.37 hours for H" to reach H# . Ä Ä 3. Torque œ ¹ PQ ‚ F¹ Ê 15 ft-lb œ ¹ PQ ¹ kFk sin 1# œ 34 ft † kFk Ê kFk œ 20 lb 4. Let a œ i j k be the vector from O to A and b œ i 3j 2k be the vector from O to B. The vector v orthogonal to a and b Ê v is parallel to b ‚ a (since the rotation is clockwise). Now b ‚ a œ i j 2k; proja b œ ˆ aa††ba ‰a œ 2i 2j 2k Ê a2, 2, 2b is the center of the circular path a1, 3, 2b takes Ê radius œ É12 a1b2 02 œ È2 Ê arc length per a second covered by the point is 3# È2 units/sec œ kvk (velocity is constant). A unit vector in the direction of v is kbb‚ ‚a k È3 È3 È 3k 2 i 2 j a 3È œ È"6 i È"6 j È26 k Ê v œ kvkŠ kbb‚ 2Š È"6 i È"6 j È26 k‹ œ ‚a k ‹ œ # 4 3 5. (a) By the Law of Cosines we have cos ! œ 3 2a35ba œ 53 and cos " œ 4 2a45ba œ 54 Ê sin ! œ 45 and sin " œ 35 5b 5b 2 2 2 2 2 2 Ê F1 œ ¢lF1 lcos !, lF1 lsin !£ œ ¢ 35 lF1 l, 45 lF1 l£, F2 œ ¢lF2 lcos " , lF2 lsin " £ œ ¢ 45 lF2 l, 35 lF2 l£, and w œ 0, 100¡. Since F1 F2 œ 0, 100¡ Ê ¢ 35 lF1 l 45 lF2 l, 45 lF1 l 35 lF2 l£ œ 0, 100¡ Ê 35 lF1 l 45 lF2 l œ 0 and 45 lF1 l 35 lF2 l œ 100. Solving the first equation for lF2 l results in: lF2 l œ 34 lF1 l. Substituting this result into the 9 second equation gives us: 45 lF1 l 20 lF1 l œ 100 Ê lF1 l œ 80 lb. Ê lF2 l œ 60 lb. Ê F1 œ 48, 64¡ and F2 œ 48, 36¡ , and ! œ tan1 ˆ 43 ‰ and " œ tan1 ˆ 34 ‰ 12 5 12 5 (b) By the Law of Cosines we have cos ! œ 5 2a13 œ 13 and cos " œ 122a1213ba13b 5 œ 13 Ê sin ! œ 12 5ba13b 13 and sin " œ 13 2 2 2 2 2 2 5 12 5 Ê F1 œ ¢lF1 lcos !, lF1 lsin !£ œ ¢ 13 lF1 l, 12 13 lF1 l£, F2 œ ¢lF2 lcos " , lF2 lsin " £ œ ¢ 13 lF2 l, 13 lF2 l£, and 5 12 5 ¡ w œ 0, 200¡. Since F1 F2 œ 0, 200¡ Ê ¢ 13 lF1 l 13 lF2 l, 12 13 lF1 l 13 lF2 l£ œ 0, 200 5 12 5 5 Ê 13 lF1 l 13 lF2 l œ 0 and 12 13 lF1 l 13 lF2 l œ 200. Solving the first equation for lF2 l results in: lF2 l œ 12 lF1 l. 25 2400 Substituting this result into the second equation gives us: 12 13 lF1 l 156 lF1 l œ 200 Ê lF1 l œ 13 ¸ 184.615 lb. 12000 28800 12000 5000 ¡ Ê lF2 l œ 1000 13 ¸ 76.923 lb. Ê F1 œ ¢ 1169 , 1169 £ ¸ 71.006, 170.414 and F2 œ ¢ 1169 , 1169 £ ¸ 71.006, 29.586¡. 6. (a) T1 œ ¢lT1 lcos !, lT1 lsin !£, T2 œ ¢lT2 lcos " , lT2 lsin " £, and w œ 0, w¡. Since T1 T2 œ 0, w¡ Ê ¢lT1 lcos ! lT2 lcos " , lT1 lsin ! lT2 lsin " £ œ 0, w¡ Ê lT1 lcos ! lT2 lcos " œ 0 and ! lT1 lsin ! lT2 lsin " œ w. Solving the first equation for lT2 l results in: lT2 l œ cos cos " lT1 l. Substituting this result into Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 12 Additional and Advanced Exercises 755 w cos " w cos " ! sin " the second equation gives us: lT1 lsin ! coscos " lT1 l œ w Ê lT1 l œ sin ! cos " cos ! sin " œ sin a! " b and ! lT2 l œ sinwacos ! "b (b) w cos " cos a! " b d w cos " d d ; d! alT1 lb œ 0 Ê w cos " cos a! " b œ 0 Ê cos a! " b œ 0 d! alT1 lb œ d! Š sin a! " b ‹ œ sin2 a! " b " cos a! " b Ê ! " œ 12 Ê ! œ 12 " ; dd!2 alT1 lb œ dd! Š w cos ‹œ sin2 a! " b 2 w cos " ˆcos2 a! " b 1‰ ; sin3 a! " b d2 œ w cos " 0 Ê local minimum when ! œ 12 " d!2 alT1 lbº !œ 12 " (c) w cos ! cos a! " b d d d w cos ! ; d" alT2 lb œ 0 Ê w cos ! cos a! " b œ 0 Ê cos a! " b œ 0 d" alT2 lb œ d" Š sin a! " b ‹ œ sin2 a! " b ! cos a! " b Ê ! " œ 12 Ê " œ 12 !; dd"2 alT2 lb œ dd" Š w cos ‹œ sin2 a! " b 2 w cos ! ˆcos2 a! " b 1‰ ; sin3 a! " b d2 œ w cos ! 0 Ê local minimum when " œ 12 ! d!2 alT2 lbº " œ 12 ! 7. (a) If P(xß yß z) is a point in the plane determined by the three points P" (x" ß y" ß z" ), P# (x# ß y# ß z# ) and Ä Ä Ä Ä Ä Ä P$ (x$ ß y$ ß z$ ), then the vectors PP" , PP# and PP$ all lie in the plane. Thus PP" † (PP# ‚ PP$ ) œ 0 â â â x" x y" y z" z â â â Ê â x# x y# y z# z â œ 0 by the determinant formula for the triple scalar product in Section 12.4. â â â x$ x y$ y z$ z â (b) Subtract row 1 from rows 2, 3, and 4 and evaluate the resulting determinant (which has the same value as the given determinant) by cofactor expansion about column 4. This expansion is exactly the determinant in part (a) so we have all points P(xß yß z) in the plane determined by P" (x" ß y" ß z" ), P# (x# ß y# ß z# ), and P$ (x$ ß y$ ß z$ ). 8. Let L" : x œ a" s b" , y œ a# s b# , z œ a$ s b$ and L# : x œ c" t d" , y œ c# t d# , z œ c$ t d$ . If L" ² L# , â â â â â a" c" b" d" â â kc" c" b" d" â â â â â then for some k, ai œ kci , i œ 1, 2, 3 and the determinant â a# c# b# d# â œ â kc# c# b# d# â œ 0, â â â â â a$ c$ b$ d$ â â kc$ c$ b$ d$ â since the first column is a multiple of the second column. The lines L" and L# intersect if and only if the Ú a s c t (b d ) œ 0 " " " " system Û a# s c# t (b# d# ) œ 0 has a nontrivial solution Í the determinant of the coefficients is zero. Ü a$ s c$ t (b$ d$ ) œ 0 9. (a) Place the tetrahedron so that A is at a0, 0, 0b, the point P is on the y-axis, and ÃBC lies in the xy-plane. Since ÃBC is an equilateral triangle, all the angles in the triangle are 60‰ and since AP bisects BC Ê ÃBP is a 30‰ - 60‰ - 90‰ trinagle. Thus the coordinates of P are Š0, È3, 0‹, the coordinates of B are Š1, È3, 0‹, and the coordinates of C are Š1, È3, 0‹. Let the coordinates of D be given by aa, b, cb. Since all of the faces are equilateral Ä Ä È ab 3 †AB 1 trinagles Ê all the angles in each of the triangles are 60‰ Ê cosanDABb œ cosa60‰ b œ AD Ä Ä œ a2ba2b œ 2 Ä Ä lADllABl È a b 3 †AC 1 È Ê a bÈ3 œ 2 and cosanDACb œ cosa60‰ b œ AD Ä Ä œ a2ba2b œ 2 Ê a b 3 œ 2. Add the two equations lADllACl to obtain: 2bÈ3 œ 4 Ê b œ È23 . Substituting this value for b in the first equation gives us: a Š È23 ‹È3 œ 2 2 Ä È Ê a œ 0. Since lAD l œ Èa2 b2 c2 œ 2 Ê 02 Š È23 ‹ c2 œ 4 Ê c œ 2È32 . Thus the coordinates of D are È Ä Ä †AP 2 1 1 ‰ Š0, È23 , 2È32 ‹. cos ) œ cosanDAPb œ AD Ä Ä œ È Ê ) œ cos Š È ‹ Ê 57.74 . 2 3 3 lADllAPl (b) Since ÃBC lies in the xy-plane Ê the normal to the face given by ÃBC is n1 œ k. The face given by ˜BCD is an Ä Ä È È adjacent face. The vectors DB œ i È13 j 2È32 k and DC œ i È13 j 2È32 k both lie in the plane containing Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 756 Chapter 12 Vectors and the Geometry of Space â i â â â ˜BCD. The normal to this plane is given by n2 œ â 1 â â 1 â j 1 È3 1 È3 adjacent faces is given by cos ) œ cosanDAPb œ lnn11ll†nn22 l œ k â â È â 2È32 ââ œ 4È2 j 2 k . The angle ) between two È3 È3 â 2È 2 â È3 â 2/È3 Ê ) œ cos1 ˆ 13 ‰ ¸ 70.53‰ . a1bŠ6/È3‹ Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä 10. Extend CD to CG so that CD œ DG. Then CG œ t CF œ CB BG and t CF œ 3 CE CA, since ACBG is a Ä Ä Ä parallelogram. If t CF 3 CE CA œ 0, then t 3 1 œ 0 Ê t œ 4, since F, E, and A are collinear. Ä Ä Ä Ä Therefore, CG œ 4 CF Ê CD œ 2 CF Ê F is the midpoint of CD. Ä 11. If Q(xß y) is a point on the line ax by œ c, then P" Q œ (x x" )i (y y" )j , and n œ ai bj is normal to the Ä x" ) b(y y" )k line. The distance is ¹projn P" Q¹ œ ¹ [(x x" )i È(y# y"#)j]†(ai bj) ¹ œ ka(x È # # a b a b " by" ck œ kaxÈ , since c œ ax by. a# b# Ä 12. (a) Let Q(xß yß z) be any point on Ax By Cz D œ 0. Let QP" œ (x x" )i (y y" )j (z z" )k, and Ä B j Ck B j Ck n œ ÈAAi . The distance is ¹projn QP" ¹ œ ¹((x x" )i (y y" )j (z z" )k) † Š ÈAAi ‹¹ # B# C# # B# C# " (Ax By Cz)k " By" Cz" Dk œ kAx" By"È Cz œ kAxÈ . # # # # # # A B C A B C (b) Since both tangent planes are parallel, one-half of the distance between them is equal to the radius of the sphere, i.e., r œ " k3 9k œ È3 (see also Exercise 12a). Clearly, the points (1ß 2ß 3) and (1ß 2ß 3) # È1 1 1 are on the line containing the sphere's center. Hence, the line containing the center is x œ 1 2t, y œ 2 4t, z œ 3 6t. The distance from the plane x y z 3 œ 0 to the center is È3 (2 4t) (3 6t) 3k Ê k(1 2t) È œ È3 from part (a) Ê t œ 0 Ê the center is at (1ß 2ß 3). Therefore 111 an equation of the sphere is (x 1)# (y 2)# (z 3)# œ 3. 13. (a) If (x" ß y" ß z" ) is on the plane Ax By Cz œ D" , then the distance d between the planes is D# k " Cz" D# k d œ kAx"È By œ kAikD"Bj # # # Ckk , since Ax" By" Cz" œ D" , by Exercise 12(a). (b) (c) A B C k12 6k d œ È4 9 1 œ È614 k2(3) (1)(2) 2(1) 4k 2(1) Dk œ k2(3) (1)(2) È14 È14 Ê D œ 8 or 4 Ê the desired plane is 2x y 2x œ 8 Dk (d) Choose the point (2ß 0ß 1) on the plane. Then k3È œ 5 Ê D œ 3 „ 5È6 Ê the desired planes are 6 x 2y z œ 3 5È6 and x 2y z œ 3 5È6. Ä Ä 14. Let n œ AB ‚ BC and D(xß yß z) be any point in the plane determined by A, B and C. Then the point D lies in Ä Ä Ä Ä this plane if and only if AD † n œ 0 Í AD † (AB ‚ BC ) œ 0. â âi â 15. n œ i 2j 6k is normal to the plane x 2y 6z œ 6; v ‚ n œ â 1 â â1 â kâ â 1 â œ 4i 5j k is parallel to the â 6â â â j kâ âi â â plane and perpendicular to the plane of v and n Ê w œ n ‚ (v ‚ n) œ â 1 2 6 â œ 32i 23j 13k is a â â â 4 5 1 â j 1 2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 12 Additional and Advanced Exercises 757 vector parallel to the plane x 2y 6z œ 6 in the direction of the projection vector projP v. Therefore, w v †w 42 " 32 23 13 ˆ 32 23 13 ‰ projP v œ projw v œ Šv † kw wk ‹ kwk œ Š kwk # ‹ w œ 32# 23# 13# w œ 1722 w œ 41 w œ 41 i 41 j 41 k 16. projz w œ projz v and w projz w œ v projz v Ê w œ (w projz w) projz w œ (v projz v) projz w œ v 2 projz v œ v 2 Š kvz†kz# ‹ z 17. (a) u ‚ v œ 2i ‚ 2j œ 4k Ê (u ‚ v) ‚ w œ 0 ; (u † w)v (v † w)u œ 0v 0u œ 0; v ‚ w œ 4i Ê u ‚ (v ‚ w) œ 0; (u † w)v (u † v)w œ 0v 0w œ 0 â â â â j k â j k â âi â i â â â â (b) u ‚ v œ â 1 1 1 â œ i 4j 3k Ê (u ‚ v) ‚ w œ â 1 4 3 â œ 10i 2j 6k ; â â â â â 2 1 2 â â 1 2 1 â (u † w)v (v † w)u œ 4(2i j 2k) 2(i j k) œ 10i 2j 6k; â â â â j k â j kâ â i âi â â â â v ‚ w œ â 2 1 2 â œ 3i 4j 5k Ê u ‚ (v ‚ w) œ â 1 1 1 â œ 9i 2j 7k; â â â â â 1 2 1 â â3 4 5â (u † w)v (u † v)w œ 4(2i j 2k) (1)(i 2j k) œ 9i 2j 7k â â â â j kâ j k â âi âi â â â â (c) u ‚ v œ â 2 1 0 â œ i 2j 4k Ê (u ‚ v) ‚ w œ â 1 2 4 â œ 4i 6j 2k ; â â â â 2 â â 2 1 1 â â1 0 (u † w)v (v † w)u œ 2(2i j k) 4(2i j) œ 4i 6j 2k; â â â â j kâ j kâ âi â i â â â â 1 0 â œ i 2 j 4k; v ‚ w œ â 2 1 1 â œ 2 i 3 j k Ê u ‚ ( v ‚ w ) œ â 2 â â â â â1 0 2â â 2 3 1 â (u † w)v (u † v)w œ 2(2i j k) 3(i 2k) œ i 2j 4k â â â j k â j â i â i â â â (d) u ‚ v œ â 1 1 2 â œ i 3j k Ê (u ‚ v) ‚ w œ â 1 3 â â â â 1 0 1 â â 2 4 â k â â 1 â œ 10i 10k ; â 2 â (u † w)v (v † w)u œ 10(i k) 0(i j 2k) œ 10i 10k; â â â â j k â j k â â i âi â â â â v ‚ w œ â 1 0 1 â œ 4i 4j 4k Ê u ‚ (v ‚ w) œ â 1 1 2 â œ 12i 4j 8k; â â â â â 2 4 2 â â 4 4 4 â (u † w)v (u † v)w œ 10(i k) 1(2i 4j 2k) œ 12i 4j 8k 18. (a) u ‚ (v ‚ w) v ‚ (w ‚ u) w ‚ (u ‚ v) œ (u † w)v (u † v)w (v † u)w (v † w)u (w † v)u (w † u)v œ 0 (b) [u † (v ‚ i)]i [(u † (v ‚ j)]j [(u † (v ‚ k)]k œ [(u ‚ v) † i]i [(u ‚ v) † j]j [(u ‚ v) † k]k œ u ‚ v u†w v†w (c) (u ‚ v) † (w ‚ r) œ u † [v ‚ (w ‚ r)] œ u † [(v † r)w (v † w)r] œ (u † w)(v † r) (u † r)(v † w) œ º u†r v†r º 19. The formula is always true; u ‚ [u ‚ (u ‚ v)] † w œ u ‚ [(u † v)u (u † u)v] † w œ [(u † v)u ‚ u (u † u)u ‚ v] † w œ kuk # u ‚ v † w œ kuk # u † v ‚ w 20. If u œ (cos B)i (sin B)j and v œ (cos A)i (sin A)j , where A B, then u ‚ v œ ckuk kvk sin (A B)d k â â j kâ â i â â œ â cos B sin B 0 â œ (cos B sin A sin B cos A)k Ê sin (A B) œ cos B sin A sin B cos A, since â â â cos A sin A 0 â kuk œ 1 and kvk œ 1. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 758 Chapter 12 Vectors and the Geometry of Space 21. If u œ ai bj and v œ ci dj , then u † v œ kuk kvk cos ) Ê ac bd œ Èa# b# Èc# d# cos ) Ê (ac bd)# œ aa# b# b ac# d# b cos# ) Ê (ac bd)# Ÿ aa# b# b ac# d# b , since cos# ) Ÿ 1. 22. If u œ ai bj ck, then u † u œ a# b# c# 0 and u † u œ 0 iff a œ b œ c œ 0. 23. ku vk # œ (u v) † (u v) œ u † u 2u † v v † v Ÿ kuk # 2 kuk kvk kvk # œ akuk kvkb# Ê ku vk Ÿ kuk kvk 24. Let ! denote the angle between w and u, and " the angle between w and v. Let a œ kuk and b œ kvk . Then # cos ! œ kwwk†kuuk œ (avkwk bkuuk)†u œ (av†kuwk kubuk †u) œ (av†kuwk kubuk †u) œ aav†kuwk aba b œ v†ukwk ba , and likewise, cos " œ u†vkwk ba . Since the angle between u and v is always Ÿ 1# and cos ! œ cos " , we have that ! œ " Ê w bisects the angle between u and v . 25. (kukv kvku) † (kvku kukv) œ kukv † kvku kvku † kvku kukv † kukv kvku † kukv œ kvku † kukv kvk# u † u kuk# v † v kvku † kukv œ kvk# kuk# kuk# kvk# œ 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. CHAPTER 13 VECTOR-VALUED FUNCTIONS AND MOTION IN SPACE 13.1 CURVES IN SPACE AND THEIR TANGENTS 1. x œ t 1 and y œ t# 1 Ê y œ (x 1)# 1 œ x# 2x; v œ ddtr œ i 2tj Ê a œ ddtv œ 2j Ê v œ i 2j and a œ 2j at t œ 1 1 2. x œ t t 1 and y œ 1t Ê x œ 1 y 1 œ 1 1 y Ê y œ x1 1; v œ ddtr œ at 11b2 i t12 j Ê a œ ddtv œ at 21b3 i t23 j y Ê v œ 4i 4j and a œ 16i 16j at t œ "# 3. x œ et and y œ 29 e2t Ê y œ 29 x# ; v œ ddtr œ et i 49 e2t j Ê a œ et i 89 e2t j Ê v œ 3i 4j and a œ 3i 8j at t œ ln 3 4. x œ cos 2t and y œ 3 sin 2t Ê x# "9 y# œ 1; v œ ddtr œ (2 sin 2t)i (6 cos 2t)j Ê a œ ddtv œ (4 cos 2t)i (12 sin 2t)j Ê v œ 6j and a œ 4i at t œ 0 5. v œ ddtr œ (cos t)i (sin t)j and a œ ddtv œ (sin t)i (cos t)j Ê for t œ 14 , v ˆ 14 ‰ œ È È2 È2 # i # j and È a ˆ 14 ‰ œ #2 i #2 j ; for t œ 1# , v ˆ 1# ‰ œ j and a ˆ 1# ‰ œ i 6. v œ ddtr œ ˆ2 sin #t ‰ i ˆ2 cos #t ‰ j and a œ ddtv œ ˆ cos #t ‰ i ˆ sin #t ‰ j Ê for t œ 1, v(1) œ 2i and a(1) œ j ; for t œ 3#1 , v ˆ 3#1 ‰ œ È2 i È2 j and a ˆ 3#1 ‰ œ È2 È2 # i # j 7. v œ ddtr œ (1 cos t)i (sin t)j and a œ ddtv œ (sin t)i (cos t)j Ê for t œ 1, v(1) œ 2i and a(1) œ j ; for t œ 3#1 , v ˆ 3#1 ‰ œ i j and a ˆ 3#1 ‰ œ i 8. v œ ddtr œ i 2tj and a œ ddtv œ 2j Ê for t œ 1, v(1) œ i 2j and a(1) œ 2j ; for t œ 0, v(0) œ i and a(0) œ 2j ; for t œ 1, v(1) œ i 2j and a(1) œ 2j Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 760 Chapter 13 Vector-Valued Functions and Motion in Space # 9. r œ (t 1)i at# 1b j 2tk Ê v œ ddtr œ i 2tj 2k Ê a œ ddt#r œ 2j ; Speed: kv(1)k œ È1# (2(1))# 2# œ 3; i 2(1)j 2k Direction: kvv(1) œ 3" i 32 j 32 k Ê v(1) œ 3 ˆ 3" i 32 j 32 k‰ (1)k œ 3 # $ # 10. r œ (1 t)i Èt 2 j t3 k Ê v œ ddtr œ i È2t2 j t# k Ê a œ ddt#r œ È22 j 2tk ; Speed: kv(1)k # v(1) # # œ Ê1# Š 2(1) È2 ‹ (1 ) œ 2; Direction: kv(1)k œ i È j (1# )k 2(1) 2 # œ "# i È"2 j "# k Ê v(1) œ 2 Š "# i È"2 j "# k‹ # 11. r œ (2 cos t)i (3 sin t)j 4tk Ê v œ ddtr œ (2 sin t)i (3 cos t)j 4k Ê a œ ddt#r œ (2 cos t)i (3 sin t)j ; 1 vˆ ‰ # # Speed: ¸v ˆ 1# ‰¸ œ Éˆ2 sin 1# ‰ ˆ3 cos 12 ‰ 4# œ 2È5; Direction: ¸v ˆ 1# ‰¸ # 2 3 4 œ Š #È sin 1# ‹ i Š #È cos 1# ‹ j #È k œ È"5 i È25 k Ê v ˆ 1# ‰ œ 2È5 Š È"5 i È25 k‹ 5 5 5 # 12. r œ (sec t)i (tan t)j 43 tk Ê v œ ddtr œ (sec t tan t)i asec# tb j 43 k Ê a œ ddt#r # # # œ asec t tan# t sec$ tb i a2 sec# t tan tb j ; Speed: ¸v ˆ 16 ‰¸ œ Éˆsec 16 tan 16 ‰ ˆsec# 16 ‰ ˆ 43 ‰ œ 2; v ˆ1‰ Direction: ¸v ˆ 16 ‰¸ œ 6 ˆsec 1 tan 1 ‰ i ˆsec# 1 ‰ j 4 k 6 6 6 3 œ 3" i 32 j 32 k # Ê v ˆ 16 ‰ œ 2 ˆ 3" i 32 j 32 k‰ # # 13. r œ (2 ln (t 1))i t# j t2 k Ê v œ ddtr œ ˆ t 2 1 ‰ i 2tj tk Ê a œ ddt#r œ ’ (t 21)# “ i 2j k ; 2 Š " 1 ‹ i 2(1)j (1)k # Speed: kv(1)k œ Éˆ 1 2 1 ‰ (2(1))# 1# œ È6; Direction: kvv(1) È6 (1)k œ œ È"6 i È26 j È16 k Ê v(1) œ È6 Š È"6 i È26 j È"6 k‹ # 14. r œ aet b i (2 cos 3t)j (2 sin 3t)k Ê v œ ddtr œ aet b i (6 sin 3t)j (6 cos 3t)k Ê a œ ddt#r œ aet b i (18 cos 3t)j (18 sin 3t)k ; Speed: kv(0)k œ Éae! b# [6 sin 3(0)]# [6 cos 3(0)]# œ È37; ! ae b i 6 sin 3(0)j 6 cos 3(0)k Direction: kvv(0) œ È"37 i È637 k Ê v(0) œ È37 Š È"37 i È637 k‹ È37 (0)k œ # 15. v œ 3i È3 j 2tk and a œ 2k Ê v(0) œ 3i È3 j and a(0) œ 2k Ê kv(0)k œ Ê3# ŠÈ3‹ 0# œ È12 and ka(0)k œ È2# œ 2; v(0) † a(0) œ 0 Ê cos ) œ 0 Ê ) œ 1# 16. v œ È2 È2 # i Š # 32t‹ j and a œ 32j Ê v(0) œ È2 È2 # i # j and a(0) œ 32j È # È Ê kv(0)k œ ÊŠ #2 ‹ Š #2 ‹ È È # È 16 2 œ 1 and ka(0)k œ È(32)# œ 32; v(0) † a(0) œ Š #2 ‹ (32) œ 16È2 Ê cos ) œ 1(32) œ #2 Ê ) œ 341 1 ‰ # 17. v œ ˆ t# 2t 1 ‰ i ˆ t# 1 j tat 1b "Î# # 2t 2 k and a œ ’ “ i ’ at# 2t 1b# “ j ’ # " $Î# “ k Ê v(0) œ j and at# 1b# at 1b a(0) œ 2i k Ê kv(0)k œ 1 and ka(0)k œ È2# 1# œ È5; v(0) † a(0) œ 0 Ê cos ) œ 0 Ê ) œ 12 18. v œ 23 (1 t)"Î# i 23 (1 t)"Î# j 3" k and a œ "3 (1 t)"Î# i 3" (1 t)"Î# j Ê v(0) œ 32 i 32 j 3" k and # # # # # a(0) œ "3 i 3" j Ê kv(0)k œ Éˆ 32 ‰ ˆ 32 ‰ ˆ 3" ‰ œ 1 and ka(0)k œ Éˆ "3 ‰ ˆ 3" ‰ œ È2 2 2 3 ; v(0) † a(0) œ 9 9 œ 0 Ê cos ) œ 0 Ê ) œ 1# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.1 Curves in Space and Their Tangents 761 19. r(t) œ (sin t)i at# cos tb j et k Ê v(t) œ (cos t)i (2t sin t)j et k ; t! œ 0 Ê v(t0 ) œ i k and r(t0 ) œ P! œ (0ß 1ß 1) Ê x œ 0 t œ t, y œ 1, and z œ 1 t are parametric equations of the tangent line 20. r(t) œ t2 i a2t 1b j t3 k Ê v(t) œ 2t i 2 j 3t2 k ; t! œ 2 Ê va2b œ 4 i 2 j 12k and r(t0 ) œ P! œ a4ß 3ß 8b Ê x œ 4 4t, y œ 3 2t, and z œ 8 12t are parametric equations of the tangent line 1 1 3 1 21. r(t) œ aln tbi tt 2 j at ln tbk Ê v(t) œ t i at 2b2 j aln t 1bk ; t! œ 1 Ê va1b œ i 3 j k and r(t0 ) œ P! œ a0ß 0ß 0b Ê x œ 0 t œ t, y œ 0 13 t œ 13 t, and z œ 0 t œ t are parametric equations of the tangent line 22. r(t) œ (cos t)i asin tb j (sin 2t)k Ê v(t) œ ( sin t)i (cos t)j (2 cos 2t)k ; t! œ 1# Ê v(t0 ) œ i 2k and r(t0 ) œ P! œ (0ß 1ß 0) Ê x œ 0 t œ t, y œ 1, and z œ 0 2t œ 2t are parametric equations of the tangent line 23. (a) v(t) œ (sin t)i (cos t)j Ê a(t) œ (cos t)i (sin t)j ; (i) kv(t)k œ È(sin t)# (cos t)# œ 1 Ê constant speed; (ii) v † a œ (sin t)(cos t) (cos t)(sin t) œ 0 Ê yes, orthogonal; (iii) counterclockwise movement; (iv) yes, r(0) œ i 0j (b) v(t) œ (2 sin 2t)i (2 cos 2t)j Ê a(t) œ (4 cos 2t)i (4 sin 2t)j; (i) kv(t)k œ È4 sin# 2t 4 cos# 2t œ 2 Ê constant speed; (ii) v † a œ 8 sin 2t cos 2t 8 cos 2t sin 2t œ 0 Ê yes, orthogonal; (iii) counterclockwise movement; (iv) yes, r(0) œ i 0j (c) v(t) œ sin ˆt 1# ‰ i cos ˆt 1# ‰ j Ê a(t) œ cos ˆt 1# ‰ i sin ˆt 1# ‰ j ; (i) kv(t)k œ Ésin# ˆt 1# ‰ cos# ˆt 1# ‰ œ 1 Ê constant speed; (ii) v † a œ sin ˆt 1# ‰ cos ˆt 1# ‰ cos ˆt 1# ‰ sin ˆt 1# ‰ œ 0 Ê yes, orthogonal; (iii) counterclockwise movement; (iv) no, r(0) œ 0i j instead of i 0j (d) v(t) œ (sin t)i (cos t)j Ê a(t) œ (cos t)i (sin t)j ; (i) kv(t)k œ È(sin t)# ( cos t)# œ 1 Ê constant speed; (ii) v † a œ (sin t)(cos t) (cos t)(sin t) œ 0 Ê yes, orthogonal; (iii) clockwise movement; (iv) yes, r(0) œ i 0j (e) v(t) œ (2t sin t)i (2t cos t)j Ê a(t) œ (2 sin t 2t cos t)i (2 cos t 2t sin t)j ; (i) kv(t)k œ Éc a2t sin tb d# (2t cos t)# œ È4t# asin# t cos# tb œ 2ktk œ 2t, t 0 Ê variable speed; (ii) v † a œ 4 at sin# t t# sin t cos tb 4 at cos# t t# cos t sin tb œ 4t Á 0 in general Ê not orthogonal in general; (iii) counterclockwise movement; (iv) yes, r(0) œ i 0j 24. Let p œ 2i 2j k denote the position vector of the point a2, 2, 1b and let, u œ È"2 i È"2 j and v œ È"3 i È"3 j È"3 k. Then r(t) œ p (cos t)u (sin t)v. Note that (2ß 2ß 1) is a point on the plane and n œ i j 2k is normal to the plane. Moreover, u and v are orthogonal unit vectors with u † n œ v † n œ 0 Ê u and v are parallel to the plane. Therefore, r(t) identifies a point that lies in the plane for each t. Also, for each t, (cos t)u (sin t)v is a unit vector. Starting at the point Š2 È12 , 2 È12 , 1‹ the vector ratb traces out a circle of radius 1 and center (2ß 2ß 1) in the plane x y 2z œ 2. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 762 Chapter 13 Vector-Valued Functions and Motion in Space 25. The velocity vector is tangent to the graph of y# œ 2x at the point (#ß #), has length 5, and a positive i dy component. Now, y# œ 2x Ê 2y dy dx œ 2 Ê dx ¹ vector i "# j Ê the velocity vector is v œ Ð#ß#Ñ œ #2†2 œ "# Ê the tangent vector lies in the direction of the 5 î "# j‰ œ È5 î "# j‰ œ 2È5i È5j 5 É1 "4 Œ # 26. (a) (b) v œ (1 cos t)i (sin t)j and a œ (sin t)i (cos t)j ; kvk# œ (1 cos t)# sin# t œ 2 2 cos t Ê kvk# is at a max when cos t œ 1 Ê t œ 1, 31, 51, etc., and at these values of t, kvk# œ 4 Ê max kvk œ È4 œ 2; kvk# is at a min when cos t œ 1 Ê t œ 0, 21, 41, etc., and at these values of t, kvk# œ 0 Ê min kvk œ 0; kak# œ sin# t cos# t œ 1 for every t Ê max kak œ min kak œ È1 œ 1 27. dtd (r † r) œ r † ddtr ddtr † r œ 2r † ddtr œ 2 † 0 œ 0 Ê r † r is a constant Ê krk œ Èr † r is constant 28. (a) (b) d du d du dw ‰ ˆ dv dt (u † v ‚ w) œ dt † (v ‚ w) u † dt (v ‚ w) œ dt † (v ‚ w) u † dt ‚ w v ‚ dt œ ddtu † (v ‚ w) u † ddtv ‚ w u † v ‚ ddtw d dr d# r dr dr d# r d# r d# r dr d$ r dr d$ r dt ’r † Š dt ‚ dt# ‹“ œ dt † Š dt ‚ dt# ‹ r † Š dt# ‚ dt# ‹ r † Š dt ‚ dt$ ‹ œ r † Š dt ‚ dt$ ‹ , since A † (A ‚ B) œ 0 and A † (B ‚ B) œ 0 for any vectors A and B dg dh 29. (a) u œ f(t)i g(t)j h(t)k Ê cu œ cf(t)i cg(t)j ch(t)k Ê dtd (cu) œ c df dt i c dt j c dt k dg dh du œ c Š df dt i dt j dt k‹ œ c dt df dg df dh (b) f u œ f f(t)i f g(t)j f h(t)k Ê dtd (f u) œ ’ ddtf f(t) f df dt “ i ’ dt g(t) f dt “ j ’ dt h(t) f dt “ k dg df dh du œ ddtf [f(t)i g(t)j h(t)k] f ’ df dt i dt j dt k“ œ dt u f dt 30. Let u œ f" (t)i f# (t)j f$ (t)k and v œ g" (t)i g# (t)j g$ (t)k. Then u v œ [f" (t) g" (t)]i [f# (t) g# (t)]j [f$ (t) g$ (t)]k Ê dtd (u v) œ [f"w (t) gw" (t)]i [f#w (t) gw# (t)]j [f$w (t) gw$ (t)]k œ [f"w (t)i f#w (t)j f$w (t)k] [gw" (t)i gw# (t)j gw$ (t)k] œ ddtu ddtv ; u v œ [f" (t) g" (t)]i [f# (t) g# (t)]j [f$ (t) g$ (t)]k Ê dtd (u v) œ [f"w (t) gw" (t)]i [f#w (t) gw# (t)]j [f$w (t) gw$ (t)]k œ [f"w (t)i f#w (t)j f$w (t)k] [gw" (t)i gw# (t)j gw$ (t)k] œ ddtu ddtv 31. Suppose r is continuous at t œ t! . Then lim r(t) œ r(t! ) Í lim [f(t)i g(t)j h(t)k] t Ä t! t Ä t! œ f(t! )i g(t! )j h(t! )k Í lim f(t) œ f(t! ), lim g(t) œ g(t! ), and lim h(t) œ h(t! ) Í f, g, and h are t Ä t! t Ä t! t Ä t! continuous at t œ t! . Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.1 Curves in Space and Their Tangents â â i â 32. lim [r" (t) ‚ r# (t)] œ lim â f" (t) t Ä t! t Ä t! â â g" (t) j f# (t) g# (t) â â i k â ââ â â lim f" (t) f$ (t) â œ â t Ä t! â g$ (t) â ââ lim g" (t) tÄt ! j lim f# (t) t Ä t! lim g# (t) t Ä t! â â lim f$ (t) ââ t Ä t! â lim g$ (t) ââ t Ä t! k œ lim r" (t) ‚ lim r# (t) œ A ‚ B t Ä t! t Ä t! 33. rw (t! ) exists Ê f w (t! )i gw (t! )j hw (t! )k exists Ê f w (t! ), gw (t! ), hw (t! ) all exist Ê f, g, and h are continuous at t œ t! Ê r(t) is continuous at t œ t! db dc 34. u œ C œ ai bj ck with a, b, c real constants Ê ddtu œ da dt i dt j dt k œ 0i 0j 0k œ 0 35-38. Example CAS commands: Maple: > with( plots ); r := t -> [sin(t)-t*cos(t),cos(t)+t*sin(t),t^2]; t0 := 3*Pi/2; lo := 0; hi := 6*Pi; P1 := spacecurve( r(t), t=lo..hi, axes=boxed, thickness=3 ): display( P1, title="#35(a) (Section 13.1)" ); Dr := unapply( diff(r(t),t), t ); # (b) Dr(t0); # (c) q1 := expand( r(t0) + Dr(t0)*(t-t0) ); T := unapply( q1, t ); P2 := spacecurve( T(t), t=lo..hi, axes=boxed, thickness=3, color=black ): display( [P1,P2], title="#35(d) (Section 13.1)" ); 39-40. Example CAS commands: Maple: a := 'a'; b := 'b'; r := (a,b,t) -> [cos(a*t),sin(a*t),b*t]; Dr := unapply( diff(r(a,b,t),t), (a,b,t) ); t0 := 3*Pi/2; q1 := expand( r(a,b,t0) + Dr(a,b,t0)*(t-t0) ); T := unapply( q1, (a,b,t) ); lo := 0; hi := 4*Pi; P := NULL: for a in [ 1, 2, 4, 6 ] do P1 := spacecurve( r(a,1,t), t=lo..hi, thickness=3 ): P2 := spacecurve( T(a,1,t), t=lo..hi, thickness=3, color=black ): P := P, display( [P1,P2], axes=boxed, title=sprintf("#39 (Section 13.1)\n a=%a",a) ); end do: display( [P], insequence=true ); 35-40. Example CAS commands: Mathematica: (assigned functions, parameters, and intervals will vary) The x-y-z components for the curve are entered as a list of functions of t. The unit vectors i, j, k are not inserted. If a graph is too small, highlight it and drag out a corner or side to make it larger. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 763 764 Chapter 13 Vector-Valued Functions and Motion in Space Only the components of r[t] and values for t0, tmin, and tmax require alteration for each problem. Clear[r, v, t, x, y, z] r[t_]={ Sin[t] t Cos[t], Cos[t] t Sin[t], t^2} t0= 31 / 2; tmin= 0; tmax= 61; ParametricPlot3D[Evaluate[r[t]], {t, tmin, tmax}, AxesLabel Ä {x, y, z}]; v[t_]= r'[t] tanline[t_]= v[t0] t r[t0] ParametricPlot3D[Evaluate[{r[t], tanline[t]}], {t, tmin, tmax}, AxesLabel Ä {x, y, z}]; For 39 and 40, the curve can be defined as a function of t, a, and b. Leave a space between a and t and b and t. Clear[r, v, t, x, y, z, a, b] r[t_,a_,b_]:={Cos[a t], Sin[a t], b t} t0= 31 / 2; tmin= 0; tmax= 41; v[t_,a_,b_]= D[r[t, a, b], t] tanline[t_,a_,b_]=v[t0, a, b] t r[t0, a, b] pa1=ParametricPlot3D[Evaluate[{r[t, 1, 1], tanline[t, 1, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}]; pa2=ParametricPlot3D[Evaluate[{r[t, 2, 1], tanline[t, 2, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}]; pa4=ParametricPlot3D[Evaluate[{r[t, 4, 1], tanline[t, 4, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}]; pa6=ParametricPlot3D[Evaluate[{r[t, 6, 1], tanline[t, 6, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}]; Show[GraphicsRow[{pa1, pa2, pa4, pa6}]] 13.2 INTEGRALS OF VECTOR FUNCTIONS; PROJECTILE MOTION 1. '01 ct$ i 7j (t 1)kd dt œ ’ t4 “ " i [7t] !" j ’ t2 t“ " k œ 4" i 7j 3# k 2. '12 (6 6t)i 3Èt j ˆ t4 ‰ k‘ dt œ c6t 3t# d #" i 2t$Î# ‘ #" j c4t" d #" k œ 3i Š4È2 2‹ j 2k 3. ' 11Î%Î% c(sin t)i (1 cos t)j asec# tb kd dt œ c cos td 1Î%1Î% i ct sin td 1Î%1Î% j ctan td 1Î%1Î% k œ Š 1 #2È2 ‹ j 2k 4. '01Î3 casec t tan tbi atan tbj a2 sin t cos tb kd dt œ '01Î3 [asec t tan tbi atan tbj asin 2tbk] dt % # ! ! # 1Î$ 1Î$ 1Î$ œ csec td ! i c ln acos tbd ! j "# cos 2t‘ ! k œ i (ln 2)j 34 k 5. '14 ˆ "t i 5 " t j #"t k‰ dt œ œ cln td %" i c ln (5 t)d %" j "# ln t‘ %" k œ (ln 4)i (ln 4)j (ln 2)k 6. '01 Š È 2 7. '01 Štet i et j k‹ dt œ ’ 12 et “ " i cet d "! j ctd !" k œ e 2 1 i e e 1 i k 8. '1ln 3 atet i et j ln t kb dt œ ctet et d ln1 3 i cet d ln1 3 j ct ln t td 1ln 3 k 1 t# 2 " È " È i 1 3t# k‹ dt œ c2 sin" td ! i ’È3 tan" t“ k œ 1i 1 4 3 k ! 2 ! œ 3a ln 3 1bi a3 ebi aln 3alnaln 3b 1b 1b k 9. '01Î2 cacos tbi asin 2tbj asin2 tb kd dt œ '01Î2 acos tbi asin 2tbj ˆ 12 12 cos 2t‰ k‘ dt œ 1 Î2 1 Î2 1 Î2 œ csin td ! i 12 cos t‘ ! j 12 t 14 sin 2t‘ ! k œ i j 14 k Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.2 Integrals of Vector Functions; Projectile Motion 1/4 765 1/4 10. '0 casec tbi atan2 tbj at sin tb kd dt œ '0 casec tbi asec2 t 1bj at sin tb kd dt 1 œ clnasec t tan tbd 1!/4 i ctan t td 1!/4 j ct cos t sin td !1/4 k œ lnŠ1 È2‹i ˆ1 14 ‰j Š È È1 ‹k 4 2 2 11. r œ ' (ti tj tk) dt œ t# i t# j t# k C ; r(0) œ 0i 0j 0k C œ i 2j 3k Ê C œ i 2j 3k # # # # # # Ê r œ Š t# 1‹ i Š t# 2‹ j Š t# 3‹ k 16 $‰ # # $‘ 12. r œ ' c(180t)i a180t 16t# b jd dt œ 90t# i ˆ90t# 16 3 t j C ; r(0) œ 90(0) i 90(0) 3 (0) j C $ ‰ œ 100j Ê C œ 100j Ê r œ 90t# i ˆ90t# 16 3 t 100 j 13. r œ ' ˆ 3# (t 1)"Î# ‰ i e t j ˆ t " 1 ‰ k‘ dt œ (t 1)$Î# i e t j ln (t 1)k C ; r(0) œ (0 1)$Î# i e ! j ln (0 1)k C œ k Ê C œ i j k Ê r œ (t 1)$Î# 1‘ i a1 e t b j [1 ln (t 1)]k 14. r œ ' cat$ 4tb i tj 2t# kd dt œ Š t4 2t# ‹ i t2 j 2t3 k C ; r(0) œ ’ 04 2(0)# “ i 02 j 2(0) 3 kC % # % $ % # $ # $ œ i j Ê C œ i j Ê r œ Š t4 2t# 1‹ i Š t# 1‹ j 2t3 k 15. ddtr œ ' (32k) dt œ 32tk C" ; ddtr (0) œ 8i 8j Ê 32(0)k C" œ 8i 8j Ê C" œ 8i 8j Ê dr œ 8i 8j 32tk ; r œ ' (8i 8j 32tk) dt œ 8ti 8tj 16t# k C# ; r(0) œ 100k dt Ê 8(0)i 8(0)j 16(0)# k C# œ 100k Ê C# œ 100k Ê r œ 8ti 8tj a100 16t# b k 16. ddtr œ ' (i j k) dt œ (ti tj tk) C" ; ddtr (0) œ 0 Ê (0i 0j 0k) C" œ 0 Ê C" œ 0 # # # Ê ddtr œ (ti tj tk) ; r œ ' (ti tj tk) dt œ Š t# i t# j t# k‹ C# ; r(0) œ 10i 10j 10k # # # Ê Š 0# i 0# j 0# k‹ C# œ 10i 10j 10k Ê C# œ 10i 10j 10k # # # Ê r œ Š t# 10‹ i Š t# 10‹ j Š t# 10‹ k 17. ddtv œ a œ 3i j k Ê v(t) œ 3ti tj tk C" ; the particle travels in the direction of the vector (4 1)i (1 2)j (4 3)k œ 3i j k (since it travels in a straight line), and at time t œ 0 it has speed 2 Ê v(0) œ È9 21 1 (3i j k) œ C" Ê ddtr œ v(t) œ Š3t È6 ‹ i Št È2 ‹ j Št È2 ‹ k 11 Ê 11 11 r(t) œ Š 3# t# È611 t‹ i Š #" t# È211 t‹ j Š #" t# È211 t‹ k C# ; r(0) œ i 2j 3k œ C# Ê r(t) œ Š 3# t# È611 t 1‹ i Š "# t# È211 t 2‹ j Š #" t# È211 t 3‹ k œ Š "# t# È211 t‹ (3i j k) (i 2j 3k) 18. ddtv œ a œ 2i j k Ê v(t) œ 2ti tj tk C" ; the particle travels in the direction of the vector (3 1)i (0 (1))j (3 2)k œ 2i j k (since it travels in a straight line), and at time t œ 0 it has speed 2 Ê v(0) œ È4 21 1 (2i j k) œ C" Ê ddtr œ v(t) œ Š2t È46 ‹ i Št È26 ‹ j Št È26 ‹ k Ê r(t) œ Št# È46 t‹ i Š "# t# È26 t‹ j Š "# t# È26 t‹ k C# ; r(0) œ i j 2k œ C# Ê r(t) œ Št# È46 t 1‹ i Š "# t# È26 t 1‹ j Š "# t# È26 t 2‹ k œ Š "# t# È26 t‹ (2i j k) (i j 2k) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 766 Chapter 13 Vector-Valued Functions and Motion in Space m m‰ 19. x œ (v! cos !)t Ê (21 km)ˆ 1000 œ (840 m/s)(cos 60°)t Ê t œ (84021,000 1 km m/s)(cos 60°) œ 50 seconds v# v# ! 20. R œ g! sin 2! and maximum R occurs when ! œ 45° Ê 24.5 km œ Š 9.8 m/s # ‹ (sin 90°) Ê v! œ È(9.8)(24,500) m# /s# œ 490 m/s v# # m/s)(sin 45°) m/s) 21. (a) t œ 2v! gsin ! œ 2(5009.8 ¸ 72.2 seconds; R œ g! sin 2! œ (500 m/s# 9.8 m/s# (sin 90°) ¸ 25,510.2 m 5000 m (b) x œ (v! cos !)t Ê 5000 m œ (500 m/s)(cos 45°)t Ê t œ (500 m/s)(cos 45°) ¸ 14.14 s; thus, y œ (v! sin !)t "# gt# Ê y ¸ (500 m/s)(sin 45°)(14.14 s) "# a9.8 m/s# b (14.14 s)# ¸ 4020 m # # !) 45°)) (c) ymax œ (v! sin œ ((5002 m/s)(sin ¸ 6378 m 2g a9.8 m/s# b 22. y œ y! (v! sin !)t "# gt# Ê y œ 32 ft (32 ft/sec)(sin 30°)t "# a32 ft/sec# b t# Ê y œ 32 16t 16t# ; the ball hits the ground when y œ 0 Ê 0 œ 32 16t 16t# Ê t œ 1 or t œ 2 Ê t œ 2 sec since t 0; thus, È x œ (v! cos !) t Ê x œ (32 ft/sec)(cos 30°)t œ 32 Š #3 ‹ (2) ¸ 55.4 ft v# v# # # # ! 23. (a) R œ g! sin 2! Ê 10 m œ Š 9.8 m/s Ê v! ¸ 9.9 m/s; # ‹ (sin 90°) Ê v! œ 98 m s # m/s) (b) 6m ¸ (9.9 9.8 m/s# (sin 2!) Ê sin 2! ¸ 0.59999 Ê 2! ¸ 36.87° or 143.12° Ê ! ¸ 18.4° or 71.6° 24. v! œ 5 ‚ 10' m/s and x œ 40 cm œ 0.4 m; thus x œ (v! cos !)t Ê 0.4m œ a5 ‚ 10' m/sb (cos 0°)t Ê t œ 0.08 ‚ 10' s œ 8 ‚ 10) s; also, y œ y! (v! sin !)t "# gt# # Ê y œ a5 ‚ 10' m/sb (sin 0°) a8 ‚ 10) sb "# a9.8 m/s# b a8 ‚ 10) sb œ 3.136 ‚ 10"% m or 3.136 ‚ 10"# cm. Therefore, it drops 3.136 ‚ 10"# cm. v# # m/s) 25. R œ g! sin 2! Ê 16,000 m œ (400 9.8 m/s# sin 2! Ê sin 2! œ 0.98 Ê 2! ¸ 78.5° or 2! ¸ 101.5° Ê ! ¸ 39.3° or 50.7° # 26. (a) R œ (2vg! ) sin 2! œ 4v#! v#! g sin 2! œ 4 Š g sin !‹ or 4 times the original range. v# (b) Now, let the initial range be R œ g! sin 2!. Then we want the factor p so that pv! will double the range v# # Ê (pvg! ) sin 2! œ 2 Š g! sin 2!‹ Ê p# œ 2 Ê p œ È2 or about 141%. The same percentage will approximately 2 2 !b !b double the height: apv0 sin œ 2av0 sin Ê p# œ 2 Ê p œ È2. 2g 2g 27. The projectile reaches its maximum height when its vertical component of velocity is zero Ê dy dt œ v0 sin ! gt œ 0 2 2 2 2 av0 sin !b !b !b ! ! v0 sin ! " Ê t œ v0 sin Ê ymax œ av0 sin !bŠ v0 sin av0 sin œ av0 sin . To find the flight time g g ‹ # gŠ g ‹ œ g 2g 2g ! we find the time when the projectile lands: av0 sin !bt "# g t2 œ 0 Ê tˆv0 sin ! "# g t‰ œ 0 Ê t œ 0 or t œ 2v0 sin . g ! t œ 0 is the time when the projectile is fired, so t œ 2v0 sin is the time when the projectile strikes the ground. The range is g v2 v2 ! ! the value of the horizontal component when t œ 2v0 sin Ê R œ x œ av0 cos !bŠ 2v0 sin ‹ œ g0 a2 sin !cos !b œ g0 sin 2!. g g The range is largest when sin 2! œ 1 Ê ! œ 45‰ . R 28. When marble A is located R units downrange, we have x œ (v! cos !)t Ê R œ (v! cos !)t Ê t œ v! cos ! . At R R " that time the height of marble A is y œ y! (v! sin !)t "# gt# œ (v! sin !) Š v! cos ! ‹ # g Š v! cos ! ‹ # R R Ê y œ R tan ! "# g Š v# cos # ! ‹ . The height of marble B at the same time t œ v cos ! seconds is ! ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. # Section 13.2 Integrals of Vector Functions; Projectile Motion 767 # R h œ R tan ! "# gt# œ R tan ! "# g Š v# cos # ! ‹ . Since the heights are the same, the marbles collide regardless ! of the initial velocity v! . 29. ddtr œ ' (gj) dt œ gtj C" and ddtr (0) œ (v! cos !)i (v! sin !)j Ê g(0)j C" œ (v! cos !)i (v! sin !)j Ê C" œ (v! cos !)i (v! sin !)j Ê ddtr œ (v! cos !)i (v! sin ! gt)j ; r œ ' [(v! cos !)i (v! sin ! gt)j] dt œ (v! t cos !)i ˆv! t sin ! "# gt# ‰ j C# and r(0) œ x! i y! j Ê [v! (0) cos !]i v! (0) sin ! "# g(0)# ‘ j C# œ x! i y! j Ê C# œ x! i y! j Ê r œ (x! v! t cos !)i ˆy! v! t sin ! "# gt# ‰ j Ê x œ x! v! t cos ! and y œ y! v! t sin ! "# gt# v# # !) 30. The maximum height is y œ (v! sin and this occurs for x œ #g! sin 2! œ #g v#! sin ! cos ! . g These equations describe parametrically the points on a curve in the xy-plane associated with the maximum heights on the parabolic trajectories in terms of the parameter (launch angle) !. Eliminating the parameter !, we have x# œ œ v%! sin# ! v% sin% ! v# ! g# œ g! (2y) (2y)# g# v# # v% Ê x# 4 Šy 4g! ‹ œ 4g!# , where x 2v# ˆv% sin# !‰ a1 sin# !b v%! sin# ! cos# ! œ ! g# g# v# v% v% Ê x# 4y# Š g ! ‹ y œ 0 Ê x# 4 ’y# Š 2g! ‹ y 16g! # “ œ 4g!# 0. 31. (a) At the time t when the projectile hits the line OR we have tan " œ yx ; x œ [v! cos (! " )]t and y œ [v! sin (! " )]t "# gt# 0 since R is below level ground. Therefore let kyk œ "# gt# [v! sin (! " )]t 0 "# gt# (v! sin (! " ))t‘ " gt v sin (! " )‘ œ # v! cos! (! ") [v! cos (! " )]t v! cos (! " ) tan " œ "# gt v! sin (! " ) t œ 2v! sin (! ") 2vg ! cos (! ") tan " , which is the time so that tan " œ Ê Ê when the projectile hits the downhill slope. Therefore, x œ [v! cos (! " )] ’ 2v! sin (! ") 2vg ! cos (! ") tan " “ œ 2v#! # g ccos (! " ) tan " sin (! " ) cos (! " )d . If x is 2v# maximized, then OR is maximized: ddx! œ g ! [ sin 2(! " ) tan " cos 2(! " )] œ 0 Ê sin 2(! " ) tan " cos 2(! " ) œ 0 Ê tan " œ cot 2(! " ) Ê 2(! " ) œ 90° " Ê ! " œ "# (90° " ) Ê ! œ "# (90° " ) œ "# of nAOR. (b) At the time t when the projectile hits OR we have tan " œ yx ; x œ [v! cos (! " )]t and y œ [v! sin (! " )]t "# gt# v sin (! " ) " gt‘ cv! sin (! " )d t "# gt# œ ! v! cos (! ")# [v! cos (! " )]t v! cos (! " ) tan " œ v! sin (! " ) "# gt t œ 2v! sin (! ") 2vg ! cos (! ") tan " , which is the time Ê tan " œ Ê Ê when the projectile hits the uphill slope. Therefore, x œ [v! cos (! " )] ’ 2v! sin (! ") 2vg ! cos (! ") tan " “ œ maximized, then OR is maximized: ddx! œ 2v#! g 2v#! # g csin (! " ) cos (! " ) cos (! " ) tan " d . If x is [cos 2(! " ) sin 2(! " ) tan " ] œ 0 Ê cos 2(! " ) sin 2(! " ) tan " œ 0 Ê cot 2(! " ) tan " œ ! Ê cot 2(! " ) œ tan " œ tan (" ) Ê 2(! " ) œ 90° (" ) œ 90° " Ê ! œ "# (90° " ) œ "# of nAOR. Therefore v! would bisect nAOR for maximum range uphill. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 768 Chapter 13 Vector-Valued Functions and Motion in Space 32. v! œ 116 ft/sec, ! œ 45°, and x œ (v! cos !)t Ê 369 œ (116 cos 45°)t Ê t ¸ 4.50 sec; also y œ (v! sin !)t "# gt# Ê y œ (116 sin 45°)(4.50) "# (32)(4.50)# ¸ 45.11 ft. It will take the ball 4.50 sec to travel 369 ft. At that time the ball will be 45.11 ft in the air and will hit the green past the pin. 33. (a) (Assuming that "x" is zero at the point of impact:) ratb œ axatbbi ayatbbj; where xatb œ a35 cos 27‰ bt and yatb œ 4 a35 sin 27‰ bt 16t2 . ‰ 2 2 !b ! 27 (b) ymax œ av0 sin 4 œ a35sin6427 b 4 ¸ 7.945 feet, which is reached at t œ v0 sin œ 35sin ¸ 0.497 seconds. 2g g 32 ‰ (c) For the time, solve y œ 4 a35 sin 27‰ bt 16t2 œ 0 for t, using the quadratic formula tœ 35 sin 27‰ Éa35 sin 27‰ b2 256 ¸ 1.201 sec. Then the range is about xa1.201b œ a35 cos 27‰ ba1.201b ¸ 37.453 feet. 32 ‰ 2 (d) For the time, solve y œ 4 a35 sin 27 bt 16t œ 7 for t, using the quadratic formula tœ 35 sin 27‰ Éa35 sin 27‰ b2 192 ¸ 0.254 and 0.740 seconds. At those times the ball is about 32 ‰ ‰ xa0.254b œ a35 cos 27 ba0.254b ¸ 7.921 feet and xa0.740b œ a35 cos 27 ba0.740b ¸ 23.077 feet the impact point, or about 37.453 7.921 ¸ 29.532 feet and 37.453 23.077 ¸ 14.376 feet from the landing spot. (e) Yes. It changes things because the ball won't clear the net (ymax ¸ 7.945). 34. x œ x! (v! cos !)t œ 0 (v! cos 40°)t ¸ 0.766 v! t and y œ y! (v! sin !)t "# gt# œ 6.5 (v! sin 40°)t 16t# ¸ 6.5 0.643 v! t 16t# ; now the shot went 73.833 ft Ê 73.833 œ 0.766 v! t Ê t ¸ 96.383 sec; the shot lands when y œ 0 v! # 148,635 Ê 0 œ 6.5 (0.643)(96.383) 16 Š 96.383 Ê v! ¸ É 148,635 v! ‹ Ê 0 ¸ 68.474 68.474 ¸ 46.6 ft/sec, the shot's initial v# ! speed 35. Flight time œ 1 sec and the measure of the angle of elevation is about 64° (using a protractor) so that t œ 2v! gsin ! # v# # sin 64°) 64° Ê 1 œ 2v! sin Ê v! ¸ 17.80 ft/sec. Then ymax œ (17.802(32) ¸ 4.00 ft and R œ g! sin 2! Ê R œ (17.80) sin 128° 32 32 ¸ 7.80 ft Ê the engine traveled about 7.80 ft in 1 sec Ê the engine velocity was about 7.80 ft/sec 36. (a) ratb œ axatbbi ayatbbj; where xatb œ a145 cos 23‰ 14bt and yatb œ 2.5 a145 sin 23‰ bt 16t2 . ‰ 2 2 !b 23 b ! 23 (b) ymax œ av0 sin 2.5 œ a145sin 2.5 ¸ 52.655 feet, which is reached at t œ v0 sin œ 145sin ¸ 1.771 seconds. 2g 64 g 32 ‰ (c) For the time, solve y œ 2.5 a145 sin 23‰ bt 16t2 œ 0 for t, using the quadratic formula tœ 145 sin 23‰ Éa145 sin 23‰ b2 160 ¸ 3.585 sec. Then the range at t ¸ 3.585 is about x œ a145 cos 23‰ 14ba3.585b 32 ¸ 428.311 feet. (d) For the time, solve y œ 2.5 a145 sin 23‰ bt 16t2 œ 20 for t, using the quadratic formula tœ 145 sin 23‰ Éa145 sin 23‰ b2 1120 ¸ 0.342 and 3.199 seconds. At those times the ball is about 32 ‰ xa0.342b œ a145 cos 23 14ba0.342b ¸ 40.860 feet from home plate and xa3.199b œ a145 cos 23‰ 14ba3.199b ¸ 382.195 feet from home plate. (e) Yes. According to part (d), the ball is still 20 feet above the ground when it is 382 feet from home plate. 37. ddt2r k ddtr œ gj Ê Patb œ k and Qatb œ gj Ê ' Patb dt œ kt Ê vatb œ e' Patb dt œ ekt Ê ddtr œ va1tb ' vatb Qatb dt 2 œ gekt ' ekt j dt œ gekt ek j C1 ‘ œ gk j Cekt , where C œ gC1 ; apply the initial condition: kt g dr ˆg ‰ dt ¹tœ0 œ av0 cos !bi av0 sin !bj œ k j C Ê C œ av0 cos !bi k v0 sin ! j Ê ddtr œ ˆv0 ekt cos !‰i ˆ gk ekt ˆ gk v0 sin !‰‰j, r œ ' c ˆv0 ekt cos !‰i ˆ gk ekt ˆ gk v0 sin !‰‰j ddt Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.2 Integrals of Vector Functions; Projectile Motion 769 ckt gt g v œ ˆ k0 ekt cos !‰i Š k e k ˆ k v0 sin !‰‹j C2 ; apply the initial condition: !‰ !‰ j C2 Ê C2 œ ˆ vk0 cos !‰i ˆ kg2 v0 sin j ra0b œ 0 œ ˆ vk0 cos !‰i ˆ kg2 v0 sin k k Ê ratb œ ˆ vk0 ˆ1 ekt ‰cos !‰i ˆ vk0 ˆ1 ekt ‰sin ! kg2 ˆ1 kt ekt ‰‰j 152 ‰ 38. (a) ratb œ axatbbi ayatbbj; where xatb œ ˆ 0.12 a1 e0.12t bacos 20‰ b and 152 ‰ 32 ‰ 0.12t yatb œ 3 ˆ 0.12 a1 e0.12t basin 20‰ b ˆ 0.12 b 2 a1 0.12t e (b) Solve graphically using a calculator or CAS: At t ¸ 1.484 seconds the ball reaches a maximum height of about 40.435 feet. (c) Use a graphing calculator or CAS to find that y œ 0 when the ball has traveled for ¸ 3.126 seconds. The range is 152 ‰ˆ about xa3.126b œ ˆ 0.12 1 e0.12a3.126b ‰acos 20‰ b ¸ 372.311 feet. (d) Use a graphing calculator or CAS to find that y œ 30 for t ¸ 0.689 and 2.305 seconds, at which times the ball is about xa0.689b ¸ 94.454 feet and xa2.305b ¸ 287.621 feet from home plate. (e) Yes, the batter has hit a home run since a graph of the trajectory shows that the ball is more than 14 feet above the ground when it passes over the fence. 39. (a) 'a kr(t) dt œ 'a [kf(t)i kg(t)j kh(t)k] dt œ 'a [kf(t)] dt i 'a [kg(t)] dt j 'a [kh(t)] dt k b b b b b œ k Œ'a f(t) dt i 'a g(t) dt j 'a h(t) dt k œ k 'a r(t) dt b b b b (b) 'a [r" (t) „ r# (t)] dt œ 'a acf" (t)i g" (t)j h" (t)kd „ cf# (t)i g# (t)j h# (t)kdb dt b b œ 'a acf" (t) „ f# (t)d i [g" (t) „ g# (tb] j [h" (t) „ h# (t)] k) dt b œ 'a cf" (t) „ f# (t)d dt i 'a cg" (t) „ g# (t)d dt j 'a ch" (t) „ h# (t)d dt k b b b œ ”'a f" (t) dt i „ 'a f# (t) dt i• ”'a g" (t) dt j „ 'a g# (t) dt j • ”'a h" (t) dt k „ 'a h# (t) dt k• b b b b b b œ 'a r" (t) dt „ 'a r# (t) dt b b (c) Let C œ c" i c# j c$ k. Then 'a C † r(t) dt œ 'a cc" f(t) c# g(t) c$ h(t)d dt b b œ c" 'a f(t) dt c# 'a g(t) dt c$ 'a h(t) dt = C † 'a r(t) dt; b b b b 'ab C ‚ r(t) dt œ 'ab cc# h(t) c$ g(t)d i cc$ f(t) c" h(t)d j cc" g(t) c# f(t)d k dt œ ”c# 'a h(t) dt c$ 'a g(t) dt• i ”c$ 'a f(t) dt c" 'a h(t) dt• j ”c" 'a g(t) dt c# 'a f(t) dt• k b b b b b b œ C ‚ 'a r(t) dt b 40. (a) Let u and r be continuous on [aß b]. Then lim u(t)r(t) œ lim [u(t)f(t)i u(t)g(t)j u(t)h(t)k] t Ä t! t Ä t! œ u(t! )f(t! )i u(t! )g(t! )j u(t! )h(t! )k œ u(t! )r(t! ) Ê ur is continuous for every t! in [aß b]. (b) Let u and r be differentiable. Then dtd (ur) œ dtd [u(t)f(t)i u(t)g(t)j u(t)h(t)k] dg df ‰ du dh ‰ ˆ du œ ˆ du dt f(t) u(t) dt i Š dt g(t) u(t) dt ‹j dt h(t) u(t) dt k dg df dh du dr œ [f(t)i g(t)j h(t)k] du dt u(t) Š dt i dt j dt k‹ œ r dt u dt 41. (a) If R" (t) and R# (t) have identical derivatives on I, then ddtR" œ dfdt" i dgdt" j dhdt" k œ dfdt# i dgdt# j dhdt# k œ ddtR# Ê dfdt" œ dfdt# , dgdt" œ dgdt# , dhdt" œ dhdt# Ê f" (t) œ f# (t) c" , g" (t) œ g# (t) c# , h" (t) œ h# (t) c$ Ê f" (t)i g" (t)j h" (t)k œ [f# (t) c" ]i [g# (t) c# ]j [h# (t) c$ ]k Ê R" (t) œ R# (t) C, where C œ c" i c# j c$ k. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 770 Chapter 13 Vector-Valued Functions and Motion in Space (b) Let R(t) be an antiderivative of r(t) on I. Then Rw (t) œ r(t). If U(t) is an antiderivative of r(t) on I, then Uw (t) œ r(t). Thus Uw (t) œ Rw (t) on I Ê U(t) œ R(t) C. 42. dtd 'a r(7 ) d7 œ dtd 'a [f(7 )i g(7 )j h(7 )k] d7 œ dtd 'a f(7 ) d7 i dtd 'a g(7 ) d7 j dtd 'a h(7 ) d7 k t t t t t œ f(t)i g(t)j h(t)k œ r(t). Since dtd 'a r(7 ) d7 œ r(t), we have that 'a r(7 ) d7 is an antiderivative of t t r. If R is any antiderivative of r, then R(t) œ 'a r(7 ) d7 C by Exercise 41(b). Then R(a) œ 'a r(7 ) d7 C t a œ 0 C Ê C œ R(a) Ê 'a r(7 ) d7 œ R(t) C œ R(t) R(a) Ê 'a r(7 ) d7 œ R(b) R(a). t b 1 ‰ 43. (a) ratb œ axatbbi ayatbbj; where xatb œ ˆ 0.08 a1 e0.08t ba152 cos 20‰ 17.6b and 152 ‰ 32 ‰ 0.08t yatb œ 3 ˆ 0.08 a1 e0.08t basin 20‰ b ˆ 0.08 b 2 a1 0.08t e (b) Solve graphically using a calculator or CAS: At t ¸ 1.527 seconds the ball reaches a maximum height of about 41.893 feet. (c) Use a graphing calculator or CAS to find that y œ 0 when the ball has traveled for ¸ 3.181 seconds. The range is 1 ‰ˆ about xa3.181b œ ˆ 0.08 1 e0.08a3.181b ‰a152 cos 20‰ 17.6b ¸ 351.734 feet. (d) Use a graphing calculator or CAS to find that y œ 35 for t ¸ 0.877 and 2.190 seconds, at which times the ball is about xa0.877b ¸ 106.028 feet and xa2.190b ¸ 251.530 feet from home plate. (e) No; the range is less than 380 feet. To find the wind needed for a home run, first use the method of part (d) to find that 1 ‰ˆ y œ 20 at t ¸ 0.376 and 2.716 seconds. Then define xawb œ ˆ 0.08 1 e0.08a2.716b ‰a152 cos 20‰ wb, and solve xawb œ 380 to find w ¸ 12.846 ft/sec. # # # sin !) sin !) !) 44. ymax œ (v! sin Ê 34 ymax œ 3(v! 8g and y œ (v! sin !)t "# gt# Ê 3(v! 8g œ (v! sin !)t "# gt# 2g Ê 3(v! sin !)# œ (8gv! sin !)t 4g# t# Ê 4g# t# (8gv! sin !)t 3(v! sin !)# œ 0 Ê 2gt 3v! sin ! œ 0 or sin ! ! 2gt v! sin ! œ 0 Ê t œ 3v!2gsin ! or t œ v! 2g . Since the time it takes to reach ymax is tmax œ v! sin , g sin ! then the time it takes the projectile to reach 34 of ymax is the shorter time t œ v! 2g or half the time it takes to reach the maximum height. 13.3 ARC LENGTH IN SPACE 1. r œ (2 cos t)i (2 sin t)j È5tk Ê v œ (2 sin t)i (2 cos t)j È5k # Ê kvk œ Ê(2 sin t)# (2 cos t)# ŠÈ5‹ œ È4 sin# t 4 cos# t 5 œ 3; T œ kvvk 1 1 œ ˆ 23 sin t‰ i ˆ 23 cos t‰ j 35 k and Length œ '0 kvk dt œ '0 3 dt œ c3td 1! œ 31 È 2. r œ (6 sin 2t)i (6 cos 2t)j 5tk Ê v œ (12 cos 2t)i (12 sin 2t)j 5k Ê kvk œ È(12 cos 2t)# (12 sin 2t)# 5# œ È144 cos# 2t 144 sin# 2t 25 œ 13; T œ v 1 5 ' ' ‰ ˆ 12 ‰ œ ˆ 12 13 cos 2t i 13 sin 2t j 13 k and Length œ 0 kvk dt œ 1 kv k 13 dt œ c13td 1! œ 131 0 # È 3. r œ ti 23 t$Î# k Ê v œ i t"Î# k Ê kvk œ É1# at"Î# b œ È1 t ; T œ kvvk œ È1" t i È1 t t k ) and Length œ '0 È1 t dt œ 23 (1 t)$Î# ‘ ! œ 52 3 8 4. r œ (2 t)i (t 1)j tk Ê v œ i j k Ê kvk œ È1# (1)# 1# œ È3 ; T œ kvvk œ È"3 i È13 j È"3 k $ and Length œ '0 È3 dt œ ’È3t“ œ 3È3 3 ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.3 Arc Length in Space 771 5. r œ acos$ tb j asin$ tb k Ê v œ a3 cos# t sin tb j a3 sin# t cos tb k Ê kvk œ Éa3 cos# t sin tb# a3 sin# t cos tb# œ Èa9 cos# t sin# tb acos# t sin# tb œ 3 kcos t sin tk ; # # cos t sin t 3 sin t cos t 1 T œ kvvk œ 33kcos t sin tk j 3 kcos t sin tk k œ ( cos t)j (sin t)k , if 0 Ÿ t Ÿ # , and 1Î2 Length œ '0 1Î2 1Î2 3 kcos t sin tk dt œ '0 3 cos t sin t dt œ '0 3 3 ‘ 1Î# œ 3# # sin 2t dt œ 4 cos 2t ! 6. r œ 6t$ i 2t$ j 3t$ k Ê v œ 18t# i 6t# j 9t# k Ê kvk œ Éa18t# b# a6t# b# a9t# b# œ È441t% œ 21t# ; "8t 6t 9t 6 2 3 '1 21t# dt œ c7t$ d " œ 49 T œ kvvk œ 21t # i 21t# j 21t# k œ 7 i 7 j 7 k and Length œ # # 2 # # È 7. r œ (t cos t)i (t sin t)j 2 3 2 t$Î# k Ê v œ (cos t t sin t)i (sin t t cos t)j ŠÈ2 t"Î# ‹ k # Ê kvk œ Ê(cos t t sin t)# (sin t t cos t)# ŠÈ2 t‹ œ È1 t# 2t œ È(t 1)# œ kt 1k œ t 1, if t 1 1 0; T œ kvvk œ ˆ cos tt t1sin t ‰ i ˆ sin ttt1cos t ‰ j Š t 2t1 ‹ k and Length œ '0 (t 1) dt œ ’ t2 t“ œ 12 1 È "Î# # # ! 8. r œ (t sin t cos t)i (t cos t sin t)j Ê v œ (sin t t cos t sin t)i (cos t t sin t cos t)j œ (t cos t)i (t sin t)j Ê kvk œ È(t cos t)# (t sin t)# œ Èt# œ ktk œ t if È2 Ÿ t Ÿ 2; T œ v kvk t‰ t‰ œ ˆ t cos i ˆ t sin j œ (cos t)i (sin t)j and Length œ 'È2 t dt œ ’ t2 “ t t 2 # # È# œ1 9. Let P(t! ) denote the point. Then v œ (5 cos t)i (5 sin t)j 12k and 261 œ '0 È25 cos# t 25 sin# t 144 dt t! œ '0 13 dt œ 13t! Ê t! œ 21, and the point is P(21) œ (5 sin 21ß 5 cos 21ß 241) œ (0ß 5ß 241) t! 10. Let P(t! ) denote the point. Then v œ (12 cos t)i (12 sin t)j 5k and 131 œ '0 È144 cos# t 144 sin# t 25 dt œ '0 13 dt œ 13t! Ê t! œ 1, and the point is t! t! P(1) œ (12 sin (1)ß 12 cos (1)ß 51) œ (0ß 12ß 51) 11. r œ (4 cos t)i (4 sin t)j 3tk Ê v œ (4 sin t)i (4 cos t)j 3k Ê kvk œ È(4 sin t)# (4 cos t)# 3# œ È25 œ 5 Ê s(t) œ '0 5 d7 œ 5t Ê Length œ s ˆ 1# ‰ œ 5#1 t 12. r œ (cos t t sin t)i (sin t t cos t)j Ê v œ (sin t sin t t cos t)i (cos t cos t t sin t)j t # œ (t cos t)i (t sin t)j Ê kvk œ È(t cos t)# (t cos t)# œ œ Èt# œ t, since 1 Ÿ t Ÿ 1 Ê s(t) œ ' 7 d7 œ t # Ê # 0 ˆ 1 ‰# # Length œ s(1) s ˆ 1# ‰ œ 1# ## œ 381 # 13. r œ aet cos tb i aet sin tb j et k Ê v œ aet cos t et sin tb i aet sin t et cos tb j et k Ê kvk œ Éaet cos t et sin tb# aet sin t et cos tb# aet b# œ œ È3e2t œ È3 et Ê s(t) œ '0 È3 e7 d7 t È œ È3 et È3 Ê Length œ s(0) s( ln 4) œ 0 ŠÈ3 e ln 4 È3‹ œ 3 4 3 14. r œ (1 2t)i (1 3t)j (6 6t)k Ê v œ 2i 3j 6k Ê kvk œ È2# 3# (6)# œ 7 Ê s(t) œ '0 7 d7 œ 7t t Ê Length œ s(0) s(1) œ 0 (7) œ 7 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 772 Chapter 13 Vector-Valued Functions and Motion in Space # # 15. r œ ŠÈ2t‹ i ŠÈ2t‹ j a1 t# b k Ê v œ È2i È2j 2tk Ê kvk œ ÊŠÈ2‹ ŠÈ2‹ (2t)# œ È4 4t# " œ 2È1 t# Ê Length œ '0 2È1 t# dt œ ’2 Š 2t È1 t# "# ln Št È1 t# ‹‹“ œ È2 ln Š1 È2‹ 1 ! 16. Let the helix make one complete turn from t œ 0 to t œ 21. Note that the radius of the cylinder is 1 Ê the circumference of the base is 21. When t œ 21, the point P is (cos 21ß sin 21ß 21) œ (1ß 0ß 21) Ê the cylinder is 21 units high. Cut the cylinder along PQ and flatten. The resulting rectangle has a width equal to the circumference of the cylinder œ 21 and a height equal to 21, the height of the cylinder. Therefore, the rectangle is a square and the portion of the helix from t œ 0 to t œ 21 is its diagonal. 17. (a) r œ (cos t)i (sin t)j (" cos t)k, 0 Ÿ t Ÿ 21 Ê x œ cos t, y œ sin t, z œ 1 cos t Ê x# y# œ cos# t sin# t œ 1, a right circular cylinder with the z-axis as the axis and radius œ 1. Therefore P(cos tß sin tß 1 cos t) lies on the cylinder x# y# œ 1; t œ 0 Ê P(1ß 0ß 0) is on the curve; t œ 1# Ê Q(!ß 1ß 1) Ä Ä is on the curve; t œ 1 Ê R(1ß 0ß 2) is on the curve. Then PQ œ i j k and PR œ 2i 2k j k× Ô i Ä Ä Ê PQ ‚ PR œ 1 " " œ 2i 2k is a vector normal to the plane of P, Q, and R. Then the Õ 2 0 2 Ø plane containing P, Q, and R has an equation 2x 2z œ 2(1) 2(0) or x z œ 1. Any point on the curve will satisfy this equation since x z œ cos t (1 cos t) œ 1. Therefore, any point on the curve lies on the intersection of the cylinder x# y# œ 1 and the plane x z œ 1 Ê the curve is an ellipse. (b) v œ ( sin t)i (cos t)j (sin t)k Ê kvk œ Èsin# t cos# t sin# t œ È1 sin# t Ê T œ v kvk (cos t)j (sin t)k k œ ( sin t)iÈ Ê T(0) œ j , T ˆ 1# ‰ œ Èi2k , T(1) œ j , T ˆ 3#1 ‰ œ iÈ 1 sin# t 2 (c) a œ ( cos t)i (sin t)j (cos t)k ; n œ i k is normal to the plane x z œ 1 Ê n † a œ cos t cos t œ 0 Ê a is orthogonal to n Ê a is parallel to the plane; a(0) œ i k , a ˆ 1# ‰ œ j , a a1b œ i k , ‰œj a ˆ 31 # 21 (d) kvk œ È1 sin# t (See part (b) Ê L œ '0 È1 sin# t dt (e) L ¸ 7.64 (by Mathematica) 18. (a) r œ (cos 4t)i (sin 4t)j 4tk Ê v œ (4 sin 4t)i (4 cos 4t)j 4k Ê kvk œ È(4 sin 4t)# (4 cos 4t)# 4# 1Î2 œ È32 œ 4È2 Ê Length œ '0 4È2 dt œ ’4È2 t“ 1Î# ! œ 21È2 (b) r œ ˆcos #t ‰ i ˆsin #t ‰ j #t k Ê v œ ˆ #" sin #t ‰ i ˆ #" cos #t ‰ j #" k # # # Ê kvk œ Éˆ "# sin #t ‰ ˆ #" cos #t ‰ ˆ #" ‰ œ É 4" 4" œ È2 # 41 È Ê Length œ '0 # 2 È dt œ ’ 22 t“ %1 ! œ 21 È 2 (c) r œ (cos t)i (sin t)j tk Ê v œ ( sin t)i (cos t)j k Ê kvk œ È( sin t)# ( cos t)# (1)# œ È1 1 œ È2 Ê Length œ 'c21 È2 dt œ ’È2 t“ 0 ! #1 œ 21 È 2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.4 Curvature and Normal Vectors of a Curve 773 19. nPQB œ nQOB œ t and PQ œ arc (AQ) œ t since PQ œ length of the unwound string œ length of arc (AQ); thus x œ OB BC œ OB DP œ cos t t sin t, and y œ PC œ QB QD œ sin t t cos t 20. r œ acos t t sin tbi asin t t cos tbj Ê v œ asin t t cos t sin tbi acos t atasin tb cos tbbj œ at cos tbi at sin tbj Ê kvk œ Éat cos tb2 at sin tb2 œ Èt2 œ ktk œ t, t t t sin t 0 Ê T œ kvvk œ t cos t i t j œ cos t i sin t j 21. v œ dtd ax0 t u1 bi dtd ay0 t u2 bj dtd az0 t u3 bk œ u1 i u2 j u3 k œ u, so satb œ '0 lvldt œ '0 luld7 œ '0 1 d7 œ t t t t 22. ratb œ t i t2 j t3 k Ê vatb œ i 2t j 3t2 k Ê lvatbl œ Éa1b2 a2tb2 a3t2 b2 œ È1 4t2 9t4 . a0, 0, 0b Ê t œ 0 and a2, 4, 8b Ê t œ 2. Thus L œ '0 lvatbl dt œ '0 È1 4t2 9t4 dt. Using Simpson's rule with n œ 10 and 2 2 0 ?x œ 2 10 œ 0.2 Ê L ¸ 0.2 3 Šlva0bl 4lva0.2bl 2lva0.4bl 4lva0.6bl 2lva0.8bl 4lva1bl 2lva1.2bl 4lva1.4bl 2lva1.6bl 4lva1.8bl lva2bl‹ ¸ 0.2 3 Š1 4a1.0837b 2a1.3676b 4a1.8991b 2a2.6919b 4a3.7417b 2a5.0421b 4a6.5890b 2a8.3800b 4a10.4134b 12.6886‹ œ 0.2 3 a143.5594b ¸ 9.5706 13.4 CURVATURE AND NORMAL VECTORS OF A CURVE sin t ‰ È1# ( tan t)# œ Èsec# t œ ksec tk œ sec t, since 1. r œ ti ln (cos t)j Ê v œ i ˆ cos t j œ i (tan t)j Ê kvk œ tan t ‰ dT 1# t 1# Ê T œ kvvk œ ˆ sec" t ‰ i ˆ sec t j œ (cos t)i (sin t)j ; dt œ ( sin t)i (cos t)j ˆ dT ‰ Ê ¸ ddtT ¸ œ È( sin t)# ( cos t)# œ 1 Ê N œ ¸ ddtT ¸ œ ( sin t)i (cos t)j ; , œ k1vk † ¸ ddtT ¸ œ sec" t † 1 œ cos t. dt t tan t ‰ 2. r œ ln (sec t)i tj Ê v œ ˆ secsec i j œ (tan t)i j Ê kvk œ È( tan t)# 1# œ Èsec# t œ ksec tk œ sec t, t 1 1 v tan since # t # Ê T œ kvk œ ˆ sec tt ‰ i ˆ sec1 t ‰ j œ (sin t)i (cos t)j ; ddtT œ (cos t)i (sin t)j ˆ dT ‰ Ê ¸ ddtT ¸ œ È(cos t)# ( sin t)# œ 1 Ê N œ ¸ ddtT ¸ œ (cos t)i (sin t)j ; , œ k1vk † ¸ ddtT ¸ œ sec" t † 1 œ cos t. dt 3. r œ (2t 3)i a5 t# b j Ê v œ 2i 2tj Ê kvk œ È2# (2t)# œ 2È1 t# Ê T œ kvvk œ È 2 # i È2t # j 2 1t 2 1t Í # # Í Í t " " ¸ ddtT ¸ œ t $ œ È " # i È t # j ; ddtT œ $i $ j Ê $ 1 t 1 t # # È È Š 1t ‹ Š 1t ‹ ŠÈ1 t# ‹ Ì ŠÈ1 t# ‹ ˆ dT ‰ œ É a1 "t# b# œ 1" t# Ê N œ ¸ ddtT ¸ œ È t # i È " dt 1t 1 t# j ; , œ k1vk † ¸ ddtT ¸ œ " † " # œ # a1 " t# b3/2 #È1 t# 1 t 4. r œ (cos t t sin t)i (sin t t cos t)j Ê v œ (t cos t)i (t sin t)j Ê kvk œ È( t cos t)# (t sin t)# œ Èt# œ ktk œ t, since t 0 Ê T œ v œ (t cos t)i(t sin t)j œ (cos t)i (sin t)j ; dT œ ( sin t)i (cos t)j Ê ¸ dT ¸ œ È( sin t)# (cos t)# kvk t dt ˆ ddtT ‰ 1 ¸ dT ¸ œ 1 Ê N œ ¸ dT ¸ œ ( sin t)i (cos t)j ; , œ kvk † dt œ "t † 1 œ "t dt dt Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 774 Chapter 13 Vector-Valued Functions and Motion in Space 5. (a) ,axb œ kva1xbk † ¹ dTdtaxb ¹. Now, v œ i f w axbj Ê kvaxbk œ É1 c f w axb d2 Ê T œ kvvk 1Î2 œ Š1 c f w axb d2 ‹ Í Í 1Î2 i f w axbŠ1 c f w axb d2 ‹ 2 f axbf axb Í xb Ê ¹ dTa ” Š1 c f axb d2 ‹3 2 • dt ¹ œ w 2 ww f ax b ww Î w Ì Thus ,axb œ j. Thus ddtT axb œ 3Î2 2 Š1 c f axb d ‹ w œË f axbf axb w ww i 3Î2 2 Š1 c f axb d ‹ w c f axb d2 Š1 c f axb d2 ‹ ww w 3 2 Š1 c f axb d ‹ w œ f ax b ww 3Î2 2 Š1 c f axb d ‹ j w kf axbk ww 2 ¹1 c f axb d ¹ w kf axbk 1 † k f a x bk 2 œ 3 2 2 a1 Òf axbÓ2 b1 2 k1 Òf axbÓ k Š1 c f axb d ‹ ww w ww w Î Î w # d y # ˆ " ‰ (b) y œ ln (cos x) Ê dy dx œ cos x ( sin x) œ tan x Ê dx# œ sec x Ê , œ # k sec# xk x œ ksec sec$ xk c1 ( tan x)# d$Î# œ sec" x œ cos x, since 1# x 1# (c) Note that f ww (x) œ 0 at an inflection point. Þ Þ Þ Þ Þ Þ 6. (a) r œ f(t)i g(t)j œ xi yj Ê v œ xi yj Ê kvk œ Èx# y# Ê T œ kvvk œ È Þ #x y Þ i ÈÞ# Þ# j x y# x y Þ Þ ÞÞ Þ ÞÞ Þ Þ ÞÞ Þ ÞÞ Þ Þ ÞÞ Þ ÞÞ Þ Þ ÞÞ Þ ÞÞ Þ Þ Þ ÞÞ Þ ÞÞ yay x x yb 2 xax y y xb 2 yay x x yb xax y y x b ay# x# bay x x yb2 dT dT ¸ ¸ œ i j Ê œ œ Ê’ “ ’ “ 3/2 Þ Þ 3/2 Þ Þ Ê 3/2 3/2 Þ Þ 3 Þ Þ Þ Þ # # # # dt dt ax y b ax y b ax# y# b ax# y# b ax# y# b Þ ÞÞ Þ ÞÞ k y x x yk Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ k y x x yk lxyyx l Þ # † kxÞ # yÞ # k œ Þ # Þ # 3/2 . x y ax y b œ kxÞ # yÞ # k ; , œ kv1k † ¸ ddtT ¸ œ È Þ #1 Þ ÞÞ Þ ÞÞ t # (b) r(t) œ ti ln (sin t)j , 0 t 1 Ê x œ t and y œ ln (sin t) Ê x œ 1, x œ 0; y œ cos sin t œ cot t, y œ csc t # t Ê , œ k csc #t 0$Î#k œ csc csc$ t œ sin t # a1 cot t)b " Þ t " (sinh t)i ln (cosh t)j Ê x œ tan" (sinh t) and y œ ln (cosh t) Ê x œ 1 cosh sinh# t œ cosh t $ # ÞÞ Þ ÞÞ ksech t sech t tanh tk sinh t # œ sech t, x œ sech t tanh t; y œ cosh œ ksech tk œ sech t asech# t tanh# tb t œ tanh t, y œ sech t Ê , œ (c) r(t) œ tan 7. (a) r(t) œ f(t)i g(t)j Ê v œ f w (t)i gw (t)j is tangent to the curve at the point (f(t)ß g(t)); n † v œ c gw (t)i f w (t)jd † cf w (t)i gw (t)jd œ gw (t)f w (t) f w (t)gw (t) œ 0; n † v œ (n † v) œ 0; thus, n and n are both normal to the curve at the point (b) r(t) œ ti e2t j Ê v œ i 2e2t j Ê n œ 2e2t i j points toward the concave side of the curve; N œ knnk and knk œ È4e4t 1 Ê N œ È2e 4t i È " 2t 1 4e 1 4e4t j (c) r(t) œ È4 t# i tj Ê v œ È t # i j Ê n œ i È t 4 t# 4t # j points toward the concave side of the curve; Ê N œ "# ŠÈ4 t# i tj‹ N œ knnk and knk œ É1 4 t t# œ È 2 4 t# 8. (a) r(t) œ ti "3 t$ j Ê v œ i t# j Ê n œ t# i j points toward the concave side of the curve when t 0 and n œ t# i j points toward the concave side when t 0 Ê N œ È " 1 t% Nœ È" 1 t% at# i jb for t 0 (b) From part (a), kvk œ È1 t% Ê T œ È " 1 t% œ 12ktkt% ; at# i jb for t 0 and # i Èt 1 t% j Ê ddtT œ 2t$ 2t $Î# i $Î# j Ê a1 t% b a1 t% b ¸ ddtT ¸ œ É 4t6 %4t$2 a1 t b ˆ dT ‰ % $ t$ 2t N œ ¸ ddtT ¸ œ 12ktkt Š 2t% $Î# i i È t % j; t Á 0. N does not exist at t œ 0, where the $Î# j‹ œ ktkÈ1 t% ktk 1 t a1 t b a1 t% b dt dt ¸ curve has a point of inflection; ddtT ¸ tœ0 œ 0 so the curvature , œ ¸ ddsT ¸ œ ¸ ddtT † ds œ 0 at t œ 0 Ê N œ ," ddsT is undefined. Since x œ t and y œ "3 t$ Ê y œ 3" x$ , the curve is the cubic power curve which is concave down for x œ t 0 and concave up for x œ t 0. 9. r œ (3 sin t)i (3 cos t)j 4tk Ê v œ (3 cos t)i (3 sin t)j 4k Ê kvk œ È(3 cos t)# (3 sin t)# 4# œ È25 œ 5 Ê T œ kvvk œ ˆ 35 cos t‰ i ˆ 35 sin t‰ j 45 k Ê ddtT œ ˆ 53 sin t‰ i ˆ 53 cos t‰ j Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.4 Curvature and Normal Vectors of a Curve # # 775 ˆ dT ‰ 3 Ê ¸ ddtT ¸ œ Éˆ 35 sin t‰ ˆ 35 cos t‰ œ 35 Ê N œ ¸ ddtT ¸ œ ( sin t)i (cos t)j ; , œ 15 † 35 œ 25 dt 10. r œ (cos t t sin t)i (sin t t cos t)j 3k Ê v œ (t cos t)i (t sin t)j Ê kvk œ È(t cos t)# (t sin t)# œ Èt# œ ktk œ t, if t 0 Ê T œ kvvk œ (cos t)i (sin t)j , t 0 Ê ddtT œ ( sin t)i (cos t)j ˆ dT ‰ Ê ¸ ddtT ¸ œ È( sin t)# (cos t)# œ 1 Ê N œ ¸ ddtT ¸ œ ( sin t)i (cos t)j ; , œ "t † 1 œ "t dt 11. r œ aet cos tb i aet sin tb j 2k Ê v œ aet cos t et sin tb i aet sin t et cos tb j Ê kvk œ Éaet cos t et sin tb# aet sin t et cos tb# œ È2e2t œ et È2 ; cos t t sin t t sin t T œ kvvk œ Š cos È ‹ i Š sin tÈ ‹ j Ê ddtT œ Š sinÈt 2 cos t ‹ i Š cos È ‹j 2 2 2 # # ˆ dT ‰ t sin t Ê ¸ ddtT ¸ œ ÊŠ sinÈt cos t ‹ Š cos È ‹ œ 1 Ê N œ ¸ ddtT ¸ œ Š cosÈt sin t ‹ i Š sinÈt cos t ‹ j ; 2 2 2 dt 2 1 1 , œ kv1k † ¸ ddtT ¸ œ et È † 1 œ et È 2 2 12. r œ (6 sin 2t)i (6 cos 2t)j 5tk Ê v œ (12 cos 2t)i (12 sin 2t)j 5k Ê kvk œ È(12 cos 2t)# (12 sin 2t)# 5# 5 dT ‰ ˆ 12 ‰ ˆ 24 ‰ ˆ 24 ‰ œ È169 œ 13 Ê T œ kvvk œ ˆ 12 13 cos 2t i 13 sin 2t j 13 k Ê dt œ 13 sin 2t i 13 cos 2t j # # ˆ dT ‰ 24 ‰ ˆ 24 ‰ œ 13 Ê ¸ ddtT ¸ œ Éˆ 24 Ê N œ ¸ ddtT ¸ œ ( sin 2t)i (cos 2t)j ; 13 sin 2t 13 cos 2t dt 1 24 , œ kv1k † ¸ ddtT ¸ œ 13 † 24 13 œ 169 . $ # 13. r œ Š t3 ‹ i Š t# ‹ j , t 0 Ê v œ t# i tj Ê kvk œ Èt% t# œ tÈt# 1, since t 0 Ê T œ kvvk œ Èt#t t i Èt#1 1 j Ê ddtT œ 1 i # t $Î# j at# 1b$Î# at 1 b ˆ dT ‰ # 1 t " dt œ É at1# œ t# 1 Ê N œ ¸ dT ¸ œ È # 1 b$ t 1 dt i È #t t 1 # Ê ¸ ddtT ¸ œ ÊŠ # " $Î# ‹ Š # t $Î# ‹ at 1 b j ; , œ kv1k † ¸ ddtT ¸ œ È 1# t # at 1 b t 1 1 † t# 1 œ " . t at# 1b$Î# 14. r œ acos$ tb i asin$ tb j , 0 t 1# Ê v œ a3 cos# t sin tb i a3 sin# t cos tb j Ê kvk œ Éa3 cos# t sin tb# a3 sin# t cos tb# œ È9 cos% t sin# t 9 sin% t cos# t œ 3 cos t sin t, since 0 t 1# ˆ dT ‰ Ê T œ kvvk œ ( cos t)i (sin t)j Ê ddtT œ (sin t)i (cos t)j Ê ¸ ddtT ¸ œ Èsin# t cos# t œ 1 Ê N œ ¸ ddtT ¸ dt œ (sin t)i (cos t)j; , œ kv1k † ¸ ddtT ¸ œ 3 cos1t sin t † 1 œ 3 cos1t sin t . 15. r œ ti â cosh at ‰ j , a 0 Ê v œ i ˆsinh at ‰ j Ê kvk œ É1 sinh# ˆ at ‰ œ Écosh# ˆ at ‰ œ cosh at Ê T œ kvvk œ ˆsech at ‰ i ˆtanh at ‰ j Ê ddtT œ ˆ "a sech at tanh at ‰ i ˆ "a sech# at ‰ j ˆ dT ‰ Ê ¸ ddtT ¸ œ É a"# sech# ˆ at ‰ tanh# ˆ at ‰ a"# sech% ˆ at ‰ œ "a sech ˆ at ‰ Ê N œ ¸ ddtT ¸ œ ˆ tanh at ‰ i ˆsech at ‰ j ; dt 1 " t " # t , œ k1vk † ¸ ddtT ¸ œ cosh t † a sech ˆ a ‰ œ a sech ˆ a ‰. a 16. r œ (cosh t)i (sinh t)j tk Ê v œ (sinh t)i (cosh t)j k Ê kvk œ Èsinh# t ( cosh t)# 1 œ È2 cosh t Ê T œ kvvk œ Š È" tanh t‹ i È" j Š È" sech t‹ k Ê ddtT œ Š È" sech# t‹ i Š È" sech t tanh t‹ k 2 Ê 2 2 ¸ ddtT ¸ œ É "# sech% t "# sech# t tanh# t œ È" sech t 2 , œ k1vk † ¸ ddtT ¸ œ È 1 2 cosh t Ê 2 2 ˆ ddtT ‰ N œ ¸ dT ¸ œ (sech t)i (tanh t)k ; dt † È" sech t œ "# sech# t. 2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 776 Chapter 13 Vector-Valued Functions and Motion in Space $Î# k2ak œ k2ak a1 4a# x# b a1 4a# x# b$Î# 17. y œ ax# Ê yw œ 2ax Ê yww œ 2a; from Exercise 5(a), ,(x) œ Ê ,w (x) œ 3# k2ak a1 4a# x# b &Î# a8a# xb ; thus, ,w (x) œ 0 Ê x œ 0. Now, ,w (x) 0 for x 0 and ,w (x) 0 for x 0 so that ,(x) has an absolute maximum at x œ 0 which is the vertex of the parabola. Since x œ 0 is the only critical point for ,(x), the curvature has no minimum value. 18. r œ (a cos t)i (b sin t)j Ê v œ (a sin t)i (b cos t)j Ê a œ (a cos t)i (b sin t)j Ê v ‚ a â â i j kâ â â â œ â a sin t b cos t 0 â œ abk Ê kv ‚ ak œ kabk œ ab, since a b 0; , (t) œ kvkv‚k$ak â â â a cos t b sin t 0 â $Î# œ ab aa# sin# t b# cos# tb &Î# ; ,w (t) œ #3 (ab) aa# sin# t b# cos# tb œ 3# (ab) aa# b# b (sin 2t) aa# sin# t b# cos# tb &Î# a2a# sin t cos t 2b# sin t cos tb ; thus, ,w (t) œ 0 Ê sin 2t œ 0 Ê t œ 0, 1 identifying points on the major axis, or t œ 1# , 3#1 identifying points on the minor axis. Furthermore, ,w (t) 0 for 0 t 1# and for 1 t 3#1 ; ,w (t) 0 for 1# t 1 and 3#1 t 21. Therefore, the points associated with t œ 0 and t œ 1 on the major axis give absolute maximum curvature and the points associated with t œ 1# and t œ 31 # on the minor axis give absolute minimum curvature. # # 19. , œ a# a b# Ê dda, œ aa#ab#bb# ; dda, œ 0 Ê a# b# œ 0 Ê a œ „ b Ê a œ b since a, b 0. Now, dda, 0 if " a b and dda, 0 if a b Ê , is at a maximum for a œ b and ,(b) œ b# b b# œ 2b is the maximum value of ,. 20. (a) From Example 5, the curvature of the helix r(t) œ (a cos t)i (a sin t)j btk, a, b 0 is , œ a# a b# ; also kvk œ Èa# b# . For the helix r(t) œ (3 cos t)i (3 sin t)j tk, 0 Ÿ t Ÿ 41, a œ 3 and b œ 1 Ê , œ # 3 # œ 3 41 and kvk œ È10 Ê K œ '0 3 1 %1 3 È 121 10 dt œ ’ È310 t“ œ È 10 10 ! 10 (b) y œ x# Ê x œ t and y œ t# , _ t _ Ê r(t) œ ti t# j Ê v œ i 2tj Ê kvk œ È1 4t# ; 4 2 T œ È1 1 4t# i È12t 4t# j; ddtT œ a1 4t4t# b3/2 i a1 4t j; ¸ ddtT ¸ œ É a16t œ 1 24t# . Thus # b3/2 1 4t# b3 2 , œ È1 1 4t# † 1 24t# œ ' œaÄ lim _ a 0 _ 2 3 ŠÈ1 4t# ‹ 2 14t# dt b lim Ä_ œaÄ lim a tan" 2ab lim _ . Then K œ 'c_ 2 _ $ ŠÈ1 4t# ‹ ŠÈ1 4t# ‹ dt œ ' _ 124t# dt '0 1 24t dt œ a Älim_ ctan" 2td !a lim bÄ_ b # a tan" 2bb œ 1# 1# œ 1 ctan" 2td 0 b bÄ_ 21. r œ ti (sin t)j Ê v œ i (cos t)j Ê kvk œ È1# (cos t)# œ È1 cos# t Ê ¸v ˆ 1# ‰¸ œ É1 cos# ˆ 1# ‰ œ 1; T œ kvvk œ Èi cos t j2 Ê ddtT œ 1 cos t sin t cos t i sin2t 3/2 j Ê a1 cos2 tb3/2 a1 cos tb ksin tk ¸ ddtT ¸ œ 1 ¸ ddtT ¸ cos2 t ; tœ 12 ¸sin 1 ¸ œ 1 cos22ˆ 1 ‰ œ 11 œ 1. Thus ,ˆ 12 ‰ œ 11 † 1 œ 1 2 # Ê 3 œ "1 œ 1 and the center is ˆ 1# ß 0‰ Ê ˆx 1# ‰ y# œ 1 2 1 22. r œ (2 ln t)i ˆt "t ‰ j Ê v œ ˆ 2t ‰ i ˆ1 t"# ‰ j Ê kvk œ É t42 ˆ1 t12 ‰ œ t t2 1 Ê T œ kvvk œ t2 2t 1 i tt2 1 j; 2 2 2 ˆt 1 ‰ 4at2 1b 16t2 dT 4t 2 1 ¸ dT ¸ t2 2 2t2 2 ¸ dT ¸ œ t2 dt œ at2 1b2 i at2 1b2 j Ê dt œ Ê 1 . Thus , œ kvk † dt œ t2 1 † t2 1 œ at2 1b2 Ê , a1b œ 22 at2 1b4 2 2 œ "# Ê 3 œ ," œ 2. The circle of curvature is tangent to the curve at P(0ß 2) Ê circle has same tangent as the curve Ê v(1) œ 2i is tangent to the circle Ê the center lies on the y-axis. If t Á 1 (t 0), then (t 1)# 0 # Ê t# 2t 1 0 Ê t# 1 2t Ê t t 1 2 since t 0 Ê t "t 2 Ê ˆt "t ‰ 2 Ê y 2 on both sides of (0ß 2) Ê the curve is concave down Ê center of circle of curvature is (0ß 4) Ê x# (y 4)# œ 4 is an equation of the circle of curvature Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.4 Curvature and Normal Vectors of a Curve 23. y œ x# Ê f w (x) œ 2x and f ww (x) œ 2 Ê ,œ k2 k 2 œ a1 (2x)# b$Î# a1 4x# b$Î# % 24. y œ x4 Ê f w (x) œ x$ and f ww (x) œ 3x# Ê ,œ k3x# k # $Î# Š1 ax$ b ‹ œ 3x# a1 x' b$Î# 25. y œ sin x Ê f w (x) œ cos x and f ww (x) œ sin x Ê ,œ k sin xk ksin xk œ a1 cos# xb$Î# a1 cos# xb$Î# 26. y œ ex Ê f w (x) œ ex and f ww (x) œ ex Ê ,œ kex k $Î# # Š1 aex b ‹ ex $Î# 1 e2x ‰ œˆ 27-34. Example CAS commands: Maple: with( plots ); r := t -> [3*cos(t),5*sin(t)]; lo := 0; hi := 2*Pi; t0 := Pi/4; P1 := plot( [r(t)[], t=lo..hi] ): display( P1, scaling=constrained, title="#27(a) (Section 13.4)" ); CURVATURE := (x,y,t) ->simplify(abs(diff(x,t)*diff(y,t,t)-diff(y,t)*diff(x,t,t))/(diff(x,t)^2+diff(y,t)^2)^(3/2)); kappa := eval(CURVATURE(r(t)[],t),t=t0); UnitNormal := (x,y,t) ->expand( [-diff(y,t),diff(x,t)]/sqrt(diff(x,t)^2+diff(y,t)^2) ); N := eval( UnitNormal(r(t)[],t), t=t0 ); C := expand( r(t0) + N/kappa ); OscCircle := (x-C[1])^2+(y-C[2])^2 = 1/kappa^2; evalf( OscCircle ); Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 777 778 Chapter 13 Vector-Valued Functions and Motion in Space P2 := implicitplot( (x-C[1])^2+(y-C[2])^2 = 1/kappa^2, x=-7..4, y=-4..6, color=blue ): display( [P1,P2], scaling=constrained, title="#27(e) (Section 13.4)" ); Mathematica: (assigned functions and parameters may vary) In Mathematica, the dot product can be applied either with a period "." or with the word, "Dot". Similarly, the cross product can be applied either with a very small "x" (in the palette next to the arrow) or with the word, "Cross". However, the Cross command assumes the vectors are in three dimensions For the purposes of applying the cross product command, we will define the position vector r as a three dimensional vector with zero for its z-component. For graphing, we will use only the first two components. Clear[r, t, x, y] r[t_]={3 Cos[t], 5 Sin[t] } t0= 1 /4; tmin= 0; tmax= 21; r2[t_]= {r[t][[1]], r[t][[2]]} pp=ParametricPlot[r2[t], {t, tmin, tmax}]; mag[v_]=Sqrt[v.v] vel[t_]= r'[t] speed[t_]=mag[vel[t]] acc[t_]= vel'[t] curv[t_]= mag[Cross[vel[t],acc[t]]]/speed[t]3 //Simplify unittan[t_]= vel[t]/speed[t]//Simplify unitnorm[t_]= unittan'[t] / mag[unittan'[t]] ctr= r[t0] + (1 / curv[t0]) unitnorm[t0] //Simplify {a,b}= {ctr[[1]], ctr[[2]]} To plot the osculating circle, load a graphics package and then plot it, and show it together with the original curve. <<GraphicsÌmplicitPlot` pc=ImplicitPlot[(x a)2 + (y b)2 == 1/curv[t0]2 , {x, 8, 8},{y, 8, 8}] radius=Graphics[Line[{{a, b}, r2[t0]}]] Show[pp, pc, radius, AspectRatio Ä 1] 13.5 TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION 1. r œ (a cos t)i (a sin t)j btk Ê v œ (a sin t)i (a cos t)j bk Ê kvk œ È(a sin t)# (a cos t)# b# œ Èa# b# Ê aT œ d kvk œ 0; a œ (a cos t)i (a sin t)j Ê kak œ È(a cos t)# (a sin t)# œ Èa# œ kak dt # Ê aN œ Ékak a#T œ Ékak# 0# œ kak œ kak Ê a œ (0)T kak N œ kak N 2. r œ (1 3t)i (t 2)j 3tk Ê v œ 3i j 3k Ê kvk œ È3# 1# (3)# œ È19 Ê aT œ dtd kvk œ 0; a œ 0 Ê aN œ Ékak# a#T œ 0 Ê a œ (0)T (0)N œ 0 3. r œ (t 1)i 2tj t# k Ê v œ i 2j 2tk Ê kvk œ È1# 2# (2t)# œ È5 4t# Ê aT œ "# a5 4t# b # "Î# œ 4t a5 4t b È # "Î# (8t) # Ê aT (1) œ È49 œ 34 ; a œ 2k Ê a(1) œ 2k Ê ka(1)k œ 2 Ê aN œ Ékak a#T œ É2# ˆ 34 ‰ È 2 5 œ É 20 Ê a(1) œ 43 T 2 3 5 N 9 œ 3 4. r œ (t cos t)i (t sin t)j t# k Ê v œ (cos t t sin t)i (sin t t cos t)j 2tk "Î# Ê kvk œ È(cos t t sin t)# (sin t t cos t)# (2t)# œ È5t# 1 Ê aT œ " a5t# 1b (10t) # Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.5 Tangential and Normal Components of Acceleration œ È5t5t# 1 Ê aT (0) œ 0; a œ (2 sin t t cos t)i (2 cos t t sin t)j 2k Ê a(0) œ 2j 2k Ê ka(0)k # œ È2# 2# œ 2È2 Ê aN œ Ékak# a#T œ ÊŠ2È2‹ 0# œ 2È2 Ê a(0) œ (0)T 2È2N œ 2È2N 5. r œ t# i ˆt "3 t$ ‰ j ˆt "3 t$ ‰ k Ê v œ 2ti a1 t# b j a1 t# b k Ê kvk œ É(2t)# a1 t# b# a1 t# b# œ È2 at% 2t# 1b œ È2 a1 t# b Ê aT œ 2tÈ2 Ê aT (0) œ 0; a œ 2i 2tj 2tk Ê a(0) œ 2i Ê ka(0)k œ 2 Ê aN œ Ékak# a#T œ È2# 0# œ 2 Ê a(0) œ (0)T 2N œ 2N 6. r œ aet cos tb i aet sin tb j È2et k Ê v œ aet cos t et sin tb i aet sin t et cos tb j È2et k # Ê kvk œ Êaet cos t et sin tb# aet sin t et cos tb# ŠÈ2et ‹ œ È4e2t œ 2et Ê aT œ 2et Ê aT (0) œ 2; a œ aet cos t et sin t et sin t et cos tb i aet sin t et cos t et cos t et sin tb j È2et k # œ a2et sin tb i a2et cos tb j È2et k Ê a(0) œ 2j È2k Ê ka(0)k œ Ê2# ŠÈ2‹ œ È6 # Ê aN œ Ékak# a#T œ ÊŠÈ6‹ 2# œ È2 Ê a(0) œ 2T È2N 7. r œ (cos t)i (sin t)j k Ê v œ ( sin t)i (cos t)j Ê kvk œ È( sin t)# (cos t)# œ 1 Ê T œ kvvk È È œ ( sin t)i (cos t)j Ê T ˆ 14 ‰ œ #2 i #2 j ; ddtT œ ( cos t)i (sin t)j Ê ¸ ddtT ¸ œ È( cos t)# ( sin t)# â â j kâ â i ˆ ddtT ‰ È2 È2 â â 1 œ 1 Ê N œ ¸ dT ¸ œ ( cos t)i (sin t)j Ê N ˆ 4 ‰ œ # i # j ; B œ T ‚ N œ â sin t cos t 0 â œ k â â dt â cos t sin t 0 â Ê B ˆ 14 ‰ œ k , the normal to the osculating plane; r ˆ 14 ‰ œ È È2 È2 # i # jk È È Ê P œ Š #2 ß #2 ß 1‹ lies on the È osculating plane Ê 0 Šx #2 ‹ 0 Šy #2 ‹ (z (1)) œ 0 Ê z œ 1 is the osculating plane; T is normal È È È È È È to the normal plane Ê Š #2 ‹ Šx #2 ‹ Š #2 ‹ Šy #2 ‹ 0(z (1)) œ 0 Ê #2 x #2 y œ 0 Ê x y œ 0 is the normal plane; N is normal to the rectifying plane È È È È È È Ê Š #2 ‹ Šx #2 ‹ Š #2 ‹ Šy #2 ‹ 0(z (1)) œ 0 Ê #2 x #2 y œ 1 Ê x y œ È2 is the rectifying plane 8. r œ (cos t)i (sin t)j tk Ê v œ ( sin t)i (cos t)j k Ê kvk œ Èsin# t cos# t 1 œ È2 Ê T œ kvvk œ Š È"2 sin t‹ i Š È"2 cos t‹ j È"2 k Ê ddtT œ Š È"2 cos t‹ i Š È"2 sin t‹ j Ê ¸ ddtT ¸ ˆ dT ‰ œ É "# cos# t "# sin# t œ È" Ê N œ ¸ ddtT ¸ œ ( cos t)i (sin t)j ; thus T(0) œ È" j È" k and N(0) œ i 2 2 2 dt â â â i j k â â " " ââ " " Ê B(0) œ ââ 0 È2 È2 â œ È2 j È2 k , the normal to the osculating plane; r(0) œ i Ê P(1ß 0ß 0) lies on â 1 0 0 ââ â the osculating plane Ê 0(x 1) È"2 (y 0) È"2 (z 0) œ 0 Ê y z œ 0 is the osculating plane; T is normal to the normal plane Ê 0(x 1) È"2 (y 0) È"2 (z 0) œ 0 Ê y z œ 0 is the normal plane; N is normal to the rectifying plane Ê 1(x 1) 0(y 0) 0(z 0) œ 0 Ê x œ 1 is the rectifying plane. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 779 780 Chapter 13 Vector-Valued Functions and Motion in Space 9. By Exercise 9 in Section 13.4, T œ ˆ 35 cos t‰ i ˆ 35 sin t‰ j 45 k and N œ ( sin t)i (cos t)j so that B œ T ‚ N â i j k ââ â â3 â œ â 5 cos t 35 sin t 45 â œ ˆ 45 cos t‰ i ˆ 45 sin t‰ j 35 k . Also v œ (3 cos t)i (3 sin t)j 4k â â â sin t cos t 0 â â â i j kâ â â â Ê a œ (3 sin t)i (3 cos t)j Ê ddta œ (3 cos t)i (3 sin t)j and v ‚ a œ â 3 cos t 3 sin t 4 â â â â 3 sin t 3 cos t 0 â œ (12 cos t)i (12 sin t)j 9k Ê kv ‚ ak2 œ a12 cos tb# a12 sin tb# a9b# œ 225. Thus 7œ â â 3 cos t â â 3 sin t â â 3 cos t â 3 sin t 4 â â 3 sin t 0 â â # 3 sin t 0 â t9 cos# tb 36 4 œ 4†a9 sin225 œ 225 œ 25 225 10. By Exercise 10 in Section 13.4, T œ (cos t)i (sin t)j and N œ ( sin t)i (cos t)j ; thus B œ T ‚ N â â j kâ â i â â œ â cos t sin t 0 â œ acos# t sin# tb k œ k. Also v œ (t cos t)i (t sin t)j â â â sin t cos t 0 â Ê a œ atasin tb cos tbi at cos t sin tbj Ê ddta œ at cos t sin t sin tbi at sin t cos t cos tbj â â i j kâ â â â t cos t t sin t 0â œ at cos t 2 sin tbi a2 cos t t sin tbj. Thus v ‚ a œ â â â â at sin t cos tb at cos t sin tb 0 â # œ [(t cos t)(t cos t sin t) (t sin t)(t sin t cos t)]k œ t# k Ê kv ‚ ak2 œ at# b œ t4 . Thus â â t cos t t sin t 0â â â â sin t t cos t 0 â â cos t t sin t â â â 2 sin t t cos t 2 cos t t sin t 0 â 7œ œ t04 œ 0 t4 t sin t 11. By Exercise 11 in Section 13.4, T œ Š cos È ‹ i Š sin tÈ2cos t ‹ j and N œ Š cosÈt 2 sin t ‹ i Š sinÈt 2 cos t ‹ j ; Thus 2 â i j k ââ â â cos t sin t sin t cos t 0 ââ œ ’Š cos# t 2 cos t sin t sin# t ‹ Š sin# t 2 sin t cos t cos# t ‹“ k È2 È2 B œ T ‚ N œ ââ 2 2 â â cos t sin t sin t cos t 0 â â â È2 È2 a2tb a2tb œ ’Š 1 sin ‹ Š 1 sin ‹“ k œ k . Also, v œ aet cos t et sin tb i aet sin t et cos tb j 2 2 Ê a œ c et asin t cos tb et acos t sin tb d i c et acos t sin tb et asin t cos tb d j=a2et sin tb i a2et cos tb j â â i j kâ â â â Ê ddta œ 2et acos t sin tb i 2et asin t cos tbj. Thus v ‚ a œ â et acos t sin tb et asin t cos tb 0 â œ 2e2t k â â t t 2e cos t 0â â 2e sin t # Ê kv ‚ ak2 œ a2e2t b œ 4e4t . Thus 7 œ â t â e acos t sin tb â â 2et sin t â â 2et acos t sin tb et asin t cos tb 0 ââ â 2et cos t 0â â 2et asin t cos tb 0 â œ0 4e4t 5 ‰ ˆ 12 ‰ 12. By Exercise 12 in Section 13.4, T œ ˆ 12 13 cos 2t i 13 sin 2t j 13 k and N œ ( sin 2t)i (cos 2t)j so â â i j kâ â â ˆ 12 12 5 â 12 ‰ ˆ ‰ ˆ5 ‰ ˆ5 ‰ B œ T ‚ N œ â 13 cos 2t 13 sin 2t 13 ââ œ 13 cos 2t i 13 sin 2t j 13 k . Also, â â a sin 2tb a cos 2tb 0â v œ (12 cos 2t)i (12 sin 2t)j 5k Ê a œ (24 sin 2t)i (24 cos 2t)j and ddta œ (48 cos 2t)i (48 sin 2t)j â â i j kâ â â â v ‚ a œ â 12 cos 2t 12 sin 2t 5 â œ (120 cos 2t)i (120 sin 2t)j 288k Ê kv ‚ ak2 â â â 24 sin 2t 24 cos 2t 0 â œ (120 cos 2t)# (120 sin 2t)# (288)# œ 120# acos# 2t sin# 2tb 288# œ 97344. Thus Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.5 Tangential and Normal Components of Acceleration 781 â â â 12 cos 2t 12 sin 2t 5 â â â â 24 sin 2t 24 cos 2t 0 â â â â 48 cos 2t 48 sin 2t 0 â 24†48b 10 7œ œ 5†a97344 œ 169 97344 13. By Exercise 13 in Section 13.4, T œ at# t1b1/2 i at# 11b1/2 j and N œ Èt#1 1 i Èt#t 1 j so that B œ T ‚ N â i â # â j k ââ â ât t 0â â t â 1 â â 0 a d # â â œ â Èt# 1 Èt# 1 â œ k. Also, v œ t i tj Ê a œ 2ti j Ê dt œ 2i so that ââ 2t " 0 ââ œ 0 Ê 7 œ 0 t â " â â 2 0 0â â Èt# 1 Èt# 1 0 â 14. By Exercise 14 in Section 13.4, T œ ( cos t)i (sin t)j and N œ (sin t)i (cos t)j so that B œ T ‚ N â â j kâ â i â â œ â cos t sin t 0 â œ k . Also, v œ a3 cos# t sin tb i a3 sin# t cos tb j â â cos t 0 â â sin t Ê a œ dtd a3 cos# t sin tb i dtd a3 sin# t cos tb j Ê ddta œ dtd ˆ dtd a3 cos# t sin tb‰ i dtd ˆ dtd a3 sin# t cos tb‰ j â â â 3 cos# t sin t 3 sin# t cos t 0â â d â d # 0 ââ œ 0 Ê 7 œ 0 Ê ââ dt a3 cos# t sin tb dt a3 sin t cos tb â d ˆ d a3 cos# t sin tb‰ d ˆ d a3 sin# t cos tb‰ 0 â â dt dt â dt dt 15. By Exercise 15 in Section 13.4, T œ kvvk œ ˆsech at ‰ i ˆtanh at ‰ j and N œ ˆ tanh at ‰ i ˆsech at ‰ j so that B œ T ‚ N â i j k ââ â â sech ˆ t ‰ t tanh ˆ a ‰ 0 ââ œ k. Also, v œ i ˆsinh at ‰ j Ê a œ ˆ "a cosh at ‰ j Ê ddta œ a"# sinh ˆ at ‰ j so that œâ a â â â tanh ˆ at ‰ sech ˆ at ‰ 0 â â â â1 sinh ˆ at ‰ 0â â â â 0 " cosh ˆ t ‰ 0 â œ 0 Ê 7 œ 0 a a â â â 0 " sinh ˆ t ‰ 0 â â â a# a 16. By Exercise 16 in Section 13.4, T œ Š È"2 tanh t‹ i È"2 j Š È"2 sech t‹ k and N œ (sech t)i (tanh t)k so that â â â â i j k â " â " " â B œ T ‚ N œ â È2 tanh t È2 È2 sech t ââ œ Š È"2 tanh t‹ i È"2 j Š È"2 sech t‹ k. Also, v œ (sinh t)i (cosh t)j k â sech t 0 tanh t ââ â â â j kâ â i â â da a œ (cosh t)i (sinh t)j Ê dt œ (sinh t)i (cosh t)j and v ‚ a œ â sinh t cosh t 1 â â â â cosh t sinh t 0 â œ (sinh t)i (cosh t)j acosh2 t sinh2 tbk œ (sinh t)i (cosh t)j k Ê kv ‚ ak# œ sinh# t cosh# t 1. Thus â â â sinh t cosh t 1 â â â â cosh t sinh t 0 â â â sinh t cosh t 0 â â " 7 œ sinh# t cosh# t1 œ sinh# t " cosh# t 1 œ # cosh# t . 17. Yes. If the car is moving along a curved path, then , Á 0 and aN œ , kvk# Á 0 Ê a œ aT T aN N Á 0 . 18. kvk constant Ê aT œ dtd kvk œ 0 Ê a œ aN N is orthogonal to T Ê the acceleration is normal to the path 19. a ¼ v Ê a ¼ T Ê aT œ 0 Ê dtd kvk œ 0 Ê kvk is constant 20. a(t) œ aT T aN N , where aT œ dtd kvk œ dtd (10) œ 0 and aN œ , kvk# œ 100, Ê a œ 0T 100,N. Now, from Exercise 5(a) Section 12.4, we find for y œ f(x) œ x# that , œ kf ww (x)k 2 2 œ ; also, $Î# œ c1 (2x)# d$Î# a1 4x# b$Î# 1 af w (x)b# ‘ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 782 Chapter 13 Vector-Valued Functions and Motion in Space r(t) œ ti t# j is the position vector of the moving mass Ê v œ i 2tj Ê kvk œ È1 4t# Ê T œ È1 " 4t# (i 2tj). At (0ß 0): T(0) œ i, N(0) œ j and ,(0) œ 2 Ê F œ ma œ m(100,)N œ 200m j ; È È 2 At ŠÈ2ß 2‹ : T ŠÈ2‹ œ "3 Ši 2È2j‹ œ 3" i 2 3 2 j , N ŠÈ2‹ œ 2 3 2 i 3" j , and , ŠÈ2‹ œ 27 Ê F œ ma È È 2 2 400 2 " 200 ‰ œ m(100,)N œ ˆ 200 27 m Š 3 i 3 j‹ œ 81 m i 81 m j 2 2 3 2 d s ds d s ‰ ˆ ds ‰ ˆ ds ‰ 21. By a œ aT T aN N we have v ‚ a œ ˆ ds dt T ‚ ’ dt2 T , dt N “ œ Š dt dt2 ‹aT ‚ Tb , dt aT ‚ Nb 3 lv‚al ds 3 ‰ œ ,ˆ ds dt B. It follows that lv ‚ al œ , ¹ dt ¹ lBl œ , lvl Ê , œ lvl3 3 22. aN œ 0 Ê , kvk# œ 0 Ê , œ 0 (since the particle is moving, we cannot have zero speed) Ê the curvature is zero so the particle is moving along a straight line 23. From Example 1, kvk œ t and aN œ t so that aN œ , kvk# Ê , œ kvaNk# œ tt# œ "t , t Á 0 Ê 3 œ ," œ t 24. r œ (x! At)i (y! Bt)j (z! Ct)k Ê v œ Ai Bj Ck Ê a œ 0 Ê v ‚ a œ 0 Ê , œ 0. Since the curve is a plane curve, 7 œ 0. 25. If a plane curve is sufficiently differentiable the torsion is zero as the following argument shows: r œ f(t)i g(t)j Ê v œ f w (t)i gw (t)j Ê a œ f ww (t)i gww (t)j Ê ddta œ f www (t)i gwww (t)j Ê â w â â f (t) gw (t) 0 â â ww â â f (t) gww (t) 0 â â www â â f (t) gwww (t) 0 â 7œ œ0 # kv ‚ a k 26. v œ aa sin tbi aa cos tbj b k and a œ aa cos tbi aa sin tbj To find the torsion: 7 œ # â â â a sin t a cos t b â â â â a cos t a sin t 0 â â â â a sin t a cos t 0 â 2 Ša È a 2 b 2 ‹ œ bâ2 cos2 t a2 sin2 t‰ a2 bˆcos2 t sin2 t‰ œ a2 aa2 b2 b œ a2 b b2 Ê a 2 aa2 b 2 b # # 7 w (b) œ aaa# bb# b# ; # 7 w (b) œ 0 Ê aaa# bb# b# œ 0 Ê a# b# œ 0 Ê b œ „ a Ê b œ a since a, b 0. Also b a Ê 7 w 0 and b a " Ê 7 w 0 so 7max occurs when b œ a Ê 7max œ a# a a# œ 2a 27. r(t) œ f(t)i g(t)j h(t)k Ê v œ f w (t)i gw (t)j hw (t)k; v † k œ 0 Ê hw (t) œ 0 Ê h(t) œ C Ê r(t) œ f(t)i g(t)j Ck and r(a) œ f(a)i g(a)j Ck œ 0 Ê f(a) œ 0, g(a) œ 0 and C œ 0 Ê h(t) œ 0. 28. From Exercise 26, v œ (a sin t)i (a cos t)j bk Ê kvk œ Èa# b# Ê T œ kvvk ˆ dT ‰ œ È #" # c(a sin t)i (a cos t)j bkd ; ddtT œ È #" # c(a cos t)i (a sin t)jd Ê N œ ¸ ddtT ¸ a b a b dt â â â i j k â â â a sin t a cos t b œ (cos t)i (sin t)j ; B œ T ‚ N œ ââ Èa# b# Èa# b# Èa# b# ââ â cos t 0 ââ sin t â t œ Èba#sin tb# i Èbacos j Èa#a b# k Ê ddtB œ Èa#" b# c(b cos t)i (b sin t)jd Ê ddtB † N œ Èa#b b# # b# Ê 7 œ k"vk ˆ ddtB † N‰ œ Š È #" # ‹ Š È #b # ‹ œ a# b b# , which is consistent with the result in Exercise 26. a b a b Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 13.5 Tangential and Normal Components of Acceleration 29-32. Example CAS commands: Maple: with( LinearAlgebra ); r := < t*cos(t) | t*sin(t) | t >; t0 := sqrt(3); rr := eval( r, t=t0 ); v := map( diff, r, t ); vv := eval( v, t=t0 ); a := map( diff, v, t ); aa := eval( a, t=t0 ); s := simplify(Norm( v, 2 )) assuming t::real; ss := eval( s, t=t0 ); T := v/s; TT := vv/ss ; q1 := map( diff, simplify(T), t ): NN := simplify(eval( q1/Norm(q1,2), t=t0 )); BB := CrossProduct( TT, NN ); kappa := Norm(CrossProduct(vv,aa),2)/ss^3; tau := simplify( Determinant(< vv, aa, eval(map(diff,a,t),t=t0) >)/Norm(CrossProduct(vv,aa),2)^3 ); a_t := eval( diff( s, t ), t=t0 ); a_n := evalf[4]( kappa*ss^2 ); Mathematica: (assigned functions and value for t0 will vary) Clear[t, v, a, t] mag[vector_]:=Sqrt[vector.vector] Print["The position vector is ", r[t_]={t Cos[t], t Sin[t], t}] Print["The velocity vector is ", v[t_]= r'[t]] Print["The acceleration vector is ", a[t_]= v'[t]] Print["The speed is ", speed[t_]= mag[v[t]]//Simplify] Print["The unit tangent vector is ", utan[t_]= v[t]/speed[t] //Simplify] Print["The curvature is ", curv[t_]= mag[Cross[v[t],a[t]]] / speed[t]3 //Simplify] Print["The torsion is ", torsion[t_]= Det[{v[t], a[t], a'[t]}] / mag[Cross[v[t],a[t]]]2 //Simplify] Print["The unit normal vector is ", unorm[t_]= utan'[t] / mag[utan'[t]] //Simplify] Print["The unit binormal vector is ", ubinorm[t_]= Cross[utan[t],unorm[t]] //Simplify] Print["The tangential component of the acceleration is ", at[t_]=a[t].utan[t] //Simplify] Print["The normal component of the acceleration is ", an[t_]=a[t].unorm[t] //Simplify] You can evaluate any of these functions at a specified value of t. t0= Sqrt[3] {utan[t0], unorm[t0], ubinorm[t0]} N[{utan[t0], unorm[t0], ubinorm[t0]}] {curv[t0], torsion[t0]} N[{curv[t0], torsion[t0]}] {at[t0], an[t0]} N[{at[t0], an[t0]}] To verify that the tangential and normal components of the acceleration agree with the formulas in the book: at[t]== speed'[t] //Simplify an[t]==curv [t] speed[t]2 //Simplify Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 783 784 Chapter 13 Vector-Valued Functions and Motion in Space 13.6 VELOCITY AND ACCELERATION IN POLAR COORDINATES 1. Þ ÞÞ . .. d) d) d) dt œ 3 œ ) Ê ) œ 0, r œ aa1 cos )b Ê r œ a sin ) dt œ 3a sin ) Ê r œ 3a cos ) dt œ 9a cos ) v œ a3a sin )bur aaa1 cos )bba3bu) œ a3a sin )bur 3aa1 cos )bu) a œ Š9a cos ) aa1 cos )ba3b2 ‹ur aaa1 cos )b † 0 2a3a sin )ba3bbu) œ a9a cos ) 9a 9a cos )bur a18a sin )bu) œ 9aa2 cos ) 1bur a18a sin )bu) 2. Þ ÞÞ . .. d) d) d) ‰ ˆ dt œ 2t œ ) Ê ) œ 2, r œ a sin 2) Ê r œ a cos 2) † 2 dt œ 4ta cos 2) Ê r œ 4ta sin 2) † 2 dt 4a cos 2) 2 œ 16t a sin 2) 4a cos 2) v œ a4ta cos 2)bur aa sin 2)ba2tbu) œ a4ta cos 2)bur a2ta sin 2)bu) a œ ’a 16t2 a sin 2) 4a cos 2)b aa sin 2)ba2tb2 “ur aa sin 2)ba2b 2a4ta cos 2)ba2tb‘u) œ ’16t2 a sin 2) 4a cos 2) 4t2 a sin 2)“ur 2a sin 2) 16t2 a cos 2)‘u) œ ’20t2 a sin 2) 4a cos 2)“ur 2a sin 2) 16t2 a cos 2)‘u) œ 4aacos 2) 5t2 sin 2)bur 2aasin 2) 8t2 cos 2)bu) 3. Þ ÞÞ . .. d) a) Ê r œ ea ) † a ddt) œ 2a ea ) Ê r œ 2a ea ) † a ddt) œ 4a2 ea ) dt œ 2 œ ) Ê ) œ 0, r œ e v œ ˆ2a ea ) ‰ur êa ) ‰a2bu) œ ˆ2a ea ) ‰ur ˆ2ea ) ‰u) a œ ’ˆ 4a2 ea ) ‰ êa ) ‰a2b2 “ur ’êa ) ‰a0b 2ˆ2a ea ) ‰a2b“u) œ ’4a2 ea ) 4ea ) “ur ’0 8a ea ) “u) œ 4ea ) aa2 1bur ˆ8a ea ) ‰u) Þ ÞÞ . .. 4. ) œ 1 et Ê ) œ et Ê ) œ et , r œ aa1 sin tb Ê r œ a cos t Ê r œ a sin t v œ aa cos tbur aaa1 sin tbbaet bu) œ aa cos tbur a et a1 sin tbu) a œ ’a a sin tb aaa1 sin tbbaet b “ur ’aaa1 sin tbbaet b 2aa cos tbaet b“u) 2 œ ’a sin t a e2t a1 sin tb“ur ’a et a1 sin tb 2a et cos t“u) œ aasin t e2t a1 sin tbbur a et aa1 sin tb 2cos tbu) œ aasin t e2t a1 sin tbbur a et a2cos t 1 sin tbu) Þ ÞÞ . .. 5. ) œ 2t Ê ) œ 2 Ê ) œ 0, r œ 2 cos 4t Ê r œ 8 sin 4t Ê r œ 32 cos 4t v œ a8 sin 4tbur a2 cos 4tba2bu) œ 8asin 4tbur 4acos 4tbu) a œ Ša32 cos 4tb a2 cos 4tba2b2 ‹ur aa2 cos 4tb † 0 2a8 sin 4tba2bbu) œ a32 cos 4t 8 cos 4tbur a0 32sin 4tbu) œ 40acos 4tbur 32asin 4tbu) r v# ! ! 6. e œ GM 1 Ê v#! œ GM(er! 1) Ê v! œ É GM(er! 1) ; Circle: e œ 0 Ê v! œ É GM r! 2GM Ellipse: 0 e 1 Ê É GM r! v! É r! Parabola: e œ 1 Ê v! œ É 2GM r! Hyperbola: e 1 Ê v! É 2GM r! GM # 7. r œ GM Ê v œ É GM v# Ê v œ r r which is constant since G, M, and r (the radius of orbit) are constant Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 13 Practice Exercises ?t) r(t) r(t) 8. ?A œ "# kr(t ?t) ‚ r(t)k Ê ??At œ "# ¹ r(t ‚ r(t)¹ œ "# ¹ r(t ?t) ? ‚ r(t)¹ ?t t " r(t ?t) r(t) œ "# ¹ r(t ??t)t r(t) ‚ r(t) ?"t r(t) ‚ r(t)¹ œ #" ¹ r(t ??t)t r(t) ‚ r(t)¹ Ê dA ‚ r(t)¹ ¹ dt œ ?tlim ?t Ä0 # Þ œ "# ¸ ddtr ‚ r(t)¸ œ "# ¸r(t) ‚ ddtr ¸ œ "# kr ‚ rk # # % # r v# # % ! ! 9. T œ Š 2r!1va! ‹ È1 e# Ê T# œ Š 4r1# va# ‹ a1 e# b œ Š 4r1# va# ‹ ”1 Š GM 1‹ • (from Equation 5) ! ! ! ! # % ˆ41# a% ‰ a2GM r! v#! b r# v % r! v!# 2GMr! v!# r!# v!% œ Š 4r1# va# ‹ ’ G!# M!# 2 Š GM ‹“ œ Š 4r1# va# ‹ ’ “œ G# M# r! G# M# ! ! ! ! # % œ a41# a% b Š 2GM r! v#! # % ˆ " ‰ˆ 2 ‰ ˆ 2 ‰ 2r! GM ‹ GM œ a41 a b 2a GM (from Equation 10) # $ # # 1 a 41 Ê T# œ 4GM Ê Ta$ œ GM minutes seconds 7 10. r œ 365.256 days œ 365.256 days ‚ 24 hours day ‚ 60 hour ‚ 60 minute œ 31,558,118.4 seconds ¸ 3.16 ‚ 10 , 41 G œ 6.6726 ‚ 1011 Nkg†m# , and the mass of the sun M œ 1.99 ‚ 1030 kg. Ta3 œ GM Ê a3 œ T2 GM 41 2 2 Ê a3 œ a3.16 ‚ 107 b 2 2 2 ˆ6.6726‚10c11 ‰ˆ1.99‚1030 ‰ 3 ¸ 3.35863335 ‚ 1033 Ê a œ È 3.35863335 ‚ 1033 41 2 ¸ 149757138111 m ¸ 149.757 billion km CHAPTER 13 PRACTICE EXERCISES 1. r(t) œ (4 cos t)i ŠÈ2 sin t‹ j Ê x œ 4 cos t # # x and y œ È2 sin t Ê 16 y# œ 1; v œ (4 sin t)i ŠÈ2 cos t‹ j and a œ (4 cos t)i ŠÈ2 sin t‹ j ; r(0) œ 4 i , v(0) œ È2j , a(0) œ 4i ; r ˆ 14 ‰ œ 2È2i j , v ˆ 14 ‰ œ 2È2i j , a ˆ 1 ‰ œ 2È2i j ; kvk œ È16 sin# t 2 cos# t 4 Ê aT œ dtd kvk œ È 14 sin# tcos t # ; at t œ 0: aT œ 0, aN œ Ékak# 0 œ 4, a œ 0T 4N œ 4N, , œ kavNk# œ 42 œ 2; 16 sin t 2 cos t È È È 4 2 4 2 4 2 aN 7 at t œ 14 : aT œ È871 œ 73 , aN œ É9 49 9 œ 3 , a œ 3 T 3 N, , œ kvk# œ 27 # # 2. r(t) œ ŠÈ3 sec t‹ i ŠÈ3 tan t‹ j Ê x œ È3 sec t and y œ È3 tan t Ê x3 y3 œ sec# t tan# t œ 1; Ê x# y# œ 3; v œ ŠÈ3 sec t tan t‹ i ŠÈ3 sec# t‹ j and a œ ŠÈ3 sec t tan# t È3 sec$ t‹ i Š2È3 sec# t tan t‹ j ; r(0) œ È3i , v(0) œ È3j , a(0) œ È3i ; kvk œ È3 sec# t tan# t 3 sec% t # $ % t tan t 18 sec t tan t Ê aT œ dtd kvk œ 6 sec ; # # % È 2 3 sec t tan t 3 sec t at t œ 0: aT œ 0, aN œ Ékak# 0 œ È3, a œ 0T È3N œ È3N, , œ kavNk# œ È3 " 3 œ È3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 785 786 Chapter 13 Vector-Valued Functions and Motion in Space 3. r œ È1" t# i È1t t# j Ê v œ t a1 t# b $Î# i a 1 t# b $Î# # # j Ê kvk œ Ê’t a1 t# b$Î# “ ’a1 t# b$Î# “ œ 1 " t# . We want to maximize kvk : ddtkvk œ a1 2tt# b# and ddtkvk œ 0 Ê a1 2tt# b# œ 0 Ê t œ 0. For t 0, a1 2tt# b# 0; for t 0, a1 2tt# b# 0 Ê kvk max occurs when t œ 0 Ê kvk max œ 1 4. r œ aet cos tb i aet sin tb j Ê v œ aet cos t et sin tb i aet sin t et cos tb j Ê a œ aet cos t et sin t et sin t et cos tb i aet sin t et cos t et cos t et sin tb j œ a2et sin tb i a2et cos tb j . Let ) be the angle between r and a . Then ) œ cos" Š krrk†kaak ‹ œ cos" 2e2t sin t cos t2e2t sin t cos t œ cos" Š 2e02t ‹ œ cos" 0 œ 1# for all t Éaet cos tb# aet sin tb# Éa2et sin tb# a2et cos tb# â â â i j kâ â â 5. v œ 3i 4j and a œ 5i 15j Ê v ‚ a œ â 3 4 0 â œ 25k Ê kv ‚ ak œ 25; kvk œ È3# 4# œ 5 â â â 5 "5 0 â " Ê , œ kvkv‚k$ak œ 25 5$ œ 5 6. , œ kyww k x 2x $Î# b $Î# œ e a1 e 1 ayw b# ‘ œ ex a1 e2x b $Î# Ê ddx, œ ex a1 e2x b 3e3x a1 e2x b &Î# œ ex a1 e2x b $Î# ex ’ 3# a1 e2x b &Î# a2e2x b“ &Î# &Î# ca1 e2x b 3e2x d œ ex a1 e2x b a1 2e2x b ; d, " " 2x 2x È2 Ê y œ " ; therefore , is at a È dx œ 0 Ê a1 2e b œ 0 Ê e œ # Ê 2x œ ln 2 Ê x œ # ln 2 œ ln 2 maximum at the point Š ln È2ß È"2 ‹ dy dx 7. r œ xi yj Ê v œ dx dt i dt j and v † i œ y Ê dt œ y. Since the particle moves around the unit circle dy dy dy dy x dx x dx x# y# œ 1, 2x dx dt 2y dt œ 0 Ê dt œ y dt Ê dt œ y (y) œ x. Since dt œ y and dt œ x, we have v œ yi xj Ê at (1ß 0), v œ j and the motion is clockwise. dy " # dx # dx 8. 9y œ x$ Ê 9 dy dt œ 3x dt Ê dt œ 3 x dt . If r œ xi yj , where x and y are differentiable functions of t, dy dy dx " # dx " # then v œ dx dt i dt j. Hence v † i œ 4 Ê dt œ 4 and v † j œ dt œ 3 x dt œ 3 (3) (4) œ 12 at (3ß 3). Also, # # # # # # ‰ ˆ 3" x# ‰ ddt#x . Hence a † i œ 2 Ê ddt#x œ 2 and a œ ddt#x i ddt#y j and ddt#y œ ˆ 32 x‰ ˆ dx dt # a † j œ ddt#y œ 23 (3)(4)# "3 (3)# (2) œ 26 at the point (xß y) œ (3ß 3). 9. dr dt orthogonal to r Ê 0 œ ddtr † r œ "# ddtr † r "# r † ddtr œ "# dtd (r † r) Ê r † r œ K, a constant. If r œ xi yj , where x and y are differentiable functions of t, then r † r œ x# y# Ê x# y# œ K, which is the equation of a circle centered at the origin. 10. (a) (b) v œ (1 1 cos 1t)i (1 sin 1t)j Ê a œ a1# sin 1tb i a1# cos 1tb j ; v(0) œ 0 and a(0) œ 1# j ; v(1) œ 21i and a(1) œ 1# j ; v(2) œ 0 and a(2) œ 1# j ; v(3) œ 21i and a(3) œ 1# j Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 13 Practice Exercises 787 (c) Forward speed at the topmost point is kv(1)k œ kv(3)k œ 21 ft/sec; since the circle makes "# revolution per second, the center moves 1 ft parallel to the x-axis each second Ê the forward speed of C is 1 ft/sec. 11. y œ y! (v! sin !)t "# gt# Ê y œ 6.5 (44 ft/sec)(sin 45°)(3 sec) "# a32 ft/sec# b (3 sec)# œ 6.5 66È2 144 ¸ 44.16 ft Ê the shot put is on the ground. Now, y œ 0 Ê 6.5 22È2t 16t# œ 0 Ê t ¸ 2.13 sec (the positive root) Ê x ¸ (44 ft/sec)(cos 45°)(2.13 sec) ¸ 66.27 ft or about 66 ft, 3 in. from the stopboard # # 45°)] !) 12. ymax œ y! (v! sin œ 7 ft [(80(2)ft/sec)(sin ¸ 57 ft #g a32 ft/sec# b (v! sin !)t "# gt# (v sin !) " gt œ ! v! cos ! # (v! cos !)t t œ 2v! sin ! 2vg ! cos ! tan 9 , which is the time when the golf ball 13. x œ (v! cos !)t and y œ (v! sin !)t "# gt# Ê tan 9 œ yx œ Ê v! cos ! tan 9 œ v! sin ! "# gt Ê hits the upward slope. At this time x œ (v! cos !) Š 2v! sin ! 2vg ! cos ! tan 9 ‹ œ Š g2 ‹ av#! sin ! cos ! v#! cos# ! tan 9b . v# sin ! cos ! v#! cos# ! tan 9 ‹ cos 9 Now OR œ cosx 9 Ê OR œ Š g2 ‹ Š ! œŠ 2v#! cos ! cos ! tan 9 sin ! ‹ Š cos g 9 cos 9 ‹ œŠ 2v#! cos ! 9 cos ! sin 9 ‹ Š sin ! cos cos ‹ #9 g 2v# cos ! œ Š g !cos# 9 ‹ [sin (! 9)]. The distance OR is maximized when x is maximized: 2v#! dx d! œ Š g ‹(cos 2! sin 2! tan 9) œ 0 Ê (cos 2! sin 2! tan 9) œ 0 Ê cot 2! tan 9 œ 0 Ê cot 2! œ tan (9) Ê 2! œ 1# 9 Ê ! œ 9# 14 5 14. (a) x œ v! (cos 40°)t and y œ 6.5 v! (sin 40°)t "# gt# œ 6.5 v! (sin 40°)t 16t# ; x œ 262 12 ft and y œ 0 ft 5 262.4167 262.4167 # # Ê 262 12 œ v! (cos 40°)t or v! œ (cos 40°)t and 0 œ 6.5 ’ (cos 40°)t “ (sin 40°)t 16t Ê t œ 14.1684 262.4167 Ê t ¸ 3.764 sec. Therefore, 262.4167 ¸ v! (cos 40°)(3.764 sec) Ê v! ¸ (cos 40°)(3.764 sec) Ê v! ¸ 91 ft/sec # 2 40°)b !) (b) ymax œ y! (v! sin ¸ 6.5 a(91)(sin ¸ 60 ft 2g (2)(32) 15. r œ (2 cos t)i (2 sin t)j t# k Ê v œ (2 sin t)i (2 cos t)j 2tk Ê kvk œ È(2 sin t)# (2 cos t)# (2t)# œ 2È1 t# Ê Length œ '0 2È1 t# dt œ ’tÈ1 t# ln ¹t È1 t# ¹“ 1Î4 1Î% ! # # 1 1 œ 14 É1 16 ln Š 14 É1 16 ‹ 16. r œ (3 cos t)i (3 sin t)j 2t$Î# k Ê v œ (3 sin t)i (3 cos t)j 3t"Î# k Ê kvk œ É(3 sin t)# (3 cos t)# a3t"Î# b œ È9 9t œ 3È1 t Ê Length œ '0 3È1 t dt œ 2(1 t)$Î# ‘ ! œ 14 3 $ 17. r œ 49 (1 t)$Î# i 49 (1 t)$Î# j "3 tk Ê v œ 32 (1 t)"Î# i 32 (1 t)"Î# j 3" k # # # Ê kvk œ É 23 (1 t)"Î# ‘ 23 (1 t)"Î# ‘ ˆ 3" ‰ œ 1 Ê T œ 32 (1 t)"Î# i 32 (1 t)"Î# j 3" k Ê T(0) œ 32 i 32 j 3" k ; ddtT œ 3" (1 t)"Î# i 3" (1 t)"Î# j Ê ddtT (0) œ 3" i 3" j Ê ¸ ddtT (0)¸ œ â â j kâ â i â 2 2 " â " " 4 â Ê N(0) œ È"2 i È"2 j ; B(0) œ T(0) ‚ N(0) œ ââ 3 3 3 â œ È i È j È k; 3 2 3 2 3 2 " " â â â È2 È2 0 â a œ "3 (1 t)"Î# i "3 (1 t)"Î# j Ê a(0) œ "3 i "3 j and v(0) œ 23 i 23 j "3 k Ê v(0) ‚ a(0) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. È2 3 # 788 Chapter 13 Vector-Valued Functions and Motion in Space â i â â œ â 23 â " â 3 k ââ È Š 32 ‹ È2 È2 " â k v ‚a k " " 4 i j k k v a k œ Ê ‚ œ Ê , (0) œ œ â 3 9 9 9 3 1$ œ 3 ; kv k $ â 0â j 23 " 3 Þ Þ a œ "6 (1 t)$Î# i "6 (1 t)$Î# j Ê a(0) œ "6 i "6 j Ê 7 (0) œ â 2 â 3 â â " â 3 â " â 6 23 " 3 " 6 kv‚ak # " ââ 3 â 0 ââ 0 ââ 2 ‰ ˆ 3" ‰ ˆ 18 œ È # Š 32 ‹ œ 6" 18. r œ aet sin 2tb i aet cos 2tb j 2et k Ê v œ aet sin 2t 2et cos 2tb i aet cos 2t 2et sin 2tb j 2et k Ê kvk œ Éaet sin 2t 2et cos 2tb# aet cos 2t 2et sin 2tb# a2et b# œ 3et Ê T œ kvvk œ ˆ "3 sin 2t 23 cos 2t‰ i ˆ 3" cos 2t 23 sin 2t‰ j 23 k Ê T(0) œ 32 i 3" j 32 k ; Ê ddtT (0) œ 32 i 34 j Ê ¸ ddtT (0)¸ œ 32 È5 â â j kâ â i â â ˆ 23 i 43 j‰ 2 1 2 4 2 5 â Ê N(0) œ 2È5 œ È"5 i È25 j ; B(0) œ T(0) ‚ N(0) œ ââ 3 3 3 â œ È i È j È k; 3 5 3 5 3 5 Š 3 ‹ â " 2 â 0â È5 â È5 dT 4 4 ˆ2 ‰ ˆ 2 ‰ dt œ 3 cos 2t 3 sin 2t i 3 sin 2t 3 cos 2t j a œ a4et cos 2t 3et sin 2tb i a3et cos 2t 4et sin 2tb j 2et k Ê a(0) œ 4i 3j 2k and v(0) œ 2i j 2k â â j kâ âi â â Ê v(0) ‚ a(0) œ â 2 " 2 â œ 8i 4j 10k Ê kv ‚ ak œ È64 16 100 œ 6È5 and kv(0)k œ 3 â â â 4 3 2 â È È Ê ,(0) œ 6 3$ 5 œ 2 9 5 ; Þ a œ a4et cos 2t 8et sin 2t 3et sin 2t 6et cos 2tb i a3et cos 2t 6et sin 2t 4et sin 2t 8et cos 2tb j 2et k Þ œ a2et cos 2t 11et sin 2tb i a11 et cos 2t 2et sin 2tb j 2et k Ê a(0) œ 2i 11j 2k Ê 7 (0) œ â â 2 â â 4 â â 2 1 3 11 kv‚ak# â 2â â 2â â 2â 80 œ 180 œ 49 19. r œ ti "# e2t j Ê v œ i e2t j Ê kvk œ È1 e4t Ê T œ È " 4t i È e 2t 1e dT 2e4t 2e2t dt œ ˆ1 e4t ‰$Î# i ˆ1 e4t ‰$Î# j 1 e4t j Ê T (ln 2) œ È"17 i È417 j ; 8 Ê ddtT (ln 2) œ 17È3217 i 17È j Ê N (ln 2) œ È417 i È"17 j ; 17 â i j k ââ â â " 4 0 ââ œ k ; a œ 2e2t j Ê a(ln 2) œ 8j and v(ln 2) œ i 4j È17 B (ln 2) œ T(ln 2) ‚ N(ln 2) œ ââ È17 â " â 4 â â È17 È17 0 â â â â i j kâ â â Þ 8 Ê v(ln 2) ‚ a(ln 2) œ â " 4 0 â œ 8k Ê kv ‚ ak œ 8 and kv(ln 2)k œ È17 Ê ,(ln 2) œ 17È ; a œ 4e2t j 17 â â â0 8 0â Þ Ê a(ln 2) œ 16j Ê 7 (ln 2) œ â â1 â â0 â â0 4 8 16 k v ‚a k # â 0â â 0â â 0â œ0 20. r œ (3 cosh 2t)i (3 sinh 2t)j 6tk Ê v œ (6 sinh 2t)i (6 cosh 2t)j 6k Ê kvk œ È36 sinh# 2t 36 cosh# 2t 36 œ 6È2 cosh 2t Ê T œ kvvk œ Š È"2 tanh 2t‹ i È"2 j Š È"2 sech 2t‹ k " 8 dT 2 2 dT # Ê T(ln 2) œ 1715 È2 i È2 j 17È2 k ; dt œ Š È2 sech 2t‹ i Š È2 sech 2t tanh 2t‹ k Ê dt (ln 2) # # # È 8 2 8 ‰ 8 ‰ ˆ 15 ‰ 128 240 240 ¸ ddtT (ln 2)¸ œ ÊŠ 128 œ Š È22 ‹ ˆ 17 i Š È22 ‹ˆ 17 ‹ Š 289 È2 ‹ œ 17 17 k œ 289È2 i 289È2 k Ê 289È2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 13 Practice Exercises â â j k â â i â 15 8 â " 8 15 8 " Ê N(ln 2) œ 17 i 17 k ; B(ln 2) œ T(ln 2) ‚ N(ln 2) œ ââ 17È2 È2 17È2 ââ œ 1715 È2 i È2 j 17È2 k ; â 8 15 â 0 17 â â 17 51 45 ‰ ˆ 15 ‰ a œ (12 cosh 2t)i (12 sinh 2t)j Ê a(ln 2) œ 12 ˆ 17 8 i 12 8 j œ # i # j and â i j k ââ â â 45 51 45 51 ‰ ˆ 17 ‰ 6 ââ v(ln 2) œ 6 ˆ 15 4 8 i 6 8 j 6k œ 4 i 4 j 6k Ê v(ln 2) ‚ a(ln 2) œ ââ 4 â 45 â 51 0â 2 # È2 Ê ,(ln 2) œ œ 135i 153j 72k Ê kv ‚ ak œ 153È2 and kv(ln 2)k œ 51 4 Þ Þ a œ (24 sinh 2t)i (24 cosh 2t)j Ê a(ln 2) œ 45i 51j Ê 153È2 $ È 2‹ Š 51 4 32 œ 867 ; â 45 51 â 6â â 4 4 â 5" 45 â â 2 â 0 2 â â â 45 51 0 â 32 7 (ln 2) œ kv‚ak# œ 867 21. r œ a2 3t 3t# b i a4t 4t# b j (6 cos t)k Ê v œ (3 6t)i (4 8t)j (6 sin t)k Ê kvk œ È(3 6t)# (4 8t)# (6 sin t)# œ È25 100t 100t# 36 sin# t "Î# Ê ddtkvk œ "# a25 100t 100t# 36 sin# tb (100 200t 72 sin t cos t) Ê aT (0) œ ddtkvk (0) œ 10; a œ 6i 8j (6 cos t)k Ê kak œ È6# 8# (6 cos t)# œ È100 36 cos# t Ê ka(0)k œ È136 Ê aN œ Ékak# a#T œ È136 10# œ È36 œ 6 Ê a(0) œ 10T 6N 22. r œ (2 t)i at 2t# b j a1 t# b k Ê v œ i (1 4t)j 2tk Ê kvk œ È1# (1 4t)# (2t)# "Î# œ È2 8t 20t# Ê d kvk œ " a2 8t 20t# b (8 40t) Ê aT œ d kvk (0) œ 2È2; a œ 4j 2k dt # dt # Ê kak œ È4# 2# œ È20 Ê aN œ Ékak# a#T œ Ê20 Š2È2‹ œ È12 œ 2È3 Ê a(0) œ 2È2T 2È3N 23. r œ (sin t)i ŠÈ2 cos t‹ j (sin t)k Ê v œ (cos t)i ŠÈ2 sin t‹ j (cos t)k # Ê kvk œ Ê(cos t)# ŠÈ2 sin t‹ (cos t)# œ È2 Ê T œ kvvk œ Š È"2 cos t‹ i (sin t)j Š È"2 cos t‹ k ; dT " " dt œ Š È2 sin t‹ i (cos t)j Š È2 sin t‹ k # # Ê ¸ ddtT ¸ œ ÊŠ È" sin t‹ ( cos t)# Š È" sin t‹ œ 1 2 2 â â i j k â â â â d T " " ˆ dt ‰ cos t sin t cos t " " â â È È Ê N œ ¸ dT ¸ œ Š È sin t‹ i (cos t)j Š È sin t‹ k ; B œ T ‚ N œ â 2 2 â 2 2 dt â " sin t cos t " sin t â â È2 â È2 â i â j k â â â â " " È È œ È2 i È2 k ; a œ ( sin t)i Š 2 cos t‹ j (sin t)k Ê v ‚ a œ â cos t 2 sin t cos t â â â â sin t È2 cos t sin t â Þ œ È2 i È2 k Ê kv ‚ ak œ È4 œ 2 Ê , œ kvkv‚k$ak œ 2 $ œ È"2 ; a œ ( cos t)i ŠÈ2 sin t‹ j (cos t)k Ê 7œ â â cos t â â sin t â â â cos t ŠÈ2‹ â È2 sin t cos t ââ È2 cos t sin t ââ (cos t) ŠÈ2‹ ŠÈ2 sin t‹ (0) (cos t) ŠÈ2‹ È2 sin t cos t ââ œ œ0 # 4 k v ‚a k 24. r œ i (5 cos t)j (3 sin t)k Ê v œ (5 sin t)j (3 cos t)k Ê a œ (5 cos t)j (3 sin t)k Ê v † a œ 25 sin t cos t 9 sin t cos t œ 16 sin t cos t; v † a œ 0 Ê 16 sin t cos t œ 0 Ê sin t œ 0 or cos t œ 0 Ê t œ 0, 1# or 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 789 790 Chapter 13 Vector-Valued Functions and Motion in Space 25. r œ 2i ˆ4 sin #t ‰ j ˆ3 1t ‰ k Ê 0 œ r † (i j) œ 2(1) ˆ4 sin #t ‰ (1) Ê 0 œ 2 4 sin #t Ê sin #t œ "# Ê #t œ 16 Ê t œ 13 (for the first time) 26. r(t) œ ti t# j t$ k Ê v œ i 2tj 3t# k Ê kvk œ È1 4t# 9t% Ê kv(1)k œ È14 Ê T(1) œ È"14 i È214 j È314 k , which is normal to the normal plane Ê È"14 (x 1) È214 (y 1) È314 (z 1) œ 0 or x 2y 3z œ 6 is an equation of the normal plane. Next we calculate N(1) which is normal to the rectifying plane. Now, a œ 2j 6tk Ê a(1) œ 2j 6k Ê v(1) ‚ a(1) â â â i j kâ # È76 È19 â â œ â " 2 3 â œ 6i 6j 2k Ê kv(1) ‚ a(1)k œ È76 Ê ,(1) œ œ kv(t)k Ê ddt#s ¹ ; ds $ œ È dt 7 14 È â â tœ1 Š 14‹ â0 2 6â œ "# a1 4t# 9t% b "Î# È a8t 36t$ b¹ # # tœ1 # ‰ N Ê 2j 6k œ È2214 , so a œ ddt#s T , ˆ ds dt È 2j3k 8 9 ‰ 11 8 9 ˆ 11 œ È2214 Š iÈ ‹ 7È19 ŠÈ14‹ N Ê N œ 2È14 7 i 7 j 7 k Ê 7 (x 1) 7 (y 1) 7 (z 1) 19 14 14 œ 0 or 11x 8y 9z œ 10 is an equation of the rectifying plane. Finally, B(1) œ T(1) ‚ N(1) â â j kâ â i È14 â â 2 3 â œ È" (3i 3j k) Ê 3(x 1) 3(y 1) (z 1) œ 0 or 3x 3y z œ Š 2È19 ‹ Š È" ‹ ˆ "7 ‰ â " 19 14 â â â 11 8 9 â œ 1 is an equation of the osculating plane. " ‰ 27. r œ et i (sin t)j ln (1 t)k Ê v œ et i (cos t)j ˆ 1 t k Ê v(0) œ i j k ; r(0) œ i Ê (1ß 0ß 0) is on the line Ê x œ 1 t, y œ t, and z œ t are parametric equations of the line 28. r œ ŠÈ2 cos t‹ i ŠÈ2 sin t‹ j tk Ê v œ ŠÈ2 sin t‹ i ŠÈ2 cos t‹ j k Ê v ˆ 14 ‰ œ ŠÈ2 sin 14 ‹ i ŠÈ2 cos 14 ‹ j k œ i j k is a vector tangent to the helix when t œ 14 Ê the tangent line is parallel to v ˆ 14 ‰ ; also r ˆ 14 ‰ œ ŠÈ2 cos 14 ‹ i ŠÈ2 sin 14 ‹ j 14 k Ê the point ˆ1ß 1ß 14 ‰ is on the line Ê x œ 1 t, y œ 1 t, and z œ 14 t are parametric equations of the line # # 29. x# œ av#! cos# !b t# and ˆy "# gt# ‰ œ av#! sin# !b t# Ê x# ˆy "# gt# ‰ œ v#! t# Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ ax x y yb# ÞÞ ÞÞ Þ Þ ÞÞ ÞÞ ÞÞ ÞÞ x x y y 30. s œ dtd Èx# y# œ È Þ # Þ # Ê x# y# s # œ x# y# xÞ # yÞ # x y ÞÞ ÞÞ Þ Þ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ ax# y# b ax# y# b ax# x# 2x x y y y# y# b ax y y x b# x# y# y# x # 2x x y y œ œ œ Þ# Þ# Þ# Þ# Þ Þ x y x y x# y# Þ ÞÞ Þ ÞÞ Þ # Þ # $Î# Þ Þ # # kx y y xk ax y b ÞÞ ÞÞ ÞÞ x y Ê È x# y# s # œ È Þ # Þ # Ê ÈÞÞ# ÞÞ# ÞÞ# œ kxÞ ÞÞy yÞ ÞÞxk œ ," œ 3 x y x y s 31. s œ a) Ê ) œ as Ê 9 œ as 1# Ê dds9 œ "a Ê , œ ¸ "a ¸ œ "a since a 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 13 Additional and Advanced Exercises y! OT 6380 32. (1) ?SOT ¸ ?TOD Ê DO OT œ SO Ê 6380 œ 6380437 # Ê y! œ 6380 6817 Ê y! ¸ 5971 km; (2) VA œ '5971 21x Ê1 Š dx dy ‹ dy # 6380 6380 œ 21'5971 È6380# y# Š È6380 # y# ‹ dy 6817 œ 21 '5971 6380 dy œ 21 c6380yd ')"( &*(" 6817 œ 16,395,469 km# ¸ 1.639 ‚ 10( km# ; # km (3) percentage visible ¸ 16,395,469 41(6380 km)# ¸ 3.21% CHAPTER 13 ADDITIONAL AND ADVANCED EXERCISES 1. (a) r()) œ (a cos ))i (a sin ))j b)k Ê ddtr œ [(a sin ))i (a cos ))j bk] ddt) ; kvk œ È2gz œ ¸ ddtr ¸ ) d) ¸ É a#2gb É a#41gbb# œ 2É a#1gbb# œ Èa# b# ddt) Ê ddt) œ É a#2gz b# œ b# Ê dt )œ#1 œ (b) ) d) É a#2gb b# dt œ Ê 2) "Î# d) "Î# Ê È œ É a#2gb œ É a#2gb b# dt Ê 2) b# t C; t œ 0 Ê ) œ 0 Ê C œ 0 ) œ É a#2gb b# t # # # gb t Ê ) œ 2 aagbt # b# b ; z œ b) Ê z œ 2 aa# b# b (c) v(t) œ ddtr œ [(a sin ))i (a cos ))j bk] ddt) œ [(a sin ))i (a cos ))j bk] Š a# gbt b# ‹ , from part (b) i (a cos ))j bk Ê v(t) œ ’ (a sin ))È T; “ Š È gbt ‹ œ È gbt # # # # # # a b a b a b # d# r ˆ d) ‰# [(a sin ))i (a cos ))j bk] ddt#) dt# œ [(a cos ))i (a sin ))j] dt # gb œ Š a# gbt b# ‹ [(a cos ))i (a sin ))j] [(a sin ))i (a cos ))j bk] Š a# b# ‹ # i (a cos ))j bk œ ’ (a sin ))È “ Š È #gb # ‹ a Š a# gbt # # b# ‹ [( cos ))i (sin ))j] a b a œ Èa#gb b# T a Š a# gbt b# ‹ # b N (there is no component in the direction of B). 2. (a) r()) œ (a) cos ))i (a) sin ))j b)k Ê ddtr œ [(a cos ) a) sin ))i (a sin ) a) cos ))j bk] ddt) ; kvk œ È2gz œ ¸ ddtr ¸ œ aa# a# )# b# b (b) s œ '0 kvk dt œ '0 aa# a# )# b# b t t "Î# È2gb) a a# ) # b# ˆ ddt) ‰ Ê ddt) œ È # "Î# d) dt dt œ '0t aa# a# )# b# b"Î# d) œ '0 aa# a# u# b# b"Î# du ) œ '0 aÉ a a# b u# du œ a '0 Èc# u# du, where c œ ) # ) # # È a# b # k ak ) Ê s œ a ’ u# Èc# u# c# ln ¹u Èc# u# ¹“ œ #a Š)Èc# )# c# ln ¹) Èc# )# ¹ c# ln c‹ ! e)r! (1 e)r! (e sin )) dr (1 e)r! (e sin )) dr 3. r œ 1(1e cos ) Ê d) œ (1 e cos ))# ; d) œ 0 Ê (1 e cos ))# œ 0 Ê (1 e)r! (e sin )) œ 0 Ê sin ) œ 0 Ê ) œ 0 or 1. Note that ddr) 0 when sin ) 0 and ddr) 0 when sin ) 0. Since sin ) 0 on e)r! 1 ) 0 and sin ) 0 on 0 ) 1, r is a minimum when ) œ 0 and r(0) œ 1(1e cos 0 œ r! 4. (a) f(x) œ x 1 "# sin x œ 0 Ê f(0) œ 1 and f(2) œ 2 1 "# sin 2 " # since ksin 2k Ÿ 1; since f is continuous on [0ß 2], the Intermediate Value Theorem implies there is a root between 0 and 2 (b) Root ¸ 1.4987011335179 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 791 792 Chapter 13 Vector-Valued Functions and Motion in Space . . . . . . . 5. (a) v œ x i y j and v œ r ur r ) u) œ arb [(cos ))i (sin ))j] ˆr )‰ [( sin ))i (cos ))j] Ê v † i œ x and . . . . . . . . v † i œ r cos ) r ) sin ) Ê x œ r cos ) r ) sin ); v † j œ y and v † j œ r sin ) r ) cos ) . . . Ê y œ r sin ) r ) cos ) . . (b) ur œ (cos ))i (sin ))j Ê v † ur œ x cos ) y sin ) . . . . œ ˆr cos ) r ) sin )‰ (cos )) ˆr sin ) r ) cos )‰ (sin )) by part (a), . . . . Ê v † ur œ r ; therefore, r œ x cos ) y sin ); . . u) œ (sin ))i (cos ))j Ê v † u) œ x sin ) y cos ) . . . . . œ ˆr cos ) r ) sin )‰ ( sin )) ˆr sin ) r ) cos )‰ (cos )) by part (a) Ê v † u) œ r ) ; . . . therefore, r ) œ x sin ) y cos ) # # # d) d r d) w ww ˆ d) ‰ f w ()) ddt#) ; v œ dr 6. r œ f()) Ê dr dt œ f ()) dt Ê dt# œ f ()) dt dt ur r dt u) "Î# "Î# d) ‰ dr d) ‰ ˆ ˆ dr ‰# r# ˆ ddt) ‰# “ œ ’af w b# f # “ œ ˆcos ) dr dt r sin ) dt i sin ) dt r cos ) dt j Ê kvk œ ’ dt Þ ÞÞ Þ ÞÞ d) dr kv ‚ ak œ kx y y xk , where x œ r cos ) and y œ r sin ). Then dx dt œ (r sin )) dt (cos )) dt # # # ˆ ddt) ‰ ; # d) dr ˆ d) ‰ (r sin )) ddt#) (cos )) ddt#r ; dy Ê ddt#x œ (2 sin )) ddt) dr dt (r cos )) dt dt œ (r cos )) dt (sin )) dt # # # # ˆ d) ‰ (r cos )) ddt#) (sin )) ddt#r . Then kv ‚ ak Ê ddt#y œ (2 cos )) ddt) dr dt (r sin )) dt $ # # # $ d) d r d) ˆ dr ‰ œ (after much algebra) r# ˆ ddt) ‰ r ddt#) dr œ ˆ ddt) ‰ Šf 2 f † f ww 2af w b2 ‹ dt r dt dt# 2 dt dt ww w 2 fb Ê , œ kvk‚vkak œ f wf†#f #2a$Î# 2 af b f ‘ # # d) d r d ) 7. (a) Let r œ 2 t and ) œ 3t Ê dr dt œ 1 and dt œ 3 Ê dt# œ dt# œ 0. The halfway point is (1ß 3) Ê t œ 1; # # # d) d r d) ˆ d) ‰ “ ur ’r ddt#) 2 dr v œ dr dt ur r dt u) Ê v(1) œ ur 3u) ; a œ ’ dt# r dt dt dt “ u) Ê a(1) œ 9ur 6u) (b) It takes the beetle 2 min to crawl to the origin Ê the rod has revolved 6 radians # # # Ê L œ '0 É[f())]# cf w ())d# d) œ '0 Éˆ2 3) ‰ ˆ 3" ‰ d) œ '0 É4 43) )9 9" d) 6 6 6 œ '0 É 37 129 ) ) d) œ 3" '0 È() 6)# 1 d) œ 3" ’ ()#6) È() 6)# 1 #" ln ¸) 6 È() 6)# 1¸“ 6 6 # œ È37 " ln ŠÈ37 6‹ ¸ 6.5 in. 6 8. (a) x œ r cos ) Ê dx œ cos ) dr r sin ) d); y œ r sin ) Ê dy œ sin ) dr r cos ) d); thus dx# œ cos# ) dr# 2r sin ) cos ) dr d) r# sin# ) d)# and dy# œ sin# ) dr# 2r sin ) cos ) dr d) r# cos# ) d)# Ê ds2 œ dx# dy# dz# œ dr# r# d)# dz# (b) (c) r œ e) Ê dr œ e) d) Ê L œ '0 ln 8 Èdr# r# d)# dz# œ '0 Èe#) e#) e#) d) ln 8 œ '0 È3e) d) œ ’È3 e) “ ln 8 ln 8 0 œ 8È3 È3 œ 7È3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. ' ! Chapter 13 Additional and Advanced Exercises â â i â 9. (a) ur ‚ u) œ â cos ) â â sin ) (b) j sin ) cos ) 793 â kâ â 0 â œ k Ê a right-handed frame of unit vectors â 0â du) dur d) œ ( sin ))i (cos ))j œ u) and d) œ ( cos ))i (sin ))j œ ur Þ ÞÞ Þ Þ Þ ÞÞ ÞÞ ÞÞ Þ Þ ÞÞ (c) From Eq. (7), v œ rur r)u) zk Ê a œ v œ a r ur r ur b ˆr )u) r) u) r) u) ‰ z k Þ# ÞÞ ÞÞ ÞÞ ÞÞ œ Š r r) ‹ ur ˆr) 2r )‰ u) z k # 10. L(t) œ r(t) ‚ mv(t) Ê ddtL œ ˆ ddtr ‚ mv‰ Šr ‚ m ddt#r ‹ Ê ddtL œ (v ‚ mv) (r ‚ ma) œ r ‚ ma ; F œ ma Ê krck$ r œ ma Ê ddtL œ r ‚ ma œ r ‚ Š krck$ r‹ œ krck$ (r ‚ r) œ 0 Ê L œ constant vector Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 794 Chapter 13 Vector-Valued Functions and Motion in Space NOTES: Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. CHAPTER 14 PARTIAL DERIVATIVES 14.1 FUNCTIONS OF SEVERAL VARIABLES 1. (a) fa0, 0b œ 0 (d) fa3, 2b œ 33 2. (a) fˆ2, 16 ‰ œ È3 2 (b) fa1, 1b œ 0 (c) fa2, 3b œ 58 1‰ (b) fˆ3, 12 œ È12 (c) fˆ1, 14 ‰ œ È12 (b) fˆ1, 12 , 14 ‰ œ 85 (c) fˆ0, 13 , 0‰ œ 3 (b) fa2, 3, 6b œ 0 (c) fa1, 2, 3b œ È35 (d) fˆ 12 , 7‰ œ 1 3. (a) fa3, 1, 2b œ 45 (d) fa2, 2, 100b œ 0 4. (a) fa0, 0, 0b œ 7 (d) fŠ È42 , È52 , È62 ‹ œ É 21 2 5. Domain: all points axß yb on or above the line yœx2 6. Domain: all points axß yb outside the circle x2 y2 œ 4 7. Domain: all points axß yb not liying on the graph of y œ x or y œ x3 8. Domain: all points axß yb not liying on the graph of x2 y2 œ 25 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 796 Chapter 14 Partial Derivatives 9. Domain: all points axß yb satisfying x2 1 Ÿ y Ÿ x2 1 10. Domain: all points axß yb satisfying ax 1bay 1b 0 11. Domain: all points axß yb satisfying ax 2bax 2bay 3bay 3b 0 12. Domain: all points axß yb inside the circle x2 y2 œ 4 such that x2 y2 Á 3 13. 14. 15. 16. 17. (a) Domain: all points in the xy-plane (b) Range: all real numbers Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.1 Functions of Several Variables (c) level curves are straight lines y x œ c parallel to the line y œ x (d) no boundary points (e) both open and closed (f) unbounded 18. (a) Domain: set of all axß yb so that y x 0 Ê y x (b) Range: z 0 (c) level curves are straight lines of the form y x œ c where c (d) boundary is Èy x œ 0 Ê y œ x, a straight line 0 (e) closed (f) unbounded 19. (a) Domain: all points in the xy-plane (b) Range: z 0 (c) level curves: for f(xß y) œ 0, the origin; for faxß yb œ c 0, ellipses with center a0ß 0b and major and minor axes along the x- and y-axes, respectively (d) no boundary points (e) both open and closed (f) unbounded 20. (a) Domain: all points in the xy-plane (b) Range: all real numbers (c) level curves: for faxß yb œ 0, the union of the lines y œ „ x; for faxß yb œ c Á 0, hyperbolas centered at a0ß 0b with foci on the x-axis if c 0 and on the y-axis if c 0 (d) no boundary points (e) both open and closed (f) unbounded 21. (a) Domain: all points in the xy-plane (b) Range: all real numbers (c) level curves are hyperbolas with the x- and y-axes as asymptotes when faxß yb Á 0, and the x- and y-axes when f(xß y) œ 0 (d) no boundary points (e) both open and closed (f) unbounded 22. (a) Domain: all axß yb Á a0ß yb (b) Range: all real numbers (c) level curves: for faxß yb œ 0, the x-axis minus the origin; for faxß yb œ c Á 0, the parabolas y œ c x# minus the origin (d) boundary is the line x œ 0 (e) open (f) unbounded 23. (a) Domain: all axß yb satisfying x# y# 16 (b) Range: z "4 (c) level curves are circles centered at the origin with radii r 4 (d) boundary is the circle x# y# œ 16 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 797 798 Chapter 14 Partial Derivatives (e) open (f) bounded 24. (a) Domain: all axß yb satisfying x# y# Ÿ 9 (b) Range: 0 Ÿ z Ÿ 3 (c) level curves are circles centered at the origin with radii r Ÿ 3 (d) boundary is the circle x# y# œ 9 (e) closed (f) bounded 25. (a) Domain: axß yb Á a0ß 0b (b) Range: all real numbers (c) level curves are circles with center a0ß 0b and radii r 0 (d) boundary is the single point a0ß 0b (e) open (f) unbounded 26. (a) Domain: all points in the xy-plane (b) Range: 0 z Ÿ 1 (c) level curves are the origin itself and the circles with center a0ß 0b and radii r 0 (d) no boundary points (e) both open and closed (f) unbounded 27. (a) Domain: all axß yb satisfying 1 Ÿ y x Ÿ 1 (b) Range: 1# Ÿ z Ÿ 1# (c) level curves are straight lines of the form y x œ c where 1 Ÿ c Ÿ 1 (d) boundary is the two straight lines y œ 1 x and y œ 1 x (e) closed (f) unbounded 28. (a) Domain: all axß yb, x Á 0 (b) Range: 1# z 1# (c) level curves are the straight lines of the form y œ c x, c any real number and x Á 0 (d) boundary is the line x œ 0 (e) open (f) unbounded 29. (a) Domain: all points axß yb outside the circle x# y# œ 1 (b) Range: all reals (c) Circles centered ar the origin with radii r 1 (d) Boundary: the cricle x# y# œ 1 (e) open (f) unbounded 30. (a) Domain: all points axß yb inside the circle x# y# œ 9 (b) Range: z ln 9 (c) Circles centered ar the origin with radii r 9 (d) Boundary: the cricle x# y# œ 9 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.1 Functions of Several Variables (e) open (f) bounded 31. f 32. e 33. a 34. c 35. d 36. b 37. (a) (b) 38. (a) (b) 39. (a) (b) 40. (a) (b) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 799 800 Chapter 14 Partial Derivatives 41. (a) (b) 42. (a) (b) 43. (a) (b) 44. (a) (b) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.1 Functions of Several Variables 45. (a) (b) 46. (a) (b) 47. (a) (b) 48. (a) (b) # # 801 49. faxß yb œ 16 x# y# and Š2È2ß È2‹ Ê z œ 16 Š2È2‹ ŠÈ2‹ œ 6 Ê 6 œ 16 x# y# Ê x# y# œ 10 50. faxß yb œ Èx# 1 and a1ß 0b Ê z œ È1# 1 œ 0 Ê x# 1 œ 0 Ê x œ 1 or x œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 802 Chapter 14 Partial Derivatives 51. faxß yb œ Èx y2 3 and a3, 1b Ê z œ É3 a1b2 3 œ 1 Ê x y2 3 œ 1 Ê x y2 œ 4 #a 1 b a 1 b 2y x x 52. faxß yb œ x 2y y 1 and a1ß 1b Ê z œ a1b 1 + 1 œ 3 Ê 3 œ x y 1 Ê y œ 4x 3 53. 54. 55. 56. 57. 58. 59. 60. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.1 Functions of Several Variables 803 61. faxß yß zb œ Èx y ln z at a3ß 1ß 1b Ê w œ Èx y ln z; at a3ß 1ß 1b Ê w œ È3 a1b ln 1 œ 2 Ê Èx y ln z œ 2 62. faxß yß zb œ ln ax# y z# b at a"ß #ß "b Ê w œ ln ax# y z# b ; at a"ß #ß "b Ê w œ ln a1 2 1b œ ln 4 Ê ln 4 œ ln ax# y z# b Ê x# y z# œ 4 63. gaxß yß zb œ Èx# y2 z# at Š1ß 1ß È2‹ Ê w œ Èx# y2 z# ; at Š1ß 1ß È2‹ Ê w œ Ê1# a1b2 ŠÈ2‹ # œ 2 Ê 2 œ Èx# y2 z# Ê x# y2 z# œ 4 1 0 a2b xyz xyz xyz 1 1 64. gaxß yß zb œ 2x y z at a1ß 0ß 2b Ê w œ 2x y z ; at a1ß 0ß 2b Ê w œ 2a1b 0 a2b œ 4 Ê 4 œ 2x y z Ê 2x y z œ 0 _ n 65. faxß yb œ ! Š xy ‹ œ n œ0 1 œ y y x for 1 Š xy ‹ ¹ xy ¹ 1 Ê Domain: all points ax, yb satisfying lxl lyl; at a1, 2b Ê since ¹ 12 ¹ 1 Ê z œ 2 2 1 œ 2 Ê y y x œ 2 Ê y œ 2x _ 66. gaxß yß zb œ ! n œ0 (x b y)n ÐxyÑÎz Ê Domain: all points ax, y, zb satisfying z Á 0; at aln 4ß ln 9ß 2b n! zn œ e Ê w œ eÐln 4 ln 9ÑÎ2 œ eÐln 36ÑÎ2 œ eln 6 œ 6 Ê 6 œ eÐxyÑÎz Ê x z y œ ln 6 67. faxß yb œ 'x È d) # œ sin1 y sin1 x Ê Domain: all points y 1) ax, yb satisfying 1 Ÿ x Ÿ 1 and 1 Ÿ y Ÿ 1; at a0, 1b Ê sin1 1 sin1 0 œ 12 Ê sin1 y sin1 x œ 12 . Since 12 Ÿ sin1 y Ÿ 12 and 12 Ÿ sin1 x Ÿ 12 , in order for sin1 y sin1 x to equal 12 , 0 Ÿ sin1 y Ÿ 12 and 12 Ÿ sin1 x Ÿ 0; that is 0 Ÿ y Ÿ 1 and 1 Ÿ x Ÿ 0. Thus y œ sinˆ 1 sin1 x‰ œ È1 x2 , x Ÿ 0 2 68. gaxß yß zb œ 'x 1 dt t# '0 È d) # œ tan1 y tan1 x sin1 ˆ 2z ‰ Ê Domain: all points ax, y, zb satisfying 2 Ÿ z Ÿ 2; y z 4) È at Š0ß 1ß È3‹ Ê tan1 1 tan1 0 sin1 Š 3 ‹ œ 71 Ê tan1 y tan1 x sin1 ˆ z ‰ œ 71 . Since 1 Ÿ sin1 ˆ z ‰ Ÿ 1 , 2 12 2 12 2 1 131 131 1 1 1 1 ‰ 1 1 1 ˆ 71 12 Ÿ tan y tan x Ÿ 12 Ê z œ 2 sin 12 tan y tan x , 12 Ÿ tan y tan x Ÿ 12 69-72. Example CAS commands: Maple: with( plots ); f := (x,y) -> x*sin(y/2) + y*sin(2*x); xdomain := x=0..5*Pi; ydomain := y=0..5*Pi; x0,y0 := 3*Pi,3*Pi; plot3d( f(x,y), xdomain, ydomain, axes=boxed, style=patch, shading=zhue, title="#69(a) (Section 14.1)" ); Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 2 2 804 Chapter 14 Partial Derivatives plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, orientation=[-90,0], title="#69(b) (Section 14.1)" ); # (b) L := evalf( f(x0,y0) ); # (c) plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, contours=[L], orientation=[-90,0], title="#45(c) (Section 13.1)" ); 73-76. Example CAS commands: Maple: eq := 4*ln(x^2+y^2+z^2)=1; implicitplot3d( eq, x=-2..2, y=-2..2, z=-2..2, grid=[30,30,30], axes=boxed, title="#73 (Section 14.1)" ); 77-80. Example CAS commands: Maple: x := (u,v) -> u*cos(v); y := (u,v) ->u*sin(v); z := (u,v) -> u; plot3d( [x(u,v),y(u,v),z(u,v)], u=0..2, v=0..2*Pi, axes=boxed, style=patchcontour, contours=[($0..4)/2], shading=zhue, title="#77 (Section 14.1)" ); 69-60. Example CAS commands: Mathematica: (assigned functions and bounds will vary) For 69 - 72, the command ContourPlot draws 2-dimensional contours that are z-level curves of surfaces z = f(x,y). Clear[x, y, f] f[x_, y_]:= x Sin[y/2] y Sin[2x] xmin= 0; xmax= 51; ymin= 0; ymax= 51; {x0, y0}={31, 31}; cp= ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading Ä False]; cp0= ContourPlot[[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, Contours Ä {f[x0,y0]}, ContourShading Ä False, PlotStyle Ä {RGBColor[1,0,0]}]; Show[cp, cp0] For 73 - 76, the command ContourPlot3D will be used. Write the function f[x, y, z] so that when it is equated to zero, it represents the level surface given. For 73, the problem associated with Log[0] can be avoided by rewriting the function as x2 + y2 +z2 - e1/4 Clear[x, y, z, f] f[x_, y_, z_]:= x2 y2 z2 Exp[1/4] ContourPlot3D[f[x, y, z], {x, 5, 5}, {y, 5, 5}, {z, 5, 5}, PlotPoints Ä {7, 7}]; For 77 - 80, the command ParametricPlot3D will be used. To get the z-level curves here, we solve x and y in terms of z and either u or v (v here), create a table of level curves, then plot that table. Clear[x, y, z, u, v] ParametricPlot3D[{u Cos[v], u Sin[v], u}, {u, 0, 2}, {v, 0, 2p}]; zlevel= Table[{z Cos[v], z sin[v]}, {z, 0, 2, .1}]; ParametricPlot[Evaluate[zlevel],{v, 0, 21}]; 14.2 LIMITS AND CONTINUITY IN HIGHER DIMENSIONS 3x# y# 5 lim # # Ðxß yÑ Ä Ð0ß 0Ñ x y 2 2. lim œ È œ0 4 Ðxß yÑ Ä Ð0ß 4Ñ Èy x # # 0 5 5 œ 3(0) 0# 0# 2 œ # 1. 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.2 Limits and Continuity in Higher Dimensions 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. lim Ðxß yÑ Ä Ð3ß 4Ñ Èx# y# 1 œ È3# 4# 1 œ È24 œ 2È6 # # # lim " Š x" y" ‹ œ #" ˆ "3 ‰‘ œ ˆ 6" ‰ œ 36 lim sec x tan y œ (sec 0) ˆtan 14 ‰ œ (1)(1) œ 1 lim cos Š xxy y 1 ‹ œ cos Š 000 0 1 ‹ œ cos 0 œ 1 Ðxß yÑ Ä Ð2ß 3Ñ Ðxß yÑ Ä ˆ0ß 14 ‰ Ðxß yÑ Ä Ð0ß 0Ñ lim # lim # exy œ e0 ln 2 œ eln ˆ 2 ‰ œ "# ln k1 x# y# k œ ln k1 (1)# (1)# k œ ln 2 ey sin x œ lim x Ðxß yÑ Ä Ð0ß 0Ñ Ðxß yÑ Ä Ð0ß 0Ñ lim lim $ 1 Ðxß yÑ Ä Ð0ß ln 2Ñ Ðxß yÑ Ä Ð1ß 1Ñ $ Ðxß yÑ Ä Ð1Î27ß 13 Ñ aey b ˆ sinx x ‰ œ e! † lim ˆ sinx x ‰ œ 1 † 1 œ 1 xÄ0 3 1 ‰ 3 3 xy œ cos É ˆ 27 cos È 1 œ cos ˆ 13 ‰ œ 21 1†sinˆ 16 ‰ 1Î2 x sin y 1 #1 œ x 1# 1 œ 2 œ 4 Ðxß yÑ Ä Ð1ß 1Î6Ñ lim lim acos 0b " cos y 1 11 y sin x œ 0 sin ˆ 1# ‰ œ 1 œ 2 lim x# 2xy y# (x y)# œ lim œ lim xy Ðx ß y Ñ Ä Ð 1 ß 1Ñ x y Ðxß yÑ Ä Ð1ß 1Ñ (x y) œ (" 1) œ 0 lim x# y# (x y)(x y) œ lim x y œ Ðxß yÑlim xy Ä Ð1ß 1Ñ Ðxß yÑ Ä Ð1ß 1Ñ (x y) œ (1 1) œ 2 lim xy y 2x 2 (x 1)(y 2) œ lim œ lim x1 x1 Ðx ß y Ñ Ä Ð 1 ß 1Ñ Ðxß yÑ Ä Ð1ß 1Ñ Ðxß yÑ Ä ˆ 12 ß 0‰ Ðxß yÑ Ä Ð1ß 1Ñ xÁy Ðxß yÑ Ä Ð1ß 1Ñ xÁy Ðxß yÑ Ä Ð1ß 1Ñ xÁ1 (y 2) œ (1 2) œ 1 xÁ1 y4 y4 " lim Ðxß yÑ Ä Ð0ß 0Ñ xÁy 1 ˆÈ x È y ‰ ˆ È x È y 2 ‰ x y 2È x 2È y œ lim œ lim Èx Èy Èx Èy Ðx ß y Ñ Ä Ð 0 ß 0Ñ Ðxß yÑ Ä Ð0ß 0Ñ ˆÈ x È y 2 ‰ xÁy œ ŠÈ0 È0 2‹ œ 2 Note: (xß y) must approach (0ß 0) through the first quadrant only with x Á y. 18. " lim œ lim œ lim œ #(2 1) œ # # # Ðxß yÑ Ä Ð2ß 4Ñ x y xy 4x 4x Ðxß yÑ Ä Ð2ß 4Ñ x(x 1)(y 4) Ðxß yÑ Ä Ð2ß 4Ñ x(x 1) y Á 4, x Á x# y Á 4, x Á x# x Á x# xy4 lim œ lim Ðxß yÑ Ä Ð2ß 2Ñ Èx y 2 Ðxß yÑ Ä Ð2ß 2Ñ xyÁ4 xyÁ4 ˆÈx y 2‰ ˆÈx y 2‰ œ Èx y 2 lim Ðxß yÑ Ä Ð2ß 2Ñ xyÁ4 ˆÈ x y 2 ‰ œ ŠÈ2 2 2‹ œ 2 2 œ 4 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 805 806 Chapter 14 Partial Derivatives 19. lim È2x y 2 È2x y 2 " œ lim 2x y 4 œ Ðxß yÑlim Ä Ð2ß 0Ñ ˆÈ2x y 2‰ ˆÈ2x y 2‰ Ðxß yÑ Ä Ð2ß 0Ñ È2x y # Ðxß yÑ Ä Ð2ß 0Ñ 2x y Á 4 2x y Á 4 œ È(2)(2)" 0 # œ 2 " 2 œ 4" 20. È x È y 1 Èx Èy 1 " œ lim x y 1 œ Ðxß yÑlim Ä Ð4ß 3Ñ ˆÈx Èy 1‰ ˆÈx Èy 1‰ Ðxß yÑ Ä Ð4ß 3Ñ Èx Èy 1 lim Ðxß yÑ Ä Ð4ß 3Ñ xyÁ1 xyÁ1 " œ È4 È œ 2 " 2 œ 4" 31 21. 22. lim # sinax# y# b a r# b œ lim sinra#r b œ lim 2r†cos œ lim x# y# 2r rÄ0 rÄ0 rÄ0 lim 1 cosaxyb œ lim 1 ucos u œ lim sin1 u œ 0 xy uÄ0 uÄ0 Ðxß yÑ Ä Ð0ß 0Ñ Ðxß yÑ Ä Ð0ß 0Ñ cosar# b œ 1 23. ax ybˆx2 xy y2 ‰ x3 y3 œ lim œ lim x y xy Ðxß yÑ Ä Ð1ß "Ñ Ðxß yÑ Ä Ð1ß "Ñ Ðxß yÑ Ä Ð1ß "Ñ 24. lim lim lim 4 4 œ 2 2 œ 2 2 œ a2 2ba22 22 b œ 32 Ðx ß y Ñ Ä Ð 2 ß 2 Ñ x y Ðxß yÑ Ä Ð2ß 2Ñ ax ybax ybax y b Ðxß yÑ Ä Ð2ß 2Ñ ax ybax y b 25. 26. 27. 28. 29. 30. lim xy lim T Ä Ð1ß 3ß 4Ñ lim lim xy 1 1 2xy yz 2(1)(1) (1)(1) œ 121" œ #" x # z# œ 1# (1)# asin# x cos# y sec# zb œ asin# 3 cos# 3b sec# 0 œ 1 1# œ 2 lim T Ä ˆ 14 ß 12 ß 2‰ tan" (xyz) œ tan" ˆ "4 † 1# † 2‰ œ tan" ˆ 14 ‰ lim ze 2y cos 2x œ 3e 2Ð0Ñ cos 21 œ (3)(1)(1) œ 3 lim ln Èx# y# z# œ ln È2# (3)# 6# œ ln È49 œ ln 7 T Ä Ð1ß 0ß 3Ñ 1 Š "x y" "z ‹ œ 1" 3" 4" œ 12 124 3 œ 19 12 T Ä Ð 1 ß 1 ß 1Ñ T Ä Ð3ß 3ß 0Ñ ax2 xy y2 b œ Š12 a1ba1b a1b2 ‹ œ 3 T Ä Ð2 ß 3 ß 6 Ñ 31. (a) All axß yb (b) All axß yb except a0ß 0b 32. (a) All axß yb so that x Á y (b) All axß yb 33. (a) All axß yb except where x œ 0 or y œ 0 (b) All axß yb 34. (a) All axß yb so that x# 3x 2 Á 0 Ê ax 2bax 1b Á 0 Ê x Á 2 and x Á 1 (b) All axß yb so that y Á x# 35. (a) All axß yß zb (b) All axß yß zb except the interior of the cylinder x# y# œ 1 36. (a) All axß yß zb so that xyz 0 (b) All axß yß zb 37. (a) All axß yß zb with z Á 0 (b) All axß yß zb with x# z# Á 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.2 Limits and Continuity in Higher Dimensions 38. (a) All axß yß zb except axß 0ß 0b (b) All axß yß zb except a0ß yß 0b or axß 0ß 0b 39. (a) All axß yß zb such that z x2 y2 1 (b) All axß yß zb such that z Á Èx2 y2 807 40. (a) All axß yß zb such that x2 y2 z2 Ÿ 4 (b) All axß yß zb such that x2 y2 z2 9 except when x2 y2 z2 œ 25 41. lim Èx#x y# œ lim b Èx#x x# œ lim b È2x kxk œ lim b Èx2 x œ lim b È"2 œ È"2 ; xÄ0 xÄ0 xÄ0 xÄ0 lim Èx#x y# œ lim c È2x kxk œ lim c È2(xx) œ lim c Ðxß yÑ Ä Ð0ß 0Ñ along y œ x x0 Ðxß yÑ Ä Ð0ß 0Ñ along y œ x x0 42. lim Ðxß yÑ Ä Ð0ß 0Ñ along y œ 0 xÄ0 xÄ0 % x% œ lim x% x 0# œ 1; lim x% y# xÄ0 Ðxß yÑ Ä Ð0ß 0Ñ along y œ x# % " x% x% œ lim x% xax# b# œ lim 2x % œ # x% y# xÄ0 xÄ0 # 43. 44. 45. 46. 47. 48. 49. 50. " " È2 œ È2 xÄ0 lim % % akx# b x% y# k# x% k# œ lim xx% œ lim xx% œ 11 k# x% y# k# x% akx# b# xÄ0 xÄ0 lim # xy x(kx) k œ lim kkx ; if k 0, the limit is 1; but if k 0, the limit is 1 # œ lim kxyk œ xlim Ä 0 kx(kx)k x Ä 0 kx k x Ä 0 kkk lim xy x kx k œ 11 x y œ xlim k Ä 0 x kx lim x2 y x2 kx k œ lim 1x œ 1kk x y œ xlim Ä 0 x kx x Ä 0 k lim # x# y kx# œ lim x kx œ 1 k k # y xÄ0 lim x# y kx4 k 4 2 4 œ 1 k2 x4 y2 œ xlim Ä 0 x k x lim xy2 1 y2 1 œ lim ay 1b œ 2; lim y 1 œ ylim Ä 1 y1 yÄ1 Ðx ß y Ñ Ä Ð 1 ß 1Ñ Ðxß yÑ Ä Ð0ß 0Ñ along y œ kx# Ðxß yÑ Ä Ð0ß 0Ñ along y œ kx kÁ0 Ðx ß y Ñ Ä Ð 0 ß 0 Ñ along y œ kx k Á 1 Ðxß yÑ Ä Ð0ß 0Ñ along y œ kx kÁ1 Ðxß yÑ Ä Ð0ß 0Ñ along y œ kx# kÁ0 Ðxß yÑ Ä Ð0ß 0Ñ along y œ kx# Ðxß yÑ Ä Ð1ß 1Ñ along x œ 1 lim Ðxß yÑ Ä Ð1ß 1Ñ along y œ 1 Ê different limits for different values of k Ê different limits for different values of k, k Á 1 Ê different limits for different values of k, k Á 1 Ê different limits for different values of k, k Á 0 Ê different limits for different values of k along y œ x xy2 1 y3 1 œ lim ay2 y 1b œ 3 y 1 œ ylim yÄ1 Ä 1 y1 xy 1 x 1 œ lim x11 œ 21 ; lim 2 x2 y2 œ xlim Ä 1 x 1 xÄ1 Ð x ß y Ñ Ä Ð 1 ß 1Ñ along y œ x2 xy 1 x 3 1 x2 x 1 2 x4 œ lim a x x 1bax2 1b x2 y2 œ xlim xÄ1 Ä1 œ 32 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 808 Chapter 14 Partial Derivatives Ú 1 if y x% 51. fax, yb œ Û 1 if y Ÿ 0 Ü 0 otherwise (a) (b) (c) x% lim fax, yb œ 1 since any path through a0, 1b that is close to a0, 1b satisfies y lim fax, yb œ 0 since any path through a2, 3b that is close to a2, 3b does not satisfiy either y lim fax, yb œ 1 and Ðxß yÑ Ä Ð0ß 1Ñ Ðx ß y Ñ Ä Ð 0 ß 0 Ñ along x œ 0 x2 52. fax, yb œ œ 3 x (a) lim fax, yb œ 0 Ê lim Ðx ß y Ñ Ä Ð 0 ß 0Ñ along y œ x2 fax, yb does not exist lim Ðxß yÑ Ä Ð0ß 0Ñ if x 0 if x 0 fax, yb œ 32 œ 9 since any path through a3, 2b that is close to a3, 2b satisfies x 0 Ðxß yÑ Ä Ð3ß 2Ñ (b) (c) x% or y Ÿ 0 Ðxß yÑ Ä Ð2ß 3Ñ lim fax, yb œ a2b3 œ 8 since any path through a2, 1b that is close to a2, 1b satisfies x 0 lim fax, yb œ 0 since the limit is 0 along any path through a0, 0b with x 0 and the limit is also zero along Ðxß yÑ Ä Ð2ß 1Ñ Ðxß yÑ Ä Ð0ß 0Ñ any path through a0, 0b with x 0 53. First consider the vertical line x œ 0 Ê 2 2 a0 b y 2x2 y œ lim 4 y2 œ lim 4 x y Ä 0 a0b y2 yÄ0 Ðxß yÑ Ä Ð0ß 0Ñ lim 0 œ 0. Now consider any nonvertical along x œ 0 through a0, 0b. The equation of any line through a0, 0b is of the form y œ mx Ê lim Ðxß yÑ Ä Ð0ß 0Ñ along y œ mx faxß yb œ 2x2 y 4 y2 x Ðxß yÑ Ä Ð0ß 0Ñ lim along y œ mx 2 3 b 2mx3 2mx œ lim x42x aamx œ lim x4 2mx lim 2 2 œ lim x2 ax2 m2 b œ lim ax2 m2 b œ 0. Thus mxb2 xÄ0 x Ä 0 m x xÄ0 xÄ0 Ðxß yÑ Ä Ð0ß 0Ñ 2x2 y x4 y2 œ 0. any line though a0, 0b 54. If f is continuous at (x! ß y! ), then lim Ðxß yÑ Ä Ðx! ß y! Ñ f(xß y) must equal f(x! ß y! ) œ 3. If f is not continuous at (x! ß y! ), the limit could have any value different from 3, and need not even exist. 55. # # lim Ðxß yÑ Ä Ð0ß 0Ñ Š1 x 3y ‹ œ 1 and lim Ðx ß y Ñ Ä Ð 0 ß 0Ñ 1œ1 Ê # # 56. If xy 0, lim 2 kxyk Š x 6y ‹ Ðx ß y Ñ Ä Ð 0 ß 0Ñ kxyk tan " xy œ 1, by the Sandwich Theorem xy lim Ðxß yÑ Ä Ð0ß 0Ñ # # œ 2xy Š x 6y ‹ lim Ðxß yÑ Ä Ð0ß 0Ñ xy œ lim Ðxß yÑ Ä Ð0ß 0Ñ ˆ2 xy ‰ 6 œ 2 and # # 2 kxyk œ lim Ðxß yÑ Ä Ð0ß 0Ñ kxyk Ðxß yÑ Ä Ð0ß 0Ñ lim œ lim Ðx ß y Ñ Ä Ð 0 ß 0 Ñ ˆ2 xy ‰ 6 œ 2 and 2 œ 2; if xy 0, lim Ðx ß y Ñ Ä Ð 0 ß 0Ñ lim 2 kxyk Š x 6y ‹ kxyk Ðxß yÑ Ä Ð0ß 0Ñ 2 kxyk kxyk œ 2 Ê lim Ðxß yÑ Ä Ð0ß 0Ñ # # œ lim 2xy Š x 6y ‹ xy Ðxß yÑ Ä Ð0ß 0Ñ 4 4 cos Èkxyk œ 2, by the Sandwich Theorem kxyk 57. The limit is 0 since ¸sin ˆ "x ‰¸ Ÿ 1 Ê 1 Ÿ sin ˆ x" ‰ Ÿ 1 Ê y Ÿ y sin ˆ x" ‰ Ÿ y for y 0, and y y sin ˆ "x ‰ y Ÿ 0. Thus as (xß y) Ä (!ß !), both y and y approach 0 Ê y sin ˆ "x ‰ Ä 0, by the Sandwich Theorem. 58. The limit is 0 since ¹cos Š "y ‹¹ Ÿ 1 Ê 1 Ÿ cos Š y" ‹ Ÿ 1 Ê x Ÿ x cos Š y" ‹ Ÿ x for x 0, and x x cos Š y" ‹ for x Ÿ 0. Thus as (xß y) Ä (!ß !), both x and x approach 0 Ê x cos Š "y ‹ Ä 0, by the Sandwich Theorem. ) 59. (a) f(xß y)k yœmx œ 1 2mm# œ 1 2tan tan# ) œ sin 2). The value of f(xß y) œ sin 2) varies with ), which is the line's angle of inclination. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. y for x Section 14.2 Limits and Continuity in Higher Dimensions (b) Since f(xß y)k yœmx œ sin 2) and since 1 Ÿ sin 2) Ÿ 1 for every ), lim Ðxß yÑ Ä Ð0ß 0Ñ 809 f(xß y) varies from 1 to 1 along y œ mx. 60. kxy ax# y# bk œ kxyk kx# y# k Ÿ kxk kyk kx# y# k œ Èx# Èy# kx# y# k Ÿ Èx# y# Èx# y# kx# y# k # # # # # # # # œ ax# y# b Ê ¹ xyxa#xy#y b ¹ Ÿ axx#yy#b œ x# y# Ê ax# y# b Ÿ xyxa#xy#y b Ÿ ax# y# b Ê 61. 62. 63. # lim Ðxß yÑ Ä Ð0ß 0Ñ # y Šxy xx# y# ‹ œ 0 by the Sandwich Theorem, since x$ xy# œ lim lim # # Ðxß yÑ Ä Ð0ß 0Ñ x y rÄ0 $ lim Ðx ß y Ñ Ä Ð 0 ß 0Ñ „ ax# y# b œ 0; thus, define fa0ß 0b œ 0 $ ) sin# )b r$ cos$ ) (r cos )) ar# sin# )b œ lim r acos ) cos œ0 r# cos# ) r# sin# ) 1 rÄ0 $ $ $ lim r acos ) sin )b y r cos ) r sin ) cos Š xx# “ œ cos 0 œ 1 y# ‹ œ lim cos Š r# cos# ) r# sin# ) ‹ œ lim cos ’ 1 lim y# r# sin# ) œ lim x# y# œ r lim r# Ä0 rÄ0 Ðxß yÑ Ä Ð0ß 0Ñ Ðxß yÑ Ä Ð0ß 0Ñ $ $ $ $ rÄ0 rÄ0 asin# )b œ sin# ); the limit does not exist since sin# ) is between 0 and 1 depending on ) 64. 65. 2r cos ) 2 cos ) 2 cos ) lim œ lim r cos ) œ cos ) ; the limit does not exist for cos ) œ 0 # # œ lim # r Ä 0 r r cos ) rÄ0 Ðxß yÑ Ä Ð0ß 0Ñ x x y 2x lim Ðxß yÑ Ä Ð0ß 0Ñ ky k krk akcos )k ksin )kb " kr cos )k kr sin )k tan" ’ kxx#k ’ “ œ lim tan" ’ “; y# “ œ lim tan r# r# rÄ0 if r Ä 0 , then rÄ0 lim b tan" ’ krk akcos )rk# ksin )kb “ œ rÄ! lim r Ä !b tan" ’ kcos )k r ksin )k “ œ 1# ; if r Ä 0 , then lim tan" ’ krk akcos )rk# ksin )kb “ œ lim c tan" Š kcos )kr ksin )k ‹ œ 1# Ê the limit is 1# rÄ! r Ä !c 66. x# y# lim # # œ lim rÄ0 Ðxß yÑ Ä Ð0ß 0Ñ x y r# cos# ) r# sin# ) œ lim r# rÄ0 acos# ) sin# )b œ lim (cos 2)) which ranges between rÄ0 1 and 1 depending on ) Ê the limit does not exist 67. lim Ðx ß y Ñ Ä Ð 0 ß 0Ñ # # # # ln Š 3x x#xy y# 3y ‹ œ lim ln Š 3r cos ) r cos r#) sin ) 3r sin ) ‹ # # % # # # # rÄ0 œ lim ln a3 r# cos# ) sin# )b œ ln 3 Ê define f(0ß 0) œ ln 3 rÄ0 68. lim Ðxß yÑ Ä Ð0ß 0Ñ (3r cos )) ar# sin# )b 3xy# œ lim x# y# œ r lim r# Ä0 rÄ0 3r cos ) sin# ) œ 0 Ê define f(0ß 0) œ 0 69. Let $ œ 0.1. Then Èx# y# $ Ê Èx# y# 0.1 Ê x# y# 0.01 Ê kx# y# 0k 0.01 Ê kf(xß y) f(!ß !)k 0.01 œ %. 70. Let $ œ 0.05. Then kxk $ and kyk $ Ê kfaxß yb fa0ß 0bk œ ¸ x# y 1 0¸ œ ¸ x# y 1 ¸ Ÿ kyk 0.05 œ %. 71. Let $ œ 0.005. Then kxk $ and kyk $ Ê kfaxß yb fa0ß 0bk œ ¸ xx#y1 0¸ œ ¸ xx#y1 ¸ Ÿ kx yk kxk kyk 0.005 0.005 œ 0.01 œ %. kx yk " 72. Let $ œ 0.01. Since 1 Ÿ cos x Ÿ 1 Ê 1 Ÿ 2 cos x Ÿ 3 Ê "3 Ÿ #cos Ÿ ¸ 2 x cosy x ¸ Ÿ kx yk x Ÿ 1 Ê 3 Ÿ kxk kyk . Then kxk $ and kyk $ Ê kfaxß yb fa0ß 0bk œ ¸ 2 x cosy x 0¸ œ ¸ 2 x cosy x ¸ Ÿ kxk kyk 0.01 0.01 œ 0.02 œ %. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 810 Chapter 14 Partial Derivatives 73. Let $ œ 0.04. Since y2 Ÿ x2 y2 Ê x2 y y2 Ÿ 1 Ê xl2xlyy2 Ÿ lxl œ Èx2 Ÿ Èx2 y2 $ Ê kfaxß yb fa0ß 0bk 2 2 2 œ ¹ x2xy y2 0¹ 0.04 œ %. 74. Let $ œ 0.01. If lyl Ÿ 1, then y2 Ÿ lyl œ Èy2 Ÿ Èx2 y2 , so lxl œ Èx2 Ÿ Èx2 y2 Ê lxl y2 Ÿ 2Èx2 y2 . Since y l y x 2 2 x2 Ÿ x2 y2 Ê x2 x y2 Ÿ 1 and y2 Ÿ x2 y2 Ê x2 y y2 Ÿ 1. Then lxx2 y2 Ÿ x2 y2 lxl x2 y2 y Ÿ lxl y 2$ 3 2 2 4 2 2 y Ê kfaxß yb fa0ß 0bk œ ¹ xx2 y2 0¹ 2a0.01b œ 0.002 œ % . 3 4 75. Let $ œ È0.015. Then Èx# y# z# $ Ê kf(xß yß z) f(!ß 0ß 0)k œ kx# y# z# 0k œ kx# y# z# k # # œ ŠÈx# t# x# ‹ ŠÈ0.015‹ œ 0.015 œ %. 76. Let $ œ 0.2. Then kxk $ , kyk $ , and kzk $ Ê kf(xß yß z) f(!ß 0ß 0)k œ kxyz 0k œ kxyzk œ kxk kyk kzk (0.2)$ œ 0.008 œ %. 77. Let $ œ 0.005. Then kxk $ , kyk $ , and kzk $ Ê kf(xß yß z) f(!ß 0ß 0)k œ ¹ x# x y# yz#z 1 0¹ œ ¹ x# x y# yz#z 1 ¹ Ÿ kx y zk Ÿ kxk kyk kzk 0.005 0.005 0.005 œ 0.015 œ %. 78. Let $ œ tan" (0.1). Then kxk $ , kyk $ , and kzk $ Ê kf(xß yß z) f(!ß 0ß 0)k œ ktan# x tan# y tan# zk Ÿ ktan# xk ktan# yk ktan# zk œ tan# x tan# y tan# z tan# $ tan# $ tan# $ œ 0.01 0.01 0.01 œ 0.03 œ %. 79. lim Ðxß yß zÑ Ä Ðx! ß y! ß z! Ñ f(xß yß z) œ lim (x y z) œ x! y! z! œ f(x! ß y! ß z! ) Ê f is continuous at lim ax# y# z# b œ x!# y!# z!# œ f(x! ß y! ß z! ) Ê f is continuous at Ðxß yß zÑ Ä Ðx! ß y! ß z! Ñ every (x! ß y! ß z! ) 80. lim Ðxß yß zÑ Ä Ðx! ß y! ß z! Ñ f(xß yß z) œ Ðxß yß zÑ Ä Ðx! ß y! ß z! Ñ every point (x! ß y! ß z! ) 14.3 PARTIAL DERIVATIVES 1. `f `f ` x œ 4x, ` y œ 3 2. `f `f ` x œ 2x y, ` y œ x 2y 3. `f `f # ` x œ 2x(y 2), ` y œ x 1 4. `f `f ` x œ 5y 14x 3, ` y œ 5x 2y 6 5. `f `f ` x œ 2y(xy 1), ` y œ 2x(xy 1) 6. `f # `f # ` x œ 6(2x 3y) , ` y œ 9(2x 3y) 7. y `f x `f ` x œ È x# y# , ` y œ È x# y# 8. `f 2x# , ` f œ $ $" y `x œ É $ $ x ˆ y ‰ `y 3É x ˆ ‰ 9. # # `f " ` " `f " ` " ` x œ (x y)# † ` x (x y) œ (x y)# , ` y œ (x y)# † ` y (x y) œ (x y)# # # # # # # 10. `` xf œ ax ayx# b(1)y#b#x(2x) œ axy# yx# b# , `` yf œ ax ayx#b(0)y#b#x(2y) œ ax# 2xy y # b# # (x y)(y) y 1 (xy ")(1) (x y)(x) `f x " 11. `` xf œ œ (xy 1)(1) œ (xy œ (xy (xy 1)# 1)# , ` y œ (xy 1)# 1)# # Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.3 Partial Derivatives 12. `` xf œ y y " ` ˆy‰ `f " ` ˆy‰ 1 x # † `x x œ x# ’1 ˆ y ‰# “ œ x# y# , ` y œ 1 ˆ y ‰# † ` y x œ x ’1 ˆ y ‰# “ œ x# y# 1 ˆ xy ‰ x x x 13. `` xf œ eÐxy1Ñ † ``x (x y 1) œ eÐxy1Ñ , `` yf œ eÐxy1Ñ † ``y (x y 1) œ eÐxy1Ñ 14. `` xf œ ex sin (x y) ex cos (x y), `` yf œ ex cos (x y) 15. `` xf œ x " y † ``x (x y) œ x " y , `` yf œ x " y † ``y (x y) œ x " y 16. `` xf œ exy † ``x (xy) † ln y œ yexy ln y, `` yf œ exy † ``y (xy) † ln y exy † "y œ xexy ln y ey xy 17. `` xf œ 2 sin (x 3y) † ``x sin (x 3y) œ 2 sin (x 3y) cos (x 3y) † ``x (x 3y) œ 2 sin (x 3y) cos (x 3y), `f ` ` ` y œ 2 sin (x 3y) † ` y sin (x 3y) œ 2 sin (x 3y) cos (x 3y) † ` y (x 3y) œ 6 sin (x 3y) cos (x 3y) 18. `` xf œ 2 cos a3x y# b † ``x cos a3x y# b œ 2 cos a3x y# b sin a3x y# b † ``x a3x y# b œ 6 cos a3x y# b sin a3x y# b , `f ` ` # # # # # ` y œ 2 cos a3x y b † ` y cos a3x y b œ 2 cos a3x y b sin a3x y b † ` y a3x y b œ 4y cos a3x y# b sin a3x y# b 19. `` xf œ yxyc1 , `` yf œ xy ln x x `f " `f ln x 20. f(xß y) œ ln ln y Ê ` x œ x ln y and ` y œ y(ln y)# 21. `` xf œ g(x), `` yf œ g(y) _ 22. f(xß y) œ ! (xy)n , kxyk 1 Ê f(xß y) œ 1 " xy Ê `` xf œ (1 "xy)# † ``x (1 xy) œ (1 yxy)# and n œ0 `f " ` x ` y œ (1 xy)# † ` y (1 xy) œ (1 xy)# 23. fx œ y# , fy œ 2xy, fz œ 4z 24. fx œ y z, fy œ x z, fz œ y x 25. fx œ 1, fy œ Èy#y z# , fz œ Èy#z z# 26. fx œ x ax# y# z# b $Î# , fy œ y ax# y# z# b $Î# , fz œ z ax# y# z# b $Î# 27. fx œ È1 yzx# y# z# , fy œ È1 xzx# y# z# , fz œ È1 xyx# y# z# " 28. fx œ kx yzk È(x , fy œ kx yzk È(xz yz)# 1 , fz œ kx yzk È(xy yz)# 1 yz)# 1 29. fx œ x 2y" 3z , fy œ x 2y2 3z , fz œ x 2y3 3z yz yz " ` ` 30. fx œ yz † xy † ``x (xy) œ (yz)(y) xy œ x , fy œ z ln (xy) yz † ` y ln (xy) œ z ln (xy) xy † ` y (xy) œ z ln (xy) z, fz œ y ln (xy) yz † ``z ln (xy) œ y ln (xy) # # # # # # # # # 31. fx œ 2xe ax y z b , fy œ 2ye ax y z b , fz œ 2ze ax y z b 32. fx œ yzexyz , fy œ xzexyz , fz œ xyexyz Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 811 812 Chapter 14 Partial Derivatives 33. fx œ sech# (x 2y 3z), fy œ 2 sech# (x 2y 3z), fz œ 3 sech# (x 2y 3z) 34. fx œ y cosh axy z# b , fy œ x cosh axy z# b , fz œ 2z cosh axy z# b 35. ``ft œ 21 sin (21t !), ``!f œ sin (21t !) Ð2uÎvÑ ` g Ð2uÎvÑ ‰ ‰ 36. `` gu œ v# eÐ2uÎvÑ † `ù ˆ 2u , ` v œ 2veÐ2uÎvÑ v# eÐ2uÎvÑ † ``v ˆ 2u 2ueÐ2uÎvÑ v œ 2ve v œ 2ve 37. `` 3h œ sin 9 cos ), ``9h œ 3 cos 9 cos ), `` h) œ 3 sin 9 sin ) 38. ``gr œ 1 cos ), `` g) œ r sin ), `` gz œ 1 # # # 2V$ v V$ v V$ v 39. Wp œ V, Wv œ P $2gv , W$ œ Vv 2g , Wv œ 2g œ g , Wg œ #g# A h 40. `Àc œ m, `Àh œ #q , `Àk œ mq , `` m œ qk c, `Àq œ km q# # # # # # 41. `` xf œ 1 y, `` yf œ 1 x, `` xf# œ 0, `` yf# œ 0, ``y`fx œ ``x`fy œ 1 # # # # 42. `` xf œ y cos xy, `` yf œ x cos xy, `` xf# œ y# sin xy, `` yf# œ x# sin xy, ``y`fx œ ``x`fy œ cos xy xy sin xy # # # # 43. `` gx œ 2xy y cos x, `` yg œ x# sin y sin x, `` xg# œ 2y y sin x, `` yg# œ cos y, ``y`gx œ ``x`gy œ 2x cos x # # # # 44. `` hx œ ey , `` hy œ xey 1, `` xh# œ 0, `` yh# œ xey , ``y`hx œ ``x`hy œ ey # # # # ` r " ` r ` r " 45. ``xr œ x" y , ``yr œ x" y , ``xr# œ (x" y)# , ` y# œ (xy)# , ` y` x œ ` x` y œ (xy)# 46. ``xs œ ” y " ` ˆy‰ " `s " ` ˆy‰ " x ˆ y‰ ˆ1‰ # • † `x x œ x # ” 1 ˆ y ‰# • œ x # y # , ` y œ ” 1 ˆ y ‰ # • † ` y x œ x ” 1 ˆ y ‰ # • œ x # y # , 1 ˆ xy ‰ x x x # # ax# y# b (1) y(2y) y(2x) 2xy 2xy x(2y) ` #s ` #s ` #s ` #s œ axy# yx# b# ` x# œ ax# y# b# œ ax# y# b# , ` y# œ ax# y# b# œ ax# y# b# , ` y` x œ ` x` y œ ax # y # b # 47. ``wx œ 2x tanaxyb x2 sec2 axyb † y œ 2x tanaxyb x2 y sec2 axyb, ``wy œ x2 sec2 axyb † x œ x3 sec2 axyb, ` #w 2 2 2 ` x# œ 2tanaxyb 2x sec axyb † y 2xy sec axyb x y a2secaxybsecaxyb tanaxyb † yb # œ 2tanaxyb 4xy sec2 axyb 2x2 y2 sec2 axyb tanaxyb, `` yw# œ x3 a2secaxybsecaxyb tanaxyb † xb œ 2x4 sec2 axyb tanaxyb ` #w ` #w 2 2 3 2 2 3 2 ` y` x œ ` x` y œ 3x sec axyb x a2secaxybsecaxyb tanaxyb † yb œ 3x sec axyb x y sec axyb tanaxyb 48. ``wx œ yex y † 2x œ 2xy ex y , ``wy œ a1bex y yex y † a1b œ ex y a1 yb , 2 2 2 2 2 # 2 2 2 2 ` #w x2 y 2xyŠex y † 2x‹ œ 2yex y a1 2x2 b, `` yw# œ Šex y † a1b‹a1 yb ex y a1b ` x# œ 2y e # # œ ex y ay 2b, ``y`wx œ ``x`wy œ Šex y † 2x‹a1 yb œ 2x ex y a1 yb 2 2 2 49. ``wx œ sinax2 yb x cosax2 yb † 2xy œ sinax2 yb 2x2 ycosax2 yb, ``wy œ x cosax2 yb † x2 œ x3 cosax2 yb, ` #w 2 2 2 2 2 3 2 2 ` x# œ cosax yb † 2xy 4xy cosax yb 2x y sinax yb † 2xy œ 6xy cosax yb 4x y sinax yb, ` #w ` #w ` #w 3 2 2 5 2 2 2 3 2 2 2 4 2 ` y# œ x sinax yb † x œ x sinax yb, ` y` x œ ` x` y œ 3x cosax yb x sinax yb † 2xy œ 3x cosax yb 2x y sinax yb Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.3 Partial Derivatives 50. ``wx œ 2 2 ˆx2 y‰ ax yba2xb ˆx2 y‰a1b ax yb y `w œ xax2 2xy , `y œ œ ax2xybx2 , y b2 ax2 yb2 ax2 yb2 2 ˆx y‰ a2x 2yb ˆx 2xy y‰2ˆx y‰a2xb 2ˆx 3x y 3 xy y ‰ ` #w œ , 2 ` x# œ 2 ax 2 y b 3 ’ax2 yb “ 2 2 2 2 3 2 2 2 2 ˆx y‰ † 0 ˆx x‰2ˆx y‰† 1 # # ˆx y‰ a2x 1b ˆx 2xy y‰2ˆx y‰† 1 ` #w œ a2xx2 y2xb3 , ``y`wx œ ``x`wy œ 2 2 ` y# œ 2 2 ’ax2 yb “ ’ax2 yb “ 2 2 2 2 2 2 2xy y œ 2x a3x x2 yb3 3 2 # # 6 ` w 6 51. ``wx œ 2x 2 3y , ``wy œ 2x 3 3y , ``y`wx œ (2x 3y)# , and ` x` y œ (2x 3y)# # # 52. ``wx œ ex ln y yx , ``wy œ yx ln x, ``y`wx œ œ "y x" , and ``x`wy œ "y "x # # 53. ``wx œ y# 2xy$ 3x# y% , ``wy œ 2xy 3x# y# 4x$ y$ , ``y`wx œ 2y 6xy# 12x# y$ , and ``x`wy œ 2y 6xy# 12x# y$ # # 54. ``wx œ sin y y cos x y, ``wy œ x cos y sin x x, ``y`wx œ cos y cos x 1, and ``x`wy œ cos y cos x 1 55. (a) x first (b) y first 56. (a) y first three times 57. fx a1ß 2b œ lim hÄ0 œ lim hÄ0 hÄ0 (d) x first (b) y first three times (c) y first twice (e) y first (f) y first (d) x first twice # # f(1 hß 2) f(1ß 2) œ lim c1 (1 h) 2 h6(1 h) d (2 6) œ lim h 6 a1 h2h h b 6 h hÄ0 hÄ0 13h 6h# œ lim h hÄ0 fy (1ß 2) œ lim (c) x first (13 6h) œ 13, f(1ß 2 h) f(1ß 2) (2 6) œ lim c1 1 (2 h) h3(2 h)d (2 6) œ lim (2 6 2h) h h hÄ0 hÄ0 œ lim (2) œ 2 hÄ0 58. fx a2ß 1b œ lim hÄ0 œ lim hÄ0 fa2 hß 1b fa2ß 1b œ lim c4 2a2 hb 3 ha2 hbd a3 2b h hÄ0 a2h 1 hb 1 œ lim h hÄ0 1 œ 1, 4 4 3a1 hb 2a1 hb# ‘ a3 2b fa2ß 1 hb fa2ß 1b œ lim h h hÄ0 hÄ0 a3 3h 2 4h 2h# b 1 h 2h# œ lim œ lim œ lim a1 2hb œ 1 h h hÄ0 hÄ0 hÄ0 fy a2ß 1b œ lim 59. fx a2ß 3b œ lim hÄ0 È 2 a 2 h b 9 1 È 4 9 1 fa2 hß 3b fa2ß 3b œ lim h h hÄ0 È2h 4 2 œ lim h hÄ0 hÄ0 œ lim fy a2ß 3b œ lim hÄ0 œ lim hÄ0 Š È2h 4 2 È2h 4 2 2 œ 12 , È2h 4 2 ‹ œ hlim h Ä 0 È2h 4 2 È 4 3 a3 h b 1 È 4 9 1 fa2ß 3 hb fa2ß 3b œ lim h h hÄ0 È3h 4 2 œ lim h hÄ0 Š È3h 4 2 È3h 4 2 3 œ 34 È3h 4 2 ‹ œ hlim h Ä 0 È2h 4 2 fa0 hß 0b fa0ß 0b œ lim h hÄ0 hÄ0 60. fx a0ß 0b œ lim fa0ß 0 hb fa0ß 0b œ lim h hÄ0 hÄ0 fy a0ß 0b œ lim sinŠh3 b 0‹ 0 h2 b 0 h sinŠ0 b h4 ‹ 0 0 b h2 h œ lim sin h3 h3 œ 1 œ lim 4 sin h4 Šh † sinh4h ‹ œ 0 † 1 œ 0 h3 œ hlim Ä0 hÄ0 hÄ0 61. (a) In the plane x œ 2 Ê fy axß yb œ 3 Ê fy a2ß 1b œ 3 Ê m œ 3 (b) In the plane y œ 1 Ê fx axß yb œ 2 Ê fy a2ß 1b œ 2 Ê m œ 2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 813 814 Chapter 14 Partial Derivatives 62. (a) In the plane x œ 1 Ê fy axß yb œ 3y2 Ê fy a1ß 1b œ 3a1b2 œ 3 Ê m œ 3 (b) In the plane y œ 1 Ê fx axß yb œ 2x Ê fy a1ß 1b œ 2a1b œ 2 Ê m œ 2 63. fz ax! ß y! ß z! b œ lim hÄ0 fz a1ß 2ß 3b œ lim hÄ0 fax! ß y! ß z! hb fax! , y! ß z! b ; h # # fa1ß 2ß 3 hb fa1, 2ß 3b œ lim 2a3 hbh 2a9b œ lim "2h h 2h œ lim a12 2hb œ 12 h hÄ0 hÄ0 hÄ0 64. fy ax! ß y! ß z! b œ lim hÄ0 fax! ß y! hß z! b fax! , y! ß z! b ; h # fa1ß hß 3b fa1, 0ß 3b œ lim a2h h9hb 0 œ lim h hÄ0 hÄ0 hÄ0 fy a1ß 0ß 3b œ lim a2h 9b œ 9 65. y ˆ3z# `` xz ‰ x z$ 2y `` xz œ 0 Ê a3xz# 2yb `` xz œ y z$ Ê at (1ß 1ß 1) we have (3 2) `` xz œ 1 1 or `` xz œ 2 66. ˆ `` xz ‰ z x ˆ yx ‰ `` xz 2x `` xz œ 0 Ê ˆz xy 2x‰ `` xz œ x Ê at (1ß 1ß 3) we have (3 1 2) `` xz œ 1 or `` xz œ "6 À a 67. a# œ b# c# 2bc cos A Ê 2a œ a2bc sin Ab `Àa Ê `Àa œ bc sin A ; also 0 œ 2b 2c cos A a2bc sin Ab ` b cos A b Ê 2c cos A 2b œ a2bc sin Ab `Àb Ê `Àb œ c bc sin A (sin A) ` a a cos A cos A À 68. sina A œ sinb B Ê œ 0 Ê asin Ab `` xa a cos A œ 0 Ê `Àa œ asin sin# A A ; also " ` a ` a ˆ sin A ‰ ` B œ ba csc B cot Bb Ê ` B œ b csc B cot B sin A 69. Differentiating each equation implicitly gives 1 œ vx ln u ˆ vu ‰ ux and 0 œ ux ln v ˆ uv ‰ vx or " v º 0 lnu v º aln ub vx ˆ vu ‰ ux œ 1 Ê œ œ aln ubalnlnvvb 1 v v x ln u ˆ uv ‰ vx aln vb ux œ 0 Ÿ u º u º ln v v 70. Differentiating each equation implicitly gives 1 œ a2xbxu a2ybyu and 0 œ a2xbxu yu or " 2y 1 º œ 2x1 4xy œ 2x 1 4xy 2y º 2x 1 º0 a2xbxu a2ybyu œ 1 Ê xu œ 2x a2xbxu yu œ 0 º " 0º yu œ #x 4xy œ 2x2x4xy œ 2x 2x4xy œ 1 1 2y ; next s œ x# y# 2x º 2x and Ê `ùs œ 2x `` ux 2y `` uy 2y œ 2x Š 2x " 4xy ‹ 2y Š 1 " 2y ‹ œ 1 " #y 1 2y2y œ 11 2y 71. fx axß yb œ œ 0 0 if y 0 Ê fx axß yb œ 0 for all points ax, yb; at y œ 0, fy axß 0b œ lim fax, 0 hhb fax, 0b œ lim fax, hhb 0 if y 0 h Ä0 h Ä0 œ lim fax,h hb œ 0 because h Ä0 lim fax,h hb œ hÄ0c 3 lim h œ 0 and h Ä0 c h limb fax,h hb œ h Ä0 2 limb hh œ 0 Ê fy axß yb œ œ h Ä0 3y2 2y if y 0 ; if y 0 fyx axß yb œ fxy axß yb œ 0 for all points ax, yb 72. At x œ 0, fx a0ß yb œ lim fa0 h, yhb fa0, yb œ lim fah, yhb 0 œ lim fah,h yb which does not exist because h Ä0 œ 2 limc hh œ 0 and hÄ0 fy axß yb œ œ limb fah,h yb œ h Ä0 h Ä0 È limb hh œ h Ä0 hÄ0 1 lim 1 œ _ Ê fx axß yb œ 2Èx h Ä0 b È h 2x if x 0 if x 0 lim fah,h yb hÄ0c ; 0 if x 0 Ê fy axß yb œ 0 for all points ax, yb; fyx axß yb œ 0 for all points ax, yb, while fxy axß yb œ 0 for all 0 if x 0 points ax, yb such that x Á 0. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.3 Partial Derivatives # # # # # 815 # 73. `` xf œ 2x, `` yf œ 2y, `` zf œ 4z Ê `` xf# œ 2, `` yf# œ 2, `` zf# œ 4 Ê `` xf# `` yf# `` zf# œ 2 2 (4) œ 0 # # # # # # 74. `` xf œ 6xz, `` yf œ 6yz, `` zf œ 6z# 3 ax# y# b , `` xf# œ 6z, `` yf# œ 6z, `` zf# œ 12z Ê `` xf# `` yf# `` zf# œ 6z 6z 12z œ 0 75. `` xf œ 2ec2y sin 2x, `` yf œ 2ec2y cos 2x, `` xf# œ 4ec2y cos 2x, `` yf# œ 4ec2y cos 2x Ê `` xf# `` yf# # # # # œ 4ec2y cos 2x 4ec2y cos 2x œ 0 # # # # # # # # 76. `` xf œ x# x y# , `` yf œ x# y y# , `` xf# œ axy# yx# b# , `` yf# œ axx# yy# b# Ê `` xf# `` yf# œ axy# yx# b# axx# yy# b# œ 0 # # # # # # # # 77. `` xf œ 3 , `` yf œ 2 , `` xf# œ 0 , `` yf# œ 0 Ê `` xf# `` yf# œ 0 0 œ 0 78. `` xf œ 1 Îy # 1 Š xy ‹ œ y# y x# , `` yf œ x Îy 2 # # # # œ y#xx# , `` xf# œ ay ayx# b†0x#b2y†2x œ ay#2xy , ` f œ ay axy#b†0 x#ab2 xb†2y œ ay# 2xy x# b2 ` y# x # b2 # # 1 Š xy ‹ # Ê `` xf# `` yf# œ ay#2xy ay# 2xy œ0 x# b2 x# b2 # # 79. `` xf œ "# ax# y# z# b $Î# $Î# ` f $Î# , ` y œ "# ax# y# z# b a2yb $Î# $Î# # # # # a2xb œ x ax# y# z# b $Î# ` f , ` z œ "# ax# y# z b a2zb œ z ax y z b ; # &Î# ` f # # # $Î# # # # # &Î# ` # f # # # $Î# 3x ax y z b , ` y # œ ax y z b 3y# ax# y# z# b , ` x # œ ax y z b # # # # $Î# &Î# ` f # # # 3z# ax# y# z# b Ê `` xf# `` yf# `` zf# ` z # œ ax y z b œ y ax# y# z# b $Î# 3x# ax# y# z# b $Î# 3z# ax# y# z# b œ ’ ax# y# z# b ’ ax# y# z# b &Î# &Î# $Î# 3y# ax# y# z# b $Î# a3x# 3y# 3z# b ax# y# z# b “ ’ ax# y# z# b “ œ 3 ax# y# z# b &Î# # “ &Î# œ0 # 80. `` xf œ 3e3x4y cos 5z, `` yf œ 4e3x4y cos 5z, `` zf œ 5e3x4y sin 5z; `` xf# œ 9e3x4y cos 5z, `` yf# œ 16e3x4y cos 5z, ` #f 3x4y cos 5z ` z# œ 25e # # # Ê `` xf# `` yf# `` zf# œ 9e3x4y cos 5z 16e3x4y cos 5z 25e3x4y cos 5z œ 0 # # # # 81. ``wx œ cos (x ct), ``wt œ c cos (x ct); `` xw# œ sin (x ct), `` tw# œ c# sin (x ct) Ê `` tw# œ c# [ sin (x ct)] œ c# `` xw# # # 82. ``wx œ 2 sin (2x 2ct), ``wt œ 2c sin (2x 2ct); `` xw# œ 4 cos (2x 2ct), `` tw# œ 4c# cos (2x 2ct) # # Ê `` tw# œ c# [4 cos (2x 2ct)] œ c# `` xw# 83. ``wx œ cos (x ct) 2 sin (2x 2ct), ``wt œ c cos (x ct) 2c sin (2x 2ct); ` #w ` #w # # ` x# œ sin (x ct) 4 cos (2x 2ct), ` t# œ c sin (x ct) 4c cos (2x 2ct) # # Ê `` tw# œ c# [ sin (x ct) 4 cos (2x 2ct)] œ c# `` xw# # # # # # 1 ` w c ` w " # # ` w 84. ``wx œ x " ct , ``wt œ x c ct ; `` xw# œ (x ct)# , ` t# œ (x ct)# Ê ` t# œ c ’ (x ct)# “ œ c ` x# # 85. ``wx œ 2 sec# (2x 2ct), ``wt œ 2c sec# (2x 2ct); `` xw# œ 8 sec# (2x 2ct) tan (2x 2ct), ` #w # # ` t# œ 8c sec (2x 2ct) tan (2x 2ct) # # Ê ux `` tw# œ c# [8 sec# (2x 2ct) tan (2x 2ct)] œ c# `` xw# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 816 Chapter 14 Partial Derivatives 86. ``wx œ 15 sin (3x 3ct) exbct , ``wt œ 15c sin (3x 3ct) cexbct ; `` xw# œ 45 cos (3x 3ct) exbct , # ` #w # # xbct ` t# œ 45c cos (3x 3ct) c e Ê `` tw# œ c# c45 cos (3x 3ct) exbct d œ c# `` xw# # # # # # # # 87. ``wt œ `` uf `ùt œ `` uf (ac) Ê `` tw# œ (ac) Š `` uf# ‹ (ac) œ a# c# `` uf# ; ``wx œ `` uf `` ux œ `` uf † a Ê `` xw# œ Ša `` uf# ‹ † a # # # # # œ a# `` uf# Ê `` tw# œ a# c# `` uf# œ c# Ša# `` uf# ‹ œ c# `` xw# 88. If the first partial derivatives are continuous throughout an open region R, then by Theorem 3 in this section of the text, f(xß y) œ f(x! ß y! ) fx (x! ß y! ) ?x fy (x! ß y! ) ?y %" ?x %# ?y, where %" , %# Ä 0 as ?x, ?y Ä 0. Then as (xß y) Ä (x! ß y! ), ?x Ä 0 and ?y Ä 0 Ê lim f(xß y) œ f(x! ß y! ) Ê f is continuous at every point (x! ß y! ) in R. Ðxß yÑ Ä Ðx! ß y! Ñ 89. Yes, since fxx , fyy , fxy , and fyx are all continuous on R, use the same reasoning as in Exercise 76 with fx (xß y) œ fx (x! ß y! ) fxx (x! ß y! ) ?x fxy (x! ß y! ) ?y %" ?x %# ?y and fy (xß y) œ fy (x! ß y! ) fyx (x! ß y! ) ?x fyy (x! ß y! ) ?y s%" ?x s%# ?y. Then lim fx (xß y) œ fx (x! ß y! ) Ðxß yÑ Ä Ðx! ß y! Ñ and lim Ðxß yÑ Ä Ðx! ß y! Ñ fy (xß y) œ fy (x! ß y! ). 90. To find ! and " so that ut œ uxx Ê ut œ " sina! xbe" t and ux œ ! cosa! xbe" t Ê uxx œ !2 sina! xbe" t ; then ut œ uxx Ê " sina! xbe" t œ !2 sina! xbe" t , thus ut œ uxx only if " œ !2 h†02 91. fx a0, 0b œ lim fa0 hß 0hb fa0ß 0b œ lim h2 0h4 h Ä0 0 0†h2 œ lim 0h œ 0; fy a0, 0b œ lim fa0ß 0 hhb fa0ß 0b œ lim 02 hh4 hÄ0 hÄ0 hÄ0 ˆky2 ‰y2 ky4 k lim f ax, yb œ lim aky2 b2 y4 œ lim k2 y4 y4 œ lim k2 1 œ k2 k 1 Ê yÄ0 yÄ0 yÄ0 Ðxß yÑ Ä Ð0ß 0Ñ hÄ0 0 œ lim 0h œ 0; hÄ0 different limits for different along x œ ky2 values of k Ê lim Ðxß yÑ Ä Ð0ß 0Ñ f ax, yb does not exist Ê f ax, yb is not continuous at a0, 0b Ê by Theorem 4, f ax, yb is not differentiable at a0, 0b. 92. fx a0, 0b œ lim fa0 hß 0hb fa0ß 0b œ lim fahß 0hb 1 œ lim 1 h 1 œ 0; fy a0, 0b œ lim fa0ß 0 hhb fa0ß 0b œ lim fa0ß hhb 1 œ lim 1 h 1 œ 0; h Ä0 lim Ðxß yÑ Ä Ð0ß 0Ñ along y œ x2 h Ä0 f ax, yb œ lim 0 œ 0 but yÄ0 h Ä0 lim Ðxß yÑ Ä Ð0ß 0Ñ along y œ 1.5x2 h Ä0 f ax, yb œ lim 1 œ 1 Ê yÄ0 h Ä0 lim Ðxß yÑ Ä Ð0ß 0Ñ h Ä0 f ax, yb does not exist Ê f ax, yb is not continuous at a0, 0b Ê by Theorem 4, f ax, yb is not differentiable at a0, 0b. 14.4 THE CHAIN RULE 1. (a) (b) 2. (a) dy `w `w dx dw ` x œ 2x, ` y œ 2y, dt œ sin t, dt œ cos t Ê dt œ 2x sin t 2y cos t œ 2 cos t sin t 2 sin t cos t œ 0; w œ x# y# œ cos# t sin# t œ 1 Ê dw dt œ 0 dw dt (1) œ 0 dy `w `w dx ` x œ 2x, ` y œ 2y, dt œ sin t cos t, dt œ sin t cos t Ê dw dt œ (2x)( sin t cos t) (2y)( sin t cos t) œ 2(cos t sin t)(cos t sin t) 2(cos t sin t)(sin t cos t) œ a2 cos# t 2 sin# tb a2 cos# t 2 sin# tb œ 0; w œ x# y# œ (cos t sin t)# (cos t sin t)# œ 2 cos# t 2 sin# t œ 2 Ê dw dt œ 0 (b) dw dt (0) œ 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.4 The Chain Rule 3. (a) (x y) dx `w " `w " `w dz " , dt œ 2 cos t sin t, dy `x œ z , `y œ z , `z œ z# dt œ 2 sin t cos t, dt œ t# x y y 2 2 cos# t sin# t x cos# t sin# t Ê dw dt œ z cos t sin t z sin t cos t z# t# œ Š " ‹ at# b œ 1; w œ z z œ Š " ‹ Š " ‹ œ t t# (b) 4. (a) (b) 5. (a) t t Ê dw dt œ 1 dw dt (3) œ 1 2y dy `w 2x `w `w 2z dx dz "Î# ` x œ x# y# z# , ` y œ x# y# z# , ` z œ x# y# z# , dt œ sin t, dt œ cos t, dt œ 2t 2 cos t sin t 2 sin t cos t 4 ˆ4t"Î# ‰ t "Î# 2y cos t 2x sin t 4zt "Î# Ê dw dt œ x# y# z# x# y# z# x# y# z# œ cos# t sin# t 16t 16 16 # # # # # œ 116t ; w œ ln ax y z b œ ln acos t sin t 16tb œ ln (1 16t) Ê dw dt œ 1 16t dw 16 dt (3) œ 49 dy `w " dx 2t " dz x `w x `w t ` x œ 2ye , ` y œ 2e , ` z œ z , dt œ t# 1 , dt œ t# 1 , dt œ e " # # x x t 4yte 2e e Ê dw dt œ t# 1 t# 1 z œ (4t) atant# tb1at 1b 2 at#t 11b eet œ 4t tan" t 1; w œ 2yex ln z œ a2 tan" tb at# 1b t " ˆ 2 ‰ # Ê dw tb (2t) 1 œ 4t tan" t 1 dt œ t# 1 at 1b a2 tan (b) 6. (a) (b) 7. (a) t dw ˆ1‰ dt (1) œ (4)(1) 4 1 œ 1 1 dy x cos xy `w `w `w dx " dz tc1 Ê dw etc1 ` x œ y cos xy, ` y œ x cos xy, ` z œ 1, dt œ 1, dt œ t , dt œ e dt œ y cos xy t œ (ln t)[cos (t ln t)] t cos (tt ln t) etc1 œ (ln t)[cos (t ln t)] cos (t ln t) etc1 ; w œ z sin xy tc1 œ etc1 sin (t ln t) Ê dw [cos (t ln t)] ln t t ˆ "t ‰‘ œ etc1 (1 ln t) cos (t ln t) dt œ e dw dt (1) œ 1 (1 0)(1) œ 0 x 4ex ln y `z `z `x `z `y 4ex x ˆ cos v ‰ 4e ysin v ` u œ ` x ` u ` y ` u œ a4e ln yb u cos v Š y ‹ (sin v) œ u v)(sin v) œ 4(u cos v)uln (u sin v) 4(u cos œ (4 cos v) ln (u sin v) 4 cos v; u sin v `z `z `x `z `y 4ex 4ex u cos v x x ˆ u sin v ‰ ` v œ ` x ` v ` y ` v œ a4e ln yb u cos v Š y ‹ (u cos v) œ a4e ln yb (tan v) y # v)(u cos v) cos v œ [4(u cos v) ln (u sin v)](tan v) 4(u cosu sin œ (4u sin v) ln (u sin v) 4usin v v ; `z sin x z œ 4e ln y œ 4(u cos v) ln (u sin v) Ê ` u œ (4 cos v) ln (u sin v) 4(u cos v) ˆ u sinvv ‰ v‰ œ (4 cos v) ln (u sin v) 4 cos v; also `` vz œ (4u sin v) ln (u sin v) 4(u cos v) ˆ uu cos sin v # cos v œ (4u sin v) ln (u sin v) 4usin v (b) At ˆ2ß 14 ‰ : `` uz œ 4 cos 14 ln ˆ2 sin 14 ‰ 4 cos 14 œ 2È2 ln È2 2È2 œ È2 (ln 2 2); (4)(2) ˆcos# 14 ‰ `z 1 1‰ ˆ œ 4È2 ln È2 4È2 œ 2È2 ln 2 4È2 ˆsin 1 ‰ ` v œ (4)(2) sin 4 ln 2 sin 4 4 8. (a) `z ù œ – `z `v œ – Š "y ‹ # Š xy ‹ Š #x ‹ — cos v – Š x ‹# 1 — sin v œ x# y# x# y# œ 1 y cos v y x sin v (u sin v)(cos v) (u cos v)(sin v) œ 0; u# y Š"‹ Š #x ‹ — (u sin v) – Š x ‹# 1 — u cos v œ x# y# x# y# œ Šx‹ 1 y y # y yu sin v xu cos v (u sin v)(u sin v) (u cos v)(u cos v) u# y " # ‰ œ sin# v cos# v œ 1; z œ tan" Š xy ‹ œ tan" (cot v) Ê `` uz œ 0 and `` vz œ ˆ 1 cot # v a csc vb œ sin# v" cos# v œ 1 (b) At ˆ1.3ß 16 ‰ : `` uz œ 0 and `` vz œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 817 818 9. (a) Chapter 14 Partial Derivatives `w `w `x `w `y `w `z ` u œ ` x ` u ` y ` u ` z ` u œ (y z)(1) (x z)(1) (y x)(v) œ x y 2z v(y x) œ (u v) (u v) 2uv v(2u) œ 2u 4uv; ``wv œ ``wx `` xv ``wy `` vy ``wz `` vz # œ (y z)(1) (x z)(1) (y x)(u) œ y x (y x)u œ 2v (2u)u œ 2v 2u ; w œ xy yz xz œ au# v# b au# v uv# b au# v uv# b œ u# v# 2u# v Ê ``wu œ 2u 4uv and (b) 10. (a) `w # ` v œ 2v 2u # At ˆ "# ß 1‰ : ``wu œ 2 ˆ "# ‰ 4 ˆ "# ‰ (1) œ 3 and ``wv œ 2(1) 2 ˆ "# ‰ œ #3 2y `w 2x 2z v v v v v ` u œ Š x# y# z# ‹ ae sin u ue cos ub Š x# y# z# ‹ ae cos u ue sin ub Š x# y# z# ‹ ae b v u ‰ aev sin u uev cos ub œ ˆ u# e2v sin# u 2ueu# esin 2v cos# u u# e2v v cos u v ‰ v ˆ u# e2v sin# u 2ue u# e2v cos# u u# e2v ae cos u ue sin ub v ‰ aev b œ 2u ; ˆ u# e2v sin# u u2ue # e2v cos# u u# e2v 2y `w 2x 2z v v v ` v œ Š x# y# z# ‹ aue sin ub Š x# y# z# ‹ aue cos ub Š x# y# z# ‹ aue b v u ‰ auev sin ub œ ˆ u# e2v sin# u 2ueu# esin 2v cos# u u# e2v v cos u ‰ v ˆ u# e2v sin# u 2ue u# e2v cos# u u# e2v aue cos ub ‰ auev b œ 2; w œ ln au# e2v sin# u u# e2v cos# u u# e2v b œ ln a2u# e2v b ˆ u# e2v sin# u u2ue # e2v cos# u u# e2v v œ ln 2 2 ln u 2v Ê ``wu œ 2u and ``wv œ 2 2 (b) At a2ß 0b: ``wu œ # œ 1 and ``wv œ 2 11. (a) rp pq qrrppq ù ù `p ù `q ù `r " œ 0; ` x œ ` p ` x ` q ` x ` r ` x œ q r (q r)# (q r)# œ (q r)# rp pq qrrppq 2p 2r ù ù `p ù `q ù `r " œ (q ` y œ ` p ` y ` q ` y ` r ` y œ q r (q r)# (q r)# œ (q r)# r)# (2x 2y 2z) (2x 2y 2z) z ù ù `p ù `q ù `r œ œ (z y)# ; ` z œ ` p ` z ` q ` z ` r ` z (2z 2y)# rp pq ppq 2p 4y y " œ q r (q r)# (q r)# œ q r (qr œ 2q r)# (q r)# œ (2z 2y)# œ (z y)# ; y (z y) y(1) y(1) ù ù u œ pq qr œ 2z 2y œ (z z y)# , and `` uz œ (z (zy)(0) 2y œ z y Ê ` x œ 0, ` y œ (z y)# y)# œ (zyy)# 2 (b) At ŠÈ3ß 2ß 1‹ : `` ux œ 0, `` uy œ (1 " 2)# œ 1, and `` uz œ (1 2)# œ 2 12. (a) z ln y ù eqr eqr cos x qr " pb (0) aqeqr sin" pb (0) œ È œ Èe cos#x œ yz if 1# x 1# ; ` x œ È1 p# (cos x) are sin 1 p# 1 sin x # " # " qr # z ˆz‰y x ù eqr qr " pb Š zy ‹ aqeqr sin" pb (0) œ z re ysin p œ œ xzyz1 ; ` y œ È1 p# (0) are sin y z qr "p ù eqr qr " pb (2z ln y) aqeqr sin" pb ˆ z"# ‰ œ a2zreqr sin" pb (ln y) qe sin ` z œ È1 p# (0) are sin z# # z œ (2z) ˆ "z ‰ ayz x ln yb az ln yz#b ay b x œ xyz ln y; u œ ez ln y sin" (sin x) œ xyz if 1# Ÿ x Ÿ 1# Ê `` ux œ yz , ù z1 , and `` uz œ ` y œ xzy œ xyz ln y from direct calculations (b) At ˆ 14 ß "# ß "# ‰ : `` xu œ ˆ "# ‰ "Î# œ È2, `` yu œ ˆ 14 ‰ ˆ "# ‰ ˆ "# ‰ Ð"Î#Ñ" È œ 1 4 2 , `` uz œ ˆ 14 ‰ ˆ "# ‰ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. "Î# ln ˆ "# ‰ œ 1 È2 ln 2 4 Section 14.4 The Chain Rule ` z dx ` z dy 13. dz dt œ ` x dt ` y dt ` z du ` z dv ` x dw 14. dz dt œ ` u dt ` v dt ` w dt 15. ``wu œ ``wx `` xu ``wy `` yu ``wz `` uz `w `w `x `w `y `w `z `v œ `x `v `y `v `z `v 16. ``wx œ ``wr ``xr ``ws ``xs ``wt ``xt `w `w `r `w `s `w `t `y œ `r `y `s `y `t `y 17. ``wu œ ``wx `` xu ``wy `` yu `w `w `x `w `y `v œ `x `v `y `v Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 819 820 Chapter 14 Partial Derivatives 18. ``wx œ ``wu `` ux ``wv `` vx `w `w ù `w `v `y œ ù `y `v `y 19. ``zt œ `` xz ``xt `` yz ``yt `z `z `x `z `y `s œ `x `s `y `s dy ` u 20. ``yr œ du `r ù 21. ``ws œ dw du ` s `w dw ` u ` t œ du ` t 22. ``wp œ ``wx `` px ``wy `` py ``wz `` pz ``wv `` vp dy ` w dy ` w dx 23. ``wr œ ``wx dx dr ` y dr œ ` x dr since dr œ 0 `w ` w dx ` w dy ` w dy dx ` s œ ` x ds ` y ds œ ` y ds since ds œ 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.4 The Chain Rule 821 25. Let F(xß y) œ x$ 2y# xy œ 0 Ê Jx (xß y) œ 3x# y 24. ``ws œ ``wx ``xs ``wy ``ys # 3x y Fx and Fy (xß y) œ 4y x Ê dy dx œ Fy œ (4y x) 4 Ê dy dx (1ß 1) œ 3 y3 Fx 26. Let F(xß y) œ xy y# 3x 3 œ 0 Ê Fx (xß y) œ y 3 and Fy (xß y) œ x 2y Ê dy dx œ Fy œ x 2y Ê dy dx (1ß 1) œ 2 2x y Fx 27. Let F(xß y) œ x# xy y# 7 œ 0 Ê Fx (xß y) œ 2x y and Fy (xß y) œ x 2y Ê dy dx œ Fy œ x 2y 4 Ê dy dx (1ß 2) œ 5 28. Let F(xß y) œ xey sin xy y ln 2 œ 0 Ê Fx (xß y) œ ey y cos xy and Fy (xß y) œ xey x sin xy 1 e y cos xy dy Fx Ê dy dx œ Fy œ xey x sin xy 1 Ê dx (!ß ln 2) œ (2 ln 2) y 29. Let F(xß yß z) œ z$ xy yz y$ 2 œ 0 Ê Fx (xß yß z) œ y, Fy (xß yß z) œ x z 3y# , Fz (xß yß z) œ 3z# y # Ê `` xz œ FFxz œ 3z# y y œ 3z#y y Ê `` xz (1ß 1ß 1) œ "4 ; `` yz œ Fyz œ x3z#zy3y œ x 3zz# 3y y F # Ê `` yz (1ß 1ß 1) œ 34 30. Let F(xß yß z) œ "x y" "z 1 œ 0 Ê Fx (xß yß z) œ x"# , Fy (xß yß z) œ y"# , Fz (xß yß z) œ z"# Ê `` xz œ FFxz œ Š x"# ‹ Š z"# ‹ # œ xz# Ê `` xz (2ß 3ß 6) œ 9; `` yz œ Fyz œ F Š y"# ‹ Š z"# ‹ # œ yz# Ê `` yz (2ß 3ß 6) œ 4 31. Let F(xß yß z) œ sin (x y) sin (y z) sin (x z) œ 0 Ê Fx (xß yß z) œ cos (x y) cos (x z), Fy (xß yß z) œ cos (x y) cos (y z), Fz (xß yß z) œ cos (y z) cos (x z) Ê `` xz œ FFxz (x y) cos (x z) cos (x y) cos (y z) y `z `z `z œ cos cos (y z) cos (x z) Ê ` x (1ß 1ß 1) œ 1; ` y œ Fz œ cos (y z) cos (x z) Ê ` y (1ß 1ß 1) œ 1 F 32. Let F(xß yß z) œ xey yez 2 ln x 2 3 ln 2 œ 0 Ê Fx (xß yß z) œ ey 2x , Fy (xß yß z) œ xey ez , Fz (xß yß z) œ yez Ê `` xz œ FFxz œ êy 2x ‰ yez Ê `` xz (1ß ln 2ß ln 3) œ 3 ln4 2 ; `` yz œ Fyz œ xeyez e Ê `` yz (1ß ln 2ß ln 3) œ 3 ln5 2 F y z 33. ``wr œ ``wx ``xr ``wy ``yr ``wz ``zr œ 2(x y z)(1) 2(x y z)[ sin (r s)] 2(x y z)[cos (r s)] œ 2(x y z)[1 sin (r s) cos (r s)] œ 2[r s cos (r s) sin (r s)][1 sin (r s) cos (r s)] Ê ``wr ¸ rœ1ßsœ1 œ 2(3)(2) œ 12 `w ¸ ‰ ˆ"‰ ˆ 2v ‰ v ˆ4‰ ˆ4‰ 34. ``wv œ ``wx `` xv ``wy `` vy ``wz `` vz œ y ˆ 2v u x(1) z (0) œ (u v) u u Ê ` v uœ1ßvœ2 œ (1) 1 1 œ 8 # " 35. ``wv œ ``wx `` xv ``wy `` yv œ ˆ2x xy# ‰ (2) ˆ "x ‰ (1) œ ’2(u 2v 1) (u2u2vv1)2 # “ (2) u 2v 1 Ê ``wv ¸ uœ0ßvœ0 œ 7 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 822 Chapter 14 Partial Derivatives 36. `` uz œ `` xz `` xu `` yz `` uy œ (y cos xy sin y)(2u) (x cos xy x cos y)(v) œ cuv cos au$ v uv$ b sin uvd (2u) cau# v# b cos au$ v uv$ b au# v# b cos uvd (v) Ê `` uz ¸ uœ0ßvœ1 œ 0 (cos 0 cos 0)(1) œ 2 dz ` x 5 `z ¸ 5 u ˆ 5 ‰ u 37. `` uz œ dx ` u œ 1 x# e œ ’ 1 aeu ln vb# “ e Ê ` u uœln 2ßvœ1 œ ’ 1 (2)# “ (2) œ 2; `z dz ` x 5 ˆ 5 ‰ ˆ "v ‰ œ ’ “ ˆ "v ‰ ` v œ dx ` v œ 1 x# 1 aeu ln vb# È Ê `` vz ¸ uœln 2ßvœ1 œ ’ 1 5(2)# “ (1) œ 1 È v3 v3 dz ` q 2 " " " `z ¸ " 38. `` uz œ dq ` u œ Š q ‹ Š 1 u# ‹ œ Š Èv 3 tan " u ‹ Š 1 u# ‹ œ atan " ub a1 u# b Ê ` u uœ1ßvœ2 œ atan " 1b a1 1# b œ 1 ; dz ` q tan " u tan " u `z " " " ` v œ dq ` v œ Š q ‹ Š 2Èv 3 ‹ œ Š Èv 3 tan " u ‹ Š 2Èv 3 ‹ œ #(v 3) Ê `` vz ¸ uœ1ßvœ2 œ "# `x `x 2 2 s t ` w s t w w 39. Let x œ s3 t2 Ê w œ fas3 t2 b œ faxb Ê ``ws œ dw , ` t œ dw dx ` s œ f axb † 3s œ 3s e dx ` t œ f axb † 2t œ 2t e 3 2 3 40. Let x œ t s2 and y œ st Ê w œ fˆt s2 ß st ‰ œ faxß yb Ê ``ws œ ``wx ``xs ``wy ``ys œ fx ax, yb † 2t s fy ax, yb † 1t œ at s2 bˆ st ‰ † 2t s ˆts2 ‰2 1 s4 t 5s4 t ` w `w `x `w `y s 4 2 2 † t œ 2s t 2 œ 2 ; ` t œ ` x ` t ` y ` t œ fx ax, yb † s fy ax, yb † t2 ˆ 2 ‰2 ts œ at s2 bˆ st ‰ † s2 2 5 5 † ˆ ts2 ‰ œ s5 s2 œ s2 ` V dI ` V dR dI dR 41. V œ IR Ê ``VI œ R and `` VR œ I; dV dt œ ` I dt ` R dt œ R dt I dt Ê 0.01 volts/sec dI œ (600 ohms) dI dt (0.04 amps)(0.5 ohms/sec) Ê dt œ 0.00005 amps/sec ` V da ` V db ` V dc da db dc 42. V œ abc Ê dV dt œ ` a dt ` b dt ` c dt œ (bc) dt (ac) dt (ab) dt ¸ Ê dV œ (2 m)(3 m)(1 m/sec) (1 m)(3 m)(1 m/sec) (1 m)(2 m)(3 m/sec) œ 3 m$ /sec dt aœ1ßbœ2ßcœ3 ` S da ` S db ` S dc and the volume is increasing; S œ 2ab 2ac 2bc Ê dS dt œ ` a dt ` b dt ` c dt db dc dS ¸ œ 2(b c) da dt 2(a c) dt 2(a b) dt Ê dt aœ1ßbœ2ßcœ3 œ 2(5 m)(1 m/sec) 2(4 m)(1 m/sec) 2(3 m)(3 m/sec) œ 0 m# /sec and the surface area is not changing; ` D da ` D db ` D dc " db dc ‰ â da ¸ D œ Èa# b# c# Ê dD Ê dD # # dt œ ` a dt ` b dt ` c dt œ È # dt b dt c dt dt aœ1ßbœ2ßcœ3 a b c " œ Š È14 ‹ [(1 m)(1 m/sec) (2 m)(1 m/sec) (3 m)(3 m/sec)] œ È614 m/sec 0 m Ê the diagonals are decreasing in length 43. `` xf œ `` uf `` xu `` vf `` xv ``wf ``wx œ `` uf (1) `` vf (0) ``wf (1) œ `` uf ``wf , `f `f ù `f `v `f `w `f `f `f `f `f ` y œ ` u ` y ` v ` y ` w ` y œ ` u (1) ` v (1) ` w (0) œ ` u ` v , and `f `f ù `f `v `f `w `f `f `f `f `f `f `f `f ` z œ ` u ` z ` v ` z ` w ` z œ ` u (0) ` v (1) ` w (1) œ ` v ` w Ê ` x ` y ` z œ 0 44. (a) (b) (c) `y `w `x `w " `w ` r œ fx ` r fy ` r œ fx cos ) fy sin ) and ` ) œ fx (r sin )) fy (r cos )) Ê r ` ) œ fx sin ) fy cos ) `w # ˆ cosr ) ‰ ``w) œ fx sin ) cos ) fy cos# ) ` r sin ) œ fx sin ) cos ) fy sin ) and Ê fy œ (sin )) ``wr ˆ cosr ) ‰ ``w) ; then ``wr œ fx cos ) (sin )) ``wr ˆ cosr ) ‰ ``w) ‘ (sin )) Ê fx cos ) œ ``wr asin# )b ``wr ˆ sin ) rcos ) ‰ ``w) œ a1 sin# )b ``wr ˆ sin ) rcos ) ‰ ``w) Ê fx œ (cos )) ``wr ˆ sinr ) ‰ ``w) # # # afx b# œ acos# )b ˆ ``wr ‰ ˆ 2 sin )r cos ) ‰ ˆ ``wr ``w) ‰ Š sinr# ) ‹ ˆ ``w) ‰ and # # # afy b# œ asin# )b ˆ ``wr ‰ ˆ 2 sin )r cos ) ‰ ˆ ``wr ``w) ‰ Š cosr# ) ‹ ˆ ``w) ‰ Ê afx b# afy b# œ ˆ ``wr ‰ r"# ˆ ``w) ‰ # Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. # 2 Section 14.4 The Chain Rule 823 45. wx œ ``wx œ ``wu `` xu ``wv `` xv œ x ``wu y ``wv Ê wxx œ ``wu x ``x ˆ ``wu ‰ y ``x ˆ ``wv ‰ # # # # # # # # œ ``wu x Š `` uw# `` xu ``v`wu `` xv ‹ y Š `ù`wv `` xu `` vw# `` xv ‹ œ ``wu x Šx `` uw# y ``v`wu ‹ y Šx `ù`wv y `` vw# ‹ # # # œ ``wu x# `` uw# 2xy ``v`wu y# `` vw# ; wy œ ``wy œ ``wu `` uy ``wv `` vy œ y ``wu x ``wv # # # # Ê wyy œ ``wu y Š `` uw# `` uy ``v`wu `` yv ‹ x Š `ù`wv `` yu `` vw# `` yv ‹ # # # # # # # œ ``wu y Šy `` uw# x ``v`wu ‹ x Šy `ù`wv x `` vw# ‹ œ ``wu y# `` uw# 2xy ``v`wu x# `` vw# ; thus # # wxx wyy œ ax# y# b `` uw# ax# y# b `` vw# œ ax# y# b (wuu wvv ) œ 0, since wuu wvv œ 0 46. ``wx œ f w (u)(1) gw (v)(1) œ f w (u) gw (v) Ê wxx œ f ww (u)(1) gww (v)(1) œ f ww (u) gww (v); `w w w ` y œ f (u)(i) g (v)(i) Ê wyy œ f ww (u) ai# b gww (v) ai# b œ f ww (u) gww (v) Ê wxx wyy œ 0 ` f dx ` f dy ` f dz 47. fx (xß yß z) œ cos t, fy (xß yß z) œ sin t, and fz (xß yß z) œ t# t 2 Ê df dt œ ` x dt ` y dt ` z dt # œ (cos t)( sin t) (sin t)(cos t) at# t 2b(1) œ t# t 2; df dt œ 0 Ê t t 2 œ 0 Ê t œ 2 or t œ 1; t œ 2 Ê x œ cos (2), y œ sin (2), z œ 2 for the point (cos (2)ß sin (2)ß 2); t œ 1 Ê x œ cos 1, y œ sin 1, z œ 1 for the point (cos 1ß sin 1ß 1) ` w dx ` w dy ` w dz 2y # 2y # 2y ˆ " ‰ 48. dw dt œ ` x dt ` y dt ` z dt œ a2xe cos 3zb ( sin t) a2x e cos 3zb t# a3x e sin 3zb (1) # 2y 3z œ 2xe2y cos 3z sin t 2x et cos 3x# e2y sin 3z; at the point on the curve z œ 0 Ê t œ z œ 0 # # 2(1) (4)(1) ¸ Ê dw 0œ4 # dt Ð1ßln 2ß0Ñ œ 0 49. (a) `T `T ` x œ 8x 4y and ` y œ 8y 4x ` T dx ` T dy Ê dT dt œ ` x dt ` y dt œ (8x 4y)( sin t) (8y 4x)(cos t) # œ (8 cos t 4 sin t)( sin t) (8 sin t 4 cos t)(cos t) œ 4 sin# t 4 cos# t Ê ddtT# œ 16 sin t cos t; dT dt œ 0 Ê 4 sin# t 4 cos# t œ 0 Ê sin# t œ cos# t Ê sin t œ cos t or sin t œ cos t Ê t œ 14 , 541 , 341 , 741 on the interval 0 Ÿ t Ÿ 21; È d# T 1 1 dt# ¹ tœ 1 œ 16 sin 4 cos 4 0 È Ê T has a minimum at (xß y) œ Š #2 ß #2 ‹ ; 4 È È d# T 31 31 dt# ¹ tœ 31 œ 16 sin 4 cos 4 0 Ê T has a maximum at (xß y) œ Š #2 ß #2 ‹ ; d# T 51 51 dt# ¹ tœ 51 œ 16 sin 4 cos 4 0 Ê T has a minimum at (xß y) œ Š #2 ß #2 ‹ ; d# T 71 71 dt# ¹ tœ 71 œ 16 sin 4 cos 4 0 Ê T has a maximum at (xß y) œ Š #2 ß #2 ‹ 4 4 4 È È È È (b) T œ 4x# 4xy 4y# Ê ``Tx œ 8x 4y, and ``Ty œ 8y 4x so the extreme values occur at the four points È È È È found in part (a): T Š #2 ß #2 ‹ œ T Š #2 ß #2 ‹ œ 4 ˆ "# ‰ 4 ˆ "# ‰ 4 ˆ "# ‰ œ 6, the maximum and È È È È T Š #2 ß #2 ‹ œ T Š #2 ß #2 ‹ œ 4 ˆ #" ‰ 4 ˆ #" ‰ 4 ˆ #" ‰ œ 2, the minimum 50. (a) `T `T ` x œ y and ` y œ x ` T dx ` T dy È È Ê dT dt œ ` x dt ` y dt œ y Š2 2 sin t‹ x Š 2 cos t‹ œ ŠÈ2 sin t‹ Š2È2 sin t‹ Š2È2 cos t‹ ŠÈ2 cos t‹ œ 4 sin# t 4 cos# t œ 4 sin# t 4 a1 sin# tb # " " # # œ 4 8 sin# t Ê ddtT# œ 16 sin t cost t; dT dt œ 0 Ê 4 8 sin t œ 0 Ê sin t œ # Ê sin t œ „ È 31 51 71 4 , 4 , 4 on the interval 0 Ÿ t Ÿ 21; d# T ˆ1‰ dt# ¹ tœ 1 œ 8 sin 2 4 œ 8 Ê T has a maximum at (xß y) œ (2ß 1); d# T ˆ 31 ‰ œ 8 dt# ¹ tœ 31 œ 8 sin 2 4 Ê T has a minimum at (xß y) œ (2ß 1); 4 4 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 2 Ê t œ 14 , 824 Chapter 14 Partial Derivatives d# T ˆ 51 ‰ œ 8 dt# ¹ tœ 51 œ 8 sin 2 4 Ê T has a maximum at (xß y) œ (2ß 1); 4 d# T ˆ 71 ‰ œ 8 dt# ¹ tœ 71 œ 8 sin 2 4 Ê T has a minimum at (xß y) œ (2ß 1) 4 (b) T œ xy 2 Ê ``Tx œ y and ``Ty œ x so the extreme values occur at the four points found in part (a): T(2ß 1) œ T(2ß 1) œ 0, the maximum and T(2ß 1) œ T(2ß 1) œ 4, the minimum ` G du ` G dx w ' 51. G(uß x) œ 'a g(tß x) dt where u œ f(x) Ê dG dx œ ` u dx ` x dx œ g(uß x)f (x) a gx (tß x) dt; thus u u F(x) œ '0 Èt% x$ dt Ê Fw (x) œ Éax# b% x$ (2x) '0 x# x# ` È % t x$ dt œ 2xÈx) x$ `x 52. Using the result in Exercise 51, F(x) œ 'x# Èt$ x# dt œ '1 Èt$ x# dt Ê Fw (x) x# 1 œ ’ Éax# b$ x# x# ' x# # ` È $ t x# dt “ œ x Èx' x# 1 `x 'x1 È x # t$ x# dt 14.5 DIRECTIONAL DERIVATIVES AND GRADIENT VECTORS 1. `f `f ` x œ 1, ` y œ 1 Ê ™ f œ i j ; f(2ß 1) œ 1 Ê 1 œ y x is the level curve 2. 2y `f 2x `f `f ` x œ x# y# Ê ` x ("ß ") œ 1; ` y œ x# y# Ê `` yf ("ß ") œ 1 Ê ™ f œ i j ; f(1ß 1) œ ln 2 # # # # Ê ln 2 œ ln ax y b Ê 2 œ x y is the level curve 3. `g 2 `x œ y Ê `` gx a2ß 1b œ 1; `` gy œ 2x y Ê `` gx a2ß 1b œ 4; Ê ™ g œ i 4j ; ga2ß 1b œ 2 Ê x œ y2# is the level curve 4. `g `x œ x Ê `` gx ŠÈ2ß "‹ œ È2; `` gy œ y Ê `` gy ŠÈ2ß "‹ œ 1 Ê ™ g œ È2 i j ; # # g ŠÈ2ß "‹ œ "# Ê "# œ x# y# or 1 œ x# y# is the level curve Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. '0x È3x # 2 # t% x$ dt Section 14.5 Directional Derivatives and Gradient Vectors 5. `f 1 ` x œ È2x 3y Ê `` xf (1ß 2) œ 21 ; `` yf œ 2È2x3 3y Ê `` xf (1ß 2) œ 43 ; Ê ™ f œ 12 i 34 j ; f(1ß 2) œ 2 Ê 4 œ 2x 3y is the level curve 6. y `f ` x œ 2y2 Èx 2x3Î2 1 Ê `` xf a4ß 2b œ 16 ; Èx `f `f 1 1 1 ` y œ 2y2 x Ê ` y a4ß 2b œ 4 Ê ™ f œ 16 i 4 j ; f a4ß 2b œ 14 Ê y œ Èx is the level curve 7. `f z ` x œ 2x x Ê `` xf (1ß 1ß 1) œ 3; `` yf œ 2y Ê `` yf ("ß "ß ") œ 2; `` zf œ 4z ln x Ê `` zf ("ß "ß ") œ 4; thus ™ f œ 3i 2j 4k 8. 9. `f z `f 11 ` f ` x œ 6xz x# z# 1 Ê ` x (1ß "ß ") œ # ; ` y œ 6yz " Ê `` zf ("ß "ß ") œ "# ; thus ™ f œ 11 # i 6j # k Ê `` yf ("ß "ß ") œ 6; `` zf œ 6z# 3 ax# y# b x# z#x 1 `f x " ` x œ ax# y# z# b$Î# x `f Ê `` xf (1ß 2ß 2) œ 26 27 ; ` y œ y y" Ê `` yf (1ß 2ß 2) œ 23 54 ; ax# y# z# b$Î# `f 23 26 23 23 ` z (1ß 2ß 2) œ 54 ; thus ™ f œ 27 i 54 j 54 k `f z " ` z œ ax# y# z# b$Î# z Ê 10. `` xf œ exy cos z Èy 1 # Ê `` xf ˆ!ß !ß 16 ‰ œ 1x `f x y sin z ` z œ e È3 `f x y cos z sin" x # 1; ` y œ e Ê `` zf ˆ!ß !ß 16 ‰ œ #" ; thus ™ f œ Š Ê `` yf ˆ0ß 0ß 16 ‰ œ È3 # ; È 3 2 È3 " # ‹i # j # k 4i 3j 4 3 11. u œ kA Ak œ È4# 3# œ 5 i 5 j ; fx (xß y) œ 2y Ê fx (5ß 5) œ 10; fy (xß y) œ 2x 6y Ê fy (5ß 5) œ 20 Ê ™ f œ 10i 20j Ê (Du f)P! œ ™ f † u œ 10 ˆ 45 ‰ 20 ˆ 35 ‰ œ 4 3i 4j 3 4 12. u œ kA Ak œ È3# (4)# œ 5 i 5 j ; fx (xß y) œ 4x Ê fx (1ß 1) œ 4; fy (xß y) œ 2y Ê fy (1ß 1) œ 2 8 Ê ™ f œ 4i 2j Ê (Du f)P! œ ™ f † u œ 12 5 5 œ 4 12i 5j y 2 5 x 2 13. u œ kA œ 12 # Ak œ È # 13 i 13 j ; gx axß yb œ axy 2b2 Ê gx a1ß 1b œ 3; gy axß yb œ axy 2b2 Ê gy a1ß 1b œ 3 2 2 12 5 15 21 Ê ™ g œ 3i 3j Ê aDu gbP! œ ™ g † u œ 36 13 13 œ 13 Š #y ‹ 3i 2j x 3 2 14. u œ kA Ak œ È3# (2)# œ È13 i È13 j ; hx (xß y) œ ˆ y ‰# x ˆ"‰ hy (xß y) œ ˆ y ‰#x x 1 ˆ #x ‰ È3 x# y# Ê1 Š 4 ‹ 1 ˆ #y ‰ È3 x# y# Ê1 Š 4 ‹ Ê hx (1ß 1) œ "# ; Ê hy (1ß 1) œ #3 Ê ™ h œ "# i #3 j Ê (Du h)P! œ ™ h † u œ 2È313 2È613 œ 2È313 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 825 826 Chapter 14 Partial Derivatives 3 i 6 j #k 3 6 2 15. u œ kA Ak œ È3# 6# (2)# œ 7 i 7 j 7 k ; fx (xß yß z) œ y z Ê fx (1ß 1ß 2) œ 1; fy (xß yß z) œ x z Ê fy (1ß 1ß 2) œ 3; fz (xß yß z) œ y x Ê fz (1ß 1ß 2) œ 0 Ê ™ f œ i 3j Ê (Du f)P! œ ™ f † u œ 37 18 7 œ 3 ijk 1 1 1 16. u œ kA Ak œ È1# 1# 1# œ È3 i È3 j È3 k ; fx (xß yß z) œ 2x Ê fx (1ß 1ß 1) œ 2; fy (xß yß z) œ 4y Ê fy (1ß 1ß 1) œ 4; fz (xß yß z) œ 6z Ê fz (1ß 1ß 1) œ 6 Ê ™ f œ 2i 4j 6k Ê (Du f)P! œ ™ f † u œ 2 Š È"3 ‹ 4 Š È"3 ‹ 6 Š È"3 ‹ œ 0 2i j 2k 2 1 2 x x 17. u œ kA Ak œ È2# 1# (2)# œ 3 i 3 j 3 k ; gx (xß yß z) œ 3e cos yz Ê gx (0ß 0ß 0) œ 3; gy (xß yß z) œ 3ze sin yz Ê gy (0ß 0ß 0) œ 0; gz (xß yß z) œ 3yex sin yz Ê gz (0ß 0ß 0) œ 0 Ê ™ g œ 3i Ê (Du g)P! œ ™ g † u œ 2 i 2j 2k 1 2 2 " "‰ ˆ 18. u œ kA Ak œ È1# 2# 2# œ 3 i 3 j 3 k ; hx (xß yß z) œ y sin xy x Ê hx 1ß 0ß # œ 1; hy (xß yß z) œ x sin xy zeyz Ê hy ˆ"ß !ß #" ‰ œ #" ; hz (xß yß z) œ yeyz "z Ê hz ˆ"ß !ß #" ‰ œ 2 Ê ™ h œ i #" j 2k Ê (Du h)P! œ ™ h † u œ "3 "3 43 œ 2 f i j " " 19. ™ f œ (2x y) i (x 2y) j Ê ™ f(1ß 1) œ i j Ê u œ k™ ™f k œ È(1)# 1# œ È i È j ; f increases 2 2 most rapidly in the direction u œ È"2 i È"2 j and decreases most rapidly in the direction u œ È"2 i È"2 j ; (Du f)P! œ ™ f † u œ k ™ f k œ È2 and (Du f)P! œ È2 f 20. ™ f œ a2xy yexy sin yb i ax# xexy sin y exy cos yb j Ê ™ f(1ß 0) œ 2j Ê u œ k™ ™f k œ j ; f increases most rapidly in the direction u œ j and decreases most rapidly in the direction u œ j ; (Du f)P! œ ™ f † u œ k ™ f k œ 2 and (Du f)P! œ 2 f i 5j k 5 " " 21. ™ f œ "y i Š yx# z‹ j yk Ê ™ f(4ß "ß ") œ i 5j k Ê u œ k™ ™f k œ È1# (5)# (1)# œ 3È3 i 3È3 j 3È3 k ; " 5 " f increases most rapidly in the direction of u œ 3È i 3È j 3È k and decreases most rapidly in the direction 3 3 3 " 5 " u œ 3È i 3È j 3È k ; (Du f)P! œ ™ f † u œ k ™ f k œ 3È3 and (Du f)P! œ 3È3 3 3 3 g 2i 2j k " 2 2 22. ™ g œ ey i xey j 2zk Ê ™ g ˆ1ß ln 2ß "# ‰ œ 2i 2j k Ê u œ k™ ™gk œ È # # # œ 3 i 3 j 3 k ; 2 2 1 2 2 " g increases most rapidly in the direction u œ 3 i 3 j 3 k and decreases most rapidly in the direction u œ 23 i 23 j 3" k ; (Du g)P! œ ™ g † u œ k ™ gk œ 3 and (Du g)P! œ 3 f " " " 23. ™ f œ ˆ "x x" ‰ i Š y" y" ‹ j ˆ "z "z ‰ k Ê ™ f("ß "ß ") œ 2i 2j 2k Ê u œ k™ ™f k œ È3 i È3 j È3 k ; f increases most rapidly in the direction u œ È"3 i È"3 j È"3 k and decreases most rapidly in the direction u œ È"3 i È"3 j È"3 k; (Du f)P! œ ™ f † u œ k ™ f k œ 2È3 and (Du f)P! œ 2È3 ™h 2y 2i 3 j 6k 24. ™ h œ Š x# 2x y# 1 ‹ i Š x# y# 1 1‹ j 6k Ê ™ h("ß "ß 0) œ 2i 3j 6k Ê u œ k™hk œ È2# 3# 6# œ 27 i 37 j 67 k ; h increases most rapidly in the direction u œ 27 i 37 j 67 k and decreases most rapidly in the direction u œ 27 i 37 j 67 k ; (Du h)P! œ ™ h † u œ k ™ hk œ 7 and (Du h)P! œ 7 Copyright © 2010 Pearson Education Inc. 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Section 14.5 Directional Derivatives and Gradient Vectors 827 25. ™ f œ 2xi 2yj Ê ™ f ŠÈ2ß È2‹ œ 2È2 i 2È2 j Ê Tangent line: 2È2 Šx È2‹ 2È2 Šy È2‹ œ 0 Ê È2x È2y œ 4 26. ™ f œ 2xi j Ê ™ f ŠÈ2ß 1‹ œ 2È2 i j Ê Tangent line: 2È2 Šx È2‹ (y 1) œ 0 Ê y œ 2È2x 3 27. ™ f œ yi xj Ê ™ f(2ß 2) œ 2i 2j Ê Tangent line: 2(x 2) 2(y 2) œ 0 Ê yœx4 28. ™ f œ (2x y)i (2y x)j Ê ™ f(1ß 2) œ 4i 5j Ê Tangent line: 4(x 1) 5(y 2) œ 0 Ê 4x 5y 14 œ 0 29. ™ f œ a2x ybi ax 2y 1bj (a) ™ fa1, 1b œ 3i 4j Ê l ™ fa1, 1bl œ 5 Ê Du fa1, 1b œ 5 in the direction of u œ 35 i 45 j (b) ™ fa1, 1b œ 3i 4j Ê l ™ fa1, 1bl œ 5 Ê Du fa1, 1b œ 5 in the direction of u œ 35 i 45 j (c) Du fa1, 1b œ 0 in the direction of u œ 45 i 35 j or u œ 45 i 35 j (d) Let u œ u1 i u2 j Ê lul œ Èu12 u22 œ 1 Ê u12 u22 œ 1; Du fa1, 1b œ ™ fa1, 1b † u œ a3i 4jb † au1 i u2 jb 2 25 2 24 œ 3u1 4u2 œ 4 Ê u2 œ 43 u1 1 Ê u12 ˆ 43 u1 1‰ œ 1 Ê 16 u1 23 u1 œ 0 Ê u1 œ 0 or u1 œ 25 ; 7 24 7 u1 œ 0 Ê u2 œ 1 Ê u œ j, or u1 œ 24 25 Ê u2 œ 25 Ê u œ 25 i 25 j (e) Let u œ u1 i u2 j Ê lul œ Èu12 u22 œ 1 Ê u12 u22 œ 1; Du fa1, 1b œ ™ fa1, 1b † u œ a3i 4jb † au1 i u2 jb 2 8 24 2 œ 3u1 4u2 œ 3 Ê u1 œ 43 u2 1 Ê ˆ 43 u2 1‰ u22 œ 1 Ê 25 9 u2 3 u2 œ 0 Ê u2 œ 0 or u2 œ 25 ; 7 7 24 u2 œ 0 Ê u1 œ 1 Ê u œ i, or u2 œ 24 25 Ê u2 œ 25 Ê u œ 25 i 25 j . 30. ™ f œ ax 2yyb2 i ax 2xyb2 j (a) ™ fˆ 21 , 23 ‰ œ 3i j Ê l ™ fˆ 21 , 23 ‰l œ È10 Ê Du fˆ 21 , 23 ‰ œ È10 in the direction of u œ È310 i È110 j (b) ™ fˆ 21 , 23 ‰ œ 3i j Ê l ™ fˆ 21 , 23 ‰l œ È10 Ê Du fa1, 1b œ È10 in the direction of u œ È310 i È110 j Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 828 Chapter 14 Partial Derivatives (c) Du fˆ 12 , 23 ‰ œ 0 in the direction of u œ È110 i È310 j or u œ È110 i È310 j (d) Let u œ u1 i u2 j Ê lul œ Èu12 u22 œ 1 Ê u12 u22 œ 1; Du fˆ 12 , 32 ‰ œ ™ fˆ 12 , 32 ‰ † u œ a3i jb † au1 i u2 jb È 6 œ 3u1 u2 œ 2 Ê u2 œ 3u1 2 Ê u12 a3u1 2b2 œ 1 Ê 10u12 12u1 3 œ 0 Ê u1 œ 6 „ 10 È 6 u1 œ 6 Ê u2 œ 2 103 10 È 6 Ê u œ 6 i 2 103 10 È6 È 6 Ê u œ 6 i 2 103 10 È6 È 6 j, or u1 œ 6 Ê u2 œ 2 103 10 È6 È6 j (e) Let u œ u1 i u2 j Ê lul œ Èu12 u22 œ 1 Ê u12 u22 œ 1; Du fˆ 12 , 32 ‰ œ ™ fˆ 12 , 32 ‰ † u œ a3i jb † au1 i u2 jb œ 3u1 u2 œ 1 Ê u2 œ 1 3u1 Ê u12 a1 3u1 b2 œ 1 Ê 10u12 6u1 œ 0 Ê u1 œ 0 or u1 œ 35 ; u1 œ 0 Ê u2 œ 1 Ê u œ j, or u1 œ 35 Ê u2 œ 45 Ê u œ 35 i 45 j 31. ™ f œ yi (x 2y)j Ê ™ f(3ß 2) œ 2i 7j ; a vector orthogonal to ™ f is v œ 7i 2j Ê u œ kvvk œ È77#i(2j 2)# œ È753 i È253 j and u œ È753 i È253 j are the directions where the derivative is zero # # 32. ™ f œ ax#4xy i ax#4x yy# b# j Ê ™ f("ß ") œ i j ; a vector orthogonal to ™ f is v œ i j y # b# Ê u œ kvvk œ È1i #j 1# œ È12 i È12 j and u œ È12 i È12 j are the directions where the derivative is zero 33. ™ f œ (2x 3y)i (3x 8y)j Ê ™ f(1ß 2) œ 4i 13j Ê k ™ f(1ß 2)k œ È(4)# (13)# œ È185 ; no, the maximum rate of change is È185 14 34. ™ T œ 2yi (2x z)j yk Ê ™ T(1ß 1ß 1) œ 2i j k Ê k ™ T(1ß 1ß 1)k œ È(2)# 1# 1# œ È6 ; no, the minimum rate of change is È6 3 35. ™ f œ fx ("ß #)i fy ("ß #)j and u" œ È1i #j 1# œ È"2 i È"2 j Ê (Du" f)(1ß 2) œ fx (1ß 2) Š È"2 ‹ fy (1ß 2) Š È"2 ‹ œ 2È2 Ê fx (1ß 2) fy (1ß 2) œ 4; u# œ j Ê (Du# f)(1ß 2) œ fx (1ß 2)(0) fy (1ß 2)(1) œ 3 Ê fy (1ß 2) œ 3 Ê fy (1ß 2) œ 3; then fx (1ß 2) 3 œ 4 Ê fx (1ß 2) œ 1; thus ™ f(1ß 2) œ i 3j and u œ kvvk œ È(1)i#2(j2)# œ È15 i È25 j Ê (Du f)P! œ ™ f † u œ È"5 È65 œ È75 f 36. (a) (Du f)P œ 2È3 Ê k ™ f k œ 2È3; u œ kvvk œ È1# i1j#k(1)# œ È13 i È13 j È"3 k; thus u œ k™ ™f k Ê ™ f œ k ™ f k u Ê ™ f œ 2È3 Š È"3 i È"3 j È"3 k‹ œ 2i 2j 2k (b) v œ i j Ê u œ kvvk œ È1i #j 1# œ È"2 i È"2 j Ê (Du f)P! œ ™ f † u œ 2 Š È"2 ‹ 2 Š È"2 ‹ 2(0) œ 2È2 37. The directional derivative is the scalar component. With ™ f evaluated at P! , the scalar component of ™ f in the direction of u is ™ f † u œ (Du f)P! . 38. Di f œ ™ f † i œ (fx i fy j fz k) † i œ fx ; similarly, Dj f œ ™ f † j œ fy and Dk f œ ™ f † k œ fz 39. If (xß y) is a point on the line, then T(xß y) œ (x x! )i (y y! )j is a vector parallel to the line Ê T † N œ 0 Ê A(x x! ) B(y y! ) œ 0, as claimed. ` (kf) ` (kf) `f `f `f `f ˆ `f ‰ ˆ `f ‰ 40. (a) ™ (kf) œ ``(kf) x i ` y j ` z k œ k ` x i k Š ` y ‹ j k ` z k œ k Š ` x i ` y j ` z k‹ œ k ™ f Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.6 Tangent Planes and Differentials 829 (b) ™ (f g) œ ` (f`x g) i ` (f`y g) j ` (f`z g) k œ Š `` xf `` xg ‹ i Š `` yf `` gy ‹ j Š `` zf `` gz ‹ k œ `` xf i `` gx i `` yf j `` gy j `` zf k `` gz k œ Š `` xf i `` yf j `` zf k‹ Š `` gx i `` gy j `` gz k‹ œ ™ f ™ g (c) ™ (f g) œ ™ f ™ g (Substitute g for g in part (b) above) ` (fg) ` (fg) `g `g `g `f `f `f (d) ™ (fg) œ ``(fg) x i `y j `z k œ Š `x g `x f ‹ i Š `y g `y f ‹ j Š `z g `z f ‹ k œ ˆ `` xf g‰ i Š `` xg f ‹ i Š `` yf g‹ j Š `` gy f ‹ j ˆ `` zf g‰ k Š `` gz f ‹ k œ f Š `` gx i `` gy j `` gz k‹ g Š `` xf i `` yf j `` zf k‹ œ f ™ g g ™ f (e) ™ Š gf ‹ œ ` Š gf ‹ `x i ` Š gf ‹ `y j ` Š gf ‹ `z g `f f `g g `f f `g g `f f `g k œ Š ` x g # ` x ‹ i Œ ` y g # ` y j Š ` z g# ` z ‹ k `f `f `f `g `g `g g Š ` x i ` y j ` z k‹ f Š ` x i ` y j ` z k‹ g ` f i g ``yf j g `` fz k f `` gx i f `` gy j f ``gz k Œ œ g# g# g# g# œ Œ `x f f™g g™f f™g œ g™ g# g# œ g# 14.6 TANGENT PLANES AND DIFFERENTIALS 1. (a) ™ f œ 2xi 2yj 2zk Ê ™ f(1ß 1ß 1) œ 2i 2j 2k Ê Tangent plane: 2(x 1) 2(y 1) 2(z 1) œ 0 Ê x y z œ 3; (b) Normal line: x œ 1 2t, y œ 1 2t, z œ 1 2t 2. (a) ™ f œ 2xi 2yj 2zk Ê ™ f(3ß 5ß 4) œ 6i 10j 8k Ê Tangent plane: 6(x 3) 10(y 5) 8(z 4) œ 0 Ê 3x 5y 4z œ 18; (b) Normal line: x œ 3 6t, y œ 5 10t, z œ 4 8t 3. (a) ™ f œ 2xi 2k Ê ™ f(2ß 0ß 2) œ 4i 2k Ê Tangent plane: 4(x 2) 2(z 2) œ 0 Ê 4x 2z 4 œ 0 Ê 2x z 2 œ 0; (b) Normal line: x œ 2 4t, y œ 0, z œ 2 2t 4. (a) ™ f œ (2x 2y)i (2x 2y)j 2zk Ê ™ f(1ß 1ß 3) œ 4j 6k Ê Tangent plane: 4(y 1) 6(z 3) œ 0 Ê 2y 3z œ 7; (b) Normal line: x œ 1, y œ 1 4t, z œ 3 6t 5. (a) ™ f œ a1 sin 1x 2xy zexz b i ax# zb j axexz yb k Ê ™ f(0ß 1ß 2) œ 2i 2j k Ê Tangent plane: 2(x 0) 2(y 1) 1(z 2) œ 0 Ê 2x 2y z 4 œ 0; (b) Normal line: x œ 2t, y œ 1 2t, z œ 2 t 6. (a) ™ f œ (2x y)i (x 2y)j k Ê ™ f(1ß 1ß 1) œ i 3j k Ê Tangent plane: 1(x 1) 3(y 1) 1(z 1) œ 0 Ê x 3y z œ 1; (b) Normal line: x œ 1 t, y œ 1 3t, z œ 1 t 7. (a) ™ f œ i j k for all points Ê ™ f(0ß 1ß 0) œ i j k Ê Tangent plane: 1(x 0) 1(y 1) 1(z 0) œ 0 Ê x y z 1 œ 0; (b) Normal line: x œ t, y œ 1 t, z œ t 8. (a) ™ f œ (2x 2y 1)i (2y 2x 3)j k Ê ™ f(2ß 3ß 18) œ 9i 7j k Ê Tangent plane: 9(x 2) 7(y 3) 1(z 18) œ 0 Ê 9x 7y z œ 21; (b) Normal line: x œ 2 9t, y œ 3 7t, z œ 18 t Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 830 Chapter 14 Partial Derivatives 2y 9. z œ f(xß y) œ ln ax# y# b Ê fx (xß y) œ x# 2x y# and fy (xß y) œ x# y# Ê fx (1ß 0) œ 2 and fy (1ß 0) œ 0 Ê from Eq. (4) the tangent plane at (1ß 0ß 0) is 2(x 1) z œ 0 or 2x z 2 œ 0 # # # # # # 10. z œ f(xß y) œ e ax y b Ê fx (xß y) œ 2xe ax y b and fy (xß y) œ 2ye ax y b Ê fx (0ß 0) œ 0 and fy (!ß !) œ 0 Ê from Eq. (4) the tangent plane at (0ß 0ß 1) is z 1 œ 0 or z œ 1 11. z œ f(Bß y) œ Èy x Ê fx (xß y) œ "# (y x)"Î# and fy (xß y) œ "# (y x)"Î# Ê fx (1ß 2) œ "# and fy ("ß #) œ "# Ê from Eq. (4) the tangent plane at (1ß 2ß 1) is "# (x 1) "# (y 2) (z 1) œ 0 Ê x y 2z 1 œ 0 12. z œ f(Bß y) œ 4x# y# Ê fx (xß y) œ 8x and fy (xß y) œ #y Ê fx (1ß 1) œ 8 and fy ("ß 1) œ # Ê from Eq. (4) the tangent plane at (1ß 1ß 5) is 8(x 1) 2(y 1) (z 5) œ 0 or 8x 2y z 5 œ 0 13. ™ f œ i 2yj 2k Ê ™ f(1ß 1ß 1) œ i 2j 2k and ™ g œ i for all points; v œ ™ f ‚ ™ g â â â i j kâ â â Ê v œ â " 2 2 â œ 2j 2k Ê Tangent line: x œ 1, y œ 1 2t, z œ 1 2t â â â" 0 0â 14. ™ f œ yzi xzj xyk Ê ™ f(1ß 1ß 1) œ i j k; ™ g œ 2xi 4yj 6zk Ê ™ g(1ß 1ß 1) œ 2i 4j 6k ; â â â i j kâ â â Ê v œ ™ f ‚ ™ g Ê â " 1 1 â œ 2i 4j 2k Ê Tangent line: x œ 1 2t, y œ 1 4t, z œ 1 2t â â â2 4 6â 15. ™ f œ 2xi 2j 2k Ê ™ f ˆ1ß 1ß "# ‰ œ 2i 2j 2k and ™ g œ j for all points; v œ ™ f ‚ ™ g â â â i j kâ â â Ê v œ â 2 2 2 â œ 2i 2k Ê Tangent line: x œ 1 2t, y œ 1, z œ "# 2t â â â0 1 0â 16. ™ f œ i 2yj k Ê ™ f ˆ "# ß 1ß "# ‰ œ i 2j k and ™ g œ j for all points; v œ ™ f ‚ ™ g â â â i j kâ â â Ê v œ â 1 2 1 â œ i k Ê Tangent line: x œ "# t, y œ 1, z œ "# t â â â0 1 0â 17. ™ f œ a3x# 6xy# 4yb i a6x# y 3y# 4xb j 2zk Ê ™ f(1ß 1ß 3) œ 13i 13j 6k ; ™ g œ 2xi 2yj 2zk â â j k â â i â â Ê ™ g("ß "ß $) œ 2i 2j 6k ; v œ ™ f ‚ ™ g Ê v œ â "3 13 6 â œ 90i 90j Ê Tangent line: â â 2 6 â â 2 x œ 1 90t, y œ 1 90t, z œ 3 18. ™ f œ 2xi 2yj Ê ™ f ŠÈ2ß È2ß 4‹ œ 2È2 i 2È2 j ; ™ g œ 2xi 2yj k Ê ™ g ŠÈ2ß È2ß 4‹ â i j k ââ â â â œ 2È2i 2È2j k ; v œ ™ f ‚ ™ g Ê v œ â 2È2 2È2 0 â œ 2È2 i 2È2 j Ê Tangent line: â â â 2È2 2È2 1 â x œ È2 2È2 t, y œ È2 2È2 t, z œ 4 3 4 12 19. ™ f œ Š x# yx# z# ‹ i Š x# yy# z# ‹ j Š x# yz# z# ‹ k Ê ™ f(3ß 4ß 12) œ 169 i 169 j 169 k; 2k 9 9 ‰ u œ kvvk œ È33#i 66#j œ 37 i 76 j 27 k Ê ™ f † u œ 1183 and df œ ( ™ f † u) ds œ ˆ 1183 (0.1) ¸ 0.0008 (2)# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.6 Tangent Planes and Differentials 831 2k 20. ™ f œ aex cos yzb i azex sin yzb j ayex sin yzb k Ê ™ f(0ß 0ß 0) œ i ; u œ kvvk œ È22#i 22#j (2)# œ È13 i È13 j È13 k Ê ™ f † u œ È13 and df œ ( ™ f † u) ds œ È13 (0.1) ¸ 0.0577 Ä 21. ™ g œ (1 cos z)i (1 sin z)j (x sin z y cos z)k Ê ™ g(2ß 1ß 0) œ 2i j k; A œ P! P" œ 2i 2j 2k i 2 j 2k 1 1 1 Ê u œ kvvk œ È(22) # 2# 2# œ È3 i È3 j È3 k Ê ™ g † u œ 0 and dg œ ( ™ g † u) ds œ (0)(0.2) œ 0 22. ™ h œ c1y sin (1xy) z# d i c1x sin (1xy)d j 2xzk Ê ™ h(1ß 1ß 1) œ (1 sin 1 1)i (1 sin 1)j 2k Ä k œ i 2k ; v œ P! P" œ i j k where P" œ (!ß !ß !) Ê u œ kvvk œ È1i#j1# œ È13 i È13 j È13 k 1# Ê ™ h † u œ È33 œ È3 and dh œ ( ™ h † u) ds œ È3(0.1) ¸ 0.1732 È 23. (a) The unit tangent vector at Š "# ß #3 ‹ in the direction of motion is u œ È3 " # i # j; È È ™ T œ (sin 2y)i (2x cos 2y)j Ê ™ T Š "# ß #3 ‹ œ Šsin È3‹ i Šcos È3‹ j Ê Du T Š "# ß #3 ‹ œ ™ T † u œ È3 È3 " cos È3 ¸ 0.935° C/ft # sin # ` T dx ` T dy (b) r(t) œ (sin 2t)i (cos 2t)j Ê v(t) œ (2 cos 2t)i (2 sin 2t)j and kvk œ 2; dT dt œ ` x dt ` y dt È œ ™ T † v œ Š ™ T † kvvk ‹ kvk œ (Du T) kvk , where u œ kvvk ; at Š "# ß #3 ‹ we have u œ È3 " # i # j from part (a) È 3 È3 " cos È3‹ † 2 œ È3 sin È3 cos È3 ¸ 1.87° C/sec Ê dT dt œ Š # sin # 24. (a) ™ T œ (4x yz)i xzj xyk Ê ™ T(8ß 6ß 4) œ 56i 32j 48k ; r(t) œ 2t# i 3tj t# k Ê the particle is at the point P()ß 6ß 4) when t œ 2; v(t) œ 4ti 3j 2tk Ê v(2) œ 8i 3j 4k Ê u œ kvvk 736 œ È889 i È389 j È489 k Ê Du T(8ß 6ß 4) œ ™ T † u œ È"89 [56 † 8 32 † 3 48 † (4)] œ È ° C/m 89 (b) dT ` T dx ` T dy dt œ ` x dt ` y dt œ 736 È89 œ 736° C/sec ¸ ™ T † v œ ( ™ T † u) kvk Ê at t œ 2, dT dt œ Du T tœ2 v(2) œ Š È89 ‹ 25. (a) f(!ß 0) œ 1, fx (xß y) œ 2x Ê fx (0ß 0) œ 0, fy (xß y) œ 2y Ê fy (0ß 0) œ 0 Ê L(xß y) œ 1 0(x 0) 0(y 0) œ 1 (b) f(1ß 1) œ 3, fx (1ß 1) œ 2, fy (1ß 1) œ 2 Ê L(xß y) œ 3 2(x 1) 2(y 1) œ 2x 2y 1 26. (a) f(!ß 0) œ 4, fx (xß y) œ 2(x y 2) Ê fx (0ß 0) œ 4, fy (xß y) œ 2(x y 2) Ê fy (0ß 0) œ 4 Ê L(xß y) œ 4 4(x 0) 4(y 0) œ 4x 4y 4 (b) f(1ß 2) œ 25, fx (1ß 2) œ 10, fy (1ß 2) œ 10 Ê L(xß y) œ 25 10(x 1) 10(y 2) œ 10x 10y 5 27. (a) f(0ß 0) œ 5, fx (xß y) œ 3 for all (xß y), fy (xß y) œ 4 for all (xß y) Ê L(xß y) œ 5 3(x 0) 4(y 0) œ 3x 4y 5 (b) f(1ß 1) œ 4, fx (1ß 1) œ 3, fy (1ß 1) œ 4 Ê L(xß y) œ 4 3(x 1) 4(y 1) œ 3x 4y 5 28. (a) f(1ß 1) œ 1, fx (xß y) œ 3x# y% Ê fx (1ß 1) œ 3, fy (xß y) œ 4x$ y$ Ê fy (1ß 1) œ 4 Ê L(xß y) œ 1 3(x 1) 4(y 1) œ 3x 4y 6 (b) f(0ß 0) œ 0, fx (!ß 0) œ 0, fy (0ß 0) œ 0 Ê L(xß y) œ 0 29. (a) f(0ß 0) œ 1, fx (xß y) œ ex cos y Ê fx (0ß 0) œ 1, fy (xß y) œ ex sin y Ê fy (0ß 0) œ 0 Ê L(xß y) œ 1 1(x 0) 0(y 0) œ x 1 (b) f ˆ0ß 1# ‰ œ 0, fx ˆ0ß 1# ‰ œ 0, fy ˆ0ß 1# ‰ œ 1 Ê L(xß y) œ 0 0(x 0) 1 ˆy 1# ‰ œ y 1# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 832 Chapter 14 Partial Derivatives 30. (a) f(0ß 0) œ 1, fx (xß y) œ e2yx Ê fx (!ß !) œ 1, fy (xß y) œ 2e2yx Ê fy (0ß 0) œ 2 Ê L(xß y) œ 1 1(x 0) 2(y 0) œ x 2y 1 (b) f(1ß 2) œ e$ , fx (1ß 2) œ e$ , fy (1ß 2) œ 2e$ Ê L(xß y) œ e$ e$ (x 1) 2e$ (y 2) œ e$ x 2e$ y 2e$ 31. (a) Wa20, 25b œ 11‰ F; Wa30, 10b œ 39‰ F; Wa15, 15b œ 0‰ F (b) Wa10, 40b œ 65.5‰ F; Wa50, 40b œ 88‰ F; Wa60, 30b œ 10.2‰ F; 5.72 0.0684t `W `W 0.16 (c) Wa25, 5b œ 17.4088‰ F; ``W V œ v0.84 v0.84 Ê ` V a25, 5b œ 0.36; ` T œ 0.6215 0.4275v Ê ``W T a25, 5b œ 1.3370 Ê LaV, Tb œ 17.4088 0.36aV 25b 1.337aT 5b œ 1.337T 0.36V 15.0938 (d) i) Wa24, 6b ¸ La24, 6b œ 15.7118 ¸ 15.7‰ F ii) Wa27, 2b ¸ La27, 2b œ 22.1398 ¸ 22.1‰ F ii) Wa5, 10b ¸ La5, 10b œ 30.2638 ¸ 30.2‰ F This value is very different because the point a5, 10b is not close to the point a25, 5b. 5.72 0.0684t `W `W 0.16 32. Wa50, 20b œ 59.5298‰ F; ``W V œ v0.84 v0.84 Ê ` V a50, 20b œ 0.2651; ` T œ 0.6215 0.4275v Ê ``W T a50, 20b œ 1.4209 Ê LaV, Tb œ 59.5298 0.2651aV 50b 1.4209aT 20b œ 1.4209T 0.2651V 17.8568 (a) Wa49, 22b ¸ La49, 22b œ 62.1065 ¸ 62.1‰ F (b) Wa53, 19b ¸ La53, 19b œ 58.9042 ¸ 58.9‰ F (c) Wa60, 30b ¸ La60, 30b œ 76.3898 ¸ 76.4‰ F 33. f(2ß 1) œ 3, fx (xß y) œ 2x 3y Ê fx (2ß 1) œ 1, fy (xß y) œ 3x Ê fy (2ß 1) œ 6 Ê L(xß y) œ 3 1(x 2) 6(y 1) œ 7 x 6y; fxx (xß y) œ 2, fyy (xß y) œ 0, fxy (xß y) œ 3 Ê M œ 3; thus kE(xß y)k Ÿ ˆ "# ‰ (3) akx 2k ky 1kb# Ÿ ˆ 3# ‰ (0.1 0.1)# œ 0.06 34. f(2ß 2) œ 11, fx (xß y) œ x y 3 Ê fx (2ß 2) œ 7, fy (xß y) œ x y# 3 Ê fy (2ß 2) œ 0 Ê L(xß y) œ 11 7(x 2) 0(y 2) œ 7x 3; fxx (xß y) œ 1, fyy (xß y) œ "# , fxy (xß y) œ 1 Ê M œ 1; thus kE(xß y)k Ÿ ˆ "# ‰ (1) akx 2k ky 2kb# Ÿ ˆ #1 ‰ (0.1 0.1)# œ 0.02 35. f(0ß 0) œ 1, fx (xß y) œ cos y Ê fx (0ß 0) œ 1, fy (xß y) œ 1 x sin y Ê fy (0ß 0) œ 1 Ê L(xß y) œ 1 1(x 0) 1(y 0) œ x y 1; fxx (xß y) œ 0, fyy (xß y) œ x cos y, fxy (xß y) œ sin y Ê Q œ 1; thus kE(xß y)k Ÿ ˆ "# ‰ (1) akxk kykb# Ÿ ˆ #1 ‰ (0.2 0.2)# œ 0.08 36. f("ß #) œ 6, fx (xß y) œ y# y sin (x 1) Ê fx (1ß 2) œ 4, fy (xß y) œ 2xy cos (x 1) Ê fy (1ß 2) œ 5 Ê L(xß y) œ 6 4(x 1) 5(y 2) œ 4x 5y 8; fxx (xß y) œ y cos (x 1), fyy (xß y) œ 2x, fxy (xß y) œ 2y sin (x 1); kx 1k Ÿ 0.1 Ê 0.9 Ÿ x Ÿ 1.1 and ky 2k Ÿ 0.1 Ê 1.9 Ÿ y Ÿ 2.1; thus the max of kfxx (xß y)k on R is 2.1, the max of kfyy (xß y)k on R is 2.2, and the max of kfxy (xß y)k on R is 2(2.1) sin (0.9 1) Ÿ 4.3 Ê M œ 4.3; thus kE(xß y)k Ÿ ˆ "# ‰ (4.3) akx 1k ky 2kb# Ÿ (2.15)(0.1 0.1)# œ 0.086 37. f(0ß 0) œ 1, fx (xß y) œ ex cos y Ê fx (0ß 0) œ 1, fy (xß y) œ ex sin y Ê fy (0ß 0) œ 0 Ê L(xß y) œ 1 1(x 0) 0(y 0) œ 1 x; fxx (xß y) œ ex cos y, fyy (xß y) œ ex cos y, fxy (xß y) œ ex sin y; kxk Ÿ 0.1 Ê 0.1 Ÿ x Ÿ 0.1 and kyk Ÿ 0.1 Ê 0.1 Ÿ y Ÿ 0.1; thus the max of kfxx (xß y)k on R is e0Þ1 cos (0.1) Ÿ 1.11, the max of kfyy (xß y)k on R is e0Þ1 cos (0.1) Ÿ 1.11, and the max of kfxy (xß y)k on R is e0Þ1 sin (0.1) Ÿ 0.12 Ê M œ 1.11; thus kE(xß y)k Ÿ ˆ "# ‰ (1.11) akxk kykb# Ÿ (0.555)(0.1 0.1)# œ 0.0222 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.6 Tangent Planes and Differentials 833 38. f(1ß 1) œ 0, fx (xß y) œ "x Ê fx (1ß 1) œ 1, fy (xß y) œ y" Ê fy (1ß 1) œ 1 Ê L(xß y) œ 0 1(x 1) 1(y 1) œ x y 2; fxx (xß y) œ x"# , fyy (xß y) œ y"# , fxy (xß y) œ 0; kx 1k Ÿ 0.2 Ê 0.98 Ÿ x Ÿ 1.2 so the max of " kfxx (xß y)k on R is (0.98) # Ÿ 1.04; ky 1k Ÿ 0.2 Ê 0.98 Ÿ y Ÿ 1.2 so the max of kfyy (xß y)k on R is " (0.98)# Ÿ 1.04 Ê M œ 1.04; thus kE(xß y)k Ÿ ˆ #" ‰ (1.04) akx 1k ky 1kb# Ÿ (0.52)(0.2 0.2)# œ 0.0832 39. (a) f("ß "ß ") œ 3, fx (1ß 1ß 1) œ y zkÐ1ß1ß1Ñ œ 2, fy (1ß 1ß 1) œ x zkÐ1ß1ß1Ñ œ 2, fz (1ß 1ß 1) œ y xkÐ1ß1ß1Ñ œ 2 Ê L(xß yß z) œ 3 2(x 1) 2(y 1) 2(z 1) œ 2x 2y 2z 3 (b) f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ 0, fy (1ß 0ß 0) œ 1, fz (1ß 0ß 0) œ 1 Ê L(xß yß z) œ 0 0(x 1) (y 0) (z 0) œ y z (c) f(0ß 0ß 0) œ 0, fx (0ß 0ß 0) œ 0, fy (0ß 0ß 0) œ 0, fz (0ß 0ß 0) œ 0 Ê L(xß yß z) œ 0 40. (a) f(1ß 1ß 1) œ 3, fx (1ß 1ß 1) œ 2xkÐ"ß"ß"Ñ œ 2, fy (1ß 1ß 1) œ 2ykÐ"ß"ß"Ñ œ 2, fz (1ß 1ß 1) œ 2zkÐ"ß"ß"Ñ œ 2 Ê L(xß yß z) œ 3 2(x 1) 2(y 1) 2(z 1) œ 2x 2y 2z 3 (b) f(0ß 1ß 0) œ 1, fx (0ß 1ß 0) œ 0, fy (!ß 1ß 0) œ 2, fz (0ß 1ß 0) œ 0 Ê L(xß yß z) œ 1 0(x 0) 2(y 1) 0(z 0) œ 2y 1 (c) f(1ß 0ß 0) œ 1, fx (1ß 0ß 0) œ 2, fy (1ß 0ß 0) œ 0, fz (1ß 0ß 0) œ 0 Ê L(xß yß z) œ 1 2(x 1) 0(y 0) 0(z 0) œ 2x 1 41. (a) f(1ß 0ß 0) œ 1, fx (1ß 0ß 0) œ Èx# xy# z# ¹ fz (1ß 0ß 0) œ Èx# zy# z# ¹ Ð1ß0ß0Ñ Ð1ß0ß0Ñ œ 1, fy (1ß 0ß 0) œ Èx# yy# z# ¹ Ð1 ß0 ß0 Ñ œ 0, œ 0 Ê L(xß yß z) œ 1 1(x 1) 0(y 0) 0(z 0) œ x (b) f(1ß 1ß 0) œ È2, fx (1ß 1ß 0) œ È"2 , fy (1ß 1ß 0) œ È"2 , fz (1ß 1ß 0) œ 0 Ê L(xß yß z) œ È2 È"2 (x 1) È"2 (y 1) 0(z 0) œ È"2 x È"2 y (c) f(1ß 2ß 2) œ 3, fx (1ß 2ß 2) œ "3 , fy (1ß 2ß 2) œ 23 , fz (1ß 2ß 2) œ 23 Ê L(xß yß z) œ 3 "3 (x 1) 23 (y 2) 23 (z 2) œ "3 x 32 y 32 z 42. (a) f ˆ 12 ß 1ß 1‰ œ 1, fx ˆ 1# ß 1ß 1‰ œ y cosz xy ¸ ˆ 1 ß"ß"‰ œ 0, fy ˆ 1# ß 1ß 1‰ œ x cosz xy ¸ ˆ 1 ß"ß"‰ œ 0, # xy fz ˆ 1# ß 1ß 1‰ œ sin z# ¹ ˆ 1 ß"ß"‰ œ 1 Ê # # L(xß yß z) œ 1 0 ˆx 1# ‰ 0(y 1) 1(z 1) œ 2 z (b) f(2ß 0ß 1) œ 0, fx (2ß 0ß 1) œ 0, fy (2ß 0ß 1) œ 2, fz (2ß 0ß 1) œ 0 Ê L(xß yß z) œ 0 0(x 2) 2(y 0) 0(z 1) œ 2y 43. (a) f(0ß 0ß 0) œ 2, fx (0ß 0ß 0) œ ex k Ð!ß!ß!Ñ œ 1, fy (0ß 0ß 0) œ sin (y z)k Ð!ß!ß!Ñ œ 0, fz (0ß 0ß 0) œ sin (y z)k Ð!ß!ß!Ñ œ 0 Ê L(xß yß z) œ 2 1(x 0) 0(y 0) 0(z 0) œ 2 x (b) f ˆ0ß 1# ß 0‰ œ 1, fx ˆ0ß 1# ß 0‰ œ 1, fy ˆ0ß 1# ß 0‰ œ 1, fz ˆ0ß 1# ß 0‰ œ 1 Ê L(xß yß z) œ 1 1(x 0) 1 ˆy 12 ‰ 1(z 0) œ x y z 1# 1 (c) f ˆ0ß 14 ß 14 ‰ œ 1, fx ˆ0ß 14 ß 14 ‰ œ 1, fy ˆ0ß 14 ß 14 ‰ œ 1, fz ˆ0ß 14 ß 14 ‰ œ 1 Ê L(xß yß z) œ 1 1(x 0) 1 ˆy 14 ‰ 1 ˆz 14 ‰ œ x y z 1# 1 44. (a) f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ (xyz)yz# 1 ¹ fz (1ß 0ß 0) œ (xyz)xy# 1 ¹ Ð"ß!ß!Ñ Ð"ß!ß!Ñ œ 0, fy (1ß 0ß 0) œ (xyz)xz# 1 ¹ Ð"ß!ß!Ñ œ 0, œ 0 Ê L(xß yß z) œ 0 (b) f(1ß 1ß 0) œ 0, fx (1ß 1ß 0) œ 0, fy (1ß 1ß 0) œ 0, fz (1ß 1ß 0) œ 1 Ê L(xß yß z) œ 0 0(x 1) 0(y 1) 1(z 0) œ z (c) f(1ß 1ß 1) œ 14 , fx (1ß 1ß 1) œ "# , fy (1ß 1ß 1) œ "# , fz (1ß 1ß 1) œ "# Ê L(xß yß z) œ 14 "# (x 1) "# (y 1) "# (z 1) œ "# x "# y "# z 14 #3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 834 Chapter 14 Partial Derivatives 45. f(xß yß z) œ xz 3yz 2 at P! (1ß 1ß 2) Ê f(1ß 1ß 2) œ 2; fx œ z, fy œ 3z, fz œ x 3y Ê L(xß yß z) œ 2 2(x 1) 6(y 1) 2(z 2) œ 2x 6y 2z 6; fxx œ 0, fyy œ 0, fzz œ 0, fxy œ 0, fyz œ 3 Ê M œ 3; thus, kE(xß yß z)k Ÿ ˆ "# ‰ (3)(0.01 0.01 0.02)# œ 0.0024 46. f(xß yß z) œ x# xy yz "4 z# at P! (1ß 1ß 2) Ê f(1ß 1ß 2) œ 5; fx œ 2x y, fy œ x z, fz œ y "# z Ê L(xß yß z) œ 5 3(x 1) 3(y 1) 2(z 2) œ 3x 3y 2z 5; fxx œ 2, fyy œ 0, fzz œ "# , fxy œ 1, fxz œ 0, fyz œ 1 Ê M œ 2; thus kE(xß yß z)k Ÿ ˆ "# ‰ (2)(0.01 0.01 0.08)# œ 0.01 47. f(xß yß z) œ xy 2yz 3xz at P! (1ß 1ß 0) Ê f(1ß 1ß 0) œ 1; fx œ y 3z, fy œ x 2z, fz œ 2y 3x Ê L(xß yß z) œ 1 (x 1) (y 1) (z 0) œ x y z 1; fxx œ 0, fyy œ 0, fzz œ 0, fxy œ 1, fxz œ 3, fyz œ 2 Ê M œ 3; thus kE(xß yß z)k Ÿ ˆ "# ‰ (3)(0.01 0.01 0.01)# œ 0.00135 48. f(xß yß z) œ È2 cos x sin (y z) at P! ˆ0ß 0ß 14 ‰ Ê f ˆ0ß 0ß 14 ‰ œ 1; fx œ È2 sin x sin (y z), fy œ È2 cos x cos (y z), fz œ È2 cos x cos (y z) Ê L(xß yß z) œ 1 0(x 0) (y 0) ˆz 14 ‰ œ y z 14 1; fxx œ È2 cos x sin (y z), fyy œ È2 cos x sin (y z), fzz œ È2 cos x sin (y z), fxy œ È2 sin x cos (y z), fxz œ È2 sin x cos (y z), fyz œ È2 cos x sin (y z). The absolute value of each of these second partial derivatives is bounded above by È2 Ê M œ È2; thus kE(xß yß z)k Ÿ ˆ " ‰ ŠÈ2‹ (0.01 0.01 0.01)# œ 0.000636. # 49. Tx (xß y) œ ey ey and Ty (xß y) œ x aey ey b Ê dT œ Tx (xß y) dx Ty (xß y) dy œ aey ey b dx x aey ey b dy Ê dTkÐ#ßln 2Ñ œ 2.5 dx 3.0 dy. If kdxk Ÿ 0.1 and kdyk Ÿ 0.02, then the maximum possible error in the computed value of T is (2.5)(0.1) (3.0)(0.02) œ 0.31 in magnitude. # 21rh dr 1r dh 50. Vr œ 21rh and Vh œ 1r# Ê dV œ Vr dr Vh dh Ê dV œ 2r dr "h dh; now ¸ drr † 100¸ Ÿ 1 and 1 r# h V œ ¸ dh ¸ ¸ dV ¸ ¸ˆ2 drr ‰ (100) ˆ dh ‰ ¸ ¸ dr ¸ ¸ dh ¸ h † 100 Ÿ 1 Ê V † 100 Ÿ h (100) Ÿ 2 r † 100 h † 100 Ÿ 2(1) 1 œ 3 Ê 3% dy 51. dx x Ÿ 0.02, y Ÿ 0.03 dy 2 (a) S œ 2x2 4xy Ê dS œ a4x 4ybdx 4x dy œ a4x2 4xyb dx x 4xy y Ÿ a4x 4xyba0.02b a4xyba0.03b œ 0.04a2x2 b 0.05a4xyb Ÿ 0.05a2x2 b 0.05a4xyb œ a0.05ba2x2 4xyb œ 0.05S 2 dy 2 2 2 (b) V œ x2 y Ê dV œ 2xy dx x2 dy œ 2x2 y dx x x y y Ÿ a2x yba0.02b ax yba0.03b œ 0.07ax yb=0.07V 52. V œ 431 r3 1 r2 h Ê dV œ a41 r2 21 rhbdr 1 r2 dh; r œ 10, h œ 15, dr œ 12 and dh œ 0 Ê dV œ Š41a10b2 21 a10ba15b‹ˆ 12 ‰ 1 a10b2 a0b œ 3501 cm3 53. Vr œ 21rh and Vh œ 1r# Ê dV œ Vr dr Vh dh Ê dV œ 21rh dr 1r# dh Ê dVkÐ5ß12Ñ œ 1201 dr 251 dh; kdrk Ÿ 0.1 cm and kdhk Ÿ 0.1 cm Ê dV Ÿ (1201)(0.1) (251)(0.1) œ 14.51 cm$ ; V(5ß 12) œ 3001 cm$ 1 Ê maximum percentage error is „ 14.5 3001 ‚ 100 œ „ 4.83% 54. (a) # " " " R œ R" R# # Ê R"# dR œ R"# dR" R"# dR# Ê dR œ Š RR" ‹ dR" Š RR# ‹ dR# " # " " (b) dR œ R# ’Š R"# ‹ dR" Š R"# ‹ dR# “ Ê dRk Ð100 400Ñ œ R# ’ (100) # dR" (400)# dR# “ Ê R will be more " # ß " " sensitive to a variation in R" since (100) # (400)# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.6 Tangent Planes and Differentials # 835 # (c) From part (a), dR œ Š RR" ‹ dR" Š RR# ‹ dR# so that R" changing from 20 to 20.1 ohms Ê dR" œ 0.1 ohm and R# changing from 25 to 24.9 ohms Ê dR# œ 0.1 ohms; R" œ R"" R"# Ê R œ 100 9 ohms ˆ 100 ‰# ˆ 100 ‰# dR ¸ 9 9 Ê dRk Ð20 25Ñ œ (20) ‚ 100 # (0.1) (25)# (0.1) ¸ 0.011 ohms Ê percentage change is R Ð20 25Ñ ß ß œ 0.011 ˆ 100 ‰ ‚ 100 ¸ 0.1% 9 55. A œ xy Ê dA œ x dy y dx; if x y then a 1-unit change in y gives a greater change in dA than a 1-unit change in x. Thus, pay more attention to y which is the smaller of the two dimensions. 56. (a) fx (xß y) œ 2x(y 1) Ê fx (1ß 0) œ 2 and fy (xß y) œ x# Ê fy (1ß 0) œ 1 Ê df œ 2 dx 1 dy Ê df is more sensitive to changes in x dx " (b) df œ 0 Ê 2 dx dy œ 0 Ê 2 dx dy 1 œ 0 Ê dy œ # 57. (a) r# œ x# y# Ê 2r dr œ 2x dx 2y dy Ê dr œ xr dx yr dy Ê dr|Ð$ß%Ñ œ ˆ 35 ‰ a „ 0.01b ˆ 45 ‰ a „ 0.01b Š y #‹ x 1 x ¸ drr ‚ 100¸ œ ¸ „ 0.014 ¸ œ „ 0.07 5 œ „ 0.014 Ê 5 ‚ 100 œ 0.28%; d) œ ˆ y ‰# Š"‹ dx ˆ y ‰#x 0.04 3 ‰ œ y#yx# dx y# x x# dy Ê d)|Ð$ß%Ñ œ ˆ 254 ‰ a „ 0.01b ˆ 25 a „ 0.01b œ …25 „#0.03 5 x 1 dy Ê maximum change in d) occurs when dx and dy have opposite signs (dx œ 0.01 and dy œ 0.01 or vice „0.0028 " ˆ 4 ‰ ¸ d)) ‚ 100¸ œ ¸ 0.927255218 versa) Ê d) œ „#0.07 ‚ 100¸ 5 ¸ „ 0.0028; ) œ tan 3 ¸ 0.927255218 Ê ¸ 0.30% (b) the radius r is more sensitive to changes in y, and the angle ) is more sensitive to changes in x 58. (a) V œ 1r# h Ê dV œ 21rh dr 1r# dh Ê at r œ 1 and h œ 5 we have dV œ 101 dr 1 dh Ê the volume is about 10 times more sensitive to a change in r " (b) dV œ 0 Ê 0 œ 21rh dr 1r# dh œ 2h dr r dh œ 10 dr dh Ê dr œ 10 dh; choose dh œ 1.5 Ê dr œ 0.15 Ê h œ 6.5 in. and r œ 0.85 in. is one solution for ?V ¸ dV œ 0 59. f(aß bß cß d) œ º a b œ ad bc Ê fa œ d, fb œ c, fc œ b, fd œ a Ê df œ d da c db b dc a dd; since c dº kak is much greater than kbk , kck , and kdk , the function f is most sensitive to a change in d. 60. ux œ ey , uy œ xey sin z, uz œ y cos z Ê du œ ey dx axey sin zb dy (y cos z) dz Ê duk ˆ2ßln 3ß 12 ‰ œ 3 dx 7 dy 0 dz œ 3 dx 7 dy Ê magnitude of the maximum possible error Ÿ 3(0.2) 7(0.6) œ 4.8 ‰ 61. QK œ "# ˆ 2KM h "Î# " ˆ 2KM ‰"Î# ˆ 2KM ‰ ˆ 2M ‰ , QM œ "# ˆ 2KM ‰"Î# ˆ 2K ‰ h h h , and Qh œ # h h# ‰ Ê dQ œ "# ˆ 2KM h ‰ œ "# ˆ 2KM h "Î# ˆ 2M ‰ dK "# ˆ 2KM ‰"Î# ˆ 2K ‰ dM "# ˆ 2KM ‰"Î# ˆ 2KM ‰ dh h h h h h# 2K 2KM 2M ‘ h dK h dM h# dh Ê dQk Ð2ß20ß0Þ0.05Ñ "Î# œ "# ’ (2)(2)(20) 0.05 “ "Î# (2)(2) (2)(2)(20) ’ (2)(20) 0.05 dK 0.05 dM (0.05)# dh“ œ (0.0125)(800 dK 80 dM 32,000 dh) Ê Q is most sensitive to changes in h 62. A œ "# ab sin C Ê Aa œ "# b sin C, Ab œ "# a sin C, Ac œ "# ab cos C Ê dA œ ˆ "# b sin C‰ da ˆ "# a sin C‰ db ˆ "# ab cos C‰ dC; dC œ k2°k œ k0.0349k radians, da œ k0.5k ft, db œ k0.5k ft; at a œ 150 ft, b œ 200 ft, and C œ 60°, we see that the change is approximately dA œ "# (200)(sin 60°) k0.5k "# (150)(sin 60°) k0.5k "# (200)(150)(cos 60°) k0.0349k œ „ 338 ft# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 836 Chapter 14 Partial Derivatives 63. z œ f(xß y) Ê g(xß yß z) œ f(xß y) z œ 0 Ê gx (xß yß z) œ fx (xß y), gy (xß yß z) œ fy (xß y) and gz (xß yß z) œ 1 Ê gx (x! ß y! ß f(x! ß y! )) œ fx (x! ß y! ), gy (x! ß y! ß f(x! ß y! )) œ fy (x! ß y! ) and gz (x! ß y! ß f(x! ß y! )) œ 1 Ê the tangent plane at the point P! is fx (x! ß y! )(x x! ) fy (x! ß y! )(y y! ) [z f(x! ß y! )] œ 0 or z œ fx (x! ß y! )(x x! ) fy (x! ß y! )(y y! ) f(x! ß y! ) 64. ™ f œ 2xi 2yj œ 2(cos t t sin t)i 2(sin t t cos t)j and v œ (t cos t)i (t sin t)j Ê u œ kvvk t)i (t sin t)j œ È(t(tcos œ (cos t)i (sin t)j since t 0 Ê (Du f)P! œ ™ f † u cos t)# (t sin t)# œ 2(cos t t sin t)(cos t) 2(sin t t cos t)(sin t) œ 2 65. ™ f œ 2xi 2yj 2zk œ (2 cos t)i (2 sin t)j 2tk and v œ ( sin t)i (cos t)j k Ê u œ kvvk sin t)i (cos t)j k t " œ È((sin œ Š Èsin t ‹ i Š cos È ‹ j È k Ê (Du f)P! œ ™ f † u t)# (cos t)# 1# 2 2 2 t " 2t œ (2 cos t) Š Èsin2 t ‹ (2 sin t) Š cos È2 ‹ (2t) Š È2 ‹ œ È2 (Du f) ˆ 14 ‰ œ Ê (Du f) ˆ 41 ‰ œ 2È12 , (Du f)(0) œ 0 and 1 2È 2 66. r œ Èti Ètj 4" (t 3)k Ê v œ #" t"Î# i "# t"Î# j 4" k ; t œ 1 Ê x œ 1, y œ 1, z œ 1 Ê P! œ (1ß 1ß 1) and v(1) œ "# i "# j "4 k ; f(xß yß z) œ x# y# z 3 œ 0 Ê ™ f œ 2xi 2yj k Ê ™ f(1ß 1ß 1) œ 2i 2j k ; therefore v œ "4 ( ™ f) Ê the curve is normal to the surface 67. r œ Èti Ètj (2t 1)k Ê v œ "# t"Î# i "# t"Î# j 2k ; t œ 1 Ê x œ 1, y œ 1, z œ 1 Ê P! œ (1ß 1ß 1) and v(1) œ "# i "# j 2k ; f(xß yß z) œ x# y# z 1 œ 0 Ê ™ f œ 2xi 2yj k Ê ™ f(1ß 1ß 1) œ 2i 2j k ; now va1b † ™ fa1ß 1ß 1b œ 0, thus the curve is tangent to the surface when t œ 1 14.7 EXTREME VALUES AND SADDLE POINTS 1. fx (xß y) œ 2x y 3 œ 0 and fy (xß y) œ x 2y 3 œ 0 Ê x œ 3 and y œ 3 Ê critical point is (3ß 3); # œ 3 0 and fxx 0 Ê local minimum of fxx (3ß 3) œ 2, fyy (3ß 3) œ 2, fxy (3ß 3) œ 1 Ê fxx fyy fxy f(3ß 3) œ 5 2. fx (xß y) œ 2y 10x 4 œ 0 and fy (xß y) œ 2x 4y 4 œ 0 Ê x œ 23 and y œ 43 Ê critical point is ˆ 23 ß 43 ‰ ; # œ 36 0 and fxx 0 Ê local maximum of fxx ˆ 23 ß 43 ‰ œ 10, fyy ˆ 23 ß 43 ‰ œ 4, fxy ˆ 23 ß 43 ‰ œ 2 Ê fxx fyy fxy f ˆ 23 ß 43 ‰ œ 0 3. fx (xß y) œ 2x y 3 œ 0 and fy (xß y) œ x 2 œ 0 Ê x œ 2 and y œ 1 Ê critical point is (2ß 1); # œ 1 0 Ê saddle point fxx (2ß 1) œ 2, fyy (2ß 1) œ 0, fxy (2ß 1) œ 1 Ê fxx fyy fxy ˆ 6 69 ‰ 4. fx (xß y) œ 5y 14x 3 œ 0 and fy (xß y) œ 5x 6 œ 0 Ê x œ 65 and y œ 69 #5 Ê critical point is 5 ß 25 ; # ‰ ˆ 6 69 ‰ ˆ 6 69 ‰ fxx ˆ 65 ß 69 25 œ 14, fyy 5 ß 25 œ 0, fxy 5 ß 25 œ 5 Ê fxx fyy fxy œ 25 0 Ê saddle point 5. fx (xß y) œ 2y 2x 3 œ 0 and fy (xß y) œ 2x 4y œ 0 Ê x œ 3 and y œ 3# Ê critical point is ˆ3ß 32 ‰ ; # œ 4 0 and fxx 0 Ê local maximum of fxx ˆ3ß 32 ‰ œ 2, fyy ˆ3ß 32 ‰ œ 4, fxy ˆ3ß 32 ‰ œ 2 Ê fxx fyy fxy f ˆ3ß 3# ‰ œ 17 # 6. fx (xß y) œ 2x 4y œ 0 and fy (xß y) œ 4x 2y 6 œ 0 Ê x œ 2 and y œ 1 Ê critical point is (2ß 1); # œ 12 0 Ê saddle point fxx (2ß 1) œ 2, fyy (2ß 1) œ 2, fxy (2ß 1) œ 4 Ê fxx fyy fxy Copyright © 2010 Pearson Education Inc. 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Section 14.7 Extreme Values and Saddle Points 837 7. fx (xß y) œ 4x 3y 5 œ 0 and fy (xß y) œ 3x 8y 2 œ 0 Ê x œ 2 and y œ 1 Ê critical point is (2ß 1); # œ 23 0 and fxx 0 Ê local minimum of f(2ß 1) œ 6 fxx (2ß 1) œ 4, fyy (2ß 1) œ 8, fxy (2ß 1) œ 3 Ê fxx fyy fxy 8. fx (xß y) œ 2x 2y 2 œ 0 and fy (xß y) œ 2x 4y 2 œ 0 Ê x œ 1 and y œ 0 Ê critical point is (1ß 0); # œ 4 0 and fxx 0 Ê local minimum of f(1ß 0) œ 0 fxx (1ß 0) œ 2, fyy (1ß 0) œ 4, fxy (1ß 0) œ 2 Ê fxx fyy fxy 9. fx (xß y) œ 2x 2 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and y œ 2 Ê critical point is (1ß 2); fxx (1ß 2) œ 2, # œ 4 0 Ê saddle point fyy (1ß 2) œ 2, fxy (1ß 2) œ 0 Ê fxx fyy fxy 10. fx (xß y) œ 2x 2y œ 0 and fy (xß y) œ 2x œ 0 Ê x œ 0 and y œ 0 Ê critical point is (0ß 0); fxx (0ß 0) œ 2, # œ 4 0 Ê saddle point fyy (0ß 0) œ 0, fxy (0ß 0) œ 2 Ê fxx fyy fxy 8y 8x ‰ 11. fx axß yb œ È56x2 112x 8 œ 0 and fy axß yb œ È56x2 8y œ 0 Ê critical point is ˆ 16 2 16x 31 7 ß0 ; 8y2 16x 31 8 8 64 # ‰ ˆ 16 ‰ ˆ 16 ‰ fxx ˆ 16 7 ß 0 œ 15 , fyy 7 ß 0 œ 15 , fxy 7 ß 0 œ 0 Ê fxx fyy fxy œ 225 0 and fxx 0 Ê local maximum of 16 ‰ fˆ 16 7 ß0 œ 7 12. fx axß yb œ 2x œ 0 and fy axß yb œ 2 2y2 2Î3 œ 0 Ê there are no solutions to the system fx axß yb œ 0 and 3ax2 y2 b2Î3 3ax y b fy axß yb œ 0, however, we must also consider where the partials are undefined, and this occurs when x œ 0 and y œ 0 Ê critical point is a0ß 0b. Note that the partial derivatives are defined at every other point other than a0ß 0b. We cannot use the second derivative test, but this is the only possible local maximum, local minimum, or saddle point. faxß yb has a local 3 maximum of fa0ß 0b œ 1 at a0ß 0b since faxß yb œ 1 È x2 y2 Ÿ 1 for all axß yb other than a0ß 0b. 13. fx (xß y) œ 3x# 2y œ 0 and fy (xß y) œ 3y# 2x œ 0 Ê x œ 0 and y œ 0, or x œ 23 and y œ 23 Ê critical points are (0ß 0) and ˆ 23 ß 23 ‰ ; for (0ß 0): fxx (0ß 0) œ 6xk Ð0ß0Ñ œ 0, fyy (0ß 0) œ 6yk Ð0ß0Ñ œ 0, fxy (0ß 0) œ 2 # Ê fxx fyy fxy œ 4 0 Ê saddle point; for ˆ 32 ß 32 ‰ : fxx ˆ 32 ß 32 ‰ œ 4, fyy ˆ 32 ß 32 ‰ œ 4, fxy ˆ 32 ß 32 ‰ œ 2 # Ê fxx fyy fxy œ 12 0 and fxx 0 Ê local maximum of f ˆ 23 ß 32 ‰ œ 170 27 14. fx (xß y) œ 3x# 3y œ 0 and fy (xß y) œ 3x 3y# œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1 Ê critical points # are (0ß 0) and (1ß 1); for (!ß !): fxx (0ß 0) œ 6xk Ð0ß0Ñ œ 0, fyy (0ß 0) œ 6yk Ð0ß0Ñ œ 0, fxy (0ß 0) œ 3 Ê fxx fyy fxy # œ 9 0 Ê saddle point; for (1ß 1): fxx (1ß 1) œ 6, fyy (1ß 1) œ 6, fxy (1ß 1) œ 3 Ê fxx fyy fxy œ 27 0 and fxx 0 Ê local maximum of f(1ß 1) œ 1 15. fx (xß y) œ 12x 6x# 6y œ 0 and fy (xß y) œ 6y 6x œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1 Ê critical # points are (0ß 0) and (1ß 1); for (!ß !): fxx (0ß 0) œ 12 12xk Ð0ß0Ñ œ 12, fyy (0ß 0) œ 6, fxy (0ß 0) œ 6 Ê fxx fyy fxy œ 36 0 and fxx 0 Ê local minimum of f(0ß 0) œ 0; for (1ß 1): fxx (1ß 1) œ 0, fyy (1ß 1) œ 6, # fxy (1ß 1) œ 6 Ê fxx fyy fxy œ 36 0 Ê saddle point 16. fx (xß y) œ 3x# 6x œ 0 Ê x œ 0 or x œ 2; fy (xß y) œ 3y# 6y œ 0 Ê y œ 0 or y œ 2 Ê the critical points are (0ß 0), (0ß 2), (2ß 0), and (2ß 2); for (!ß !): fxx (0ß 0) œ 6x 6k Ð0ß0Ñ œ 6, fyy (0ß 0) œ 6y 6k Ð0ß0Ñ œ 6, # fxy (0ß 0) œ 0 Ê fxx fyy fxy œ 36 0 Ê saddle point; for (0ß 2): fxx (0ß 2) œ 6, fyy (0ß 2) œ 6, fxy (0ß 2) œ 0 # Ê fxx fyy fxy œ 36 0 and fxx 0 Ê local minimum of f(0ß 2) œ 12; for (2ß 0): fxx (2ß 0) œ 6, # fyy (2ß 0) œ 6, fxy (2ß 0) œ 0 Ê fxx fyy fxy œ 36 0 and fxx 0 Ê local maximum of f(2ß 0) œ 4; # for (2ß 2): fxx (2ß 2) œ 6, fyy (2ß 2) œ 6, fxy (2ß 2) œ 0 Ê fxx fyy fxy œ 36 0 Ê saddle point Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 838 Chapter 14 Partial Derivatives 17. fx axß yb œ 3x2 3y2 15 œ 0 and fy axß yb œ 6x y 3y2 15 œ 0 Ê critical points are a2ß 1b, a2ß 1b, Š0ß È5‹, and Š0ß È5‹; for a2ß 1b: fxx a2ß 1b œ 6xk a2ß1b œ 12, fyy a2ß 1b œ a6x 6ybk a2ß1b œ 18, fxy a2ß 1b œ 6yk a2ß1b œ 6 # Ê fxx fyy fxy œ 180 0 and fxx 0 Ê local minimum of fa2ß 1b œ 30; for a2ß 1b: fxx a2ß 1b œ 6xk a2ß1b # œ 12, fyy a2ß 1b œ a6x 6ybk a2ß1b œ 18, fxy a2ß 1b œ 6yk a2ß1b œ 6 Ê fxx fyy fxy œ 180 0 and fxx 0 Ê local maximum of fa2ß 1b œ 30; for Š0ß È5‹: fxx Š0ß È5‹ œ 6x¹ œ a6x 6ybk Š0ßÈ5‹ œ 6È5, fxy Š0ß È5‹ œ 6y¹ for Š0ß È5‹: fxx Š0ß È5‹ œ 6x¹ fxy Š0ß È5‹ œ 6y¹ Š0ßÈ5‹ Š0ßÈ5‹ Š0ßÈ5‹ Š0ßÈ5‹ œ 0, fyy Š0ß È5‹ # œ 6È5 Ê fxx fyy fxy œ 180 0 Ê saddle pointà œ 0, fyy Š0ß È5‹ œ a6x 6ybk Š0ßÈ5‹ œ 6È5, # œ 6È5 Ê fxx fyy fxy œ 180 0 Ê saddle point. 18. fx (xß y) œ 6x# 18x œ 0 Ê 6x(x 3) œ 0 Ê x œ 0 or x œ 3; fy (xß y) œ 6y# 6y 12 œ 0 Ê 6(y 2)(y 1) œ 0 Ê y œ 2 or y œ 1 Ê the critical points are (0ß 2), (0ß 1), (3ß 2), and (3ß 1); fxx (xß y) œ 12x 18, fyy (xß y) œ 12y 6, and fxy (xß y) œ 0; for (!ß 2): fxx (0ß 2) œ 18, fyy (0ß 2) œ 18, fxy (0ß 2) œ 0 # Ê fxx fyy fxy œ 324 0 and fxx 0 Ê local maximum of f(0ß 2) œ 20; for (0ß 1): fxx (0ß 1) œ 18, # fyy (0ß 1) œ 18, fxy (0ß 1) œ 0 Ê fxx fyy fxy œ 324 0 Ê saddle point; for (3ß 2): fxx (3ß 2) œ 18, # fyy (3ß 2) œ 18, fxy (3ß 2) œ 0 Ê fxx fyy fxy œ 324 0 Ê saddle point; for (3ß 1): fxx (3ß 1) œ 18, # fyy (3ß 1) œ 18, fxy (3ß 1) œ 0 Ê fxx fyy fxy œ 324 0 and fxx 0 Ê local minimum of f(3ß 1) œ 34 19. fx (xß y) œ 4y 4x$ œ 0 and fy (xß y) œ 4x 4y$ œ 0 Ê x œ y Ê x a1 x# b œ 0 Ê x œ 0, 1, 1 Ê the critical points are (0ß 0), (1ß 1), and (1ß 1); for (!ß !): fxx (0ß 0) œ 12x# k Ð0ß0Ñ œ 0, fyy (0ß 0) œ 12y# k Ð0ß0Ñ œ 0, # fxy (0ß 0) œ 4 Ê fxx fyy fxy œ 16 0 Ê saddle point; for (1ß 1): fxx (1ß 1) œ 12, fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 # Ê fxx fyy fxy œ 128 0 and fxx 0 Ê local maximum of f(1ß 1) œ 2; for (1ß 1): fxx (1ß 1) œ 12, # fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 Ê fxx fyy fxy œ 128 0 and fxx 0 Ê local maximum of f(1ß 1) œ 2 20. fx (xß y) œ 4x$ 4y œ 0 and fy (xß y) œ 4y$ 4x œ 0 Ê x œ y Ê x$ x œ 0 Ê x a1 x# b œ 0 Ê x œ 0, 1, 1 Ê the critical points are (0ß 0), (1ß 1), and (1ß 1); fxx (xß y) œ 12x# , fyy (xß y) œ 12y# , and fxy (xß y) œ 4; # for (!ß 0): fxx (0ß 0) œ 0, fyy (0ß 0) œ 0, fxy (0ß 0) œ 4 Ê fxx fyy fxy œ 16 0 Ê saddle point; for (1ß 1): # fxx (1ß 1) œ 12, fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 Ê fxx fyy fxy œ 128 0 and fxx 0 Ê local minimum of # f("ß 1) œ 2; for (1ß 1): fxx (1ß 1) œ 12, fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 Ê fxx fyy fxy œ 128 0 and fxx 0 Ê local minimum of f(1ß 1) œ 2 21. fx (xß y) œ ax# y2x œ 0 and fy (xß y) œ ax# y2y œ 0 Ê x œ 0 and y œ 0 Ê the critical point is (!ß 0); # 1b# # 1b# # # # # fxx œ a4xx# y2y# 1b2$ , fyy œ ax2x# y#4y 1b$2 , fxy œ ax# 8xy ; fxx (!ß !) œ 2, fyy (0ß 0) œ 2, fxy (0ß 0) œ 0 y# 1b$ # Ê fxx fyy fxy œ 4 0 and fxx 0 Ê local maximum of f(0ß 0) œ 1 22. fx (xß y) œ x1# y œ 0 and fy (xß y) œ x y1# œ 0 Ê x œ 1 and y œ 1 Ê the critical point is (1ß 1); fxx œ x2$ , fyy œ y2$ , # fxy œ 1; fxx (1ß 1) œ 2, fyy (1ß 1) œ 2, fxy (1ß 1) œ 1 Ê fxx fyy fxy œ 3 0 and fxx 2 Ê local minimum of f(1ß 1) œ 3 23. fx (xß y) œ y cos x œ 0 and fy (xß y) œ sin x œ 0 Ê x œ n1, n an integer, and y œ 0 Ê the critical points are (n1ß 0), n an integer (Note: cos x and sin x cannot both be 0 for the same x, so sin x must be 0 and y œ 0); fxx œ y sin x, fyy œ 0, fxy œ cos x; fxx (n1ß 0) œ 0, fyy (n1ß 0) œ 0, fxy (n1ß 0) œ 1 if n is even and fxy (n1ß 0) œ 1 # if n is odd Ê fxx fyy fxy œ 1 0 Ê saddle point. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.7 Extreme Values and Saddle Points 839 24. fx (xß y) œ 2e2x cos y œ 0 and fy (xß y) œ e2x sin y œ 0 Ê no solution since e2x Á 0 for any x and the functions cos y and sin y cannot equal 0 for the same y Ê no critical points Ê no extrema and no saddle points 25. fx axß yb œ a2x 4bex y 4x œ 0 and fy axß yb œ 2yex y 4x œ 0 Ê critical point is a2ß 0b; fxx a2ß 0b œ e24 , fxy a2ß 0b œ 0, 2 2 2 2 # fyy a2ß 0b œ e24 Ê fxx fyy fxy œ e48 0 and fxx 0 Ê local mimimum of fa2ß 0b œ e14 26. fx axß yb œ yex œ 0 and fy axß yb œ ey ex œ 0 Ê critical point is a0ß 0b; fxx a2ß 0b œ 0, fxy a2ß 0b œ 1, fyy a2ß 0b œ 1 # Ê fxx fyy fxy œ 1 0 Ê saddle point 27. fx axß yb œ 2xey œ 0 and fy axß yb œ 2yey ey ax2 y2 b œ 0 Ê critical points are a0ß 0b and a0ß 2b; for a0ß 0b: fxx a0ß 0b œ 2ey k a0ß0b œ 2, fyy a0ß 0b œ a2ey 4yey ey ax2 y2 bbk a0ß0b œ 2, fxy a0ß 0b œ 2xey k a0ß0b œ 0 # Ê fxx fyy fxy œ 4 0 and fxx 0 Ê local mimimum of fa0ß 0b œ 0; for a0ß 2b: fxx a0ß 2b œ 2ey k a0ß2b œ e22 , # fyy a0ß 2b œ a2ey 4yey ey ax2 y2 bbk a0ß2b œ e22 , fxy a0ß 2b œ 2xey k a0ß2b œ 0 Ê fxx fyy fxy œ e44 0 Ê saddle point 28. fx axß yb œ ex ax2 2x y2 b œ 0 and fy axß yb œ 2yex œ 0 Ê critical points are a0ß 0b and a2ß 0b; for a0ß 0b: fxx a0ß 0b œ ex ax2 4x 2 y2 bk a0ß0b œ 2, fyy a0ß 0b œ 2ex k a0ß0b œ 2, fxy a0ß 0b œ 2yex k a0ß0b œ 0 # Ê fxx fyy fxy œ 4 0 and fxx 0 Ê saddle point; for a2ß 0b: fxx a2ß 0b œ ex ax2 4x 2 y2 bk a2ß0b œ e22 , # fyy a2ß 0b œ 2ex k a2ß0b œ e22 , fxy a2ß 0b œ 2yex k a2ß0b œ 0 Ê fxx fyy fxy œ e44 0 and fxx 0 Ê local maximum of fa2ß 0b œ e42 29. fx axß yb œ 4 x2 œ 0 and fy axß yb œ 1 y1 œ 0 Ê critical point is ˆ 21 , 1‰ ; fxx ˆ 21 , 1‰ œ 8, fyy ˆ 12 , 1‰ œ 1, # œ 8 0 and fxx 0 Ê local maximum of fˆ 12 , 1‰ œ 3 2ln 2 fxy ˆ 12 , 1‰ œ 0 Ê fxx fyy fxy 30. fx axß yb œ 2x x 1 y œ 0 and fy axß yb œ 1 x 1 y œ 0 Ê critical point is ˆ 21 , 23 ‰ ; fxx ˆ 21 , 23 ‰ œ 1, fyy ˆ 12 , 32 ‰ œ 1, # œ 2 0 Ê saddle point fxy ˆ 12 , 32 ‰ œ 1 Ê fxx fyy fxy On OA, f(xß y) œ f(0ß y) œ y# 4y 1 on 0 Ÿ y Ÿ 2; f w (0ß y) œ 2y 4 œ 0 Ê y œ 2; f(0ß 0) œ 1 and f(!ß #) œ 3 (ii) On AB, f(xß y) œ f(xß 2) œ 2x# 4x 3 on 0 Ÿ x Ÿ 1; f w (xß 2) œ 4x 4 œ 0 Ê x œ 1; f(0ß 2) œ 3 and f(1ß #) œ 5 (iii) On OB, f(xß y) œ f(xß 2x) œ 6x# 12x 1 on 0 Ÿ x Ÿ 1; endpoint values have been found above; f w (xß 2x) œ 12x 12 œ 0 Ê x œ 1 and y œ 2, but ("ß #) is not an interior point of OB (iv) For interior points of the triangular region, fx (xß y) œ 4x 4 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and y œ 2, but (1ß 2) is not an interior point of the region. Therefore, the absolute maximum is 1 at (0ß 0) and the absolute minimum is 5 at ("ß #). 31. (i) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 840 Chapter 14 Partial Derivatives On OA, D(xß y) œ D(0ß y) œ y# 1 on 0 Ÿ y Ÿ 4; Dw (0ß y) œ 2y œ 0 Ê y œ 0; D(!ß !) œ 1 and D(!ß %) œ 17 (ii) On AB, D(xß y) œ D(xß 4) œ x# 4x 17 on 0 Ÿ x Ÿ 4; Dw (xß 4) œ 2x 4 œ 0 Ê x œ 2 and (2ß 4) is an interior point of AB; D(#ß %) œ 13 and D(%ß %) œ D(!ß %) œ 17 (iii) On OB, D(xß y) œ D(xß x) œ x# 1 on 0 Ÿ x Ÿ 4; Dw (xß x) œ 2x œ 0 Ê x œ 0 and y œ 0, which is not an interior point of OB; endpoint values have been found above (iv) For interior points of the triangular region, fx (xß y) œ 2x y œ 0 and fy (xß y) œ x 2y œ 0 Ê x œ 0 and y œ 0, which is not an interior point of the region. Therefore, the absolute maximum is 17 at (!ß %) and (%ß %), and the absolute minimum is 1 at (0ß 0). 32. (i) On OA, f(xß y) œ f(!ß y) œ y# on 0 Ÿ y Ÿ 2; f w (0ß y) œ 2y œ 0 Ê y œ 0 and x œ 0; f(0ß 0) œ 0 and f(0ß #) œ 4 (ii) On OB, f(xß y) œ f(xß 0) œ x# on 0 Ÿ x Ÿ 1; f w (xß 0) œ 2x œ 0 Ê x œ 0 and y œ 0; f(0ß 0) œ 0 and f(1ß 0) œ 1 (iii) On AB, f(xß y) œ f(xß 2x 2) œ 5x# 8x 4 on 0 Ÿ x Ÿ 1; f w (xß 2x 2) œ 10x 8 œ 0 Ê x œ 45 and y œ 25 ; f ˆ 45 ß 25 ‰ œ 45 ; endpoint values have been found above. 33. (i) (iv) For interior points of the triangular region, fx (xß y) œ 2x œ 0 and fy (xß y) œ 2y œ 0 Ê x œ 0 and y œ 0, but (!ß 0) is not an interior point of the region. Therefore the absolute maximum is 4 at (0ß 2) and the absolute minimum is 0 at (0ß 0). 34. (i) (ii) On AB, T(xß y) œ T(!ß y) œ y# on 3 Ÿ y Ÿ 3; Tw (0ß y) œ 2y œ 0 Ê y œ 0 and x œ 0; T(0ß 0) œ 0, T(!ß 3) œ 9, and T(!ß 3) œ 9 On BC, T(xß y) œ T(xß 3) œ x# 3x 9 on 0 Ÿ x Ÿ 5; Tw (xß 3) œ 2x 3 œ 0 Ê x œ 3# and y œ 3; T ˆ 3# ß 3‰ œ 27 4 and T(5ß 3) œ 19 (iii) On CD, T(xß y) œ T(5ß y) œ y# 5y 5 on 3 Ÿ y Ÿ 3;Tw (5ß y) œ 2y 5 œ 0 Ê y œ 5# and x œ 5;T ˆ5ß 5# ‰ œ 45 4 , T(&ß 3) œ 11 and T(5ß 3) œ 19 (iv) On AD, T(xß y) œ T(xß 3) œ x# 9x 9 on 0 Ÿ x Ÿ 5; Tw (xß 3) œ 2x 9 œ 0 Ê x œ 9# and y œ 3; T ˆ 9# ß 3‰ œ 45 4 , T(!ß 3) œ 9 and T(&ß 3) œ 11 (v) For interior points of the rectangular region, Tx (xß y) œ 2x y 6 œ 0 and Ty (xß y) œ x 2y œ 0 Ê x œ 4 and y œ 2 Ê (4ß 2) is an interior critical point with T(4ß 2) œ 12. Therefore the absolute maximum is 19 at (5ß 3) and the absolute minimum is 12 at (4ß 2). Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.7 Extreme Values and Saddle Points 35. (i) (ii) 841 On OC, T(xß y) œ T(xß 0) œ x# 6x 2 on 0 Ÿ x Ÿ 5; Tw (xß 0) œ 2x 6 œ 0 Ê x œ 3 and y œ 0; T(3ß 0) œ 7, T(0ß 0) œ 2, and T(5ß 0) œ 3 On CB, T(xß y) œ T(5ß y) œ y# 5y 3 on 3 Ÿ y Ÿ 0; Tw (5ß y) œ 2y 5 œ 0 Ê y œ 5# and x œ 5; T ˆ5ß 5# ‰ œ 37 4 and T(5ß 3) œ 9 (iii) On AB, T(xß y) œ T(xß 3) œ x# 9x 11 on 0 Ÿ x Ÿ 5; Tw (xß 3) œ 2x 9 œ 0 Ê x œ 9# and y œ 3; T ˆ 9# ß 3‰ œ 37 4 and T(!ß 3) œ 11 (iv) On AO, T(xß y) œ T(!ß y) œ y# 2 on 3 Ÿ y Ÿ 0; Tw (0ß y) œ 2y œ 0 Ê y œ 0 and x œ 0, but (0ß 0) is not an interior point of AO (v) For interior points of the rectangular region, Tx (xß y) œ 2x y 6 œ 0 and Ty (xß y) œ x 2y œ 0 Ê x œ 4 and y œ 2, an interior critical point with T(%ß 2) œ 10. Therefore the absolute maximum is 11 at (!ß 3) and the absolute minimum is 10 at (4ß 2). 36. (i) (ii) On OA, f(xß y) œ f(!ß y) œ 24y# on 0 Ÿ y Ÿ 1; f w (0ß y) œ 48y œ 0 Ê y œ 0 and x œ 0, but (0ß 0) is not an interior point of OA; f(!ß 0) œ 0 and f(!ß 1) œ 24 On AB, f(xß y) œ f(xß 1) œ 48x 32x$ 24 on 0 Ÿ x Ÿ 1; f w (xß 1) œ 48 96x# œ 0 Ê x œ È"2 and y œ 1, or x œ È"2 and y œ 1, but Š È"2 ß 1‹ is not in the interior of AB; f Š È"2 ß 1‹ œ 16È2 24 and f(1ß 1) œ 8 (iii) On BC, f(xß y) œ f("ß y) œ 48y 32 24y# on 0 Ÿ y Ÿ 1; f w ("ß y) œ 48 48y œ 0 Ê y œ 1 and x œ 1, but ("ß ") is not an interior point of BC; f("ß 0) œ 32 and f("ß ") œ 8 (iv) On OC, f(xß y) œ f(xß 0) œ 32x$ on 0 Ÿ x Ÿ 1; f w (xß 0) œ 96x# œ 0 Ê x œ 0 and y œ 0, but (0ß 0) is not an interior point of OC; f(!ß 0) œ 0 and f("ß 0) œ 32 (v) For interior points of the rectangular region, fx (xß y) œ 48y 96x# œ 0 and fy (xß y) œ 48x 48y œ 0 Ê x œ 0 and y œ 0, or x œ "# and y œ "# , but (0ß 0) is not an interior point of the region; f ˆ "# ß "# ‰ œ 2. Therefore the absolute maximum is 2 at ˆ "# ß "# ‰ and the absolute minimum is 32 at (1ß 0). 37. (i) On AB, f(xß y) œ f(1ß y) œ 3 cos y on 14 Ÿ y Ÿ 14 ; f w (1ß y) œ 3 sin y œ 0 Ê y œ 0 and x œ 1; È È f("ß 0) œ 3, f ˆ1ß 14 ‰ œ 3 # 2 , and f ˆ1ß 14 ‰ œ 3 # 2 (ii) On CD, f(xß y) œ f($ß y) œ 3 cos y on 14 Ÿ y Ÿ 14 ; f w (3ß y) œ 3 sin y œ 0 Ê y œ 0 and x œ 3; È È f(3ß 0) œ 3, f ˆ3ß 14 ‰ œ 3 # 2 and f ˆ3ß 14 ‰ œ 3 # 2 (iii) On BC, f(xß y) œ f ˆxß 14 ‰ œ È2 # # a4x x b on È 1 Ÿ x Ÿ 3; f w ˆxß 14 ‰ œ È2(2 x) œ 0 Ê x œ 2 and y œ 14 ; f ˆ2ß 14 ‰ œ 2È2, f ˆ1ß 14 ‰ œ 3 # 2 , and È f ˆ3ß 14 ‰ œ 3 # 2 È2 1‰ # wˆ È2(2 x) œ 0 # a4x x b on 1 Ÿ x Ÿ 3; f xß 4 œ È È f ˆ2ß 14 ‰ œ 2È2, f ˆ1ß 14 ‰ œ 3 # 2 , and f ˆ3ß 14 ‰ œ 3 # 2 (iv) On AD, f(xß y) œ f ˆxß 14 ‰ œ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Ê x œ 2 and y œ 14 ; 842 Chapter 14 Partial Derivatives (v) For interior points of the region, fx (xß y) œ (4 2x) cos y œ 0 and fy (xß y) œ a4x x# b sin y œ 0 Ê x œ 2 and y œ 0, which is an interior critical point with f(2ß 0) œ 4. Therefore the absolute maximum is 4 at È (2ß 0) and the absolute minimum is 3 # 2 at ˆ3ß 14 ‰ , ˆ3ß 14 ‰ , ˆ1ß 14 ‰ , and ˆ1ß 14 ‰ . On OA, f(xß y) œ f(!ß y) œ 2y 1 on 0 Ÿ y Ÿ 1; f w (0ß y) œ 2 Ê no interior critical points; f(0ß 0) œ 1 and f(0ß 1) œ 3 (ii) On OB, f(xß y) œ f(xß 0) œ 4x 1 on 0 Ÿ x Ÿ 1; f w (xß 0) œ 4 Ê no interior critical points; f(1ß 0) œ 5 (iii) On AB, f(xß y) œ f(xß x 1) œ 8x# 6x 3 on 0 Ÿ x Ÿ 1; f w (xß x 1) œ 16x 6 œ 0 Ê x œ 38 and y œ 58 ; f ˆ 38 ß 58 ‰ œ 15 8 , f(0ß 1) œ 3, and f("ß 0) œ 5 38. (i) (iv) For interior points of the triangular region, fx (xß y) œ 4 8y œ 0 and fy (xß y) œ 8x 2 œ 0 Ê y œ "# and x œ 4" which is an interior critical point with f ˆ 4" ß #" ‰ œ 2. Therefore the absolute maximum is 5 at (1ß 0) and the absolute minimum is 1 at (0ß 0). 39. Let F(aß b) œ 'a a6 x x# b dx where a Ÿ b. The boundary of the domain of F is the line a œ b in the ab-plane, and b F(aß a) œ 0, so F is identically 0 on the boundary of its domain. For interior critical points we have: `F `F # # ` a œ a6 a a b œ 0 Ê a œ 3, 2 and ` b œ a6 b b b œ 0 Ê b œ 3, 2. Since a Ÿ b, there is only one interior critical point (3ß 2) and F(3ß 2) œ 'c3 a6 x x# b dx gives the area under the parabola y œ 6 x x# that is 2 above the x-axis. Therefore, a œ 3 and b œ 2. 40. Let F(aß b) œ 'a a24 2x x# b b "Î$ dx where a Ÿ b. The boundary of the domain of F is the line a œ b and on this line F is identically 0. For interior critical points we have: ``Fa œ a24 2a a# b `F # "Î$ œ0 ` b œ a24 2b b b "Î$ œ 0 Ê a œ 4, 6 and Ê b œ 4, 6. Since a Ÿ b, there is only one critical point (6ß 4) and F(6ß 4) œ 'c6 a24 2x x# b dx gives the area under the curve y œ a24 2x x# b 4 "Î$ that is above the x-axis. Therefore, a œ 6 and b œ 4. 41. Tx (xß y) œ 2x 1 œ 0 and Ty (xß y) œ 4y œ 0 Ê x œ "# and y œ 0 with T ˆ "# ß 0‰ œ 4" ; on the boundary È x# y# œ 1: T(xß y) œ x# x 2 for 1 Ÿ x Ÿ 1 Ê Tw (xß y) œ 2x 1 œ 0 Ê x œ "# and y œ „ #3 ; È È È T Š "# ß #3 ‹ œ 94 , T Š #" ß #3 ‹ œ 94 , T(1ß 0) œ 2, and T("ß 0) œ 0 Ê the hottest is 2 4" ° at Š #" ß #3 ‹ and È Š "# ß #3 ‹ ; the coldest is "4 ° at ˆ "# ß 0‰ . 42. fx (xß y) œ y 2 2x œ 0 and fy (xß y) œ x "y œ 0 Ê x œ "# and y œ 2; fxx ˆ "# ß 2‰ œ x2# ¸ ˆ 1 ß2‰ œ 8, 2 fyy ˆ #" ß 2‰ œ y"# ¹ 1 œ 4" , fxy ˆ "# ß 2‰ œ 1 ˆ ß2‰ 2 Ê # fxx fyy fxy œ 1 0 and fxx 0 Ê a local minimum of f ˆ "# ß 2‰ œ 2 ln "# œ 2 ln 2 43. (a) fx (xß y) œ 2x 4y œ 0 and fy (xß y) œ 2y 4x œ 0 Ê x œ 0 and y œ 0; fxx (0ß 0) œ 2, fyy (0ß 0) œ 2, # œ 12 0 Ê saddle point at (0ß 0) fxy (0ß 0) œ 4 Ê fxx fyy fxy (b) fx (xß y) œ 2x 2 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and y œ 2; fxx (1ß 2) œ 2, fyy (1ß 2) œ 2, # œ 4 0 and fxx 0 Ê local minimum at ("ß #) fxy (1ß 2) œ 0 Ê fxx fyy fxy Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.7 Extreme Values and Saddle Points 843 (c) fx (xß y) œ 9x# 9 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ „ 1 and y œ 2; fxx (1ß 2) œ 18xk Ð1ß2Ñ œ 18, # œ 36 0 and fxx 0 Ê local minimum at ("ß #); fyy (1ß 2) œ 2, fxy (1ß 2) œ 0 Ê fxx fyy fxy # fxx (1ß 2) œ 18, fyy ("ß 2) œ 2, fxy ("ß 2) œ 0 Ê fxx fyy fxy œ 36 0 Ê saddle point at ("ß 2) 44. (a) Minimum at (0ß 0) since f(xß y) 0 for all other (xß y) (b) Maximum of 1 at (!ß !) since f(xß y) 1 for all other (xß y) (c) Neither since f(xß y) 0 for x 0 and f(xß y) 0 for x 0 (d) Neither since f(xß y) 0 for x 0 and f(xß y) 0 for x 0 (e) Neither since f(xß y) 0 for x 0 and y 0, but f(xß y) 0 for x 0 and y 0 (f) Minimum at (0ß 0) since f(xß y) 0 for all other (xß y) 45. If k œ 0, then f(xß y) œ x# y# Ê fx (xß y) œ 2x œ 0 and fy (xß y) œ 2y œ 0 Ê x œ 0 and y œ 0 Ê (0ß 0) is the only critical point. If k Á 0, fx (xß y) œ 2x ky œ 0 Ê y œ 2k x; fy (xß y) œ kx 2y œ 0 Ê kx 2 ˆ 2k x‰ œ 0 4‰ ˆ ˆ 2‰ Ê kx 4x k œ 0 Ê k k x œ 0 Ê x œ 0 or k œ „ 2 Ê y œ k (0) œ 0 or y œ „ x; in any case (0ß 0) is a critical point. # 46. (See Exercise 45 above): fxx (xß y) œ 2, fyy (xß y) œ 2, and fxy (xß y) œ k Ê fxx fyy fxy œ 4 k# ; f will have a saddle point at (0ß 0) if 4 k# 0 Ê k 2 or k 2; f will have a local minimum at (0ß 0) if 4 k# 0 Ê 2 k 2; the test is inconclusive if 4 k# œ 0 Ê k œ „ 2. 47. No; for example f(xß y) œ xy has a saddle point at (aß b) œ (0ß 0) where fx œ fy œ 0. # 48. If fxx (aß b) and fyy (aß b) differ in sign, then fxx (aß b) fyy (aß b) 0 so fxx fyy fxy 0. The surface must therefore have a saddle point at (aß b) by the second derivative test. 49. We want the point on z œ 10 x# y# where the tangent plane is parallel to the plane x 2y 3z œ 0. To find a normal vector to z œ 10 x# y# let w œ z x# y# 10. Then ™ w œ 2xi 2yj k is normal to z œ 10 x# y# at (xß y). The vector ™ w is parallel to i 2j 3k which is normal to the plane x 2y 3z œ 0 if " ‰ 6xi 6yj 3k œ i 2j 3k or x œ "6 and y œ "3 . Thus the point is ˆ "6 ß "3 ß 10 36 9" ‰ or ˆ 6" ß 3" ß 355 36 . 50. We want the point on z œ x# y# 10 where the tangent plane is parallel to the plane x 2y z œ 0. Let w œ z x# y# 10, then ™ w œ 2xi 2yj k is normal to z œ x# y# 10 at (xß y). The vector ™ w is parallel ‰ to i 2j k which is normal to the plane if x œ "# and y œ 1. Thus the point ˆ "# ß 1ß 4" 1 10‰ or ˆ #" ß 1ß 45 4 is the point on the surface z œ x# y# 10 nearest the plane x 2y z œ 0. 51. daxß yß zb œ Éax 0b2 ay 0b2 az 0b2 Ê we can minimize daxß yß zb by minimizing Daxß yß zb œ x2 y2 z2 ; 3x 2y z œ 6 Ê z œ 6 3x 2y Ê Daxß yb œ x2 y2 a6 3x 2yb2 Ê Dx axß yb œ 2x 6a6 3x 2yb œ 0 and Dy axß yb œ 2y 4a6 3x 2yb œ 0 Ê critical point is ˆ 97 , 67 ‰ Ê z œ 37 ; Dxx ˆ 97 , 67 ‰ œ 20, Dyy ˆ 12 , 1‰ œ 10, È Dxy ˆ 12 , 1‰ œ 12 Ê Dxx Dyy D#xy œ 56 0 and Dxx 0 Ê local minimum of dˆ 97 , 67 , 37 ‰ œ 3 714 52. daxß yß zb œ Éax 2b2 ay 1b2 az 1b2 Ê we can minimize daxß yß zb by minimizing Daxß yß zb œ ax 2b2 ay 1b2 az 1b2 ; x y z œ 2 Ê z œ x y 2 Ê Daxß yb œ ax 2b2 ay 1b2 ax y 3b2 Ê Dx axß yb œ 2ax 2b 2ax y 3b œ 0 and Dy axß yb œ 2ay 1b 2ax y 3b œ 0 Ê critical point is ˆ 83 , 13 ‰ Ê z œ 13 ; Dxx ˆ 83 , 13 ‰ œ 4, Dyy ˆ 83 , 13 ‰ œ 4, Dxy ˆ 83 , 13 ‰ œ 2 Ê Dxx Dyy D#xy œ 12 0 and Dxx 0 Ê local minimum of dˆ 83 , 13 , 13 ‰ œ È2 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 844 Chapter 14 Partial Derivatives 53. saxß yß zb œ x2 y2 z2 ; x y z œ 9 Ê z œ 9 x y Ê saxß yb œ x2 y2 a9 x yb2 Ê sx axß yb œ 2x 2a9 x yb œ 0 and sy axß yb œ 2y 2a9 x yb œ 0 Ê critical point is a3, 3b Ê z œ 3; sxx a3, 3b œ 4, syy a3, 3b œ 4, sxy a3, 3b œ 2 Ê sxx syy s#xy œ 12 0 and sxx 0 Ê local minimum of sa3, 3, 3b œ 27 54. paxß yß zb œ xyz; x y z œ 3 Ê z œ 3 x y Ê paxß yb œ x ya3 x yb œ 3x y x2 y x y2 Ê px axß yb œ 3y 2xy y2 œ 0 and py axß yb œ 3x x2 2xy œ 0 Ê critical points are a0, 0b, a0, 3b, a3, 0b, and a1, 1b; for a0, 0b Ê z œ 3; pxx a0, 0b œ 0, pyy a0, 0b œ 0, pxy a0, 0b œ 3 Ê pxx pyy p#xy œ 9 0 Ê saddle point; for a0, 3b Ê z œ 0; pxx a0, 3b œ 6, pyy a0, 3b œ 0, pxy a0, 3b œ 3 Ê pxx pyy p#xy œ 9 0 Ê saddle point; for a3, 0b Ê z œ 0; pxx a3, 0b œ 0, pyy a3, 0b œ 6, pxy a3, 0b œ 3 Ê pxx pyy p#xy œ 9 0 Ê saddle point; for a1, 1b Ê z œ 1; pxx a1, 1b œ 2, pyy a1, 1b œ 2, pxy a1, 1b œ 1 Ê pxx pyy p#xy œ 3 0 and pxx 0 Ê local maximum of pa1, 1, 1b œ 1 55. saxß yß zb œ xy yz xz; x y z œ 6 Ê z œ 6 x y Ê saxß yb œ xy ya6 x yb xa6 x yb œ 6x 6y xy x2 y2 Ê sx axß yb œ 6 2x y œ 0 and sy axß yb œ 6 x 2y œ 0 Ê critical point is a2, 2b Ê z œ 2; sxx a2, 2b œ 2, syy a2, 2b œ 2, sxy a2, 2b œ 1 Ê sxx syy s#xy œ 3 0 and sxx 0 Ê local maximum of sa2, 2, 2b œ 12 56. daxß yß zb œ Éax 6b2 ay 4b2 az 0b2 Ê we can minimize daxß yß zb by minimizing Daxß yß zb œ ax 6b2 ay 4b2 z2 ; z œ Èx2 y2 Ê Daxß yb œ ax 6b2 ay 4b2 x2 y2 œ 2x2 2y2 12x 8y 52 Ê Dx axß yb œ 4x 12 œ 0 and Dy axß yb œ 4y 8 œ 0 Ê critical point is a3, 2b Ê z œ È13; Dxx a3, 2b œ 4, Dyy a3, 2b œ 4, Dxy a3, 2b œ 0 Ê Dxx Dyy D# œ 16 0 and Dxx 0 Ê local xy minimum of dŠ3, 2, È13‹ œ È26 57. Vaxß yß zb œ a2xba2yba2zb œ 8xyz; x2 y2 z2 œ 4 Ê z œ È4 x2 y2 Ê Vaxß yb œ 8xyÈ4 x2 y2 , x 0 and y y 8y y 8x 0 Ê Vx axß yb œ 32yÈ4 16x œ 0 and Vy axß yb œ 32xÈ416x œ 0 Ê critical points are x2 y2 x2 y2 2 3 2 3 a0, 0b, Š È#3 , È#3 ‹, Š È#3 , È#3 ‹, Š È#3 , È#3 ‹, and Š È#3 , È#3 ‹. Only a0, 0b and Š È#3 , È#3 ‹ satisfy x 0 and y 0 2 2 È Va0ß 0b œ 0 and VŠ È#3 , È#3 ‹ œ 364 È3 ; On x œ 0, 0 Ÿ y Ÿ 2 Ê Va0ß yb œ 8a0by 4 0 y œ 0, no critical points, Va0ß 0b œ 0, Va0ß 2b œ 0; On y œ 0, 0 Ÿ x Ÿ 2 Ê Vaxß 0b œ 8xa0bÈ4 x2 02 œ 0, no critical points, Va0ß 0b œ 0, 2 Va0ß 2b œ 0; On y œ È4 x2 , 0 Ÿ x Ÿ 2 Ê VŠxß È4 x2 ‹ œ 8xÈ4 x2 Ê4 x2 ŠÈ4 x2 ‹ œ 0 # # # no critical points, Va0ß 2b œ 0, Va2ß 0b œ 0. Thus, there is a maximum volume of 364 È3 if the box is È3 ‚ È3 ‚ È3 . 27 27 27 54 58. Saxß yß zb œ 2xy 2yz 2xz; xyz œ 27 Ê z œ xy Ê Saxß yß zb œ 2xy 2yŠ xy ‹ 2xŠ xy ‹ œ 2xy 54 x y , x 0, 54 y 0; Sx axß yb œ 2y 54 x2 œ 0 and Sy axß yb œ 2x y2 œ 0 Ê Critical point is a3, 3b Ê z œ 3; Sxx a3, 3b œ 4, Syy a3, 3b œ 4, Dxy a3, 3b œ 2 Ê Dxx Dyy D#xy œ 12 0 and Dxx 0 Ê local minimum of Sa3ß 3ß 3b œ 54 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.7 Extreme Values and Saddle Points 845 59. Let x œ height of the box, y œ width, and z œ length, cut out squares of length x from corner of the material See diagram at right. Fold along the dashed lines to form the box. From the diagram we see that the length of the material is 2x y and the width is 2x z. Thus a2x yba2x zb œ 12 2ˆ6 2 x2 xy‰ . Since Vax, y, zb œ x y z 2x y ˆ 2x y 6 2 x2 xy‰ Ê Vax, yb œ , where x 0, y 0. 2x y 2 3 2 2 ˆ 4 3y 4x y 4x y xy3 ‰ Vx ax, yb œ œ 0 and a2x yb2 Êzœ Vy ax, yb œ 2ˆ12x2 4 x4 4x3 y x2 y2 ‰ œ 0 Ê critical points are ŠÈ3, 0‹, ŠÈ3, 0‹, Š È13 , È43 ‹, a2x yb2 and Š È13 , È43 ‹. Only ŠÈ3, 0‹ and Š È13 , È43 ‹ satisfy x 0 and y 0. For ŠÈ3, 0‹: z œ 0; Vxx ŠÈ3, 0‹ œ 0, # Vyy ŠÈ3, 0‹ œ 2È3, Vxy ŠÈ3, 0‹ œ 4È3 Ê Vxx Vyy Vxy œ 48 0 Ê saddle point. For Š È13 , È43 ‹: z œ È43 ; 1 4 2 1 4 4 16 # Vxx Š È13 , È43 ‹ œ 380 È3 , Vyy Š È3 , È3 ‹ œ 3È3 , Vxy Š È3 , È3 ‹ œ 3È3 Ê Vxx Vyy Vxy œ 3 0 and 16 Vxx 0 Ê local maximum of VŠ È13 , È43 , È43 ‹ œ 3È 3 60. (a) (i) On x œ 0, f(xß y) œ f(0ß y) œ y# y 1 for 0 Ÿ y Ÿ 1; f w (0ß y) œ 2y 1 œ 0 Ê y œ "# and x œ 0; f ˆ0ß "# ‰ œ 34 , f(0ß 0) œ 1, and f(0ß 1) œ 1 (ii) On y œ 1, f(xß y) œ f(xß 1) œ x# x 1 for 0 Ÿ x Ÿ 1; f w (xß 1) œ 2x 1 œ 0 Ê x œ "# and y œ 1, but ˆ "# ß 1‰ is outside the domain; f(0ß 1) œ 1 and f("ß ") œ 3 (iii) On x œ 1, f(xß y) œ f("ß y) œ y# y 1 for 0 Ÿ y Ÿ 1; f w (1ß y) œ 2y 1 œ 0 Ê y œ "# and x œ 1, but ˆ1ß "# ‰ is outside the domain; f(1ß 0) œ 1 and f("ß ") œ 3 (iv) On y œ 0, f(xß y) œ f(xß 0) œ x# x 1 for 0 Ÿ x Ÿ 1; f w (xß 0) œ 2x 1 œ 0 Ê x œ "# and y œ 0; f ˆ "# ß 0‰ œ 34 ; f(0ß 0) œ 1, and f("ß 0) œ 1 (v) On the interior of the square, fx (xß y) œ 2x 2y 1 œ 0 and fy (xß y) œ 2y 2x 1 œ 0 Ê 2x 2y œ 1 Ê (x y) œ "# . Then f(xß y) œ x# y# 2xy x y 1 œ (x y)# (x y) 1 œ 34 is the absolute minimum value when 2x 2y œ 1. (b) The absolute maximum is f("ß ") œ 3. 61. (a) (i) dy df ` f dx ` f dy dx dt œ ` x dt ` y dt œ dt dt œ 2 sin t 2 cos t œ 0 On the semicircle x# y# œ 4, y Ê cos t œ sin t Ê x œ y 0, we have t œ 14 and x œ y œ È2 Ê f ŠÈ2ß È2‹ œ 2È2. At the endpoints, f(2ß 0) œ 2 and f(#ß !) œ 2. Therefore the absolute minimum is f(2ß 0) œ 2 when t œ 1; the absolute maximum is f ŠÈ2ß È2‹ œ 2È2 when t œ 1 . 4 (ii) On the quartercircle x# y# œ 4, x 0 and y 0, the endpoints give f(!ß 2) œ 2 and f(#ß 0) œ 2. Therefore the absolute minimum is f(2ß 0) œ 2 and f(!ß 2) œ 2 when t œ 0, 1# respectively; the absolute maximum is f ŠÈ2ß È2‹ œ 2È2 when t œ 14 . (b) (i) dg ` g dx ` g dy dy dx # # dt œ ` x dt ` y dt œ y dt x dt œ 4 sin t 4 cos t œ 0 On the semicircle x# y# œ 4, y Ê cos t œ „ sin t Ê x œ „ y. 0, we obtain x œ y œ È2 at t œ 14 and x œ È2, y œ È2 at t œ 341 . Then g ŠÈ2ß È2‹ œ 2 and g ŠÈ2ß È2‹ œ 2. At the endpoints, g(2ß 0) œ g(#ß 0) œ 0. Therefore the absolute minimum is g ŠÈ2ß È2‹ œ 2 when t œ 341 ; the absolute maximum is g ŠÈ2 ß È2‹ œ 2 when t œ 14 . Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 846 Chapter 14 Partial Derivatives (ii) On the quartercircle x# y# œ 4, x 0 and y 0, the endpoints give g(!ß 2) œ 0 and g(#ß 0) œ 0. Therefore the absolute minimum is g(2ß 0) œ 0 and g(!ß 2) œ 0 when t œ 0, 1# respectively; the absolute maximum is g ŠÈ2ß È2‹ œ 2 when t œ 14 . (c) (i) (ii) dy dh ` h dx ` h dy dx dt œ ` x dt ` y dt œ 4x dt 2y dt œ (8 cos t)(2 sin t) (4 sin t)(2 cos t) œ 8 cos t sin t œ 0 Ê t œ 0, 1# , 1 yielding the points (2ß 0), (0ß 2) for 0 Ÿ t Ÿ 1. # # On the semicircle x y œ 4, y 0 we have h(2ß 0) œ 8, h(0ß 2) œ 4, and h(2ß 0) œ 8. Therefore, the absolute minimum is h(!ß 2) œ 4 when t œ 1# ; the absolute maximum is h(2ß 0) œ 8 and h(2ß 0) œ 8 when t œ 0, 1 respectively. On the quartercircle x# y# œ 4, x 0 and y absolute maximum is h(2ß 0) œ 8 when t œ 0. 62. (a) (i) 0 the absolute minimum is h(0ß 2) œ 4 when t œ 1# ; the dy df ` f dx ` f dy dx dt œ ` x dt ` y dt œ 2 dt 3 dt œ 6 sin t 6 cos t œ 0 È # # Ê sin t œ cos t Ê t œ 14 for 0 Ÿ t Ÿ 1. È 0, f(xß y) œ 2x 3y œ 6 cos t 6 sin t œ 6 Š #2 ‹ 6 Š #2 ‹ œ 6È2 On the semi-ellipse, x9 y4 œ 1, y at t œ 14 . At the endpoints, f(3ß 0) œ 6 and f(3ß 0) œ 6. The absolute minimum is f(3ß 0) œ 6 when È t œ 1; the absolute maximum is f Š 3 # 2 ß È2‹ œ 6È2 when t œ 14 . (ii) On the quarter ellipse, at the endpoints f(0ß 2) œ 6 and f(3ß 0) œ 6. The absolute minimum is f(3ß 0) œ 6 È and f(0ß 2) œ 6 when t œ 0, 1 respectively; the absolute maximum is f Š 3 2 ß È2‹ œ 6È2 when t œ 1 . # # 4 ` g dx ` g dy dy dx # # (b) dg dt œ ` x dt ` y dt œ y dt x dt œ (2 sin t)(3 sin t) (3 cos t)(2 cos t) œ 6 acos t sin tb œ 6 cos 2t œ 0 Ê t œ 14 , 341 for 0 Ÿ t Ÿ 1. È (i) On the semi-ellipse, g(xß y) œ xy œ 6 sin t cos t. Then g Š 3 # 2 ß È2‹ œ 3 when t œ 14 , and È g Š 3 # 2 ß È2‹ œ 3 when t œ 341 . At the endpoints, g(3ß 0) œ g($ß 0) œ 0. The absolute minimum is È È g Š 3 # 2 ß È2‹ œ 3 when t œ 341 ; the absolute maximum is g Š 3 # 2 ß È2‹ œ 3 when t œ 14 . (ii) On the quarter ellipse, at the endpoints g(!ß 2) œ 0 and g($ß 0) œ 0. The absolute minimum is g(3ß 0) œ 0 È and g(0ß 2) œ 0 at t œ 0, 1 respectively; the absolute maximum is g Š 3 2 ß È2‹ œ 3 when t œ 1 . # (c) (i) (ii) # 4 dy dh ` h dx ` h dy dx dt œ ` x dt ` y dt œ 2x dt 6y dt œ (6 cos t)(3 sin t) (12 sin t)(2 cos t) œ 6 sin t cos t œ 0 Ê t œ 0, 1# , 1 for 0 Ÿ t Ÿ 1, yielding the points (3ß 0), (0ß 2), and (3ß 0). On the semi-ellipse, y 0 so that h(3ß 0) œ 9, h(0ß 2) œ 12, and h(3ß 0) œ 9. The absolute minimum is h(3ß 0) œ 9 and h(3ß 0) œ 9 when t œ 0, 1 respectively; the absolute maximum is h(!ß 2) œ 12 when t œ 1# . On the quarter ellipse, the absolute minimum is h(3ß 0) œ 9 when t œ 0; the absolute maximum is h(!ß 2) œ 12 when t œ 1# . dy ` f dx ` f dy dx 63. df dt œ ` x dt ` y dt œ y dt x dt (i) (ii) " " x œ 2t and y œ t 1 Ê df dt œ (t 1)(2) (2t)(1) œ 4t 2 œ 0 Ê t œ # Ê x œ 1 and y œ # with f ˆ1ß "# ‰ œ "# . The absolute minimum is f ˆ1ß "# ‰ œ "# when t œ "# ; there is no absolute maximum. For the endpoints: t œ 1 Ê x œ 2 and y œ 0 with f(2ß 0) œ 0; t œ 0 Ê x œ 0 and y œ 1 with f(!ß 1) œ 0. The absolute minimum is f ˆ1ß "# ‰ œ "# when t œ "# ; the absolute maximum is f(0ß 1) œ 0 and f(#ß 0) œ 0 when t œ 1, 0 respectively. (iii) There are no interior critical points. For the endpoints: t œ 0 Ê x œ 0 and y œ 1 with f(0ß 1) œ 0; t œ 1 Ê x œ 2 and y œ 2 with f(2ß 2) œ 4. The absolute minimum is f(0ß 1) œ 0 when t œ 0; the absolute maximum is f(2ß 2) œ 4 when t œ 1. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.7 Extreme Values and Saddle Points dy df ` f dx ` f dy dx dt œ ` x dt ` y dt œ 2x dt 2y dt 4 4 2 (i) x œ t and y œ 2 2t Ê df dt œ (2t)(1) 2(2 2t)(2) œ 10t 8 œ 0 Ê t œ 5 Ê x œ 5 and y œ 5 with 4 f ˆ 45 ß 25 ‰ œ "#65 25 œ 45 . The absolute minimum is f ˆ 45 ß 25 ‰ œ 45 when t œ 45 ; there is no absolute 64. (a) (ii) maximum along the line. For the endpoints: t œ 0 Ê x œ 0 and y œ 2 with f(0ß 2) œ 4; t œ 1 Ê x œ 1 and y œ 0 with f(1ß 0) œ 1. The absolute minimum is f ˆ 45 ß 25 ‰ œ 45 at the interior critical point when t œ 45 ; the absolute maximum is f(0ß 2) œ 4 at the endpoint when t œ 0. (b) (i) dg ` g dx ` g dy 2y dy 2x dx dt œ ` x dt ` y dt œ ’ ax# y# b# “ dt ’ ax# y# b# “ dt # # x œ t and y œ 2 2t Ê x# y# œ 5t# 8t 4 Ê dg dt œ a5t 8t 4b [(2t)(1) (2)(2 2t)(2)] # œ a5t# 8t 4b (10t 8) œ 0 Ê t œ 45 Ê x œ 45 and y œ 25 with g ˆ 45 ß 25 ‰ œ ˆ "4 ‰ œ 54 . The absolute 5 maximum is g ˆ 45 ß 25 ‰ œ 54 when t œ 45 ; there is no absolute minimum along the line since x and y can be (ii) as large as we please. For the endpoints: t œ 0 Ê x œ 0 and y œ 2 with g(0ß 2) œ "4 ; t œ 1 Ê x œ 1 and y œ 0 with g(1ß 0) œ 1. The absolute minimum is g(0ß 2) œ "4 when t œ 0; the absolute maximum is g ˆ 45 ß 52 ‰ œ 45 when t œ 54 . 65. w œ am x1 b y1 b2 am x2 b y2 b2 â am xn b yn b2 w Ê `` m œ 2am x1 b y1 bax1 b 2am x2 b y2 bax2 b â 2am xn b yn baxn b Ê ``wb œ 2am x1 b y1 ba1b 2am x2 b y2 ba1b â 2am xn b yn ba1b `w ‘ ` m œ 0 Ê 2 am x1 b y1 bax1 b am x2 b y2 bax2 b â am xn b yn baxn b œ 0 Ê m x21 b x1 x1 y1 m x#2 b x2 x2 y2 â m xn2 b xn xn yn œ 0 Ê max21 x2# â xn2 b bax1 x2 â xn b ax1 y1 x2 y2 â xn yn b œ 0 n n n Ê m! ax2k b b! xk ! axk yk b œ 0 k œ1 k œ1 k œ1 `w ‘ ` b œ 0 Ê 2 am x1 b y1 b am x2 b y2 b â am xn b yn b œ 0 Ê m x1 b y1 m x2 b y2 â m xn b yn œ 0 Ê max1 x2 â xn b ab b â bb ay1 y2 â yn b œ 0 n n n n n k œ1 k œ1 k œ1 k œ1 k œ1 n n kœ1 n kœ 1 n n n k œ1 k œ1 k œ1 Ê m ! xk b ! 1 ! yk œ 0 Ê m ! xk bn ! yk œ 0 Ê b œ 1n Œ ! yk m! xk . n w Substituting for b in the equation obtained for `` m we get m ! ax2k b 1n Œ ! yk m! xk ! xk ! axk yk b œ 0. n n k œ1 n n k œ1 n k œ1 k œ1 kœ1 Multiply both sides by n to obtain m n ! ax2k b Œ ! yk m! xk ! xk n ! axk yk b œ 0 k œ1 n n n k œ1 k œ1 n n k œ1 2 n n Ê m n ! ax2k b Œ ! xk Œ ! yk mŒ ! xk n ! axk yk b œ 0 k œ1 2 k œ1 kœ1 n n n k œ1 k œ1 k œ1 n n n Ê m n ! ax2k b mŒ ! xk œ n ! axk yk b Œ ! xk Œ ! yk kœ1 kœ1 n 2 n Ê m–n! ax2k b Œ ! xk — œ n ! axk yk b Œ ! xk Œ ! yk k œ1 k œ1 n Êmœ k œ1 n n n ! axk yk bŒ ! xk Œ ! yk kœ1 kœ1 n n kœ1 kœ1 kœ1 2 n! ax2k bŒ ! xk k œ1 n œ n kœ1 n Œ ! xk Œ ! yk n ! axk yk b kœ1 kœ1 n 2 kœ1 n Œ ! x k n ! ax k b kœ1 2 kœ1 To show that these values for m and b minimize the sum of the squares of the distances, use second derivative test. n n ` 2w 2 2 2 ! ax2k b; ` 2 w œ 2 x1 2 x2 â 2 xn œ 2! xk ; ` 2 w2 œ 2 2 â 2 œ 2 n œ 2 x 2 x â 2 x œ 2 2 n # 1 `m `m `b `b k œ1 k œ1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 847 848 Chapter 14 Partial Derivatives 2 n n kœ1 kœ1 # n n kœ1 k œ1 2 The discriminant is: Š `` mw2 ‹Š `` bw2 ‹ Š ``m w` b ‹ œ ”2 ! ax2k b•a2 nb ”2 ! xk • œ 4–n ! ax2k b Œ ! xk —. 2 n 2 2 2 n Now, n ! ax2k b Œ ! xk œ nax12 x#2 â x#n b ax1 x2 â xn bax1 x2 â xn b k œ1 kœ1 2 2 œ n x1 n x# â n x#n x21 x1 x2 â x1 xn x2 x1 x2# â x2 xn xn x1 xn x2 â x#n œ an 1b x21 an 1b x2# â an 1b x#n 2 x1 x2 2 x1 x3 â 2 x1 xn 2 x2 x3 â 2 x2 xn â 2 xn1 xn œ a x21 2 x1 x2 x#2 b a x12 2 x1 x3 x23 b â ax21 2 x1 xn xn# b ax2# 2 x2 x3 x23 b â a x#2 2 x2 xn xn# b â ax2n1 2 xn1 xn x#n b œ ax1 x2 b2 ax1 x3 b2 â ax1 xn b2 ax2 x3 b2 â ax2 xn b2 â axn1 xn b2 0. 2 n n 2 2 2 2 Thus we have : Š `` mw2 ‹Š `` bw2 ‹ Š ``m w` b ‹ œ 4–n ! ax2k b Œ ! xk — 4a0b œ 0. If x1 œ x2 œ â œ xn then kœ1 k œ1 2 n Š `` mw2 ‹Š `` bw2 ‹ Š ``m w` b ‹ œ 0. Also, `` mw2 œ 2 ! ax2k b 2 2 2 2 k œ1 0. If x1 œ x2 œ â œ xn œ 0, then `` mw2 œ 0. 2 2 Provided that at least one xi is nonzero and different from the rest of xj , j Á i, then Š `` mw2 ‹Š `` bw2 ‹ Š ``m w` b ‹ 0 and 2 ` 2w ` m2 0 2 2 Ê the values given above for m and b minimize w. 3(6) 3 66. m œ (0)(5) (0)# 3(8) œ 4 and b œ "3 5 34 (0)‘ œ 53 Ê y œ 34 x 53 ; y¸ xœ4 œ 14 3 1) 3("4) 67. m œ (2)((2) œ 20 # 3(10) 13 and 9 ‰ ‘ b œ "3 1 ˆ 20 13 (2) œ 13 9 71 ¸ Ê y œ 20 13 x 13 ; y xœ4 œ 13 3(8) 3 68. m œ (3)(5) (3)# 3(5) œ 2 and b œ "3 5 32 (3)‘ œ 16 Ê y œ 32 x 16 ; y¸ xœ4 œ 37 6 k 1 2 3 D xk 2 0 2 0 yk 0 2 3 5 x#k 4 0 4 8 xk yk 0 0 6 6 k 1 2 3 D xk 1 0 3 2 yk 2 1 4 1 x#k 1 0 9 10 xk yk 2 0 12 14 k 1 2 3 D xk 0 1 2 3 yk 0 2 3 5 x#k 0 1 4 5 xk yk 0 2 6 8 69-74. Example CAS commands: Maple: f := (x,y) -> x^2+y^3-3*x*y; x0,x1 := -5,5; y0,y1 := -5,5; plot3d( f(x,y), x=x0..x1, y=y0..y1, axes=boxed, shading=zhue, title="#69(a) (Section 14.7)" ); plot3d( f(x,y), x=x0..x1, y=y0..y1, grid=[40,40], axes=boxed, shading=zhue, style=patchcontour, title="#69(b) (Section 14.7)" ); fx := D[1](f); # (c) fy := D[2](f); crit_pts := solve( {fx(x,y)=0,fy(x,y)=0}, {x,y} ); fxx := D[1](fx); # (d) fxy := D[2](fx); Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.8 Lagrange Multipliers 849 fyy := D[2](fy); discr := unapply( fxx(x,y)*fyy(x,y)-fxy(x,y)^2, (x,y) ); for CP in {crit_pts} do # (e) eval( [x,y,fxx(x,y),discr(x,y)], CP ); end do; # (0,0) is a saddle point # ( 9/4, 3/2) is a local minimum Mathematica: (assigned functions and bounds will vary) Clear[x,y,f] f[x_,y_]:= x2 y3 3x y xmin= 5; xmax= 5; ymin= 5; ymax= 5; Plot3D[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, AxesLabel Ä {x, y, z}] ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading Ä False, Contours Ä 40] fx= D[f[x,y], x]; fy= D[f[x,y], y]; critical=Solve[{fx==0, fy==0},{x, y}] fxx= D[fx, x]; fxy= D[fx, y]; fyy= D[fy, y]; discriminant= fxx fyy fxy2 {{x, y}, f[x, y], discriminant, fxx} /.critical 14.8 LAGRANGE MULTIPLIERS 1. ™ f œ yi xj and ™ g œ 2xi 4yj so that ™ f œ - ™ g Ê yi xj œ -(2xi 4yj) Ê y œ 2x- and x œ 4yÈ Ê x œ 8x-# Ê - œ „ 42 or x œ 0. CASE 1: If x œ 0, then y œ 0. But (0ß 0) is not on the ellipse so x Á 0. # È CASE 2: x Á 0 Ê - œ „ 42 Ê x œ „ È2y Ê Š „ È2y‹ 2y# œ 1 Ê y œ „ "# . È È Therefore f takes on its extreme values at Š „ 22 ß "# ‹ and Š „ 22 ß "# ‹ . The extreme values of f on the ellipse È are „ #2 . 2. ™ f œ yi xj and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê yi xj œ -(2xi 2yj) Ê y œ 2x- and x œ 2yÊ x œ 4x-# Ê x œ 0 or - œ „ 12 . CASE 1: If x œ 0, then y œ 0. But (0ß 0) is not on the circle x# y# 10 œ 0 so x Á 0. CASE 2: x Á 0 Ê - œ „ 12 Ê y œ 2x ˆ „ "# ‰ œ „ x Ê x# a „ xb# 10 œ 0 Ê x œ „ È5 Ê y œ „ È5. Therefore f takes on its extreme values at Š „ È5ß È5‹ and Š „ È5ß È5‹ . The extreme values of f on the circle are 5 and 5. 3. ™ f œ 2xi 2yj and ™ g œ i 3j so that ™ f œ - ™ g Ê 2xi 2yj œ -(i 3j) Ê x œ -# and y œ 3#Ê ˆ -# ‰ 3 ˆ 3#- ‰ œ 10 Ê - œ 2 Ê x œ 1 and y œ 3 Ê f takes on its extreme value at (1ß 3) on the line. The extreme value is f("ß $) œ 49 1 9 œ 39. 4. ™ f œ 2xyi x# j and ™ g œ i j so that ™ f œ - ™ g Ê 2xyi x# j œ -(i j) Ê 2xy œ - and x# œ Ê 2xy œ x# Ê x œ 0 or 2y œ x. CASE 1: If x œ 0, then x y œ 3 Ê y œ 3. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 850 Chapter 14 Partial Derivatives CASE 2: If x Á 0, then 2y œ x so that x y œ 3 Ê 2y y œ 3 Ê y œ 1 Ê x œ 2. Therefore f takes on its extreme values at (!ß 3) and (2ß "). The extreme values of f are f(0ß 3) œ 0 and f(2ß 1) œ 4. 5. We optimize f(xß y) œ x# y# , the square of the distance to the origin, subject to the constraint g(xß y) œ xy# 54 œ 0. Thus ™ f œ 2xi 2yj and ™ g œ y# i 2xyj so that ™ f œ - ™ g Ê 2xi 2yj œ - ay# i 2xyjb Ê 2x œ -y# and 2y œ 2-xy. CASE 1: If y œ 0, then x œ 0. But (0ß 0) does not satisfy the constraint xy# œ 54 so y Á 0. CASE 2: If y Á 0, then 2 œ 2-x Ê x œ -" Ê 2 ˆ -" ‰ œ -y# Ê y# œ -2# . Then xy# œ 54 Ê ˆ -" ‰ ˆ -2# ‰ œ 54 Ê -$ œ " Ê - œ " Ê x œ 3 and y# œ 18 Ê x œ 3 and y œ „ 3È2. 27 3 Therefore Š$ß „ 3È2‹ are the points on the curve xy# œ 54 nearest the origin (since xy# œ 54 has points increasingly far away as y gets close to 0, no points are farthest away). 6. We optimize f(xß y) œ x# y# , the square of the distance to the origin subject to the constraint g(xß y) œ x# y 2 œ 0. Thus ™ f œ 2xi 2yj and ™ g œ 2xyi x# j so that ™ f œ - ™ g Ê 2x œ 2xy- and 2y œ x# - Ê - œ 2y x# , since 2y x œ 0 Ê y œ 0 (but g(0ß 0) Á 0). Thus x Á 0 and 2x œ 2xy ˆ x# ‰ Ê x# œ 2y# Ê a2y# b y 2 œ 0 Ê y œ 1 (since y 0) Ê x œ „ È2 . Therefore Š „ È2ß 1‹ are the points on the curve x# y œ 2 nearest the origin (since x# y œ 2 has points increasingly far away as x gets close to 0, no points are farthest away). 7. (a) ™ f œ i j and ™ g œ yi xj so that ™ f œ - ™ g Ê i j œ -(yi xj) Ê 1 œ -y and 1 œ -x Ê y œ -" and x œ -" Ê -"# œ 16 Ê - œ „ 4" . Use - œ "4 since x 0 and y 0. Then x œ 4 and y œ 4 Ê the minimum value is 8 at the point (4ß 4). Now, xy œ 16, x 0, y 0 is a branch of a hyperbola in the first quadrant with the x-and y-axes as asymptotes. The equations x y œ c give a family of parallel lines with m œ 1. As these lines move away from the origin, the number c increases. Thus the minimum value of c occurs where x y œ c is tangent to the hyperbola's branch. (b) ™ f œ yi xj and ™ g œ i j so that ™ f œ - ™ g Ê yi xj œ -(i j) Ê y œ - œ x y y œ 16 Ê y œ 8 Ê x œ 8 Ê f()ß )) œ 64 is the maximum value. The equations xy œ c (x 0 and y 0 or x 0 and y 0 to get a maximum value) give a family of hyperbolas in the first and third quadrants with the x- and y-axes as asymptotes. The maximum value of c occurs where the hyperbola xy œ c is tangent to the line x y œ 16. 8. Let f(xß y) œ x# y# be the square of the distance from the origin. Then ™ f œ 2xi 2yj and ™ g œ (2x y)i (2y x)j so that ™ f œ - ™ g Ê 2x œ -(2x y) and 2y œ -(2y x) Ê 2y2yx œ Ê 2x œ Š 2y2yx ‹ (2x y) Ê x(2y x) œ y(2x y) Ê x# œ y# Ê y œ „ x. CASE 1: y œ x Ê x# x(x) x# 1 œ 0 Ê x œ „ È"3 and y œ x. CASE 2: y œ x Ê x# x(x) (x)# 1 œ 0 Ê x œ „ 1 and y œ x. Thus f Š È"3 ß È"3 ‹ œ 23 œ f Š È"3 ß È"3 ‹ and f(1ß 1) œ 2 œ f(1ß 1). Therefore the points (1ß 1) and (1ß 1) are the farthest away; Š È"3 ß È"3 ‹ and Š È"3 ß È"3 ‹ are the closest points to the origin. 9. V œ 1r# h Ê 161 œ 1r# h Ê 16 œ r# h Ê g(rß h) œ r# h 16; S œ 21rh 21r# Ê ™ S œ (21h 41r)i 21rj and ™ g œ 2rhi r# j so that ™ S œ - ™ g Ê (21rh 41r)i 21rj œ - a2rhi r# jb Ê 21rh 41r œ 2rh- and 21r œ -r# Ê r œ 0 or - œ 2r1 . But r œ 0 gives no physical can, so r Á 0 Ê - œ 2r1 Ê 21h 41r œ 2rh ˆ 2r1 ‰ Ê 2r œ h Ê 16 œ r# (2r) Ê r œ 2 Ê h œ 4; thus r œ 2 cm and h œ 4 cm give the only extreme surface area of 241 cm# . Since r œ 4 cm and h œ 1 cm Ê V œ 161 cm$ and S œ 401 cm# , which is a larger surface area, then 241 cm# must be the minimum surface area. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.8 Lagrange Multipliers 851 10. For a cylinder of radius r and height h we want to maximize the surface area S œ 21rh subject to the constraint # g(rß h) œ r# ˆ h# ‰ a# œ 0. Thus ™ S œ 21hi 21rj and ™ g œ 2ri h# j so that ™ S œ - ™ g Ê 21h œ 2-r and # 21r œ -#h Ê 1rh œ - and 21r œ ˆ 1rh ‰ ˆ #h ‰ Ê 4r# œ h# Ê h œ 2r Ê r# 4r4 œ a# Ê 2r# œ a# Ê r œ Èa Ê h œ aÈ2 Ê 2 S œ 21 Š Èa2 ‹ ŠaÈ2‹ œ 21a# . # # x 11. A œ (2x)(2y) œ 4xy subject to g(xß y) œ 16 y9 1 œ 0; ™ A œ 4yi 4xj and ™ g œ x8 i 2y 9 j so that ™ A 2y 2y 32y x x ‰ ˆ 32y ‰ œ - ™ g Ê 4yi 4xj œ - ˆ 8 i 9 j‰ Ê 4y œ ˆ 8 ‰ - and 4x œ ˆ 9 ‰ - Ê - œ x and 4x œ ˆ 2y 9 x ˆ „43 x‰# œ 1 Ê x# œ 8 Ê x œ „ 2È2 . We use x œ 2È2 since x represents distance. 9 È Then y œ 34 Š2È2‹ œ 3 # 2 , so the length is 2x œ 4È2 and the width is 2y œ 3È2. # x Ê y œ „ 34 x Ê 16 # # 2y 12. P œ 4x 4y subject to g(xß y) œ xa# yb# 1 œ 0; ™ P œ 4i 4j and ™ g œ 2x a# i b# j so that ™ P œ - ™ g # # # # 2a b x ‰ ˆ 2y ‰ ˆ 2y ‰ 2a Ê 4 œ ˆ 2x a# - and 4 œ b# - Ê - œ x and 4 œ b# Š x ‹ Ê y œ Š a# ‹ x Ê a# œ 1 Ê aa# b# b x# œ a% Ê x œ È #a # a b# # , since x 0 Ê y œ Š ba# ‹ x œ È #b # a b# # # # # Š ba# ‹ x# b# # # # œ 1 Ê xa# bax% Ê width œ 2x œ È 2a # # a b# # 4a 4b and height œ 2y œ Èa2b Ê perimeter is P œ 4x 4y œ È œ 4Èa# b# # b# a# b# 13. ™ f œ 2xi 2yj and ™ g œ (2x 2)i (2y 4)j so that ™ f œ - ™ g œ 2xi 2yj œ -[(2x 2)i (2y 4)j] 2# # Ê 2x œ -(2x 2) and 2y œ -(2y 4) Ê x œ - 1 and y œ -1 , - Á 1 Ê y œ 2x Ê x 2x (2x) 4(2x) œ 0 Ê x œ 0 and y œ 0, or x œ 2 and y œ 4. Therefore f(0ß 0) œ 0 is the minimum value and f(2ß 4) œ 20 is the maximum value. (Note that - œ 1 gives 2x œ 2x 2 or ! œ 2, which is impossible.) 3 3 ‰ 14. ™ f œ 3i j and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê 3 œ 2-x and 1 œ 2-y Ê - œ 2x and 1 œ 2 ˆ 2x y # Ê y œ x3 Ê x# ˆ x3 ‰ œ 4 Ê 10x# œ 36 Ê x œ „ È610 Ê x œ È610 and y œ È210 , or x œ È610 and y œ È210 . Therefore f Š È610 ß È210 ‹ œ È2010 6 œ 2È10 6 ¸ 12.325 is the maximum value, and f Š È610 ß È210 ‹ œ 2È10 6 ¸ 0.325 is the minimum value. 15. ™ T œ (8x 4y)i (4x 2y)j and g(xß y) œ x# y# 25 œ 0 Ê ™ g œ 2xi 2yj so that ™ T œ - ™ g Ê (8x 4y)i (4x 2y)j œ -(2xi 2yj) Ê 8x 4y œ 2-x and 4x 2y œ 2-y Ê y œ -2x1 , - Á 1 Ê 8x 4 ˆ -2x1 ‰ œ 2-x Ê x œ 0, or - œ 0, or - œ 5. CASE 1: x œ 0 Ê y œ 0; but (0ß 0) is not on x# y# œ 25 so x Á 0. CASE 2: - œ 0 Ê y œ 2x Ê x# (2x)# œ 25 Ê x œ „ È5 and y œ 2x. # CASE 3: - œ 5 Ê y œ 42x œ #x Ê x# ˆ #x ‰ œ 25 Ê x œ „ 2È5 Ê x œ 2È5 and y œ È5, or x œ 2È5 and y œ È5 . Therefore T ŠÈ5ß 2È5‹ œ 0° œ T ŠÈ5ß 2È5‹ is the minimum value and T Š2È5ß È5‹ œ 125° œ T Š2È5ß È5‹ is the maximum value. (Note: - œ 1 Ê x œ 0 from the equation 4x 2y œ 2-y; but we found x Á 0 in CASE 1.) 16. The surface area is given by S œ 41r# 21rh subject to the constraint V(rß h) œ 43 1r$ 1r# h œ 8000. Thus ™ S œ (81r 21h)i 21rj and ™ V œ a41r# 21rhb i 1r# j so that ™ S œ - ™ V œ (81r 21h)i 21rj œ - ca41r# 21rhb i 1r# jd Ê 81r 21h œ - a41r# 21rhb and 21r œ -1r# Ê r œ 0 or 2 œ r-. But r Á 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 852 Chapter 14 Partial Derivatives so 2 œ r- Ê - œ 2r Ê 4r h œ 2r a2r# rhb Ê h œ 0 Ê the tank is a sphere (there is no cylindrical part) and 4 $ 3 1r œ 8000 Ê r œ 10 ˆ 16 ‰ "Î$ . 17. Let f(xß yß z) œ (x 1)# (y 1)# (z 1)# be the square of the distance from (1ß 1ß 1). Then ™ f œ 2(x 1)i 2(y 1)j 2(z 1)k and ™ g œ i 2j 3k so that ™ f œ - ™ g Ê 2(x 1)i 2(y 1)j 2(z 1)k œ -(i 2j 3k) Ê 2(x 1) œ -, 2(y 1) œ 2-, 2(z 1) œ 3Ê 2(y 1) œ 2[2(x 1)] and 2(z 1) œ 3[2(x 1)] Ê x œ y # 1 Ê z 2 œ 3 ˆ y # 1 ‰ or z œ 3y # 1 ; thus y1 ˆ 3y # 1 ‰ 13 œ 0 Ê y œ 2 Ê x œ 3# and z œ #5 . Therefore the point ˆ #3 ß 2ß 5# ‰ is closest (since no # 2y 3 point on the plane is farthest from the point (1ß 1ß 1)). 18. Let f(xß yß z) œ (x 1)# (y 1)# (z 1)# be the square of the distance from (1ß 1ß 1). Then ™ f œ 2(x 1)i 2(y 1)j 2(z 1)k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê x 1 œ -x, y 1 œ -y # ‰# ˆ 1 " - ‰# œ 4 and z 1 œ -z Ê x œ 1 " - , y œ 1 " - , and z œ 1" - for - Á 1 Ê ˆ 1 " - ‰ ˆ 1" Ê " " - œ „ È23 Ê x œ È23 , y œ È23 , z œ È23 or x œ È23 , y œ È23 , z œ È23 . The largest value of f occurs where x 0, y 0, and z 0 or at the point Š È23 ß È23 ß È23 ‹ on the sphere. 19. Let f(xß yß z) œ x# y# z# be the square of the distance from the origin. Then ™ f œ 2xi 2yj 2zk and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê 2xi 2yj 2zk œ -(2xi 2yj 2zk) Ê 2x œ 2x-, 2y œ 2y-, and 2z œ 2z- Ê x œ 0 or - œ 1. CASE 1: - œ 1 Ê 2y œ 2y Ê y œ 0; 2z œ 2z Ê z œ 0 Ê x# 1 œ 0 Ê x# 1 œ 0 Ê x œ „ 1 and y œ z œ 0. CASE 2: x œ 0 Ê y# z# œ 1, which has no solution. Therefore the points on the unit circle x# y# œ 1, are the points on the surface x# y# z# œ 1 closest to the originÞ The minimum distance is 1. 20. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk and ™ g œ yi xj k so that ™ f œ - ™ g Ê 2xi 2yj 2zk œ -(yi xj k) Ê 2x œ -y, 2y œ -x, and 2z œ Ê x œ -#y Ê 2y œ - Š -#y ‹ Ê y œ 0 or - œ „ 2. CASE 1: y œ 0 Ê x œ 0 Ê z 1 œ 0 Ê z œ 1. CASE 2: - œ 2 Ê x œ y and z œ 1 Ê x# (1) 1 œ 0 Ê x# 2 œ 0, so no solution. CASE 3: - œ 2 Ê x œ y and z œ 1 Ê (y)y 1 1 œ 0 Ê y œ 0, again. Therefore (0ß 0ß 1) is the point on the surface closest to the origin since this point gives the only extreme value and there is no maximum distance from the surface to the origin. 21. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk and ™ g œ yi xj 2zk so that ™ f œ - ™ g Ê 2xi 2yj 2zk œ -(yi xj 2zk) Ê 2x œ y-, 2y œ x-, and 2z œ 2z- Ê - œ 1 or z œ 0. CASE 1: - œ 1 Ê 2x œ y and 2y œ x Ê y œ 0 and x œ 0 Ê z# 4 œ 0 Ê z œ „ 2 and x œ y œ 0. # # CASE 2: z œ 0 Ê xy 4 œ 0 Ê y œ 4x . Then 2x œ 4x - Ê - œ x# , and x8 œ x- Ê x8 œ x Š x# ‹ Ê x% œ 16 Ê x œ „ 2. Thus, x œ 2 and y œ 2, or x = 2 and y œ 2. Therefore we get four points: (#ß 2ß 0), (2ß 2ß 0), (0ß 0ß 2) and (!ß 0ß 2). But the points (!ß 0ß 2) and (!ß !ß 2) are closest to the origin since they are 2 units away and the others are 2È2 units away. 22. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk and ™ g œ yzi xzj xyk so that ™ f œ - ™ g Ê 2x œ -yz, 2y œ -xz, and 2z œ -xy Ê 2x# œ -xyz and 2y# œ -yxz Ê x# œ y# Ê y œ „ x Ê z œ „ x Ê x a „ xb a „ xb œ 1 Ê x œ „ 1 Ê the points are (1ß 1ß 1), ("ß 1ß 1), ("ß "ß "), and (1ß 1, 1). Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.8 Lagrange Multipliers 853 23. ™ f œ i 2j 5k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê i 2j 5k œ -(2xi 2yj 2zk) Ê 1 œ 2x-, 2 œ 2y-, and 5 œ 2z- Ê x œ #"- , y œ -" œ 2x, and z œ #5- œ 5x Ê x# (2x)# (5x)# œ 30 Ê x œ „ 1. Thus, x œ 1, y œ 2, z œ 5 or x œ 1, y œ 2, z œ 5. Therefore f(1ß 2ß 5) œ 30 is the maximum value and f(1ß 2ß 5) œ 30 is the minimum value. 24. ™ f œ i 2j 3k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê i 2j 3k œ -(2xi 2yj 2zk) Ê 1 œ 2x-, 2 œ 2y-, and 3 œ 2z- Ê x œ #"- , y œ -" œ 2x, and z œ #3- œ 3x Ê x# (2x)# (3x)# œ 25 Ê x œ „ È514 . Thus, x œ È514 , y œ È1014 , z œ È1514 or x œ È514 , y œ È1014 , z œ È1514 . Therefore f Š È514 ß È1014 ß È1514 ‹ œ 5È14 is the maximum value and f Š È514 ß È1014 , È1514 ‹ œ 5È14 is the minimum value. 25. f(xß yß z) œ x# y# z# and g(xß yß z) œ x y z 9 œ 0 Ê ™ f œ 2xi 2yj 2zk and ™ g œ i j k so that ™ f œ - ™ g Ê 2xi 2yj 2zk œ -(i j k) Ê 2x œ -, 2y œ -, and 2z œ - Ê x œ y œ z Ê x x x 9 œ 0 Ê x œ 3, y œ 3, and z œ 3. 26. f(xß yß z) œ xyz and g(xß yß z) œ x y z# 16 œ 0 Ê ™ f œ yzi xzj xyk and ™ g œ i j 2zk so that ™ f œ - ™ g Ê yzi xzj xyk œ -(i j 2zk) Ê yz œ -, xz œ -, and xy œ 2z- Ê yz œ xz Ê z œ 0 or y œ x. But z 0 so that y œ x Ê x# œ 2z- and xz œ -. Then x# œ 2z(xz) Ê x œ 0 or x œ 2z# . But x 0 so that 32 x œ 2z# Ê y œ 2z# Ê 2z# 2z# z# œ 16 Ê z œ „ È45 . We use z œ È45 since z 0. Then x œ 32 5 and y œ 5 32 4 4096 which yields f Š 32 5 ß 5 ß È5 ‹ œ 25È5 . 27. V œ xyz and g(xß yß z) œ x# y# z# 1 œ 0 Ê ™ V œ yzi xzj xyk and ™ g œ 2xi 2yj 2zk so that ™ V œ - ™ g Ê yz œ -x, xz œ -y, and xy œ -z Ê xyz œ -x# and xyz œ -y# Ê y œ „ x Ê z œ „ x Ê x# x# x# œ 1 Ê x œ È"3 since x 0 Ê the dimensions of the box are È13 by È13 by È13 for maximum volume. (Note that there is no minimum volume since the box could be made arbitrarily thin.) 28. V œ xyz with xß yß z all positive and xa yb cz œ 1; thus V œ xyz and g(xß yß z) œ bcx acy abz abc œ 0 Ê ™ V œ yzi xzj xyk and ™ g œ bci acj abk so that ™ V œ - ™ g Ê yz œ -bc, xz œ -ac, and xy œ -ab Ê xyz œ -bcx, xyz œ -acy, and xyz œ -abz Ê - Á 0. Also, -bcx œ -acy œ -abz Ê bx œ ay, cy œ bz, and a cx œ az Ê y œ ba x and z œ ac x. Then xa by zc œ 1 Ê xa b" ˆ ba x‰ "c ˆ ca x‰ œ 1 Ê 3x a œ 1 Ê xœ 3 Ê y œ ˆ ba ‰ ˆ 3a ‰ œ b3 and z œ ˆ ca ‰ ˆ 3a ‰ œ 3c Ê V œ xyz œ ˆ 3a ‰ ˆ b3 ‰ ˆ 3c ‰ œ abc 27 is the maximum volume. (Note that there is no minimum volume since the box could be made arbitrarily thin.) 29. ™ T œ 16xi 4zj (4y 16)k and ™ g œ 8xi 2yj 8zk so that ™ T œ - ™ g Ê 16xi 4zj (4y 16)k œ -(8xi 2yj 8zk) Ê 16x œ 8x-, 4z œ 2y-, and 4y 16 œ 8z- Ê - œ 2 or x œ 0. CASE 1: - œ 2 Ê 4z œ 2y(2) Ê z œ y. Then 4z 16 œ 16z Ê z œ 43 Ê y œ 43 . Then # # 4x# ˆ 43 ‰ 4 ˆ 43 ‰ œ 16 Ê x œ „ 43 . 2z # # # # # CASE 2: x œ 0 Ê - œ 2z y Ê 4y 16 œ 8z Š y ‹ Ê y 4y œ 4z Ê 4(0) y ay 4yb 16 œ 0 Ê y# 2y 8 œ 0 Ê (y 4)(y 2) œ 0 Ê y œ 4 or y œ 2. Now y œ 4 Ê 4z# œ 4# 4(4) Ê z œ 0 and y œ 2 Ê 4z# œ (2)# 4(2) Ê z œ „ È3. ° ° The temperatures are T ˆ „ 43 ß 43 ß 43 ‰ œ 642 23 , T(0ß 4ß 0) œ 600°, T Š0ß 2ß È3‹ œ Š600 24È3‹ , and ° T Š0ß 2ß È3‹ œ Š600 24È3‹ ¸ 641.6°. Therefore ˆ „ 43 ß 43 ß 43 ‰ are the hottest points on the space probe. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 854 Chapter 14 Partial Derivatives 30. ™ T œ 400yz# i 400xz# j 800xyzk and ™ g œ 2xi 2yj 2zk so that ™ T œ - ™ g Ê 400yz# i 400xz# j 800xyzk œ -(2xi 2yj 2zk) Ê 400yz# œ 2x-, 400xz# œ 2y-, and 800xyz œ 2z-. È Solving this system yields the points a!ß „ 1ß 0b , a „ 1ß 0ß 0b , and Š „ "# ß „ "# ß „ #2 ‹ . The corresponding È temperatures are T a!ß „ 1ß 0b œ 0, T a „ 1ß 0ß 0b œ 0, and T Š „ "# ß „ "# ß „ #2 ‹ œ „ 50. Therefore 50 is the È È maximum temperature at Š "# ß "# ß „ #2 ‹ and Š "# ß "# ß „ #2 ‹ ; 50 is the minimum temperature at È È Š "# ß "# ß „ #2 ‹ and Š #" ß #" ß „ #2 ‹ . 31. ™ U œ (y 2)i xj and ™ g œ 2i j so that ™ U œ - ™ g Ê (y 2)i xj œ -(2i j) Ê y # œ 2- and x œ - Ê y 2 œ 2x Ê y œ 2x 2 Ê 2x (2x 2) œ 30 Ê x œ 8 and y œ 14. Therefore U(8ß 14) œ $128 is the maximum value of U under the constraint. 32. ™ M œ (6 z)i 2yj xk and ™ g œ 2xi 2yj 2zk so that ™ M œ - ™ g Ê (6 z)i 2yj xk œ -(2xi 2yj 2zk) Ê 6 z œ 2x-, 2y œ 2y-, x œ 2z- Ê - œ 1 or y œ 0. CASE 1: - œ 1 Ê 6 z œ 2x and x œ 2z Ê 6 z œ 2(2z) Ê z œ 2 and x œ 4. Then (4)# y# 2# 36 œ 0 Ê y œ „ 4. x x ‰ CASE 2: y œ 0, 6 z œ 2x-, and x œ 2z- Ê - œ 2z Ê 6 z œ 2x ˆ 2z Ê 6z z# œ x# Ê a6z z# b 0# z# œ 36 Ê z œ 6 or z œ 3. Now z œ 6 Ê x# œ 0 Ê x œ 0; z œ 3 Ê x# œ 27 Ê x œ „ 3È3. Therefore we have the points Š „ 3È3ß 0ß 3‹ , (0ß 0ß 6), and a4ß „ 4ß 2b . Then M Š3È3ß 0ß 3‹ œ 27È3 60 ¸ 106.8, M Š3È3ß 0ß 3‹ œ 60 27È3 ¸ 13.2, M(0ß 0ß 6) œ 60, and M(4ß 4ß 2) œ 12 œ M(4ß 4ß 2). Therefore, the weakest field is at a4ß „ 4ß 2b . 33. Let g" (xß yß z) œ 2x y œ 0 and g# (xß yß z) œ y z œ 0 Ê ™ g" œ 2i j , ™ g# œ j k , and ™ f œ 2xi 2j 2zk so that ™ f œ - ™ g" . ™ g# Ê 2xi 2j 2zk œ -(2i j) .(j k) Ê 2xi 2j 2zk œ 2-i (. -)j .k Ê 2x œ 2-, 2 œ . -, and 2z œ . Ê x œ -. Then 2 œ 2z x Ê x œ 2z 2 so that 2x y œ 0 Ê 2(2z 2) y œ 0 Ê 4z 4 y œ 0. This equation coupled with y z œ 0 implies z œ 43 and y œ 43 . Then # # x œ 23 so that ˆ 23 ß 43 ß 43 ‰ is the point that gives the maximum value f ˆ 23 ß 43 ß 43 ‰ œ ˆ 23 ‰ 2 ˆ 43 ‰ ˆ 43 ‰ œ 43 . 34. Let g" (xß yß z) œ x 2y 3z 6 œ 0 and g# (xß yß z) œ x 3y 9z 9 œ 0 Ê ™ g" œ i 2j 3k , ™ g# œ i 3j 9k , and ™ f œ 2xi 2yj 2zk so that ™ f œ - ™ g" . ™ g# Ê 2xi 2yj 2zk œ -(i 2j 3k) .(i 3j 9k) Ê 2x œ - ., 2y œ 2- 3., and 2z œ 3- 9.. Then 0 œ x 2y 3z 6 ‰ œ "# (- .) (2- 3.) ˆ 9# - 27 # . 6 Ê 7- 17. œ 6; 0 œ x 3y 9z 9 " 9 27 81 Ê # (- .) ˆ3- # .‰ ˆ # - # .‰ 9 Ê 34- 91. œ 18. Solving these two equations for - and . gives -. 2- 3. 3- 9. 78 81 9 - œ 240 œ 123 œ 59 . The minimum value is 59 and . œ 59 Ê x œ # œ 59 , y œ # 59 , and z œ # 21,771 81 123 9 369 f ˆ 59 ß 59 ß 59 ‰ œ 59# œ 59 . (Note that there is no maximum value of f subject to the constraints because at least one of the variables x, y, or z can be made arbitrary and assume a value as large as we please.) 35. Let f(xß yß z) œ x# y# z# be the square of the distance from the origin. We want to minimize f(xß yß z) subject to the constraints g" (xß yß z) œ y 2z 12 œ 0 and g# (xß yß z) œ x y 6 œ 0. Thus ™ f œ 2xi 2yj 2zk , ™ g" œ j 2k, and ™ g# œ i j so that ™ f œ - ™ g" . ™ g# Ê 2x œ ., 2y œ - ., and 2z œ 2-. Then 0 œ y 2z 12 œ ˆ -# .# ‰ 2- 12 Ê #5 - "# . œ 12 Ê 5- . œ 24; 0 œ x y 6 œ .# ˆ -# .# ‰ 6 Ê "# - . œ 6 Ê - #. œ 12. Solving these two equations for - and . gives - œ 4 and . œ 4 Ê x œ .# œ 2, y œ - # . œ 4, and z œ - œ 4. The point (2ß 4ß 4) on the line of intersection is closest to the origin. (There is no maximum distance from the origin since points on the line can be arbitrarily far away.) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.8 Lagrange Multipliers 855 36. The maximum value is f ˆ 23 ß 43 ß 43 ‰ œ 43 from Exercise 33 above. 37. Let g" (xß yß z) œ z 1 œ 0 and g# (xß yß z) œ x# y# z# 10 œ 0 Ê ™ g" œ k , ™ g# œ 2xi 2yj 2zk , and ™ f œ 2xyzi x# zj x# yk so that ™ f œ - ™ g" . ™ g# Ê 2xyzi x# zj x# yk œ -(k) .(2xi 2yj 2zk) Ê 2xyz œ 2x., x# z œ 2y., and x# y œ 2z. - Ê xyz œ x. Ê x œ 0 or yz œ . Ê . œ y since z œ 1. CASE 1: x œ 0 and z œ 1 Ê y# 9 œ 0 (from g# ) Ê y œ „ 3 yielding the points a0ß „ 3ß 1b. CASE 2: . œ y Ê x# z œ 2y# Ê x# œ 2y# (since z œ 1) Ê 2y# y# 1 10 œ 0 (from g# ) Ê 3y# 9 œ 0 # Ê y œ „ È3 Ê x# œ 2 Š „ È3‹ Ê x œ „ È6 yielding the points Š „ È6ß „ È3ß "‹ . Now f a!ß „ 3ß 1b œ 1 and f Š „ È6ß „ È3ß "‹ œ 6 Š „ È3‹ 1 œ 1 „ 6È3. Therefore the maximum of f is 1 6È3 at Š „ È6ß È3ß 1‹, and the minimum of f is 1 6È3 at Š „ È6ß È3ß "‹ . 38. (a) Let g" (xß yß z) œ x y z 40 œ 0 and g# (xß yß z) œ x y z œ 0 Ê ™ g" œ i j k , ™ g# œ i j k , and ™ w œ yzi xzj xyk so that ™ w œ - ™ g" . ™ g# Ê yzi xzj xyk œ -(i j k) .(i j k) Ê yz œ - ., xz œ - ., and xy œ - . Ê yz œ xz Ê z œ 0 or y œ x. CASE 1: z œ 0 Ê x y œ 40 and x y œ 0 Ê no solution. CASE 2: x œ y Ê 2x z 40 œ 0 and 2x z œ 0 Ê z œ 20 Ê x œ 10 and y œ 10 Ê w œ (10)(10)(20) œ 2000 â â âi j k â â â " â œ 2i 2j is parallel to the line of intersection Ê the line is x œ 2t 10, (b) n œ â " " â â â " " " â y œ 2t 10, z œ 20. Since z œ 20, we see that w œ xyz œ (2t 10)(2t 10)(20) œ a4t# 100b (20) which has its maximum when t œ 0 Ê x œ 10, y œ 10, and z œ 20. 39. Let g" (Bß yß z) œ y x œ 0 and g# (xß yß z) œ x# y# z# 4 œ 0. Then ™ f œ yi xj 2zk , ™ g" œ i j , and ™ g# œ 2xi 2yj 2zk so that ™ f œ - ™ g" . ™ g# Ê yi xj 2zk œ -(i j) .(2xi 2yj 2zk) Ê y œ - 2x., x œ - 2y., and 2z œ 2z. Ê z œ 0 or . œ 1. CASE 1: z œ 0 Ê x# y# 4 œ 0 Ê 2x# 4 œ 0 (since x œ y) Ê x œ „ È2 and y œ „ È2 yielding the points Š „ È2ß „ È2ß !‹ . CASE 2: . œ 1 Ê y œ - 2x and x œ - 2y Ê x y œ 2(x y) Ê 2x œ 2(2x) since x œ y Ê x œ 0 Ê y œ 0 Ê z# 4 œ 0 Ê z œ „ 2 yielding the points a!ß !ß „ 2b . Now, f a!ß !ß „ 2b œ 4 and f Š „ È2ß „ È2ß !‹ œ 2. Therefore the maximum value of f is 4 at a!ß !ß „ 2b and the minimum value of f is 2 at Š „ È2ß „ È2ß !‹ . 40. Let f(xß yß z) œ x# y# z# be the square of the distance from the origin. We want to minimize f(xß yß z) subject to the constraints g" (xß yß z) œ 2y 4z 5 œ 0 and g# (xß yß z) œ 4x# 4y# z# œ 0. Thus ™ f œ 2xi 2yj 2zk , ™ g" œ 2j 4k , and ™ g# œ 8xi 8yj 2zk so that ™ f œ - ™ g" . ™ g# Ê 2xi 2yj 2zk œ -(2j 4k) .(8xi 8yj 2zk) Ê 2x œ 8x., 2y œ 2- 8y., and 2z œ 4- 2z. Ê x œ 0 or . œ "4 . CASE 1: x œ 0 Ê 4(0)# 4y# z# œ 0 Ê z œ „ 2y Ê 2y 4(2y) 5 œ 0 Ê y œ "# , or 2y 4(2y) 5 œ 0 Ê y œ 56 yielding the points ˆ!ß "# ß "‰ and ˆ!ß 56 ß 53 ‰ . CASE 2: . œ "4 Ê y œ - y Ê - œ 0 Ê 2z œ 4(0) 2z ˆ 4" ‰ Ê z œ 0 Ê 2y 4(0) œ 5 Ê y œ 5# and # (0)# œ 4x# 4 ˆ #5 ‰ Ê no solution. " ˆ " ‰ Then f ˆ!ß "# ß 1‰ œ 54 and f ˆ!ß 56 ß 35 ‰ œ 25 ˆ 36 "9 ‰ œ 125 36 Ê the point !ß # ß 1 is closest to the origin. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 856 Chapter 14 Partial Derivatives 41. ™ f œ i j and ™ g œ yi xj so that ™ f œ - ™ g Ê i j œ -(yi xj) Ê 1 œ y- and 1 œ x- Ê y œ x Ê y# œ 16 Ê y œ „ 4 Ê (4ß 4) and (%ß 4) are candidates for the location of extreme values. But as x Ä _, y Ä _ and f(xß y) Ä _; as x Ä _, y Ä 0 and f(xß y) Ä _. Therefore no maximum or minimum value exists subject to the constraint. 4 42. Let f(Aß Bß C) œ ! (Axk Byk C zk )# œ C# (B C 1)# (A B C 1)# (A C 1)# . We want k œ1 to minimize f. Then fA (Aß Bß C) œ 4A 2B 4C, fB (Aß Bß C) œ 2A 4B 4C 4, and fC (Aß Bß C) œ 4A 4B 8C 2. Set each partial derivative equal to 0 and solve the system to get A œ "# , B œ 3# , and C œ "4 or the critical point of f is ˆ #" ß 3# ß "4 ‰ . 43. (a) Maximize f(aß bß c) œ a# b# c# subject to a# b# c# œ r# . Thus ™ f œ 2ab# c# i 2a# bc# j 2a# b# ck and ™ g œ 2ai 2bj 2ck so that ™ f œ - ™ g Ê 2ab# c# œ 2a-, 2a# bc# œ 2b-, and 2a# b# c œ 2cÊ 2a# b# c# œ 2a# - œ 2b# - œ 2c# - Ê - œ 0 or a# œ b# œ c# . CASE 1: - œ 0 Ê a# b# c# œ 0. # $ CASE 2: a# œ b# œ c# Ê f(aß bß c) œ a# a# a# and 3a# œ r# Ê f(aß bß c) œ Š r3 ‹ is the maximum value. (b) The point ŠÈaß Èbß Èc‹ is on the sphere if a b c œ r# . Moreover, by part (a), abc œ f ŠÈaß Èbß Èc‹ # $ # Ÿ Š r3 ‹ Ê (abc)"Î$ Ÿ r3 œ a b3 c , as claimed. n 44. Let f(x" ß x# ß á ß xn ) œ ! ai xi œ a" x" a# x# á an xn and g(x" ß x# ß á ß xn ) œ x"# x## á xn# 1. Then we i œ1 a# a# # want ™ f œ - ™ g Ê a" œ -(2x" ), a# œ -(2x# ), á , an œ -(2xn ), - Á 0 Ê xi œ 2a-i Ê 4-"# 4-## á 4a-n# œ 1 n n iœ1 i œ1 "Î# Ê 4-# œ ! a#i Ê 2- œ Œ! a#i n n n n i œ1 i œ1 i œ1 i œ1 "Î# Ê f(x" ß x# ß á ß xn ) œ ! ai xi œ ! ai ˆ #a-i ‰ œ #"- ! a#i œ Œ! a#i the maximum value. 45-50. Example CAS commands: Maple: f := (x,y,z) -> x*y+y*z; g1 := (x,y,z) -> x^2+y^2-2; g2 := (x,y,z) -> x^2+z^2-2; h := unapply( f(x,y,z)-lambda[1]*g1(x,y,z)-lambda[2]*g2(x,y,z), (x,y,z,lambda[1],lambda[2]) ); hx := diff( h(x,y,z,lambda[1],lambda[2]), x ); hy := diff( h(x,y,z,lambda[1],lambda[2]), y ); hz := diff( h(x,y,z,lambda[1],lambda[2]), z ); hl1 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[1] ); hl2 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[2] ); sys := { hx=0, hy=0, hz=0, hl1=0, hl2=0 }; q1 := solve( sys, {x,y,z,lambda[1],lambda[2]} ); q2 := map(allvalues,{q1}); for p in q2 do eval( [x,y,z,f(x,y,z)], p ); ``=evalf(eval( [x,y,z,f(x,y,z)], p )); end do; Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. # (a) #(b) # (c) # (d) is Section 14.9 Taylor's Formula for Two Variables Mathematica: (assigned functions will vary) Clear[x, y, z, lambda1, lambda2] f[x_,y_,z_]:= x y y z g1[x_,y_,z_]:= x2 y2 2 g2[x_,y_,z_]:= x2 z2 2 h = f[x, y, z] lambda1 g1[x, y, z] lambda2 g2[x, y, z]; hx= D[h, x]; hy= D[h, y]; hz= D[h,z]; hL1=D[h, lambda1]; hL2= D[h, lambda2]; critical=Solve[{hx==0, hy==0, hz==0, hL1==0, hL2==0, g1[x,y,z]==0, g2[x,y,z]==0}, {x, y, z, lambda1, lambda2}]//N {{x, y, z}, f[x, y, z]}/.critical 14.9 TAYLOR'S FORMULA FOR TWO VARIABLES 1. f(xß y) œ xey Ê fx œ ey , fy œ xey , fxx œ 0, fxy œ ey , fyy œ xey Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d œ 0 x † 1 y † 0 "# ax# † 0 2xy † 1 y# † 0b œ x xy quadratic approximation; fxxx œ 0, fxxy œ 0, fxyy œ ey , fyyy œ xey Ê f(xß y) ¸ quadratic "6 cx$ fxxx (!ß !) 3x# yfxxy (0ß 0) 3xy# fxyy (!ß !) y$ fyyy (0ß 0)d œ x xy "6 ax$ † 0 3x# y † 0 3xy# † 1 y$ † 0b œ x xy "# xy# , cubic approximation 2. f(xß y) œ ex cos y Ê fx œ ex cos y, fy œ ex sin y, fxx œ ex cos y, fxy œ ex sin y, fyy œ ex cos y Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (!ß 0) "# cx# fxx (!ß !) 2xyfxy (!ß !) y# fyy (0ß 0)d œ 1 x † 1 y † 0 "# cx# † 1 2xy † 0 y# † (1)d œ 1 x "# ax# y# b , quadratic approximation; fxxx œ ex cos y, fxxy œ ex sin y, fxyy œ ex cos y, fyyy œ ex sin y Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d œ 1 x "# ax# y# b 6" cx$ † 1 3x# y † 0 3xy# † (1) y$ † 0d œ 1 x "# ax# y# b 6" ax$ 3xy# b , cubic approximation 3. f(xß y) œ y sin x Ê fx œ y cos x, fy œ sin x, fxx œ y sin x, fxy œ cos x, fyy œ 0 Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (!ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d œ 0 x † 0 y † 0 "# ax# † 0 2xy † 1 y# † 0b œ xy, quadratic approximation; fxxx œ y cos x, fxxy œ sin x, fxyy œ 0, fyyy œ 0 Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d œ xy "6 ax$ † 0 3x# y † 0 3xy# † 0 y$ † 0b œ xy, cubic approximation 4. f(xß y) œ sin x cos y Ê fx œ cos x cos y, fy œ sin x sin y, fxx œ sin x cos y, fxy œ cos x sin y, fyy œ sin x cos y Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d œ 0 x † 1 y † 0 "# ax# † 0 2xy † 0 y# † 0b œ x, quadratic approximation; fxxx œ cos x cos y, fxxy œ sin x sin y, fxyy œ cos x cos y, fyyy œ sin x sin y Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d œ x "6 cx$ † (1) 3x# y † 0 3xy# † (1) y$ † 0d œ x 6" ax$ 3xy# b, cubic approximation x x x 5. f(xß y) œ ex ln (1 y) Ê fx œ ex ln (1 y), fy œ 1 e y , fxx œ ex ln (1 y), fxy œ 1 e y , fyy œ (1 e y)# Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d œ 0 x † 0 y † 1 "# cx# † 0 2xy † 1 y# † (1)d œ y "# a2xy y# b , quadratic approximation; x x x fxxx œ ex ln (1 y), fxxy œ 1 e y , fxyy œ (1 e y)# , fyyy œ (1 2e y)$ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 857 858 Chapter 14 Partial Derivatives Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d œ y "2 a2xy y# b 6" cx$ † 0 3x# y † 1 3xy# † (1) y$ † 2d œ y "# a2xy y# b 6" a3x# y 3xy# 2y$ b , cubic approximation 6. f(xß y) œ ln (2x y 1) Ê fx œ 2x 2y 1 , fy œ #x "y 1 , fxx œ (2x y4 1)# , fxy œ (2x y2 1)# , " # # fyy œ (2x " y 1)# Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) # cx fxx (0ß 0) 2xyfxy (0ß 0) y fyy (0ß 0)d œ 0 x † 2 y † 1 "# cx# † (4) 2xy † (2) y# † (1)d œ 2x y "# a4x# 4xy y# b œ (2x y) "# (2x y)# , quadratic approximation; fxxx œ (2x 16y 1)$ , fxxy œ (2x 8y 1)$ , fxyy œ (2x 4y 1)$ , fyyy œ (2x 2y 1)$ Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d œ (2x y) "# (2x y)# 6" ax$ † 16 3x# y † 8 3xy# † 4 y$ † 2b œ (2x y) "# (2x y)# 3" a8x$ 12x# y 6xy# y# b œ (2x y) "# (2x y)# 3" (2x y)$ , cubic approximation 7. f(xß y) œ sin ax# y# b Ê fx œ 2x cos ax# y# b , fy œ 2y cos ax# y# b , fxx œ 2 cos ax# y# b 4x# sin ax# y# b , fxy œ 4xy sin ax# y# b , fyy œ 2 cos ax# y# b 4y# sin ax# y# b Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d œ 0 x † 0 y † 0 "# ax# † 2 2xy † 0 y# † 2b œ x# y# , quadratic approximation; fxxx œ 12x sin ax# y# b 8x$ cos ax# y# b , fxxy œ 4y sin ax# y# b 8x# y cos ax# y# b , fxyy œ 4x sin ax# y# b 8xy# cos ax# y# b , fyyy œ 12y sin ax# y# b 8y$ cos ax# y# b Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d œ x# y# "6 ax$ † 0 3x# y † 0 3xy# † 0 y$ † 0b œ x# y# , cubic approximation 8. f(xß y) œ cos ax# y# b Ê fx œ 2x sin ax# y# b , fy œ 2y sin ax# y# b , fxx œ 2 sin ax# y# b 4x# cos ax# y# b , fxy œ 4xy cos ax# y# b , fyy œ 2 sin ax# y# b 4y# cos ax# y# b Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d œ 1 x † 0 y † 0 "# cx# † 0 2xy † 0 y# † 0d œ 1, quadratic approximation; fxxx œ 12x cos ax# y# b 8x$ sin ax# y# b , fxxy œ 4y cos ax# y# b 8x# y sin ax# y# b , fxyy œ 4x cos ax# y# b 8xy# sin ax# y# b , fyyy œ 12y cos ax# y# b 8y$ sin ax# y# b Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d œ 1 "6 ax$ † 0 3x# y † 0 3xy# † 0 y$ † 0b œ 1, cubic approximation 9. f(xß y) œ 1 x" y Ê fx œ (1 x" y)# œ fy , fxx œ (1 x2 y)$ œ fxy œ fyy Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d œ 1 x † 1 y † 1 "# ax# † 2 2xy † 2 y# † 2b œ 1 (x y) ax# 2xy y# b œ 1 (x y) (x y)# , quadratic approximation; fxxx œ (1 x6 y)% œ fxxy œ fxyy œ fyyy Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d œ 1 (x y) (x y)# "6 ax$ † 6 3x# y † 6 3xy# † 6 y$ † 6b œ 1 (x y) (x y)# ax$ 3x# y 3xy# y$ b œ 1 (x y) (x y)# (x y)$ , cubic approximation # y) 10. f(xß y) œ 1 x " y xy Ê fx œ (1 x1yy xy)# , fy œ (" x1yx xy)# , fxx œ (1 2(1 x y xy)$ , # fxy œ (" x 1y xy)# , fyy œ (1 2(x"yx) xy)$ Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d œ 1 x † 1 y † 1 "# ax# † 2 2xy † 1 y# † 2b œ 1 x y x# xy y# , quadratic approximation; Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.10 Partial Derivatives with Constrained Variables $ 6(1 y)(1 x)](1 y) y) fxxx œ (1 6(1 , fxxy œ [4(1 x y (1 xy)x , x y xy)% y xy)% $ x)(1 y)](1 x) x) fxyy œ [4(1 x y (1 xy)x 6(1 , fyyy œ (1 6(1 y xy)% x y xy)% Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d œ 1 x y x# xy y# "6 ax$ † 6 3x# y † 2 3xy# † 2 y$ † 6b œ 1 x y x# xy y# x$ x# y xy# y$ , cubic approximation 11. f(xß y) œ cos x cos y Ê fx œ sin x cos y, fy œ cos x sin y, fxx œ cos x cos y, fxy œ sin x sin y, fyy œ cos x cos y Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d # # œ 1 x † 0 y † 0 "# cx# † (1) 2xy † 0 y# † (1)d œ 1 x# y# , quadratic approximation. Since all partial derivatives of f are products of sines and cosines, the absolute value of these derivatives is less than or equal to 1 Ê E(xß y) Ÿ "6 c(0.1)$ 3(0.1)$ 3(0.1)$ 0.1)$ d Ÿ 0.00134. 12. f(xß y) œ ex sin y Ê fx œ ex sin y, fy œ ex cos y, fxx œ ex sin y, fxy œ ex cos y, fyy œ ex sin y Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d œ 0 x † 0 y † 1 "# ax# † 0 2xy † 1 y# † 0b œ y xy , quadratic approximation. Now, fxxx œ ex sin y, fxxy œ ex cos y, fxyy œ ex sin y, and fyyy œ ex cos y. Since kxk Ÿ 0.1, kex sin yk Ÿ ke0Þ1 sin 0.1k ¸ 0.11 and kex cos yk Ÿ ke0Þ1 cos 0.1k ¸ 1.11. Therefore, E(xß y) Ÿ "6 c(0.11)(0.1)$ 3(1.11)(0.1)$ 3(0.11)(0.1)$ (1.11)(0.1)$ d Ÿ 0.000814. 14.10 PARTIAL DERIVATIVES WITH CONSTRAINED VARIABLES 1. w œ x# y# z# and z œ x# y# : Î x œ x(yß z) Ñ y yœy Ä w Ê Š ``wy ‹ œ ``wx `` xy ``wy `` yy ``wz `` yz ; `` yz œ 0 and `` yz œ 2x `` yx 2y `` yy (a) Œ Ä z z Ï zœz Ò œ 2x `` xy 2y Ê 0 œ 2x `` xy 2y Ê `` xy œ xy Ê Š ``wy ‹ œ (2x) ˆ xy ‰ (2y)(1) (2z)(0) œ 2y 2y œ 0 z xœx Ñ x y œ y(xß z) Ä w Ê ˆ ``wz ‰x œ ``wx `` xz ``wy `` yz ``wz `` zz ; `` xz œ 0 and `` zz œ 2x `` xz 2y `` yz (b) Œ Ä z Ï zœz Ò Î " Ê 1 œ 2y `` yz Ê `` yz œ #"y Ê ˆ ``wz ‰x œ (2x)(0) (2y) Š 2y ‹ (2z)(1) œ 1 2z Î x œ x(yß z) Ñ y yœy Ä w Ê ˆ ``wz ‰y œ ``wx `` xz ``wy `` yz ``wz `` zz ; `` yz œ 0 and `` zz œ 2x `` xz 2y `` yz (c) Œ Ä z Ï zœz Ò 1 Ê 1 œ 2x `` xz Ê `` xz œ 2x Ê ˆ ``wz ‰ œ (2x) ˆ #"x ‰ (2y)(0) (2z)(1) œ 1 2z y 2. w œ x# y z sin t and x y œ t: Î xœx Ñ ÎxÑ Ð yœy Ó y Ä Ð Ä w Ê Š ``wy ‹ œ ``wx `` xy ``wy `` yy ``wz `` yz ``wt ``yt ; `` xy œ 0, `` yz œ 0, and (a) Ó œ z z xz ÏzÒ Ït œ x yÒ ß `t `y œ 1 Ê Š ``wy ‹ œ (2x)(0) (1)(1) (1)(0) (cos t)(1) œ 1 cos t œ 1 cos (x y) xßt Îx œ t yÑ ÎyÑ Ð yœy Ó z Ä Ð Ä w Ê Š ``wy ‹ œ ``wx `` xy ``wy `` yy ``wz `` yz ``wt ``yt ; `` yz œ 0 and ``yt œ 0 (b) Ó z z œ zt ÏtÒ Ï tœt Ò ß Ê `` xy œ ``yt `` yy œ 1 Ê Š ``wy ‹ œ (2x)(1) (1)(1) (1)(0) (cos t)(0) œ 1 2at yb œ 1 2y 2t zßt Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 859 860 Chapter 14 Partial Derivatives Î xœx Ñ ÎxÑ Ð yœy Ó y Ä Ð Ä w Ê ˆ ``wz ‰x y œ ``wx `` xz ``wy `` yz ``wz `` zz ``wt ``zt ; `` xz œ 0 and `` yz œ 0 (c) Ó œ z z ÏzÒ Ït œ x yÒ ß Ê ˆ ``wz ‰x y œ (2x)(0) (1)(0) (1)(1) (cos t)(0) œ 1 ß Îx œ t yÑ ÎyÑ Ð yœy Ó z Ä Ð Ä w Ê ˆ ``wz ‰y t œ ``wx `` xz ``wy `` yz ``wz `` zz ``wt ``zt ; `` yz œ 0 and ``zt œ 0 (d) Ó zœz ÏtÒ Ï tœt Ò ` w Ê ˆ ` z ‰ œ (2x)(0) (1)(0) (1)(1) (cos t)(0) œ 1 ß yßt Î xœx Ñ ÎxÑ Ð y œ t xÓ z Ä Ð Ä w Ê ˆ ``wt ‰x z œ ``wx ``xt ``wy ``yt ``wz ``zt ``wt `` tt ; ``xt œ 0 and ``zt œ 0 (e) Ó zœz ÏtÒ Ï tœt Ò Ê ˆ ``wt ‰ œ (2x)(0) (1)(1) (1)(0) (cos t)(1) œ 1 cos t ß xßz Îx œ t yÑ ÎyÑ Ð yœy Ó z Ä Ð Ä w Ê ˆ ``wt ‰y z œ ``wx ``xt ``wy ``yt ``wz ``zt ``wt `` tt ; ``yt œ 0 and ``zt œ 0 (f) Ó œ z z ÏtÒ Ï tœt Ò Ê ˆ ``wt ‰ œ (2x)(1) (1)(0) (1)(0) (cos t)(1) œ cos t 2x œ cos t 2(t y) ß yßz 3. U œ f(Pß Vß T) and PV œ nRT Î PœP Ñ P `V Ù `T Ù V ‰ ˆ Ù ‰ ˆ ` U ‰ ˆ nR VœV Ä U Ê ˆ `ÙP ‰V œ `ÙP `` PP `` U (a) Œ Ä V ` P ` T ` P œ ` P ` V (0) ` T V PV ÏT œ Ò nR V ‰ œ `ÙP ˆ `ÙT ‰ ˆ nR ÎP œ V Ñ V U `V Ù `T Ù ˆ ` U ‰ ˆ nR ‰ ˆ Ù ‰ Ä U Ê ˆ `ÙT ‰V œ `ÙP `` TP `` V (b) Œ Ä VœV `T `T `T œ `P V ` V (0) ` T T Ï TœT Ò ‰ Ù œ ˆ `ÙP ‰ ˆ nR V `T nRT 4. w œ x# y# z# and y sin z z sin x œ 0 Î xœx Ñ x yœy Ä w Ê ˆ ``wx ‰y œ ``wx `` xx ``wy `` yx ``wz `` xz ; `` yx œ 0 and (a) Œ Ä y Ï z œ z(xß y) Ò z cos x `z 1 (y cos z) `` xz (sin x) `` xz z cos x œ 0 Ê `` xz œ y cos z sin x . At (0ß 1ß 1), ` x œ 1 œ 1 Ê ˆ ``wx ‰yk (0ß1ß1) œ (2x)(1) (2y)(0) (2z)(1)k Ð0ß1ß1Ñ œ 21# Î x œ x(yß z) Ñ y yœy Ä w Ê ˆ ``wz ‰y œ ``wx `` xz ``wy `` yz ``wz `` zz œ (2x) `` xz (2y)(0) (2z)(1) (b) Œ Ä z Ï zœz Ò œ (2x) `` xz 2z. Now (sin z) `` yz y cos z sin x (z cos x) `` xz œ 0 and `` yz œ 0 z sin x Ê y cos z sin x (z cos x) `` xz œ 0 Ê `` xz œ y cos . At (!ß "ß 1), `` xz œ (11)(1)0 œ 1" z cos x Ê ˆ ``wz ‰Ck (!,"ß1Ñ œ 2(0) ˆ 1" ‰ 21 œ 21 5. w œ x# y# yz z$ and x# y# z# œ 6 Î xœx Ñ x yœy (a) Œ Ä Ä w Ê Š ``wy ‹ œ ``wx `` yx ``wy `` yy ``wz `` yz y x Ï z œ z(xß y) Ò Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 14.10 Partial Derivatives with Constrained Variables œ a2xy# b (0) a2x# y zb (1) ay 3z# b `` yz œ 2x# y z ay 3z# b `` yz . Now (2x) `` xy 2y (2z) `` yz œ 0 and `x `y œ 0 Ê 2y (2z) `` yz œ 0 Ê `` yz œ yz . At (wß xß yß z) œ (4ß 2ß 1ß 1), `` yz œ "1 œ 1 Ê Š ``wy ‹ ¹ x (4ß2ß1ßc1) œ c(2)(2)# (1) (1)d c1 3(1)# d (1) œ 5 Î x œ x(yß z) Ñ y yœy (b) Œ Ä Ä w Ê Š ``wy ‹ œ ``wx `` xy ``wy `` yy ``wz `` yz z z Ï zœz Ò œ a2xy# b `` yx a2x# y zb (1) ay 3z# b (0) œ a2x# yb `` yx 2x# y z. Now (2x) `` yx 2y (2z) `` yz œ 0 and `z `y œ 0 Ê (2x) `` xy 2y œ 0 Ê `` xy œ yx . At (wß xß yß z) œ (4ß 2ß 1ß 1), `` yx œ "2 Ê Š ``wy ‹ ¹ œ (2)(2)(1) z "# ‰ (2)(2)# (1) (1) œ 5 (4ß2ß1ßc1) #ˆ 6. y œ uv Ê 1 œ v `` uy u `` vy ; x œ u# v# and `` xy œ 0 Ê 0 œ 2u `` uy 2v `` vy Ê `` vy œ ˆ uv ‰ `` uy Ê 1 # # œ v `` uy u Š uv `` uy ‹ œ Š v v u ‹ `` yu Ê `` yu œ v# v u# . At (uß v) œ ŠÈ2ß 1‹ , `` uy œ " # 1# ŠÈ2‹ œ 1 Ê Š `` uy ‹ œ 1 x r x œ r cos ) 7. Œ Ä Œ Ê ˆ ``xr ‰) œ cos ); x# y# œ r# Ê 2x 2y `` xy œ 2r ``xr and `` xy œ 0 Ê 2x œ 2r ``xr ) y œ r sin ) Ê ``xr œ xr Ê ˆ ``xr ‰ œ È #x # y x y 8. If x, y, and z are independent, then ˆ ``wx ‰y z œ ``wx `` xx ``wy `` xy ``wz `` xz ``wt ``xt ß œ (2x)(1) (2y)(0) (4)(0) (1) ˆ ``xt ‰ œ 2x ``xt . Thus x 2z t œ 25 Ê 1 0 ``xt œ 0 Ê ``xt œ 1 Ê ˆ ``wx ‰ œ 2x 1. On the other hand, if x, y, and t are independent, then ˆ ``wx ‰ yßz yßt œ ``wx `` xx ``wy `` yx ``wz `` xz ``wt ``xt œ (2x)(1) (2y)(0) 4 `` xz (1)(0) œ 2x 4 `` xz . Ê 1 2 `` xz 0 œ 0 Ê `` xz œ #" Ê ˆ ``wx ‰yßt œ 2x 4 ˆ #" ‰ œ 2x 2. Thus, x 2z t œ 25 9. If x is a differentiable function of y and z, then f(xß yß z) œ 0 Ê `` xf `` xx `` yf `` yx `` zf `` xz œ 0 Ê `` xf `` yf `` yx œ 0 Ê Š `` xy ‹ œ `` f/f/`` yz . Similarly, if y is a differentiable function of x and z, Š `` yz ‹ œ `` f/f/`` xz and if z is a z x differentiable function of x and y, ˆ `` xz ‰ œ `` f/f/`` xy . Then Š `` xy ‹ Š `` yz ‹ ˆ `` xz ‰ y z x y œ Š `` f/f/`` yz ‹ ˆ `` f/f/`` xz ‰ Š `` f/f/`` xy ‹ œ 1. df ` u df `z df ` u df `z `z 10. z œ z f(u) and u œ xy Ê `` xz œ 1 du ` x œ 1 y du ; also ` y œ 0 du ` y œ x du so that x ` x y ` y df ‰ df ‰ œ x ˆ1 y du y ˆx du œx 11. If x and y are independent, then g(xß yß z) œ 0 Ê `` gx `` xy `` yg `` yy `` gz `` yz œ 0 and `` xy œ 0 Ê `` yg `` gz `` yz œ 0 `y Ê Š `` yz ‹ œ `` g/ g/` z , as claimed. x 12. Let x and y be independent. Then f(xß yß zß w) œ 0, g(xß yß zß w) œ 0 and `` yx œ 0 Ê `` xf `` xx `` yf `` yx `` zf `` xz ``wf ``wx œ `` xf `` zf `` xz ``wf ``wx œ 0 and `g `x `g `y `g `z `g `w `g `g `z `g `w ` x ` x ` y ` x ` z ` x ` w ` x œ ` x ` z ` x ` w ` x œ 0 imply Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 861 862 Chapter 14 Partial Derivatives `f `z `f `w `f `z `x `w `x œ `x `g `z `g `w œ `g `z `x `w `x `x Ê ˆ `` xz ‰y œ c ``xf » c `g `x `f `z » `g `z `f `w `g » `w `f `w `g » `w œ ``xf ``wg `` gx ``wf `g `f `f `g `z `w `z `w `f `g `f `g `z `w `w `z œ ``xf ``wg ``wf ``gx , as claimed. Likewise, f(xß yß zß w) œ 0, g(xß yß zß w) œ 0 and `` yx œ 0 Ê `` xf `` yx `` yf `` yy `` zf `` yz ``wf ``wy œ `` yf `` zf `` yz ``wf ``wy œ 0 and (similarly) `` gy `` gz `` yz ``wg ``wy œ 0 imply `f `z `f `w `f `z `y `w `y œ `y `g `z `g `w œ `g `z `y `w `y `y Ê Š ``wy ‹ œ x `f `z » `g `z `f `z » `g `z c ``yf c `` gy » `f `w `g » `w `f `g `g `f `f `g `f `g `z `w `z `w `w `z œ `f`z`g`y `g`z ``f y œ ``fz ``gy ``yf ``zg , as claimed. `z `w CHAPTER 14 PRACTICE EXERCISES 1. Domain: All points in the xy-plane Range: z 0 Level curves are ellipses with major axis along the y-axis and minor axis along the x-axis. 2. Domain: All points in the xy-plane Range: 0 z _ Level curves are the straight lines x y œ ln z with slope 1, and z 0. 3. Domain: All (xß y) such that x Á 0 and y Á 0 Range: z Á 0 Level curves are hyperbolas with the x- and y-axes as asymptotes. 4. Domain: All (xß y) so that x# y Range: z 0 0 Level curves are the parabolas y œ x# c, c 0. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 14 Practice Exercises 5. Domain: All points (xß yß z) in space Range: All real numbers Level surfaces are paraboloids of revolution with the z-axis as axis. 6. Domain: All points (xß yß z) in space Range: Nonnegative real numbers Level surfaces are ellipsoids with center (0ß 0ß 0). 7. Domain: All (xß yß z) such that (xß yß z) Á (0ß !ß 0) Range: Positive real numbers Level surfaces are spheres with center (0ß 0ß 0) and radius r 0. 8. Domain: All points (xß yß z) in space Range: (0ß 1] Level surfaces are spheres with center (0ß 0ß 0) and radius r 0. 9. lim Ðxß yÑ Ä Ð1ß ln 2Ñ ey cos x œ eln 2 cos 1 œ (2)(1) œ 2 2y 20 10. lim œ 0 cos 0 œ 2 Ðxß yÑ Ä Ð0ß 0Ñ x cos y 11. lim lim œ lim œ 11 œ # # # œ Ðxß yÑ Ä Ð1ß 1Ñ x y Ðxß yÑ Ä Ð1ß 1Ñ (x y)(x y) Ðxß yÑ Ä Ð1ß 1Ñ x y xÁ „y xÁ „y 12. 13. 14. xy xy 1 (xy 1) ax# y# xy 1b x$ y$ 1 œ lim œ lim xy 1 xy 1 Ðxß yÑ Ä Ð1ß 1Ñ Ðx ß y Ñ Ä Ð 1 ß 1Ñ Ðxß yÑ Ä Ð1ß 1Ñ lim " " ax# y# xy 1b œ 1# † 1# 1 † 1 1 œ 3 lim ln kx y zk œ ln k1 (1) ek œ ln e œ 1 lim tan" (x y z) œ tan" (1 (1) (1)) œ tan" (1) œ 14 P Ä Ð1ß 1ß eÑ P Ä Ð1 ß 1 ß 1 Ñ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 863 864 Chapter 14 Partial Derivatives 15. Let y œ kx# , k Á 1. Then kx# lim œ lim # # # œ 1 k# which gives different limits for Ðxß yÑ Ä Ð0ß 0Ñ x y axß kx# b Ä Ð0ß 0Ñ x kx # yÁx y k different values of k Ê the limit does not exist. 16. Let y œ kx, k Á 0. Then # x# y# x# (kx)# œ lim œ 1 k k which gives different limits for xy x(kx) Ðxß yÑ Ä Ð0ß 0Ñ (xß kxÑ Ä Ð0ß 0Ñ lim xy Á 0 different values of k Ê the limit does not exist. 17. Let y œ kx. Then x# y# x# k# x# 1 k# lim # # œ x# k# x# œ 1 k# which gives different limits for different values Ðxß yÑ Ä Ð0ß 0Ñ x y of k Ê the limit does not exist so f(0ß 0) cannot be defined in a way that makes f continuous at the origin. 18. Along the x-axis, y œ 0 and lim Ðxß yÑ Ä Ð0ß 0Ñ 1, x 0 sin (x y) sin x œœ , so the kxkkyk œ xlim ", x 0 Ä 0 k xk limit fails to exist Ê f is not continuous at (0ß 0). 19. ``gr œ cos ) sin ), `` g) œ r sin ) r cos ) 20. `` xf œ #" Š x# 2x y# ‹ y ‹ x# y # 1 ˆx‰ Š y 2y `f " œ x# x y# x# y y# œ xx# y# , ` y œ # Š x # y # ‹ Š 1x ‹ y # 1 ˆx‰ y œ x# y y# x# x y# œ xx# y# 21. ``Rf" œ R"# , ``Rf# œ R"# , ``Rf$ œ R"# " # $ 22. hx (xß yß z) œ 21 cos (21x y 3z), hy (xß yß z) œ cos (21x y 3z), hz (xß yß z) œ 3 cos (21x y 3z) `P nT ` P nR ` P nRT 23. `` Pn œ RT V , ` R œ V , ` T œ V , ` V œ V# 24. fr (rß jß Tß w) œ 2r"# j É 1Tw , fj (rß jß Tß w) œ #r"j# É 1Tw , fT (rß jß Tß w) œ ˆ #"rj ‰ Š È"1w ‹ Š 2È" T ‹ œ 4r"j É T1"w œ 4r"jT É 1Tw , fw (rß jß Tß w) œ ˆ #"rj ‰ É T1 ˆ "# w$Î# ‰ œ 4r"jw É 1Tw # # # # ` g ` g " 25. `` gx œ "y , `` yg œ 1 yx# Ê `` xg# œ 0, `` yg# œ 2x y $ , ` y ` x œ ` x` y œ y # 26. gx (xß y) œ ex y cos x, gy (xß y) œ sin x Ê gxx (xß y) œ ex y sin x, gyy (xß y) œ 0, gxy (xß y) œ gyx (xß y) œ cos x # # # # # 27. `` xf œ 1 y 15x# x#2x 1 , `` yf œ x Ê `` xf# œ 30x ax2#2x1b# , `` yf# œ 0, ``y`fx œ ``x`fy œ 1 28. fx (xß y) œ 3y, fy (xß y) œ 2y 3x sin y 7ey Ê fxx (xß y) œ 0, fyy (xß y) œ 2 cos y 7ey , fxy (xß y) œ fyx (xß y) œ 3 " t dy 29. ``wx œ y cos (xy 1), ``wy œ x cos (xy 1), dx dt œ e , dt œ t 1 t ˆ " ‰ Ê dw dt œ [y cos (xy 1)]e [x cos (xy 1)] t1 ; t œ 0 Ê x œ 1 and y œ 0 ¸ œ 0 † 1 [1 † (1)] ˆ 0" 1 ‰ œ 1 Ê dw dt t œ0 "Î# dy 30. ``wx œ ey , ``wy œ xey sin z, ``wz œ y cos z sin z, dx , dt œ 1 "t , dz dt œ t dt œ 1 y "Î# Ê dw axey sin zb ˆ1 "t ‰ (y cos z sin z)1; t œ 1 Ê x œ 2, y œ 0, and z œ 1 dt œ e t ¸ œ 1 † 1 (2 † 1 0)(2) (0 0)1 œ 5 Ê dw dt t œ1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 14 Practice Exercises 31. ``wx œ 2 cos (2x y), ``wy œ cos (2x y), ``xr œ 1, ``xs œ cos s, ``yr œ s, ``ys œ r Ê ``wr œ [2 cos (2x y)](1) [ cos (2x y)](s); r œ 1 and s œ 0 Ê x œ 1 and y œ 0 Ê ``wr ¸ œ (2 cos 21) (cos 21)(0) œ 2; ``ws œ [2 cos (2x y)](cos s) [ cos (2x y)](r) Ð1ß0Ñ Ê `w ¸ ` s Ð1ß0Ñ œ (2 cos 21)(cos 0) (cos 21)(1) œ 2 1 `x " ‰ `w ¸ 2 u ˆ x ˆ2 "‰ 32. ``wu œ dw dx ` u œ 1 x# x# 1 a2e cos vb ; u œ v œ 0 Ê x œ 2 Ê ` u Ð0ß0Ñ œ 5 5 (2) œ 5 ; `w dw ` x " ‰ u ˆ x ` v œ dx ` v œ 1 x# x# 1 a2e sin vb Ê ``wv ¸ Ð0ß0Ñ œ ˆ 52 5" ‰ (0) œ 0 dy dz 33. `` xf œ y z, `` yf œ x z, `` zf œ y x, dx dt œ sin t, dt œ cos t, dt œ 2 sin 2t Ê df dt œ (y z)(sin t) (x z)(cos t) 2(y x)(sin 2t); t œ 1 Ê x œ cos 1, y œ sin 1, and z œ cos 2 df ¸ Ê dt œ (sin 1 cos 2)(sin 1) (cos 1 cos 2)(cos 1) 2(sin 1 cos 1)(sin 2) t œ1 `s dw `w dw ` s dw dw `w `w dw dw 34. ``wx œ dw ds ` x œ (5) ds and ` y œ ds ` y œ (1) ds œ ds Ê ` x 5 ` y œ 5 ds 5 ds œ 0 1 y cos xy Fx 35. F(xß y) œ 1 x y# sin xy Ê Fx œ 1 y cos xy and Fy œ 2y x cos xy Ê dy dx œ Fy œ 2y x cos xy œ 12yy xcoscosxyxy Ê at (xß y) œ (!ß 1) we have dy dx ¹ Ð0ß1Ñ œ 12" œ 1 xby 2y e Fx 36. F(xß y) œ 2xy exy 2 Ê Fx œ 2y exy and Fy œ 2x exy Ê dy dx œ Fy œ 2x exby Ê at (xß y) œ (!ß ln 2) we have dy dx ¹ Ð0ßln 2Ñ 2 œ 2 ln0 2 2 œ (ln 2 1) # # 37. ™ f œ ( sin x cos y)i (cos x sin y)j Ê ™ f k ˆ 14 14 ‰ œ "# i "# j Ê k ™ f k œ Éˆ "# ‰ ˆ "# ‰ œ È" œ ß È È 2 È È2 # ; È f 2 2 2 2 u œ k™ ™f k œ # i # j Ê f increases most rapidly in the direction u œ # i # j and decreases most È2 È2 È2 È2 # i # j ; (Du f)P! œ k ™ f k œ # and (Dcu f)P! œ # ; 7 u" œ kvvk œ È33i #4j4# œ 53 i 54 j Ê (Du" f)P! œ ™ f † u" œ ˆ "# ‰ ˆ 35 ‰ ˆ "# ‰ ˆ 45 ‰ œ 10 rapidly in the direction u œ f 1 1 38. ™ f œ 2xec2y i 2x# ec2y j Ê ™ f k Ð1ß0Ñ œ #i #j Ê k ™ f k œ È2# (2)# œ 2È2; u œ k™ ™f k œ È2 i È2 j Ê f increases most rapidly in the direction u œ È12 i È12 j and decreases most rapidly in the direction u œ È12 i È12 j ; (Du f)P! œ k ™ f k œ 2È2 and (Dcu f)P! œ 2È2 ; u" œ kvvk œ È1i #j 1# œ È12 i È12 j Ê (Du" f)P! œ ™ f † u" œ (2) Š È"2 ‹ (2) Š È"2 ‹ œ 0 2 3 6 39. ™ f œ Š 2x 3y 6z ‹ i Š 2x 3y 6z ‹ j Š 2x 3y 6z ‹ k Ê ™ f k Ð 1ß 1ß1Ñ œ 2i 3j 6k ; f 2i 3j 6k 2 3 6 2 3 6 u œ k™ ™f k œ È2# 3# 6# œ 7 i 7 j 7 k Ê f increases most rapidly in the direction u œ 7 i 7 j 7 k and decreases most rapidly in the direction u œ 27 i 37 j 67 k ; (Du f)P! œ k ™ f k œ 7, (Du f)P! œ 7; u" œ kvvk œ 27 i 37 j 67 k Ê (Du" f)P! œ (Du f)P! œ 7 f 2 " 40. ™ f œ (2x 3y)i (3x 2)j (1 2z)k Ê ™ f k Ð0ß0ß0Ñ œ 2j k ; u œ k™ ™f k œ È5 j È5 k Ê f increases most rapidly in the direction u œ È25 j È"5 k and decreases most rapidly in the direction u œ È25 j È"5 k ; k (Du f)P! œ k ™ f k œ È5 and (Du f)P! œ È5 ; u" œ kvvk œ È1i#j1# œ È"3 i È"3 j È"3 k 1# Ê (Du" f)P! œ ™ f † u" œ (0) Š È"3 ‹ (2) Š È"3 ‹ (1) Š È"3 ‹ œ È33 œ È3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 865 866 Chapter 14 Partial Derivatives 41. r œ (cos 3t)i (sin 3t)j 3tk Ê v(t) œ (3 sin 3t)i (3 cos 3t)j 3k Ê v ˆ 13 ‰ œ 3j 3k Ê u œ È"2 j È"2 k ; f(xß yß z) œ xyz Ê ™ f œ yzi xzj xyk ; t œ 13 yields the point on the helix (1ß 0ß 1) Ê ™ f k Ð 1ß0ß1Ñ œ 1j Ê ™ f † u œ (1j) † Š È"2 j È"2 k‹ œ È12 42. f(xß yß z) œ xyz Ê ™ f œ yzi xzj xyk ; at (1ß 1ß 1) we get ™ f œ i j k Ê the maximum value of Du f k œ k ™ f k œ È3 Ð1ß1ß1Ñ 43. (a) Let ™ f œ ai bj at (1ß 2). The direction toward (2ß 2) is determined by v" œ (2 1)i (2 2)j œ i œ u so that ™ f † u œ 2 Ê a œ 2. The direction toward (1ß 1) is determined by v# œ (1 1)i (1 2)j œ j œ u so that ™ f † u œ 2 Ê b œ 2 Ê b œ 2. Therefore ™ f œ 2i 2j ; fx a1, 2b œ fy a1, 2b œ 2. (b) The direction toward (4ß 6) is determined by v$ œ (4 1)i (6 2)j œ 3i 4j Ê u œ 35 i 45 j Ê ™ f † u œ 14 5 . 44. (a) True (b) False (c) True (d) True 45. ™ f œ 2xi j 2zk Ê ™ f k Ð0ß 1ß 1Ñ œ j 2k , ™ f k Ð0ß0ß0Ñ œ j , ™ f k Ð0ß 1ß1Ñ œ j 2k 46. ™ f œ 2yj 2zk Ê ™ f k Ð2ß2ß0Ñ œ 4j , ™ f k Ð2ß 2ß0Ñ œ 4j , ™ f k Ð2ß0ß2Ñ œ 4k , ™ f k Ð2ß0ß 2Ñ œ 4k 47. ™ f œ 2xi j 5k Ê ™ f k Ð2ß 1ß1Ñ œ 4i j 5k Ê Tangent Plane: 4(x 2) (y 1) 5(z 1) œ 0 Ê 4x y 5z œ 4; Normal Line: x œ 2 4t, y œ 1 t, z œ 1 5t 48. ™ f œ 2xi 2yj k Ê ™ f k Ð1ß1ß2Ñ œ 2i 2j k Ê Tangent Plane: 2(x 1) 2(y 1) (z 2) œ 0 Ê 2x 2y z 6 œ 0; Normal Line: x œ 1 2t, y œ 1 2t, z œ 2 t 2y `z ¸ `z `z 49. `` xz œ x# 2x y# Ê ` x Ð0ß1ß0Ñ œ 0 and ` y œ x# y# Ê ` y ¹ Ð0ß1ß0Ñ œ 2; thus the tangent plane is 2(y 1) (z 0) œ 0 or 2y z 2 œ 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 14 Practice Exercises 50. `` xz œ 2x ax# y# b # Ê `` xz ¸ ˆ1 1 1 ‰ œ #" and `` yz œ 2y ax# y# b ß ß # 2 Ê `` yz ¹ ˆ1ß1ß 1 ‰ 2 867 œ "# ; thus the tangent plane is "# (x 1) "# (y 1) ˆz "# ‰ œ 0 or x y 2z 3 œ 0 51. ™ f œ ( cos x)i j Ê ™ f k Ð1ß1Ñ œ i j Ê the tangent line is (x 1) (y 1) œ 0 Ê x y œ 1 1; the normal line is y 1 œ 1(x 1) Ê y œ x 1 1 52. ™ f œ xi yj Ê ™ f k Ð1ß2Ñ œ i 2j Ê the tangent line is (x 1) 2(y 2) œ 0 Ê y œ "# x #3 ; the normal line is y 2 œ 2(x 1) Ê y œ 2x 4 53. Let f(xß yß z) œ x# 2y 2z 4 and g(xß yß z) œ y 1. Then ™ f œ 2xi 2j 2kk a1 1 12 b œ 2i 2j 2k â â â i j kâ â â and ™ g œ j Ê ™ f ‚ ™ g œ â 2 2 2 â œ 2i 2k Ê the line is x œ 1 2t, y œ 1, z œ "# 2t â â â0 " 0â ß ß 54. Let f(xß yß z) œ x y# z 2 and g(xß yß z) œ y 1. Then ™ f œ i 2yj kk a 12 1 12 b œ i 2j k and â â â i j kâ â â ™ g œ j Ê ™ f ‚ ™ g œ â 1 2 1 â œ i k Ê the line is x œ "# t, y œ 1, z œ "# t â â â0 " 0â ß ß 55. f ˆ 14 ß 14 ‰ œ "# , fx ˆ 14 ß 14 ‰ œ cos x cos yk Ð1Î4ß1Î4Ñ œ "# , fy ˆ 14 ß 14 ‰ œ sin x sin yk Ð1Î4ß1Î4Ñ œ "# Ê L(xß y) œ "# "# ˆx 14 ‰ "# ˆy 14 ‰ œ "# "# x "# y; fxx (xß y) œ sin x cos y, fyy (xß y) œ sin x cos y, and fxy (xß y) œ cos x sin y. Thus an upper bound for E depends on the bound M used for kfxx k , kfxy k , and kfyy k . With M œ È2 È 1¸ 1 ¸ ‰# " È2 ˆ¸ ¸ Ÿ 42 (0.2)# Ÿ 0.0142; # we have kE(xß y)k Ÿ # Š # ‹ x 4 y 4 # with M œ 1, kE(xß y)k Ÿ "# (1) ˆ¸x 14 ¸ ¸y 14 ¸‰ œ "# (0.2)# œ 0.02. 56. f(1ß 1) œ 0, fx (1ß 1) œ yk Ð1ß1Ñ œ 1, fy (1ß 1) œ x 6yk Ð1ß1Ñ œ 5 Ê L(xß y) œ (x 1) 5(y 1) œ x 5y 4; fxx (xß y) œ 0, fyy (xß y) œ 6, and fxy (xß y) œ 1 Ê maximum of kfxx k , kfyy k , and kfxy k is 6 Ê M œ 6 Ê kE(xß y)k Ÿ "# (6) akx 1k ky 1kb# œ "# (6)(0.1 0.2)# œ 0.27 57. f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ y 3zk Ð1ß0ß0Ñ œ 0, fy (1ß 0ß 0) œ x 2zk Ð1ß0ß0Ñ œ 1, fz (1ß 0ß 0) œ 2y 3xk Ð1ß0ß0Ñ œ 3 Ê L(xß yß z) œ 0(x 1) (y 0) 3(z 0) œ y 3z; f(1ß 1ß 0) œ 1, fx (1ß 1ß 0) œ 1, fy (1ß 1ß 0) œ 1, fz ("ß "ß !) œ 1 Ê L(xß yß z) œ 1 (x 1) (y 1) 1(z 0) œ x y z 1 58. f ˆ0ß !ß 14 ‰ œ 1, fx ˆ!ß 0ß 14 ‰ œ È2 sin x sin (y z)¹ ˆ0ß0ß 1 ‰ œ 0, fy ˆ!ß 0ß 14 ‰ œ È2 cos x cos (y z)¹ 4 fz ˆ!ß 0ß 14 ‰ œ È2 cos x cos (y z)¹ ˆ0ß0ß 1 ‰ œ 1, 4 œ 1 Ê L(xß yß z) œ 1 1(y 0) 1 ˆz 14 ‰ œ 1 y z 14 ; 4 È2 È2 È2 È2 ˆ1 1 ‰ ˆ1 1 ‰ ˆ1 1 ‰ # , fx 4 ß 4 ß 0 œ # , fy 4 ß 4 ß 0 œ # , fz 4 ß 4 ß 0 œ # È È È È È È È È L(xß yß z) œ #2 #2 ˆx 14 ‰ #2 ˆy 14 ‰ #2 (z 0) œ #2 #2 x #2 y #2 z f ˆ 14 ß 14 ß 0‰ œ Ê ˆ0ß0ß 1 ‰ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 868 Chapter 14 Partial Derivatives 59. V œ 1r# h Ê dV œ 21rh dr 1r# dh Ê dVk Ð1Þ5ß5280Ñ œ 21(1.5)(5280) dr 1(1.5)# dh œ 15,8401 dr 2.251 dh. You should be more careful with the diameter since it has a greater effect on dV. 60. df œ (2x y) dx (x 2y) dy Ê df k Ð1ß2Ñ œ 3 dy Ê f is more sensitive to changes in y; in fact, near the point (1ß 2) a change in x does not change f. " 24 ¸ 61. dI œ R" dV RV# dR Ê dI¸ Ð24ß100Ñ œ 100 dV 100 # dR Ê dI dVœ1ßdRœ20 œ 0.01 (480)(.0001) œ 0.038, " ‰ 20 ‰ or increases by 0.038 amps; % change in V œ (100) ˆ 24 ¸ 4.17%; % change in R œ ˆ 100 (100) œ 20%; 24 I œ 100 œ 0.24 Ê estimated % change in I œ dII ‚ 100 œ 0.038 0.24 ‚ 100 ¸ 15.83% Ê more sensitive to voltage change. 62. A œ 1ab Ê dA œ 1b da 1a db Ê dAk Ð10ß16Ñ œ 161 da 101 db; da œ „ 0.1 and db œ „ 0.1 ¸ ¸ 2.61 ¸ Ê dA œ „ 261(0.1) œ „ 2.61 and A œ 1(10)(16) œ 1601 Ê ¸ dA A ‚ 100 œ 1601 ‚ 100 ¸ 1.625% 63. (a) y œ uv Ê dy œ v du u dv; percentage change in u Ÿ 2% Ê kduk Ÿ 0.02, and percentage change in v Ÿ 3% dy v du u dv dv dv ¸ du ¸ ¸ du ¸ ¸ dv ¸ Ê kdvk Ÿ 0.03; dy œ du y œ uv u v Ê ¹ y ‚ 100¹ œ u ‚ 100 v ‚ 100 Ÿ u ‚ 100 v ‚ 100 Ÿ 2% 3% œ 5% dv du dv du dv (b) z œ u v Ê dzz œ duu v œ u v u v Ÿ u v (since u 0, v 0) dy dv ¸ Ê ¸ dzz ‚ 100¸ Ÿ ¸ du u ‚ 100 v ‚ 100 œ ¹ y ‚ 100¹ (0.425)(7) (0.725)(7) 64. C œ 71.84w07425 h0 725 Ê Cw œ 71.84w 1 425 h0 725 and Ch œ 71.84w0 425 h1 725 Þ Ê Þ Þ Þ Þ Þ 2.975 5.075 dC œ 71.84w 1 425 h0 725 dw 71.84w0 425 h1 725 dh; thus when w œ 70 and h œ 180 we have Þ Þ Þ Þ dCk Ð70ß180Ñ ¸ (0.00000225) dw (0.00000149) dh Ê 1 kg error in weight has more effect 65. fx (xß y) œ 2x y 2 œ 0 and fy (xß y) œ x 2y 2 œ 0 Ê x œ 2 and y œ 2 Ê (2ß 2) is the critical point; # œ 3 0 and fxx 0 Ê local minimum value fxx (2ß 2) œ 2, fyy (#ß 2) œ 2, fxy (#ß 2) œ 1 Ê fxx fyy fxy of f(#ß 2) œ 8 66. fx (xß y) œ 10x 4y 4 œ 0 and fy (xß y) œ 4x 4y 4 œ 0 Ê x œ 0 and y œ 1 Ê (0ß 1) is the critical point; # œ 56 0 Ê saddle point with f(0ß 1) œ 2 fxx (0ß 1) œ 10, fyy (0ß 1) œ 4, fxy (0ß 1) œ 4 Ê fxx fyy fxy 67. fx (xß y) œ 6x# 3y œ 0 and fy (xß y) œ 3x 6y# œ 0 Ê y œ 2x# and 3x 6 a4x% b œ 0 Ê x a1 8x$ b œ 0 Ê x œ 0 and y œ 0, or x œ "# and y œ "# Ê the critical points are (0ß 0) and ˆ "# ß "# ‰ . For (!ß !): # fxx (!ß !) œ 12xk Ð0ß0Ñ œ 0, fyy (!ß !) œ 12yk Ð0ß0Ñ œ 0, fxy (!ß 0) œ 3 Ê fxx fyy fxy œ 9 0 Ê saddle point with # f(0ß 0) œ 0. For ˆ "# ß "# ‰: fxx œ 6, fyy œ 6, fxy œ 3 Ê fxx fyy fxy œ 27 0 and fxx 0 Ê local maximum " " " value of f ˆ # ß # ‰ œ 4 68. fx (xß y) œ 3x# 3y œ 0 and fy (xß y) œ 3y# 3x œ 0 Ê y œ x# and x% x œ 0 Ê x ax$ 1b œ 0 Ê the critical points are (0ß 0) and (1ß 1) . For (!ß !): fxx (!ß !) œ 6xk Ð0ß0Ñ œ 0, fyy (!ß !) œ 6yk Ð0ß0Ñ œ 0, fxy (!ß 0) œ 3 # Ê fxx fyy fxy œ 9 0 Ê saddle point with f(0ß 0) œ 15. For (1ß 1): fxx (1ß 1) œ 6, fyy (1ß 1) œ 6, fxy (1ß 1) œ 3 # Ê fxx fyy fxy œ 27 0 and fxx 0 Ê local minimum value of f(1ß 1) œ 14 69. fx (xß y) œ 3x# 6x œ 0 and fy (xß y) œ 3y# 6y œ 0 Ê x(x 2) œ 0 and y(y 2) œ 0 Ê x œ 0 or x œ 2 and y œ 0 or y œ 2 Ê the critical points are (0ß 0), (0ß 2), (2ß 0), and (2ß 2) . For (!ß !): fxx (!ß !) œ 6x 6k Ð0ß0Ñ # œ 6, fyy (!ß !) œ 6y 6k Ð0ß0Ñ œ 6, fxy (!ß 0) œ 0 Ê fxx fyy fxy œ 36 0 Ê saddle point with f(0ß 0) œ 0. For # (0ß 2): fxx (!ß 2) œ 6, fyy (0ß #) œ 6, fxy (!ß 2) œ 0 Ê fxx fyy fxy œ 36 0 and fxx 0 Ê local minimum value of Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 14 Practice Exercises 869 # f(!ß 2) œ 4. For (#ß 0): fxx (2ß 0) œ 6, fyy (#ß 0) œ 6, fxy (2ß 0) œ 0 Ê fxx fyy fxy œ 36 0 and fxx 0 Ê local maximum value of f(2ß 0) œ 4. For (2ß 2): fxx (2ß 2) œ 6, fyy (2ß 2) œ 6, fxy (2ß 2) œ 0 # Ê fxx fyy fxy œ 36 0 Ê saddle point with f(2ß 2) œ 0. 70. fx (xß y) œ 4x$ 16x œ 0 Ê 4x ax# 4b œ 0 Ê x œ 0, 2, 2; fy (xß y) œ 6y 6 œ 0 Ê y œ 1. Therefore the critical points are (0ß 1), (2ß 1), and (2ß 1). For (!ß 1): fxx (!ß 1) œ 12x# 16k Ð0ß1Ñ œ 16, fyy (!ß 1) œ 6, fxy (!ß 1) œ 0 # Ê fxx fyy fxy œ 96 0 Ê saddle point with f(0ß 1) œ 3. For (2ß 1): fxx (2ß 1) œ 32, fyy (2ß 1) œ 6, # fxy (2ß 1) œ 0 Ê fxx fyy fxy œ 192 0 and fxx 0 Ê local minimum value of f(2ß 1) œ 19. For (#ß 1): # fxx (2ß 1) œ 32, fyy (#ß 1) œ 6, fxy (2ß 1) œ 0 Ê fxx fyy fxy œ 192 0 and fxx 0 Ê local minimum value of f(2ß 1) œ 19. 71. (i) On OA, f(xß y) œ f(0ß y) œ y# 3y for 0 Ÿ y Ÿ 4 Ê f w (!ß y) œ 2y 3 œ 0 Ê y œ 3# . But ˆ!ß 3# ‰ is not in the region. Endpoints: f(0ß 0) œ 0 and f(0ß 4) œ 28. (ii) On AB, f(xß y) œ f(xß x 4) œ x# 10x 28 for 0 Ÿ x Ÿ 4 Ê f w (xß x 4) œ 2x 10 œ 0 Ê x œ 5, y œ 1. But (5ß 1) is not in the region. Endpoints: f(4ß 0) œ 4 and f(!ß 4) œ 28. (iii) On OB, f(xß y) œ f(xß 0) œ x# 3x for 0 Ÿ x Ÿ 4 Ê f w (xß 0) œ 2x 3 Ê x œ 3# and y œ 0 Ê ˆ 3# ß 0‰ is a critical point with f ˆ 3# ß !‰ œ 94 . Endpoints: f(0ß 0) œ 0 and f(%ß 0) œ 4. (iv) For the interior of the triangular region, fx (xß y) œ 2x y 3 œ 0 and fy (xß y) œ x 2y 3 œ 0 Ê x œ 3 and y œ 3. But (3ß 3) is not in the region. Therefore the absolute maximum is 28 at (0ß 4) and the absolute minimum is 94 at ˆ 3# ß !‰ . On OA, f(xß y) œ f(0ß y) œ y# 4y 1 for 0 Ÿ y Ÿ 2 Ê f w (!ß y) œ 2y 4 œ 0 Ê y œ 2 and x œ 0. But (0ß 2) is not in the interior of OA. Endpoints: f(0ß 0) œ 1 and f(0ß 2) œ 5. (ii) On AB, f(xß y) œ f(xß 2) œ x# 2x 5 for 0 Ÿ x Ÿ 4 Ê f w (xß 2) œ 2x 2 œ 0 Ê x œ 1 and y œ 2 Ê (1ß 2) is an interior critical point of AB with f(1ß 2) œ 4. Endpoints: f(4ß 2) œ 13 and f(!ß 2) œ 5. (iii) On BC, f(xß y) œ f(4ß y) œ y# 4y 9 for 0 Ÿ y Ÿ 2 Ê f w (4ß y) œ 2y 4 œ 0 Ê y œ # and x œ 4. But (4ß 2) is not in the interior of BC. Endpoints: f(4ß 0) œ 9 and f(%ß 2) œ 13. (iv) On OC, f(xß y) œ f(xß 0) œ x# 2x 1 for 0 Ÿ x Ÿ 4 Ê f w (xß 0) œ 2x 2 œ 0 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of OC with f(1ß 0) œ 0. Endpoints: f(0ß 0) œ 1 and f(4ß 0) œ 9. (v) For the interior of the rectangular region, fx (xß y) œ 2x 2 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and y œ 2. But (1ß 2) is not in the interior of the region. Therefore the absolute maximum is 13 at (4ß 2) and the absolute minimum is 0 at (1ß 0). 72. (i) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 870 Chapter 14 Partial Derivatives 73. (i) On AB, f(xß y) œ f(2ß y) œ y# y 4 for 2 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2y 1 Ê y œ "# and x œ 2 Ê ˆ2ß "# ‰ is an interior critical point in AB with f ˆ2ß "# ‰ œ 17 4 . Endpoints: f(2ß 2) œ 2 and f(2ß 2) œ 2. On BC, f(xß y) œ f(xß 2) œ 2 for 2 Ÿ x Ÿ 2 Ê f w (xß 2) œ 0 Ê no critical points in the interior of BC. Endpoints: f(2ß 2) œ 2 and f(2ß 2) œ 2. (iii) On CD, f(xß y) œ f(2ß y) œ y# 5y 4 for 2 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2y 5 œ 0 Ê y œ 5# and x œ 2. But ˆ#ß 5# ‰ is not in the region. (ii) Endpoints: f(2ß 2) œ 18 and f(2ß 2) œ 2. (iv) On AD, f(xß y) œ f(xß 2) œ 4x 10 for 2 Ÿ x Ÿ 2 Ê f w (xß 2) œ 4 Ê no critical points in the interior of AD. Endpoints: f(2ß 2) œ 2 and f(2ß 2) œ 18. (v) For the interior of the square, fx (xß y) œ y 2 œ 0 and fy (xß y) œ 2y x 3 œ 0 Ê y œ 2 and x œ 1 Ê (1ß 2) is an interior critical point of the square with f(1ß 2) œ 2. Therefore the absolute maximum "‰ ˆ is 18 at (2ß 2) and the absolute minimum is 17 4 at #ß # . On OA, f(xß y) œ f(0ß y) œ 2y y# for 0 Ÿ y Ÿ 2 Ê f w (!ß y) œ 2 2y œ 0 Ê y œ 1 and x œ 0 Ê (!ß 1) is an interior critical point of OA with f(0ß 1) œ 1. Endpoints: f(0ß 0) œ 0 and f(0ß 2) œ 0. (ii) On AB, f(xß y) œ f(xß 2) œ 2x x# for 0 Ÿ x Ÿ 2 Ê f w (xß 2) œ 2 2x œ 0 Ê x œ 1 and y œ 2 Ê (1ß 2) is an interior critical point of AB with f(1ß 2) œ 1. Endpoints: f(0ß 2) œ 0 and f(2ß 2) œ 0. (iii) On BC, f(xß y) œ f(2ß y) œ 2y y# for 0 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2 2y œ 0 Ê y œ 1 and x œ 2 Ê (2ß 1) is an interior critical point of BC with f(2ß 1) œ 1. Endpoints: f(2ß 0) œ 0 and f(2ß 2) œ 0. (iv) On OC, f(xß y) œ f(xß 0) œ 2x x# for 0 Ÿ x Ÿ 2 Ê f w (xß 0) œ 2 2x œ 0 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of OC with f(1ß 0) œ 1. Endpoints: f(0ß 0) œ 0 and f(0ß 2) œ 0. (v) For the interior of the rectangular region, fx (xß y) œ 2 2x œ 0 and fy (xß y) œ 2 2y œ 0 Ê x œ 1 and y œ 1 Ê (1ß 1) is an interior critical point of the square with f(1ß 1) œ 2. Therefore the absolute maximum is 2 at (1ß 1) and the absolute minimum is 0 at the four corners (0ß 0), (0ß 2), (2ß 2), and (2ß 0). 74. (i) On AB, f(xß y) œ f(xß x 2) œ 2x 4 for 2 Ÿ x Ÿ 2 Ê f w (xß x 2) œ 2 œ 0 Ê no critical points in the interior of AB. Endpoints: f(2ß 0) œ 8 and f(2ß 4) œ 0. (ii) On BC, f(xß y) œ f(2ß y) œ y# 4y for 0 Ÿ y Ÿ 4 Ê f w (2ß y) œ 2y 4 œ 0 Ê y œ 2 and x œ 2 Ê (2ß 2) is an interior critical point of BC with f(2ß 2) œ 4. Endpoints: f(2ß 0) œ 0 and f(2ß 4) œ 0. (iii) On AC, f(xß y) œ f(xß 0) œ x# 2x for 2 Ÿ x Ÿ 2 Ê f w (xß 0) œ 2x 2 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of AC with f(1ß 0) œ 1. Endpoints: f(2ß 0) œ 8 and f(2ß 0) œ 0. (iv) For the interior of the triangular region, fx (xß y) œ 2x 2 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and y œ 2 Ê (1ß 2) is an interior critical point of the region with f(1ß 2) œ 3. Therefore the absolute maximum is 8 at (2ß 0) and the absolute minimum is 1 at (1ß 0). 75. (i) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 14 Practice Exercises 76. (i) (ii) 871 On AB, faxß yb œ faxß xb œ 4x# 2x% 16 for 2 Ÿ x Ÿ 2 Ê f w axß xb œ 8x 8x$ œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1, or x œ 1 and y œ 1 Ê a0ß 0b, a1ß 1b, a1ß 1b are all interior points of AB with fa0ß 0b œ 16, fa1ß 1b œ 18, and fa1ß 1b œ 18. Endpoints: fa2ß 2b œ 0 and fa2ß 2b œ 0. On BC, faxß yb œ fa2ß yb œ 8y y% for 2 Ÿ y Ÿ 2 3 Ê f w a2ß yb œ 8 4y$ œ 0 Ê y œ È 2 and x œ 2 3 Ê Š2ß È 2‹ is an interior critical point of BC with 3 3 f Š2ß È 2‹ œ 6 È 2. Endpoints: fa2ß 2b œ 32 and fa2ß 2b œ 0. 3 (iii) On AC, faxß yb œ faxß 2b œ 8x x% for 2 Ÿ x Ÿ 2 Ê f w axß 2b œ 8 4x$ œ 0 Ê x œ È 2 and y œ 2 3 3 3 Ê ŠÈ 2ß 2‹ is an interior critical point of AC with f ŠÈ 2ß 2‹ œ 6 È 2. Endpoints: fa2ß 2b œ 0 and fa2ß 2b œ 32. (iv) For the interior of the triangular region, fx axß yb œ 4y 4x$ œ 0 and fy axß yb œ 4x 4y$ œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1 or x œ 1 and y œ 1. But neither of the points a0ß 0b and a1ß 1b, or a1ß 1b are interior to the region. Therefore the absolute maximum is 18 at (1ß 1) and (1ß 1), and the absolute minimum is 32 at a2ß 2b. On AB, f(xß y) œ f(1ß y) œ y$ 3y# 2 for 1 Ÿ y Ÿ 1 Ê f w (1ß y) œ 3y# 6y œ 0 Ê y œ 0 and x œ 1, or y œ 2 and x œ 1 Ê (1ß 0) is an interior critical point of AB with f(1ß 0) œ 2; (1ß 2) is outside the boundary. Endpoints: f(1ß 1) œ 2 and f(1ß 1) œ 0. (ii) On BC, f(xß y) œ f(xß 1) œ x$ 3x# 2 for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 6x œ 0 Ê x œ 0 and y œ 1, or x œ 2 and y œ 1 Ê (0ß 1) is an interior critical point of BC with f(!ß 1) œ 2; (2ß 1) is outside the boundary. Endpoints: f("ß 1) œ 0 and f("ß 1) œ 2. (iii) On CD, f(xß y) œ f("ß y) œ y$ 3y# 4 for 1 Ÿ y Ÿ 1 Ê f w ("ß y) œ 3y# 6y œ 0 Ê y œ 0 and x œ 1, or y œ 2 and x œ 1 Ê ("ß 0) is an interior critical point of CD with f("ß 0) œ 4; (1ß 2) is outside the boundary. Endpoints: f(1ß 1) œ 2 and f("ß 1) œ 0. (iv) On AD, f(xß y) œ f(xß 1) œ x$ 3x# 4 for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 6x œ 0 Ê x œ 0 and y œ 1, or x œ 2 and y œ 1 Ê (0ß 1) is an interior point of AD with f(0ß 1) œ 4; (#ß 1) is outside the boundary. Endpoints: f(1ß 1) œ 2 and f("ß 1) œ 0. (v) For the interior of the square, fx (xß y) œ 3x# 6x œ 0 and fy (xß y) œ 3y# 6y œ 0 Ê x œ 0 or x œ 2, and y œ 0 or y œ 2 Ê (0ß 0) is an interior critical point of the square region with f(!ß 0) œ 0; the points (0ß 2), (2ß 0), and (2ß 2) are outside the region. Therefore the absolute maximum is 4 at (1ß 0) and the absolute minimum is 4 at (0ß 1). 77. (i) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 872 Chapter 14 Partial Derivatives On AB, f(xß y) œ f(1ß y) œ y$ 3y for 1 Ÿ y Ÿ 1 Ê f w (1ß y) œ 3y# 3 œ 0 Ê y œ „ 1 and x œ 1 yielding the corner points (1ß 1) and (1ß 1) with f(1ß 1) œ 2 and f(1ß 1) œ 2. (ii) On BC, f(xß y) œ f(xß 1) œ x$ 3x 2 for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 3 œ 0 Ê no solution. Endpoints: f("ß 1) œ 2 and f("ß 1) œ 6. (iii) On CD, f(xß y) œ f("ß y) œ y$ 3y 2 for 1 Ÿ y Ÿ 1 Ê f w ("ß y) œ 3y# 3 œ 0 Ê no solution. Endpoints: f(1ß 1) œ 6 and f("ß 1) œ 2. (iv) On AD, f(xß y) œ f(xß 1) œ x$ 3x for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 3 œ 0 Ê x œ „ 1 and y œ 1 yielding the corner points (1ß 1) and (1ß 1) with f(1ß 1) œ 2 and f(1ß 1) œ 2 (v) For the interior of the square, fx (xß y) œ 3x# 3y œ 0 and fy (xß y) œ 3y# 3x œ 0 Ê y œ x# and x% x œ 0 Ê x œ 0 or x œ 1 Ê y œ 0 or y œ 1 Ê (!ß 0) is an interior critical point of the square region with f(0ß 0) œ 1; (1ß 1) is on the boundary. Therefore the absolute maximum is 6 at ("ß 1) and the absolute minimum is 2 at (1ß 1) and (1ß 1). 78. (i) 79. ™ f œ 3x# i 2yj and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê 3x# i 2yj œ -(2xi 2yj) Ê 3x# œ 2x- and 2y œ 2y- Ê - œ 1 or y œ 0. CASE 1: - œ 1 Ê 3x# œ 2x Ê x œ 0 or x œ 23 ; x œ 0 Ê y œ „ 1 yielding the points (0ß 1) and (!ß 1); x œ 23 È È È Ê y œ „ 35 yielding the points Š 32 ß 35 ‹ and Š 32 ß 35 ‹ . CASE 2: y œ 0 Ê x# 1 œ 0 Ê x œ „ 1 yielding the points (1ß 0) and (1ß 0). È Evaluations give f a!ß „ 1b œ 1, f Š 23 ß „ 35 ‹ œ 23 27 , f("ß 0) œ 1, and f("ß 0) œ 1. Therefore the absolute maximum is 1 at a!ß „ 1b and (1ß 0), and the absolute minimum is 1 at ("ß !). 80. ™ f œ yi xj and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê yi xj œ -(2xi 2yj) Ê y œ 2-x and xy œ 2-y Ê x œ 2-(2-x) œ 4-# x Ê x œ 0 or 4-# œ 1. CASE 1: x œ 0 Ê y œ 0 but (0ß 0) does not lie on the circle, so no solution. CASE 2: 4-# œ 1 Ê - œ "# or - œ "# . For - œ "# , y œ x Ê 1 œ x# y# œ 2x# Ê x œ C œ „ È"2 yielding the points Š È"2 ß È"2 ‹ and Š È"2 , È"2 ‹ . For - œ #" , y œ x Ê 1 œ x# y# œ 2x# Ê x œ „ È"2 and y œ x yielding the points Š È"2 ß È"2 ‹ and Š È"2 , È"2 ‹ . Evaluations give the absolute maximum value f Š È"2 ß È"2 ‹ œ f Š È"2 ß È"2 ‹ œ #" and the absolute minimum value f Š È"2 ß È"2 ‹ œ f Š È"2 ß È"2 ‹ œ #" . 81. (i) f(xß y) œ x# 3y# 2y on x# y# œ 1 Ê ™ f œ 2xi (6y 2)j and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê 2xi (6y 2)j œ -(2xi 2yj) Ê 2x œ 2x- and 6y 2 œ 2y- Ê - œ 1 or x œ 0. È È CASE 1: - œ 1 Ê 6y 2 œ 2y Ê y œ "# and x œ „ #3 yielding the points Š „ #3 ß #" ‹ . CASE 2: x œ 0 Ê y# œ 1 Ê y œ „ 1 yielding the points a!ß „ 1b . È Evaluations give f Š „ #3 ß "# ‹ œ "# , f(0ß 1) œ 5, and f(!ß 1) œ 1. Therefore "# and 5 are the extreme values on the boundary of the disk. (ii) For the interior of the disk, fx (xß y) œ 2x œ 0 and fy (xß y) œ 6y 2 œ 0 Ê x œ 0 and y œ "3 Ê ˆ!ß 13 ‰ is an interior critical point with f ˆ!ß 3" ‰ œ 3" . Therefore the absolute maximum of f on the disk is 5 at (0ß 1) and the absolute minimum of f on the disk is "3 at ˆ!ß 3" ‰ . Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 14 Practice Exercises 873 82. (i) f(xß y) œ x# y# 3x xy on x# y# œ 9 Ê ™ f œ (2x 3 y)i (2y x)j and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê (2x 3 y)i (2y x)j œ -(2xi 2yj) Ê 2x 3 y œ 2x- and 2y x œ 2yx x Ê 2x(" -) y œ 3 and x 2y(1 -) œ 0 Ê 1 - œ 2y and (2x) Š 2y ‹ y œ 3 Ê x# y# œ 3y Ê x# œ y# 3y. Thus, 9 œ x# y# œ y# 3y y# Ê 2y# 3y 9 œ 0 Ê (2y 3)(y 3) œ 0 Ê y œ 3, 3# . For y œ 3, x# y# œ 9 Ê x œ 0 yielding the point (0ß 3). For y œ 3# , x# y# œ 9 È È È 3 3 3 3 3 27 3 Ê x# 94 œ 9 Ê x# œ 27 4 Ê x œ „ # . Evaluations give f(0ß 3) œ 9, f Š # ß # ‹ œ 9 4 È È ¸ 20.691, and f Š 3 # 3 , 3# ‹ œ 9 274 3 ¸ 2.691. (ii) For the interior of the disk, fx (xß y) œ 2x 3 y œ 0 and fy (xß y) œ 2y x œ 0 Ê x œ 2 and y œ 1 Ê (2ß 1) is an interior critical point of the disk with f(2ß 1) œ 3. Therefore, the absolute maximum of f on È È the disk is 9 274 3 at Š 3 # 3 ß 3# ‹ and the absolute minimum of f on the disk is 3 at (2ß 1). 83. ™ f œ i j k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê i j k œ -(2xi 2yj 2zk) Ê 1 œ 2x-, 1 œ 2y-, 1 œ 2z- Ê x œ y œ z œ -" . Thus x# y# z# œ 1 Ê 3x# œ 1 Ê x œ „ È"3 yielding the points Š È"3 ß È"3 , È"3 ‹ and Š È"3 , È"3 , È"3 ‹ . Evaluations give the absolute maximum value of f Š È"3 ß È"3 ß È"3 ‹ œ È33 œ È3 and the absolute minimum value of f Š È"3 ß È"3 ß È"3 ‹ œ È3. 84. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin and g(xß yß z) œ x# zy 4. Then ™ f œ 2xi 2yj 2zk and ™ g œ 2xi zj yk so that ™ f œ - ™ g Ê 2x œ 2-x, 2y œ -z, and 2z œ -y Ê x œ 0 or - œ 1. CASE 1: x œ 0 Ê zy œ 4 Ê z œ 4y and y œ 4z Ê 2 Š y4 ‹ œ -y and 2 ˆ 4z ‰ œ -z Ê -8 œ y# and 8 # - œz Ê y# œ z# Ê y œ „ z. But y œ x Ê z# œ 4 leads to no solution, so y œ z Ê z# œ 4 Ê z œ „ 2 yielding the points (0ß 2ß 2) and (0ß 2ß 2). CASE 2: - œ 1 Ê 2z œ y and 2y œ z Ê 2y œ ˆ y# ‰ Ê 4y œ y Ê y œ 0 Ê z œ 0 Ê x# 4 œ 0 Ê x œ „ 2 yielding the points (2ß 0ß 0) and (2ß !ß 0). Evaluations give f(0ß 2ß 2) œ f(0ß 2ß 2) œ 8 and f(2ß 0ß 0) œ f(2ß !ß 0) œ 4. Thus the points (2ß 0ß 0) and (2ß !ß 0) on the surface are closest to the origin. 85. The cost is f(xß yß z) œ 2axy 2bxz 2cyz subject to the constraint xyz œ V. Then ™ f œ - ™ g Ê 2ay 2bz œ -yz, 2ax 2cz œ -xz, and 2bx 2cy œ -xy Ê 2axy 2bxz œ -xyz, 2axy 2cyz œ -xyz, and 2bxz 2cyz œ -xyz Ê 2axy 2bxz œ 2axy 2cyz Ê y œ ˆ bc ‰ x. Also 2axy 2bxz œ 2bxz 2cyz Ê z œ ˆ ca ‰ x. # # Then x ˆ bc x‰ ˆ ca x‰ œ V Ê x$ œ cabV Ê width œ x œ Š cabV ‹ # Height œ z œ ˆ ac ‰ Š cabV ‹ "Î$ # œ Š abcV ‹ "Î$ "Î$ # , Depth œ y œ ˆ bc ‰ Š cabV ‹ "Î$ # œ Š bacV ‹ "Î$ , and . 86. The volume of the pyramid in the first octant formed by the plane is V(aß bß c) œ "3 ˆ "# ab‰ c œ "6 abc. The point (2ß 1ß 2) on the plane Ê 2a "b 2c œ 1. We want to minimize V subject to the constraint 2bc ac 2ab œ abc. ac ab Thus, ™ V œ bc 6 i 6 j 6 k and ™ g œ (c 2b bc)i (2c 2a ac)j (2b a ab)k so that ™ V œ - ™ g ac ab abc Ê bc 6 œ -(c 2b bc), 6 œ -(2c 2a ac), and 6 œ -(2b a ab) Ê 6 œ -(ac 2ab abc), abc abc 6 œ -(2bc 2ab abc), and 6 œ -(2bc ac abc) Ê -ac œ 2-bc and 2-ab œ 2-bc. Now - Á 0 since a Á 0, b Á 0, and c Á 0 Ê ac œ 2bc and ab œ bc Ê a œ 2b œ c. Substituting into the constraint equation gives y 2 2 2 x z a a a œ 1 Ê a œ 6 Ê b œ 3 and c œ 6. Therefore the desired plane is 6 3 6 œ 1 or x 2y z œ 6. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 874 Chapter 14 Partial Derivatives 87. ™ f œ (y z)i xj xk , ™ g œ 2xi 2yj , and ™ h œ zi xk so that ™ f œ - ™ g . ™ h Ê (y z)i xj xk œ -(2xi 2yj) .(zi xk) Ê y z œ 2-x .z, x œ 2-y, x œ .x Ê x œ 0 or . œ 1. CASE 1: x œ 0 which is impossible since xz œ 1. CASE 2: . œ 1 Ê y z œ 2-x z Ê y œ 2-x and x œ 2-y Ê y œ (2-)(2-y) Ê y œ 0 or 4-# œ 1. If y œ 0, then x# œ 1 Ê x œ „ 1 so with xz œ 1 we obtain the points (1ß 0ß 1) and (1ß 0ß 1). If 4-# œ 1, then - œ „ "# . For - œ "# , y œ x so x# y# œ 1 Ê x# œ "# Ê x œ „ È"2 with xz œ 1 Ê z œ „ È2, and we obtain the points Š È"2 ß È"2 ß È2‹ and Š È"2 ß È"2 ß È2‹ . For - œ "# , y œ x Ê x# œ "# Ê x œ „ È"2 with xz œ 1 Ê z œ „ È2, and we obtain the points Š È"2 ß È"2 , È2‹ and Š È"2 ß È"2 ß È2‹ . Evaluations give f(1ß 0ß 1) œ 1, f(1ß 0ß 1) œ 1, f Š È"2 ß È"2 ß È2‹ œ "# , f Š È"2 ß È"2 , È2‹ œ "# , f Š È"2 ß È"2 ß È2‹ œ 3# , and f Š È"2 ß È"2 ß È2‹ œ 3# . Therefore the absolute maximum is 3# at Š È"2 ß È"2 ß È2‹ and Š È"2 ß È"2 ß È2‹ , and the absolute minimum is "# at Š È"2 ß È"2 ß È2‹ and Š È"2 ß È"2 ß È2‹ . 88. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk , ™ g œ i j k , and ™ h œ 4xi 4yj 2zk so that ™ f œ - ™ g . ™ h Ê 2x œ - 4x., 2y œ - 4y., and 2z œ - 2z. Ê - œ 2x(1 2.) œ 2y(1 2.) œ 2z(1 2.) Ê x œ y or . œ "# . CASE 1: x œ y Ê z# œ 4x# Ê z œ „ 2x so that x y z œ 1 Ê x x 2x œ 1 or x x 2x œ 1 (impossible) Ê x œ "4 Ê y œ "4 and z œ "# yielding the point ˆ "4 ß "4 ß "# ‰ . CASE 2: . œ "# Ê - œ 0 Ê 0 œ 2z(1 1) Ê z œ 0 so that 2x# 2y# œ 0 Ê x œ y œ 0. But the origin (!ß 0ß 0) fails to satisfy the first constraint x y z œ 1. Therefore, the point ˆ "4 ß 4" ß "# ‰ on the curve of intersection is closest to the origin. 89. (a) y, z are independent with w œ x# eyz and z œ x# y# Ê ``wy œ ``wx `` xy ``wy `` yy ``wz `` yz œ a2xeyz b `` xy azx# eyz b (1) ayx# eyz b (0); z œ x# y# Ê 0 œ 2x `` xy 2y Ê `` xy œ yx ; therefore, Š ``wy ‹ œ a2xeyz b ˆ xy ‰ zx# eyz œ a2y zx# b eyz z (b) z, x are independent with w œ x# eyz and z œ x# y# Ê ``wz œ ``wx `` xz ``wy `` yz ``wz `` zz œ a2xeyz b (0) azx# eyz b `` yz ayx# eyz b (1); z œ x# y# Ê 1 œ 0 2y `` yz Ê `` yz œ #"y ; therefore, 1 z ˆ ``wz ‰ œ azx# eyz b Š 2y ‹ yx# eyz œ x# eyz Šy 2y ‹ x (c) z, y are independent with w œ x# eyz and z œ x# y# Ê ``wz œ ``wx `` xz ``wy `` yz ``wz `` zz œ a2xeyz b `` xz azx# eyz b (0) ayx# eyz b (1); z œ x# y# Ê 1 œ 2x `` xz 0 Ê `` xz œ #"x ; therefore, 1 ‰ ˆ ``wz ‰ œ a2xeyz b ˆ 2x yx# eyz œ a1 x# yb eyz y `V Ù `T 90. (a) T, P are independent with U œ f(Pß Vß T) and PV œ nRT Ê `ÙT œ `ÙP `` TP `` U V `T `T `T ‰ ˆ ``VT ‰ ˆ `ÙT ‰ (1); PV œ nRT Ê P ``VT œ nR Ê ``VT œ nR œ ˆ `ÙP ‰ (0) ˆ `` U V P ; therefore, ˆ `ÙT ‰ œ ˆ `` U ‰ ˆ nR ‰ Ù V P `T P Ù `P Ù `V Ù `T (b) V, T are independent with U œ f(Pß Vß T) and PV œ nRT Ê `` U V œ `P `V `V `V `T `V U‰ œ ˆ `ÙP ‰ ˆ ``VP ‰ ˆ `` V (1) ˆ `ÙT ‰ (0); PV œ nRT Ê V ``VP P œ (nR) ˆ ``VT ‰ œ 0 Ê ``VP œ VP ; therefore, U ˆ `` U ‰ œ ˆ `ÙP ‰ ˆ VP ‰ `` V V T Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 14 Practice Exercises 875 91. Note that x œ r cos ) and y œ r sin ) Ê r œ Èx# y# and ) œ tan" ˆ yx ‰ . Thus, `w `w `r `w `) x ˆ `w ‰ ˆ ` w ‰ y ` x œ ` r ` x ` ) ` x œ ` r Š È x # y # ‹ ` ) Š x # y# ‹ œ (cos )) ``wr ˆ sinr ) ‰ ``w) ; y `w `w `r `w `) x ˆ `w ‰ ˆ `w ‰ ` y œ ` r ` y ` ) ) y œ ` r Š È x # y# ‹ ` ) Š x# y# ‹ œ (sin )) ``wr ˆ cosr ) ‰ ``w) 92. zx œ fu `` ux fv `` vx œ afu afv , and zy œ fu `` uy fv `` vy œ bfu bfv ù dw `w dw ` u dw " `w dw " `w dw 93. `` uy œ b and `` ux œ a Ê ``wx œ dw du ` x œ a du and ` y œ du ` y œ b du Ê a ` x œ du and b ` y œ du Ê "a ``wx œ b" ``wy Ê b ``wx œ a ``wy 2(r s) 2(r s) 2y 2(r s) " `w rs 94. ``wx œ x# 2x y# 2z œ (r s)# (r s)# 4rs - œ 2 ar# 2rs s# b œ r s , ` y œ x# y# 2z œ #(r s)# œ (r s)# , and ``wz œ x# y2# 2z œ (r " s)# Ê ``wr œ ``wx ``xr ``wy ``yr ``wz ``zr œ r " s (rrs)s # ’ (r " s)# “ (2s) œ (r2rs)2s# œ r 2 s and ``ws œ ``wx ``xs ``wy ``ys ``wz `` zs œ r " s (rrs)s # ’ (r " s)# “ (2r) œ r 2 s 95. eu cos v x œ 0 Ê aeu cos vb `` ux aeu sin vb `` vx œ 1; eu sin v y œ 0 Ê aeu sin vb `` ux aeu cos vb `` vx œ 0. Solving this system yields `` ux œ eu cos v and `` vx œ eu sin v. Similarly, eu cos v x œ 0 Ê aeu cos vb `` uy aeu sin vb `` vy œ 0 and eu sin v y œ 0 Ê aeu sin vb `` uy aeu cos vb `` vy œ 1. Solving this second system yields `` uy œ eu sin v and `` vy œ eu cos v. Therefore Š `` ux i `` uy j‹ † Š `` vx i `` vy j‹ œ caeu cos vb i aeu sin vb jd † caeu sin vb i aeu cos vb jd œ 0 Ê the vectors are orthogonal Ê the angle between the vectors is the constant 1# . 96. `` g) œ `` xf `` x) `` yf `` y) œ (r sin )) `` xf (r cos )) `` yf # Ê `` )g# œ (r sin )) Š `` xf# `` x) ``y`fx `` y) ‹ (r cos )) `` xf (r cos )) Š ``x`fy `` x) `` yf# `` y) ‹ (r sin )) `` yf # # # # œ (r sin )) Š `` x) `` y) ‹ (r cos )) (r cos )) Š `` x) `` y) ‹ (r sin )) œ (r sin ) r cos ))(r sin ) r cos )) (r cos ) r sin )) œ (2)(2) (0 2) œ 4 2 œ 2 at (rß )) œ ˆ2ß 1# ‰ . 97. (y z)# (z x)# œ 16 Ê ™ f œ 2(z x)i 2(y z)j 2(y 2z x)k ; if the normal line is parallel to the yz-plane, then x is constant Ê `` xf œ 0 Ê 2(z x) œ 0 Ê z œ x Ê (y z)# (z z)# œ 16 Ê y z œ „ 4. Let x œ t Ê z œ t Ê y œ t „ 4. Therefore the points are (tß t „ 4ß t), t a real number. 98. Let f(xß yß z) œ xy yz zx x z# œ 0. If the tangent plane is to be parallel to the xy-plane, then ™ f is perpendicular to the xy-plane Ê ™ f † i œ 0 and ™ f † j œ 0. Now ™ f œ (y z 1)i (x z)j (y x 2z)k so that ™ f † i œ y z 1 œ 0 Ê y z œ 1 Ê y œ 1 z, and ™ f † j œ x z œ 0 Ê x œ z. Then z(1 z) (" z)z z(z) (z) z# œ 0 Ê z 2z# œ 0 Ê z œ "# or z œ 0. Now z œ "# Ê x œ "# and y œ "# Ê ˆ "# ß "# ß "# ‰ is one desired point; z œ 0 Ê x œ 0 and y œ 1 Ê (0ß 1ß 0) is a second desired point. 99. ™ f œ -(xi yj zk) Ê `` xf œ -x Ê f(xß yß z) œ "# -x# g(yß z) for some function g Ê -y œ `` yf œ `` yg Ê g(yß z) œ "# -y# h(z) for some function h Ê -z œ `` zf œ `` gz œ hw (z) Ê h(z) œ #" -z# C for some arbitrary constant C Ê g(yß z) œ "# -y# ˆ "# -z# C‰ Ê f(xß yß z) œ "# -x# "# -y# "# -z# C Ê f(0ß 0ß a) œ "# -a# C and f(0ß 0ß a) œ "# -(a)# C Ê f(0ß 0ß a) œ f(0ß 0ß a) for any constant a, as claimed. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 876 Chapter 14 Partial Derivatives œ lim f(0 su" ß 0 su# ßs0 su$ )f(0ß 0ß 0) , s 0 sÄ0 ‰ 100. ˆ df ds u (0 0 0) ß ß ß És# u#" s# u## s# u#$ 0 œ lim sÄ0 s ,s0 sÉu#" u## u#$ œ lim œ lim kuk œ 1; s sÄ0 sÄ0 however, ™ f œ Èx# xy# z# i Èx# yy# z# j Èx# zy# z# k fails to exist at the origin (0ß 0ß 0) 101. Let f(xß yß z) œ xy z 2 Ê ™ f œ yi xj k . At (1ß 1ß 1), we have ™ f œ i j k Ê the normal line is x œ 1 t, y œ 1 t, z œ 1 t, so at t œ 1 Ê x œ 0, y œ 0, z œ 0 and the normal line passes through the origin. 102. (b) f(xß yß z) œ x# y# z# œ 4 Ê ™ f œ 2xi 2yj 2zk Ê at (2ß 3ß 3) the gradient is ™ f œ 4i 6j 6k which is normal to the surface (c) Tangent plane: 4x 6y 6z œ 8 or 2x 3y 3z œ 4 Normal line: x œ 2 4t, y œ 3 6t, z œ 3 6t CHAPTER 14 ADDITIONAL AND ADVANCED EXERCISES 1. By definition, fxy (!ß 0) œ lim hÄ0 fx (0ß h) fx (0ß 0) so we need to calculate the first partial derivatives in the h numerator. For (xß y) Á (0ß 0) we calculate fx (xß y) by applying the differentiation rules to the formula for # # $ # # # # $ # $ ax y b (2x) f(xß y): fx (xß y) œ xx#yyy# (xy) ax y b (2x) œ xx#yyy# ax#4x yy# b# Ê fx (0ß h) œ hh# œ h. ax # y # b # For (xß y) œ (0ß 0) we apply the definition: fx (!ß 0) œ lim hÄ0 fxy (0ß 0) œ lim hÄ0 $ h 0 œ 1. h Similarly, fyx (0ß 0) œ lim $ # $ # hÄ0 $ f(hß 0) f(0ß0) œ lim 0 h 0 œ 0. h hÄ0 Then by definition fy (hß 0) fy (!ß 0) , so for (xß y) Á (0ß 0) we have h 4x y h fy (xß y) œ xx#xy y# ax# y# b# Ê fy (hß 0) œ h# œ h; for (xß y) œ (0ß 0) we obtain fy (0ß 0) œ lim œ lim hÄ0 2. 00 h œ 0. `w x ` x œ 1 e cos y # Then by definition fyx (0ß 0) œ lim hÄ0 hÄ0 h0 h œ 1. f(0ß h) f(!ß 0) h Note that fxy (0ß 0) Á fyx (0ß 0) in this case. Ê w œ x ex cos y g(y); ``wy œ ex sin y gw (y) œ 2y ex sin y Ê gw (y) œ 2y Ê g(y) œ y C; w œ ln 2 when x œ ln 2 and y œ 0 Ê ln 2 œ ln 2 eln 2 cos 0 0# C Ê 0 œ 2 C Ê C œ 2. Thus, w œ x ex cos y g(y) œ x ex cos y y# 2. 3. Substitution of u u(x) and v œ v(x) in g(uß v) gives g(u(x)ß v(x)) which is a function of the independent ` g du ` g dv ` ' du ` ' dv variable x. Then, g(uß v) œ 'u f(t) dt Ê dg dx œ ` u dx ` v dx œ Š ` u u f(t) dt‹ dx Š ` v u f(t) dt‹ dx v v v ` ' dv du dv dv du œ Š `ù 'v f(t) dt‹ du dx Š ` v u f(t) dt‹ dx œ f(u(x)) dx f(v(x)) dx œ f(v(x)) dx f(u(x)) dx u v # # # # # # `r d f ˆ `r ‰ ` r d f `r df ` r 4. Applying the chain rules, fx œ df df dr ` x Ê fxx œ Š dr# ‹ ` x dr ` x# . Similarly, fyy œ Š dr# ‹ Š ` y ‹ dr ` y# and # # # # ` r `r x ` r fzz œ Š ddr#f ‹ ˆ ``zr ‰ df dr ` z# . Moreover, ` x œ Èx# y# z# Ê ` x# œ # Ê ``yr# œ x# z# `r z 3 ; and ` z œ È # x y # z# ˆÈx# y# z# ‰ # Ê ``z#r œ y # z# y `r 3 ; `y œ È # x y # z# ˆÈx# y# z# ‰ x# y# 3 . ˆÈ x # y # z # ‰ Next, fxx fyy fzz œ 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 14 Additional and Advanced Exercises # y z ‰ Ê Š ddr#f ‹ Š x# xy# z# ‹ ˆ df dr Œ ˆÈ # # # # # # x y ‰ Š ddr#f ‹ Š x# yz # z# ‹ ˆ df dr Œ ˆÈ # # # # y# d# f x y z# ‰ # 3 3 x # z# Š dr# ‹ Š x# y# z# ‹ ˆ dr ‰ Œ ˆÈx# y# z# ‰3 df d# f x y z# ‰ 877 d# f œ 0 Ê dr# Š Èx# y# z# ‹ dr œ 0 Ê dr# r dr œ 0 2 df 2 df df 2 dr Ê drd (f w ) œ ˆ 2r ‰ f w , where f w œ df Ê ln f w œ 2 ln r ln C Ê f w œ Cr# , or dr Ê f œ r w w df # dr œ Cr Ê f(r) œ Cr b œ ar b for some constants a and b (setting a œ C) 5. (a) Let u œ tx, v œ ty, and w œ f(uß v) œ f(u(tß x)ß v(tß y)) œ f(txß ty) œ tn f(xß y), where t, x, and y are independent variables. Then ntnc1 f(xß y) œ ``wt œ ``wu `ùt ``wv ``vt œ x ``wu y ``wv . Now, `w `w ù `w `v `w `w ˆ `w ‰ ˆ `w ‰ ˆ " ‰ ˆ ``wx ‰ . Likewise, ` x œ ` u ` x ` v ` x œ ` u (t) ` v (0) œ t ` u Ê ` u œ t `w `w ù `w `v ˆ `w ‰ ˆ `w ‰ ` y œ ` u ` y ` v ` y œ ` u (0) ` v (t) Ê ``wv œ ˆ "t ‰ Š ``wy ‹ . Therefore, ntnc1 f(xß y) œ x ``wu y ``wv œ ˆ xt ‰ ˆ ``wx ‰ ˆ yt ‰ Š ``wy ‹. When t œ 1, u œ x, v œ y, and w œ f(xß y) Ê ``wx œ `` xf and ``wy œ `` xf Ê nf(xß y) œ x `` xf y `` yf , as claimed. (b) From part (a), ntnc1 f(xß y) œ x ``wu y ``wv . Differentiating with respect to t again we obtain n(n 1)tnc2 f(xß y) œ x `` uw# `ùt x ``v`wu ``vt y `ù`wv `ùt y `` vw# ``vt œ x# `` uw# 2xy `ù`wv y# `` vw# . # # # # # # # # # # # # Also from part (a), `` xw# œ ``x ˆ ``wx ‰ œ ``x ˆt ``wu ‰ œ t `` uw# `` ux t ``v`wu `` vx œ t# `` uw# , `` yw# œ ``y Š ``wy ‹ # # # # # # œ ``y ˆt ``wv ‰ œ t `ù`wv `` yu t `` vw# `` yv œ t# `` vw# , and ``y`wx œ ``y ˆ ``wx ‰ œ ``y ˆt ``wu ‰ œ t `` uw# `` yu t ``v`wu `` yv # # # # # # # œ t# ``v`wu Ê ˆ t"# ‰ `` xw# œ `` uw# , ˆ t"# ‰ `` yw# œ `` vw# , and ˆ t"# ‰ ``y`wx œ ``v`wu ‰ Š ``y`wx ‹ Š yt# ‹ Š `` yw# ‹ for t Á 0. When t œ 1, w œ f(xß y) and Ê n(n 1)tnc2 f(xß y) œ Š xt# ‹ Š `` xw# ‹ ˆ 2xy t# # # # # # # # # we have n(n 1)f(xß y) œ x# Š `` xf# ‹ 2xy Š ``x`fy ‹ y# Š `` yf# ‹ as claimed. 6. (a) lim rÄ0 sin 6r sin t œ 1, where t œ 6r 6r œ t lim Ä0 t ˆ sin6h6h ‰ 1 f(0 hß 0) f(0ß 0) 6h 6h 6 œ lim œ lim sin 6h œ lim 6 cos12h h h 6h# hÄ0 hÄ0 hÄ0 hÄ0 œ lim 3612sin 6h œ 0 (applying l'Hopital's rule twice) s hÄ0 (b) fr (0ß 0) œ lim ˆ sin6r6r ‰ ˆ sin6r6r ‰ f(rß ) h) f(rß )) œ lim œ lim h0 œ 0 h h hÄ0 hÄ0 hÄ0 (c) f) (rß )) œ lim 7. (a) r œ xi yj zk Ê r œ krk œ Èx# y# z# and ™ r œ Èx# xy# z# i Èx# yy# z# j Èx# zy# z# k œ rr (b) rn œ ˆÈx# y# z# ‰ n Ê ™ arn b œ nx ax# y# z# b ÐnÎ2Ñ 1 ÐnÎ2Ñ 1 ÐnÎ2Ñ 1 i ny ax# y# z# b j nz ax# y# z# b k œ nrn 2 r # (c) Let n œ 2 in part (b). Then "# ™ ar# b œ r Ê ™ ˆ "# r# ‰ œ r Ê r# œ #" ax# y# z# b is the function. (d) dr œ dxi dyj dzk Ê r † dr œ x dx y dy z dz, and dr œ rx dx ry dy rz dz œ xr dx yr dy zr dz Ê r dr œ x dx y dy z dz œ r † dr (e) A œ ai bj ck Ê A † r œ ax by cz Ê ™ (A † r) œ ai bj ck œ A dy dy ` f dx ` f dy `f `f dx dx 8. f(g(t)ß h(t)) œ c Ê 0 œ df dt œ ` x dt ` y dt œ Š ` x i ` y j‹ † Š dt i dt j‹ , where dt i dt j is the tangent vector Ê ™ f is orthogonal to the tangent vector 9. f(xß yß z) œ xz# yz cos xy 1 Ê ™ f œ az# y sin xyb i (z x sin xy)j (2xz y)k Ê ™ f(0ß 0ß 1) œ i j Ê the tangent plane is x y œ 0; r œ (ln t)i (t ln t)j tk Ê rw œ ˆ "t ‰ i (ln t 1)j k ; x œ y œ 0, z œ 1 Ê t œ 1 Ê rw (1) œ i j k . Since (i j k) † (i j) œ rw (1) † ™ f œ 0, r is parallel to the plane, and r(1) œ 0i 0j k Ê r is contained in the plane. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 878 Chapter 14 Partial Derivatives 10. Let f(xß yß z) œ x$ y$ z$ xyz Ê ™ f œ a3x# yzb i a3y# xzb j a3z# xyb k Ê ™ f(0ß 1ß 1) œ i 3j 3k $ Ê the tangent plane is x 3y 3z œ 0; r œ Š t4 2‹ i ˆ 4t 3‰ j (cos (t 2)) k # Ê rw œ Š 3t4 ‹ i ˆ t4# ‰ j (sin (t 2)) k ; x œ 0, y œ 1, z œ 1 Ê t œ 2 Ê rw (2) œ 3i j . Since rw (2) † ™ f œ 0 Ê r is parallel to the plane, and r(2) œ i k Ê r is contained in the plane. # 11. `` xz œ 3x# 9y œ 0 and `` yz œ 3y# 9x œ 0 Ê y œ "3 x# and 3 ˆ "3 x# ‰ 9x œ 0 Ê "3 x% 9x œ 0 Ê x ax$ 27b œ 0 Ê x œ 0 or x œ 3. Now x œ 0 Ê y œ 0 or (!ß 0) and x œ 3 Ê y œ 3 or (3ß 3). Next ` #z ` #z ` #z ` x# œ 6x, ` y# œ 6y, and ` x` y œ 9. # # # # # # # For (!ß 0), `` xz# `` yz# Š ``x`zy ‹ œ 81 Ê no extremum (a saddle point), # # and for (3ß 3), `` xz# `` yz# Š ``x`zy ‹ œ 243 0 and `` xz# œ 18 0 Ê a local minimum. 12. f(xß y) œ 6xyeÐ2x3yÑ Ê fx (xß y) œ 6y(1 2x)eÐ2x3yÑ œ 0 and fy (xß y) œ 6x(1 3y)eÐ2x3yÑ œ 0 Ê x œ 0 and y œ 0, or x œ "# and y œ 3" . The value f(0ß 0) œ 0 is on the boundary, and f ˆ "# ß "3 ‰ œ e"2 . On the positive y-axis, f(0ß y) œ 0, and on the positive x-axis, f(xß 0) œ 0. As x Ä _ or y Ä _ we see that f(xß y) Ä 0. Thus the absolute maximum of f in the closed first quadrant is e"2 at the point ˆ #" ß 3" ‰ . # # # 2y 2z 13. Let f(xß yß z) œ xa# yb# cz# 1 Ê ™ f œ 2x a# i b# j c# k Ê an equation of the plane tangent at the point # # # 2x! 2y! 2z! !‰ ˆ 2y! ‰ ˆ 2z! ‰ ˆ x! ‰ ˆ y! ‰ ˆ z! ‰ P! (x! ß y! ß y! ) is ˆ 2x a# x b# y c# z œ a# b# c# œ 2 or a# x b# y c# z œ 1. # # # The intercepts of the plane are Š xa! ß 0ß 0‹ , Š0ß by! ß 0‹ and Š!ß !ß zc! ‹ . The volume of the tetrahedron formed by the # # # # " plane and the coordinate planes is V œ ˆ "3 ‰ ˆ #" ‰ Š xa! ‹ Š by! ‹ Š cz! ‹ Ê we need to maximize V(xß yß z) œ (abc) 6 (xyz) # # # # # (abc) 2y " 2x " subject to the constraint f(xß yß z) œ xa# by# cz# œ 1. Thus, ’ (abc) 6 “ Š x# yz ‹ œ a# -, ’ 6 “ Š xy# z ‹ œ b# -, # " 2z # # # and ’ (abc) 6 “ Š xyz# ‹ œ c# -. Multiply the first equation by a yz, the second by b xz, and the third by c xy. Then equate the first and second Ê a# y# œ b# x# Ê y œ ba x, x 0; equate the first and third Ê a# z# œ c# x# Ê z œ ca x, x 0; substitute into f(xß yß z) œ 0 Ê x œ Èa3 Ê y œ Èb3 Ê z œ Èc3 Ê V œ È3 # abc. 14. 2(x u) œ -, 2(y v) œ -, 2(x u) œ ., and 2(y v) œ 2.v Ê x u œ v y, x u œ .# , and y v œ .v Ê x u œ .v œ .# Ê v œ "# or . œ 0. CASE 1: . œ 0 Ê x œ u, y œ v, and - œ 0; then y œ x 1 Ê v œ u 1 and v# œ u Ê v œ v# 1 È Ê v# v 1 œ 0 Ê v œ 1 „ #1 4 Ê no real solution. CASE 2: v œ #" and u œ v# Ê u œ 4" ; x 4" œ #" y and y œ x 1 Ê x 4" œ x #" Ê 2x œ 4" Ê x œ 8" # # # Ê y œ 7 . Then f ˆ " ß 7 ß " ß " ‰ œ ˆ " " ‰ ˆ 7 " ‰ œ 2 ˆ 3 ‰ Ê the minimum distance is 3 È2. 8 8 8 4 # 8 4 8 # 8 8 (Notice that f has no maximum value.) 15. Let (x! ß y! ) be any point in R. We must show lim Ðhß kÑ Ä Ð0ß 0Ñ lim Ðxß yÑ Ä Ðx! ß y! Ñ f(xß y) œ f(x! ß y! ) or, equivalently that kf(x! hß y! k) f(x! ß y! )k œ 0. Consider f(x! hß y! k) f(x! ß y! ) œ [f(x! hß y! k) f(x! ß y! k)] [f(x! ß y! k) f(x! ß y! )]. Let F(x) œ f(xß y! k) and apply the Mean Value Theorem: there exists 0 with x! 0 x! h such that Fw (0 )h œ F(x! h) F(x! ) Ê hfx (0ß y! k) œ f(x! hß y! k) f(x! ß y! k). Similarly, k fy (x! ß () œ f(x! ß y! k) f(x! ß y! ) for some ( with y! ( y! k. Then kf(x! hß y! k) f(x! ß y! )k Ÿ khfx (0ß y! k)k kkfy (x! ß ()k . If M, N are positive real numbers such that kfx k Ÿ M and kfy k Ÿ N for all (xß y) in the xy-plane, then kf(x! hß y! k) f(x! ß y! )k Ÿ M khk N kkk . As (hß k) Ä 0, kf(x! hß y! k) f(x! ß y! )k Ä 0 Ê lim kf(x! hß y! k) f(x! ß y! )k Ðhß kÑ Ä Ð0ß 0Ñ œ 0 Ê f is continuous at (x! ß y! ). Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 14 Additional and Advanced Exercises 879 dr 16. At extreme values, ™ f and v œ ddtr are orthogonal because df dt œ ™ f † dt œ 0 by the First Derivative Theorem for Local Extreme Values. 17. `` xf œ 0 Ê f(xß y) œ h(y) is a function of y only. Also, `` yg œ `` xf œ 0 Ê g(xß y) œ k(x) is a function of x only. Moreover, `` yf œ `` xg Ê hw (y) œ kw (x) for all x and y. This can happen only if hw (y) œ kw (x) œ c is a constant. Integration gives h(y) œ cy c" and k(x) œ cx c# , where c" and c# are constants. Therefore f(xß y) œ cy c" and g(xß y) œ cx c# . Then f(1ß 2) œ g(1ß 2) œ 5 Ê 5 œ 2c c" œ c c# , and f(0ß 0) œ 4 Ê c" œ 4 Ê c œ "# Ê c# œ 9# . Thus, f(xß y) œ "# y 4 and g(xß y) œ "# x #9 . 18. Let g(xß y) œ Du f(xß y) œ fx (xß y)a fy (xß y)b. Then Du g(xß y) œ gx (xß y)a gy (xß y)b œ fxx (xß y)a# fyx (xß y)ab fxy (xß y)ba fyy (xß y)b# œ fxx (xß y)a# 2fxy (xß y)ab fyy (xß y)b# . 19. Since the particle is heat-seeking, at each point (xß y) it moves in the direction of maximal temperature increase, that is in the direction of ™ T(xß y) œ aec2y sin xb i a2ec2y cos xb j . Since ™ T(xß y) is parallel to c2y cos x the particle's velocity vector, it is tangent to the path y œ f(x) of the particle Ê f w (x) œ 2eec2y sin x œ 2 cot x. Integration gives f(x) œ 2 ln ksin xk C and f ˆ 14 ‰ œ 0 Ê 0 œ 2 ln ¸sin 14 ¸ C Ê C œ 2 ln # È2 2 # œ ln Š È2 ‹ œ ln 2. Therefore, the path of the particle is the graph of y œ 2 ln ksin xk ln 2. 20. The line of travel is x œ t, y œ t, z œ 30 5t, and the bullet hits the surface z œ 2x# 3y# when 30 5t œ 2t# 3t# Ê t# t 6 œ 0 Ê (t 3)(t 2) œ 0 Ê t œ 2 (since t 0). Thus the bullet hits the surface at the point (2ß 2ß 20). Now, the vector 4xi 6yj k is normal to the surface at any (xß yß z), so that n œ 8i 12j k is normal to the surface at (2ß 2ß 20). If v œ i j 5k , then the velocity of the particle †25 ‰ 600 50 ‰ after the ricochet is w œ v 2 projn v œ v Š 2knvk†#n ‹ n œ v ˆ 2209 n œ (i j 5k) ˆ 400 209 i 209 j 209 k 391 995 œ 191 209 i 209 j 209 k . 21. (a) k is a vector normal to z œ 10 x# y# at the point (!ß 0ß 10). So directions tangential to S at (!ß 0ß 10) will be unit vectors u œ ai bj . Also, ™ T(xß yß z) œ (2xy 4) i ax# 2yz 14b j ay# 1b k Ê ™ T(!ß 0ß 10) œ 4i 14j k . We seek the unit vector u œ ai bj such that Du T(0ß 0ß 10) œ (4i 14j k) † (ai bj) œ (4i 14j) † (ai bj) is a maximum. The maximum will occur when ai bj has the same direction as 4i 14j , or u œ È"53 (2i 7j). (b) A vector normal to S at (1ß 1ß 8) is n œ 2i 2j k . Now, ™ T(1ß 1ß 8) œ 6i 31j 2k and we seek the unit vector u such that Du T(1ß 1ß 8) œ ™ T † u has its largest value. Now write ™ T œ v w , where v is parallel to ™ T and w is orthogonal to ™ T. Then Du T œ ™ T † u œ (v w) † u œ v † u w † u œ w † u. Thus Du T(1ß 1ß 8) is a maximum when u has the same direction as w . Now, w œ ™ T Š ™knTk#†n ‹ n 62 2 ‰ ‰ i ˆ31 152 ‰ j ˆ2 76 ‰ œ (6i 31j 2k) ˆ 124 (2i 2j k) œ ˆ6 152 41 9 9 9 k 127 58 w " œ 98 9 i 9 j 9 k Ê u œ kwk œ È29,097 (98i 127j 58k). 22. Suppose the surface (boundary) of the mineral deposit is the graph of z œ f(xß y) (where the z-axis points up into the air). Then `` xf i `` yf j k is an outer normal to the mineral deposit at (xß y) and `` xf i `` yf j points in the direction of steepest ascent of the mineral deposit. This is in the direction of the vector `` xf i `` yf j at (0ß 0) (the location of the 1st borehole) that the geologists should drill their fourth borehole. To approximate this vector we use the fact that (0ß 0ß 1000), (0ß 100ß 950), and (100ß !ß 1025) lie on the graph of z œ f(xß y). The plane containing these three points is a good â â j k â â i â â "00 50 â approximation to the tangent plane to z œ f(xß y) at the point (0ß 0ß 0). A normal to this plane is â 0 â â 25 â â "00 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 880 Chapter 14 Partial Derivatives œ 2500i 5000j 10,000k, or i 2j 4k. So at (0ß 0) the vector `` xf i `` yf j is approximately i 2j . Thus the geologists should drill their fourth borehole in the direction of È"5 (i 2j) from the first borehole. 23. w œ ert sin 1x Ê wt œ rert sin 1x and wx œ 1ert cos 1x Ê wxx œ 1# ert sin 1x; wxx œ c"# wt , where c# is the positive constant determined by the material of the rod Ê 1# ert sin 1x œ c"# arert sin 1xb # # Ê ar c# 1# b ert sin 1x œ 0 Ê r œ c# 1# Ê w œ ec 1 t sin 1x 24. w œ ert sin kx Ê wt œ rert sin kx and wx œ kert cos kx Ê wxx œ k# ert sin kx; wxx œ c"# wt # # Ê k# ert sin kx œ c"# arert sin kxb Ê ar c# k# b ert sin kx œ 0 Ê r œ c# k# Ê w œ ec k t sin kx. # # # # # # Now, w(Lß t) œ 0 Ê ec k t sin kL œ 0 Ê kL œ n1 for n an integer Ê k œ nL1 Ê w œ ec n 1 tÎL sin ˆ nL1 x‰ . # # # # As t Ä _, w Ä 0 since ¸sin ˆ nL1 x‰¸ Ÿ 1 and ec n 1 tÎL Ä 0. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. CHAPTER 15 MULTIPLE INTEGRALS 15.1 DOUBLE AND ITERATED INTEGRALS OVER RECTANGLES 1. '12 '04 2xy dy dx œ '12 cx y# d 40 dx œ '12 16x dx œ c8 x# d 21 œ 24 2. '02 'c11 ax yb dy dx œ '02 xy 12 y# ‘ "" dx œ '02 2x dx œ c x# d 20 œ 4 3. ! 'c01 'c11 (x y 1) dx dy œ 'c01 ’ x2 yx x“" dy œ 'c01 (2y 2) dy œ cy# 2yd " œ1 4. '01 '01 Š1 x 2 y ‹ dx dy œ '01 ’x x6 x2y “" dy œ '01 Š 65 y2 ‹dy œ ’ 56 y y6 “1 œ 23 # " # # # 3 # 3 0 0 5. '03 '02 a4 y# b dy dx œ '03 ’4y y3 “ # dx œ '03 163 dx œ 163 x‘30 œ 16 6. '03 'c02 ax# y 2xyb dy dx œ '03 ’ x 2y xy# “! dx œ '03 a4x 2x# b dx œ ’2x# 2x3 “3 œ 0 7. '01 '01 1 yx y dx dy œ '01 clnl1 x yld"0 dy œ '01 lnl1 yldy œ cy lnl1 yl y lnl1 yld 10 œ 2 ln 2 1 8. '14 '04 ˆ 2x Èy‰ dx dy œ '14 41 x2 xÈy‘ !4 dy œ '14 ˆ4 4 y1/2 ‰dy œ 4y 38 y3/2 ‘41 œ 923 9. '0ln 2 '1ln 5 e2x y dy dx œ '0ln 2 ce2x y dln" 5 dx œ '0ln 2 a5e2x e2x 1 b dx œ 52 e2x "# e2x 1 ‘0ln 2 œ 32 a5 eb $ ! # # $ # ! 10. '0 '1 x y ex dy dx œ '0 "# x y2 ex ‘" dx œ '0 32 x ex dx œ 32 x ex 32 ex ‘0 œ 32 1 2 1 2 1Î2 1 1 11. 'c1 '0 y sin x dx dy œ 'c1 cy cos xd10 Î2 dy œ 'c1 y dy œ "# y2 ‘ 1 œ 32 2 2 2 2 12. '1 '0 asin x cos yb dx dy œ '1 ccos x x cos yd01 dy œ '1 a2 1 cos yb dy œ c2y 1 sin yd121 œ 21 21 1 21 21 13. ' ' a6 y# 2 xbdA œ '0 '0 a6 y# 2 xb dy dx œ '0 c2 y3 2 x yd0 dx œ '0 a16 4 xb dx œ c16 x 2 x2 d0 œ 14 1 2 1 2 1 1 R 14. ' ' R Èx y2 dA œ '04 '12 Èy x dy dx œ '04 ’ Èyx “2 dx œ '04 "# x1Î2 dx œ 31 x3Î2 ‘40 œ 38 2 1 15. ' ' x y cos y dA œ 'c1 '0 x y cos y dy dx œ 'c1 cx y sin y x cos yd10 dx œ 'c1 a2xb dx œ cx2 dc1 œ 0 1 1 1 1 R 16. ' ' y sinax yb dA œ 'c1 '0 y sinax yb dy dx œ 'c1 cy cosax yb sinax ybd10 dx 0 1 0 R œ 'c1 asinax 1b 1 cosax 1b sin xbdx œ ccosax 1b 1 sinax 1b cos xdc0 1 œ 4 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 1 882 Chapter 15 Multiple Integrals 17. ' ' ex y dA œ '0 ln 2 R '0ln 2 ex y dy dx œ '0ln 2 cex y dln0 2 dx œ '0ln 2 aex ln 2 ex b dx œ cex ln 2 ex dln0 2 œ "# 18. ' ' x y ex y dA œ '0 '0 x y ex y dy dx œ '0 ’ "# ex y “ dx œ '0 ˆ "# ex "# ‰ dx œ "# ex "# x‘0 œ "# ae2 3b 2 2 1 2 2 2 1 2 2 0 R 19. ' ' xx2 y 1 dA œ '0 '0 xx2 y 1 dy dx œ '0 ’ 4axx2y 1b “ dx œ '0 x24x 1 dx œ c2 lnlx2 1ld0 œ 2 ln 2 3 1 2 1 3 2 4 1 1 0 R 1 1 20. ' ' x2 y2y 1 dA œ '0 '0 ax yby2 1 dx dy œ '0 ctan1 ax ybd0 dy œ '0 tan1 y dy œ y tan1 y "# lnl1 y2 l‘0 œ 14 "# ln 2 1 1 1 1 R 21. '1 '1 2 2 1 xy dy dx œ '12 x" (ln 2 ln 1) dx œ (ln 2) '12 x" dx œ (ln 2)# 22. '0 '0 y cos xy dx dy œ '0 csin xyd 1! dy œ '0 sin 1y dy œ 1" cos 1y‘ ! œ 1" (1 1) œ 12 1 1 1 1 " 23. V œ ' ' fax, yb dA œ 'c1 'c1 ax2 y2 b dy dx œ 'c1 x2 y 31 y3 ‘ 1 dx œ 'c1 ˆ2 x2 32 ‰ dx œ 32 x3 32 x‘ 1 œ 38 1 1 1 1 " " R 2 3‘ 2‰ 88 24. V œ ' ' fax, yb dA œ '0 '0 a16 x2 y2 b dy dx œ '0 16 y x2 y 13 y3 ‘0 dx œ '0 ˆ 88 3 2 x dx œ 3 x 3 x 0 2 2 2 2 2 2 R œ 160 3 25Þ V œ ' ' fax, yb dA œ '0 '0 a2 x yb dy dx œ '0 2 y x y "# y2 ‘ ! dx œ '0 ˆ 32 x‰ dx œ 32 x "# x2 ‘ ! œ 1 1 1 1 " 1 " R 26Þ V œ ' ' fax, yb dA œ '0 '0 y2 dy dx œ '0 ’ y4 “ dx œ '0 1 dx œ cxd40 œ 4 4 2 4 2 2 4 0 R 27Þ V œ ' ' fax, yb dA œ '0 1Î2 R '01Î4 2 sin x cos y dy dx œ '01Î2 c2 sin x sin yd01Î4 dx œ '01Î2 ŠÈ2 sin x‹ dx œ ’È2 cos x“1Î2 0 œ È2 16 ‰ 16 ‘ 28. V œ ' ' fax, yb dA œ '0 '0 a4 y2 b dy dx œ '0 4 y 13 y3 ‘0 dx œ '0 ˆ 16 3 dx œ 3 x ! œ 3 1 2 1 2 1 " R 15.2 DOUBLE INTEGRALS OVER GENERAL REGIONS 1. 2. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.2 Double Integrals Over General Regions 3. 4. 5. 6. 7. 8. 9. (a) '! 'x3 dy dx (b) '! '0 10. (a) '! '0 dy dx (b) '! 'yÎ2 dx dy 11. (a) '! 'x2 dy dx (b) '! 'yÎ3 dx dy 12. (a) '! '1 dy dx (b) '1 'ln y dx dy # 3 3 # 8 2x 3x ex 6 9 e2 y1Î3 dx dy 3 Èy 2 Èx 13. (a) '! '0 9 8 dy dx (b) '0 'y2 dx dy 3 9 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 883 884 Chapter 15 Multiple Integrals 14. (a) '! 1Î4 'tan1 x dy dx tanc1 y (b) '0 '0 1 15. (a) '! ln 3 dx dy 'e1c dy dx x (b) '1Î3 'ln y dx dy 1 ln 3 16. (a) '! '0 dy dx '1 'ln x dy dx 1 1 e 1 (b) '0 '0 dx dy 1 ey 17. (a) '! 'x 1 3 2x dy dx (b) '0 '0 dx dy '1 '0 1 y 18. (a) '1 'x2 2 x2 Èy 3 a 3 y bÎ 2 dx dy dy dx Èy (b) '0 'Èy dx dy '1 'y 2 dx dy 1 3 19. '0 '0 (x sin y) dy dx œ '0 c x cos yd x! dx 1 x 1 œ '0 (x x cos x) dx œ ’ x2 (cos x x sin x)“ 1 # # 1 ! œ 1# 2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.2 Double Integrals Over General Regions 20. '0 '0 1 y dy dx œ '0 ’ y2 “ 1 sin x # sin x ! dx œ '0 "# sin# x dx 1 œ "4 '0 (1 cos 2x) dx œ "4 x "2 sin 2x‘ ! œ 14 1 21. '1 ln 8 1 '0ln yexby dx dy œ '1ln 8 cexbyd !ln y dy œ '1ln 8 ayey eyb dy œ c(y 1)ey ey d 1ln 8 œ 8(ln 8 1) 8 e œ 8 ln 8 16 e 22. '1 'y dx dy œ '1 ay# yb dy œ ’ y3 y# “ 2 y# 2 $ # œ ˆ 83 2‰ ˆ "3 "# ‰ œ 37 3# œ 56 # " 23. '0 '0 3y$ exy dx dy œ '0 c3y# exy d 0 dy 1 y# 1 y# œ '0 Š3y# ey 3y# ‹ dy œ ’ey y$ “ œ e 2 1 $ " $ ! 24. È '14 '0 x 3# eyÎÈx dy dx œ '14 32 Èx eyÎÈx ‘ 0Èx dx 4 % œ 3# (e 1) '1 Èx dx œ 23 (e 1) ˆ 32 ‰ x$Î# ‘ " œ 7(e 1) 25. '1 'x 2 2x 1cx 26. '0 '0 1 x y dy dx œ '12 cx ln yd x2x dx œ (ln 2) '12 x dx œ 3# ln 2 ax# y# b dy dx œ '0 ’x# y y3 “ 1 $ " x 0 dx œ '0 ’x# (1 x) (13x) “ dx œ '0 ’x# x$ (13x) “ dx 1 1 $ $ % " $ % œ ’ x3 x4 (11#x) “ œ ˆ "3 4" 0‰ ˆ0 0 1"# ‰ œ "6 ! 1cu 27. '0 '0 1 ˆv Èu‰ dv du œ ' ’ v2 vÈu“ 0 1 # " u 0 du œ '0 ’ 12u# u Èu(1 u)“ du 1 # " œ '0 Š "# u u# u"Î# u$Î# ‹ du œ ’ u2 u# u6 32 u$Î# 25 u&Î# “ œ #" #" "6 32 25 œ #" 25 œ 10 1 # # $ " ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 885 886 Chapter 15 Multiple Integrals 28. '1 '0 es ln t ds dt œ '1 ces ln td 0ln t dt œ '1 (t ln t ln t) dt œ ’ t2 ln t t4 t ln t t“ 2 ln t 2 2 œ (2 ln 2 1 2 ln 2 2) ˆ "4 1‰ œ 4" cv # # # " 29. 'c2 'v 2 dp dv œ 2'c2 cpd vv dv œ 2'c2 2v dv 0 0 0 œ 2 cv# d c2 œ 8 0 È1cs 30. '0 '0 1 # È1cs 8t dt ds œ '0 c4t# d 0 1 # ds œ '0 4 a1 s# b ds œ 4 ’s s3 “ œ 38 1 $ " ! 1Î3 31. 'c1Î3 '0 sec t 1Î3 1Î3 3 cos t du dt œ ' 1Î3 c(3 cos t)ud 0sec t œ 'c1Î3 3 dt œ 21 3Î2 32. '0 '14 2u 4 v 2u dv du œ '03Î2 2u v 4 ‘ 14 2u du # œ '0 3Î2 Ð4 y)Î2 33. '2 '0 4 x2 34. ' 2 '0 0 $Î2 a3 2ub du œ c3u u# d ! œ 92 dx dy dy dx Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.2 Double Integrals Over General Regions 35. '0 'x# dy dx 1 x È1cy 36. '0 '1cy dx dy 1 37. '1 'ln y dx dy e 1 38. '1 '0 dy dx 2 ln x 39. '0 '0 9 40. 1 2 È9cy 16x dx dy È '04 '0 4cx y dy dx Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 887 888 Chapter 15 Multiple Integrals È1cx 41. 'c1 '0 1 È4cy 42. 'c2 '0 2 # 3y dy dx # 6x dx dy 43. '0 'ey x y dx dy 1 44. '0 1 Î2 e c1 '0sin y x y2 dx dy 45. '1 'ln x ax ybdy dx e3 46. '0 1Î3 3 È 'tan 3x Èx y dy dx Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.2 Double Integrals Over General Regions 47. '0 'x siny y dy dx œ '0 '0 siny y dx dy œ '0 sin y dy œ 2 1 1 1 1 y 48. '0 'x 2y# sin xy dy dx œ '0 '0 2y# sin xy dx dy 2 2 2 y œ '0 c2y cos xyd 0y dy œ '0 a2y cos y# 2yb dy 2 2 # œ c sin y# y# d ! œ 4 sin 4 49. '0 'y x# exy dx dy œ '0 '0 x# exy dy dx œ '0 cxexy d 0x dx 1 1 1 x 1 œ '0 axex xb dx œ ’ "2 ex x# “ œ e# 2 1 " # # # ! 4cx# 50. '0 '0 2 xe2y 4y dy dx œ œ '0 ’ #x(4ey) “ 4 51. '0 2 # 2y Èln 3 Èln 3 'y/2 Èln 3 œ '0 È4cy 0 È '04 '0 4cy 4xey dx dy dy œ '0 e# dy œ ’ e4 “ œ e 4 " 4 2y 1 2y % ) ! Èln 3 ex dx dy œ '0 # # # Èln 3 2xex dx œ cex d 0 52. '0 'ÈxÎ3 ey dy dx œ '0 '0 3 2y 1 $ 3y# '02x ex dy dx # œ eln 3 1 œ 2 $ ey dx dy œ '0 3y# ey dy œ cey d ! œ e 1 1 53. '0 1Î16 $ $ " 'y1Î2 cos a161x& b dx dy œ '01Î2 '0x cos a161x& b dy dx % "Î% 161x b œ '0 x% cos a161x& b dx œ ’ sin a80 “ 1 1Î2 & "Î# ! œ 80"1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 889 890 Chapter 15 Multiple Integrals 54. '0 'È$ x y%"1 dy dx œ '0 '0 y%"1 dx dy 8 2 y$ 2 œ '0 y%y1 dy œ "4 cln ay% 1bd ! œ ln417 2 # $ 55. ' ' ay 2x# b dA R xb1 1cx œ 'c1 'cxc1 ay 2x# b dy dx '0 'xc1 ay 2x# b dy dx 0 1 x " 1x œ 'c1 "2 y# 2x# y‘ x1 dx '0 2" y# 2x# y‘ x1 dx 0 1 œ 'c1 "# (x 1)# 2x# (x 1) "# (x 1)# 2x# (x 1)‘dx 0 '0 "# (1 x)# 2x# (1 x) "# (x 1)# 2x# (x 1)‘ dx 1 œ 4 'c1 ax$ x# b dx 4 '0 ax$ x# b dx 0 1 % $ œ 4 ’ x4 x3 “ 56. ' ' xy dA œ '0 0 2Î3 R % c1 " % $ 3 4 ‰ 8 4 ’ x4 x3 “ œ 4 ’ (41) (31) “ 4 ˆ 4" 3" ‰ œ 8 ˆ 12 12 œ 12 œ 32 $ ! 'x2x xy dy dx '21Î3 'x2 x xy dy dx 2Î3 2x 2 x œ '0 "2 xy# ‘ x dx '2Î3 2" xy# ‘ x dx 1 œ '0 ˆ2x$ "# x$ ‰ dx '2Î3 "# x(2 x)# "# x$ ‘ dx 2Î3 œ '0 2Î3 1 3 $ # x dx '21Î3 a2x x# b dx " 2Î3 2‰ 8 ‰‘ 6 ‰ ˆ 4 ˆ 2 ‰ ˆ 27 ˆ 36 16 ‰ 13 œ 38 x% ‘ 0 x# 23 x$ ‘ #Î$ œ ˆ 38 ‰ ˆ 16 œ 81 27 81 1 3 9 3 81 81 81 œ 81 2cx 57. V œ '0 'x 1 ax# y# b dy dx œ '0 ’x# y y3 “ 1 $ 2cx (2x) dx œ '0 ’2x# 7x3 (23x) “ dx œ ’ 2x3 7x 12 12 “ 1 x $ $ $ % 7 " ‰ ‰ 4 œ ˆ 23 12 12 ˆ0 0 16 12 œ 3 2cx# 58. V œ 'c2 'x 1 x# dy dx œ 'c2 cx# yd x 2cx# 1 % " ! dx œ 'c2 a2x# x% x$ b dx œ 23 x$ 15 x& 14 x% ‘ # 1 " 32 16 ‰ 63 ˆ 40 12 15 ‰ ˆ 320 384 240 ‰ 189 œ ˆ 23 "5 4" ‰ ˆ 16 3 5 4 œ 60 60 60 60 60 60 œ 60 œ 20 4cx# 59. V œ 'c4 '3x 1 4cx (x 4) dy dx œ 'c4 cxy 4yd 3x dx œ 'c4 cx a4 x# b 4 a4 x# b 3x# 12xd dx 1 1 # 625 ‰ 157 " œ 'c4 ax$ 7x# 8x 16b dx œ 41 x% 37 x$ 4x# 16x‘ % œ ˆ 4" 37 12‰ ˆ 64 3 64 œ 3 4 œ 12 1 " È 4 cx 60. V œ '0 '0 2 # (3 y) dy dx œ '0 ’3y y2 “ 2 # $ È 4c x 0 # dx œ '0 ’3È4 x# Š 4#x ‹“ dx 2 # # 918 œ ’ 32 xÈ4 x# 6 sin" ˆ x# ‰ 2x x6 “ œ 6 ˆ 1# ‰ 4 86 œ 31 16 6 œ 3 ! 61. V œ '0 '0 a4 y# b dx dy œ '0 c4x y# xd ! dy œ '0 a12 3y# b dy œ c12y y$ d ! œ 24 8 œ 16 2 3 2 $ 2 # Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.2 Double Integrals Over General Regions 4cx# 62. V œ '0 '0 2 a4 x# yb dy dx œ '0 ’a4 x# b y y2 “ 2 # # 4cx# ! dx œ '0 "# a4 x# b dx œ '0 Š8 4x# x# ‹ dx 2 2 # % " &‘ 32 48032096 œ 8x 43 x$ 10 x ! œ 16 32 œ 128 3 10 œ 30 15 2cx 63. V œ '0 '0 2 a12 3y# b dy dx œ '0 c12y y$ d ! 2 xb1 #x dx œ '0 c24 12x (2 x)$ d dx œ ’24x 6x# (24x) “ œ 20 2 % # ! 1cx 64. V œ 'c1 'cxc1 (3 3x) dy dx '0 'xc1 (3 3x) dy dx œ 6 'c1 a1 x# b dx 6 '0 (1 x)# dx œ 4 2 œ 6 0 1 1Îx 0 1 65. V œ '1 'c1Îx (x 1) dy dx œ '1 cxy yd 1Î1xÎx dx œ '1 1 "x ˆ1 x" ‰‘dx œ 2 '1 ˆ1 x" ‰ dx œ 2 cx ln xd #" 2 2 2 2 œ 2(1 ln 2) 66. V œ 4 '0 1Î3 '0sec x a1 y# b dy dx œ 4 '01Î3 ’y y3 “ sec x dx œ 4 '01Î3 Šsec x sec3 x ‹ dx $ $ 0 1Î$ œ 23 c7 ln ksec x tan xk sec x tan xd ! œ 23 ’7 ln Š2 È3‹ 2È3“ 67. 68. 69. '1_ 'ec1 x"y dy dx œ '1_ ’ lnx y “ " dx œ '1_ ˆ x x ‰ dx œ lim x $ $ $ ec x 1/ ˆ1cx ‰ È1cx 1 70. 'c1 'c1/È1cx (2y 1) dy dx œ 'c1 cy# ydº 1 1/ # 1Î# bÄ_ ˆ "b 1‰ œ 1 _ _ 'c_ ' _ ax 1b"ay 1b -dx dy œ 2 '0_ Š y 21 ‹ Š lim # 1 b c1/ a1cx b œ 4 lim c csin" b 0d œ 21 bÄ1 # x" ‘ b œ lim 1 dx œ 'c1 È 2 # dx œ 4 lim c csin" xd ! 1 x bÄ1 # 1Î# # # 71. bÄ_ # bÄ_ tan" b tan" 0‹ dy œ 21 lim bÄ_ '0b y "1 dy # œ 21 Š lim tan" b tan" 0‹ œ (21) ˆ 1# ‰ œ 1# bÄ_ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 891 892 Chapter 15 Multiple Integrals _ _ _ _ 72. '0 '0 xecÐx 2yÑ dx dy œ '0 e2y lim bÄ_ _ œ '0 ec2y dy œ "# lim bÄ_ cxex ex d b0 dy œ '0 e2y lim bÄ_ abeb eb 1b dy aec2b 1b œ "# 3 73. ' ' f(xß y) dA ¸ "4 f ˆ "# ß 0‰ 8" f(0ß 0) 8" f ˆ "4 ß 0‰ œ "4 ˆ "# ‰ 8" ˆ0 "4 ‰ œ 32 R " 128 ‰ ˆ 9 11 ‰ ˆ 7 13 ‰ ˆ 9 13 ‰ 74. ' ' f(xß y) dA ¸ "4 ’f ˆ 47 ß 11 4 f 4 ß 4 f 4 ß 4 f 4 ß 4 “ œ 16 (29 31 33 35) œ 16 œ 8 R 75. The ray ) œ 16 meets the circle x# y# œ 4 at the point ŠÈ3ß 1‹ Ê the ray is represented by the line y œ Èx3 . Thus, È3 È4cx È3 x# b ' ' f(xß y) dA œ ' ' È È4x# dy dx œ ' ’a4 x# b Èx3 È4 x# “ dx œ ”4x x3$ a4È 0 xÎ 3 0 3 3 • # $Î# R _ _ _ _ 1) ' ˆ x#3x x#3x ‰ dx œ 6 '2 76. '2 '0 ax# xb "(y1)#Î$ dy dx œ '2 ’ 3(y ax# xb “ dx œ 2 2 œ 6 lim '2b ˆ x" 1 "x ‰ dx œ 6 lim œ 6 ’ lim ln ˆ1 "b ‰ ln 2“ œ 6 ln 2 bÄ_ bÄ_ 2cx 77. V œ '0 'x 1 2 "Î$ bÄ_ 0 cln (x 1) ln xd 2b œ 6 lim bÄ_ ax# y# b dy dx œ '0 ’x# y y3 “ 1 $ 2cx $ $ $ 2 2 ' ' 1x dx x(x1) [ln (b 1) ln b ln 1 ln 2] dx % 7 ‰ 4 œ ˆ 23 12 1"# ‰ ˆ0 0 16 12 œ 3 78. '0 atan" 1x tan" xb dx œ '0 'x 0 x (2x) œ '0 ’2x# 7x3 (23x) “ dx œ ’ 2x3 7x 12 12 “ 1 È3 " 1y# dy dx œ % " ! '02 'yyÎ1 1"y dx dy '221 'y2Î1 1"y dx dy # # 21 ˆ 21 1 " ‰ y 2 y ‰ # # " ‰ œ 0 11y# dy 2 1y1# dy œ ˆ 12" y #"1 ln a1 y# b‘ 2 1 cln a1 y bd ! 2 tan œ ˆ 1211 ‰ ln 5 2 tan" 21 2"1 ln a1 41# b 2 tan" 2 #"1 ln 5 œ 2 tan" 21 2 tan" 2 2"1 ln a1 41# b ln#5 2ˆ 79. To maximize the integral, we want the domain to include all points where the integrand is positive and to exclude all points where the integrand is negative. These criteria are met by the points (xß y) such that 4 x# 2y# 0 or x# 2y# Ÿ 4, which is the ellipse x# 2y# œ 4 together with its interior. 80. To minimize the integral, we want the domain to include all points where the integrand is negative and to exclude all points where the integrand is positive. These criteria are met by the points (xß y) such that x# y# 9 Ÿ 0 or x# y# Ÿ 9, which is the closed disk of radius 3 centered at the origin. 81. No, it is not possible. By Fubini's theorem, the two orders of integration must give the same result. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. È œ 209 3 Section 15.2 Double Integrals Over General Regions 82. One way would be to partition R into two triangles with the line y œ 1. The integral of f over R could then be written as a sum of integrals that could be evaluated by integrating first with respect to x and then with respect to y: ' ' f(xß y) dA R 2cÐyÎ2Ñ œ '0 '2c2y 1 ÐyÎ2Ñ f(xß y) dx dy '1 'yc1 2 2 f(xß y) dx dy. Partitioning R with the line x œ 1 would let us write the integral of f over R as a sum of iterated integrals with order dy dx. 83. ' b ' b e x y dx dy œ ' b ' b e y e x dx dy œ ' b e y Œ' b e x dx dy œ Œ' b e x dx Œ' b e y dy b b # b # # b # b # # b # b # # b # # œ Œ'cb ecx dx œ Œ2 '0 ecx dx œ 4 Œ'0 ecx dx ; taking limits as b Ä _ gives the stated result. b # b # b # 84. '0 '0 (yx1)#Î$ dy dx œ '0 '0 (yx1)#Î$ dx dy œ '0 (y"1)#Î$ ’ x3 “ dy œ 3" '0 (ydy1)#Î$ 1 3 3 # 1 3 # $ " 3 ! œ "3 lim c '0 (ydy1)#Î$ "3 lim b 'b (ydy1)#Î$ œ lim c (y 1)"Î$ ‘ 0 lim b (y 1)"Î$ ‘ b bÄ1 bÄ1 bÄ1 bÄ1 b 3 b 3 3 3 œ ’ lim c (b 1)"Î$ (1)"Î$ “ ’ lim b (b 1)"Î$ (2)"Î$ “ œ (0 1) Š0 È 2‹ œ 1 È 2 bÄ1 bÄ1 85-88. Example CAS commands: Maple: f := (x,y) -> 1/x/y; q1 := Int( Int( f(x,y), y=1..x ), x=1..3 ); evalf( q1 ); value( q1 ); evalf( value(q1) ); 89-94. Example CAS commands: Maple: f := (x,y) -> exp(x^2); c,d := 0,1; g1 := y ->2*y; g2 := y -> 4; q5 := Int( Int( f(x,y), x=g1(y)..g2(y) ), y=c..d ); value( q5 ); plot3d( 0, x=g1(y)..g2(y), y=c..d, color=pink, style=patchnogrid, axes=boxed, orientation=[-90,0], scaling=constrained, title="#89 (Section 15.2)" ); r5 := Int( Int( f(x,y), y=0..x/2 ), x=0..2 ) + Int( Int( f(x,y), y=0..1 ), x=2..4 ); value( r5); value( q5-r5 ); 85-94. Example CAS commands: Mathematica: (functions and bounds will vary) You can integrate using the built-in integral signs or with the command Integrate. In the Integrate command, the integration begins with the variable on the right. (In this case, y going from 1 to x). Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 893 894 Chapter 15 Multiple Integrals Clear[x, y, f] f[x_, y_]:= 1 / (x y) Integrate[f[x, y], {x, 1, 3}, {y, 1, x}] To reverse the order of integration, it is best to first plot the region over which the integration extends. This can be done with ImplicitPlot and all bounds involving both x and y can be plotted. A graphics package must be loaded. Remember to use the double equal sign for the equations of the bounding curves. Clear[x, y, f] <<GraphicsÌmplicitPlot` ImplicitPlot[{x==2y, x==4, y==0, y==1},{x, 0, 4.1}, {y, 0, 1.1}]; f[x_, y_]:=Exp[x2 ] Integrate[f[x, y], {x, 0, 2}, {y, 0, x/2}] Integrate[f[x, y], {x, 2, 4}, {y, 0, 1}] To get a numerical value for the result, use the numerical integrator, NIntegrate. Verify that this equals the original. Integrate[f[x, y], {x, 0, 2}, {y, 0, x/2}] NIntegrate[f[x, y], {x, 2, 4}, {y, 0, 1}] NIntegrate[f[x, y], {y, 0, 1},{x, 2y, 4}] Another way to show a region is with the FilledPlot command. This assumes that functions are given as y = f(x). Clear[x, y, f] <<Graphics`FilledPlot` FilledPlot[{x2 , 9},{x, 0,3}, AxesLabels Ä {x, y}]; f[x_, y_]:= x Cos[y2 ] Integrate[f[x, y], {y, 0, 9}, {x, 0, Sqrt[y]}] " 85. '1 '1 xy dy dx ¸ 0.603 86. '0 '0 ec ˆx by ‰ dy dx ¸ 0.558 87. '0 '0 tan" xy dy dx ¸ 0.233 88. 3 1 x 1 1 '0 '2ye dx dy 4 # È # 'c11 '0 1cx 3È1 x# y# dy dx ¸ 3.142 # The following graph was generated using Mathematica. 89. Evaluate the integrals: 1 1 x# œ '0 '0 ex dy dx '2 '0 ex dy dx 2 x/2 # 4 1 # œ "4 4" ê4 2È1 erfia2b 2È1 erfia4b‰ ¸ 1.1494 ‚ 106 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.2 Double Integrals Over General Regions 90. Evaluate the integrals: Èy '0 'x x cosay2 bdy dx œ '0 '0 x cosay2 bdx dy 3 9 9 The following graph was generated using Mathematica. 2 œ sina481b ¸ 0.157472 91. Evaluate the integrals: '0 'y 2 4 È2y 3 Èx ax2 y xy2 bdx dy œ '0 'x2 /32 ax2 y xy2 bdy dx 8 3 The following graph was generated using Mathematica. œ 67,520 693 ¸ 97.4315 92. Evaluate the integrals: 4cy# '0 '0 2 È4cx exy dx dy œ '0 '0 4 exy dy dx The following graph was generated using Mathematica. ¸ 20.5648 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 895 896 Chapter 15 Multiple Integrals 93. Evaluate the integrals: The following graph was generated using Mathematica. '1 '0 x1 y dy dx 1 2 4 2 œ '0 '1 x1 y dx dy '1 'Èy x1 y dx dy 2 x# ‰ 1 lnˆ 27 4 ¸ 0.909543 94. Evaluate the integrals: Èx '1 'y Èx1y dx dy œ '1 '1 2 8 8 2 3 2 The following graph was generated using Mathematica. 3 1 Èx2 y2 dy dx ¸ 0.866649 15.3 AREA BY DOUBLE INTEGRATION 1. '02 '02cx dy dx œ '02 (2 x) dx œ ’2x x2 “ # œ 2, # 2 2cy 2 or '0 '0 dx dy œ '0 (2 y) dy œ 2 2. ! '02 '2x4 dy dx œ '02 (4 2x) dx œ c4x x# d 02 œ 4, 4 yÎ2 4 or '0 '0 dx dy œ '0 y# dy œ 4 'c12 'yccy2 dx dy œ 'c12 ay# y 2b dy # 3. $ # œ ’ y3 y# 2y“ " # œ ˆ "3 "# 2‰ ˆ 83 2 4‰ œ 9# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.3 Area by Double Integration '02 'cyycy dx dy œ '02 a2y y# b dy œ ’y# y3 “ # # 4. $ ! œ 4 83 œ 43 5. '0ln 2 '0e dy dx œ '0ln 2 ex dx œ cex d 0ln 2 œ 2 1 œ 1 6. '1e 'ln2 xln x dy dx œ '1e ln x dx œ cx ln x xd 1e x œ (e e) (0 1) œ 1 '01 'y2ycy dx dy œ '01 a2y 2y# b dy œ y# 23 y$ ‘ "! # 7. # œ "3 'c11 '2yy cc12 dx dy œ 'c11 ay# 1 2y# 2b dy # 8. # œ 'c1 a1 y# b dy œ ’y y3 “ 1 9. $ " " œ 34 '02 'y3y 1 dx dy œ '02 cxd 3y y dy œ '0 a2yb dy œ cy2 d0 œ 4 2 2 y 10. '1 '1 y 1 dx dy œ '0 cxdln 1 y dy 2 ln y 2 œ '1 aln y 1 yb dy œ ’y ln y 2y y2 “ 2 2 2 1 œ 2 ln 2 12 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 897 898 Chapter 15 Multiple Integrals 11. '0 'xÎ2 1 dy dx '1 'xÎ2 1 dy dx 1 2x 2 3x ' 3x œ '0 cyd2x xÎ2 dx 1 cydxÎ2 dx 1 2 œ '0 ˆ 32 x‰dx '1 ˆ3 32 x‰dx 1 2 1 2 œ 34 x2 ‘0 3x 34 x2 ‘1 œ 32 Èx Èx 12. '0 'x 1 dy dx '1 'x 2 1 dy dx 1 4 œ '0 cydxx dx '1 cydx 2 dx 1 4 È Èx œ '0 ˆÈx x‰dx '1 ˆÈx x 2‰dx 1 4 1 4 œ 23 x3Î2 12 x2 ‘0 23 x3Î2 12 x2 2x‘1 œ 13 3 13. '0 'y#Î3 dx dy œ '0 Š2y y3 ‹ dy œ ’y# y9 “ 6 2y 6 # $ ' ! œ 36 216 9 œ 12 2xcx# 14. '0 'cx 3 dy dx œ '0 a3x x# b dx œ 32 x# 3" x$ ‘ ! 3 $ 9 œ 27 # 9œ # 15. '0 'sincosx x dy dx 1Î4 œ '0 (cos x sin x) dx œ csin x cos xd 01Î4 1Î4 È È œ Š #2 #2 ‹ (0 1) œ È2 1 yb2 16. 'c1 'y# dx dy œ 'c1 ay 2 y# b dy œ ’ y2 2y y3 “ 2 2 # œ ˆ2 4 83 ‰ ˆ "# 2 3" ‰ œ 5 "# œ 9# $ 2 c1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.3 Area by Double Integration 17. 'c1 'c2x dy dx '0 'cxÎ2 dy dx 0 1 x 2 1 x œ ' 1 (1 x) dx '0 ˆ1 x# ‰ dx 0 2 # œ ’x x2 “ 0 2 # c1 ’x x4 “ œ ˆ1 "# ‰ (2 1) œ 3# 0 Èx 18. '0 'x#c4 dy dx '0 '0 dy dx 2 0 4 œ '0 a4 x# b dx '0 x"Î# dx 2 4 $ 2 32 œ ’4x x3 “ 32 x$Î# ‘ 0 œ ˆ8 38 ‰ 16 3 œ 3 4 0 19. (a) average œ 1"# '0 '0 sin (x y) dy dx œ 1"# '0 c cos (x y)d 01 dx œ 1"# '0 [ cos (x 1) cos x] dx 1 1 1 1 œ 1"# c sin (x 1) sin xd 01 œ 1"# [( sin 21 sin 1) ( sin 1 sin 0)] œ 0 " (b) average œ 1# Š#‹ '01 '01Î2 sin (x y) dy dx œ 12 '01 c cos (x y)d 1! Î# dx œ 12 '01 cos ˆx 1# ‰ cos x‘ dx # # 1 œ 12# sin ˆx 1# ‰ sin x‘ 0 œ 12# ˆ sin 3#1 sin 1‰ ˆ sin 1# sin 0‰‘ œ 14# 20. average value over the square œ '0 '0 xy dy dx œ '0 ’ xy2 “ dx œ '0 #x dx œ 4" œ 0.25; 1 1 1 È1cx average value over the quarter circle œ ˆ 1 ‰ '0 '0 1 " 4 " # 1 ! # xy dy dx œ 14 '0 ’ xy2 “ 1 # È1cx # dx 0 œ 12 '0 ax x$ b dx œ 12 ’ x2 x4 “ œ #"1 ¸ 0.159. The average value over the square is larger. 1 # 4 " ! 8 21. average height œ "4 '0 '0 ax# y# b dy dx œ 4" '0 ’x# y y3 “ dx œ "4 '0 ˆ2x# 38 ‰ dx œ "# ’ x3 4x 3 “ œ 3 2 2 2 $ # 2 $ ! 22. average œ (ln"2)# 'ln 2 2 ln 2 ' # ' 2 ln 2 23. 'c5 'c2 10,000e dy dx œ 10,000 a1 e# b 'c5 x 0 ! 'ln2 2ln 2 xy" dy dx œ (ln"2) 'ln2 2ln 2 ’ lnxy “ 2 ln 2 dx ln 2 2 ln 2 œ (ln"2)# ln 2 "x (ln 2 ln ln 2 ln ln 2) dx œ ˆ ln"2 ‰ ln 2 œ ˆ ln"# ‰ (ln 2 ln ln 2 ln ln 2) œ 1 5 # 5 y k k 1 # 2 ln 2 dx ˆ " ‰ x œ ln # cln xd ln 2 dx œ 10,000 a1 e# b ’ kxk 1 # 'c!5 1 dx '!5 1 dx “ x # x # ! & œ 10,000 a1 ec2 b 2 ln ˆ1 x# ‰‘ & 10,000 a1 ec2 b 2 ln ˆ1 x# ‰‘ ! œ 10,000 a1 e# b 2 ln ˆ1 5# ‰‘ 10,000 a1 e# b 2 ln ˆ1 5# ‰‘ œ 40,000 a1 e# b ln ˆ 72 ‰ ¸ 43,329 2ycy# 24. '0 'y# 1 100(y 1) dx dy œ '0 c100(y 1)xd y2y# cy dy œ '0 100(y 1) a2y 2y# b dy œ 200 '0 ay y$ b dy 1 # % # 1 " œ 200 ’ y2 y4 “ œ (200) ˆ "4 ‰ œ 50 ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 1 899 900 Chapter 15 Multiple Integrals 25. Let (xi ß yi ) be the location of the weather station in county i for i œ 1ß á ß 254. The average temperature 254 ! T(xi ßyi ) ?i A in Texas at time t! is approximately iœ1 A , where T(xi ß yi ) is the temperature at time t! at the weather station in county i, ?i A is the area of county i, and A is the area of Texas. 26. Let y œ faxb be a nonnegative, continuous function on Òa, bÓ, then A œ ' ' dA œ 'a '0 dy dx œ 'a cydf0axb dx œ 'a faxb dx b faxb b R 15.4 DOUBLE INTEGRALS IN POLAR FORM 1. x2 y2 œ 92 Ê r œ 9 Ê 12 Ÿ ) Ÿ 21, 0 Ÿ r Ÿ 9 2. x2 y2 œ 12 Ê r œ 1, x2 y2 œ 42 Ê r œ 4 Ê 12 Ÿ ) Ÿ 12 , 1 Ÿ r Ÿ 4 3. y œ x Ê ) œ 14 , y œ x Ê ) œ 341 , y œ 1 Ê r œ csc ) Ê 14 Ÿ ) Ÿ 341 , 0 Ÿ r Ÿ csc ) 4. x œ 1 Ê r œ sec ), y œ È3x Ê ) œ 13 Ê 0 Ÿ ) Ÿ 13 , 0 Ÿ r Ÿ sec ) 5. x2 y2 œ 12 Ê r œ 1, x œ 2È3 Ê r œ 2È3 sec ), y œ 2 Ê r œ 2 csc ); 2È3 sec ) œ 2 csc ) Ê ) œ 16 Ê 0 Ÿ ) Ÿ 1 , 1 Ÿ r Ÿ 2È3 sec ); 1 Ÿ ) Ÿ 1 , 1 Ÿ r Ÿ 2È3 csc ) 6 6 2 6. x2 y2 œ 22 Ê r œ 2, x œ 1 Ê r œ sec ); 2 œ sec ) Ê ) œ 13 or ) œ 13 Ê 13 Ÿ ) Ÿ 13 , sec ) Ÿ r Ÿ 2 7. x2 y2 œ 2x Ê r œ 2 cos ) Ê 12 Ÿ ) Ÿ 12 , 0 Ÿ r Ÿ 2 cos ) 8. x2 y2 œ 2y Ê r œ 2 sin ) Ê 0 Ÿ ) Ÿ 1, 0 Ÿ r Ÿ 2 sin ) 9. È 'c11 '0 1cx dy dx œ '01 '01 r dr d) œ "# '01 d) œ 1# # È1cy ax# y# b dx dy œ '0 '01 r$ dr d) œ 4" '01Î2 d) œ 18 È4cy ax# y# b dx dy œ '0 '02 r$ dr d) œ 4 '01Î2 d) œ 21 10. '0 '0 1 11. '0 '0 2 1Î2 # # Èa cx 1Î2 12. 'ca 'cÈa#cx# dy dx œ '0 '0 r dr d) œ a2 '0 d) œ 1a# # a 21 # a # 13. '0 '0 x dx dy œ '1Î4 '0 6 1Î2 y 14. '0 '0 y dy dx œ '0 6 csc ) 1Î4 21 r# cos ) dr d) œ 72 '1Î4 cot ) csc# ) d) œ 36 ccot# )d 1Î4 œ 36 1Î2 1Î2 '02 sec r# sin ) dr d) œ 83 '0 Î4 tan ) sec# ) d) œ 43 ) 1 2 x È3 '1x dy dx œ '1ÎÎ64 'csc 3 sec r dr d) œ ' ÎÎ64 ˆ 32 sec2 ) 12 csc2 )‰ d) œ 32 tan ) 12 cot )‘ ÎÎ64 œ 2 È3 15. '1 1 È ) ) 16. 'È2 'È4 y2 dy dx œ '1Î4 '2 2 y 1Î2 1 1 1 1 2 csc ) r dr d) œ '1Î6 a2csc2 ) 2b d) œ 2 cot ) 12 )‘ 1Î4 œ 2 12 1Î4 1Î2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. b Section 15.4 Double Integrals in Polar Form 17. 'c1 'cÈ1 c x# 1 È2x# y# dy dx œ '1 0 31Î2 0 '01 1 2r r dr d) œ 2 '131Î2 '01 ˆ1 1 " r ‰ dr d) œ 2 '131Î2 (1 ln 2) d) œ (1 ln 2)1 È1 c x 18. 'c1 'cÈ1 c x# a1 x#2 y# b# dy dx œ 4 '0 1 19. '0 ln 2 1Î2 # È '01 ' 1Î2 1 " r ‘ "! d) œ 2 '01Î2 d) œ 1 2r dr d) œ 4 0 a1 r# b# '0 (ln 2) c y eÈx y dx dy œ '01Î2 '0ln 2 rer dr d) œ '01Î2 a2 ln 2 1b d) œ 1# a2 ln 2 1b # # # # È1 c y 20. 'c1 'cÈ1 c y# ln ax# y# 1b dx dy œ 4 '0 1Î2 # 1 È 2 x# 21. '0 'x 1 # È2 ax 2yb dy dx œ '1Î4 '0 1 Î2 '01 ln ar# 1b r dr d) œ 2'01Î2 aln 4 1b d) œ 1aln 4 1b 1Î2 È È È È 1 Î2 1 Î4 È2x x# 22. '1 '0 2 1 dy dx œ ax # y # b # 3 3 0 œ '1Î4 Š 2 3 2 cos ) 4 3 2 sin )‹ d) œ ’ 2 3 2 sin ) 4 3 2 cos )“ 1 Î2 œ 1 È 1 x# È1 y# '01 '0 1 Î4 ) '01Î4 'sec2cos) ) r1 r dr d) œ '01Î4 2r1 ‘2cos d) œ œ '0 ˆ 21 cos2 ) 18 sec2 )‰ d) sec ) 4 2 x y dy dx or x y dx dy È3y 24. '1Î2 'È1 y# x dx dy or 1 È È '0 3Î2 'È1 1 x x dy dx 'È33Î2 'xÎ1 È3 x dy dx # 25. '0 '0 y2 ax2 y2 b dy dx or 2 d) 2 ˆ1 È 2 ‰ 3 1Î4 1 œ 14 ) 18 sin2 ) 18 tan )‘0 œ 16 23. '0 '0 È2 ar cos ) 2r sin )b r dr d) œ '1Î4 ’ r3 cos ) 2r3 sin )“ x '02 'y2 y2 ax2 y2 b dx dy Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 901 902 Chapter 15 Multiple Integrals 26. '0 '0 ax2 y2 b dy dx or 3 4 3 '04 '03 ax2 y2 b3 dx dy 27. '0 1Î2 È '02 2 sin 2 r dr d) œ 2 '0 Î2 (2 sin 2)) d) œ 2(1 1) ) 1 28. A œ 2 '0 '11 cos r dr d) œ '0 Î2 a2 cos ) cos# )b d) œ 8 4 1 29. A œ 2 '0 '012 cos 3 r dr d) œ 144 '0 Î6 cos# 3) d) œ 121 1Î2 1Î6 ) ) 30. A œ '0 '0 21 4)Î3 31. A œ '0 1Î2 1 1 r dr d) œ 89 '0 )# d) œ 64271 21 $ '01 sin r dr d) œ 12 '0 Î2 ˆ 3# 2 sin ) cos#2) ‰ d) œ 381 1 ) 32. A œ 4 '0 1Î2 1 '01 cos r dr d) œ 2 '0 Î2 ˆ 3# 2 cos ) cos#2) ‰ d) œ 321 4 ) 1 33. average œ 14a# '0 '0a rÈa# r# dr d) œ 314a '01Î2 a$ d) œ 2a3 34. average œ 14a# '0 '0a r# dr d) œ 314a '01Î2 a$ d) œ 2a3 1Î2 # 1Î2 # Èa cx 35. average œ 1"a# 'ca 'cÈa#cx# Èx# y# dy dx œ 1"a# '0 '0 r# dr d) œ 3a1 '0 d) œ 2a 3 # a 21 # 21 a 36. average œ 1" ' ' c(1 x)# y# d dy dx œ 1" '0 '0 c(1 r cos ))# r# sin# )d r dr d) 21 1 R œ 1" '021 '01 ar$ 2r# cos ) rb dr d) œ 1" '021 ˆ 34 2 cos3 ) ‰ d) œ 1" 34 ) 2 sin3 ) ‘ 021 œ #3 Èe Èe 37. '0 '1 Š lnrr ‹ r dr d) œ '0 '1 2 ln r dr d) œ 2'0 cr ln r rd1e d) œ 2'0 Èe ˆ "# 1‰ 1‘ d) œ 21ˆ2 Èe‰ 21 21 # 21 21 "Î# 38. '0 '1 Š lnrr ‹ dr d) œ '0 '1 ˆ 2 lnr r ‰ dr d) œ '0 c(ln r)# d 1 d) œ '0 d) œ 21 21 e 21 # 39. V œ 2 '0 1Î2 21 e 21 e '11 cos r# cos ) dr d) œ 23 '0 Î2 a3 cos# ) 3 cos$ ) cos% )b d) ) 1 1Î2 œ 23 "85) sin 2) 3 sin ) sin$ ) sin324) ‘ 0 œ 34 581 40. V œ 4 '0 1Î4 È '0 2 cos 2 rÈ2 r# dr d) œ 43 '0 Î4 (2 2 cos 2))$Î# 2$Î# ‘ d) ) 1 '0 a1 cos# )b sin ) d) œ 213 2 323 ’ cos3 ) cos )“ œ 213 2 32 3 È 1Î4 È $ 1Î4 0 œ 61 È2 40È2 64 9 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.4 Double Integrals in Polar Form _ _ 41. (a) I# œ '0 '0 e ax y b dx dy œ '0 œ "# '0 # 1Î2 lim bÄ_ t# Î 1 2 # '0_ Šer ‹ r dr d) œ '01Î2 ” lim '0b rer dr• d) # # bÄ_ Šeb 1‹ d) œ "# '0 1Î2 # _ 903 d) œ 14 Ê I œ È1 # ' 2e dt œ È21 '0 et dt œ Š È21 ‹ Š # ‹ œ 1, from part (a) (b) x lim Ä _ 0 È1 x _ _ 42. '0 '0 œ 14 lim bÄ_ dx dy œ '0 Î 1 2 " a1 x# y# b# # '0_ ˆ1 1 " b# ‰ œ 14 È1 r dr d) œ 1# lim a1 r# b# bÄ_ È3Î2 43. Over the disk x# y# Ÿ 34 : ' ' 1 x"# y# dA œ '0 '0 21 R 21 21 œ '0 ˆ "# ln "4 ‰ d) œ (ln 2) '0 d) œ 1 ln 4 '0b r dr œ 14 lim a1 r# b# bÄ_ r 1 r# dr d) œ 1 " r# ‘ b 0 '021 2" ln a1 r# b‘ 0È3Î2 d) Over the disk x# y# Ÿ 1: ' ' 1 x"# y# dA œ '0 '0 1 r r# dr d) œ '0 ’ lim c '0 1 r r# dr“ d) œ '0 21 21 21 1 aÄ1 R lim a Ä 1c x# y# Ÿ 1 a "# ln a1 a# b‘ d) œ 21 † lim c aÄ1 "# ln a1 a# b‘ œ 21 † _, so the integral does not exist over 44. The area in polar coordinates is given by A œ '! '0 r dr d) œ '! ’ r2 “ " fÐ)Ñ " # fÐ)Ñ 0 d) œ "# '! f # ()) d) œ '! "# r# d), " " where r œ f()) 45. average œ 1"a# '0 '0 c(r cos ) h)# r# sin# )d r dr d) œ 1"a# '0 '0 ar$ 2r# h cos ) rh# b dr d) 21 21 a a œ 1"a# '0 Š a4 2a h3cos ) a #h ‹ d) œ 1" '0 Š a4 2ah 3cos ) h# ‹ d) œ 1" ’ a4) 2ah 3sin ) h#) “ 21 % $ # # 21 # # # # 21 0 œ "# aa# 2h# b 46. A œ '1Î4 'csc ) r dr d) œ "# '1Î4 a4 sin# ) csc# )b d) 31Î4 2 sin ) 31Î4 œ "# c2) sin 2) cot )d 131Î4Î4 œ 1# 47-50. Example CAS commands: Maple: f := (x,y) -> y/(x^2+y^2); a,b := 0,1; f1 := x -> x; f2 := x -> 1; plot3d( f(x,y), y=f1(x)..f2(x), x=a..b, axes=boxed, style=patchnogrid, shading=zhue, orientation=[0,180], title="#47(a) (Section 15.4)" ); # (a) q1 := eval( x=a, [x=r*cos(theta),y=r*sin(theta)] ); # (b) q2 := eval( x=b, [x=r*cos(theta),y=r*sin(theta)] ); q3 := eval( y=f1(x), [x=r*cos(theta),y=r*sin(theta)] ); q4 := eval( y=f2(x), [x=r*cos(theta),y=r*sin(theta)] ); theta1 := solve( q3, theta ); Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 904 Chapter 15 Multiple Integrals theta2 := solve( q1, theta ); r1 := 0; r2 := solve( q4, r ); plot3d(0,r=r1..r2, theta=theta1..theta2, axes=boxed, style=patchnogrid, shading=zhue, orientation=[-90,0], title="#47(c) (Section 15.4)" ); fP := simplify(eval( f(x,y), [x=r*cos(theta),y=r*sin(theta)] )); # (d) q5 := Int( Int( fP*r, r=r1..r2 ), theta=theta1..theta2 ); value( q5 ); Mathematica: (functions and bounds will vary) For 47 and 48, begin by drawing the region of integration with the FilledPlot command. Clear[x, y, r, t] <<Graphics`FilledPlot` FilledPlot[{x, 1}, {x, 0, 1}, AspectRatio Ä 1, AxesLabel Ä {x,y}]; The picture demonstrates that r goes from 0 to the line y=1 or r = 1/ Sin[t], while t goes from 1/4 to 1/2. f:= y / (x2 y2 ) topolar={x Ä r Cos[t], y Ä r Sin[t]}; fp= f/.topolar //Simplify Integrate[r fp, {t, 1/4, 1/2}, {r, 0, 1/Sin[t]}] For 49 and 50, drawing the region of integration with the ImplicitPlot command. Clear[x, y] <<GraphicsÌmplicitPlot` ImplicitPlot[{x==y, x==2 y, y==0, y==1}, {x, 0, 2.1}, {y, 0, 1.1}]; The picture shows that as t goes from 0 to 1/4, r goes from 0 to the line x=2 y. Solve will find the bound for r. bdr=Solve[r Cos[t]==2 r Sin[t], r]//Simplify f:=Sqrt[x y] topolar={x Ä r Cos[t], y Ä r Sin[t]}; fp= f/.topolar //Simplify Integrate[r fp, {t, 0, 1/4}, {r, 0, bdr[[1, 1, 2]]}] 15.5 TRIPLE INTEGRALS IN RECTANGULAR COORDINATES 1. '01 '01cx 'x1bz Fax, y, zb dy dz dx œ '01 '01cx 'x1bz dy dz dx œ '01 '01cx a1 x zb dz dx œ '0 ’a1 xb xa1 xb a1 2 xb “dx œ '0 1 2. 3. 2 1 " 3 a1 xb2 dx œ ’ a1 6 xb “ œ 61 2 0 '01 '02 '03 dz dy dx œ '01 '02 3 dy dx œ '01 6 dx œ 6, '02 '01 '03 dz dx dy, '03 '02 '01 dx dy dz, '02 '03 '01 dx dz dy, '03 '01 '02 dy dx dz, '01 '03 '02 dy dz dx '01 '02c2x '03c3xc3yÎ2 dz dy dx 1 2 2x œ '0 '0 ˆ3 3x 32 y‰ dy dx 1 œ '0 3(1 x) † 2(1 x) 34 † 4(1 x)# ‘ dx 1 " œ 3 '0 (1 x)# dx œ c (1 x)$ d ! œ 1, '02 '01cyÎ2 '03 3x 3yÎ2 dz dx dy, '01 '03 3x '02 2x 2zÎ3 dy dz dx, Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.5 Triple Integrals in Rectangular Coordinates '03 '01 zÎ3 '02 2x 2zÎ3 dy dx dz, '02 '03 3yÎ2 '01 yÎ2 zÎ3 dx dz dy, '03 '02 2zÎ3 '01 yÎ2 zÎ3 dx dy dz 4. È '02 '03 '0 4 x dz dy dx œ '02 '03 È4 x# dy dx œ '02 3È4 x# dx œ 3# ’xÈ4 x# 4 sin" x# “ 2 œ 6 sin" 1 œ 31, # È4cx '0 '0 '0 3 5. 2 È È4cx dz dx dy, '0 '0 # 2 # È4cz '0 dy dz dx, '0 '0 3 2 È # È4cz '0 dy dx dz, '0 '0 '0 3 2 3 # 0 È4cz dx dy dz, '0 '0 '0 3 2 # dx dz dy 'c22 'cÈ44ccxx 'x8bcyx cy dz dy dx œ 4 '02 '0 4cx 'x8bcyx cy dz dy dx # # # # È4cx œ 4'0 '0 2 # È4cx œ 8 '0 '0 2 œ 8'0 1Î2 # # # # # c8 2 ax# y# bd dy dx # a4 x# y# b dy dx '02 a4 r# b r dr d) œ 8 '01Î2 ’2r# r4 “ # d) % ! 1Î2 œ 32 '0 d) œ 32 ˆ 1# ‰ œ 161, È # # 'c22 'cÈ44ccyy 'x8bcyx cy dz dx dy, # # # È # # # È 'c22 'y4 'cÈzzccyy dx dz dy 'c22 '48cy 'cÈ88cczzccyy dx dz dy, # # # # # # È È È È È È '0 'cÈzz 'cÈzzccyy dx dy dz '48 'cÈ88cczz 'cÈ88cczzccyy dx dy dz, 'c22 'x4 'cÈzzccxx dy dz dx 'c22 '48cx 'cÈ88cczzccxx dy dz dx, È È È È '04 'cÈzz 'cÈzzccxx dy dx dz '48 'cÈ88cczz 'cÈ88cczzccxx dy dx dz 4 # # # # # # # # # # # # # # 6. The projection of D onto the xy-plane has the boundary x# y# œ 2y Ê x# (y 1)# œ 1, which is a circle. Therefore the two integrals are: È È '02 'cÈ2y2yccyy 'x2yby dz dx dy and 'c11 '11cbÈ11ccxx 'x2yby dz dy dx # # # # # # # # 7. '01 '01 '01 ax# y# z# b dz dy dx œ '01 '01 ˆx# y# 3" ‰ dy dx œ '01 ˆx# 23 ‰ dx œ 1 8. '0 2 '03y'x8bc3yx cy dz dx dy œ '0 2 '03ya8 2x# 4y# b dx dy œ '0 2 8x 23 x$ 4xy# ‘ 03y dy È # # È2 È # È # È %‘ 2 œ '0 a24y 18y$ 12y$ b dy œ 12y# "5 2 y 0 œ 24 30 œ 6 '1e '1e '1e xyz" dx dy dz œ '1e '1e ’ lnyzx “ e dy dz œ '1e '1e yz3 dy dz œ '1e ’ lnzy “ e dz œ '1e 6z dz œ 6 2 9. 3 2 3 1 3c3x 10. '0 '0 1 2 2 1 '03c3xcy dz dy dx œ '01 '03c3x (3 3x y) dy dx œ '01 (3 3x)# "# (3 3x)# ‘ dx œ #9 '01 (1 x)# dx " œ 3# c(1 x)$ d ! œ 3# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 905 906 Chapter 15 Multiple Integrals 11. '0 1Î6 È '01 ' 3# y sin z dx dy dz œ '01Î6 '01 5y sin z dy dz œ 5# '01Î6 sin z dz œ 5Š2 4 3‹ 12. 'c1 '0 '0 ax y zb dy dx dz œ 'c1 '0 œ xy "2 y# zy‘0 dx dz œ 'c1 '0 a2x 2 2zb dx dz 1 1 2 1 1 1 2 1 œ 'c1 cx# 2x 2zxd 0 dz œ 'c1 a3 2zb dz œ c3z z# dc1 œ 6 1 È9 c x 13. '0 '0 3 È 1 È '0 9 c x dz dy dx œ '03 '0 9 c x È9 x# dy dx œ '03 a9 x# b dx œ ’9x x3 “ $ œ 18 # # # $ ! È4cy 2xby 14. '0 'cÈ4cy# '0 # 2 1 " È4cy È4cy dz dx dy œ '0 'cÈ4cy# (2x y) dx dy œ '0 cx# xyd È4 y# dy œ '0 a4 y# b # 2 # 2 c c $Î# # œ ’ 23 a4 y# b 2cx 15. '0 '0 1 2 "Î# (2y) dy “ œ 23 (4)$Î# œ 16 3 ! '02cxcydz dy dx œ '01 '02cx (2 x y) dy dx œ '01 (2 x)# "# (2 x)# ‘ dx œ "# '01 (2 x)# dx " œ "6 (2 x)$ ‘ ! œ 6" 86 œ 67 1cx# '34cx cyx dz dy dx œ '01 '01cx x a1 x# yb dy dx œ '01 x ’a1 x# b# "# a1 x# b“ dx œ '01 "# x a1 x# b# dx 16. '0 '0 1 # # $ " " " œ ’ 12 a1 x# b “ œ 12 ! 17. '0 '0 '0 cos (u v w) du dv dw œ '0 '0 [sin (w v 1) sin (w v)] dv dw 1 1 1 1 1 œ '0 [( cos (w 21) cos (w 1)) (cos (w 1) cos w)] dw 1 œ c sin (w 21) sin (w 1) sin w sin (w 1)d 1! œ 0 Èe 18. '0 '1 1 œ 19. '0 2 2 Èe 6 1Î4 œ '0 È lim b Ä _ s s 1 '01 s es ds œ 2 6Èe cs es es d 01 œ 2 6Èe '0ln sec v 'c2t_ ex dx dt dv œ '01Î4 '0ln sec v 1Î4 ae2t eb b dt dv œ '0 1Î4 '0ln sec v e2t dt dv œ '01Î4 ˆ "# e2 ln sec v "# ‰ dv # 1Î% Š sec# v #" ‹ dv œ tan2 v 2v ‘ ! œ #" 18 È4cq 20. '0 '0 '0 7 È '1e s es ln r alnttb dt dr ds œ '01 '1 e as es ln rb’ 31 aln tb3 “ e dr ds œ '01 '1 e s3e ln r dr ds œ '01 s3e cr ln r rd 1Èe ds 2 q r 1 dp dq dr œ 1cx# 21. (a) 'c1 '0 1 # ! 'x1cz dy dz dx 1cy 1 cÈz 1 y# 1 cz Èy 1cy (e) '0 'cÈy '0 1 cÈz (b) '0 '0 'c1 1 dy dz dx (d) 'c1 '0 '0 dx dz dy 0 È1cz (b) '0 'cÈ1cz 'x# 'cÈÈyy dx dz dy 22. (a) '0 '0 'c1 1 '07 '02 qÈr41 q dq dr œ '07 3(r " 1) ’ a4 q# b$Î# “ # dr œ 38 '07 r"1 dr œ 8 ln3 8 œ 8 ln 2 # (d) '0 '0 1 # 1 0 1 y# 1 y# 1 1 cÈz (c) '0 'c1 1 dy dx dz 23. V œ '0 'c1 '0 dz dy dx œ '0 'c1 y# dy dx œ 23 '0 dx œ 23 1 'cÈÈyy dx dy dz dz dx dy (e) 'c1 '0 '0 dz dx dy 1 1 cz (c) '0 '0 1 dy dx dz 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. '01 dx dy dz Section 15.5 Triple Integrals in Rectangular Coordinates 1cx 24. V œ '0 '0 1 '02c2z dy dz dx œ '01 '01cx (2 2z) dz dx œ '01 c2z z# d 01cx dx œ '01 a1 x# b dx œ ’x x3 “ " œ 32 $ ! È4cx 25. V œ '0 '0 4 È '02cy dz dy dx œ '04 '0 4cx (2 y) dy dx œ '04 ’2È4 x ˆ 4#x ‰“ dx % 20 œ 43 (4 x)$Î# "4 (4 x)# ‘ ! œ 43 (4)$Î# "4 (16) œ 32 3 4œ 3 cy 26. V œ 2 '0 'cÈ1cx# '0 dz dy dx œ 2 '0 'cÈ1cx# y dy dx œ '0 a1 x# b dx œ 23 1 0 2c2x 27. V œ '0 '0 1 1 0 1 '03c3xc3yÎ2 dz dy dx œ '01 '02 2x ˆ3 3x 3# y‰ dy dx œ '01 6(1 x)# 43 † 4(1 x)# ‘ dx œ '0 3(1 x)# dx œ c (1 x)$ d ! œ 1 1 " 1cx '0cos Ð1xÎ2Ñ dz dy dx œ '01 '01 x cos ˆ 1#x ‰ dy dx œ '01 ˆcos 1#x ‰ (1 x) dx 1 1 1Î2 " œ '0 cos ˆ 1#x ‰ dx '0 x cos ˆ 1#x ‰ dx œ 12 sin 1#x ‘ ! 14 '0 u cos u du œ 12 14 ccos u u sin ud 01Î2 28. V œ '0 '0 1 # # œ 12 14# ˆ 1# 1‰ œ 14# È1cx 29. V œ 8 '0 '0 1 4cx# 30. V œ '0 '0 2 # È È '0 1cx dz dy dx œ 8 '01 '0 1cx È1 x# dy dx œ 8 '01 a1 x# b dx œ 163 # # '04cx cydz dy dx œ '02 '04cx a4 x# yb dy dx œ '02 ’a4 x# b# "# a4 x# b# “ dx # # œ "# '0 a4 x# b dx œ '0 Š8 4x# x# ‹ dx œ 128 15 2 2 # c Î 31. V œ '0 '0 ˆÈ16 y# ‰ 2 4 % '04 ydx dz dy œ '04 '0 16 y Î2 (4 y) dz dy œ '04 È16#y (4 y) dy ˆÈ #‰ # % $Î# œ '0 2È16 y# dy "# '0 yÈ16 y# dy œ yÈ16 y# 16 sin" y4 ‘ ! ’ 6" a16 y# b “ 4 4 œ 16 ˆ 1# ‰ "6 (16)$Î# œ 81 32 3 È4cx 3cx 32. V œ 'c2 'cÈ4cx# '0 2 # È4cx dz dy dx œ 'c2 'cÈ4cx# (3 x) dy dx œ 2 'c2 (3 x)È4 x# dx 2 # 2 " œ 12 sin 2cx 33. '0 '0 2 " 1 12 sin (1) œ 12 ˆ 1# ‰ 12 ˆ 1# ‰ œ 121 # # ’ 23 a4 x# b $Î# # 'Ð24cx2xcyÑÎ2y2 dz dy dx œ '02 '02cx ˆ3 3x# 3y# ‰ dy dx œ '0 3 ˆ1 x# ‰ (2 x) 34 (2 x)# ‘ dx 2 œ '0 ’6 6x 3x# 3(24 x) “ dx 2 # # $ # œ ’6x 3x# x2 (24x) “ œ (12 12 4 0) 24 œ 2 $ ! 2 œ 3 'c2 2È4 x# dx 2 'c2 xÈ4 x# dx œ 3 ’xÈ4 x# 4 sin" x2 “ 2 % $ ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. “ # 907 908 Chapter 15 Multiple Integrals 34. V œ '0 'z 'z 4 8 8cz dx dy dz œ '0 'z (8 2z) dy dz œ '0 (8 2z)() z) dz œ '0 a64 24z 2z# b dz 4 8 4 4 % œ 64z 12z# 23 z$ ‘ ! œ 320 3 È4cx Î2 35. V œ 2 'c2 '0 2 # È '0x 2 dz dy dx œ 2 'c22 '0 4x Î2 (x 2) dy dx œ 'c22 (x 2)È4 x# dx # œ 'c2 2È4 x# dx 'c2 xÈ4 x# dx œ ’xÈ4 x# 4 sin" x# “ 2 2 œ 4 ˆ 1# ‰ 4 ˆ 1# ‰ œ 41 1cy# 36. V œ 2 '0 '0 1 # # ’ "3 a4 x# b $Î# # “ # '0x by dz dx dy œ 2 '01 '01cy ax# y# b dx dy œ 2'01 ’ x3 xy# “ 1cy dy # # # # $ 0 1 1 1 # œ 2 '0 a1 y# b ’ "3 a1 y# b y# “ dy œ 2 '0 a1 y# b ˆ 3" 3" y# 3" y% ‰ dy œ 23 '0 a1 y' b dy " ( œ 23 ’y y7 “ œ ˆ 23 ‰ ˆ 67 ‰ œ 47 ! 37. average œ "8 '0 '0 '0 ax# 9b dz dy dx œ 8" '0 '0 a2x# 18b dy dx œ 8" '0 a4x# 36b dx œ 31 3 2 2 2 2 2 2 38. average œ "2 '0 '0 '0 (x y z) dz dy dx œ 2" '0 '0 (2x 2y 2) dy dx œ 2" '0 (2x 1) dx œ 0 1 1 2 1 1 1 39. average œ '0 '0 '0 ax# y# z# b dz dy dx œ '0 '0 ˆx# y# "3 ‰ dy dx œ '0 ˆx# 23 ‰ dx œ 1 1 1 1 1 1 1 40. average œ "8 '0 '0 '0 xyz dz dy dx œ 4" '0 '0 xy dy dx œ 2" '0 x dx œ 1 2 2 2 2 ax b 41. '0 '0 '2y 4 cos dx dy dz œ '0 '0 '0 2È z 4 1 2 4 # xÎ2 2 2 2 4 cos ax# b dy dx dz œ 2È z '04 '02 x cosÈzax b dx dz œ '04 ˆ sin# 4 ‰ z"Î# dz # % œ (sin 4)z"Î# ‘ ! œ 2 sin 4 Èy 42. '0 '0 'x# 12xz ezy dy dx dz œ '0 '0 '0 1 1 1 # 1 1 12xz ezy dx dy dz œ '0 '0 6yz ezy dy dz œ '0 ’3ezy “ dz 1 # 1 # 1 # ! 1 œ 3'0 aez zb dz œ 3 cez 1d "! œ 3e 6 43. '0 'È$ z '0 1 1 ln 3 1e2x sin a1y# b dx dy dz œ y# " '01 'È1z 41 siny a1y b dy dz œ '01 '0y 41 siny a1y b dz dy $ # # $ # # œ '0 41y sin a1y# b dy œ c2 cos a1y# bd ! œ 2(1) 2(1) œ 4 1 4cx# 44. '0 '0 2 " È 2z 2z ‰ '0x sin ' 2 ' 4cx x4sin z2z dz dx œ '04 '0 4cz ˆ sin ' 4 ˆ sin 2z ‰ " 4 z dy dz dx œ 0 0 4 z x dx dz œ 0 4 z # (4 z) dz # % % # œ "4 cos 2z‘ ! œ 4" "# sin# z‘ ! œ sin# 4 4cacx# 'a4cx cydz dy dx œ 154 Ê '01 '04cacx a4 x# y ab dy dx œ 154 1 1 1 # # # 4 4 Ê '0 ’a4 a x# b "# a4 a x# b “ dx œ 15 Ê "# '0 a4 a x# b dx œ 15 Ê '0 c(4 a)# 2x# (4 a) x% d dx 45. '0 '0 1 # # & " 8 8 8 œ 15 Ê ’(4 a)# x 32 x$ (4 a) x5 “ œ 15 Ê (4 a)# 32 (4 a) "5 œ 15 Ê 15(4 a)# 10(4 a) 5 œ 0 ! Ê 3(4 a)# 2(4 a) 1 œ 0 Ê [3(4 a) 1][(4 a) 1] œ 0 Ê 4 a œ "3 or 4 a œ 1 Ê a œ 13 3 or a œ 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.6 Moments and Centers of Mass # 1 1 46. The volume of the ellipsoid xa# by# zc# œ 1 is 4abc so that 4(1)(2)(c) œ 81 Ê c œ 3. 3 3 # # 47. To minimize the integral, we want the domain to include all points where the integrand is negative and to exclude all points where it is positive. These criteria are met by the points (xß yß z) such that 4x# 4y# z# 4 Ÿ 0 or 4x# 4y# z# Ÿ 4, which is a solid ellipsoid centered at the origin. 48. To maximize the integral, we want the domain to include all points where the integrand is positive and to exclude all points where it is negative. These criteria are met by the points (xß yß z) such that 1 x# y# z# 0 or x# y# z# Ÿ 1, which is a solid sphere of radius 1 centered at the origin. 49-52. Example CAS commands: Maple: F := (x,y,z) -> x^2*y^2*z; q1 := Int( Int( Int( F(x,y,z), y=-sqrt(1-x^2)..sqrt(1-x^2) ), x=-1..1 ), z=0..1 ); value( q1 ); Mathematica: (functions and bounds will vary) Clear[f, x, y, z]; f:= x2 y2 z Integrate[f, {x,1,1}, {y,Sqrt[1 x2 ], Sqrt[1 x2 ]}, {z, 0, 1}] N[%] topolar={x Ä r Cos[t], y Ä r Sin[t]}; fp= f/.topolar //Simplify Integrate[r fp, {t, 0, 21}, {r, 0, 1},{z, 0, 1}] N[%] 15.6 MOMENTS AND CENTERS OF MASS 2cx# 1. M œ '0 'x 1 2cx# 3 dy dx œ 3'0 a2 x# xb dx œ 7# ; My œ '0 'x 1 1 2cx# œ 3'0 a2x x$ x# b dx œ 45 ; Mx œ '0 'x 1 1 3x dy dx œ 3 '0 cxyd x2cx dx 1 3y dy dx œ 3# '0 cy# d x 2cx# 1 # dx œ 3# '0 a4 5x# x% b dx œ 19 5 1 5 38 Ê x œ 14 and y œ 35 2. M œ $ '0 '0 dy dx œ $ '0 3 dx œ 9$ ; Ix œ $ '0 '0 y# dy dx œ $ '0 ’ y3 “ dx œ 27$ ; 3 3 3 Iy œ $ '0 '0 x dy dx œ $ ' 3 3 3 3 $ cx yd ! dx œ $ 0 # # 3 3 $ 0 '0 3x dx œ 27$ 3 3 # "' # ' ' 3. M œ '0 'y#Î2 dx dy œ '0 Š4 y y# ‹ dy œ 14 3 ; My œ 0 y# Î2 x dx dy œ # 0 cx d y# Î2 dy 2 4 y 2 2 # 4 y 2 4cy 4 y ' 'y#Î2 y dx dy œ '0 Š4y y# y# ‹ dy œ 103 œ "# '0 Š16 8y y# y4 ‹ dy œ 128 15 ; Mx œ 0 2 2 % 2 $ 5 Ê x œ 64 35 and y œ 7 3cx 4. M œ '0 '0 3 3cx dy dx œ '0 (3 x) dx œ 9# ; My œ '0 '0 3 3 x dy dx œ '0 cxyd 03cx dx œ '0 a3x x# b dx œ 9# 3 3 Ê x œ 1 and y œ 1, by symmetry Èa cx 5. M œ '0 '0 a # # Èa cx dy dx œ 1a4 ; My œ '0 '0 # a # # È x dy dx œ '0 cxyd 0 a cx dx œ '0 xÈa# x# dx œ a3 a # # a 4a Ê x œ y œ 31 , by symmetry Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. $ 909 910 Chapter 15 Multiple Integrals 6. M œ '0 '0 1 dy dx œ '0 sin x dx œ 2; Mx œ '0 '0 1 sin x 1 sin x œ "4 '0 (1 cos 2x) dx œ 14 Ê x œ 1# and y œ 18 1 È4cx 7. Ix œ 'c2 'cÈ4cx# y# dy dx œ 'c2 ’ y3 “ 2 # 2 $ y dy dx œ "# '0 cy# d 0 1 sin x dx œ "# '0 sin# x dx 1 È4cx 2 $Î# dx œ 32 'c2 a4 x# b dx œ 41; Iy œ 41, by symmetry; È c 4cx # # I o œ I x I y œ 81 8. Iy œ '1 '0 21 ˆsin# x‰Îx# x# dy dx œ '1 asin# x 0b dx œ "# '1 (1 cos 2x) dx œ 1# 21 21 9. M œ 'c_ '0 dy dx œ ' _ ex dx œ ex 0 œ 'b0 xex dx œ lim b Ä _ œ "# 'c_ e 0 _ b Ä _ 'b e 0 dx œ "4 bÄ_ 2 2 ycy# # Ix œ '0 'cy y (x y) dx dy œ ' 2 È3Î2 È12 4y 12. M œ 'cÈ3Î2 '4y# 2cx 13. M œ '0 'x 1 1 ycy# cy 2 È3Î2 5x dx dy œ 5 ' È3Î2 ’ x2 “ # abeb eb b œ 1; Mx œ 'c_ '0 y dy dx ex b Ä _ x"#Î2 1‘ b œ 1 e 0 y 8 dy œ '0 Š y2 2y$ 2y# ‹ dy œ ’ 10 y# 2y3 “ œ 15 ; 2 ycy# È12 4y # 4y# % & % $ # ! dy œ ' 2 ' 64 ; Š y2 2y& 2y% ‹ dy œ 105 0 È3Î2 dy œ 5# ' È3Î2 a12 4y# 16y% b dy œ 23È3 2cx (6x 3y 3) dy dx œ '0 6xy 3# y# 3y‘ x dx œ '0 a12 12x# b dx œ 8; 2cx My œ '0 'x # 0 Ê x œ 1 and y œ "4 # # ’ x 2y xy$ “ 0 cy # b Ä _ ex 0 bÄ_ 11. M œ '0 'cy (x y) dx dy œ '0 ’ x2 xy“ eb œ 1; My œ ' _ '0 x dy dx œ ' _ xex dx 0 lim lim '0b xe x#Î2 dx œ lim x dy dx œ lim ycy# b Ä _ 2x # e x Î2 10. My œ '0 '0 lim cxex ex d b0 œ 1 lim dx œ "# lim b Ä _ 2x 'b0 ex dx œ 1 0 1 1 2cx x(6x 3y 3) dy dx œ '0 a12x 12x$ b dx œ 3; Mx œ '0 'x 1 1 y(6x 3y 3) dy dx 3 17 œ '0 a14 6x 6x# 2x$ b dx œ 17 # Ê x œ 8 and y œ 16 1 2ycy# 14. M œ '0 'y# 1 2ycy# My œ '0 'y# 1 2ycy# (y 1) dx dy œ '0 a2y 2y$ b dy œ "# ; Mx œ '0 'y# 1 1 4 y(y 1) dx dy œ '0 a2y# 2y% b dy œ 15 ; 1 2ycy# 8 8 x(y 1) dx dy œ '0 a2y# 2y% b dy œ 154 Ê x œ 15 and y œ 15 ; Ix œ '0 'y# 1 1 y# (y 1) dx dy œ 2 '0 ay$ y& b dy œ "6 1 15. M œ '0 '0 (x y 1) dx dy œ '0 (6y 24) dy œ 27; Mx œ '0 '0 y(x y 1) dx dy œ '0 y(6y 24) dy œ 14; 1 6 1 1 6 1 14 ' ' # My œ '0 '0 x(x y 1) dx dy œ '0 (18y 90) dy œ 99 Ê x œ 11 3 and y œ 27 ; Iy œ 0 0 x (x y 1) dx dy 1 6 1 1 6 ‰ œ 216 '0 ˆ y3 11 6 dy œ 432 1 32 16. M œ 'c1 'x# (y 1) dy dx œ 'c1 Š x# x# 3# ‹ dx œ 15 ; Mx œ 'c1 'x# y(y 1) dy dx œ 'c1 Š 56 x3 x# ‹ dx 1 1 1 1 % 1 1 ' % 9 $ ' ' ' 3x x ' ' # œ 48 35 ; My œ c1 x# x(y 1) dy dx œ c1 Š # # x ‹ dx œ 0 Ê x œ 0 and y œ 14 ; Iy œ c1 x# x (y 1) dy dx 1 1 1 & 16 œ 'c1 Š 3x2 x2 x% ‹ dx œ 35 1 # ' Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 1 1 Section 15.6 Moments and Centers of Mass 911 13 ' ' ' 7x x 17. M œ 'c1 '0 (7y 1) dy dx œ 'c1 Š 7x# x# ‹ dx œ 31 15 ; Mx œ c1 0 y(7y 1) dy dx œ c1 Š 3 2 ‹ dx œ 15 ; x# 1 1 x# 1 % 1 ' % ' ' # My œ 'c1 '0 x(7y 1) dy dx œ 'c1 Š 7x# x$ ‹ dx œ 0 Ê x œ 0 and y œ 13 31 ; Iy œ c1 0 x (7y 1) dy dx x# 1 1 x# 1 & œ 'c1 Š 7x# x% ‹ dx œ 75 1 ' x ‰ x ‰ x ‰ 18. M œ '0 'c1 ˆ1 20 dy dx œ '0 ˆ2 10 dx œ 60; Mx œ '0 'c1 y ˆ1 20 dy dx œ '0 ’ˆ1 #x0 ‰ Š y# ‹“ 20 1 20 20 1 20 # " " dx œ 0; x ‰ x ' 'c1 y# ˆ1 20x ‰ dy dx My œ '0 'c1 x ˆ1 20 dy dx œ '0 Š2x 10 Ê x œ 100 ‹ dx œ 2000 3 9 and y œ 0; Ix œ 0 20 1 20 20 # 1 x ‰ œ 23 '0 ˆ1 20 dx œ 20 20 19. M œ '0 'cy (y 1) dx dy œ '0 a2y# 2yb dy œ 53 ; Mx œ '0 'cy y(y 1) dx dy œ 2 ' ay$ y# b dy œ 76 ; 1 y 1 1 1 y 0 7 My œ '0 'cy x(y 1) dx dy œ '0 0 dy œ 0 Ê x œ 0 and y œ 10 ; Ix œ '0 'cy y# (y 1) dx dy œ '0 a2y% 2y$ b dy 1 y 1 1 y 1 9 3 œ 10 ; Iy œ '0 'cy x# (y 1) dx dy œ 3" '0 a2y% 2y$ b dy œ 10 Ê Io œ Ix Iy œ 56 1 y 1 20. M œ '0 'cy a3x# 1b dx dy œ '0 a2y$ 2yb dy œ 3# ; Mx œ '0 'cy y a3x# 1b dx dy œ '0 a2y% 2y# b dy œ 16 15 ; 1 y 1 1 y 1 5 & $ ' ' # # ' My œ '0 'cy x a3x# 1b dx dy œ 0 Ê x œ 0 and y œ 32 45 ; Ix œ 0 cy y a3x 1b dx dy œ 0 a2y 2y b dy œ 6 ; 1 y 1 y 1 6 Iy œ '0 'cy x# a3x# 1b dx dy œ 2 '0 ˆ 35 y& 3" y$ ‰ dy œ 11 30 Ê Io œ Ix Iy œ 5 1 y 1 21. Ix œ '0 '0 '0 ay# z# b dz dy dx œ '0 '0 Šcy# c3 ‹ dy dx œ '0 Š cb3 c3b ‹ dx œ abc ab3 c b a b c a b a $ $ # $ # œ M3 ab# c# b where M œ abc; Iy œ M3 aa# c# b and Iz œ M3 aa# b# b , by symmetry Ð4 2yÑÎ3 22. The plane z œ 4 3 2y is the top of the wedge Ê Ix œ 'c3 'c2 'c4Î3 3 4 ay# z# b dz dy dx Ð4 2yÑÎ3 y) 64 ' ' ' œ 'c3 'c2 ’ 8y3 2y3 8(281 81 “ dy dx œ 'c3 104 3 dx œ 208; Iy œ c3 c2 c4Î3 3 4 # $ 3 $ 3 4 ax# z# b dz dy dx 64 ‰ œ 'c3 'c2 ’ (4 812y) x (4 3 2y) 4x3 81 “ dy dx œ 'c3 ˆ12x# 32 3 dx œ 280; 3 4 $ Ð4 2yÑÎ3 Iz œ 'c3 'c2 'c4Î3 3 4 # 3 # ' 3 ax# 2b dx œ 360 ‰ ax# y# b dz dy dx œ ' 3 ' 2 ax# y# b ˆ 83 2y 3 dy dx œ 12 3 4 3 ' ' ' 23. M œ 4 '0 '0 '4y# dz dy dx œ 4 '0 '0 a4 4y# b dy dx œ 16 '0 23 dx œ 32 3 ; Mxy œ 4 0 0 4y# z dz dy dx 1 1 4 1 1 1 1 1 4 12 '0 dx œ 128 œ 2 '0 '0 a16 16y% b dy dx œ 128 5 5 Ê z œ 5 , and x œ y œ 0, by symmetry; 1 1 1 64y 7904 % ' 1976 ‰ Ix œ 4 '0 '0 '4y# ay# z# b dz dy dx œ 4 '0 '0 ’ˆ4y# 64 3 Š4y 3 ‹“ dy dx œ 4 0 105 dx œ 105 ; 1 1 4 1 1 1 ' 64y # # ' ˆ 8 # 128 ‰ dx ‰ Iy œ 4 '0 '0 '4y# ax# z# b dz dy dx œ 4 '0 '0 ’ˆ4x# 64 3 Š4x y 3 ‹“ dy dx œ 4 0 3 x 7 1 1 4 1 1 1 ' ' ' ' # # ' ' # ## # % œ 4832 63 ; Iz œ 4 0 0 4y# ax y b dz dy dx œ 16 0 0 ax x y y y b dy dx 1 1 4 1 1 2 œ 16 '0 Š 2x3 15 ‹ dx œ 256 45 1 # ŠÈ4 x# ‹Î2 24. (a) M œ 'c2 'ŠcÈ4cx#‹Î2 '0 2 ŠÈ4 x# ‹Î2 2 x Myz œ 'c2 'ŠcÈ4cx#‹Î2 '0 2 ŠÈ4 x# ‹Î2 dz dy dx œ ' 2 'Š È4 x#‹Î2 (2 x) dy dx œ ' 2 (2 x) ŠÈ4 x# ‹ dx œ 41; 2 x 2 ŠÈ4 x# ‹Î2 2 x dz dy dx œ ' 2 'Š È4 x#‹Î2 x(2 x) dy dx œ ' 2 x(2 x) ŠÈ4 x# ‹ dx œ 21; 2 2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 912 Chapter 15 Multiple Integrals ŠÈ4 x# ‹Î2 Mxz œ 'c2 'ŠcÈ4cx#‹Î2 '0 2 2 x ŠÈ4 x# ‹Î2 y dz dy dx œ ' 2 'Š È4 x#‹Î2 y(2 x) dy dx 2 œ "# 'c2 (2 x) ’ 44x 44x “ dx œ 0 Ê x œ "# and y œ 0 2 # # ŠÈ4 x# ‹Î2 (b) Mxy œ 'c2 'ŠcÈ4cx#‹Î2 '0 2 2 x ŠÈ4 x# ‹Î2 z dz dy dx œ "# ' 2 'Š È4 x# ‹Î2 (2 x)# dy dx œ "# ' 2 (2 x)# ŠÈ4 x# ‹ dx 2 2 œ 51 Ê z œ 54 È4cx 25. (a) M œ 4 '0 '0 2 # 'x4 y dz dy dx œ 4 '01Î2 '02 'r 4 r dz dr d) œ 4 '01Î2 '02 a4r r$ b dr d) œ 4 '01Î2 4 d) œ 81; # # # '0 d) œ 6431 Ê z œ 83 , and x œ y œ 0, by symmetry Mxy œ '0 '0 'r# zr dz dr d) œ '0 '0 #r a16 r% b dr d) œ 32 3 21 2 21 4 Èc (b) M œ 81 Ê 41 œ '0 '0 21 21 2 È 'r c r dz dr d) œ '021 '0 c acr r$ b dr d) œ '021 c4 d) œ c#1 Ê c# œ 8 Ê c œ 2È2, # # # since c 0 26. M œ 8; Mxy œ 'c1 '3 'c1 z dz dy dx œ 'c1 '3 ’ z2 “ 1 5 1 1 5 # dy dx œ 0; Myz œ 'c1 '3 'c1 x dz dy dx " 1 " 5 1 œ 2 'c1 '3 x dy dx œ 4 'c1 x dx œ 0; Mxz œ 'c1 '3 'c1 y dz dy dx œ 2 'c1 '3 y dy dx œ 16 'c1 dx œ 32 1 5 1 1 5 1 1 5 1 Ê x œ 0, y œ 4, z œ 0; Ix œ 'c1 '3 'c1 ay# z# b dz dy dx œ 'c1 '3 ˆ2y# 23 ‰ dy dx œ 32 'c1 100 dx œ 400 3 ; 1 5 1 1 5 1 Iy œ 'c1 '3 'c1 ax# z# b dz dy dx œ 'c1 '3 ˆ2x# 23 ‰ dy dx œ 43 'c1 a3x# 1b dx œ 16 3 ; 1 5 1 1 5 1 400 ‰ Iz œ 'c1 '3 'c1 ax# y# b dz dy dx œ 2 'c1 '3 ax# y# b dy dx œ 2 'c1 ˆ2x# 98 3 dx œ 3 1 5 1 1 5 1 Ð2 yÑÎ2 27. The plane y 2z œ 2 is the top of the wedge Ê IL œ 'c2 'c2 'c1 2 4 c(y 6)# z# d dz dy dx 49 # œ 'c2 'c2 ’ (y 6)#(4 y) (2 24y) "3 “ dy dx; let t œ 2 y Ê IL œ 4 'c2 Š 13t 24 5t 16t 3 ‹ dt œ 1386; 2 4 # 4 $ $ M œ "# (3)(6)(4) œ 36 Ð2 yÑÎ2 28. The plane y 2z œ 2 is the top of the wedge Ê IL œ 'c2 'c2 'c1 2 4 c(x 4)# y# d dz dy dx œ "# 'c2 'c2 ax# 8x 16 y# b (4 y) dy dx œ 'c2 a9x# 72x 162b dx œ 696; M œ "# (3)(6)(4) œ 36 2 4 2 2cx '02cxcy 2x dz dy dx œ '02 '02cx a4x 2x# 2xyb dy dx œ '02 ax$ 4x# 4xb dx œ 43 2 2cx 2cxcy 2 2cx 2 8 8 (b) Mxy œ '0 '0 '0 2xz dz dy dx œ '0 '0 x(2 x y)# dy dx œ '0 x(23 x) dx œ 15 ; Mxz œ 15 by 2 2cx 2cxcy 2 2cx 2 # symmetry; Myz œ '0 '0 '0 2x# dz dy dx œ '0 '0 2x# (2 x y) dy dx œ '0 a2x x# b dx œ 16 15 29. (a) M œ '0 '0 2 $ Ê x œ 45 , and y œ z œ 25 Èx 30. (a) M œ '0 '0 2 2 # Èx (b) Myz œ '0 '0 È '04cx kxy dz dy dx œ k'02 '0 x xy a4 x# b dy dx œ k# '02 a4x# x% b dx œ 32k 15 È '04cx kx# y dz dy dx œ k '02 '0 x x# y a4 x# b dy dx œ k# '02 a4x$ x& b dx œ 8k3 # Èx Ê x œ 54 ; Mxz œ '0 '0 2 È '04cx kxy# dz dy dx œ k'02 '0 x xy# a4 x# b dy dx œ k3 '02 ˆ4x&Î# x*Î# ‰ dx # Èx œ 2562312k Ê y œ 4077 2 ; Mxy œ '0 '0 È È 2 È '04cx kxyz dz dy dx œ '02 '0 x xy a4 x# b# dy dx # 8 œ k4 '0 a16x# 8x% x' b dx œ 256k 105 Ê z œ 7 2 31. (a) M œ '0 '0 '0 (x y z 1) dz dy dx œ '0 '0 ˆx y 3# ‰ dy dx œ '0 (x 2) dx œ 5# 1 1 1 1 1 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.6 Moments and Centers of Mass 4 ‰ (b) Mxy œ '0 '0 '0 z(x y z 1) dz dy dx œ "# '0 '0 ˆx y 53 ‰ dy dx œ "# '0 ˆx 13 6 dx œ 3 1 1 1 1 1 1 8 Ê Mxy œ Myz œ Mxz œ 43 , by symmetry Ê x œ y œ z œ 15 (c) Iz œ '0 '0 '0 ax# y# b (x y z 1) dz dy dx œ '0 '0 ax# y# b ˆx y 3# ‰ dy dx 1 1 1 1 1 11 œ '0 ˆx$ 2x# "3 x 43 ‰ dx œ 11 6 Ê Ix œ Iy œ Iz œ 6 , by symmetry 1 32. The plane y 2z œ 2 is the top of the wedge. Ð2 yÑÎ2 (a) M œ 'c1 'c2 'c1 1 4 4 1 Ð2 yÑÎ2 (b) Myz œ 'c1 'c2 'c1 1 (x 1) dz dy dx œ 'c1 'c2 (x 1) ˆ2 y# ‰ dy dx œ 18 Ð2 yÑÎ2 Mxz œ 'c1 'c2 'c1 1 4 Ð2 yÑÎ2 Mxy œ 'c1 'c2 'c1 1 4 Ð2 yÑÎ2 (c) Ix œ 'c1 'c2 'c1 1 4 Ð2 yÑÎ2 Iy œ 'c1 'c2 'c1 1 4 Ð2 yÑÎ2 Iz œ 'c1 'c2 'c1 1 4 1cz Èz 4 x(x 1) dz dy dx œ 'c1 'c2 x(x 1) ˆ2 y# ‰ dy dx œ 6; 1 4 y(x 1) dz dy dx œ 'c1 'c2 y(x 1) ˆ2 y# ‰ dy dx œ 0; 1 4 z(x 1) dz dy dx œ "# 'c1 'c2 (x 1) Š y4 y‹ dy dx œ 0 Ê x œ 13 , and y œ z œ 0 1 4 # (x 1) ay# z# b dz dy dx œ 'c1 'c2 (x 1) ’2y# "3 y# 3" ˆ" 2y ‰ “ dy dx œ 45; 1 4 $ $ (x 1) ax# z# b dz dy dx œ 'c1 'c2 (x 1) ’2x# "3 x#y 3" ˆ" y2 ‰ “ dy dx œ 15; 1 4 $ # (x 1) ax# y# b dz dy dx œ 'c1 'c2 (x 1) ˆ2 y# ‰ ax# y# b dy dx œ 42 1 4 1cz 33. M œ '0 'zc1 '0 (2y 5) dy dx dz œ '0 'zc1 ˆz 5Èz‰ dx dz œ '0 2 ˆz 5Èz‰ (1 z) dz 1 1 1 $Î# œ 2 '0 ˆ5z"Î# z 5z$Î# z# ‰ dz œ 2 10 "# z# 2z&Î# 3" z$ ‘ ! œ 2 ˆ 93 3# ‰ œ 3 3 z " 1 È4cx 16c2 ˆx# by# ‰ 34. M œ 'c2 'cÈ4cx# '2 ax#by# b 2 # È ' 4cx Èx# y# c16 4 ax# y# bd dy dx c2 cÈ4cx Èx# y# dz dy dx œ ' 2 # # œ 4 '0 '0 r a4 r# b r dr d) œ 4 '0 ’ 4r3 r5 “ d) œ 4 '0 21 21 2 $ & # 21 ! 64 5121 15 d) œ 15 35. (a) x œ Myz œ 0 Ê ' ' ' x$ (xß yß z) dx dy dz œ 0 Ê Myz œ 0 M R (b) IL œ ' ' ' kv hik# dm œ ' ' ' k(x h) i yjk# dm œ ' ' ' ax# 2xh h# y# b dm D D D œ ' ' ' ax# y# b dm 2h ' ' ' x dm h# ' ' ' dm œ Ix 0 h# m œ Ic m h# m Þ D D Þ D 36. IL œ Ic m mh# œ 25 ma# ma# œ 75 ma# Þ Þ # # # 37. (a) (xß yß z) œ ˆ #a ß #b ß #c ‰ Ê Iz œ Ic m abc ŠÉ a4 b4 ‹ Ê IcÞmÞ œ Iz abc aa4 b b # Þ # # # # # Þ # # b œ abc aa3 b b abc aa4 b b œ abc aa1# b b ; RcÞmÞ œ É IcMÞmÞ œ É a 12 # # # # # (b) IL œ IcÞmÞ abc ŒÉ a4 ˆ b# 2b‰ œ abc aa12 b b abc aa 4 9b b œ abc a4a1# 28b b # # IL œ abc aa 3 7b b ; RL œ É M œ É a 37b Ð4 2yÑÎ3 38. M œ 'c3 'c2 'c4Î3 3 4 # # # # # # # dz dy dx œ ' 3 ' 2 23 (4 y) dy dx œ ' 3 23 ’4y y2 “ 3 4 # 3 # # % # dx œ 12 ' 3 dx œ 72; 3 x œ y œ z œ 0 from Exercise 22 Ê Ix œ IcÞmÞ 72 ŠÈ0# 0# ‹ œ IcÞmÞ Ê IL œ IcÞmÞ 72 ŠÉ16 16 9 ‹ ‰ œ 1488 œ 208 72 ˆ 160 9 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. # 913 914 Chapter 15 Multiple Integrals 15.7 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 1. È '021 '01 'r 2cr dz r dr d) œ '021 '01 ’r a2 r# b"Î# r# “ dr d) œ '021 ’ "3 a2 r# b$Î# r3 “ " d) # $ ! œ '0 Š 2 3 23 ‹ d) œ 21 2. È # $ % ! 91 Š8È2 7‹ 2 '021 '0 Î2 '03 24r dz r dr d) œ '02 '0 Î2 a3r 24r$ b dr d) œ '02 32 r# 6r% ‘ !Î2 d) œ 3# '02 Š 4)1 164)1 ‹ d) ) # 1 1 & œ 3# ’ 12)1# 20)1% “ È 21 ! ) 1 1 ) 1 1 # % # % œ 1751 '01 '0 Î ' 3È44 rr z dz r dr d) œ '0 '0 Î "# c9 a4 r# b a4 r# bd r dr d) œ 4 '0 '0 Î a4r r$ b dr d) ) 1 # 1 ) 1 1 ) 1 # œ 4 '0 ’2r# r4 “ 1 % '021 '01 'r 2cr ˆ 5. 3 # $ 4. 41 ŠÈ2 "‹ '021 '03 'r Î318cr dz r dr d) œ '021 '03 ’r a18 r# b"Î# r3 “ dr d) œ '021 ’ "3 a18 r# b$Î# 12r “ $ d) œ 3. $Î# #‰ "Î# ) Î1 ! œ 4 '0 Š 21)# 4)1% ‹ d) œ 37151 1 # % 3 dz r dr d) œ 3 '0 '0 ’r a2 r# b 21 1 "Î# r# “ dr d) œ 3 '0 ’ a2 r# b 21 "Î# $ " r3 “ d) 21 œ 3 '0 ŠÈ2 43 ‹ d) œ 1 Š6È2 8‹ ! 6. '021 '01 'c11ÎÎ22 ar# sin# ) z# b dz r dr d) œ '021 '01 ˆr$ sin# ) 12r ‰ dr d) œ '021 Š sin4 ) 24" ‹ d) œ 13 7. 21 z 3 '021 '03 '0zÎ3 r$ dr dz d) œ '021 '03 324 dz d) œ '0 20 d) œ 3101 8. 'c11 '021 '01bcos 4r dr d) dz œ 'c11 '02 2(1 cos ))# d) dz œ 'c11 61 d) œ 121 9. '01 '0 z '021 ar# cos# ) z# b r d) dr dz œ '01 '0 z ’ r2) r sin4 2) z# )“ #1 r dr dz œ '01 '0 z a1r$ 21rz# b dr dz # % ) 1 È È œ '0 ’ 14r 1r# z# “ 1 È4cr 10. '0 'rc2 2 % # Èz # È # dz œ '0 Š 14z 1z$ ‹ dz œ ’ 112z 14z “ œ 13 1 # $ % " ! ! ! È '021 (r sin ) 1) r d) dz dr œ '02 'rc24cr 21r dz dr œ 21'02 ’r a4 r# b"Î# r# 2r“ dr $Î# œ 21 ’ "3 a4 r# b $ # # r3 r# “ œ 21 38 4 3" (4)$Î# ‘ œ 81 ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.7 Triple Integrals in Cylindrical and Spherical Coordinates È4cr 11. (a) '0 '0 '0 21 1 È3 # dz r dr d) È (b) '0 '0 '01 r dr dz d) '021 'È23 '0 4cz r dr dz d) (c) '0 '0 # 21 È4cr 1 12. (a) '0 '0 'r 21 1 '021 r d) dz dr 2cr# dz r dr d) È2cz (b) '0 '0 '0 r dr dz d) '0 '1 '0 21 (c) '0 'r 1 z 2cr# '0 r d) dz dr 1Î2 cos ) 1Î2 1 1 # Î 19. '0 1Î4 Î Î r$ dz dr d) œ ' 1Î2 '0 r% cos ) dr d) œ "5 ' 1Î2 cos ) d) œ 25 1 2 1 2 1 '04cr sin f(rß )ß z) dz r dr d) 1cos ) '04 f(rß )ß z) dz r dr d) 21. '0 '0 '0 1 1 2 sin 9 Î 3 cos ) 1 2 Î 2 cos ) Î csc ) 18. ' 1Î2 'cos ) '0sec '02 r sin f(rß )ß z) dz r dr d) ) 1 2 16. ' 1Î2 '0 ) 17. 'c1Î2 '1 1 2 r dr dz d) 21 r cos ) 2 sin ) 2 '03r f(rß )ß z) dz r dr d) 14. 'c1Î2 '0 '0 15. '0 '0 21 1 13. 'c1Î2 '0 # 20. '1Î4 '0 ) 1 2 '05 r cos f(rß )ß z) dz r dr d) ) '03 r sin f(rß )ß z) dz r dr d) ) '02 r sin f(rß )ß z) dz r dr d) ) 3# sin 9 d3 d9 d) œ 83 '0 '0 sin% 9 d9 d) œ 83 '0 Š’ sin 94cos 9 “ 34 '0 sin# 9 d9‹ d) 1 1 1 1 $ ! 1 1 1 1 1 œ 2 '0 '0 sin# 9 d9 d) œ '0 ) sin#2) ‘ ! d) œ '0 1 d) œ 1# 22. '0 '0 21 1Î4 '02 (3 cos 9) 3# sin 9 d3 d9 d) œ '021 '01Î4 4 cos 9 sin 9 d9 d) œ '021 c2 sin# 9d 1! Î% d) œ '021 d) œ 21 Ð1 cos 9ÑÎ2 23. '0 '0 '0 21 1 1 " ' ' " ' 3# sin 9 d3 d9 d) œ 24 (1 cos 9)$ sin 9 d9 d) œ 96 c(1 cos 9)% d ! d) 0 0 0 21 1 21 1 " ' " 1 ' œ 96 a2% 0b d) œ 16 96 0 d) œ 6 (21) œ 3 0 21 24. '0 31Î2 œ 56 21 '01 '01 53$ sin$ 9 d3 d9 d) œ 54 '031Î2 '01 sin$ 9 d9 d) œ 54 '031Î2 Š’ sin 93cos 9 “ 1 23 '01 sin 9 d9‹ d) # '031Î2 c cos 9d 1! d) œ 53 '031Î2 d) œ 5#1 ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 915 916 Chapter 15 Multiple Integrals 25. '0 '0 21 1Î3 'sec2 9 33# sin 9 d3 d9 d) œ '021 '01Î3 a8 sec$ 9b sin 9 d9 d) œ '021 8 cos 9 "2 sec# 9‘ !1Î$ d) œ '0 (4 2) ˆ8 "# ‰‘ d) œ 5# '0 d) œ 51 21 21 26. '0 '0 21 1Î4 '0sec 9 3$ sin 9 cos 9 d3 d9 d) œ "4 '021 '01Î4 tan 9 sec# 9 d9 d) œ "4 '021 "2 tan# 9‘ !1Î% d) œ "8 '021 d) œ 14 1Î2 27. '0 'c1 '1Î4 3$ sin 29 d9 d) d3 œ '0 ' 1 3$ cos229 ‘ 1Î% d) d3 œ '0 ' 1 3# d) d3 œ '0 3#1 d3 œ ’ 138 “ œ 21 2 0 2 1Î# 0 2 0 2 $ $ % # ! 28. '1Î6 'csc 9 '021 3# sin 9 d) d3 d9 œ 21 '11ÎÎ63 'csc2 csc9 9 3# sin 9 d3 d9 œ 231 '11ÎÎ63 c3$ sin 9d csc2 csc9 9 d9 œ 1431 '11ÎÎ63 csc# 9 d9 œ 28È1 29. '0 '0 '0 123 sin$ 9 d9 d) d3 œ '0 '0 Œ123 ’ sin 39 cos 9 “ 1Î3 2 csc 9 1 1 3 1Î4 1 1 # 1Î% ! 83 '0 sin 9 d9 d) d3 1 Î4 23 3 103 53 # ' œ '0 '0 Š È 83 ccos 9d ! ‹ d) d3 œ '0 '0 Š83 10 È2 ‹ d) d3 œ 1 0 Š83 È2 ‹ d3 œ 1 ’43 È2 “ 2 1 1 œ 1 1 1Î% 1 # " ! Š4È2 5‹ 1 È2 1 Î2 1 Î2 1Î2 1Î2 1Î2 1 Î2 30. '1Î6 ' 1Î2 'csc 9 53% sin$ 9 d3 d) d9 œ '1Î6 ' 1Î2 a32 csc& 9b sin$ 9 d) d9 œ '1Î6 ' 1Î2 a32 sin$ 9 csc# 9b d) d9 2 œ 1 '1Î6 a32 sin$ 9 csc# 9b d9 œ 1 ’ 32 sin 39 cos 9 “ 1Î2 # È 1Î# 1Î' 6431 '1Î6 sin 9 d9 1 ccot 9d 1Î' 1Î2 1Î# È3 È È ˆ 6431 ‰ Š #3 ‹ œ 3313 3 œ 111È3 3 1 1Î# œ 1 Š 3224 3 ‹ 6431 ccos 9d 1Î' 1 ŠÈ3‹ œ 31. (a) x# y# œ 1 Ê 3# sin# 9 œ 1, and 3 sin 9 œ 1 Ê 3 œ csc 9; thus '021 '01Î6 '02 3# sin 9 d3 d9 d) '021 '11ÎÎ62 '0csc 9 3# sin 9 d3 d9 d) sin " Ð1Î3Ñ (b) '0 '1 '1Î6 21 2 3# sin 9 d9 d3 d) '0 '0 '0 21 1Î6 2 3# sin 9 d9 d3 d) 32. (a) '0 '0 '0sec 9 3# sin 9 d3 d9 d) 21 1 1Î4 (b) '0 '0 '0 3# sin 9 d9 d3 d) 21 1Î4 È2 '0 '1 21 33. V œ '0 '0 21 1Î2 'cos1Î4" Ð"Î3Ñ 3# sin 9 d9 d3 d) 'cos2 9 3# sin 9 d3 d9 d) œ "3 '021 '01Î2 a8 cos$ 9b sin 9 d9 d) œ 3" '0 ’8 cos 9 cos4 9 “ 21 % 1Î# ! 311 ‰ d) œ 3" '0 ˆ8 4" ‰ d) œ ˆ 31 12 (21) œ 6 21 '11 cos 9 3# sin 9 d3 d9 d) œ "3 '021 '01Î2 a3 cos 9 3 cos# 9 cos$ 9b sin 9 d9 d) 21 21 1Î# 111 ' 21 ˆ 11 ‰ œ 3" '0 3# cos# 9 cos$ 9 14 cos% 9‘ ! d) œ 3" '0 ˆ 32 1 "4 ‰ d) œ 11 12 0 d) œ 12 (21) œ 6 34. V œ '0 '0 21 1Î2 1ccos 9 35. V œ '0 '0 '0 21 " œ 12 (2)% 1 9) 3# sin 9 d3 d9 d) œ "3 '0 '0 (1 cos 9)$ sin 9 d9 d) œ 3" '0 ’ (" cos “ d) 4 21 1 21 % '021 d) œ 34 (21) œ 831 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 1 ! 3 Section 15.7 Triple Integrals in Cylindrical and Spherical Coordinates 1Î# 9) '01 cos 9 3# sin 9 d3 d9 d) œ "3 '021 '01Î2 (1 cos 9)$ sin 9 d9 d) œ 3" '021 ’ (" cos d) “ 4 36. V œ '0 '0 21 " œ 12 1Î2 % ! '021 d) œ 12" (21) œ 16 37. V œ '0 '1Î4 '0 21 917 1Î2 3# sin 9 d3 d9 d) œ 83 '0 '1Î4 cos$ 9 sin 9 d9 d) œ 83 '0 ’ cos4 9 “ 2 cos 9 21 1Î2 21 % " ‰' œ ˆ 83 ‰ ˆ 16 d) œ "6 (21) œ 13 0 1Î# 1Î% d) 21 38. V œ '0 '1Î3 '0 3# sin 9 d3 d9 d) œ 83 '0 '1Î3 sin 9 d9 d) œ 83 '0 c cos 9d 1Î$ d) œ 43 '0 d) œ 831 21 1Î2 39. (a) 8'0 1Î2 È4cx 2 1Î2 (c) '0 1Î2 1Î2 21 '01Î2 '02 3# sin 9 d3 d9 d) (c) 8 '0 '0 40. (a) '0 21 2 # È (b) 8'0 1Î2 '0 4cx cy dz dy dx # È È 21 1Î# È '02 '0 4cr dz r dr d) # # 1Î2 '03Î 2 'r 9 r dz r dr d) (b) '0 # '01Î4 '03 3# sin 9 d3 d9 d) È '01Î4 '03 3# sin 9 d3 d9 d) œ 9 '01Î2 '01Î4 sin 9 d9 d) œ 9 '01Î2 Š È" 1‹ d) œ 91 Š2 4 2‹ 2 41. (a) V œ '0 '0 21 1Î3 'sec2 9 3# sin 9 d3 d9 d) È3cx È3 È4cx cy (c) V œ 'cÈ3 'cÈ3cx# '1 # È3 # (d) V œ '0 '0 ’r a4 r# b 21 21 È1cr 42. (a) Iz œ '0 '0 '0 21 1 # '1 # dz r dr d) # "Î# dz dy dx r“ dr d) œ '0 ” a4 3r b 21 # $Î# # r# • '021 d) œ 531 œ 56 È3 È4cr (b) V œ '0 '0 È$ ! d) œ '0 Š 3" #3 4 3 ‹ d) 21 $Î# r# dz r dr d) (b) Iz œ '0 '0 '01 a3# sin# 9b a3# sin 9b d3 d9 d), since r# œ x# y# œ 3# sin# 9 cos# ) 3# sin# 9 sin# ) œ 3# sin# 9 (c) Iz œ '0 '0 " " $ 5 sin 9 d9 d) œ 5 21 1Î2 21 1Î2 '021 Œ’ sin 93cos 9 “ 1Î# 32 '01Î2 sin 9 d9 d) œ 152 '021 c cos 9d !1Î# d) # ! 2 œ 15 (21) œ 4151 43. V œ 4 '0 '01 'r 4 14r dz r dr d) œ 4 '01Î2 '01 a5r 4r$ r& b dr d) œ 4 '01Î2 ˆ 5# 1 "6 ‰ d) œ 4 '01Î2 d) œ 831 44. V œ 4'0 '01 ' 1Èr1 r dz r dr d) œ 4 '01Î2 '01 Šr r# rÈ1r# ‹ dr d) œ 4 '01Î2 ’ r2 r3 "3 a1 r# b$Î# “ " d) 1Î2 1Î2 œ 4 '0 # % # $ # 1Î2 ' 1Î2 ˆ "# "3 "3 ‰ d) œ 2 0 45. V œ '31Î2 '0 21 3 cos ) ! d) œ 2 ˆ 1# ‰ œ 1 '0cr sin dz r dr d) œ '321Î2 '03 cos r# sin ) dr d) œ '321Î2 a9 cos$ )b (sin )) d) œ 94 cos% )‘ #$11Î# ) 1 ) 1 œ 94 0 œ 94 c3 cos ) 46. V œ 2 '1Î2 '0 1 # '0r dz r dr d) œ 2 '1Î2 '0c3 cos r# dr d) œ 23 '1Î2 27 cos$ ) d) œ 18 Œ’ cos )3 sin ) “ 1 1 1Î# ) 1 32 '1Î2 cos ) d) œ 12 csin )d 11Î# œ 12 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 918 Chapter 15 Multiple Integrals 47. V œ '0 1Î2 È '0sin '0 1 r dz r dr d) œ '0 Î2 '0sin rÈ1r# dr d) œ '0 Î2 ’ "3 a1 r# b$Î# “ sin d) ) # 1 œ "3 '0 ’a1 sin# )b 1Î2 1Î# 1Î2 ! $Î# 1“ d) œ "3 '0 acos$ ) 1b d) œ "3 Œ’ cos )3 sin ) “ 1 Î2 È ) # 1 œ '0 ’ a1 cos )b # 1Î# œ 1# 32 ccos )d ! $Î# ) 1Î# ! 32 '0 cos ) d) 3) ‘ ! 1 Î2 1Î# 21 21Î3 ) 1 1“ d) œ '0 a1 sin 1Î2 $ ! '01Î2 sin ) d) 1Î# # )b d) œ ’) sin )3cos ) “ 23 ! œ 1# 32 œ 316 4 49. V œ '0 '1Î3 '0 3# sin 9 d3 d9 d) œ '0 '1Î3 1Î6 # '0cos '03 1 r dz r dr d) œ '0 Î2 '0cos 3rÈ1r# dr d) œ '0 Î2 ’ a1 r# b$Î# “ cos d) 1Î2 50. V œ '0 ) 1 31 16 œ 418 œ 29 csin )d ! 48. V œ '0 ) 21 a 21Î3 a$ a$ 3 sin 9 d9 d) œ 3 '021 c cos 9d #11Î$Î$ d) œ a3 '021 ˆ "# "# ‰ d) œ 213a $ '01Î2 '0a 3# sin 9 d3 d9 d) œ a3 '01Î6 '01Î2 sin 9 d9 d) œ a3 '01Î6 d) œ a181 $ 51. V œ '0 '0 21 1Î3 $ $ 'sec2 9 3# sin 9 d3 d9 d) œ "3 '0 '0 a8 sin 9 tan 9 sec# 9b d9 d) 21 1Î3 œ "3 '0 8 cos 9 "2 tan# 9‘ ! 21 1Î$ d) œ "3 '0 4 #" (3) 8‘ d) œ 3" '0 21 52. V œ 4 '0 1Î2 '0 œ 28 3 1Î2 21 5 5 51 # d) œ 6 (21) œ 3 '01Î4 'sec2 sec9 9 3# sin 9 d3 d9 d) œ 43 '01Î2 '01Î4 a8 sec$ 9 sec$ 9b sin 9 d9 d) '01Î4 sec$ 9 sin 9 d9 d) œ 283 '01Î2 '01Î4 tan 9 sec# 9 d9 d) œ 283 '01Î2 2" tan# 9‘ !1Î% d) œ 143 '01Î2 d) œ 731 53. V œ 4 '0 '01 '0r dz r dr d) œ 4 '01Î2 '01 r$ dr d) œ '01Î2 d) œ 1# 54. V œ 4 '0 '01 'r r 1 dz r dr d) œ 4 '01Î2 '01 r dr d) œ 2 '01Î2 d) œ 1 55. V œ 8 '0 '1 2 '0r dz r dr d) œ 8 '01Î2 '1 2 r# dr d) œ 8 Š 2È32" ‹ '01Î2 d) œ 41 Š2 3 2"‹ 56. V œ 8 '0 '1 2 '0 2 r dz r dr d) œ 8 '01Î2 '1 2 rÈ2 r# dr d) œ 8 '01Î2 ’ "3 a2 r# b$Î# “ # d) œ 83 '01Î2 d) œ 431 1Î2 1Î2 # # # 1Î2 1Î2 È È È È # 4cr sin ) 2 È dz r dr d) œ '0 '0 a4r r# sin )b dr d) œ 8 '0 ˆ1 sin3 ) ‰ d) œ 161 21 4cr cos )cr sin ) 58. V œ '0 '0 '0 21 È 1 57. V œ '0 '0 '0 21 È 2 21 2 dz r dr d) œ '0 '0 c4r r# (cos ) sin ))d dr d) œ 83 '0 (3 cos ) sin )) d) œ 161 21 21 2 59. The paraboloids intersect when 4x# 4y# œ 5 x# y# Ê x# y# œ 1 and z œ 4 Ê V œ 4 '0 1Î2 '01 '4r5 r dz r dr d) œ 4 '01Î2 '01 a5r 5r$ b dr d) œ 20 '01Î2 ’ r2 r4 “ " d) œ 5'01Î2 d) œ 5#1 # # # % ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. $ Section 15.7 Triple Integrals in Cylindrical and Spherical Coordinates 60. The paraboloid intersects the xy-plane when 9 x# y# œ 0 Ê x# y# œ 9 Ê V œ 4 '0 1Î2 œ 4 '0 1Î2 919 '13 '09cr dz r dr d) # '13 a9r r$ b dr d) œ 4 '01Î2 ’ 9r2 r4 “ $ d) œ 4 '01Î2 ˆ 814 174 ‰ d) œ 64 '01Î2 d) œ 321 # % " È4cr 61. V œ 8 '0 '0 '0 21 1 # dz r dr d) œ 8 '0 '0 r a4 r# b 21 œ 83 '0 ˆ3$Î# 8‰ d) œ 21 1 "Î# dr d) œ 8 '0 ’ "3 a4 r# b 21 $Î# " “ d) ! 41 Š8 3È3‹ 3 62. The sphere and paraboloid intersect when x# y# z# œ 2 and z œ x# y# Ê z# z 2 œ 0 Ê (z 2)(z 1) œ 0 Ê z œ 1 or z œ 2 Ê z œ 1 since z 0. Thus, x# y# œ 1 and the volume is given by the triple integral V œ 4 '0 1Î2 œ 4 '0 ’ "3 a2 r# b 1Î2 È '01 'r 2 r dz r dr d) œ 4 '01Î2 '01 ’r a2 r# b"Î# r$ “ dr d) # # 1 Š8 È 2 7 ‹ 7 r4 “ d) œ 4 '0 Š 2 3 2 12 ‹ d) œ $Î# % " 1Î2 È 6 ! 63. average œ #"1 '0 '0 'c1 r# dz dr d) œ #"1 '0 '0 2r# dr d) œ 3"1 '0 d) œ 23 21 1 21 1 È1cr 21 1 3 ' ' 64. average œ ˆ 4"1 ‰ '0 '0 'cÈ1cr# r# dz dr d) œ 41 2r# È1 r# dr d) 0 0 21 1 21 # 1 3 œ 231 ' " ’ "8 sin" r "8 rÈ1 r# a1 2r# b“ d) œ 1631 0 ! 21 '021 ˆ 1# 0‰ d) œ 323 '021 d) œ ˆ 323 ‰ (21) œ 3161 65. average œ ˆ 4"1 ‰ '0 '0 '0 3$ sin 9 d3 d9 d) œ 1631 '0 '0 sin 9 d9 d) œ 831 '0 d) œ 43 21 1 21 1 1 21 3 66. average œ ˆ 2"1 ‰ '0 '0 21 1Î2 3 œ 1631 '01 3$ cos 9 sin 9 d3 d9 d) œ 831 '021 '01Î2 cos 9 sin 9 d9 d) œ 831 '021 ’ sin2 9 “ 1Î# d) # ! '021 d) œ ˆ 1631 ‰ (21) œ 38 67. M œ 4 '0 '01 '0r dz r dr d) œ 4 '01Î2 '01 r# dr d) œ 43 '01Î2 d) œ 231 ; Mxy œ '021 '01 '0r z dz r dr d) 21 1 21 œ "# '0 '0 r$ dr d) œ 18 '0 d) œ 14 Ê z œ MM œ ˆ 14 ‰ ˆ 231 ‰ œ 38 , and x œ y œ 0, by symmetry 1Î2 xy 68. M œ '0 '02 '0r dz r dr d) œ '01Î2 '02 r# dr d) œ 83 '01Î2 d) œ 431 ; Myz œ '01Î2 '02 '0r x dz r dr d) 2 2 r 2 1Î2 1Î2 1Î2 1Î2 œ '0 '0 r$ cos ) dr d) œ 4 '0 cos ) d) œ 4; Mxz œ '0 '0 '0 y dz r dr d) œ '0 '0 r$ sin ) dr d) 2 r 2 1Î2 1Î2 1Î2 1Î2 M œ 4 '0 sin ) d) œ 4; Mxy œ '0 '0 '0 z dz r dr d) œ "# '0 '0 r$ dr d) œ 2 '0 d) œ 1 Ê x œ M œ 13 , 1Î2 yz M y œ MMxz œ 13 , and z œ Mxy œ 34 69. M œ 831 ; Mxy œ '0 '1Î3 '0 z3# sin 9 d3 d9 d) œ '0 '1Î3 '0 3$ cos 9 sin 9 d3 d9 d) œ 4 '0 '1Î3 cos 9 sin 9 d9 d) 21 œ 4 '0 ’ sin2 9 “ 21 # 70. M œ '0 '0 1Î# 1Î$ 1Î2 21 2 1Î2 21 2 d) œ 4 '0 ˆ "# 38 ‰ d) œ "# '0 d) œ 1 Ê z œ Mxy œ (1) ˆ 831 ‰ œ 38 , and x œ y œ 0, by symmetry 21 21 M È '0a 3# sin 9 d3 d9 d) œ a3 '021 '01Î4 sin 9 d9 d) œ a3 '021 2 #È2 d) œ 1a Š23 2‹ ; 1Î4 1Î4 21 a 21 21 a ' Mxy œ '0 '0 '0 3$ sin 9 cos 9 d3 d9 d) œ a4 '0 '0 sin 9 cos 9 d9 d) œ 16 d) œ 18a 0 21 1Î4 1Î2 $ $ $ % % % Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 920 Chapter 15 Multiple Integrals % Ê z œ Mxy œ Š 18a ‹ – M Èr 71. M œ '0 '0 '0 21 4 3 — œ ˆ 8 ‰Š 3a 1a$ Š2È2‹ 3 Š2È2‹ a 2 È2 , and x œ y œ 0, by symmetry # ‹œ 16 Èr '0 d) œ 1285 1 ; Mxy œ '0 '0 '0 z dz r dr d) dz r dr d) œ '0 '0 r$Î# dr d) œ 64 5 21 21 4 21 4 '0 d) œ 6431 Ê z œ Mxy œ 65 , and x œ y œ 0, by symmetry œ "# '0 '0 r# dr d) œ 32 3 21 21 4 1Î3 M È1 r 1Î3 1Î3 72. M œ 'c1Î3 '0 ' È1 r# dz r dr d) œ ' 1Î3 '0 2rÈ1 r# dr d) œ ' 1Î3 ’ 23 a1 r# b œ 23 ' # 1 1 $Î# " “ d) ! È ' 1Î3 '0 ' È11 rr r# cos ) dz dr d) œ 2 ' 11ÎÎ33 '01 r# È1 r# cos ) dr d) 1Î3 1Î3 d) œ ˆ 23 ‰ ˆ 231 ‰ œ 491 ; Myz œ c1Î3 # 1 # 1 Î3 1 Î3 œ 2 'c1Î3 ’ 18 sin" r "8 rÈ1 r# a1 2r# b“ cos ) d) œ 18 ' 1Î3 cos ) d) œ 18 csin )d 1Î13Î3 œ ˆ 18 ‰ Š2 † #3 ‹ œ 1 8 3 Ê " È È ! Myz 9È 3 x œ M œ 32 , and y œ z œ 0, by symmetry 73. We orient the cone with its vertex at the origin and axis along the z-axis Ê 9 œ 14 . We use the the x-axis which is through the vertex and parallel to the base of the cone Ê Ix œ '0 '0 'r ar# sin# ) z# b dz r dr d) 21 1 1 " ) ) ‘ #1 œ '0 '0 Šr$ sin# ) r% sin# ) 3r r3 ‹ dr d) œ '0 Š sin20 ) 10 sin802) 10 œ #10 15 œ 14 ‹ d) œ 40 ! 21 1 74. Iz œ '0 '0 21 1a œ 815 21 % Èa cr # a # ' cÈa cr # # # # # r$ dz dr d) œ '0 '0 2r$ Èa# r# dr d) œ 2 '0 ’Š r5 2a 15 ‹ aa r b 21 21 a # “ d) œ 2 '0 21 $Î# a # ! 2 & 15 a d) & % 75. Iz œ '0 '0 'ˆ h ‰ r ax# y# b dz r dr d) œ '0 '0 ' ' hr r$ dz dr d) œ '0 '0 Šhr$ hra ‹ dr d) œ '0 21 a 21 h a hr a a '0 d) œ 110ha œ '0 h Š a4 5a ‹ d) œ ha 20 21 h a % & 21 % 21 h 21 a % & a r h ’ r4 5a “ d) ! a % " ' 76. (a) M œ '0 '0 '0 z dz r dr d) œ '0 '0 "# r& dr d) œ 12 d) œ 16 ; Mxy œ '0 '0 '0 z# dz r dr d) 0 21 1 21 r# 21 1 21 r# 1 " ' 1 œ 3" '0 '0 r( dr d) œ 24 d) œ 12 Ê z œ #" , and x œ y œ 0, by symmetry; Iz œ '0 '0 '0 zr$ dz dr d) 0 21 21 1 21 r# 1 " ' œ "# '0 '0 r( dr d) œ 16 d) œ 18 0 21 21 1 ' '0 '0 zr# dz dr d) (b) M œ '0 '0 '0 r# dz dr d) œ '0 '0 r% dr d) œ 5" '0 d) œ 21 5 ; Mxy œ 0 21 1 21 r# 21 1 21 1 r# " ' 5 œ 2" '0 '0 r' dr d) œ 14 d) œ 17 Ê z œ 14 , and x œ y œ 0, by symmetry; Iz œ '0 '0 '0 r% dz dr d) 0 21 21 1 21 r# 1 œ '0 '0 r' dr d) œ 7" '0 d) œ 21 7 21 21 1 77. (a) M œ '0 '0 'r z dz r dr d) œ "# '0 '0 ar r$ b dr d) œ 8" '0 d) œ 14 ; Mxy œ '0 '0 'r z# dz r dr d) 21 1 21 1 21 1 21 1 1 " ' œ 3" '0 '0 ar r% b dr d) œ 10 d) œ 15 Ê z œ 45 , and x œ y œ 0, by symmetry; Iz œ '0 '0 'r zr$ dz dr d) 0 21 21 1 1 " ' œ "# '0 '0 ar$ r& b dr d) œ 24 d) œ 12 0 21 21 1 1 21 1 (b) M œ '0 '0 'r z# dz r dr d) œ 15 from part (a); Mxy œ '0 '0 'r z$ dz r dr d) œ 4" '0 '0 ar r& b dr d) 21 1 1 21 1 21 1 1 " ' œ 12 d) œ 16 Ê z œ 56 , and x œ y œ 0, by symmetry; Iz œ '0 '0 'r z# r$ dz dr d) œ "3 '0 '0 ar$ r' b dr d) 0 21 21 1 1 " ' 1 œ 28 d) œ 14 0 21 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 21 1 Section 15.7 Triple Integrals in Cylindrical and Spherical Coordinates 921 78. (a) M œ '0 '0 '0 3% sin 9 d3 d9 d) œ a5 '0 '0 sin 9 d9 d) œ 2a5 '0 d) œ 415a ; Iz œ '0 '0 '0 3' sin$ 9 d3 d9 d) 21 1 a 21 & 1 & 21 21 & 1 a '0 d) œ 8a211 œ a7 '0 '0 a1 cos# 9b sin 9 d9 d) œ a7 '0 ’ cos 9 cos3 9 “ d) œ 4a #1 21 ( 1 21 ( 1 $ ( 21 ( ! 21 a 21 21 1 1 29 ) (b) M œ '0 '0 '0 3$ sin# 9 d3 d9 d) œ a4 '0 '0 (1cos d9 d) œ 18a '0 d) œ 1 4a ; # 1 1 21 a 21 Iz œ '0 '0 '0 3& sin% 9 d3 d9 d) œ a6 '0 '0 sin% 9 d9 d) 1 21 21 21 1 1 œ a6 '0 Š’ sin 49 cos 9 “ 43 '0 sin# 9 d9‹ d) œ a8 '0 9# sin429 ‘ ! d) œ 116a '0 d) œ a 81 % % # % ' $ ' ' ' ' # ! 79. M œ '0 '0 '0 Èa cr dz r dr d) œ '0 '0 21 a 21 $ a d) œ 2ha3 1 ; Mxy œ '0 '0 '0 œ ha '0 3 h a # # 21 21 # a a '021 ’ 3" aa# r# b$Î# “ a d) h È # h # a r a r dr d) œ a Èa cr # h a # ! h ' ' z dz r dr d) œ 2a aa# r r$ b dr d) # 0 0 21 # a h ' œ 2a Š a# a4 ‹ d) œ a h4 1 Ê z œ Š 1a4h ‹ ˆ 2ha3# 1 ‰ œ 83 h, and x œ y œ 0, by symmetry # 0 21 # % % # # # # 80. Let the base radius of the cone be a and the height h, and place the cone's axis of symmetry along the z-axis with the vertex at the origin. Then M œ 1a3 h and Mxy œ '0 '0 'ˆ h ‰ r z dz r dr d) œ "# '0 21 # a 21 h '0a Šh# r ha r$ ‹ dr d) # # a œ h# '0 ’ r2 4ar # “ d) œ h# '0 Š a# a4 ‹ d) œ h 8a '0 d) œ h a4 1 Ê z œ MMxy œ Š h a4 1 ‹ ˆ 1a3# h ‰ œ 43 h, and 21 # # a % # 21 # # # # 21 # # # # ! x œ y œ 0, by symmetry Ê the centroid is one fourth of the way from the base to the vertex 81. The density distribution function is linear so it has the form $ (3) œ k3 C, where 3 is the distance from the center of the planet. Now, $ (R) œ 0 Ê kR C œ 0, and $ (3) œ k3 kR. It remains to determine the constant k: M œ '0 '0 '0 (k3 kR) 3# sin 9 d3 d9 d) œ '0 '0 ’k 34 kR 33 “ sin 9 d9 d) 21 œ' Ê 21 ' 1 21 R ' 1 1 % $ ' 21 21 R ! % % % k Š R4 R3 ‹ sin 9 d9 d) œ 0 1k# R% c cos 9d 1! d) œ 0 6k R% d) œ k13R 0 0 ‰ R œ 13M $ (3) œ 13M 3 13M R . At the center of the planet 3 œ 0 Ê $ (0) œ ˆ 13M R$ . R% R% R% Ê k œ 13M R% 82. The mass of the plant's atmosphere to an altitude h above the surface of the planet is the triple integral M(h) œ '0 '0 'R .! ecÐ3RÑ 3# sin 9 d3 d9 d) œ 'R '0 '0 .! ecÐ3RÑ 3# sin 9 d9 d) d3 21 1 h 21 h 1 œ 'R '0 .! ecÐ3RÑ 3# ( cos 9)‘ ! d) d3 œ 2 'R '0 .! ecR ec3 3# d) d3 œ 41.! ecR 'R ec3 3# d3 h 21 1 œ 41.! ecR ’ 3 ec c3 # ch # œ 41.! ecR Š h ec 23ce# c3 2hec# ch c3 2ec$ “ ch h 21 h h R (by parts) # 2ec$ R ec cR 2Rec# cR cR 2ec$ ‹ . The mass of the planet's atmosphere is therefore M œ lim hÄ_ # 2 M(h) œ 41.! Š Rc 2R c# c$ ‹ . 83. (a) A plane perpendicular to the x-axis has the form x œ a in rectangular coordinates Ê r cos ) œ a Ê r œ cosa ) Ê r œ a sec ), in cylindrical coordinates. (b) A plane perpendicular to the y-axis has the form y œ b in rectangular coordinates Ê r sin ) œ b Ê r œ sinb ) Ê r œ b csc ), in cylindrical coordinates. 84. ax by œ c Ê aar cos )b bar sin )b œ c Ê raa cos ) b sin )b œ c Ê r œ a cos ) c b sin ) . Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 922 Chapter 15 Multiple Integrals 85. The equation r œ fazb implies that the point ar, ), zb œ afazb, ), zb will lie on the surface for all ). In particular afazb, ) 1, zb lies on the surface whenever afazb, ), zb does Ê the surface is symmetric with respect to the z-axis. 86. The equation 3 œ fa9b implies that the point a3, 9, )b œ afa9b, 9, )b lies on the surface for all ). In particular, if afa9b, 9, )b lies on the surface, then afa9b, 9, ) 1b lies on the surface, so the surface is symmetric wiith respect to the z-axis. 15.8 SUBSTITUTIONS IN MULTIPLE INTEGRALS 1. (a) x y œ u and 2x y œ v Ê 3x œ u v and y œ x u Ê x œ "3 (u v) and y œ 3" (2u v); ` (xßy) ` (ußv) œ » " 3 23 " " 2 " 3 " »œ 9 9 œ 3 3 (b) The line segment y œ x from (!ß 0) to (1ß 1) is x y œ 0 Ê u œ 0; the line segment y œ 2x from (0ß 0) to (1ß 2) is 2x y œ 0 Ê v œ 0; the line segment x œ 1 from (1ß 1) to ("ß 2) is (x y) (2x y) œ 3 Ê u v œ 3. The transformed region is sketched at the right. 2. (a) x 2y œ u and x y œ v Ê 3y œ u v and x œ v y Ê y œ "3 (u v) and x œ 3" (u 2v); " ` (xßy) 3 œ »1 ` (ußv) 3 2 3 œ 9" 92 œ 3" 3" » (b) The triangular region in the xy-plane has vertices (0ß 0), (2ß 0), and ˆ 23 ß 23 ‰ . The line segment y œ x from (0ß 0) to ˆ 23 ß 23 ‰ is x y œ 0 Ê v œ 0; the line segment y œ 0 from (0ß 0) to (#ß 0) Ê u œ v; the line segment x 2y œ 2 from ˆ 23 ß 23 ‰ to (2ß 0) Ê u œ 2. The transformed region is sketched at the right. " 3. (a) 3x 2y œ u and x 4y œ v Ê 5x œ 2u v and y œ "# (u 3x) Ê x œ 5" (2u v) and y œ 10 (3v u); ` (xßy) ` (ußv) œ » 2 5 1 10 15 6 1 " 3 » œ 50 50 œ 10 10 (b) The x-axis y œ 0 Ê u œ 3v; the y-axis x œ 0 Ê v œ 2u; the line x y œ 1 " Ê "5 (2u v) 10 (3v u) œ 1 Ê 2(2u v) (3v u) œ 10 Ê 3u v œ 10. The transformed region is sketched at the right. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.8 Substitutions in Multiple Integrals 4. (a) 2x 3y œ u and x y œ v Ê x œ u 3v and y œ v x Ê x œ u 3v and y œ u 2v; " 3 ` (xßy) ` (ußv) œ º 1 2 º œ 2 3 œ 1 (b) The line x œ 3 Ê u 3v œ 3 or u 3v œ 3; x œ 0 Ê u 3v œ 0; y œ x Ê v œ 0; y œ x 1 Ê v œ 1. The transformed region is the parallelogram sketched at the right. 5. '04 'yÐÎy2Î2Ñ 1 ˆx y# ‰ dx dy œ '04 ’ x2 xy# “ " dy œ "# '04 ’ˆ y# 1‰# ˆ y# ‰# ˆ y# 1‰ y ˆ y# ‰ y“ dy # y 2 y 2 œ "# '0 (y 1 y) dy œ "# '0 dy œ "# (4) œ 2 4 6. 4 ' ' a2x# xy y# b dx dy œ ' ' (x y)(2x y) dx dy R R ßy) " '' œ ' ' uv ¹ `` (x uv du dv; (ußv) ¹ du dv œ 3 G G We find the boundaries of G from the boundaries of R, shown in the accompanying figure: xy-equations for Corresponding uv-equations Simplified for the boundary of G uv-equations the boundary of R y œ 2x 4 " 2 3 (2u v) œ 3 (u v) 4 vœ4 y œ 2x 7 2 " 3 (2u v) œ 3 (u v) 7 vœ7 yœx2 1 " 3 (2u v) œ 3 (u v) 2 uœ2 yœx1 1 " 3 (2u v) œ 3 (u v) 1 u œ 1 ' ˆ 11 ‰ u Ê "3 ' ' uv du dv œ "3 'c1 '4 uv dv du œ "3 'c1 u ’ v2 “ du œ 11 # c1 u du œ # ’ 2 “ 2 7 2 # 2 % G 7. ( # # " 33 ‰ œ ˆ 11 4 (4 1) œ 4 ' ' a3x# 14xy 8y# b dx dy R œ ' ' (3x 2y)(x 4y) dx dy R ßy) " '' œ ' ' uv ¹ `` (x uv du dv; (ußv) ¹ du dv œ 10 G G We find the boundaries of G from the boundaries of R, shown in the accompanying figure: xy-equations for the boundary of R Corresponding uv-equations Simplified for the boundary of G uv-equations y œ 3# x 1 3 " 10 (3v u) œ 10 (2u v) 1 uœ2 y œ 3# x 3 " 3 10 (3v u) œ 10 (2u v) 3 uœ6 y œ 4" x " 1 10 (3v u) œ 20 (2u v) vœ0 y œ 4" x 1 " 1 10 (3v u) œ 20 (2u v) 1 vœ4 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 923 924 Chapter 15 Multiple Integrals " '' " ' ' " ' Ê 10 uv du dv œ 10 uv dv du œ 10 u ’ v2 “ du œ 45 '2 u du œ ˆ 54 ‰ ’ u2 “ œ ˆ 54 ‰ (18 2) œ 64 5 2 0 2 6 4 6 # 6 ' # ! G 8. % # ' ' 2(x y) dx dy œ ' ' 2v ¹ `` (x(ußßy)v) ¹ du dv œ ' ' 2v du dv; the region G is sketched in Exercise 4 R G G 3c3v Ê ' ' 2v du dv œ '0 'c3v 2v du dv œ '0 2v(3 3v 3v) dv œ '0 6v dv œ c3v# d ! œ 3 1 1 1 " G v" uv# œ v" u v" u œ 2u v ; v u º y œ x Ê uv œ uv Ê v œ 1, and y œ 4x Ê v œ 2; xy œ 1 Ê u œ 1, and xy œ 9 Ê u œ 3; thus (xßy) 9. x œ uv and y œ uv Ê yx œ v# and xy œ u# ; `` (u ßv) œ J(uß v) œ º ' ' ŠÉ yx Èxy‹ dx dy œ ' ' (v u) ˆ 2uv ‰ dv du œ ' ' Š2u 2uv # ‹ dv du œ ' c2uv 2u# ln vd #" du R 3 2 3 2 3 1 1 1 1 1 œ '1 a2u 2u# ln 2b du œ u# 23 u# ln 2‘ " œ 8 23 (26)(ln 2) œ 8 52 3 (ln 2) 3 10. (a) $ " ` (xßy) ` (ußv) œ J(uß v) œ º v 0 œ u, and uº the region G is sketched at the right (b) x œ 1 Ê u œ 1, and x œ 2 Ê u œ 2; y œ 1 Ê uv œ 1 Ê v œ "u , and y œ 2 Ê uv œ 2 Ê v œ 2u ; thus, '12 '12 yx dy dx œ '12 '12ÎÎuu ˆ uvu ‰ u dv du œ '12 '12ÎÎuu uv dv du œ '12 u ’ v2 “ 2Îu du œ '12 u ˆ u2 2u" ‰ du # # 1Îu # # 3 3 œ #3 '1 u ˆ u"# ‰ du œ 3# cln ud #" œ 3# ln 2; '1 '1 yx dy dx œ '1 ’ x1 † y2 “ dx œ 3# '1 dx x œ # cln xd " œ # ln 2 2 2 2 2 2 2 2 1 ßy) 11. x œ ar cos ) and y œ ar sin ) Ê ``(x (rß)) œ J(rß ) ) œ º a cos ) b sin ) ar sin ) œ abr cos# ) abr sin# ) œ abr; br cos ) º I! œ ' ' ax# y# b dA œ '0 '0 r# aa# cos# ) b# sin# )b kJ(rß ))k dr d) œ '0 '0 abr$ aa# cos# ) b# sin# )b dr d) 21 21 1 1 R œ ab 4 '021 aa# cos# ) b# sin# )b d) œ ab4 ’ a2) a sin4 2) b2) b sin4 2) “ 21 œ ab1 aa4 b b # # # # # # ! ßy) 12. `` (x (ußv) œ J(uß v) œ º È1cu# 1 a 0 œ ab; A œ ' ' dy dx œ ' ' ab du dv œ 'c1 'cÈ1cu# ab dv du º 0 b R G œ 2ab 'c1 È1u# du œ 2ab ’ u2 È1 u# "# sin" u“ 1 " " œ ab csin" 1 sin" (1)d œ ab 1# ˆ 1# ‰‘ œ ab1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 15.8 Substitutions in Multiple Integrals 13. The region of integration R in the xy-plane is sketched in the figure at the right. The boundaries of the image G are obtained as follows, with G sketched at the right: xy-equations for Corresponding uv-equations Simplified for the boundary of G uv-equations the boundary of R xœy " 1 3 (u 2v) œ 3 (u v) vœ0 x œ 2 2y " 2 3 (u 2v) œ 2 3 (u v) uœ2 yœ0 0 œ 13 (u v) vœu ßy) " '0 Also, from Exercise 2, `` (x (ußv) œ J(uß v) œ 3 Ê 2Î3 'y2 2y (x 2y) eÐy xÑ dx dy œ '02 '0u ue v ¸ 3" ¸ dv du œ "3 '0 u cecv d !u du œ 3" '0 u a1 ecu b du œ 3" ’u au ecu b u# ecu “ œ 3" c2 a2 ec2 b 2 ec2 1d 2 2 # # ! œ "3 a3ec2 1b ¸ 0.4687 14. x œ u v# and y œ v Ê 2x y œ (2u v) v œ 2u and " ` (xßy) ` (ußv) œ J(uß v) œ º " # v º œ 1; next, u œ x # 0 " œ x y# and v œ y, so the boundaries of the region of integration R in the xy-plane are transformed to the boundaries of G: xy-equations for Corresponding uv-equations Simplified for the boundary of G uv-equations the boundary of R x œ y# x œ #y 2 u #v œ #v u #v œ #v 2 uœ2 yœ0 vœ0 vœ0 yœ2 vœ2 vœ2 ÐyÎ2Ñ 2 Ê '0 'yÎ2 2 uœ0 y$ (2x y) eÐ2xyÑ dx dy œ '0 '0 v$ (2u) e4u du dv œ '0 v$ ’ "4 e4u “ dv œ 4" '0 v$ ae16 1b dv # 2 2 2 # # # 2 ! % # œ "4 ae16 1b ’ v4 “ œ e16 1 ! v" uv# œ v" u v" u œ 2u v ; v u º y œ x Ê uv œ uv Ê v œ 1, and y œ 4x Ê v œ 2; xy œ 1 Ê u œ 1, and xy œ 4 Ê u œ 2; thus (xßy) 15. x œ uv and y œ uv Ê yx œ v# and xy œ u# ; `` (u ßv) œ J(uß v) œ º '12 '1yÎyax2 y2 b dx dy '24 'y4ÎÎ4yax2 y2 b dx dy œ '12 '12 Š uv u2 v2 ‹ ˆ 2uv ‰ du dv œ '12 '12 Š 2uv 2u3 v‹ du dv 2 3 2 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 925 926 Chapter 15 Multiple Integrals u 1 4 15 15v ‰ 15 15v 225 œ '1 ’ 2v dv œ '1 ˆ 2v dv œ ’ 4v 3 2 u v“ 3 2 2 4 “ œ 16 2 # 4 2 2 " 2 " ßy) 16. x œ u2 v2 and y œ 2uv; `` (x (ußv) œ J(uß v) œ º 2u 2v 2v œ 4u2 4v2 œ 4au2 v2 b ; 2u º y œ 2È1 x Ê y2 œ 4a1 xb Ê a2uvb2 œ 4a1 au2 v2 bb Ê u œ „ 1; y œ 0 Ê 2uv œ 0 Ê u œ 0 or v œ 0; x œ 0 Ê u2 v2 œ 0 Ê u œ v or u œ v; This gives us four triangular regions, but only the one in the quadrant where both u, v are positive maps into the region R in the xy-plane. È '01 '02 1 x Èx2 y2 dx dy œ '01 '0u Éau2 v2 b2 a2uvb2 † 4au2 v2 b dv du œ 4'01 '0u au2 v2 b2 dv du 2 u 112 1 6 ‘ 2 56 '2 5 œ 4'1 u4 v 23 u2 v3 15 v5 ‘0 du œ 112 15 1 u du œ 15 6 u " œ 45 ßy) 17. (a) x œ u cos v and y œ u sin v Ê `` (x (ußv) œ º ßy) (b) x œ u sin v and y œ u cos v Ê `` (x (ußv) œ º cos v u sin v œ u cos# v u sin# v œ u sin v u cos v º sin v u cos v œ u sin# v u cos# v œ u cos v u sin v º â â cos v â z) œ â sin v 18. (a) x œ u cos v, y œ u sin v, z œ w Ê ``(u(xßßvyßßw) â â 0 u sin v u cos v 0 â â2 â z) (b) x œ 2u 1, y œ 3v 4, z œ "# (w 4) Ê ``(u(xßßvyßßw) œ â0 â â0 â â sin 9 cos ) â 19. â sin 9 sin ) â â cos 9 œ (cos 9) º 3 cos 9 cos ) 3 cos 9 sin ) 3 sin 9 3 cos 9 cos ) 3 cos 9 sin ) 0 3 0 â 0â â 0 â œ u cos# v u sin# v œ u â "â â 0â 0 ââ œ (2)(3) ˆ #" ‰ œ 3 " â # â â 3 sin 9 sin ) â â 3 sin 9 cos ) â â 0 â 3 sin 9 sin ) sin 9 cos ) (3 sin 9) º 3 sin 9 cos ) º sin 9 sin ) 3 sin 9 sin ) 3 sin 9 cos ) º œ a3# cos 9b asin 9 cos 9 cos# ) sin 9 cos 9 sin# )b a3# sin 9b asin# 9 cos# ) sin# 9 sin# )b œ 3# sin 9 cos# 9 3# sin$ 9 œ a3# sin 9b acos# 9 sin# 9b œ 3# sin 9 w w ' ' 20. Let u œ gaxb Ê Jaxb œ du dx œ g axb Ê a faub du œ gaab fagaxbbg axb dx in accordance with Theorem 7 in gabb b Section 5.6. Note that gw axb represents the Jacobian of the transformation u œ gaxb or x œ g" aub. 21. '0 '0 'yÎ2 3 4 œ' 1 ÐyÎ2Ñ xz ˆ 2x # y 3z ‰ dx dy dz œ ' ' ’ x2 xy # 3 “ % # # ’ (y 4 1) y4 yz 3 “ ! dz œ 0 3 ' 3 4 0 0 # 3 "‰ ˆ 49 4z 3 4 dz œ 0 "ÐyÎ2Ñ yÎ2 dy dz œ '0 '0 "# (y 1) y# 3z ‘ dy dz 3 4 '0 ˆ2 4z3 ‰ dz œ ’2z 2z3 “ $ œ 12 3 # ! â â âa 0 0â # # # â â 22. J(uß vß w) œ â 0 b 0 â œ abc; the transformation takes the ellipsoid region xa# by# cz# Ÿ 1 in xyz-space â â â0 0 câ into the spherical region u# v# w# Ÿ 1 in uvw-space ˆwhich has volume V œ 43 1‰ Ê V œ ' ' ' dx dy dz œ ' ' ' abc du dv dw œ 413abc R G Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 15 Practice Exercises 927 â â âa 0 0â â â 23. J(uß vß w) œ â 0 b 0 â œ abc; for R and G as in Exercise 22, ' ' ' kxyzk dx dy dz â â R â0 0 câ œ ' ' ' a# b# c# uvw dw dv du œ 8a# b# c# '0 1Î2 G # # # œ 4a 3b c '01Î2 '01 (3 sin 9 cos ))(3 sin 9 sin ))(3 cos 9) a3# sin 9b d3 d9 d) '01Î2 '01Î2 sin ) cos ) sin$ 9 cos 9 d9 d) œ a b3 c '01Î2 sin ) cos ) d) œ a b6 c # # # # # # â 1 â â 24. u œ x, v œ xy, and w œ 3z Ê x œ u, y œ vu , and z œ "3 w Ê J(uß vß w) œ â uv# â â 0 0 " u 0 0 ââ " 0 ââ œ 3u ; " â â 3 ' ' ' ax# y 3xyzb dx dy dz œ ' ' ' u# ˆ vu ‰ 3u ˆ vu ‰ ˆ w3 ‰‘ kJ(uß vß w)k du dv dw œ "3 ' ' ' ˆv vw ‰ du dv dw u 0 0 1 3 D œ "3 2 2 G '0 '0 (v vw ln 2) dv dw œ 3" '03 (1 w ln 2) ’ v2 “ # dw œ 32 '03 (1 w ln 2) dw œ 32 ’w w2 ln 2“ $ 3 2 # # ! ! œ 23 ˆ3 #9 ln 2‰ œ 2 3 ln 2 œ 2 ln 8 # # # 25. The first moment about the xy-coordinate plane for the semi-ellipsoid, xa# by# cz# œ 1 using the transformation in Exercise 23 is, Mxy œ ' ' ' z dz dy dx œ ' ' ' cw kJ(uß vß w)k du dv dw D G œ abc# ' ' ' w du dv dw œ aabc# b † aMxy of the hemisphere x# y# z# œ 1, z G # 0b œ abc4 1 ; # 1 3 ‰ 3 the mass of the semi-ellipsoid is 2abc Ê z œ Š abc4 1 ‹ ˆ 2abc 3 1 œ 8 c 26. A solid of revolution is symmetric about the axis of revolution, therefore, the height of the solid is solely a function of r. That is, y œ faxb œ farb. Using cylindrical coordinates with x œ r cos ), y œ y and z œ r sin ), we have V œ ' ' ' r dy d) dr œ 'a '0 '0 farb 21 b G r dy d) dr œ 'a '0 c r y d0 d) dr œ 'a '0 r farb d) dr œ 'a c r)farb d201 dr b 21 farb b 21 b 'ab 21rfarbdr. In the last integral, r is a dummy or stand-in variable and as such it can be replaced by any variable name. b Choosing x instead of r we have V œ 'a 21xfaxbdx, which is the same result obtained using the shell method. CHAPTER 15 PRACTICE EXERCISES 1. '110 '01Îyyexy dx dy œ '110 cexy d !"Îy dy 10 œ '1 (e 1) dy œ 9e 9 '01 '0x eyÎx dy dx œ '01 x eyÎx ‘ !x dx $ 2. $ œ '0 Šxex x‹ dx œ ’ "2 ex x# “ œ e # 2 1 # # # " ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 928 3. Chapter 15 Multiple Integrals È # È9 4t # 9 '03Î2 ' È99 4t4t t ds dt œ '03Î2 ctsd È # 4t# dt œ '0 2tÈ9 4t# dt œ ’ "6 a9 4t# b $Î# $Î# 3Î2 “ 9 œ 6" ˆ0$Î# 9$Î# ‰ œ 27 6 œ # 4. ! '01 'È2cy Èy xy dx dy œ '01 y ’ x2 “ 2cÈy dy # Èy '0 y ˆ4 4Èy y y‰ dy 1 œ "# œ '0 ˆ2y 2y$Î# ‰ dy œ ’y# 4y5 “ œ 5" 1 " &Î# ! 'c02 '2x4 cb x4 dy dx œ 'c02 ax# 2xb dx # 5. $ œ ’ x3 x# “ ! # œ ˆ 38 4‰ œ 43 '04 'c(Èy c4 c4)/2y dx dy œ '04 ˆ y c2 4 È4 y‰ dy 4 2 œ ’ y2 2y 32 a4 yb3/2 “ œ 4 8 23 † 43/2 0 4 œ 4 16 3 œ 3 6. '01 'yÈy Èx dx dy œ '01 23 x$Î# ‘ yÈy dy œ 32 '0 ˆy$Î% y$Î# ‰ dy œ 32 47 y(Î% 52 y&Î# ‘ ! " 1 4 œ 23 ˆ 47 25 ‰ œ 35 '01 'xx Èx dy dx œ '01 x1/2 ax x2 b dx œ '01 ˆx3/2 x5/2 ‰ dx 2 1 4 œ 25 x5/2 27 x7/2 ‘0 œ 25 27 œ 35 7. È È 'c33 '0Ð1Î2Ñ 9 x y dy dx œ 'c33 ’ y2 “ Ð1Î2Ñ 9 x dx # # # ! œ' $ " x$ a9 x# b dx œ ’ 9x 8 24 “ $ c3 8 3 27 ‰ 9 ˆ 27 27 ‰ 27 œ ˆ 27 8 24 8 24 œ 6 œ # È '03Î2 'È99 4y4y y dx dy œ '03Î2 2yÈ94y# dy # # 3/2 œ "4 † 23 a94y# b3/2 º 0 9 œ 6" † 93/2 œ 27 6 œ # '02 '04 x 2x dy dx œ '02 c2xyd 04 x dx 2 2 œ '0 a2xa4 x2 bb dx œ '0 a8x 2x3 b dx 2 8. 2 4 2 œ ’4x2 x2 “ œ 16 16 2 œ 8 È4 c y '0 '0 4 ! È4 c y 2x dx dy œ '0 cx2 d 0 4 dy œ '0 a4 yb dy œ ’ 4y y2 “ œ 16 16 2 œ 8 4 2 4 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 15 Practice Exercises 9. '01 '2y2 4 cos ax# b dx dy œ '02 '0x/2 4 cos ax# b dy dx œ '02 2x cos ax# b dx œ csin ax# bd #! œ sin 4 10. '0 'yÎ2 ex dx dy œ '0 '0 ex dy dx œ '0 2xex dx œ cex d ! œ e 1 2 1 1 # 2x 1 # 11. '0 'È$ x y% " 1 dy dx œ '0 '0 8 2 y$ 2 # # " '02 y 4y 1 dy œ ln417 $ " dx dy œ "4 y% 1 % 12. '0 'È$ y 21 sinx#a1x b dx dy œ '0 '0 21 sinx#a1x b dy dx œ '0 21x sin a1x# b dx œ c cos a1x# bd ! œ (1) (1) œ 2 1 1 x$ 1 # 4 c x# 1 # " 13. A œ 'c2 '2x b 4 dy dx œ 'c2 ax# 2xb dx œ 43 0 2cx 15. V œ '0 'x 1 4 ax# y# b dy dx œ '0 ’x# y y3 “ 1 $ Èy 14. A œ '1 '2 c y dx dy œ '1 ˆÈy 2 y‰ dy œ 37 6 0 2cx x 4 dx œ '0 ’2x# (23x) 7x3 “ dx œ ’ 2x3 (212x) 7x 12 “ 1 $ $ % $ " 7 ‰ 2% œ ˆ 23 12 12 12 œ 43 6 c x# 16. V œ 'c3 'x 2 x# dy dx œ 'c3 cx# yd x " ! dx œ 'c3 a6x# x% x$ b dx œ 125 4 6 c x# 2 % 2 17. average value œ '0 '0 xy dy dx œ '0 ’ xy2 “ dx œ '0 x2 dx œ 4" 1 1 1 " # 1 ! È1 c x 18. average value œ ˆ "1 ‰ '0 '0 1 xy dy dx œ 14 '0 ’ xy2 “ # 1 4 È1 c x # È1 c x # ! dx œ 12 '0 ax x$ b dx œ #"1 1 " 19. 'c1 'cÈ1 c x# a1 x#2 y# b2 dy dx œ '0 '0 a1 2rr# b# dr d) œ '0 1 " r# ‘ ! d) œ #" '0 d) œ 1 1 21 # 21 1 È1 c y 21 20. 'c1 'cÈ1 c y# ln ax# y# 1b dx dy œ '0 '0 r ln ar# 1b dr d) œ '0 '1 1 21 # 21 1 2 " " # ln u du d) œ # '021 cu ln u ud #" d) œ "# '0 (2 ln 2 1) d) œ [ln (4) 1] 1 21 Ècos 2) 21. ax# y# b ax# y# b œ 0 Ê r% r# cos 2) œ 0 Ê r# œ cos 2) so the integral is 'c1Î4 '0 1Î4 # 1Î4 œ 'c1Î4 ’ 2 a1 " r# b “ 1Î4 Ècos 2) ! d) œ "# 1Î4 " " ' " ' 11ÎÎ44 ˆ1 1 cos ‰ ˆ1 # cos ‰ ) d) 2) d) œ # 1Î4 # œ "# 'c1Î4 Š1 sec# ) ‹ d) œ "# ) tan2 ) ‘ 1Î4 œ 14 2 22. (a) ' ' R 1Î4 # " dx dy œ a1 x # y # b # '01Î3 '0sec ) r dr d) œ a1 r# b# œ '0 ’ "# # a1 "sec# )b “ d) œ #" '0 1Î3 1Î3 œ "# ’ È"2 tan" Èu2 “ (b) ' ' R È$ ! " dx dy œ a1 x # y # b # œ '0 1Î2 œ r dr d) œ a1 r# b# " " lim ’ " # a1 b# b “d) œ # bÄ_ # ) 1 # u œ tan ) sec# ) 1 sec# ) d); ” du œ sec# ) d) • È2 " É 3 # 4 tan '01Î2 '0_ '0 Î3 ’ 2 a1 " r b “ sec d) '0 1Î2 '01Î2 lim bÄ_ ! È3 Ä #" '0 du 2 u # b ’ 2 a1 " r# b “ d) 0 d) œ 14 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. r dr d) a1 r# b# 929 930 Chapter 15 Multiple Integrals 23. '0 '0 '0 cos (x y z) dx dy dz œ '0 '0 [sin (z y 1) sin (z y)] dy dz 1 1 1 1 1 œ '0 [ cos (z 21) cos (z 1) cos z cos (z 1)] dz œ 0 1 24. 'ln 6 '0 ln 7 ln 2 'lnln45 eÐxyzÑ dz dy dx œ 'lnln67 '0ln 2 eÐxyÑ dy dx œ 'lnln67 ex dx œ 1 xby 25. '0 '0 '0 1 x# 8 (2x y z) dz dy dx œ '0 '0 Š 3x# 3y# ‹ dy dx œ '0 Š 3x# x# ‹ dx œ 35 x# 1 1 # # % ' e ' ' " ' 26. '1 '1 '0 2y z$ dy dz dx œ 1 1 z dz dx œ 1 ln x dx œ cx ln x xd 1 œ 1 e x z e 27. V œ 2 '0 1Î2 x e ' 0cos y '0 2x dz dx dy œ 2 '01Î2 ' 0cos y 2x dx dy œ 2 '01Î2 cos# y dy œ 2 ’ y2 sin42y “ 1Î2 œ 1# ! È4cx 28. V œ 4 '0 '0 2 œ ’x a4 x# b # È '04cx dz dy dx œ 4 '02 '0 4cx a4 x# b dy dx œ 4 '02 a4 x# b$Î# dx # $Î# # # 6xÈ4 x# 24 sin" x2 “ œ 24 sin" 1 œ 121 ! 29. average œ "3 '0 '0 '0 30xzÈx# y dz dy dx œ 3" '0 '0 15xÈx# y dy dx œ 3" '0 '0 15xÈx# y dx dy 1 3 œ "3 '0 ’5 ax# yb 3 1 1 3 3 1 “ dy œ "3 '0 5(1 y)$Î# 5y$Î# ‘ dy œ "3 2(1 y)&Î# 2y&Î# ‘ ! œ "3 2(4)&Î# 2(3)&Î# 2‘ $Î# " $ 3 ! œ "3 2 ˆ31 3&Î# ‰‘ 30. average œ 413a$ '0 '0 '0 3$ sin 9 d3 d9 d) œ 163a1 '0 '0 sin 9 d9 d) œ 83a1 '0 d) œ 3a 4 21 È2 È2cy 1 È4cx cy 31. (a) 'cÈ2 'cÈ2cy# 'Èx#by# # 21 a # 3 dz dx dy '02 33# sin 9 d3 d9 d) (c) '0 '0 'r 1Î4 È2 È4cr 21 # 21 # (b) '0 '0 21 1 È2 3 dz r dr d) œ 3 '0 '0 ’r a4 r# b 21 "Î# r# “ dr d) œ 3 '0 ’ "3 a4 r# b 21 $Î# œ '0 ˆ2$Î# 2$Î# 4$Î# ‰ d) œ Š8 4È2‹'0 d) œ 21 Š8 4È2‹ 21 21 1Î2 1 1Î2 1 1Î2 $ r3 “ È# ! d) 32. (a) 'c1Î2 '0 ' r# 21(r cos ))(r sin ))# dz r dr d) œ ' 1Î2 '0 ' r# 21r$ cos ) sin# ) dz r dr d) r# 1Î2 (b) 'c1Î2 '0 ' r# 21r$ cos ) sin# ) dz r dr d) œ 84 '0 r# 1 r# '01 r' sin# ) cos ) dr d) œ 12'01Î2 sin# ) cos ) d) œ 4 33. (a) '0 '0 '0sec 9 3# sin 9 d3 d9 d) 21 sec 9 21 21 21 1Î4 1Î4 1Î4 (b) '0 '0 '0 3# sin 9 d3 d9 d) œ 3" '0 '0 (sec 9)(sec 9 tan 9) d9 d) œ 3" '0 2" tan# 9‘ ! d) œ 6" '0 d) œ 13 21 1Î4 È1cx È 1 r 1Î2 '0 x y (6 4y) dz dy dx (b) '0 '0 '0 (6 4r sin )) dz r dr d) csc 9 1Î2 1Î2 (c) '0 '1Î4 '0 (6 43 sin 9 sin )) a3# sin 9b d3 d9 d) 34. (a) '0 '0 1 # # # (d) '0 '01 '0r (6 4r sin )) dz r dr d) œ '01Î2 '01 a6r# 4r$ sin )b dr d) œ '01Î2 c2r$ r% sin )d "! d) 1Î2 1Î# œ '0 (2 sin )) d) œ c2) cos )d ! œ 1 1 1Î2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 15 Practice Exercises È3cx 1 È3 È3cx È4cx cy 35. '0 'È1cx# '1 # # z# yx dz dy dx '1 # '0 # È '1 4cx cy z# yx dz dy dx # # 36. (a) Bounded on the top and bottom by the sphere x# y# z# œ 4, on the right by the right circular cylinder (x 1)# y# œ 1, on the left by the plane y œ 0 (b) '0 1Î2 È '02 cos 'cÈ44ccrr dz r dr d) ) # # È8cr 37. (a) V œ '0 '0 '2 21 œ' 2 # dz r dr d) œ '0 '0 ŠrÈ8 r# 2r‹ dr d) œ '0 ’ "3 a8 r# b 21 21 "3 (4)$Î# 4 3" (8)$Î# ‘ d) œ 0 (b) V œ '0 '0 21 1Î4 21 2 È '0 21 $Î# 1Î4 21 21 21 1Î3 21 È2 1Î% ! d) 81 Š4È2 5‹ ‹ d) œ 3 '02 (3 sin 9)# a3# sin 9b d3 d9 d) œ '021 '01Î3 '02 3% sin$ 9 d3 d9 d) '0 '0 asin 9 cos# 9 sin 9b d9 d) œ 325 '0 ’ cos 9 cos3 9 “ œ 32 5 21 È '0 d) œ 81 Š4 3 2 5‹ 21 '2 sec8 9 3# sin 9 d3 d9 d) œ 83 '021 '01Î4 Š2È2 sin 9 sec$ 9 sin 9‹ d9 d) œ 83 '0 Š2 "# 2È2‹ d) œ 83 '0 Š 5 #4 38. Iz œ '0 '0 ! 4 4 È È 3 Š2 3 2 8‹ d) œ 3 Š4 2 5‹ œ 83 '0 '0 Š2È2 sin 9 tan 9 sec# 9‹ d9 d) œ 83 '0 ’2È2 cos 9 "# tan# 9“ 21 # r # “ d) 1Î3 21 $ 1Î$ ! d) œ 831 39. With the centers of the spheres at the origin, Iz œ '0 '0 'a $ (3 sin 9)# a3# sin 9b d3 d9 d) 21 1 b œ $ ab 5 a b '0 '0 sin$ 9 d9 d) œ $ ab 5 a b '0 '0 asin 9 cos# 9 sin 9b d9 d) & 21 & 1 & 21 & 1 œ $ ab 5 a b '0 ’ cos 9 cos3 9 “ d) œ 4$ ab15 a b '0 d) œ 81$ ab15 a b & 21 & $ 1 & & 21 & & ! 1ccos ) 40. Iz œ '0 '0 '0 21 1 1ccos ) (3 sin 9)# a3# sin 9b d3 d9 d) œ '0 '0 '0 21 1 3% sin$ 9 d3 d9 d) œ "5 '0 '0 (1 cos 9)& sin$ 9 d9 d) œ '0 '0 (1 cos 9)' (1 cos 9) sin 9 d9 d); 21 1 21 1 # 21 2 21 21 u œ 1 cos 9 " " " " " 2u( u) ' ) ” du œ sin 9 d9 • Ä 5 '0 '0 u (2 u) du d) œ 5 '0 ’ 7 8 “ d) œ 5 '0 ˆ 7 8 ‰ 2 d) ! œ "5 '0 21 $ & 2 †2 56 d) œ 32 35 '0 21 d) œ 64351 41. M œ '1 '2Îx dy dx œ '1 ˆ2 2x ‰ dx œ 2 ln 4; My œ '1 '2Îx x dy dx œ '1 x ˆ2 2x ‰ dx œ 1; 2 2 2 2 2 2 Mx œ '1 '2Îx y dy dx œ '1 ˆ2 x2# ‰ dx œ 1 Ê x œ y œ # "ln 4 2 2 2 2y c y# 2y c y# 4y y 64 ' ' ' # $ 42. M œ '0 'c2y dx dy œ '0 a4y y# b dy œ 32 3 ; Mx œ 0 c2y y dx dy œ 0 a4y y b dy œ ’ 3 4 “ œ 3 ; 4 4 2y c y# 4 My œ '0 'c2y x dx dy œ ' 4 4 # ’ a2y#y b 0 # % % y& 2y# “ dy œ ’ 10 y2 “ œ 128 5 ! 4 $ 4 2 ! Ê $ 44. (a) Io œ 'c2 'c1 ax# y# b dy dx œ 'c2 ˆ2x# 23 ‰ dx œ 40 3 2 1 % Mx y 12 xœ M M œ 5 and y œ M œ 2 14x 43. Io œ '0 '2x ax# y# b (3) dy dx œ 3 '0 Š4x# 64 3 3 ‹ dx œ 104 2 % 2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 931 932 Chapter 15 Multiple Integrals 4a b ' ' # ' 2a (b) Ix œ 'ca 'cb y# dy dx œ 'ca 2b3 dx œ 4ab Ê Io œ Ix Iy 3 ; Iy œ cb ca x dx dy œ cb 3 dy œ 3 a b a # $ b $ a b $ $ # 4ab ab a b 4a b œ 4ab 3 3 œ 3 $ $ 45. M œ $ '0 '0 2xÎ3 3 ' ' dy dx œ $ '0 2x 3 dx œ 3$ ; Ix œ $ 0 0 3 2xÎ3 3 8$ ' 8$ ‰ 3 y# dy dx œ 81 x$ dx œ ˆ 81 Š 4 ‹ œ 2$ 0 3 % "3 46. M œ '0 'x# (x 1) dy dx œ '0 ax x$ b dx œ "4 ; Mx œ '0 'x# y(x 1) dy dx œ #" '0 ax$ x& x# x% b dx œ 120 ; 1 x 1 1 x 1 2 8 13 My œ '0 'x# x(x 1) dy dx œ '0 ax# x% b dx œ 15 Ê x œ 15 and y œ 30 ; Ix œ '0 'x# y# (x 1) dy dx 1 x 1 1 x Ix 17 17 1 œ "3 '0 ax% x( x$ x' b dx œ 280 Ê Rx œ É M œ É 70 ; Iy œ '0 'x# x# (x 1) dy dx œ '0 ax$ x& b dx œ 12 1 1 x 1 47. M œ 'c1 'c1 ˆx# y# 3" ‰ dy dx œ 'c1 ˆ2x# 34 ‰ dx œ 4; Mx œ 'c1 'c1 y ˆx# y# "3 ‰ dy dx œ 'c1 0 dx œ 0; 1 1 1 1 1 1 My œ 'c1 'c1 x ˆx# y# "3 ‰ dy dx œ 'c1 ˆ2x$ 34 x‰ dx œ 0 1 1 1 48. Place the ?ABC with its vertices at A(0ß 0), B(bß 0) and C(aß h). The line through the points A and C is Ða y œ ha x; the line through the points C and B is y œ a h b (x b). Thus, M œ '0 'ayÎh h Ða bÑyÎh b ' ' œ b$ '0 ˆ1 yh ‰ dy œ $bh # ; Ix œ 0 ayÎh h h 1Î3 1Î3 bÑyÎh b y# $ dx dy œ b$ '0 Šy# yh ‹ dy œ $1bh# h 1Î3 $ $ dx dy $ 1Î3 49. M œ ' 1Î3 '0 r dr d) œ 9# ' 1Î3 d) œ 31; My œ ' 1Î3 '0 r# cos ) dr d) œ 9 ' 1Î3 cos ) d) œ 9È3 Ê x œ 3 1 3 , 3 3 È and y œ 0 by symmetry 50. M œ '0 1Î2 13 '13 r dr d) œ 4 '01Î2 d) œ 21; My œ '01Î2 '13 r# cos ) dr d) œ 263 '01Î2 cos ) d) œ 263 Ê x œ 31 , and 13 y œ 31 by symmetry 51. (a) M œ 2 '0 1Î2 '11bcos r dr d) ) (b) 2) ‰ œ '0 ˆ2 cos ) 1 cos d) œ 8 4 1 ; # 1Î2 1Î2 My œ 'c1Î2 '1 1 cos ) (r cos )) r dr d) œ 'c1Î2 Šcos# ) cos$ ) cos3 ) ‹ d) 1Î2 % 1 32 œ 32 24151 Ê x œ 15 61 48 , and y œ 0 by symmetry ) 52. (a) M œ 'c! '0 r dr d) œ 'c! a# d) œ a# !; My œ 'c! '0 (r cos )) r dr d) œ 'c! a cos d) œ 2a 3sin ! 3 ! ! a ! # ! a $ ! 2a sin ! Ê x œ 2a 3sin œ0 ! , and y œ 0 by symmetry; lim c x œ lim c 3! 2a (b) x œ 51 and y œ 0 !Ä1 !Ä1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. $ Chapter 15 Additional and Advanced Exercises 53. x œ u y and y œ v Ê x œ u v and y œ v " " Ê J(uß v) œ º œ 1; the boundary of the 0 "º image G is obtained from the boundary of R as follows: xy-equations for Corresponding uv-equations Simplified the boundary of R for the boundary of G uv-equations yœx yœ0 vœuv uœ0 vœ0 _ _ _ sÐuvÑ Ê '0 '0 esx f(x yß y) dy dx œ '0 '0 e x vœ0 f(uß v) du dv # s !t $s "t 54. If s œ !x " y and t œ # x $ y where (!$ "# )# œ ac b# , then x œ !$ "# , y œ !$ "# , and J(sß t) œ (!$ " "#)# º $ # _ _ # # " œ " Ê 'c_ ' _ e as t b È " # ds dt ac b ! º !$ "# '021 '0_ rer dr d) œ È " # œÈ " ac b# # ac b# '021 d) œ È 1 ac b# . Therefore, È 1 ac b# œ 1 Ê ac b# œ 1# . CHAPTER 15 ADDITIONAL AND ADVANCED EXERCISES 6cx# 1. (a) V œ 'c3 'x 2 6cx# (c) V œ 'c3 'x 2 6cx# (b) V œ 'c3 'x 2 x# dy dx 6cx# x# dy dx œ 'c3 'x 2 '0x dz dy dx # & % a6x# x% x$ b dx œ ’2x$ x5 x4 “ # $ œ 125 4 2. Place the sphere's center at the origin with the surface of the water at z œ 3. Then 9 œ 25 x# y# Ê x# y# œ 16 is the projection of the volume of water onto the xy-plane c3 Ê V œ '0 '0 'cÈ25cr# dz r dr d) œ '0 '0 ŠrÈ25 r# 3r‹ dr d) œ '0 ’ "3 a25 r# b 21 4 21 21 4 $Î# % 3# r# “ d) ! 21 21 521 œ '0 "3 (9)$Î# 24 3" (25)$Î# ‘ d) œ '0 26 3 d) œ 3 2crÐcos ) 3. Using cylindrical coordinates, V œ '0 '0 '0 21 1 sin )Ñ dz r dr d) œ '0 '0 a2r r# cos ) r# sin )b dr d) 21 1 œ '0 ˆ1 "3 cos ) "3 sin )‰ d) œ ) "3 sin ) 3" cos )‘ ! œ 21 21 #1 4. V œ 4 '0 1Î2 œ 4 '0 1Î2 È '01 'r 2 r dz r dr d) œ 4 '01Î2 '01 ŠrÈ2 r# r$ ‹ dr d) œ 4'01Î2 ’ "3 a2 r# b$Î# r4 “ " d) # % # ! ' 1Î2 È È Š "3 "4 2 3 2 ‹ d) œ Š 8 327 ‹ 0 d) œ 1 Š8È2 7‹ 6 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 933 934 Chapter 15 Multiple Integrals 5. The surfaces intersect when 3 x# y# œ 2x# 2y# Ê x# y# œ 1. Thus the volume is È1 c x V œ 4 '0 '0 1 6. V œ 8 '0 1Î2 # '2x3cx2ycy dz dy dx œ 4 '01Î2 '01 '2r3 r dz r dr d) œ 4 '01Î2 '01 a3r 3r$ b dr d) œ 3'01Î2 d) œ 31# # # # # # # '01Î2 '02 sin 9 3# sin 9 d3 d9 d) œ 643 '01Î2 '01Î2 sin% 9 d9 d) '0 ” sin 94cos 9 ¹ œ 64 3 1Î2 $ 1Î# ! 43 '0 sin# 9 d9• d) œ 16 '0 92 sin429 ‘ ! 1Î2 1Î2 1Î# d) œ 41 '0 d) œ 21# 1Î2 7. (a) The radius of the hole is 1, and the radius of the sphere is 2. È3 È4cz (b) V œ 2 '0 '0 21 8. V œ '0 '0 1 3 sin ) '1 È # È3 r dr dz d) œ '0 '0 21 a3 z# b dz d) œ 2È3 '0 d) œ 4È31 21 '0 9cr dz r dr d) œ '0 '03 sin rÈ9 r# dr d) œ '0 ’ "3 a9 r# b$Î# “ 3 sin d) # 1 ) ) 1 ! 1 1 1 $Î# $Î# œ '0 ’ "3 a9 9 sin# )b 3" (9)$Î# “ d) œ 9'0 ’1 a1 sin# )b “ d) œ 9'0 a1 cos$ )b d) œ '0 a1 cos ) sin# ) cos )b d) œ 9 ’) sin ) sin3 ) “ œ 91 1 $ 1 ! # # 9. The surfaces intersect when x# y# œ x y# 1 Ê x# y# œ 1. Thus the volume in cylindrical coordinates is V œ 4 '0 1Î2 10. V œ '0 1Î2 '01 'r r 1 Î2 dz r dr d) œ 4 '01Î2 '01 Š #r r# ‹ dr d) œ 4'01Î2 ’ r4 r8 “ " d) œ "# '01Î2 d) œ 14 ˆ# ‰ $ % ! '12 '0r sin cos dz r dr d) œ '0 Î2 '12 r$ sin ) cos ) dr d) œ '0 Î2 ’ r4 “ # sin ) cos ) d) # ) ) 1 '0 sin ) cos ) d) œ 154 ’ sin2 ) “ œ 15 4 1Î2 # _ ecax ecbx 11. '0 # # _ 1Î# ! 1 " œ 15 8 _ dx œ '0 'a exy dy dx œ 'a '0 exy dx dy œ 'a Š lim b b b tÄ_ x % '0t exy dx‹ dy œ 'a lim ’ e y “ dy œ 'a lim Š "y e y ‹ dy œ 'a "y dy œ cln yd ab œ ln ˆ ba ‰ cxy b tÄ_ t cyt b b tÄ_ ! 12. (a) The region of integration is sketched at the right Ê '0 a sin " È 'y cota "c y ln ax# y# b dx dy # # œ '0 '0 r ln ar# b dr d); " a a# " u œ r# " ” du œ 2r dr • Ä # '0 '0 ln u du d) œ "# '0 [u ln u u] !a d) " # œ "# '0 ’2a# ln a a# lim t ln t“ d) œ a# '0 (2 ln a 1) d) œ a# " ˆln a "# ‰ " " # tÄ0 È a cos " '0(tan ")x ln ax# y# b dy dx 'aacos " '0 a cx ln ax# y# b dy dx (b) '0 # # Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 15 Additional and Advanced Exercises 13. '0 '0 emÐxtÑ f(t) dt du œ '0 't emÐxtÑ f(t) du dt œ '0 (x t)emÐxtÑ f(t) dt; also x u x x x '0x '0v '0u emÐxtÑ f(t) dt du dv œ '0x 't x 't v emÐxtÑ f(t) du dv dt œ '0x 't x (v t)emÐxtÑ f(t) dv dt x x x œ '0 "2 (v t)# emÐxtÑ f(t)‘ t dt œ '0 (x # t) emÐxtÑ f(t) dt # 14. '0 f(x) Š'0 g(xy)f(y) dy‹ dx œ '0 '0 g(xy)f(x)f(y) dy dx 1 x 1 x œ '0 'y g(xy)f(x)f(y) dx dy œ '0 f(y) Œ'y g(xy)f(x) dx dy; 1 1 1 1 '01 '01 g akxykb f(x)f(y) dx dy œ '01 '0x g(xy)f(x)f(y) dy dx '01 'x1 g(yx)f(x)f(y) dy dx 1 1 1 1 œ '0 'y g(xy)f(x)f(y) dx dy '0 'x g(yx)f(x)f(y) dy dx œ '0 'y g(xy)f(x)f(y) dx dy '0 'y g(xy)f(y)f(x) dx dy ðóóóóóóóóóóóóñóóóóóóóóóóóóò 1 1 1 1 simply interchange x and y variable names œ 2'0 'y g(xy)f(x)f(y) dx dy, and the statement now follows. 1 1 15. Io (a) œ '0 '0 xÎa# a ax# y# b dy dx œ '0 ’x# y y3 “ a $ # œ a4 1"# a# ; Iow (a) œ #" a "6 a$ œ 0 % Ê a xÎa# ! œ "3 x x x dx œ '0 Š xa# 3a ' ‹ dx œ ’ 4a# 12a' “ a Ê aœ $ $ % % " É "3 œ È . % 3 % a ! Since Io (a) œ "# #" a% 0, the ww value of a does provide a minimum for the polar moment of inertia Io (a). 64 16. Io œ '0 '2x ax# y# b (3) dy dx œ 3 '0 Š4x# 14x 3 3 ‹ dx œ 104 2 4 2 $ 17. M œ 'c) 'b sec ) r dr d) œ 'c) Š a# b# sec# )‹ d) ) ) a # # œ a# ) b# tan ) œ a# cos" ˆ ba ‰ b# Š È a# b# ‹ b œ a# cos" ˆ ba ‰ bÈa# b# ; Io œ 'c) 'b sec ) r$ dr d) ) a œ "4 'c) aa% b% sec% )b d) ) œ "4 'c) ca% b% a1 tan# )b asec# )bd d) ) % $ ) œ "4 ’a% ) b% tan ) b tan “ 3 % % % $ ) ) ) ) œ a#) b tan b tan # 6 œ "# a% cos" ˆ ba ‰ "# b$ Èa# b# 6" b$ aa# b# b 2 ˆy# Î2‰ $Î# y 18. M œ 'c2 '1cay#Î4b dx dy œ ' 2 Š1 y4 ‹ dy œ ’y 12 “ 2 œ' 2 2c ˆy# Î2‰ # ’x “ c2 2 1 ay# Î4b 2 dy œ ' # $ 2 3 3 a4 y# b dy œ 32 c2 32 # # 2 ˆy# Î2‰ œ 38 ; My œ ' 2 '1 ay#Î4b x dx dy 2 'c2 a16 8y# y% b dy œ 163 ’16y 8y3 y5 “ # 2 $ & ! My 3 ˆ 32 ‰ ˆ 3 ‰ ˆ 3215†8 ‰ œ 48 ˆ 48 ‰ ˆ 83 ‰ œ 56 , and y œ 0 by symmetry œ 16 32 64 3 5 œ 16 15 Ê x œ M œ 15 19. '0 '0 emax ab x ßa y b dy dx œ '0 '0 a b # # # # a bxÎa eb x dy dx '0 '0 # # b ayÎb # # ea y dx dy Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 935 936 Chapter 15 Multiple Integrals " " œ '0 ˆ ba x‰ eb x dx '0 ˆ ba y‰ ea y dy œ ’ 2ab eb x “ ’ 2ba ea y “ œ #"ab Šeb a 1‹ #"ab Šea b 1‹ a b # # # # # # a b # # ! # # # # ! # # " œ ab Šea b 1‹ ßy) ' ` F(xßy) ' ` F(x"ßy) ` F(x` y!ßy) “ dx œ cF(x" ß y) F(x! ß y)d yy!" 20. 'y 'x ``F(x x ` y dx dy œ y ’ ` y “ dy œ y ’ ` y y" x" ! ! # y" x" y" ! x! ! œ F(x" ß y" ) F(x! ß y" ) F(x" ß y! ) F(x! ß y! ) 21. (a) (i) (ii) (iii) (iv) (b) '0 ln 2 Fubini's Theorem Treating G(y) as a constant Algebraic rearrangement The definite integral is a constant number '01Î2 ex cos y dy dx œ Œ'0ln 2 ex dx Œ'01Î2 cos y dy œ aeln 2 e0 b ˆsin 1# sin 0‰ œ (1)(1) œ 1 (c) '1 'c1 yx# dx dy œ Œ'1 y"# dy Œ'c1 x dx œ ’ y" “ ’ x2 “ 2 1 2 # 1 # " " " œ ˆ "# 1‰ ˆ "# "# ‰ œ 0 22. (a) ™ f œ xi yj Ê Du f œ u" x u# y; the area of the region of integration is "# 1cx Ê average œ 2'0 '0 1 (u" x u# y) dy dx œ 2 '0 u" x(1 x) "# u# (1 x)# ‘ dx 1 " $ œ 2 ’u" Š x2 x3 ‹ ˆ "# u# ‰ (13x) “ œ 2 ˆ 6" u" 6" u# ‰ œ 3" (u" u# ) # $ ! (b) " average œ area _ _ 23. (a) I# œ '0 '0 œ "# '0 1Î2 _ M u" ' ' u# ' ' ' ' (u" x u# y) dA œ area x dA area y dA œ u" Š My ‹ u# ˆ MMx ‰ œ u" x u# y R R 1Î2 e ˆx y ‰ dx dy œ '0 # lim bÄ_ # R '0_ ae r b r dr d) œ '01Î2 ” lim '0b re r dr• d) # # bÄ_ aecb 1b d) œ "# '0 d) œ 14 Ê I œ 1Î2 # _ (b) > ˆ "# ‰ œ '0 t"Î# et dt œ '0 ay# b "Î# y# e _ È1 # (2y) dy œ 2 '0 ey dy œ 2 Š # ‹ œ È1, where y œ Èt È1 # 24. Q œ '0 '0 kr# (1 sin )) dr d) œ kR3 '0 (1 sin )) d) œ kR3 c) cos )d #!1 œ 21kR 3 21 R $ 21 $ Èh Èh c x $ 25. For a height h in the bowl the volume of water is V œ 'cÈh 'cÈh c x# 'x#by# dz dy dx Èh Èh c x Èh # h œ 'cÈh 'cÈh c x# ah x# y# b dy dx œ '0 '0 ah r# b r dr d) œ '0 ’ hr2 r4 “ 21 # 21 # % Èh ! d) œ '0 21 h# h# 1 4 d) œ # . Since the top of the bowl has area 101, then we calibrate the bowl by comparing it to a right circular cylinder # whose cross sectional area is 101 from z œ 0 to z œ 10. If such a cylinder contains h#1 cubic inches of water # # to a depth w then we have 101w œ h 1 Ê w œ h . So for 1 inch of rain, w œ 1 and h œ È20; for 3 inches of # 20 rain, w œ 3 and h œ È60. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 15 Additional and Advanced Exercises 26. (a) An equation for the satellite dish in standard position is z œ "# x# "# y# . Since the axis is tilted 30°, a unit vector v œ 0i aj bk normal to the plane of the È3 # È " v œ # j #3 k water level satisfies b œ v † k œ cos ˆ 16 ‰ œ Ê a œ È1 b# œ "# Ê È Ê "# (y 1) #3 ˆz "# ‰ œ 0 Ê z œ È"3 y Š "# È"3 ‹ is an equation of the plane of the water level. Therefore È3 yb 2 c È3 the volume of water is V œ ' ' ' 1 x#b 1 y# 1 1 2 R 1 dz dy dx, where R is the interior of the ellipse 2 È 3 Ê 3 4 Š È 3 1‹ 2 x# y# È23 y 1 È23 œ 0. When x œ 0, then y œ ! or y œ " , where ! œ È 2 and " œ È Ê 43 4 Š 2 1‹ 3 3 b1c 3 cy ‹ yb c È ' 'c b cÈ c ' È b Ê Vœ # "Î# " Š 2 3 y 2 ! Š È3 y# ‹ 2 3 y # 3 "Î# 2 1 1 1 # 2 x 1 2 4 2 # 1 3 1 dz dx dy 1 # 2 y dz dz (b) x œ 0 Ê z œ "# y# and dy œ y; y œ 1 Ê dy œ 1 Ê the tangent line has slope 1 or a 45° slant Ê at 45° and thereafter, the dish will not hold water. 27. The cylinder is given by x# y# œ 1 from z œ 1 to _ Ê ' ' ' z ar# z# b _ œ '0 '0 '1 21 1 '021 '01 '1a z dz r dr d) œ a lim Ä_ ar# z# b&Î# ' '0 ’ˆ "3 ‰ œ a lim Ä_ 0 21 dV D rz dz dr d) ar# z# b&Î# ' 21 ' 1 a 1 &Î# r “ dr d) œ a lim ’ˆ 3" ‰ # r # $Î# ˆ "3 ‰ # r $Î# “ dr d) Ä_ 0 0 ar# z# b$Î# 1 ar a b ar 1 b " 21 21 "Î# "Î# "Î# "Î# œ a lim 3" ar# 1b 3" ˆ2"Î# ‰ 3" aa# b 3" “ d) ’ 3" ar# a# b “ d) œ a lim ’ 3" a1 a# b Ä_ 0 Ä _ 0 ! È È "Î# œ a lim 21 ’ 3" a1 a# b 3" Š #2 ‹ 3" ˆ "a ‰ 3" “ œ 21 ’ 3" ˆ 3" ‰ #2 “ . Ä_ ' ' 28. Let's see? The length of the "unit" line segment is: L œ 2'0 dx œ 2. 1 È1 c x The area of the unit circle is: A œ 4'0 '0 1 2 dy dx œ 1. È1 c x The volume of the unit sphere is: V œ 8'0 '0 1 2 È '0 1 c x c y dz dy dx œ 43 1. 2 2 Therefore, the hypervolume of the unit 4-sphere should be: È1cx Vhyper œ 16'0 '0 1 2 È È '0 1cx cy '0 1cx cy cz dw dz dy dx. 2 2 2 2 2 Mathematica is able to handle this integral, but we'll use the brute force approach. È1cx Vhyper œ 16'0 '0 1 È È 2 È È È 2 2 2 2 2 1 1cx '0 1cx cy È1 x2 y2 É1 1 c xz c y dz dy dx œ – œ 16'0 '0 2 2 È1cx 1 È1cx œ 16'0 '0 1 2 2 2 2 2 2 2 z È1 x2 y2 œ cos ) 2 2 œ 16'0 '0 È '0 1cx cy '0 1cx cy cz dw dz dy dx œ 16'01 '0 1cx '0 1cx cy È1 x2 y2 z2 dz dy dx dz œ È1 x2 y2 sin ) d) È1cx a1 x2 y2 b'1/2 È1 cos2 ) sin ) d) dy dx œ 16'0 '0 0 1 2 — a1 x2 y2 b'1/2 sin2 ) d) dy dx 0 '1 3/2 1 2 2 2È È 2 1 x2 3" a1 x2 b ‹ dx 4 a1 x y b dy dx œ 41 0 Š 1 x x œ 41'0 È1 x2 a1 x2 b 1 $ x ‘ dx œ 83 1'0 a1 x2 b 1 1 3 3/2 dx œ ” 0 x œ cos ) œ 83 1'1/2 sin4 ) d) dx œ sin ) d) • 2) ‰ œ 83 1'1/2 ˆ 1 cos d) œ 23 1'1/2 a1 2 cos 2) cos2 2)bd) œ 23 1'1/2 ˆ #3 2 cos 2) cos2 4) ‰d) œ 12 2 0 2 0 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 2 937 938 Chapter 15 Multiple Integrals NOTES: Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. CHAPTER 16 INTEGRATION IN VECTOR FIELDS 16.1 LINE INTEGRALS 1. r œ ti a" tbj Ê x œ t and y œ 1 t Ê y œ 1 x Ê (c) 2. r œ i j tk Ê x œ 1, y œ 1, and z œ t Ê (e) 3. r œ a2 cos tbi a2 sin tbj Ê x œ 2 cos t and y œ 2 sin t Ê x# y# œ 4 Ê (g) 4. r œ ti Ê x œ t, y œ 0, and z œ 0 Ê (a) 5. r œ ti tj tk Ê x œ t, y œ t, and z œ t Ê (d) 6. r œ tj a2 2tbk Ê y œ t and z œ 2 2t Ê z œ 2 2y Ê (b) # 7. r œ at# 1b j 2tk Ê y œ t# 1 and z œ 2t Ê y œ z4 1 Ê (f) 8. r œ a2 cos tbi a2 sin tbk Ê x œ 2 cos t and z œ 2 sin t Ê x# z# œ 4 Ê (h) 9. ratb œ ti a1 tbj , 0 Ÿ t Ÿ 1 Ê ddtr œ i j Ê ¸ ddtr ¸ œ È2 j ; x œ t and y œ 1 t Ê x y œ t (" t) œ 1 Ê 'C faxß yß zb ds œ '0 fatß 1 tß 0b ¸ ddtr ¸ dt œ '0 (1) ŠÈ2‹ dt œ ’È2 t“ œ È2 1 " 1 ! 10. r(t) œ ti (1 t)j k , 0 Ÿ t Ÿ 1 Ê ddtr œ i j Ê ¸ ddtr ¸ œ È2; x œ t, y œ 1 t, and z œ 1 Ê x y z 2 œ t (1 t) 1 2 œ 2t 2 Ê 'C f(xß yß z) ds œ '0 (2t 2) È2 dt œ È2 ct# 2td ! œ È2 1 " 11. r(t) œ 2ti tj (2 2t)k , 0 Ÿ t Ÿ 1 Ê ddtr œ 2i j 2k Ê ¸ ddtr ¸ œ È4 1 4 œ 3; xy y z œ (2t)t t (2 2t) Ê 'C f(xß yß z) ds œ '0 a2t# t 2b 3 dt œ 3 23 t$ "# t# 2t‘ ! œ 3 ˆ 23 "# 2‰ œ 13 # 1 " 12. r(t) œ (4 cos t)i (4 sin t)j 3tk , 21 Ÿ t Ÿ 21 Ê ddtr œ (4 sin t)i (4 cos t)j 3k Ê ¸ ddtr ¸ œ È16 sin# t 16 cos# t 9 œ 5; Èx# y# œ È16 cos# t 16 sin# t œ 4 Ê 'C f(xß yß z) ds œ 'c21 (4)(5) dt 21 1 œ c20td ## 1 œ 801 13. r(t) œ (i 2j 3k) t(i 3j 2k) œ (1 t)i (2 3t)j (3 2t)k , 0 Ÿ t Ÿ 1 Ê ddtr œ i 3j 2k Ê ¸ ddtr ¸ œ È1 9 4 œ È14 ; x y z œ (1 t) (2 3t) (3 2t) œ 6 6t Ê 'C f(xß yß z) ds œ '0 (6 6t) È14 dt œ 6È14 ’t t2 “ œ Š6È14‹ ˆ "# ‰ œ 3È14 1 # " ! È È È 14. r(t) œ ti tj tk , 1 Ÿ t Ÿ _ Ê ddtr œ i j k Ê ¸ ddtr ¸ œ È3 ; x# y#3 z# œ t# t#3 t# œ 3t#3 _ È _ Ê 'C f(xß yß z) ds œ '1 Š 3t#3 ‹ È3 dt œ 1t ‘ " œ lim ˆ b" 1‰ œ 1 bÄ_ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 940 Chapter 16 Integration in Vector Fields 15. C" : r(t) œ ti t# j , 0 Ÿ t Ÿ 1 Ê ddtr œ i 2tj Ê ¸ ddtr ¸ œ È1 4t# ; x Èy z# œ t Èt# 0 œ t ktk œ 2t $Î# 0 Ê 'C f(xß yß z) ds œ '0 2tÈ1 4t# dt œ ’ "6 a" 4t# b “ œ 6" (5)$Î# "6 œ "6 Š5È5 1‹ ; " 1 since t ! " C# : r(t) œ i j tk, 0 Ÿ t Ÿ 1 Ê Ê ¸ ddtr ¸ œ 1; x Èy z# œ 1 È1 t# œ 2 t# dr dt œ k " Ê 'C f(xß yß z) ds œ '0 a2 t# b (1) dt œ 2t "3 t$ ‘ ! œ 2 3" œ 53 ; therefore 'C f(xß yß z) ds 1 # œ 'C f(xß yß z) ds 'C f(xß yß z) ds œ 56 È5 3# " # 16. C" : r(t) œ tk , 0 Ÿ t Ÿ 1 Ê ddtr œ k Ê ¸ ddtr ¸ œ 1; x Èy z# œ 0 È0 t# œ t# Ê 'C f(xß yß z) ds œ '0 at# b (1) dt œ ’ t3 “ œ 3" ; 1 $ " ! " C# : r(t) œ tj k, 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê ¸ ddtr ¸ œ 1; x Èy z# œ 0 Èt 1 œ Èt 1 Ê 'C f(xß yß z) ds œ '0 ˆÈt 1‰ (1) dt œ 23 t$Î# t‘ ! œ 23 1 œ 3" ; " 1 # C$ : r(t) œ ti j k , 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê ¸ ddtr ¸ œ 1; x Èy z# œ t È1 1 œ t Ê 'C f(xß yß z) ds œ '0 (t)(1) dt œ ’ t2 “ œ "# Ê 'C f(xß yß z) ds œ 'C f ds 'C f ds 'C f ds œ 3" ˆ 3" ‰ "# 1 # " ! $ œ "6 " # $ 17. r(t) œ ti tj tk , 0 a Ÿ t Ÿ b Ê ddtr œ i j k Ê ¸ ddtr ¸ œ È3 ; x#xyy#zz# œ t#ttt#tt# œ 1t Ê 'C f(xß yß z) ds œ 'a ˆ 1t ‰ È3 dt œ ’È3 ln ktk “ œ È3 ln ˆ ba ‰ , since 0 a Ÿ b b b a 18. r(t) œ aa cos tb j aa sin tb k , 0 Ÿ t Ÿ 21 Ê ddtr œ (a sin t) j (a cos t) k Ê ¸ ddtr ¸ œ Èa# sin# t a# cos# t œ kak ; kak sin t, 0 Ÿ t Ÿ 1 Èx# z# œ È0 a# sin# t œ œ Ê 'C f(xß yß z) ds œ '0 kak# sin t dt '1 kak# sin t dt kak sin t, 1 Ÿ t Ÿ 21 1 1 21 #1 œ ca# cos td ! ca# cos td 1 œ ca# (1) a# d ca# a# (1)d œ 4a# 19. (a) ratb œ ti "# tj , 0 Ÿ t Ÿ 4 Ê ddtr œ i "# j Ê ¸ ddtr ¸ œ dr dt œ i 2tj (b) ratb œ ti t j , 0 Ÿ t Ÿ 2 Ê 2 3Î2 2 1 œ ’ 12 a1 4t2 b È5 2 Ê 'C x ds œ '0 t 4 È5 È5 2 dt œ 2 '04 t dt œ ’ È45 t2 “ 4 œ 4È5 Ê ¸ ddtr ¸ œ È1 4t2 Ê 'C x ds œ '0 t È1 4t2 dt 2 È 17 " “ œ 17 12 ! 20. (a) ratb œ ti 4tj , 0 Ÿ t Ÿ 1 Ê ddtr œ i 4j Ê ¸ ddtr ¸ œ È17 Ê 'C Èx 2y ds œ '0 Èt 2a4tb È17 dt 1 œ È17'0 È9t dt œ 3È17'0 Èt dt œ ’2È17 t2Î3 “ œ 2È17 1 1 1 ! (b) C" : ratb œ ti , 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê ¸ ddtr ¸ œ 1; C2 : ratb œ i tj, 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê ¸ ddtr ¸ œ 1 'C Èx 2y ds œ 'C Èx 2y ds 'C Èx 2y ds œ '01 Èt 2a0b dt '02 È1 2atb dt 1 2 2 1 2 È È 1 œ '0 Èt dt '0 È1 2t dt œ 23 t2Î3 ‘ ! ’ 13 a1 2tb2Î3 “ œ 23 Š 5 3 5 31 ‹ œ 5 53 1 ! 21. ratb œ 4ti 3tj , 1 Ÿ t Ÿ 2 Ê ddtr œ 4i 3j Ê ¸ ddtr ¸ œ 5 Ê 'C y ex ds œ 'c1 a3tb ea4tb † 5dt 2 16t œ 15'c1 t e16t dt œ ’ 15 “ 32 e 2 2 2 2 c1 2 2 15 16 15 16 64 64 œ 15 32 e 32 e œ 32 ae e b Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. ! Section 16.1 Line Integrals 941 22. ratb œ acos tbi asin tbj , 0 Ÿ t Ÿ 21 Ê ddtr œ asin tbi acos tbj Ê ¸ ddtr ¸ œ Èsin2 t cos2 t œ 1 Ê 'C ax y 3b ds œ '0 acos t sin t 3b † 1 dt œ csin t cos t 3td 201 œ 61 21 23. ratb œ t2 i t3 j , 1 Ÿ t Ÿ 2 Ê ddtr œ 2ti 3t2 j Ê ¸ ddtr ¸ œ Éa2tb2 a3t2 b2 œ tÈ4 9t2 Ê 'C yx4Î3 ds 2 œ '1 2 ˆt2 ‰2 1 a4 9t2 b † tÈ4 9t2 dt œ '1 t È4 9t2 dt œ ’ 27 3Î2 2 2 at3 b4Î3 “ œ 80 1 È10 13È13 27 24. ratb œ t3 i t4 j , 12 Ÿ t Ÿ 1 Ê ddtr œ 3t2 i 4t3 j Ê ¸ ddtr ¸ œ Éa3t2 b2 a4t3 b2 œ t2 È9 16t2 Ê 'C 1 a9 16t2 b œ '1Î2 t3t † t2 È9 16t2 dt œ '1Î2 t È9 16t2 dt œ ’ 48 1 È4 1 3 Î2 1 “ 1Î2 œ 125 4813 Èy x ds È13 25. C" : ratb œ ti t2 j , 0 Ÿ t Ÿ 1 Ê ddtr œ i 2tj Ê ¸ ddtr ¸ œ È1 4t2 ; C2 : ratb œ a1 tbi a1 tbj, 0 Ÿ t Ÿ 1 Ê ddtr œ i j Ê ¸ ddtr ¸ œ È2 Ê 'C ˆx Èy‰ds œ 'C ˆx Èy‰ds 'C ˆx Èy‰ds 1 2 œ '0 Št Èt2 ‹È1 4t2 dt '0 Ša1 tb È1 t‹ È2dt œ '0 2tÈ1 4t2 dt '0 Š1 t È1 t‹ È2dt 1 1 œ ’ 16 a1 4t2 b 1 3 Î2 1 1 0 0 Î “ È2’t "# t2 23 a1 tb3 2 “ œ 5 1 È È5 1 È È 7 6 2 œ 5 5 76 2 1 6 26. C" : ratb œ ti , 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê ¸ ddtr ¸ œ 1; C2 : ratb œ i tj, 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê ¸ ddtr ¸ œ 1; C3 : ratb œ a1 tbi j, 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê ¸ ddtr ¸ œ 1; C4 : ratb œ a1 tbj, 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê ¸ ddtr ¸ œ 1; Ê 'C x2 y12 1 ds œ 'C x2 y12 1 ds 'C x2 y12 1 ds 'C x2 y12 1 ds 'C x2 y12 1 ds œ' 1 dt 2 0 t 1 ' 1 1 dt 2 0 t 2 ' 2 1 dt 2 0 a1 tb 2 ' 3 4 1 dt 2 0 a1 tb 1 1 1 0 0 1 œ ctan1 td 0 È12 ’tan1 Š Èt 2 ‹“ È12 ’tan1 Š tÈ ‹“ ctan1 a1 tbd 0 œ 12 È22 tan1 Š È12 ‹ 2 1 1 # # dr dr ¸ 27. r(x) œ xi yj œ xi x# j , 0 Ÿ x Ÿ 2 Ê dx œ i xj Ê ¸ dx œ È1 x# ; f(xß y) œ f Šxß x# ‹ œ œ '0 (2x)È1 x# dx œ ’ 23 a1 x# b 2 $Î# # x$ # Š x# ‹ È “ œ 23 ˆ5$Î# 1‰ œ 10 35 2 ! 28. r(t) œ a1 tbi #1 a1 tb2 j, 0 Ÿ t Ÿ 1 Ê ¸ ddtr ¸ œ É1 a1 tb# ; f(xß y) œ f Ša1 tbß #1 a1 tb2 ‹ œ Ê 'C f ds œ '0 1 a1 tb 14 a1 tb4 É1 a1 tb # œ 2x Ê 'C f ds a1 tb 14 a1 tb4 É1 a1 tb# 1 É1 a1 tb# dt œ ' Ša1 tb 14 a1 tb4 ‹ dt œ ’ "# a1 tb2 20 a1 tb5 “ 1 " 0 ! œ 0 ˆ "# #"0 ‰ œ #110 29. r(t) œ (2 cos t) i (2 sin t) j , 0 Ÿ t Ÿ 1# Ê ddtr œ (2 sin t) i (2 cos t) j Ê ¸ ddtr ¸ œ 2; f(xß y) œ f(2 cos tß 2 sin t) œ 2 cos t 2 sin t Ê 'C f ds œ '0 (2 cos t 2 sin t)(2) dt œ c4 sin t 4 cos td ! 1Î2 1Î# œ 4 (4) œ 8 30. r(t) œ (2 sin t) i (2 cos t) j , 0 Ÿ t Ÿ 14 Ê ddtr œ (2 cos t) i (2 sin t) j Ê ¸ ddtr ¸ œ 2; f(xß y) œ f(2 sin tß 2 cos t) œ 4 sin# t 2 cos t Ê 'C f ds œ '0 a4 sin# t 2 cos t b (2) dt œ c4t 2 sin 2t 4 sin td 01Î% œ 1 2Š1 È2‹ 1Î4 31. y œ x2 , 0 Ÿ x Ÿ 2 Ê ratb œ ti t2 j , 0 Ÿ t Ÿ 2 Ê ddtr œ i 2tj Ê ¸ ddtr ¸ œ È1 4t2 Ê A œ 'C fax, yb ds È 1 3 Î2 œ 'C ˆx Èy‰ds œ '0 Št Èt2 ‹È1 4t2 dt œ '0 2tÈ1 4t2 dt œ ’ 16 a1 4t2 b “ œ 17 17 6 2 2 2 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 942 Chapter 16 Integration in Vector Fields È13 3 Ê Aœ 32. 2x 3y œ 6, 0 Ÿ x Ÿ 6 Ê ratb œ ti ˆ2 23 t‰j , 0 Ÿ t Ÿ 6 Ê ddtr œ i 23 j Ê ¸ ddtr ¸ œ œ 'C a4 3x 2ybds œ '0 ˆ4 3t 2ˆ2 23 t‰‰ 313 dt œ È 6 È13 3 'C fax, yb ds '06 ˆ8 35 t‰dt œ È313 8t 65 t2 ‘ 60 œ 26È13 33. r(t) œ at# 1b j 2tk , 0 Ÿ t Ÿ 1 Ê ddtr œ 2tj 2k Ê ¸ ddtr ¸ œ 2Èt# 1; M œ 'C $ (xß yß z) ds œ '0 $ (t) Š2Èt# 1‹ dt 1 œ '0 ˆ 3# t‰ Š2Èt# 1‹ dt œ ’at# 1b “ œ 2$Î# 1 œ 2È2 1 3/2 " 1 ! 34. r(t) œ at# 1b j 2tk , 1 Ÿ t Ÿ 1 Ê ddtr œ 2tj 2k Ê ¸ dr ¸ œ 2Èt# 1; M œ ' $ (xß yß z) ds dt C œ 'c1 1 ˆ15Èat# 1b 2‰ Š2Èt# 1‹ dt œ 'c1 30 at# 1b dt œ ’30 Š t3 t‹“ 1 $ " " œ 60 ˆ 3" 1‰ œ 80; Mxz œ 'C y$ (xß yß z) ds œ 'c1 at# 1b c30 at# 1bd dt 1 œ 'c1 30 at% 1b dt œ ’30 Š t5 t‹“ 1 & " " œ 60 ˆ 5" 1‰ 48 œ 48 Ê y œ MMxz œ 80 œ 53 ; Myz œ 'C x$ (xß yß z) ds œ 'C 0 $ ds œ 0 Ê x œ 0; z œ 0 by symmetry (since $ is independent of z) Ê (xß yß z) œ ˆ!ß 35 ß 0‰ 35. r(t) œ È2t i È2t j a4 t# b k , 0 Ÿ t Ÿ 1 Ê ddtr œ È2i È2j 2tk Ê ¸ ddtr ¸ œ È2 2 4t# œ 2È1 t# ; (a) M œ 'C $ ds œ '0 (3t) Š2È1 t# ‹ dt œ ’2 a1 t# b 1 $Î# " “ œ 2 ˆ2$Î# 1‰ œ 4È2 2 ! (b) M œ 'C $ ds œ '0 a1b Š2È1 t# ‹ dt œ ’tÈ1 t# ln Št È1 t# ‹“ œ ’È2 ln Š1 È2‹“ a0 ln 1b " 1 ! œ È2 ln Š1 È2‹ 36. r(t) œ ti 2tj 23 t$Î# k , 0 Ÿ t Ÿ 2 Ê ddtr œ i 2j t"Î# k Ê ¸ ddtr ¸ œ È1 4 t œ È5 t; M œ 'C $ ds œ '0 ˆ3È5 t‰ ˆÈ5 t‰ dt œ '0 3(5 t) dt œ 32 (5 t)# ‘ ! œ 3# a7# 5# b œ 3# (24) œ 36; 2 2 # Myz œ 'C x$ ds œ '0 t[3(5 t)] dt œ '0 a15t 3t# b dt œ "25 t# t$ ‘ ! œ 30 8 œ 38; 2 2 # Mxz œ 'C y$ ds œ '0 2t[3(5 t)] dt œ 2 '0 a15t 3t# b dt œ 76; Mxy œ 'C z$ ds œ '0 23 t$Î# [3(5 t)] dt 2 2 2 È2 œ 144 È2 Ê x œ Myz œ '0 ˆ10t$Î# 2t&Î# ‰ dt œ 4t&Î# 47 t(Î# ‘ ! œ 4(2)&Î# 47 (2)(Î# œ 16È2 32 7 7 M # 2 M È 144 2 xy Mxz 19 76 19 4 È œ 38 2 36 œ 18 , y œ M œ 36 œ 9 , and z œ M œ 7†36 œ 7 dy dz 37. Let x œ a cos t and y œ a sin t, 0 Ÿ t Ÿ 21. Then dx dt œ a sin t, dt œ a cos t, dt œ 0 ‰# Š dy ˆ dz ‰# dt œ a dt; Iz œ ' ax# y# b $ ds œ ' aa# sin# t a# cos# tb a$ dt Ê Êˆ dx dt dt ‹ dt C 0 # 21 œ '0 a$ $ dt œ 21$ a$ . 21 38. r(t) œ tj (2 2t)k , 0 Ÿ t Ÿ 1 Ê ddtr œ j 2k Ê ¸ ddtr ¸ œ È5; M œ 'C $ ds œ '0 $ È5 dt œ $ È5; 1 Ix œ 'C ay# z# b $ ds œ '0 ct# (2 2t)# d $ È5 dt œ '0 a5t# 8t 4b $ È5 dt œ $ È5 53 t$ 4t# 4t‘ ! œ 53 $ È5 ; 1 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. " Section 16.1 Line Integrals 943 Iy œ 'C ax# z# b $ ds œ '0 c0# (2 2t)# d $ È5 dt œ '0 a4t# 8t 4b $ È5 dt œ $ È5 43 t$ 4t# 4t‘ ! œ 43 $ È5 ; 1 1 " Iz œ 'C ax# y# b $ ds œ '0 a0# t# b $ È5 dt œ $ È5 ’ t3 “ œ 3" $ È5 1 $ " ! 39. r(t) œ (cos t)i (sin t)j tk , 0 Ÿ t Ÿ 21 Ê ddtr œ ( sin t)i (cos t)j k Ê ¸ ddtr ¸ œ Èsin# t cos# t 1 œ È2; (a) Iz œ 'C ax# y# b $ ds œ '0 acos# t sin# tb $ È2 dt œ 21$ È2 21 (b) Iz œ 'C ax# y# b $ ds œ '0 $ È2 dt œ 41$ È2 41 È 40. r(t) œ (t cos t)i (t sin t)j 2 3 2 t$Î# k , 0 Ÿ t Ÿ 1 Ê ddtr œ (cos t t sin t)i (sin t t cos t)j È2t k Ê ¸ ddtr ¸ œ È(t 1)# œ t 1 for 0 Ÿ t Ÿ 1; M œ 'C $ ds œ '0 (t 1) dt œ "2 (t 1)# ‘ ! œ "# a2# 1# b œ #3 ; 1 " Mxy œ 'C z$ ds œ '0 Š 2 3 2 t$Î# ‹ (t 1) dt œ 2 3 2 '0 ˆt&Î# t$Î# ‰ dt œ 2 3 2 27 t(Î# 25 t&Î# ‘ ! È 1 È È 1 " ‰ 16 2 Ê z œ Mxy œ Š 1635 2 ‹ ˆ 23 ‰ œ 321052 ; Iz œ ' ax# y# b $ ds œ 2 3 2 ˆ 27 52 ‰ œ 2 3 2 ˆ 24 35 œ 35 C È È È È M È 7 œ '0 at# cos# t t# sin# tb (t 1) dt œ '0 at$ t# b dt œ ’ t4 t3 “ œ "4 "3 œ 12 1 1 % $ " ! 41. $ (xß yß z) œ 2 z and r(t) œ (cos t)j (sin t)k , 0 Ÿ t Ÿ 1 Ê M œ 21 2 as found in Example 3 of the text; also ¸ ddtr ¸ œ 1; Ix œ 'C ay# z# b $ ds œ '0 acos# t sin# tb (2 sin t) dt œ '0 (2 sin t) dt œ 21 2 1 È 1 # 42. r(t) œ ti 2 3 2 t$Î# j t# k , 0 Ÿ t Ÿ 2 Ê ddtr œ i È2 t"Î# j tk Ê ¸ ddtr ¸ œ È1 2t t# œ È(1 t)# œ 1 t for 0 Ÿ t Ÿ 2; M œ 'C $ ds œ '0 ˆ t"1 ‰ (1 t) dt œ '0 dt œ 2; Myz œ 'C x$ ds œ '0 t ˆ t"1 ‰ (1 t) dt œ ’ t2 “ œ 2; 2 Mxz œ 'C y$ ds œ ' 2 # È 2È2 $Î# 32 dt œ ’ 4152 t&Î# “ œ 15 ; Mxy œ 3 t 0 ! 2 2 'C z$ ds œ ' # $ # t dt œ ’ t6 “ œ 3% 0 # ! 2 # # ! Ê M x œ Myz œ 1, xy # ' # # ' ˆ 8 $ " % ‰ dt œ ’ 92 t% 20t “ œ 329 3220 œ 232 y œ MMxz œ 16 15 , and z œ M œ 3 ; Ix œ C ay z b $ ds œ 0 9 t 4 t 45 ; 2 M Iy œ 'C ax z b $ ds œ '0 t # 2 # ˆ# & # $ t& 64 4" t% ‰ dt œ ’ t3 20 “ œ 38 32 20 œ 15 ; Iz œ ! # ! 'C ax y b $ ds # # 56 œ '0 ˆt# 89 t$ ‰ dt œ ’ t3 29 t% “ œ 83 32 9 œ 9 2 $ # ! 43-46. Example CAS commands: Maple: f := (x,y,z) -> sqrt(1+30*x^2+10*y); g := t -> t; h := t -> t^2; k := t -> 3*t^2; a,b := 0,2; ds := ( D(g)^2 + D(h)^2 + D(k)^2 )^(1/2): 'ds' = ds(t)*'dt'; F := f(g,h,k): 'F(t)' = F(t); Int( f, s=C..NULL ) = Int( simplify(F(t)*ds(t)), t=a..b ); `` = value(rhs(%)); # (a) # (b) # (c) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 944 Chapter 16 Integration in Vector Fields Mathematica: (functions and domains may vary) Clear[x, y, z, r, t, f] f[x_,y_,z_]:= Sqrt[1 30x2 10y] {a,b}= {0, 2}; x[t_]:= t y[t_]:= t2 z[t_]:= 3t2 r[t_]:= {x[t], y[t], z[t]} v[t_]:= D[r[t], t] mag[vector_]:=Sqrt[vector.vector] Integrate[f[x[t],y[t],z[t]] mag[v[t]], {t, a, b}] N[%] 16.2 VECTOR FIELDS, WORK, CIRCULATION, AND FLUX 1. f(xß yß z) œ ax# y# z# b "Î# Ê `` xf œ #" ax# y# z# b $Î# $Î# `f # # # $Î# and `` zf œ z ax# y# z# b ` y œ y ax y z b (2x) œ x ax# y# z# b Ê $Î# ; similarly, ™ f œ # xi #yj #zk$Î# ax y z b 2. f(xß yß z) œ ln Èx# y# z# œ "# ln ax# y# z# b Ê `` xf œ "# Š x# y"# z# ‹ (2x) œ x# yx# z# ; similarly, `` yf œ x# yy# z# and `` zf œ x# yz# z# Ê ™ f œ xx#i yyj#zzk# `g 2y `g z 3. g(xß yß z) œ ez ln ax# y# b Ê `` gx œ x# 2x y# , ` y œ x# y# and ` z œ e z Ê ™ g œ Š x#2xy# ‹ i Š x# 2y y# ‹ j e k 4. g(xß yß z) œ xy yz xz Ê `` gx œ y z, `` gy œ x z, and `` gz œ y x Ê ™ g œ (y z)i (B z)j (x y)k 5. kFk inversely proportional to the square of the distance from (xß y) to the origin Ê È(M(xß y))# (N(xß y))# x œ x# k y# , k 0; F points toward the origin Ê F is in the direction of n œ Èx# i Èx#y y# j y# Ê F œ an , for some constant a 0. Then M(xß y) œ Èx#ax and N(xß y) œ Èx#ay y# y# Ê È(M(xß y))# (N(xß y))# œ a Ê a œ x# k y# Ê F œ kx i # ky# $Î# j , for any constant k 0 ax# y# b$Î# ax y b 6. Given x# y# œ a# b# , let x œ Èa# b# cos t and y œ Èa# b# sin t. Then r œ ŠÈa# b# cos t‹ i ŠÈa# b# sin t‹ j traces the circle in a clockwise direction as t goes from 0 to 21 Ê v œ ŠÈa# b# sin t‹ i ŠÈa# b# cos t‹ j is tangent to the circle in a clockwise direction. Thus, let F œ v Ê F œ yi xj and F(0ß 0) œ 0 . 7. Substitute the parametric representations for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F , and calculate 'C F † ddtr . (a) F œ 3ti 2tj 4tk and ddtr œ i j k Ê F † ddtr œ 9t Ê '0 9t dt œ 9# 1 (b) F œ 3t# i 2tj 4t% k and ddtr œ i 2tj 4t$ k Ê F † ddtr œ 7t# 16t( Ê '0 a7t# 16t( b dt œ 37 t$ 2t) ‘ ! 1 œ 73 2 œ 13 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. " Section 16.2 Vector Fields, Work, Circulation, and Flux (c) r" œ ti tj and r# œ i j tk ; F" œ 3ti 2tj and ddtr" œ i j Ê F" † ddtr" œ 5t Ê '0 5t dt œ #5 ; 1 F# œ 3i 2j 4tk and ddtr# œ k Ê F# † ddtr# œ 4t Ê '0 4t dt œ 2 Ê #5 2 œ #9 1 8. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F, and calculate 'C F † ddtr . " ‰ dr dr " " ' " (a) F œ ˆ t# td ! œ 14 1 j and dt œ i j k Ê F † dt œ t# 1 Ê 0 t# 1 dt œ ctan 1 " " ‰ dr dr 2t $ # ' 2t (b) F œ ˆ t# 1 j and dt œ i 2tj 4t k Ê F † dt œ t# 1 Ê 0 t# 1 dt œ cln at 1bd ! œ ln 2 1 " dr" d r" d r# " ‰ " " (c) r" œ ti tj and r# œ i j tk ; F" œ ˆ t# 1 j and dt œ i j Ê F" † dt œ t# 1 ; F# œ # j and dt œ k 1 " Ê F# † ddtr# œ 0 Ê '0 t# 1 dt œ 4 1 9. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F, and calculate 'C F † ddtr . " (a) F œ Èti 2tj Ètk and ddtr œ i j k Ê F † ddtr œ 2Èt 2t Ê '0 ˆ2Èt 2t‰ dt œ 43 t$Î# t# ‘ ! œ "3 1 (b) F œ t# i 2tj tk and ddtr œ i 2tj 4t$ k Ê F † ddtr œ 4t% 3t# Ê '0 a4t% 3t# b dt œ 45 t& t$ ‘ ! œ "5 1 " (c) r" œ ti tj and r# œ i j tk ; F" œ 2tj Èt k and ddtr" œ i j Ê F" † ddtr" œ 2t Ê '0 2t dt œ 1; 1 F# œ Èti 2j k and ddtr# œ k Ê F# † ddtr# œ 1 Ê '0 dt œ 1 Ê 1 1 œ 0 1 10. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F, and calculate 'C F † ddtr . (a) F œ t# i t# j t# k and ddtr œ i j k Ê F † ddtr œ 3t# Ê '0 3t# dt œ 1 1 (b) F œ t$ i t' j t& k and ddtr œ i 2tj 4t$ k Ê F † ddtr œ t$ 2t( 4t) Ê '0 at$ 2t( 4t) b dt 1 % ) " 17 œ ’ t4 t4 94 t* “ œ 18 ! (c) r" œ ti tj and r# œ i j tk ; F" œ t# i and ddtr" œ i j Ê F" † ddtr" œ t# Ê '0 t# dt œ 3" ; 1 F# œ i tj tk and ddtr# œ k Ê F# † ddtr# œ t Ê '0 t dt œ "# Ê "3 #" œ 65 1 11. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F, and calculate 'C F † ddtr . (a) F œ a3t# 3tb i 3tj k and ddtr œ i j k Ê F † ddtr œ 3t# 1 Ê '0 a3t# 1b dt œ ct$ td ! œ 2 1 " (b) F œ a3t# 3tb i 3t% j k and ddtr œ i 2tj 4t$ k Ê F † ddtr œ 6t& 4t$ 3t# 3t " Ê '0 a6t& 4t$ 3t# 3tb dt œ t' t% t$ 3# t# ‘ ! œ 3# 1 (c) r" œ ti tj and r# œ i j tk ; F" œ a3t# 3tb i k and ddtr" œ i j Ê F" † ddtr" œ 3t# 3t " Ê '0 a3t# 3tb dt œ t$ 32 t# ‘ ! œ "# ; F# œ 3tj k and ddtr# œ k Ê F# † ddtr# œ 1 Ê '0 dt œ 1 1 1 Ê "# 1 œ 12 12. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector field F, and calculate 'C F † ddtr . (a) F œ 2ti 2tj 2tk and ddtr œ i j k Ê F † ddtr œ 6t Ê '0 6t dt œ c3t# d ! œ 3 1 " Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 945 946 Chapter 16 Integration in Vector Fields (b) F œ at# t% b i at% tb j at t# b k and ddtr œ i 2tj 4t$ k Ê F † ddtr œ 6t& 5t% 3t# Ê '0 a6t& 5t% 3t# b dt œ ct' t& t$ d ! œ 3 1 " (c) r" œ ti tj and r# œ i j tk ; F" œ ti tj 2tk and ddtr" œ i j Ê F" † ddtr" œ 2t Ê '0 2t dt œ "; 1 F# œ (1 t)i (t 1)j 2k and ddtr# œ k Ê F# † ddtr# œ 2 Ê '0 2 dt œ 2 Ê " 2 œ 3 1 3 13. x œ t, y œ 2t 1, 0 Ÿ t Ÿ 3 Ê dx œ dt Ê 'C ax yb dx œ '0 at a2t 1bb dt œ '0 at 1b dt œ "# t2 t‘ ! œ 15 2 3 3 14. x œ t, y œ t2 , 1 Ÿ t Ÿ 2 Ê dy œ 2t dt Ê 'C xy dy œ '1 tt2 a2tb dt œ '1 2 dt œ c2td21 œ 2 2 2 15. C1 : x œ t, y œ 0, 0 Ÿ t Ÿ 3 Ê dy œ 0; C2 : x œ 3, y œ t, 0 Ÿ t Ÿ 3 Ê dy œ dt Ê 'C ax2 y2 b dy œ 'C ax2 y2 b dx 'C ax2 y2 b dx œ '0 at2 02 b † 0 '0 a32 t2 b dt œ '0 a9 t2 bdt œ 9t 13 t3 ‘ ! œ 36 3 1 3 3 3 2 16. C1 : x œ t, y œ 3t, 0 Ÿ t Ÿ 1 Ê dx œ dt; C2 : x œ 1 t, y œ 3, 0 Ÿ t Ÿ 1 Ê dx œ dt; C3 : x œ 0, y œ 3 t, 0 Ÿ t Ÿ 3 Ê dx œ 0 Ê 'C Èx y dx œ 'C Èx y dx 'C Èx y dx 'C Èx y dx 1 2 3 œ '0 Èt 3t dt '0 Èa1 tb 3 a1bdt '0 È0 a3 tb † 0 œ '0 2Èt dt '0 È4 t dt 1 1 3 1 1 1 1 È œ 43 t2Î3 ‘ ! ’ 23 a4 tb2Î3 “ œ 43 Š2È3 16 3 ‹œ 2 34 ! 17. ratb œ ti j t2 k , 0 Ÿ t Ÿ 1 Ê dx œ dt, dy œ 0, dz œ 2t dt (a) 'C ax y zb dx œ '0 at 1 t2 b dt œ 12 t2 t 13 t3 ‘ ! œ 56 1 1 (b) 'C ax y zb dy œ '0 at 1 t2 b † 0 œ 0 1 (c) 'C ax y zb dz œ '0 at 1 t2 b 2t dt œ '0 a2t2 2t 2t3 b dt œ œ 23 t3 t2 12 t4 ‘ ! œ 56 1 1 1 18. ratb œ acos tbi asin tbj acos tbk , 0 Ÿ t Ÿ 1 Ê dx œ sin t dt, dy œ cos t dt, dz œ sin t dt (a) 'C x z dx œ '0 acos tb acos tbasin tbdt œ '0 cos2 t sin tdt œ ’ 13 acos tb3 “ œ 23 1 1 1 ! 1 1 1 1 (b) 'C x z dy œ '0 acos tb acos tbacos tbdt œ '0 cos3 t dt œ '0 a1 sin2 tb cos t dt œ ’ 13 asin tb3 sin t“ œ 0 (c) 'C x y z dz œ '0 acos tbasin tb acos tbasin tbdt œ '0 cos t sin 1 1 1 œ 18 '0 a1 cos 4tb dt œ 18 t 32 sin 4t‘ ! œ 18 1 2 2 t dt œ 14 '0 sin 1 2 2t dt œ 41 1 19. r œ ti t# j tk , 0 Ÿ t Ÿ 1, and F œ xyi yj yzk Ê F œ t$ i t# j t$ k and ddtr œ i 2tj k Ê F † ddtr œ 2t$ Ê work œ '0 2t$ dt œ "# 1 20. r œ (cos t)i (sin t)j 6t k , 0 Ÿ t Ÿ 21, and F œ 2yi 3xj (x y)k Ê F œ (2 sin t)i (3 cos t)j (cos t sin t)k and ddtr œ ( sin t)i (cos t)j 6" k Ê F † ddtr œ 3 cos# t 2sin2 t "6 cos t 6" sin t Ê work œ '0 ˆ3 cos# t 2 sin2 t 6" cos t 6" sin t‰ dt 21 #1 œ 32 t 34 sin 2t t sin2 2t "6 sin t "6 cos t‘ ! œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. ' 1 ! 1 cos 4t dt 2 0 Section 16.2 Vector Fields, Work, Circulation, and Flux 21. r œ (sin t)i (cos t)j tk , 0 Ÿ t Ÿ 21, and F œ zi xj yk Ê F œ ti (sin t)j (cos t)k and dr dt œ (cos t)i (sin t)j k Ê F † ddtr œ t cos t sin# t cos t Ê work œ '0 at cos t sin# t cos tb dt 21 #1 œ cos t t sin t 2t sin4 2t sin t‘ ! œ 1 22. r œ (sin t)i (cos t)j 6t k , 0 Ÿ t Ÿ 21, and F œ 6zi y# j 12xk Ê F œ ti acos# tbj (12 sin t)k and dr " dt œ (cos t)i (sin t)j 6 k Ê F † ddtr œ t cos t sin t cos# t 2 sin t Ê work œ '0 at cos t sin t cos# t 2 sin tb dt œ cos t t sin t 13 cos$ t 2 cos t‘ ! œ 0 21 #1 23. x œ t and y œ x# œ t# Ê r œ ti t# j , 1 Ÿ t Ÿ 2, and F œ xyi (x y)j Ê F œ t$ i at t# b j and dr dt œ i 2tj Ê F † ddtr œ t$ a2t# 2t$ b œ 3t$ 2t# Ê 'C xy dx (x y) dy œ 'C F † ddtr dt œ 'c" a3t$ 2t# b dt # # 69 ‰ ˆ 3 2 ‰ 45 18 œ 34 t% 32 t$ ‘ " œ ˆ12 16 3 4 3 œ 4 3 œ 4 24. Along (0ß 0) to (1ß 0): r œ ti , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ ti tj and ddtr œ i Ê F † ddtr œ t; Along (1ß 0) to (0ß 1): r œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ (1 2t)i j and dr dr dt œ i j Ê F † dt œ 2t; Along (0ß 1) to (0ß 0): r œ (1 t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ (t 1)i (1 t)j and dr dr dt œ j Ê F † dt œ t 1 Ê " œ c2t# td ! œ 2 1 œ 1 25. r œ xi yj œ y# i yj , 2 'C (x y) dx (x y) dy œ '01 t dt '01 2t dt '01 (t 1) dt œ '01 (4t 1) dt dr dr 1, and F œ x# i yj œ y% i yj Ê dy œ 2yi j and F † dy œ 2y& y y c1 c1 dr 4‰ 3 63 39 Ê 'C F † T ds œ '2 F † dy dy œ '2 a2y& yb dy œ 3" y' "# y# ‘ # œ ˆ 3" #" ‰ ˆ 64 3 # œ # 3 œ # " 26. r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ 1# , and F œ yi xj Ê F œ (sin t)i (cos t)j and ddtr œ ( sin t)i (cos t)j Ê F † ddtr œ sin# t cos# t œ 1 Ê 'C F † dr œ '0 1Î2 (1) dt œ 1# 27. r œ (i j) t(i 2j) œ (1 t)i (1 2t)j , 0 Ÿ t Ÿ 1, and F œ xyi (y x)j Ê F œ a1 3t 2t# b i tj and dr dt œ i 2j Ê F † ddtr œ 1 5t 2t# Ê work œ 'C F † ddtr dt œ '0 a1 5t 2t# b dt œ t 25 t# 23 t$ ‘ ! œ 25 6 1 " 28. r œ (2 cos t)i (2 sin t)j , 0 Ÿ t Ÿ 21, and F œ ™ f œ 2(x y)i 2(x y)j Ê F œ 4(cos t sin t)i 4(cos t sin t)j and ddtr œ (2 sin t)i (2 cos t)j Ê F † ddtr œ 8 asin t cos t sin# tb 8 acos# t cos t sin tb œ 8 acos# t sin# tb œ 8 cos 2t Ê work œ 'C ™ f † dr œ 'C F † ddtr dt œ '0 8 cos 2t dt œ c4 sin 2td #!1 œ 0 21 29. (a) r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ 21, F" œ xi yj , and F# œ yi xj Ê ddtr œ ( sin t)i (cos t)j , F" œ (cos t)i (sin t)j , and F# œ ( sin t)i (cos t)j Ê F" † ddtr œ 0 and F# † ddtr œ sin# t cos# t œ 1 Ê Circ" œ '0 0 dt œ 0 and Circ# œ '0 dt œ 21; n œ (cos t)i (sin t)j Ê F" † n œ cos# t sin# t œ 1 and 21 21 F# † n œ 0 Ê Flux" œ '0 dt œ 21 and Flux# œ '0 0 dt œ 0 21 21 (b) r œ (cos t)i (4 sin t)j , 0 Ÿ t Ÿ 21 Ê ddtr œ ( sin t)i (4 cos t)j , F" œ (cos t)i (4 sin t)j , and F# œ (4 sin t)i (cos t)j Ê F" † ddtr œ 15 sin t cos t and F# † ddtr œ 4 Ê Circ" œ '0 15 sin t cos t dt 21 œ "25 sin# t‘ ! œ 0 and Circ# œ '0 4 dt œ 81; n œ Š È417 cos t‹ i Š È"17 sin t‹ j Ê F" † n #1 21 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 947 948 Chapter 16 Integration in Vector Fields œ È417 cos# t È417 sin# t and F# † n œ È1517 sin t cos t Ê Flux" œ '0 (F" † n) kvk dt œ '0 Š È417 ‹ È17 dt 21 21 # ‘ œ 81 and Flux# œ '0 (F# † n) kvk dt œ '0 Š È1517 sin t cos t‹ È17 dt œ 15 2 sin t ! œ 0 21 21 #1 30. r œ (a cos t)i (a sin t)j , 0 Ÿ t Ÿ 21, F" œ 2xi 3yj , and F# œ 2xi (x y)j Ê ddtr œ (a sin t)i (a cos t)j , F" œ (2a cos t)i (3a sin t)j , and F# œ (2a cos t)i (a cos t a sin t)j Ê n kvk œ (a cos t)i (a sin t)j , F" † n kvk œ 2a# cos# t 3a# sin# t, and F# † n kvk œ 2a# cos# t a# sin t cos t a# sin# t Ê Flux" œ '0 a2a# cos# t 3a# sin# tb dt œ 2a# 2t sin4 2t ‘ ! 3a# 2t sin4 2t ‘ ! œ 1a# , and 21 #1 #1 Flux# œ '0 a2a# cos# t a# sin t cos t a# sin# tb dt œ 2a# 2t sin4 2t ‘ ! a# csin# td ! a# 2t sin4 2t ‘ ! œ 1a# 21 #1 #1 #1 # 31. F" œ (a cos t)i (a sin t)j , ddtr" œ (a sin t)i (a cos t)j Ê F" † ddtr" œ 0 Ê Circ" œ 0; M" œ a cos t, N" œ a sin t, dx œ a sin t dt, dy œ a cos t dt Ê Flux" œ 'C M" dy N" dx œ '0 aa# cos# t a# sin# tb dt 1 œ '0 a# dt œ a# 1; 1 F# œ ti , ddtr# œ i Ê F# † ddtr# œ t Ê Circ# œ 'ca t dt œ 0; M# œ t, N# œ 0, dx œ dt, dy œ 0 Ê Flux# a œ 'C M# dy N# dx œ 'ca 0 dt œ 0; therefore, Circ œ Circ" Circ# œ 0 and Flux œ Flux" Flux# œ a# 1 a 32. F" œ aa# cos# tb i aa# sin# tb j , ddtr" œ (a sin t)i (a cos t)j Ê F" † ddtr" œ a$ sin t cos# t a$ cos t sin# t Ê Circ" œ '0 aa$ sin t cos# t a$ cos t sin# tb dt œ 2a3 ; M" œ a# cos# t, N" œ a# sin# t, dy œ a cos t dt, 1 $ dx œ a sin t dt Ê Flux" œ 'C M" dy N" dx œ '0 aa$ cos$ t a$ sin$ tb dt œ 43 a$ ; 1 F# œ t# i , ddtr# œ i Ê F# † ddtr# œ t# Ê Circ# œ 'ca t# dt œ 2a3 ; M# œ t# , N# œ 0, dy œ 0, dx œ dt a $ Ê Flux# œ 'C M# dy N# dx œ 0; therefore, Circ œ Circ" Circ# œ 0 and Flux œ Flux" Flux# œ 43 a$ 33. F" œ (a sin t)i (a cos t)j , ddtr" œ (a sin t)i (a cos t)j Ê F" † ddtr" œ a# sin# t a# cos# t œ a# Ê Circ" œ '0 a# dt œ a# 1 ; M" œ a sin t, N" œ a cos t, dx œ a sin t dt, dy œ a cos t dt 1 Ê Flux" œ 'C M" dy N" dx œ '0 aa# sin t cos t a# sin t cos tb dt œ 0; F# œ tj , ddtr# œ i Ê F# † ddtr# œ 0 1 Ê Circ# œ 0; M# œ 0, N# œ t, dx œ dt, dy œ 0 Ê Flux# œ 'C M# dy N# dx œ 'ca t dt œ 0; therefore, a Circ œ Circ" Circ# œ a# 1 and Flux œ Flux" Flux# œ 0 34. F" œ aa# sin# tb i aa# cos# tb j , ddtr" œ (a sin t)i (a cos t)j Ê F" † ddtr" œ a$ sin$ t a$ cos$ t Ê Circ" œ '0 aa$ sin$ t a$ cos$ tb dt œ 43 a$ ; M" œ a# sin# t, N" œ a# cos# t, dy œ a cos t dt, dx œ a sin t dt 1 Ê Flux" œ 'C M" dy N" dx œ '0 aa$ cos t sin# t a$ sin t cos# tb dt œ 23 a$ ; F# œ t# j , ddtr# œ i Ê F# † ddtr# œ 0 1 Ê Circ# œ 0; M# œ 0, N# œ t# , dy œ 0, dx œ dt Ê Flux# œ 'C M# dy N# dx œ 'ca t# dt œ 23 a$ ; therefore, a Circ œ Circ" Circ# œ 43 a$ and Flux œ Flux" Flux# œ 0 35. (a) r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr œ (sin t)i (cos t)j and F œ (cos t sin t)i acos# t sin# tb j Ê F † ddtr œ sin t cos t sin# t cos t Ê 'C F † T ds 1 œ '0 a sin t cos t sin# t cos tb dt œ 2" sin# t #t sin4 2t sin t‘ ! œ 1# 1 (b) r œ (1 2t)i , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr œ 2i and F œ (1 2t)i (1 2t)# j Ê F † ddtr œ 4t 2 Ê 'C F † T ds œ '0 (4t 2) dt œ c2t# 2td ! œ 0 1 " Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.2 Vector Fields, Work, Circulation, and Flux 949 (c) r" œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr" œ i j and F œ (1 2t)i a1 2t 2t# b j Ê F † ddtr" œ (2t 1) a1 2t 2t# b œ 2t# Ê Flow" œ 'C F † ddtr" œ '0 2t# dt œ 32 ; r# œ ti (t 1)j , 1 " 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr# œ i j and F œ i at# t# 2t 1b j œ i a2t# 2t 1b j Ê F † ddtr# œ 1 a2t# 2t 1b œ 2t 2t# Ê Flow# œ 'C F † ddtr# œ '0 a2t 2t# b dt 1 # " œ t# 23 t$ ‘ ! œ 3" Ê Flow œ Flow" Flow# œ 32 3" œ 1 36. From (1ß 0) to (0ß 1): r" œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr" œ i j , F œ i a1 2t 2t# b j , and n" kv" k œ i j Ê F † n" kv" k œ 2t 2t# Ê Flux" œ '0 a2t 2t# b dt 1 " œ t# 23 t$ ‘ ! œ 3" ; From (0ß 1) to (1ß 0): r# œ ti (1 t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr# œ i j , F œ (1 2t)i a1 2t 2t# b j , and n# kv# k œ i j Ê F † n# kv# k œ (2t 1) a1 2t 2t# b œ 2 4t 2t# " Ê Flux# œ '0 a2 4t 2t# b dt œ 2t 2t# 23 t$ ‘ ! œ 23 ; 1 From (1ß 0) to (1ß 0): r$ œ (1 2t)i , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr$ œ 2i , F œ (1 2t)i a1 4t 4t# b j , and n$ kv$ k œ 2j Ê F † n$ kv$ k œ 2 a1 4t 4t# b Ê Flux$ œ 2 '0 a1 4t 4t# b dt œ 2 t 2t# 43 t$ ‘ ! œ 32 Ê Flux œ Flux" Flux# Flux$ œ 3" 32 32 œ "3 1 " 37. (a) y œ 2x, 0 Ÿ x Ÿ 2 Ê ratb œ ti 2tj , 0 Ÿ t Ÿ 2 Ê ddtr œ i 2tj Ê F † ddtr œ Ša2tb2 i 2atba2tbj‹ † ai 2jb œ 4t2 8t2 œ 12t2 Ê Flow œ 'C F † ddtr dt œ '0 12t2 dt œ c4t3 d ! œ 32 2 2 2 (b) y œ x2 , 0 Ÿ x Ÿ 2 Ê ratb œ ti t2 j , 0 Ÿ t Ÿ 2 Ê ddtr œ i 2tj Ê F † ddtr œ Šat2 b i 2atbat2 bj‹ † ai 2tjb œ t4 4t4 œ 5t4 Ê Flow œ 'C F † ddtr dt œ '0 5t4 dt œ ct5 d ! œ 32 2 2 (c) answers will vary, one possible path is y œ 12 x3 , 0 Ÿ x Ÿ 2 Ê ratb œ ti "# t3 j , 0 Ÿ t Ÿ 2 Ê ddtr œ i 3t2 j Ê F † ddtr œ Šˆ "# t3 ‰ i 2atbˆ "# t3 ‰j‹ † ai 3t2 jb œ 14 t6 32 t6 œ 74 t6 Ê Flow œ 'C F † ddtr dt œ '0 74 t6 dt œ 14 t7 ‘ ! 2 2 œ 32 38. (a) C1 : ratb œ a1 tbi j , 0 Ÿ t Ÿ 2 Ê ddtr œ i Ê F † ddtr œ aa1bi aa1 tb 2a1bbjb † aib œ 1; C2 : ratb œ i a1 tbj , 0 Ÿ t Ÿ 2 Ê ddtr œ j Ê F † ddtr œ aa1 tbi aa1b 2a1 tbbjb † ajb œ 2t 1; C3 : ratb œ at 1bi j , 0 Ÿ t Ÿ 2 Ê ddtr œ i Ê F † ddtr œ aa1bi aat 1b 2a1bbjb † aib œ 1; C4 : ratb œ i at 1bj , 0 Ÿ t Ÿ 2 Ê ddtr œ j Ê F † ddtr œ aat 1bi aa1b 2at 1bbjb † ajb œ 2t 1; Ê Flow œ 'C F † ddtr dt œ 'C F † ddtr dt 'C F † ddtr dt 'C F † ddtr dt 'C F † ddtr dt 1 2 œ '0 a1b dt '0 a2t 1b dt '0 a1b dt ' 2 2 2 3 4 2 2 2 a2t 1b dt œ ctd 2! ct2 td ! ctd !2 ct2 td ! 0 œ 2 2 2 2 œ 0 (b) x2 y2 œ 4 Ê ratb œ a2cos tbi a2sin tbj , 0 Ÿ t Ÿ 21 Ê ddtr œ a2sin tbi a2cos tbj Ê F † ddtr œ aa2sin tbi a2cos t 2a2sin tbbjb † aa2sin tbi a2cos tbjb œ 4sin2 t 4cos2 t 8sin t cos t œ 4cos 2t 4sin 2t Ê Flow œ 'C F † ddtr dt œ '0 a4cos 2t 4sin 2tb dt œ c2sin 2t 2cos 2td 2!1 œ 0 21 (c) answers will vary, one possible path is: C1 : ratb œ ti , 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê F † ddtr œ aa0bi at 2a1bbjb † aib œ 0; C2 : ratb œ a1 tbi tj , 0 Ÿ t Ÿ 1 Ê ddtr œ i j Ê F † ddtr œ ati aa1 tb 2tbjb † ai jb œ 1; C3 : ratb œ a1 tbj , 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê F † ddtr œ aa1 tbi a0 2a1 tbbjb † ajb œ 2t 1; Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 2 950 Chapter 16 Integration in Vector Fields Ê Flow œ 'C F † ddtr dt œ 'C F † ddtr dt 'C F † ddtr dt 'C F † ddtr dt œ '0 a0b dt '0 a1b dt '0 a2t 1b dt 1 1 2 1 1 3 1 œ 0 ctd 1! ct2 td ! œ 1 a1b œ 0 39. F œ Èx#y y# i Èx#x y# j on x# y# œ 4; at (2ß 0), F œ j ; at (0ß 2), F œ i ; at (2ß 0), È F œ j ; at (!ß 2), F œ i ; at ŠÈ2ß È2‹ , F œ 3 i " j ; # # È at ŠÈ2ß È2‹ , F œ #3 i #" j ; at ŠÈ2ß È2‹ , È F œ #3 i #" j ; at ŠÈ2ß È2‹ , F œ È3 " # i # j 40. F œ xi yj on x# y# œ 1; at (1ß 0), F œ i ; at (1ß 0), F œ i ; at (0ß 1), F œ j ; at (0ß 1), È È F œ j ; at Š "# ß #3 ‹, F œ #" i #3 j ; È È at Š "# ß #3 ‹, F œ "# i #3 j ; È È at Š "# ß #3 ‹, F œ #" i #3 j ; È È at Š "# ß #3 ‹, F œ "# i #3 j . 41. (a) G œ P(xß y)i Q(xß y)j is to have a magnitude Èa# b# and to be tangent to x# y# œ a# b# in a counterclockwise direction. Thus x# y# œ a# b# Ê 2x 2yyw œ 0 Ê yw œ xy is the slope of the tangent line at any point on the circle Ê yw œ ba at (aß b). Let v œ bi aj Ê kvk œ Èa# b# , with v in a counterclockwise direction and tangent to the circle. Then let P(xß y) œ y and Q(xß y) œ x Ê G œ yi xj Ê for (aß b) on x# y# œ a# b# we have G œ bi aj and kGk œ Èa# b# . (b) G œ ˆÈx# y# ‰ F œ ŠÈa# b# ‹ F . 42. (a) From Exercise 41, part a, yi xj is a vector tangent to the circle and pointing in a counterclockwise direction Ê yi xj is a vector tangent to the circle pointing in a clockwise direction Ê G œ Èyxi #xjy# is a unit vector tangent to the circle and pointing in a clockwise direction. (b) G œ F 43. The slope of the line through (xß y) and the origin is yx Ê v œ xi yj is a vector parallel to that line and pointing away from the origin Ê F œ Èxxi #yjy# is the unit vector pointing toward the origin. 44. (a) From Exercise 43, Èxxi #yjy# is a unit vector through (xß y) pointing toward the origin and we want kFk to have magnitude Èx# y# Ê F œ Èx# y# Š Èxxi #yjy# ‹ œ xi yj . yj (b) We want kFk œ Èx#C y# where C Á 0 is a constant Ê F œ Èx#C y# Š Èxxi #yjy# ‹ œ C Š xx#i y# ‹. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.2 Vector Fields, Work, Circulation, and Flux 951 45. Yes. The work and area have the same numerical value because work œ 'C F † dr œ 'C yi † dr ‘ œ 'b [f(t)i] † i df dt j dt a [On the path, y equals f(t)] œ 'a f(t) dt œ Area under the curve b [because f(t) 0] dr 46. r œ xi yj œ xi f(x)j Ê dx œ i f w (x)j ; F œ Èx#k y# (xi yj) has constant magnitude k and points away †f (x) dr d È # from the origin Ê F † dx œ Èx#kx y# Èk†xy#†f(x)y# œ kxÈx#k†f(x) œ k dx x [f(x)]# , by the chain rule [f(x)]# w w dr d È # Ê 'C F † T ds œ 'C F † dx dx œ 'a k dx x [f(x)]# dx œ k Èx# [f(x)]# ‘ a b b œ k ˆÈb# [f(b)]# Èa# [f(a)]# ‰ , as claimed. 47. F œ 4t$ i 8t# j 2k and ddtr œ i 2tj Ê F † ddtr œ 12t$ Ê Flow œ '0 12t$ dt œ c3t% d ! œ 48 2 # 48. F œ 12t# j 9t# k and ddtr œ 3j 4k Ê F † ddtr œ 72t# Ê Flow œ '0 72t# dt œ c24t$ d ! œ 24 1 " 49. F œ (cos t sin t)i (cos t)k and ddtr œ ( sin t)i (cos t)k Ê F † ddtr œ sin t cos t 1 Ê Flow œ '0 ( sin t cos t 1) dt œ 2" cos# t t‘ ! œ ˆ #" 1‰ ˆ #" 0‰ œ 1 1 1 50. F œ (2 sin t)i (2 cos t)j 2k and ddtr œ (2 sin t)i (2 cos t)j 2k Ê F † ddtr œ 4 sin# t 4 cos# t 4 œ 0 Ê Flow œ 0 51. C" : r œ (cos t)i (sin t)j tk , 0 Ÿ t Ÿ 1# Ê F œ (2 cos t)i 2tj (2 sin t)k and ddtr œ ( sin t)i (cos t)j k Ê F † ddtr œ 2 cos t sin t 2t cos t 2 sin t œ sin 2t 2t cos t 2 sin t Ê Flow" œ '0 1Î2 1Î# ( sin 2t 2t cos t 2 sin t) dt œ 2" cos 2t 2t sin t 2 cos t 2 cos t‘ ! œ 1 1; C# : r œ j 1# (1 t)k , 0 Ÿ t Ÿ 1 Ê F œ 1(1 t)j 2k and ddtr œ 1# k Ê F † ddtr œ 1 Ê Flow# œ '0 1 dt œ c1td "! œ 1; 1 C$ : r œ ti (1 t)j , 0 Ÿ t Ÿ 1 Ê F œ 2ti 2(1 t)k and ddtr œ i j Ê F † ddtr œ 2t Ê Flow$ œ '0 2t dt œ ct# d ! œ 1 Ê Circulation œ (1 1) 1 1 œ 0 1 " dy dz ` f dx ` f dy ` f dz " dr d # # # 52. F † ddtr œ x dx dt y dt z dt œ ` x dt ` y dt ` z dt , where f(xß yß z) œ # ax y x b Ê F † dt œ dt afaratbbb by the chain rule Ê Circulation œ 'C F † ddtr dt œ 'a dtd afaratbbb dt œ farabbb faraabb. Since C is an entire ellipse, b rabb œ raab, thus the Circulation œ 0. 53. Let x œ t be the parameter Ê y œ x# œ t# and z œ x œ t Ê r œ ti t# j tk , 0 Ÿ t Ÿ 1 from (0ß 0ß 0) to (1ß 1ß 1) Ê ddtr œ i 2tj k and F œ xyi yj yzk œ t$ i t# j t$ k Ê F † ddtr œ t$ 2t$ t$ œ 2t$ Ê Flow œ '0 2t$ dt 1 œ "# ` f dy ` z dz df dr # $ ) 54. (a) F œ ™ axy# z$ b Ê F † ddtr œ `` xf dx dt ` y dt ` z dt œ dt , where f(xß yß z) œ xy z Ê C F † dt dt œ 'a dtd afaratbbb dt œ farabbb faraabb œ 0 since C is an entire ellipse. b Ð2ß1ß 1Ñ (b) 'C F † ddtr œ 'Ð1ß1ß1Ñ d # $ # $ Ð#ß"ß"Ñ # $ # $ dt axy z b dt œ cxy z d Ð"ß"ß"Ñ œ (2)(1) (1) (1)(1) (1) œ 2 1 œ 3 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 952 Chapter 16 Integration in Vector Fields 55-60. Example CAS commands: Maple: with( LinearAlgebra );#55 F := r -> < r[1]*r[2]^6 | 3*r[1]*(r[1]*r[2]^5+2) >; r := t -> < 2*cos(t) | sin(t) >; a,b := 0,2*Pi; dr := map(diff,r(t),t); # (a) F(r(t)); # (b) q1 := simplify( F(r(t)) . dr ) assuming t::real; # (c) q2 := Int( q1, t=a..b ); value( q2 ); Mathematica: (functions and bounds will vary): Exercises 55 and 56 use vectors in 2 dimensions Clear[x, y, t, f, r, v] f[x_, y_]:= {x y6 , 3x (x y5 2)} {a, b}={0, 21}; x[t_]:= 2 Cos[t] y[t_]:= Sin[t] r[t_]:={x[t], y[t]} v[t_]:= r'[t] integrand= f[x[t], y[t]] . v[t] //Simplify Integrate[integrand,{t, a, b}] N[%] If the integration takes too long or cannot be done, use NIntegrate to integrate numerically. This is suggested for exercises 57 - 60 that use vectors in 3 dimensions. Be certain to leave spaces between variables to be multiplied. Clear[x, y, z, t, f, r, v] f[x_, y_, z_]:= {y y z Cos[x y z], x2 x z Cos[x y z], z x y Cos[x y z]} {a, b}={0, 21}; x[t_]:= 2 Cos[t] y[t_]:= 3 Sin[t] z[t_]:= 1 r[t_]:={x[t], y[t], z[t]} v[t_]:= r'[t] integrand= f[x[t], y[t],z[t]] . v[t] //Simplify NIntegrate[integrand,{t, a, b}] 16.3 PATH INDEPENDENCE, POTENTIAL FUNCTIONS, AND CONSERVATIVE FIELDS 1. `P `N `M `P `N `M `y œ x œ `z , `z œ y œ `x , `x œ z œ `y 2. `P `N `M `P `N `M ` y œ x cos z œ ` z , ` z œ y cos z œ ` x , ` x œ sin z œ ` y 3. `P `N ` y œ 1 Á 1 œ ` z 5. `N `M `x œ 0 Á 1 œ `y 6. `P `N `M `P `N `M x ` y œ 0 œ ` z , ` z œ 0 œ ` x , ` x œ e sin y œ ` y Ê Conservative Ê Not Conservative 4. Ê Conservative `N `M ` x œ 1 Á 1 œ ` y Ê Not Conservative Ê Not Conservative Ê Conservative Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.3 Path Independence, Potential Functions, and Conservative Fields 7. `f ` x œ 2x # 953 # Ê f(xß yß z) œ x# g(yß z) Ê `` yf œ `` yg œ 3y Ê g(yß z) œ 3y# h(z) Ê f(xß yß z) œ x# 3y# h(z) # Ê `` zf œ hw (z) œ 4z Ê h(z) œ 2z# C Ê f(xß yß z) œ x# 3y# 2z# C 8. `f `x œ y z Ê f(xß yß z) œ (y z)x g(yß z) Ê `` yf œ x `` yg œ x z Ê `` yg œ z Ê g(yß z) œ zy h(z) Ê f(xß yß z) œ (y z)x zy h(z) Ê `` zf œ x y hw (z) œ x y Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ (y z)x zy C 9. `f y2z `x œ e y2z œ xe Ê f(xß yß z) œ xey2z g(yß z) Ê `` yf œ xey2z `` yg œ xey2z Ê `` yg œ 0 Ê f(xß yß z) h(z) Ê `` zf œ 2xey2z hw (z) œ 2xey2z Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xey2z C 10. `` xf œ y sin z Ê f(xß yß z) œ xy sin z g(yß z) Ê `` yf œ x sin z `` yg œ x sin z Ê `` yg œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ xy sin z h(z) Ê `` zf œ xy cos z hw (z) œ xy cos z Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xy sin z C 11. `` zf œ y# z z# Ê f(xß yß z) œ #" ln ay# z# b g(xß y) Ê `` xf œ `` xg œ ln x sec# (x y) Ê g(xß y) œ (x ln x x) tan (x y) h(y) Ê f(xß yß z) œ "# ln ay# z# b (x ln x x) tan (x y) h(y) Ê `` yf œ y# y z# sec# (x y) hw (y) œ sec# (x y) y# y z# Ê hw (y) œ 0 Ê h(y) œ C Ê f(xß yß z) œ "# ln ay# z# b (x ln x x) tan (x y) C 12. `` xf œ 1 yx# y# Ê f(xß yß z) œ tan" (xy) g(yß z) Ê `` yf œ 1 xx# y# `` yg œ 1 xx# y# È1 z y# z# Ê `` gy œ È1 z y# z# Ê g(yß z) œ sin" (yz) h(z) Ê f(xß yß z) œ tan" (xy) sin" (yz) h(z) Ê `` zf œ È1 y y# z# hw (z) œ È1 y y# z# "z Ê hw (z) œ "z Ê h(z) œ ln kzk C Ê f(xß yß z) œ tan" (xy) sin" (yz) ln kzk C 13. Let F(xß yß z) œ 2xi 2yj 2zk Ê `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 0 œ ``My Ê M dx N dy P dz is exact; `` xf œ 2x Ê f(xß yß z) œ x# g(yß z) Ê `` yf œ `` yg œ 2y Ê g(yß z) œ y# h(z) Ê f(xß yß z) œ x# y# œ h(z) Ð2ß3ß 6Ñ Ê `` zf œ hw (z) œ 2z Ê h(z) œ z# C Ê f(xß yß z) œ x# y# z# C Ê 'Ð0ß0ß0Ñ 2x dx 2y dy 2z dz œ f(2ß 3ß 6) f(!ß !ß !) œ 2# 3# (6)# œ 49 14. Let F(xß yß z) œ yzi xzj xyk Ê `` Py œ x œ ``Nz , ``Mz œ y œ `` Px , ``Nx œ z œ ``My Ê M dx N dy P dz is exact; `` xf œ yz Ê f(xß yß z) œ xyz g(yß z) Ê `` yf œ xz `` yg œ xz Ê `` yg œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ xyz h(z) Ê `` zf œ xy hw (z) œ xy Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xyz C Ð3ß5ß0Ñ Ê 'Ð1ß1ß2Ñ yz dx xz dy xy dz œ f(3ß 5ß 0) f(1ß 1ß 2) œ 0 2 œ 2 15. Let F(xß yß z) œ 2xyi ax# z# b j 2yzk Ê `` Py œ 2z œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 2x œ ``My Ê M dx N dy P dz is exact; `` xf œ 2xy Ê f(xß yß z) œ x# y g(yß z) Ê `` yf œ x# `` gy œ x# z# Ê `` gy œ z# Ê g(yß z) œ yz# h(z) Ê f(xß yß z) œ x# y yz# h(z) Ê `` zf œ 2yz hw (z) œ 2yz Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x# y yz# C Ê 'Ð0ß0ß0Ñ 2xy dx ax# z# b dy 2yz dz œ f("ß #ß $) f(!ß !ß !) œ 2 2(3)# œ 16 Ð1ß2ß3Ñ 16. Let F(xß yß z) œ 2xi y# j ˆ 1 4 z# ‰ k Ê `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 0 œ ``My $ Ê M dx N dy P dz is exact; `` xf œ 2x Ê f(xß yß z) œ x# g(yß z) Ê `` yf œ `` yg œ y# Ê g(yß z) œ y3 h(z) Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 954 Chapter 16 Integration in Vector Fields $ Ê f(xß yß z) œ x# y3 h(z) Ê `` zf œ hw (z) œ 1 4 z# Ê h(z) œ 4 tan" z C Ê f(xß yß z) œ x# y3 4 tan" z C Ê 'Ð0ß0ß0Ñ 2x dx y# dy 1 4 z# dz œ f(3ß 3ß 1) f(!ß !ß !) Ð3ß3ß1Ñ $ 1‰ œ ˆ9 27 3 4 † 4 (! ! 0) œ 1 17. Let F(xß yß z) œ (sin y cos x)i (cos y sin x)j k Ê `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ cos y cos x œ ``My Ê M dx N dy P dz is exact; `` xf œ sin y cos x Ê f(xß yß z) œ sin y sin x g(yß z) Ê `` yf œ cos y sin x `` gy œ cos y sin x Ê `` gy œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ sin y sin x h(z) Ê `` zf œ hw (z) œ 1 Ê h(z) œ z C Ê f(xß yß z) œ sin y sin x z C Ê 'Ð1ß0ß0Ñ sin y cos x dx cos y sin x dy dz œ f(0ß 1ß 1) f(1ß !ß !) Ð0ß1ß1Ñ œ (0 1) (0 0) œ 1 18. Let F(xß yß z) œ (2 cos y)i Š "y 2x sin y‹ j ˆ "z ‰ k Ê `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 2 sin y œ ``My Ê M dx N dy P dz is exact; `` xf œ 2 cos y Ê f(xß yß z) œ 2x cos y g(yß z) Ê `` yf œ 2x sin y `` gy œ y" 2x sin y Ê `` yg œ y" Ê g(yß z) œ ln kyk h(z) Ê f(xß yß z) œ 2x cos y ln kyk h(z) Ê `` zf œ hw (z) œ "z Ê h(z) œ ln kzk C Ê f(xß yß z) œ 2x cos y ln kyk ln kzk C Ê 'Ð0ß2ß1Ñ Ð1ß1Î2ß2Ñ 2 cos y dx Š "y 2x sin y‹ dy "z dz œ f ˆ1ß 1# ß 2‰ f(!ß #ß ") œ ˆ2 † 0 ln 1# ln 2‰ (0 † cos 2 ln 2 ln 1) œ ln 1# # `N `M `P `N `M 19. Let F(xß yß z) œ 3x# i Š zy ‹ j (2z ln y)k Ê `` Py œ 2z y œ `z , `z œ 0 œ `x , `x œ 0 œ `y Ê M dx N dy P dz is exact; `` xf œ 3x# Ê f(xß yß z) œ x$ g(yß z) Ê `` yf œ `` gy œ zy Ê g(yß z) œ z# ln y h(z) # Ê f(xß yß z) œ x$ z# ln y h(z) Ê `` zf œ 2z ln y hw (z) œ 2z ln y Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x$ z# ln y C Ê 'Ð1ß1ß1Ñ 3x# dx zy dy 2z ln y dz œ f(1ß 2ß 3) f("ß "ß ") Ð1ß2ß3Ñ # œ (1 9 ln 2 C) (1 0 C) œ 9 ln 2 # `M 20. Let F(xß yß z) œ (2x ln y yz)i Š xy xz‹ j (xy)k Ê `` Py œ x œ ``Nz , ``Mz œ y œ `` Px , ``Nx œ 2x y z œ `y Ê M dx N dy P dz is exact; `` xf œ 2x ln y yz Ê f(xß yß z) œ x# ln y xyz g(yß z) Ê `` yf œ xy xz `` gy # œ xy xz Ê `` yg œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ x# ln y xyz h(z) Ê `` zf œ xy hw (z) œ xy Ê hw (z) œ 0 # Ê h(z) œ C Ê f(xß yß z) œ x# ln y xyz C Ê 'Ð1ß2ß1Ñ (2x ln y yz) dx Š xy xz‹ dy xy dz Ð2ß1ß1Ñ # œ f(2ß 1ß 1) f("ß 2ß 1) œ (4 ln 1 2 C) (ln 2 2 C) œ ln 2 21. Let F(xß yß z) œ Š "y ‹ i Š 1z yx# ‹ j ˆ zy# ‰ k Ê `` Py œ z"# œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ y1# œ ``My Ê M dx N dy P dz is exact; `` xf œ "y Ê f(xß yß z) œ xy g(yß z) Ê `` yf œ yx# `` gy œ "z yx# Ê `` gy œ "z Ê g(yß z) œ yz h(z) Ê f(xß yß z) œ xy yz h(z) Ê `` zf œ zy# hw (z) œ zy# Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ yx yz C Ê 'Ð1ß1ß1Ñ "y dx Š 1z yx# ‹ dy zy# dz œ f(2ß 2ß 2) f("ß 1ß 1) œ ˆ 2# 2# C‰ ˆ "1 "1 C‰ Ð2ß2ß2Ñ œ0 22. Let F(xß yß z) œ 2xxi#2yy#j z2z# k Šand let 3# œ x# y# z# Ê `` 3x œ 3x , `` 3y œ 3y , `` 3z œ 3z ‹ Ê `` Py œ 4yz œ ``Nz , ``Mz œ 4xz œ `` Px , ``Nx œ 4xy œ ``My Ê M dx N dy P dz is exact; 3% 3% 3% `f 2x ` x œ x # y # z# `g 2y Ê f(xß yß z) œ ln ax# y# z# b g(yß z) Ê `` yf œ x# 2y y # z # ` y œ x # y # z# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.3 Path Independence, Potential Functions, and Conservative Fields 955 w Ê `` gy œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ ln ax# y# z# b h(z) Ê `` zf œ x# 2z y# z# h (z) w # # # œ x# 2z y# z# Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ ln ax y z b C Ð2ß2ß2Ñ Ê 'Ð 1ß 1ß 1Ñ 2x dxx# 2yy#dyz#2z dz œ f(2ß 2ß 2) f("ß 1ß 1) œ ln 12 ln 3 œ ln 4 23. r œ (i j k) t(i 2j 2k) œ (1 t)i (1 2t)j (1 2t)k, 0 Ÿ t Ÿ 1 Ê dx œ dt, dy œ 2 dt, dz œ 2 dt Ð2ß3ß 1Ñ Ê 'Ð1ß1ß1Ñ y dx x dy 4 dz œ '0 (2t 1) dt (t 1)(2 dt) 4(2) dt œ '0 (4t 5) dt œ c2t# 5td ! œ 3 1 1 " 24. r œ t(3j 4k), 0 Ÿ t Ÿ 1 Ê dx œ 0, dy œ 3 dt, dz œ 4 dt Ê 'Ð0ß0ß0Ñ x# dx yz dy Š y# ‹ dz Ð0ß3ß4Ñ # œ '0 a12t# b (3 dt) Š 9t# ‹ (4 dt) œ '0 54t# dt œ c18t# d ! œ 18 1 1 # " 25. `` Py œ 0 œ ``Nz , ``Mz œ 2z œ `` Px , ``Nx œ 0 œ ``My Ê M dx N dy P dz is exact Ê F is conservative Ê path independence 26. `` Py œ ˆÈ # yz# x y z# ‰ $ œ ``Nz , ``Mz œ ˆÈ # xz# x y z# ‰ $ œ `` Px , ``Nx œ ˆÈ # xy# x y z# ‰ $ œ ``My Ê M dx N dy P dz is exact Ê F is conservative Ê path independence `M 27. `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 2x y# œ ` y Ê F is conservative Ê there exists an f so that F œ ™ f; `f 2x `x œ y # # # Ê f(xß y) œ xy g(y) Ê `` yf œ xy# gw (y) œ 1 y#x Ê gw (y) œ y"# Ê g(y) œ "y C # # Ê f(xß y) œ xy y" C Ê F œ ™ Š x y 1 ‹ 28. `` Py œ cos z œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ ey œ ``My Ê F is conservative Ê there exists an f so that F œ ™ f; x `f x ` x œ e ln y Ê f(xß yß z) œ ex ln y g(yß z) Ê `` yf œ ey `` yg œ ey sin z Ê `` yg œ sin z Ê g(yß z) x x œ y sin z h(z) Ê f(xß yß z) œ ex ln y y sin z h(z) Ê `` zf œ y cos z hw (z) œ y cos z Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ ex ln y y sin z C Ê F œ ™ aex ln y y sin zb 29. `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 1 œ ``My Ê F is conservative Ê there exists an f so that F œ ™ f; `f # `x œ x y Ê f(xß yß z) œ "3 x$ xy g(yß z) Ê `` yf œ x `` gy œ y# x Ê `` yg œ y# Ê g(yß z) œ "3 y$ h(z) Ê f(xß yß z) œ "3 x$ xy 3" y$ h(z) Ê `` zf œ hw (z) œ zez Ê h(z) œ zez ez C Ê f(xß yß z) œ "3 x$ xy 3" y$ zez ez C Ê F œ ™ ˆ "3 x$ xy 3" y$ zez ez ‰ (a) work œ 'A F † ddtr dt œ 'A F † dr œ 3" x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ ˆ 3" 0 0 e e‰ ˆ 3" 0 0 1‰ B B Ð"ß!ß"Ñ œ1 (b) work œ 'A F † dr œ "3 x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ 1 B Ð"ß!ß"Ñ (c) work œ 'A F † dr œ "3 x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ 1 B Ð"ß!ß"Ñ Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 0) to (1ß 0ß 1). B 30. `` Py œ xeyz xyzeyz cos y œ ``Nz , ``Mz œ yeyz œ `` Px , ``Nx œ zeyz œ ``My Ê F is conservative Ê there exists an f so that F œ ™ f; `` xf œ eyz Ê f(xß yß z) œ xeyz g(yß z) Ê `` yf œ xzeyz `` yg œ xzeyz z cos y Ê `` gy œ z cos y Ê g(yß z) œ z sin y h(z) Ê f(xß yß z) œ xeyz z sin y h(z) Ê `` zf œ xyeyz sin y hw (z) œ xyeyz sin y Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xeyz z sin y C Ê F œ ™ axeyz z sin yb Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 956 Chapter 16 Integration in Vector Fields (a) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ B Ð"ß1Î#ß!Ñ œ (1 0) (1 0) œ 0 (b) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ B Ð"ß1Î#ß!Ñ (c) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ B Ð"ß1Î#ß!Ñ œ0 œ0 Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 1) to ˆ1ß 1# ß 0‰ . B 31. (a) F œ ™ ax$ y# b Ê F œ 3x# y# i 2x$ yj ; let C" be the path from (1ß 1) to (0ß 0) Ê x œ t 1 and y œ t 1, 0 Ÿ t Ÿ 1 Ê F œ 3(t 1)# (t 1)# i 2(t 1)$ (t 1)j œ 3(t 1)% i 2(t 1)% j and r" œ (t 1)i (t 1)j Ê dr" œ dt i dt j Ê 'C F † dr" œ '0 c3(t 1)% 2(t 1)% d dt 1 " 1 " œ '0 5(t 1)% dt œ c(t 1)& d ! œ 1; let C# be the path from (0ß 0) to (1ß 1) Ê x œ t and y œ t, 1 0 Ÿ t Ÿ 1 Ê F œ 3t% i 2t% j and r# œ ti tj Ê dr# œ dt i dt j Ê 'C F † dr# œ '0 a3t% 2t% b dt 1 œ '0 5t% dt œ 1 Ê 'C F † dr œ 'C F † dr" 'C F † dr# œ 2 " # # Ð1ß1Ñ (b) Since f(xß y) œ x$ y# is a potential function for F, 'Ð 1ß1Ñ F † dr œ f(1ß 1) f(1ß 1) œ 2 32. `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 2x sin y œ ``My Ê F is conservative Ê there exists an f so that F œ ™ f; `f ` x œ 2x cos y Ê f(xß yß z) œ x# cos y g(yß z) Ê `` yf œ x# sin y `` yg œ x# sin y Ê `` yg œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ x# cos y h(z) Ê `` zf œ hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x# cos y C Ê F œ ™ ax# cos yb (a) 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ œ 0 1 œ 1 Ð!ß"Ñ (b) 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß1Ñ œ 1 (1) œ 2 Ð"ß!Ñ (c) 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ œ 1 1 œ 0 Ð"ß!Ñ (d) 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ œ 1 1 œ 0 Ð"ß!Ñ 33. (a) If the differential form is exact, then `` Py œ ``Nz Ê 2ay œ cy for all y Ê 2a œ c, ``Mz œ `` Px Ê 2cx œ 2cx for all x, and ``Nx œ ``My Ê by œ 2ay for all y Ê b œ 2a and c œ 2a (b) F œ ™ f Ê the differential form with a œ 1 in part (a) is exact Ê b œ 2 and c œ 2 34. F œ ™ f Ê g(xß yß z) œ 'Ð0ß0ß0Ñ F † dr œ 'Ð0ß0ß0Ñ ™ f † dr œ f(xß yß z) f(0ß 0ß 0) Ê `` gx œ `` xf 0, `` yg œ `` yf 0, and ÐxßyßzÑ `g `f `z œ `z 0 ÐxßyßzÑ Ê ™ g œ ™ f œ F, as claimed 35. The path will not matter; the work along any path will be the same because the field is conservative. 36. The field is not conservative, for otherwise the work would be the same along C" and C# . 37. Let the coordinates of points A and B be axA , yA , zA b and axB , yB , zB b, respectively. The force F œ ai bj ck is conservative because all the partial derivatives of M, N, and P are zero. Therefore, the potential function is fax, y, zb œ ax by cz C, and the work done by the force in moving a particle along any path from A to B is faBb faAb œ f axB , yB , zB b faxA , yA , zA b œ aaxB byB czB Cb aaxA byA czA Cb Ä œ aaxB xA b bayB yA b cazB zA b œ F † BA Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.4 Green's Theorem in the Plane 38. (a) Let GmM œ C Ê F œ C ’ Ê x i # #y # $Î# j # #z # $Î# k“ ax# y# z# b$Î# ax y z b ax y z b 3yzC 3xyC `P `N `M 3xzC `P `N `M ` y œ ax# y# z# b&Î# œ ` z , ` z œ ax# y# z# b&Î# œ ` x , ` x œ ax# y# z# b&Î# œ ` y Ê F œ ™ f for yC `g `f ` y œ ax# y# z# b$Î# ` y some f; `` xf œ xC Ê f(xß yß z) œ # #C # "Î# g(yß z) Ê ax# y# z# b$Î# ax y z b `g yC œ # # # $Î# Ê ` y œ 0 Ê g(yß z) œ h(z) Ê `` zf œ # zC hw (z) œ # zC ax y z b ax y# z# b$Î# ax y# z# b$Î# Ê h(z) œ C" Ê f(xß yß z) œ C C" . ax# y# z# b"Î# Let C" œ 0 Ê f(xß yß z) œ GmM is a potential ax# y# z# b"Î# function for F. (b) If s is the distance of (xß yß z) from the origin, then s œ Èx# y# z# . The work done by the gravitational field F is work œ 'P F † dr œ ’ Èx#GmM “ y # z# P# T# " T" " " GmM œ GmM s# s" œ GmM Š s# s" ‹ , as claimed. 16.4 GREEN'S THEOREM IN THE PLANE 1. M œ y œ a sin t, N œ x œ a cos t, dx œ a sin t dt, dy œ a cos t dt Ê ``Mx œ 0, ``My œ 1, ``Nx œ 1, and `N ` y œ 0; Equation (3): )C M dy N dx œ '0 [(a sin t)(a cos t) (a cos t)(a sin t)] dt œ '0 0 dt œ 0; 21 21 ' ' Š ``Mx ``Ny ‹ dx dy œ ' ' 0 dx dy œ 0, Flux R R 21 21 Equation (4): )C M dx N dy œ '0 [(a sin t)(a sin t) (a cos t)(a cos t)] dt œ '0 a# dt œ 21a# ; ' ' Š ``Nx ``My ‹ dx dy œ ' ' a Èa c x # ca cc R # 2 dy dx œ 'ca 4Èa# x# dx œ 4 ’ x2 Èa# x# a# sin" xa “ a # a ca œ 2a# ˆ 1# 1# ‰ œ 2a# 1, Circulation 2. M œ y œ a sin t, N œ 0, dx œ a sin t dt, dy œ a cos t dt Ê ``Mx œ 0, ``My œ 1, ``Nx œ 0, and ``Ny œ 0; #1 Equation (3): )C M dy N dx œ '0 a# sin t cos t dt œ a# 2" sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Flux 21 R 21 #1 Equation (4): )C M dx N dy œ '0 aa# sin# tb dt œ a# 2t sin4 2t ‘ ! œ 1a# ; ' ' Š ``Nx ``My ‹ dx dy œ ' ' 1 dx dy œ '0 '0 r dr d) œ ' 21 a R 21 # a# d) œ 1a# , Circulation 0 R 3. M œ 2x œ 2a cos t, N œ 3y œ 3a sin t, dx œ a sin t dt, dy œ a cos t dt Ê ``Mx œ 2, ``My œ 0, ``Nx œ 0, and `N ` y œ 3; Equation (3): )C M dy N dx œ '0 [(2a cos t)(a cos t) (3a sin t)(a sin t)] dt 21 œ '0 a2a# cos# t 3a# sin# tb dt œ 2a# 2t sin4 2t ‘ ! 3a# 2t sin4 2t ‘ ! œ 21a# 31a# œ 1a# ; 21 #1 #1 ' ' Š ``Mx ``Ny ‹ œ ' ' 1 dx dy œ ' ' r dr d) œ ' a## d) œ 1a# , Flux 0 0 0 21 R a 21 R 21 Equation (4): )C M dx N dy œ '0 [(2a cos t)(a sin t) (3a sin t)(a cos t)] dt 21 #1 œ '0 a2a# sin t cos t 3a# sin t cos tb dt œ 5a# 12 sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Circulation R 4. M œ x# y œ a$ cos# t, N œ xy# œ a$ cos t sin# t, dx œ a sin t dt, dy œ a cos t dt Ê ``Mx œ 2xy, ``My œ x2 , ``Nx œ y# , and ``Ny œ 2xy; Equation (3): )C M dy N dx œ '0 aa% cos$ t sin t a% cos t sin$ tb œ ’ a4 cos% t a4 sin% t“ 21 % % Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. #1 ! œ 0; 957 958 Chapter 16 Integration in Vector Fields ' ' Š ``Mx ``Ny ‹ dx dy œ ' ' (2xy 2xy) dx dy œ 0, Flux R R 21 21 Equation (4): )C M dx N dy œ '0 aa% cos# t sin# t a% cos# t sin# tb dt œ '0 a2a% cos# t sin# tb dt 21 41 %1 œ '0 "# a% sin# 2t dt œ a4 '0 sin# u du œ a4 u2 sin42u ‘ ! œ 1#a ; ' ' Š ``Nx ``My ‹ dx dy œ ' ' ay# x# b dx dy % % % R 21 a 21 œ '0 '0 r# † r dr d) œ '0 a4 d) œ 1#a , Circulation % R % 5. M œ x y, N œ y x Ê ``Mx œ 1, ``My œ 1, ``Nx œ 1, ``Ny œ 1 Ê Flux œ ' ' 2 dx dy œ '0 '0 2 dx dy œ 2; 1 1 R Circ œ ' ' [1 (1)] dx dy œ 0 R 6. M œ x# 4y, N œ x y# Ê ``Mx œ 2x, ``My œ 4, ``Nx œ 1, ``Ny œ 2y Ê Flux œ ' ' (2x 2y) dx dy R 1 1 1 1 " " œ '0 '0 (2x 2y) dx dy œ '0 cx# 2xyd ! dy œ '0 (1 2y) dy œ cy y# d ! œ 2; Circ œ ' ' (1 4) dx dy R œ '0 '0 3 dx dy œ 3 1 1 7. M œ y# x# , N œ x# y# Ê ``Mx œ 2x, ``My œ 2y, ``Nx œ 2x, ``Ny œ 2y Ê Flux œ ' ' (2x 2y) dx dy œ '0 '0 (2x 2y) dy dx œ '0 a2x 3 x 3 R # $ x b dx œ "3 x$ ‘ ! œ 9; Circ œ # ' ' (2x 2y) dx dy 3 x 3 œ '0 '0 (2x 2y) dy dx œ '0 x# dx œ 9 R 8. M œ x y, N œ ax# y# b Ê ``Mx œ 1, ``My œ 1, ``Nx œ 2x, ``Ny œ 2y Ê Flux œ ' ' (1 2y) dx dy R 1 x 1 1 x œ '0 '0 (1 2y) dy dx œ '0 ax x# b dx œ "6 ; Circ œ ' ' (2x 1) dx dy œ '0 '0 (2x 1) dy dx œ '0 a2x 1 # R xb dx œ 76 9. M œ xy y2 , N œ x y Ê ``Mx œ y, ``My œ x 2y, ``Nx œ 1, ``Ny œ 1 Ê Flux œ ' ' ay a1bb dy dx R Èx ' ' a1 ax 2ybb dy dx œ '0 'x2 ay 1b dy dx œ '0 ˆ "# x Èx "# x4 x# ‰ dx œ 11 60 ; Circ œ 1 1 R Èx 7 œ '0 'x2 a1 x 2yb dy dx œ '0 ˆÈx x3Î2 x x# x3 x4 ‰ dx œ 60 1 1 10. M œ x 3y, N œ 2x y Ê ``Mx œ 1, ``My œ 3, ``Nx œ 2, ``Ny œ 1 Ê Flux œ ' ' a1 a1bb dy dx œ 0 R È2 È2 2 x Î2 Circ œ ' ' a2 3b dy dx œ 'cÈ2 'È 2 c x Î2 a1b dy dx œ È22 'cÈ2 È2 x2 dx œ 1È2 Èa R 2b a 2b 11. M œ x3 y2 , N œ "# x4 y Ê ``Mx œ 3x2 y2 , ``My œ 2x3 y, ``Nx œ 2x3 y, ``Ny œ "# x4 Ê Flux œ ' ' ˆ3x2 y2 "# x4 ‰ dy dx R ' ' a2x3 y 2x3 yb dy dx œ 0 œ '0 'x2 x ˆ3x2 y2 "# x4 ‰ dy dx œ '0 ˆ3x5 72 x6 3x7 x8 ‰ dx œ 64 9 ; Circ œ 2 x 2 R Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.4 Green's Theorem in the Plane 12. M œ 1 x y2 , N œ tan1 y Ê ``Mx œ 1 1 y2 , ``My œ a12xy2yb2 , ``Nx œ 0, ``Ny œ 1 1 y2 Ê Flux œ ' ' Š 1 1 y2 1 1 y2 ‹ dx dy R È1 y2 œ 'c1 'È1 y2 1 2 1 y2 dx dy œ ' 4È 1 y2 dx œ 41È2 41 ; Circ œ c1 1 y2 1 È1 y2 ' ' Š0 Š 2x y ‹‹ dy dx a1 y 2 b 2 R y œ 'c1 'È1 y2 Š a1 2x ‹ dy dx œ 'c1 a0b dx œ 0 y 2 b2 1 1 13. M œ x ex sin y, N œ x ex cos y Ê ``Mx œ 1 ex sin y, ``My œ ex cos y, ``Nx œ 1 ex cos y, ``Ny œ ex sin y Ècos 2) 1Î4 Ê Flux œ ' ' dx dy œ 'c1Î4 '0 R Î 1Î% r dr d) œ ' 1Î4 ˆ "# cos 2)‰ d) œ 4" sin 2)‘ 1Î% œ "# ; 1 4 Ècos 2) 1Î4 Circ œ ' ' a1 ex cos y ex cos yb dx dy œ ' ' dx dy œ 'c1Î4 '0 R R Î r dr d) œ ' 1Î4 ˆ "# cos 2)‰ d) œ "# 1 4 2y `N 14. M œ tan" yx , N œ ln ax# y# b Ê ``Mx œ x#yy# , ``My œ x# x y# , ``Nx œ x# 2x y# , ` y œ x# y# ' ' ˆ r sinr# ) ‰ r dr d) œ '0 sin ) d) œ 2; Ê Flux œ ' ' Š x#yy# x# 2y y# ‹ dx dy œ 0 1 1 R 1 2 x ' ' ˆ r cosr# ) ‰ r dr d) œ '0 cos ) d) œ 0 Circ œ ' ' Š x# 2x y# x# y# ‹ dx dy œ 0 1 1 1 2 R 15. M œ xy, N œ y# Ê ``Mx œ y, ``My œ x, ``Nx œ 0, ``Ny œ 2y Ê Flux œ ' ' (y 2y) dy dx œ '0 'x# 3y dy dx 1 x R 1 1 x 1 œ '0 Š 3x# 3x# ‹ dx œ 5" ; Circ œ ' ' x dy dx œ '0 'x x dy dx œ '0 ax# x$ b dx œ 1"# # % # R 16. M œ sin y, N œ x cos y Ê ``Mx œ 0, ``My œ cos y, ``Nx œ cos y, ``Ny œ x sin y Ê Flux œ ' ' (x sin y) dx dy œ '0 1Î2 R '01Î2 (x sin y) dx dy œ '01Î2 Š 18 sin y‹ dy œ 18 ; # Circ œ ' ' [cos y ( cos y)] dx dy œ '0 1Î2 R '01Î2 2 cos y dx dy œ '01Î2 1 cos y dy œ c1 sin yd 1Î# œ1 ! 17. M œ 3xy 1 x y# , N œ ex tan " y Ê ``Mx œ 3y 1 " y# , ``Ny œ 1 " y# aÐ1 Ê Flux œ ' ' Š3y 1 " y# 1 " y# ‹ dx dy œ ' ' 3y dx dy œ '0 '0 21 R R œ '0 a$ (1 cos ))$ (sin )) d) œ ’ a4 (1 cos ))% “ 21 # $ #1 ! cos )Ñ (3r sin )) r dr d) œ 4a$ a4a$ b œ 0 18. M œ y ex ln y, N œ ey Ê ``My œ 1 ey , ``Nx œ ey Ê Circ œ ' ' ’ ey Š1 ey ‹“ dx dy œ ' ' (1) dx dy x x x x x R R 1 3cx 1 1 œ 'c1 'x b 1 dy dx œ 'c1 ca3 x# b ax% 1bd dx œ 'c1 ax% x# 2b dx œ 44 15 # % 19. M œ 2xy$ , N œ 4x# y# Ê ``My œ 6xy# , ``Nx œ 8xy# Ê work œ )C 2xy$ dx 4x# y# dy œ ' ' a8xy# 6xy# b dx dy œ '0 '0 2xy dy dx œ ' 1 x$ # R 1 2 "! 2 x dx œ 33 0 3 20. M œ 4x 2y, N œ 2x 4y Ê ``My œ 2, ``Nx œ 2 Ê work œ )C (4x 2y) dx (2x 4y) dy œ ' ' [2 (2)] dx dy œ 4 ' ' dx dy œ 4(Area of the circle) œ 4(1 † 4) œ 161 R R Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 959 960 Chapter 16 Integration in Vector Fields 21. M œ y# , N œ x# Ê ``My œ 2y, ``Nx œ 2x Ê )C y# dx x# dy œ ' ' (2x 2y) dy dx 1cx œ '0 '0 1 R (2x 2y) dy dx œ '0 a3x 4x 1b dx œ cx 2x# xd ! œ 1 2 1 œ 0 1 # " $ 22. M œ 3y, N œ 2x Ê ``My œ 3, ``Nx œ 2 Ê )C 3y dx 2x dy œ ' ' a2 3b dx dy œ '0 '0 1 sin x a1bdy dx R œ '0 sin x dx œ 2 1 23. M œ 6y x, N œ y 2x Ê ``My œ 6, ``Nx œ 2 Ê )C (6y x) dx (y 2x) dy œ ' ' (2 6) dy dx R œ 4(Area of the circle) œ 161 24. M œ 2x y# , N œ 2xy 3y Ê ``My œ 2y, ``Nx œ 2y Ê )C a2x y# b dx (2xy 3y) dy œ ' ' (2y 2y) dx dy œ 0 R 25. M œ x œ a cos t, N œ y œ a sin t Ê dx œ a sin t dt, dy œ a cos t dt Ê Area œ "# )C x dy y dx œ "# '0 aa# cos# t a# sin# tb dt œ "# '0 a# dt œ 1a# 21 21 26. M œ x œ a cos t, N œ y œ b sin t Ê dx œ a sin t dt, dy œ b cos t dt Ê Area œ "# )C x dy y dx œ "# '0 aab cos# t ab sin# tb dt œ "# '0 ab dt œ 1ab 21 21 27. M œ x œ cos$ t, N œ y œ sin$ t Ê dx œ 3 cos# t sin t dt, dy œ 3 sin# t cos t dt Ê Area œ "# )C x dy y dx 3 ' œ "# '0 a3 sin# t cos# tb acos# t sin# tb dt œ "# '0 a3 sin# t cos# tb dt œ 38 '0 sin# 2t dt œ 16 sin# u du 0 21 21 21 41 %1 3 u sin 2u ‘ œ 16 œ 38 1 2 4 ! 28. C1 : M œ x œ t, N œ y œ 0 Ê dx œ dt, dy œ 0; C2 : M œ x œ a21 tb sina21 tb œ 21 t sin t, N œ y œ 1 cosa21 tb œ 1 cos t Ê dx œ acos t 1b dt, dy œ sin t dt Ê Area œ "# )C x dy y dx œ "# )C x dy y dx "# )C x dy y dx " œ "# 2 '021 a0bdt "# '021 ca21 t sin tbasin tb a1 cos tb acos t 1bd dt œ "# '021 a2 cos t t sin t 2 21 sin tb dt œ 12 c3 sin t t cos t 2t 21 cos td20 1 œ 31 29. (a) M œ f(x), N œ g(y) Ê ``My œ 0, ``Nx œ 0 Ê )C f(x) dx g(y) dy œ ' ' Š ``Nx ``My ‹ dx dy œ ' ' 0 dx dy œ 0 R (b) M œ ky, N œ hx Ê `M `N ` y œ k, ` x œ h R Ê )C ky dx hx dy œ ' ' Š ``Nx ``My ‹ dx dy œ ' ' (h k) dx dy œ (h k)(Area of the region) R R 30. M œ xy# , N œ x# y 2x Ê ``My œ 2xy, ``Nx œ 2xy 2 Ê )C xy# dx ax# y 2xb dy œ ' ' Š ``Nx ``My ‹ dx dy œ ' ' (2xy 2 2xy) dx dy œ 2 ' ' dx dy œ 2 times the area of the square R R R Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.4 Green's Theorem in the Plane 961 31. The integral is 0 for any simple closed plane curve C. The reasoning: By the tangential form of Green's Theorem, with M œ 4x$ y and N œ x% , )C 4x$ y dx x% dy œ ' ' ’ ``x ax% b ``y a4x$ yb“ dx dy R œ ' ' ðóóñóóò a4x$ 4x$ b dx dy œ 0. R 0 32. The integral is 0 for any simple closed curve C. The reasoning: By the normal form of Green's theorem, with ` ` $ $ M œ x$ and N œ y$ , )C y$ dy x$ dx œ ' ' ”ðñò ` x ay b ï ` y ax b • dx dy œ 0. R 0 0 33. Let M œ x and N œ 0 Ê ``Mx œ 1 and ``Ny œ 0 Ê )C M dy N dx œ ' ' Š ``Mx ``Ny ‹ dx dy Ê )C x dy R œ ' ' (1 0) dx dy Ê Area of R œ ' ' dx dy œ ) x dy; similarly, M œ y and N œ 0 Ê ``My œ 1 and R C R Ê )C M dx N dy œ ' ' Š ``Nx ``My ‹ dy dx Ê )C y dx œ ' ' (0 1) dy dx Ê )C y dx `N `x œ 0 R œ ' ' dx dy œ Area of R R R 34. 'a f(x) dx œ Area of R œ )C y dx, from Exercise 33 b ' ' x $ (xßy) dA ' ' x dA y R R 35. Let $ (xß y) œ 1 Ê x œ M M œ ' ' $ (xßy) dA œ ' ' dA œ R ' ' x dA Ê Ax œ ' ' x dA œ ' ' (x 0) dx dy R A R R R R R œ )C x# dy, Ax œ ' ' x dA œ ' ' (0 x) dx dy œ ) xy dx, and Ax œ ' ' x dA œ ' ' ˆ 23 x "3 x‰ dx dy # R œ) R " # x dy 3" xy dx C 3 Ê " # C )C x dy œ ) xy dx œ 3" C # )C x dy xy dx œ Ax # 36. If $ (xß y) œ 1, then Iy œ ' ' x# $ (xß y) dA œ ' ' x# dA œ ' ' ax# 0b dy dx œ "3 )C x$ dy, R R R ' ' x# dA œ ' ' a0 x# b dy dx œ ) x# y dx, and ' ' x# dA œ ' ' ˆ 34 x# 4" x# ‰ dy dx R C R œ) " $ x dy 4" x# y dx œ 4" C 4 R R )C x dy x y dx Ê )C x dy œ )C x# y dx œ 4" )C x$ dy x# y dx œ Iy $ " 3 # $ 37. M œ `` yf , N œ `` xf Ê ``My œ `` yf# , ``Nx œ `` xf# Ê )C `` yf dx `` xf dy œ ' ' Š `` xf# `` yf# ‹ dx dy œ 0 for such curves C # # # # R 38. M œ "4 x# y "3 y$ , N œ x Ê ``My œ 14 x# y# , ``Nx œ 1 Ê Curl œ ``Nx ``My œ 1 ˆ "4 x# y# ‰ 0 in the interior of the ellipse "4 x# y# œ 1 Ê work œ 'C F † dr œ ' ' ˆ1 4" x# y# ‰ dx dy will be maximized on the region R R œ {(xß y) | curl F} 0 or over the region enclosed by 1 œ "4 x# y# 2y 2y 2x 39. (a) ™ f œ Š x# 2x y# ‹ i Š x# y# ‹ j Ê M œ x# y# , N œ x# y# ; since M, N are discontinuous at (0ß 0), we compute 'C ™ f † n ds directly since Green's Theorem does not apply. Let x œ a cos t, y œ a sin t Ê dx œ a sin t dt, dy œ a cos t dt, M œ 2a cos t, N œ 2a sin t, 0 Ÿ t Ÿ 21, so 'C ™ f † n ds œ 'C M dy N dx œ '0 ˆ 2a cos t‰aa cos tb ˆ 2a sin t‰aa sin tb ‘dt œ '0 2acos2 t sin2 tbdt œ 41. Note that this holds for any 21 21 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 962 Chapter 16 Integration in Vector Fields a 0, so 'C ™ f † n ds œ 41 for any circle C centered at a0, 0b traversed counterclockwise and 'C ™ f † n ds œ 41 if C is traversed clockwise. (b) If K does not enclose the point (0ß 0) we may apply Green's Theorem: 'C ™ f † n ds œ 'C M dy N dx œ ' ' Š ``Mx ``Ny ‹ dx dy œ ' ' Š ax2 y2 b2 ax2 y2 b2 ‹ dx dy œ ' ' 0 dx dy œ 0. If K does enclose the point 2 ˆy 2 x 2 ‰ R 2 ˆx 2 y 2 ‰ R R (0ß 0) we proceed as follows: Choose a small enough so that the circle C centered at (0ß 0) of radius a lies entirely within K. Green's Theorem applies to the region R that lies between K and C. Thus, as before, 0 œ ' ' Š ``Mx ``Ny ‹ dx dy R œ 'K M dy N dx 'C M dy N dx where K is traversed counterclockwise and C is traversed clockwise. Hence by part (a) 0 œ ’ ' M dy N dx “ 41 Ê 41 œ ' M dy N dx œ 'K ™ f † n ds. We have shown: K K 'K ™ f † n ds œ œ 0 if (0ß 0) lies inside K if (0ß 0) lies outside K 41 40. Assume a particle has a closed trajectory in R and let C" be the path Ê C" encloses a simply connected region R" Ê C" is a simple closed curve. Then the flux over R" is )C F † n ds œ 0, since the velocity vectors F are " tangent to C" . But 0 œ )C F † n ds œ )C M dy N dx œ ' ' Š ``Mx ``Ny ‹ dx dy Ê Mx Ny œ 0, which is a " " R" contradiction. Therefore, C" cannot be a closed trajectory. 41. 'g ÐyÑ ``Nx dx dy œ N(g# (y)ß y) N(g" (y)ß y) Ê 'c 'g ÐyÑ ˆ ``Nx dx‰ dy œ 'c [N(g# (y)ß y) N(g" (y)ß y)] dy g# ÐyÑ d " g# ÐyÑ d " d d d c œ 'c N(g# (y)ß y) dy 'c N(g" (y)ß y) dy œ 'c N(g# (y)ß y) dy 'd N(g" (y)ß y) dy œ 'C N dy 'C N dy œ )C dy Ê )C N dy œ ' ' ``Nx dx dy # " R 42. The curl of a conservative two-dimensional field is zero. The reasoning: A two-dimensional field F œ Mi Nj can be considered to be the restriction to the xy-plane of a three-dimensional field whose k component is zero, and whose i and j components are independent of z. For such a field to be conservative, we must have `N `M `N `M ` x œ ` y by the component test in Section 16.3 Ê curl F œ ` x ` y œ 0. 43-46. Example CAS commands: Maple: with( plots );#43 M := (x,y) -> 2*x-y; N := (x,y) -> x+3*y; C := x^2 + 4*y^2 = 4; implicitplot( C, x=-2..2, y=-2..2, scaling=constrained, title="#43(a) (Section 16.4)" ); curlF_k := D[1](N) - D[2](M): # (b) 'curlF_k' = curlF_k(x,y); top,bot := solve( C, y ); # (c) left,right := -2, 2; q1 := Int( Int( curlF_k(x,y), y=bot..top ), x=left..right ); value( q1 ); Mathematica: (functions and bounds will vary) The ImplicitPlot command will be useful for 43 and 44, but is not needed for 43 and 44. In 44, the equation of the line from (0, 4) to (2, 0) must be determined first. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.5 Surfaces and Area 963 Clear[x, y, f] <<GraphicsÌmplicitPlot` f[x_, y_]:= {2x y, x 3y} curve= x2 4y2 ==4 ImplicitPlot[curve, {x, 3, 3},{y, 2, 2}, AspectRatio Ä Automatic, AxesLabel Ä {x, y}]; ybounds= Solve[curve, y] {y1, y2}=y/.ybounds; integrand:=D[f[x,y][[2]], x] D[f[x,y][[1]], y]//Simplify Integrate[integrand, {x, 2, 2}, {y, y1, y2}] N[%] Bounds for y are determined differently in 45 and 46. In 46, note equation of the line from (0, 4) to (2, 0). Clear[x, y, f] f[x_, y_]:= {x Exp[y], 4x2 Log[y]} ybound = 4 2x Plot[{0, ybound}, {x, 0,2. 1}, AspectRatio Ä Automatic, AxesLabel Ä {x, y}]; integrand:=D[f[x, y][[2]], x] D[f[x, y][[1]], y]//Simplify Integrate[integrand, {x, 0, 2}, {y, 0, ybound}] N[%] 16.5 SURFACES AND AREA # 1. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ ˆÈx# y# ‰ œ r# . Then r(rß )) œ (r cos ))i (r sin ))j r# k , 0 Ÿ r Ÿ 2, 0 Ÿ ) Ÿ 21. 2. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ 9 x# y# œ 9 r# . Then r(rß )) œ (r cos ))i (r sin ))j a9 r# b k ; z 0 Ê 9 r# 0 Ê r# Ÿ 9 Ê 3 Ÿ r Ÿ 3, 0 Ÿ ) Ÿ 21. But 3 Ÿ r Ÿ 0 gives the same points as 0 Ÿ r Ÿ 3, so let 0 Ÿ r Ÿ 3. È x# y# Ê z œ #r . Then r(rß )) œ (r cos ))i (r sin ))j ˆ #r ‰ k . # r For 0 Ÿ z Ÿ 3, 0 Ÿ # Ÿ 3 Ê 0 Ÿ r Ÿ 6; to get only the first octant, let 0 Ÿ ) Ÿ 1# . 3. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ 4. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ 2Èx# y# Ê z œ 2r. Then r(rß )) œ (r cos ))i (r sin ))j 2rk . For 2 Ÿ z Ÿ 4, 2 Ÿ 2r Ÿ 4 Ê 1 Ÿ r Ÿ 2, and let 0 Ÿ ) Ÿ 21. 5. In cylindrical coordinates, let x œ r cos ), y œ r sin ); since x# y# œ r2 Ê z# œ 9 ax# y# b œ 9 r# Ê z œ È9 r# , z 0. Then r(rß )) œ (r cos ))i (r sin ))j È9 r# k . Let 0 Ÿ ) Ÿ 21. For the domain # of r: z œ Èx# y# and x# y# z# œ 9 Ê x# y# ˆÈx# y# ‰ œ 9 Ê 2 ax# y# b œ 9 Ê 2r# œ 9 Ê r œ È32 Ê 0 Ÿ r Ÿ È32 . 6. In cylindrical coordinates, r(rß )) œ (r cos ))i (r sin ))j È4 r# k (see Exercise 5 above with x# y# z# œ 4, instead of x# y# z# œ 9). For the first octant, let 0 Ÿ ) Ÿ 1# . For the domain of r: z œ Èx# y# and # x# y# z# œ 4 Ê x# y# ˆÈx# y# ‰ œ 4 Ê 2 ax# y# b œ 4 Ê 2r# œ 4 Ê r œ È2. Thus, let È2 Ÿ r Ÿ 2 (to get the portion of the sphere between the cone and the xy-plane). Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 964 Chapter 16 Integration in Vector Fields 7. In spherical coordinates, x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), 3 œ Èx# y# z# Ê 3# œ 3 Ê 3 œ È3 Ê z œ È3 cos 9 for the sphere; z œ È3 È3 cos 9 # œ È È Ê cos 9 œ #" Ê 9 œ 13 ; z œ #3 Ê #3 œ È3 cos 9 Ê cos 9 œ "# Ê 9 œ 231 . Then r(9ß )) œ ŠÈ3 sin 9 cos )‹ i ŠÈ3 sin 9 sin )‹ j ŠÈ3 cos 9‹ k , 1 21 3 Ÿ 9 Ÿ 3 and 0 Ÿ ) Ÿ 21. 8. In spherical coordinates, x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), 3 œ Èx# y# z# Ê 3# œ 8 Ê 3 œ È8 œ 2È2 Ê x œ 2È2 sin 9 cos ), y œ 2È2 sin 9 sin ), and z œ 2È2 cos 9. Thus let r(9ß )) œ Š2È2 sin 9 cos )‹ i Š2È2 sin 9 sin )‹ j Š2È2 cos 9‹ k ; z œ 2 Ê 2 œ 2È2 cos 9 Ê cos 9 œ È"2 Ê 9 œ 341 ; z œ 2È2 Ê 2È2 œ 2È2 cos 9 Ê cos 9 œ 1 Ê 9 œ 0. Thus 0 Ÿ 9 Ÿ 341 and 0 Ÿ ) Ÿ 21 . 9. Since z œ 4 y# , we can let r be a function of x and y Ê r(xß y) œ xi yj a4 y# b k . Then z œ 0 Ê 0 œ 4 y# Ê y œ „ 2. Thus, let 2 Ÿ y Ÿ 2 and 0 Ÿ x Ÿ 2. 10. Since y œ x# , we can let r be a function of x and z Ê r(xß z) œ xi x# j zk . Then y œ 2 Ê x# œ 2 Ê x œ „ È2. Thus, let È2 Ÿ x Ÿ È2 and 0 Ÿ z Ÿ 3. 11. When x œ 0, let y# z# œ 9 be the circular section in the yz-plane. Use polar coordinates in the yz-plane Ê y œ 3 cos ) and z œ 3 sin ). Thus let x œ u and ) œ v Ê r(u,v) œ ui (3 cos v)j (3 sin v)k where 0 Ÿ u Ÿ 3, and 0 Ÿ v Ÿ 21. 12. When y œ 0, let x# z# œ 4 be the circular section in the xz-plane. Use polar coordinates in the xz-plane Ê x œ 2 cos ) and z œ 2 sin ). Thus let y œ u and ) œ v Ê r(u,v) œ (2 cos v)i uj (3 sin v)k where 2 Ÿ u Ÿ 2, and 0 Ÿ v Ÿ 1 (since we want the portion above the xy-plane). 13. (a) x y z œ 1 Ê z œ 1 x y. In cylindrical coordinates, let x œ r cos ) and y œ r sin ) Ê z œ 1 r cos ) r sin ) Ê r(rß )) œ (r cos ))i (r sin ))j (1 r cos ) r sin ))k , 0 Ÿ ) Ÿ 21 and 0 Ÿ r Ÿ 3. (b) In a fashion similar to cylindrical coordinates, but working in the yz-plane instead of the xy-plane, let y œ u cos v, z œ u sin v where u œ Èy# z# and v is the angle formed by (xß yß z), (xß 0ß 0), and (xß yß 0) with (xß 0ß 0) as vertex. Since x y z œ 1 Ê x œ 1 y z Ê x œ 1 u cos v u sin v, then r is a function of u and v Ê r(uß v) œ (1 u cos v u sin v)i (u cos v)j (u sin v)k , 0 Ÿ u Ÿ 3 and 0 Ÿ v Ÿ 21. 14. (a) In a fashion similar to cylindrical coordinates, but working in the xz-plane instead of the xy-plane, let x œ u cos v, z œ u sin v where u œ Èx# z# and v is the angle formed by (xß yß z), (yß 0ß 0), and (xß yß 0) with vertex (yß 0ß 0). Since x y 2z œ 2 Ê y œ x 2z 2, then r(uß v) œ (u cos v)i (u cos v 2u sin v 2)j (u sin v)k , 0 Ÿ u Ÿ È3 and 0 Ÿ v Ÿ 21. (b) In a fashion similar to cylindrical coordinates, but working in the yz-plane instead of the xy-plane, let y œ u cos v, z œ u sin v where u œ Èy# z# and v is the angle formed by (xß yß z), (xß 0ß 0), and (xß yß 0) with vertex (xß 0ß 0). Since x y 2z œ 2 Ê x œ y 2z 2, then r(uß v) œ (u cos v 2u sin v 2)i (u cos v)j (u sin v)k , 0 Ÿ u Ÿ È2 and 0 Ÿ v Ÿ 21. 15. Let x œ w cos v and z œ w sin v. Then (x 2)# z# œ 4 Ê x# 4x z# œ 0 Ê w# cos# v 4w cos v w# sin# v œ 0 Ê w# 4w cos v œ 0 Ê w œ 0 or w 4 cos v œ 0 Ê w œ 0 or w œ 4 cos v. Now w œ 0 Ê x œ 0 and y œ 0, which is a line not a cylinder. Therefore, let w œ 4 cos v Ê x œ (4 cos v)(cos v) œ 4 cos# v and z œ 4 cos v sin v. Finally, let y œ u. Then r(uß v) œ a4 cos# vb i uj (4 cos v sin v)k , 1# Ÿ v Ÿ 1# and 0 Ÿ u Ÿ 3. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.5 Surfaces and Area 16. Let y œ w cos v and z œ w sin v. Then y# (z 5)# œ 25 Ê y# z# 10z œ 0 Ê w# cos# v w# sin# v 10w sin v œ 0 Ê w# 10w sin v œ 0 Ê w(w 10 sin v) œ 0 Ê w œ 0 or w œ 10 sin v. Now w œ 0 Ê y œ 0 and z œ 0, which is a line not a cylinder. Therefore, let w œ 10 sin v Ê y œ 10 sin v cos v and z œ 10 sin# v. Finally, let x œ u. Then r(uß v) œ ui (10 sin v cos v)j a10 sin# vb k , 0 Ÿ u Ÿ 10 and 0 Ÿ v Ÿ 1. 17. Let x œ r cos ) and y œ r sin ). Then r(rß )) œ (r cos ))i (r sin ))j ˆ 2 r#sin ) ‰ k , 0 Ÿ r Ÿ 1 and 0 Ÿ ) Ÿ 21 )‰ Ê rr œ (cos ))i (sin ))j ˆ sin# ) ‰ k and r) œ (r sin ))i (r cos ))j ˆ r cos k # â â i j k â â â sin ) sin# ) ââ Ê rr ‚ r) œ â cos ) â ) ââ â r sin ) r cos ) r cos # œ Š r sin#) cos ) (sin ))(r# cos )) ‹ i Š r sin# ) r cos# ) ‹ j ar cos# ) r sin# )b k œ #r j rk # # Ê krr ‚ r) k œ É r4 r# œ È5 r # # Ê A œ '0 '0 21 1 È5 r # dr d) œ '021 ’ È45 r “ " d) œ '021 d) œ 1È# 5 # ! 18. Let x œ r cos ) and y œ r sin ) Ê z œ x œ r cos ), 0 Ÿ r Ÿ 2 and 0 Ÿ ) Ÿ 21. Then r(rß )) œ (r cos ))i (r sin ))j (r cos ))k Ê rr œ (cos ))i (sin ))j (cos ))k and r) œ (r sin ))i (r cos ))j (r sin ))k â â i j k â â â â sin ) cos ) â Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) r sin ) â œ ar sin# ) r cos# )b i (r sin ) cos ) r sin ) cos ))j ar cos# ) r sin# )b k œ ri rk Ê krr ‚ r) k œ Èr# r# œ rÈ2 Ê A œ '0 '0 rÈ2 dr d) œ '0 ’ r 2 2 “ d) œ '0 2È2 d) œ 41È2 21 21 2 #È # 21 ! 19. Let x œ r cos ) and y œ r sin ) Ê z œ 2Èx# y# œ 2r, 1 Ÿ r Ÿ 3 and 0 Ÿ ) Ÿ 21. Then r(rß )) œ (r cos ))i (r sin ))j 2rk Ê rr œ (cos ))i (sin ))j 2k and r) œ (r sin ))i (r cos ))j â â i j kâ â â â sin ) 2 â œ (2r cos ))i (2r sin ))j ar cos# ) r sin# )b k Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) 0 â œ (2r cos ))i (2r sin ))j rk Ê krr ‚ r) k œ È4r# cos# ) 4r# sin# ) r# œ È5r# œ rÈ5 Ê A œ '0 '1 rÈ5 dr d) œ '0 ’ r 2 5 “ d) œ '0 4È5 d) œ 81È5 21 21 3 #È $ 21 " È x# y# 20. Let x œ r cos ) and y œ r sin ) Ê z œ œ 3r , 3 Ÿ r Ÿ 4 and 0 Ÿ ) Ÿ 21. Then 3 r(rß )) œ (r cos ))i (r sin ))j ˆ 3r ‰ k Ê rr œ (cos ))i (sin ))j ˆ 3" ‰ k and r) œ (r sin ))i (r cos ))j â â i j kâ â â " â # # ˆ " ‰ ˆ" ‰ Ê rr ‚ r) œ â cos ) sin ) 3 ââ œ 3 r cos ) i 3 r sin ) j ar cos ) r sin )b k â â r sin ) r cos ) 0 â # È r 10 œ ˆ "3 r cos )‰ i ˆ 3" r sin )‰ j rk Ê krr ‚ r) k œ É 9" r# cos# ) 9" r# sin# ) r# œ É 10r 9 œ 3 Ê A œ '0 '3 r 310 dr d) œ '0 ’ r 6 10 “ d) œ '0 21 4 È 21 #È % $ 21 È 7È10 d) œ 71 3 10 6 21. Let x œ r cos ) and y œ r sin ) Ê r# œ x# y# œ 1, 1 Ÿ z Ÿ 4 and 0 Ÿ ) Ÿ 21. Then r(zß )) œ (cos ))i (sin ))j zk Ê rz œ k and r) œ ( sin ))i (cos ))j Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 965 966 Chapter 16 Integration in Vector Fields â â i â Ê r) ‚ rz œ â sin ) â â 0 j cos ) 0 â kâ â 0 â œ (cos ))i (sin )) j Ê kr) ‚ rz k œ Ècos# ) sin# ) œ 1 â 1â Ê A œ '0 '1 1 dr d) œ '0 3 d) œ 61 21 21 4 22. Let x œ u cos v and z œ u sin v Ê u# œ x# z# œ 10, 1 Ÿ y Ÿ 1, 0 Ÿ v Ÿ 21. Then r(yß v) œ (u cos v)i yj (u sin v)k œ ŠÈ10 cos v‹ i yj ŠÈ10 sin v‹ k â i â â È È È Ê rv œ Š 10 sin v‹ i Š 10 cos v‹ k and ry œ j Ê rv ‚ ry œ â 10 sin v â â 0 â j k â â 0 È10 cos v â â â 1 0 œ ŠÈ10 cos v‹ i ŠÈ10 sin v‹ k Ê krv ‚ ry k œ È10 Ê A œ '0 'c1 È10 du dv œ '0 ’È10u“ 21 21 1 œ '0 2È10 dv œ 41È10 21 " " dv 23. z œ 2 x# y# and z œ Èx# y# Ê z œ 2 z# Ê z# z 2 œ 0 Ê z œ 2 or z œ 1. Since z œ Èx# y# we get z œ 1 where the cone intersects the paraboloid. When x œ 0 and y œ 0, z œ 2 Ê the vertex of the paraboloid is (0ß 0ß 2). Therefore, z ranges from 1 to 2 on the “cap" Ê r ranges from 1 (when x# y# œ 1) to 0 (when x œ 0 and y œ 0 at the vertex). Let x œ r cos ), y œ r sin ), and z œ 2 r# . Then r(rß )) œ (r cos ))i (r sin ))j a2 r# b k , 0 Ÿ r Ÿ 1, 0 Ÿ ) Ÿ 21 Ê rr œ (cos ))i (sin ))j 2rk and â â i j k â â â â sin ) 2r â r) œ (r sin ))i (r cos ))j Ê rr ‚ r) œ â cos ) â â 0 â â r sin ) r cos ) œ a2r# cos )b i a2r# sin )b j rk Ê krr ‚ r) k œ È4r% cos# ) 4r% sin# ) r# œ rÈ4r# 1 " a4r# 1b Ê A œ '0 '0 rÈ4r# 1 dr d) œ '0 ’ 12 21 1 21 0, “ d) œ '0 Š 5 15# " ‹ d) œ 16 Š5È5 1‹ $Î# " 21 È ! 24. Let x œ r cos ), y œ r sin ) and z œ x# y# œ r# . Then r(rß )) œ (r cos ))i (r sin ))j r# k , 1 Ÿ r Ÿ 2, 0 Ÿ ) Ÿ 21 Ê rr œ (cos ))i (sin ))j 2rk and r) œ (r sin ))i (r cos ))j â â i j kâ â â â sin ) 2r â œ a2r# cos )b i a2r# sin )b j rk Ê krr ‚ r) k Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) 0 â " œ È4r% cos# ) 4r% sin# ) r# œ rÈ4r# 1 Ê A œ '0 '1 rÈ4r# 1 dr d) œ '0 ’ 12 a4r# 1b 21 œ '0 Š 17 21 21 2 $Î# # “ d) " È17 5È5 ‹ d) œ 16 Š17È17 5È5‹ 1# 25. Let x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), and z œ 3 cos 9 Ê 3 œ Èx# y# z# œ È2 on the sphere. Next, x# y# z# œ 2 and z œ Èx# y# Ê z# z# œ 2 Ê z# œ 1 Ê z œ 1 since z 0 Ê 9 œ 1 . For the lower 4 portion of the sphere cut by the cone, we get 9 œ 1. Then r(9ß )) œ ŠÈ2 sin 9 cos )‹ i ŠÈ2 sin 9 sin )‹ j ŠÈ2 cos 9‹ k , 1 Ÿ 9 Ÿ 1, 0 Ÿ ) Ÿ 21 4 Ê r9 œ ŠÈ2 cos 9 cos )‹ i ŠÈ2 cos 9 sin )‹ j ŠÈ2 sin 9‹ k and r) œ ŠÈ2 sin 9 sin )‹ i ŠÈ2 sin 9 cos )‹ j â â i j k â â âÈ â È È 2 cos 9 sin ) 2 sin 9 â Ê r9 ‚ r) œ â 2 cos 9 cos ) â â â È2 sin 9 sin ) È2 sin 9 cos ) â 0 # # œ a2 sin 9 cos )b i a2 sin 9 sin )b j (2 sin 9 cos 9)k Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.5 Surfaces and Area 967 Ê kr9 ‚ r) k œ È4 sin% 9 cos# ) 4 sin% 9 sin# ) 4 sin# 9 cos# 9 œ È4 sin# 9 œ 2 ksin 9k œ 2 sin 9 Ê A œ '0 '1Î4 2 sin 9 d9 d) œ '0 Š2 È2‹ d) œ Š4 2È2‹ 1 21 1 21 26. Let x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), and z œ 3 cos 9 Ê 3 œ Èx# y# z# œ 2 on the sphere. Next, z œ 1 Ê 1 œ 2 cos 9 Ê cos 9 œ "# Ê 9 œ 231 ; z œ È3 Ê È3 œ 2 cos 9 Ê cos 9 œ r(9ß )) œ (2 sin 9 cos ))i (2 sin 9 sin ))j (2 cos 9)k , 16 Ÿ 9 Ÿ 231 , 0 Ÿ ) Ÿ 21 È3 # Ê 9 œ 16 . Then Ê r9 œ (2 cos 9 cos ))i (2 cos 9 sin ))j (2 sin 9)k and r) œ (2 sin 9 sin ))i (2 sin 9 cos )) j â â i j k â â â â Ê r9 ‚ r) œ â 2 cos 9 cos ) 2 cos 9 sin ) 2 sin 9 â â â 0 â 2 sin 9 sin ) 2 sin 9 cos ) â œ a4 sin# 9 cos )b i a4 sin# 9 sin )b j (4 sin 9 cos 9)k Ê kr9 ‚ r) k œ È16 sin% 9 cos# ) 16 sin% 9 sin# ) 16 sin# 9 cos# 9 œ È16 sin# 9 œ 4 ksin 9k œ 4 sin 9 Ê A œ '0 '1Î6 4 sin 9 d9 d) œ '0 Š2 2È3‹ d) œ Š4 4È3‹ 1 21 21Î3 21 27. The parametrization r(rß )) œ (r cos ))i (r sin ))j rk at P! œ ŠÈ2ß È2ß 2‹ Ê ) œ 1 , r œ 2, 4 È È rr œ (cos ))i (sin ))j k œ #2 i #2 j k and r) œ (r sin ))i (r cos ))j œ È2i È2j â i j k ââ â âÈ â Ê rr ‚ r) œ â 2/2 È2/2 1 â â â â È 2 È 2 0 â œ È2i È2j 2k Ê the tangent plane is 0 œ ŠÈ2i È2j 2k‹ † ’Šx È2‹ i Šy È2‹ j (z 2)k“ Ê È2x È2y 2z œ 0, or x y È2z œ 0. The parametrization r(rß )) Ê x œ r cos ), y œ r sin ) and z œ r Ê x# y# œ r# œ z# Ê the surface is z œ Èx# y# . 28. The parametrization r(9ß )) œ (4 sin 9 cos ))i (4 sin 9 sin ))j (4 cos 9)k at P! œ ŠÈ2ß È2ß 2È3‹ Ê 3 œ 4 and z œ 2È3 œ 4 cos 9 Ê 9 œ 16 ; also x œ È2 and y œ È2 Ê ) œ 14 . Then r9 œ (4 cos 9 cos ))i (4 cos 9 sin ))j (4 sin 9)k œ È6i È6j 2k and r) œ (4 sin 9 sin ))i (4 sin 9 cos ))j â i j k ââ â â â œ È2i È2j at P! Ê r9 ‚ r) œ â È6 È6 2 â â â â È2 È2 0 â œ 2È2i 2È2j 4È3k Ê the tangent plane is Š2È2i 2È2j 4È3k‹ † ’Šx È2‹ i Šy È2‹ j Šz 2È3‹ k“ œ 0 Ê È2x È2y 2È3z œ 16, or x y È6z œ 8È2. The parametrization Ê x œ 4 sin 9 cos ), y œ 4 sin 9 sin ), z œ 4 cos 9 Ê the surface is x# y# z# œ 16, z 0. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 968 Chapter 16 Integration in Vector Fields 29. The parametrization r()ß z) œ (3 sin 2))i a6 sin# )b j zk È at P! œ Š 3 # 3 ß #9 ß 0‹ Ê ) œ 13 and z œ 0. Then r) œ (6 cos 2))i (12 sin ) cos ))j œ 3i 3È3j and rz œ k at P! â i j k ââ â â â Ê r) ‚ rz œ â 3 3È3 0 â œ 3È3i 3j â â â 0 0 1â Ê the tangent plane is È Š3È3i 3j‹ † ’Šx 3 3 ‹ i ˆy 9 ‰ j (z 0)k“ œ 0 # # Ê È3x y œ 9. The parametrization Ê x œ 3 sin 2) # and y œ 6 sin# ) Ê x# y# œ 9 sin# 2) a6 sin# )b œ 9 a4 sin# ) cos# )b 36 sin% ) œ 6 a6 sin# )b œ 6y Ê x# y# 6y 9 œ 9 Ê x# (y 3)# œ 9 30. The parametrization r(xß y) œ xi yj x# k at P! œ (1ß 2ß 1) Ê rx œ i 2xk œ i 2k and ry œ j at P! â â âi j k â â â Ê rx ‚ ry œ â 1 0 2 â œ 2i k Ê the tangent plane â â â0 " 0 â is (2i k) † [(x 1)i (y 2)j (z 1)k] œ 0 Ê 2x z œ 1. The parametrization Ê x œ x, y œ y and z œ x# Ê the surface is z œ x# 31. (a) An arbitrary point on the circle C is (xß z) œ (R r cos u, r sin u) Ê (xß yß z) is on the torus with x œ (R r cos u) cos v, y œ (R r cos u) sin v, and z œ r sin u, 0 Ÿ u Ÿ 21, 0 Ÿ v Ÿ 21 (b) ru œ (r sin u cos v)i (r sin u sin v)j (r cos u)k and rv œ ((R r cos u) sin v)i ((R r cos u) cos v)j â â i j k â â â â r sin u sin v r cos u â Ê ru ‚ rv œ â r sin u cos v â â 0 â â (R r cos u) sin v (R r cos u) cos v œ (R r cos u)(r cos v cos u)i (R r cos u)(r sin v cos u)j (r sin u)(R r cos u)k Ê kru ‚ rv k# œ (R r cos u)# ar# cos# v cos# u r# sin# v cos# u r# sin# ub Ê kru ‚ rv k œ r(R r cos u) Ê A œ '0 '0 21 21 arR r# cos ub du dv œ '0 21rR dv œ 41# rR 21 32. (a) The point (xß yß z) is on the surface for fixed x œ f(u) when y œ g(u) sin ˆ 1# v‰ and z œ g(u) cos ˆ 1# v‰ Ê x œ f(u), y œ g(u) cos v, and z œ g(u) sin v Ê r(uß v) œ f(u)i (g(u) cos v)j (g(u) sin v)k , 0 Ÿ v Ÿ 21, aŸuŸb (b) Let u œ y and x œ u# Ê f(u) œ u# and g(u) œ u Ê r(uß v) œ u# i (u cos v)j (u sin v)k , 0 Ÿ v Ÿ 21, 0 Ÿ u # # # 33. (a) Let w# zc# œ 1 where w œ cos 9 and zc œ sin 9 Ê xa# yb# œ cos# 9 Ê xa œ cos 9 cos ) and yb œ cos 9 sin ) Ê x œ a cos ) cos 9, y œ b sin ) cos 9, and z œ c sin 9 Ê r()ß 9) œ (a cos ) cos 9)i (b sin ) cos 9)j (c sin 9)k Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.5 Surfaces and Area (b) r) œ (a sin ) cos 9)i (b cos ) cos 9)j and r9 œ (a cos ) sin 9)i (b sin ) sin 9)j (c cos 9)k â â i j k â â â â 0 â Ê r) ‚ r9 œ â a sin ) cos 9 b cos ) cos 9 â â â a cos ) sin 9 b sin ) sin 9 c cos 9 â œ abc cos ) cos# 9b i aac sin ) cos# 9b j (ab sin 9 cos 9)k Ê kr) ‚ r9 k# œ b# c# cos# ) cos% 9 a# c# sin# ) cos% 9 a# b# sin# 9 cos# 9, and the result follows. A Ê '0 '0 kr) ‚ r9 k d9 d) œ '0 '0 c a# b# sin# 9 cos# 9 b# c# cos# ) cos% 9 a# c# sin# ) cos% 9 d d9 d) 21 1 21 1 1/2 34. (a) r()ß u) œ (cosh u cos ))i (cosh u sin ))j (sinh u)k (b) r()ß u) œ (a cosh u cos ))i (b cosh u sin ))j (c sinh u)k 35. r()ß u) œ (5 cosh u cos ))i (5 cosh u sin ))j (5 sinh u)k Ê r) œ (5 cosh u sin ))i (5 cosh u cos ))j and ru œ (5 sinh u cos ))i (5 sinh u sin ))j (5 cosh u)k â â i j k â â â â 0 Ê r) ‚ ru œ â 5 cosh u sin ) 5 cosh u cos ) â â â 5 sinh u sin ) 5 cosh u â â 5 sinh u cos ) œ a25 cosh# u cos )b i a25 cosh# u sin )b j (25 cosh u sinh u)k. At the point (x! ß y! ß 0), where x#0 y#0 œ 25 we have 5 sinh u œ 0 Ê u œ 0 and x! œ 25 cos ), y! œ 25 sin ) Ê the tangent plane is 5(x! i y! j) † [(x x! )i (y y! )j zk] œ 0 Ê x! x x0# y! y y0# œ 0 Ê x! x y! y œ 25 # # # 36. Let zc# w# œ 1 where zc œ cosh u and w œ sinh u Ê w# œ xa# by# Ê xa œ w cos ) and by œ w sin ) Ê x œ a sinh u cos ), y œ b sinh u sin ), and z œ c cosh u Ê r()ß u) œ (a sinh u cos ))i (b sinh u sin ))j (c cosh u)k , 0 Ÿ ) Ÿ 21, _ u _ 37. p œ k , ™ f œ 2xi 2yj k Ê k ™ f k œ È(2x)# (2y)# (1)# œ È4x# 4y# 1 and k ™ f † pk œ 1; z œ 2 Ê x# y# œ 2; thus S œ ' ' k™f k dA œ ' ' È4x# 4y# 1 dx dy R k™f†pk R È2 $Î# " a4r# 1b “ œ ' ' È4r# cos# ) 4r# sin# ) 1 r dr d) œ '0 '0 È4r# 1 r dr d) œ '0 ’ 12 21 21 R œ '0 21 È# ! d) 13 13 6 d) œ 3 1 38. p œ k , ™ f œ 2xi 2yj k Ê k ™ f k œ È4x# 4y# 1 and k ™ f † pk œ 1; 2 Ÿ x# y# Ÿ 6 È6 k™f k ' ' È4x# 4y# 1 dx dy œ ' ' È4r# 1 r dr d) œ ' 'È È4r# 1 r dr d) Ê S œ ' ' k™ f†pk dA œ 0 2 21 R " œ '0 ’ 12 a4r# 1b 21 R $Î# R È' “ È d) œ '0 21 # 49 49 6 d) œ 3 1 39. p œ k , ™ f œ i 2j 2k Ê k ™ f k œ 3 and k ™ f † pk œ 2; x œ y# and x œ 2 y# intersect at (1ß 1) and (1ß 1) 2 c y# k™f k ' ' 3# dx dy œ ' ' # Ê S œ ' ' k™ f†pk dA œ c1 y 1 R R 3 # dx dy œ 'c11 a3 3y# b dy œ 4 40. p œ k , ™ f œ 2xi 2k Ê k ™ f k œ È4x# 4 œ 2Èx# 1 and k ™ f † pk œ 2 Ê S œ ' ' œ'' R 2È x# 1 dx dy œ # È3 È3 '0 '0 Èx# 1 dy dx œ '0 x R k™f k k™f†pk dA È$ $Î# xÈx# 1 dx œ ’ 3" ax# 1b “ œ 3" (4)$Î# 3" œ 37 ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 969 970 Chapter 16 Integration in Vector Fields 41. p œ k , ™ f œ 2xi 2j 2k Ê k ™ f k œ È(2x)# (2)# (2)# œ È4x# 8 œ 2Èx# 2 and k ™ f † pk œ 2 Ê Sœ'' R k™f k k™f†pk dA œ ' ' 2Èx##2 dx dy œ ' ' Èx# 2 dy dx œ ' 3xÈx# 2 dx œ ’ax# 2b$Î# “ 0 0 0 2 3x 2 R # ! œ 6È 6 2È 2 42. p œ k , ™ f œ 2xi 2yj 2zk Ê k ™ f k œ È4x# 4y# 4z# œ È8 œ 2È2 and k ™ f † pk œ 2z; x# y# z# œ 2 and z œ Èx# y# Ê x# y# œ 1; thus, S œ ' ' œ È2 ' ' " È2 È2 ax# y# b dA œ R '0 ' 21 k™f k k™f†pk dA œ R 1 r dr d) œ È2 0 È 2 r# '0 2 È2 ' ' " dA ' ' 2È #z dA œ z 21 R R Š1 È2‹ d) œ 21 Š2 È2‹ 43. p œ k , ™ f œ ci k Ê k ™ f k œ Èc# 1 and k ™ f † pk œ 1 Ê S œ ' ' R 21 1 21 È c 1 œ '0 '0 Èc# 1 r dr d) œ '0 d) œ 1Èc# 1 # k™f k k™f†pk dA œ ' ' Èc# 1 dx dy R # 44. p œ k , ™ f œ 2xi 2zj Ê k ™ f k œ È(2x)# (2z)# œ 2 and k ™ f † pk œ 2z for the upper surface, z Ê Sœ'' R k™f k k™f†pk dA œ '' R 2 #z dA œ '' R " È1 x# dy dx œ 2 1Î2 1Î2 'c1Î2 '0 1 È1 x# dy dx œ ' 1Î2 0 " dx 1Î2 È1 x# "Î# œ csin" xd "Î# œ 16 ˆ 16 ‰ œ 13 45. p œ i , ™ f œ i 2yj 2zk Ê k ™ f k œ È1# (2y)# (2z)# œ È1 4y# 4z# and k ™ f † pk œ 1; 1 Ÿ y# z# Ÿ 4 k™f k ' ' È1 4y# 4z# dy dz œ ' ' È1 4r# cos# ) 4r# sin# ) r dr d) Ê S œ ' ' k™ f†pk dA œ 0 1 21 R R " a1 4r# b œ '0 '1 È1 4r# r dr d) œ '0 ’ 12 21 21 2 2 “ d) œ '0 $Î# # 21 " 1 " È È È È 1# Š17 17 5 5‹ d) œ 6 Š17 17 5 5‹ 46. p œ j , ™ f œ 2xi j 2zk Ê k ™ f k œ È4x# 4z# 1 and k ™ f † pk œ 1; y œ 0 and x# y z# œ 2 Ê x# z# œ 2; k™f k ' ' È4x# 4z# 1 dx dz œ ' thus, S œ ' ' k™ f†pk dA œ 0 21 R R È '0 2 È4r# 1 r dr d) œ '021 136 d) œ 133 1 # # # 47. p œ k , ™ f œ ˆ2x 2x ‰ i È15 j k Ê k ™ f k œ Êˆ2x 2x ‰ ŠÈ15‹ (1)# œ É4x# 8 x4# œ Éˆ2x 2x ‰ k™f k ' ' a2x 2x" b dx dy œ 2x 2x , on 1 Ÿ x Ÿ 2 and k ™ f † pk œ 1 Ê S œ ' ' k™ f†pk dA œ R R œ '0 '1 a2x 2x" b dx dy œ '0 cx# 2 ln xd " dy œ '0 (3 2 ln 2) dy œ 3 2 ln 2 1 2 1 1 # 48. p œ k , ™ f œ 3Èx i 3Èy j 3k Ê k ™ f k œ È9x 9y 9 œ 3Èx y 1 and k ™ f † pk œ 3 k™f k ' ' Èx y 1 dx dy œ ' ' Èx y 1 dx dy œ ' 23 (x y 1)$Î# ‘ ! dy Ê S œ ' ' k™ f†pk dA œ 0 0 0 1 R 1 1 " R 1 " 4 4 4 œ '0 23 (y 2)$Î# 23 (y 1)$Î# ‘ dy œ 15 (y 2)&Î# 15 (y 1)&Î# ‘ ! œ 15 (3)&Î# (2)&Î# (2)&Î# 1‘ 4 œ 15 Š9È3 8È2 1‹ 49. fx (xß y) œ 2x, fy (xß y) œ 2y Ê Éfx# fy# 1 œ È4x# 4y# 1 Ê Area œ ' ' È4x# 4y# 1 dx dy È3 œ '0 '0 21 R È4r# 1 r dr d) œ 1 Š13È13 1‹ 6 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.5 Surfaces and Area 971 50. fy (yß z) œ 2y, fz (yß z) œ 2z Ê Éfy# fz# 1 œ È4y# 4z# 1 Ê Area œ ' ' È4y# 4z# 1 dy dz R 21 1 œ '0 '0 È4r# 1 r dr d) œ 16 Š5È5 1‹ 51. fx (xß y) œ Èx#x y# , fy (xß y) œ Èx#y y# Ê Éfx# fy# 1 œ É x# x y# x# y y# 1 œ È2 Ê Area œ ' ' È2 dx dy # # Rxy œ È2aArea between the ellipse and the circleb œ È2a61 1b œ 51È2 52. Over Rxy : z œ 2 23 x 2y Ê fx (xß y) œ 23 , fy (xß y) œ 2 Ê Éfx# fy# 1 œ É 49 4 1 œ 73 Ê Area œ ' ' Rxy 7 7 ˆ 7 ‰ ˆ #3 ‰ œ #7 . 3 dA œ 3 (Area of the shadow triangle in the xy-plane) œ 3 Over Rxz : y œ 1 13 x "# z Ê fx (xß z) œ 13 , fz (xß z) œ "# Ê Èfx# fz# 1 œ É 19 "4 1 œ 76 Ê Area œ ' ' 76 dA œ 76 (Area of the shadow triangle in the xz-plane) œ ˆ 76 ‰ (3) œ 7# . Rxz Over Ryz : x œ 3 3y 3# z Ê fy (yß z) œ 3, fz (yß z) œ 3# Ê Éfy# fz# 1 œ É9 94 1 œ 7# Ê Area œ ' ' 72 dA œ 72 (Area of the shadow triangle in the yz-plane) œ ˆ 72 ‰ (1) œ 7# . Ryz 16 2 $Î# 53. y œ 23 z$Î# Ê fx (xß z) œ 0, fz (xß z) œ z"Î# Ê Èfx# fz# 1 œ Èz 1 ; y œ 16 Ê zœ4 3 Ê 3 œ 3 z Ê Area œ '0 '0 Èz 1 dx dz œ '0 Èz 1 dz œ 23 Š5È5 1‹ 4 1 4 4 c z# 54. y œ 4 z Ê fx (xß z) œ 0, fz (xß z) œ 1 Ê Èfx# fz# 1 œ È2 Ê Area œ ' ' È2 dA œ '0 '0 2 È2 dx dz Rxz œ È2 '0 a4 z# b dz œ 163 2 È 2 55. rax, yb œ x i y j fax, yb k Ê rx ax, yb œ i fx ax, yb k, ry ax, yb œ j fy ax, ybk â â k â âi j â â Ê rx ‚ ry œ â 1 0 fx ax, yb â œ fx ax, yb i fy ax, yb j k â â â 0 1 fy ax, yb â Ê lrx ‚ ry l œ Éafx ax, ybb2 afy ax, ybb2 12 œ Éfx ax, yb2 fy ax, yb2 1 Ê d5 œ Éfx ax, yb2 fy ax, yb2 1 dA 56. S is obtained by rotating y œ faxb, a Ÿ x Ÿ b about the x-axis where faxb 0 (a) Let ax, y, zb be a point on S. Consider the cross section when x œ x‡ , the cross section is a circle with radius r œ fax‡ b. The set of parametric equations for this circle are given by ya)b œ r cos ) œ fax‡ b cos ) and za)b œ r sin ) œ fax‡ b sin ) where 0 Ÿ ) Ÿ 21. Since x can take on any value between a and b we have xax, )b œ x, yax, )b œ faxb cos ), zax, )b œ faxb sin ) where a Ÿ x Ÿ b and 0 Ÿ ) Ÿ 21. Thus rax, )b œ x i faxb cos ) j faxb sin ) k (b) rx ax, )b œ i f w axb cos ) j f w axb sin ) k and r) ax, )b œ faxb sin ) j faxb cos ) k â â j k âi â â â Ê rx ‚ r) œ â 1 f w axb cos ) f w axb sin ) â œ faxb † f w axb i faxb cos ) j faxb sin ) k â â â 0 faxb sin ) faxb cos ) â Ê lrx ‚ r) l œ Éafaxb † f w axbb2 afaxb cos )b2 afaxb sin )b2 œ faxbÉ1 af w axbb2 21 A œ 'a '0 faxbÉ1 af w axbb2 d) dx œ 'a ”ŒfaxbÉ1 af w axbb2 )• dx œ 'a 2 1 faxbÉ1 af w axbb2 dx b 21 b b 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 972 Chapter 16 Integration in Vector Fields 16.6 SURFACE INTEGRALS â âi â 1. Let the parametrization be r(xß z) œ xi x# j zk Ê rx œ i 2xj and rz œ k Ê rx ‚ rz œ â 1 â â0 j 2x 0 â kâ â 0â â "â " œ 2xi j Ê krx ‚ rz k œ È4x# 1 Ê ' ' G(xß yß z) d5 œ '0 '0 xÈ4x# 1 dx dz œ '0 ’ 12 a4x# 1b 3 œ' 2 3 S $Î# # “ dz ! È " " Š17È17 1‹ dz œ 17 17 4 0 1# 3 2. Let the parametrization be r(xß y) œ xi yj È4 y# k , 2 Ÿ y Ÿ 2 Ê rx œ i and ry œ j È4y y# k âi j â k â â # â1 0 â 0 Ê rx ‚ ry œ â â œ y j k Ê krx ‚ ry k œ É 4 y y# 1 œ È42 y# â 0 1 y â È 4 y# â È 4 y# â Ê ' ' G(xß yß z) d5 œ '1 'c2 È4 y# Š È42 y# ‹ dy dx œ 24 4 2 S 3. Let the parametrization be r(9ß )) œ (sin 9 cos ))i (sin 9 sin ))j (cos 9)k (spherical coordinates with 3 œ 1 on the sphere), 0 Ÿ 9 Ÿ 1, 0 Ÿ ) Ÿ 21 Ê r9 œ (cos 9 cos ))i (cos 9 sin ))j (sin 9)k and â â i j k â â â â r) œ ( sin 9 sin ))i (sin 9 cos ))j Ê r9 ‚ r) œ â cos 9 cos ) cos 9 sin ) sin 9 â â â 0 â â sin 9 sin ) sin 9 cos ) œ asin# 9 cos )b i asin# 9 sin )b j (sin 9 cos 9)k Ê kr9 ‚ r) k œ Èsin% 9 cos# ) sin% 9 sin# ) sin# 9 cos# 9 œ sin 9; x œ sin 9 cos ) Ê G(xß yß z) œ cos# ) sin# 9 Ê ' ' G(xß yß z) d5 œ '0 '0 acos# ) sin# 9b (sin 9) d9 d) 21 1 S 21 c1 u œ cos 9 œ '0 '0 acos )b a1 cos 9b (sin 9) d9 d); ” Ä '0 '1 acos# )b au# 1b du d) • du œ sin 9 d9 21 1 # # œ '0 acos# )b ’ u3 u“ 21 $ " " d) œ 34 '0 cos# ) d) œ 34 2) sin42) ‘ ! œ 431 21 #1 4. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i (a sin 9 sin ))j (a cos 9)k (spherical coordinates with 3 œ a, a 0, on the sphere), 0 Ÿ 9 Ÿ 1# (since z 0), 0 Ÿ ) Ÿ 21 Ê r9 œ (a cos 9 cos ))i (a cos 9 sin ))j (a sin 9)k and â â i j k â â â â r) œ (a sin 9 sin ))i (a sin 9 cos ))j Ê r9 ‚ r) œ â a cos 9 cos ) a cos 9 sin ) a sin 9 â â â 0 â a sin 9 sin ) a sin 9 cos ) â œ aa# sin# 9 cos )b i aa# sin# 9 sin )b j (a# sin 9 cos 9)k Ê kr9 ‚ r) k œ Èa% sin% 9 cos# ) a% sin% 9 sin# ) a% sin# 9 cos# 9 œ a# sin 9; z œ a cos 9 Ê G(xß yß z) œ a# cos# 9 Ê ' ' G(xß yß z) d5 œ '0 '0 21 S 1Î2 aa# cos# 9b aa# sin 9b d9 d) œ 23 1a% 5. Let the parametrization be r(xß y) œ xi yj (4 x y)k Ê rx œ i k and ry œ j k â â âi j k â 1 1 â â Ê rx ‚ ry œ â 1 0 1 â œ i j k Ê krx ‚ ry k œ È3 Ê ' ' F(xß yß z) d5 œ '0 '0 (4 x y) È3 dy dx â â S â 0 1 " â œ '0 È3 ’4y xy y2 “ dx œ '0 È3 ˆ #7 x‰ dx œ È3 ’ 27 x x# “ œ 3È3 1 # " ! 1 # " ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.6 Surface Integrals 973 6. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j rk , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â sin ) " â Ê rr œ (cos ))i (sin ))j k and r) œ (r sin ))i (r cos ))j Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) 0 â œ (r cos ))i (r sin ))j rk Ê krr ‚ r) k œ È(r cos ))# (r sin ))# r# œ rÈ2; z œ r and x œ r cos ) Ê F(xß yß z) œ r r cos ) Ê ' ' F(xß yß z) d5 œ '0 '0 (r r cos )) ŠrÈ2‹ dr d) œ È2 '0 '0 (1 cos )) r# dr d) 21 21 1 1 S È œ 213 2 7. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j a1 r# b k , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21 â â i j k â â â â sin ) 2r â Ê rr œ (cos ))i (sin ))j 2rk and r) œ (r sin ))i (r cos ))j Ê rr ‚ r) œ â cos ) â â 0 â â r sin ) r cos ) œ a2r# cos )b i a2r# sin )b j rk Ê krr ‚ r) k œ Éa2r# cos )b# a2r# sin )b r# œ rÈ1 4r# ; z œ 1 r# and x œ r cos ) Ê H(xß yß z) œ ar# cos# )b È1 4r# Ê ' ' H(xß yß z) d5 S 21 1 21 1 œ '0 '0 ar# cos# )b ŠÈ1 4r# ‹ ŠrÈ1 4r# ‹ dr d) œ '0 '0 r$ a1 4r# b cos# ) dr d) œ 11121 8. Let the parametrization be r(9ß )) œ (2 sin 9 cos ))i (2 sin 9 sin ))j (2 cos 9)k (spherical coordinates with 3 œ 2 on the sphere), 0 Ÿ 9 Ÿ 1 ; x# y# z# œ 4 and z œ Èx# y# Ê z# z# œ 4 Ê z# œ 2 Ê z œ È2 (since 4 È2 # Ê 9 œ 14 , 0 Ÿ ) Ÿ 21; r9 œ (2 cos 9 cos ))i (2 cos 9 sin ))j (2 sin 9)k â â i j k â â â â and r) œ (2 sin 9 sin ))i (2 sin 9 cos ))j Ê r9 ‚ r) œ â 2 cos 9 cos ) 2 cos 9 sin ) 2 sin 9 â â â 0 â 2 sin 9 sin ) 2 sin 9 cos ) â 0) Ê 2 cos 9 œ È2 Ê cos 9 œ z œ a4 sin# 9 cos )b i a4 sin# 9 sin )b j (4 sin 9 cos 9)k Ê kr9 ‚ r) k œ È16 sin% 9 cos# ) 16 sin% 9 sin# ) 16 sin# 9 cos# 9 œ 4 sin 9; y œ 2 sin 9 sin ) and z œ 2 cos 9 Ê H(xß yß z) œ 4 cos 9 sin 9 sin ) Ê ' ' H(xß yß z) d5 œ '0 '0 21 œ '0 '0 16 sin# 9 cos 9 sin ) d9 d) œ 0 21 1Î4 1Î4 (4 cos 9 sin 9 sin ))(4 sin 9) d9 d) S 9. The bottom face S of the cube is in the xy-plane Ê z œ 0 Ê G(xß yß 0) œ x y and f(xß yß z) œ z œ 0 Ê p œ k and ™ f œ k Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dy Ê ' ' G d5 œ ' ' (x y) dx dy œ '0 '0 (x y) dx dy œ ' a a S a # Š a# ay‹ dy œ a$ . 0 R Because of symmetry, we also get a$ over the face of the cube in the xz-plane and a$ over the face of the cube in the yz-plane. Next, on the top of the cube, G(xß yß z) œ G(xß yß a) œ x y a and f(xß yß z) œ z œ a Ê p œ k and ™ f œ k Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dy ' ' G d5 œ ' ' (x y a) dx dy œ ' ' (x y a) dx dy œ ' ' (x y) dx dy ' ' a dx dy œ 2a$ . 0 0 0 0 0 0 a S a a a a a R Because of symmetry, the integral is also 2a$ over each of the other two faces. Therefore, ' ' (x y z) d5 œ 3 aa$ 2a$ b œ 9a$ . cube 10. On the face S in the xz-plane, we have y œ 0 Ê f(xß yß z) œ y œ 0 and G(xß yß z) œ G(xß 0ß z) œ z Ê p œ j and ™ f œ j Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dz Ê ' ' G d5 œ ' ' (y z) d5 œ '0 '0 z dx dz œ '0 2z dz œ 1. 1 S 2 1 S On the face in the xy-plane, we have z œ 0 Ê f(xß yß z) œ z œ 0 and G(xß yß z) œ G(xß yß 0) œ y Ê p œ k and ™ f œ k Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 974 Chapter 16 Integration in Vector Fields Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dy Ê ' ' G d5 œ ' ' y d5 œ '0 '0 y dx dy œ 1. 1 S 2 S On the triangular face in the plane x œ 2 we have f(xß yß z) œ x œ 2 and G(xß yß z) œ G(2ß yß z) œ y z Ê p œ i and 1cy ™ f œ i Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dz dy Ê ' ' G d5 œ ' ' (y z) d5 œ '0 '0 1 S œ '0 "# a1 y# b dy œ 3" . (y z) dz dy S 1 On the triangular face in the yz-plane, we have x œ 0 Ê f(xß yß z) œ x œ 0 and G(xß yß z) œ G(0ß yß z) œ y z Ê p œ i and ™ f œ i Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dz dy Ê ' ' G d5 œ ' ' (y z) d5 S 1 1cy œ '0 '0 (y z) dz dy œ "3 . S Finally, on the sloped face, we have y z œ 1 Ê f(xß yß z) œ y z œ 1 and G(xß yß z) œ y z œ 1 Ê p œ k and ™ f œ j k Ê k ™ f k œ È2 and k ™ f † pk œ 1 Ê d5 œ È2 dx dy Ê ' ' G d5 œ ' ' (y z) d5 œ '0 '0 È2 dx dy œ 2È2. Therefore, ' ' 1 2 wedge S S G(xß yß z) d5 œ 1 1 "3 3" 2È2 œ 38 2È2 11. On the faces in the coordinate planes, G(xß yß z) œ 0 Ê the integral over these faces is 0. On the face x œ a, we have f(xß yß z) œ x œ a and G(xß yß z) œ G(aß yß z) œ ayz Ê p œ i and ™ f œ i Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dy dz Ê ' ' G d5 œ ' ' ayz d5 œ '0 '0 ayz dy dz œ ab4c . c S b # # S On the face y œ b, we have f(xß yß z) œ y œ b and G(xß yß z) œ G(xß bß z) œ bxz Ê p œ j and ™ f œ j Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dz Ê ' ' G d5 œ ' ' bxz d5 œ '0 '0 bxz dx dz œ a bc 4 . c S a # # S On the face z œ c, we have f(xß yß z) œ z œ c and G(xß yß z) œ G(xß yß c) œ cxy Ê p œ k and ™ f œ k Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dy dx Ê ' ' G d5 œ ' ' cxy d5 œ '0 '0 cxy dx dy œ a 4b c . Therefore, b ' ' G(xß yß z) d5 œ abc(ab 4ac bc) . S a # # S S 12. On the face x œ a, we have f(xß yß z) œ x œ a and G(xß yß z) œ G(aß yß z) œ ayz Ê p œ i and ™ f œ i Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dz dy Ê ' ' G d5 œ ' ' ayz d5 œ 'cb 'cc ayz dz dy œ 0. Because of the symmetry b S c S of G on all the other faces, all the integrals are 0, and ' ' G(xß yß z) d5 œ 0. S 13. f(xß yß z) œ 2x 2y z œ 2 Ê ™ f œ 2i 2j k and G(xß yß z) œ x y (2 2x 2y) œ 2 x y Ê p œ k , k ™ f k œ 3 and k ™ f † pk œ 1 Ê d5 œ 3 dy dx; z œ 0 Ê 2x 2y œ 2 Ê y œ 1 x Ê ' ' G d5 œ ' ' (2 x y) d5 S 1 1cx 1 1 œ 3 '0 '0 (2 x y) dy dx œ 3 '0 (2 x)(1 x) "# (1 x)# ‘ dx œ 3 '0 Š #3 2x x# ‹ dx œ 2 S # 14. f(xß yß z) œ y# 4z œ 16 Ê ™ f œ 2yj 4k Ê k ™ f k œ È4y# 16 œ 2Èy# 4 and p œ k Ê k ™ f † pk œ 4 Ê d5 œ œ' 2È y# 4 dx dy 4 Èy 4 Ê ' ' G d5 œ 'c4 '0 ˆxÈy# 4‰ Š # ‹ dx dy œ 'c4 '0 x ay # 4b dx dy 4 1 # 4 1 # S % " " y$ # ‰ 56 a y 4 b dy œ 4y œ #" ˆ 64 ’ “ # 4 3 3 16 œ 3 c4 ! 4 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.6 Surface Integrals 975 15. fax, y, zb œ x y2 z œ 0 Ê ™ f œ i 2yj k Ê k ™ f k œ È4y# 2 œ È2È2y# 1 and p œ k Ê k ™ f † pk œ 1 È2È2y# 1 dx dy Ê 1 Ê d5 œ œ È2'0 y 1 3È ' ' G d5 œ ' ' ax y2 xb È2È2y# 1 dx dy œ È2' ' y2 È2y# 1 dx dy S 1 y 1 y 0 0 0 0 È È 2y# 1 dy œ 6 630 2 16. fax, y, zb œ x2 y z œ 0 Ê ™ f œ 2xi j k Ê k ™ f k œ È4x# 2 œ È2È2x# 1 and p œ k Ê k ™ f † pk œ 1 È2È2x# 1 dx dy Ê 1 Ê d5 œ È È œ 3 6 6 2 ' ' ' G d5 œ ' ' x È2È2x# 1 dx dy œ È2' ' x È2x# 1 dx dy S 1 1 1 1 c1 0 c1 0 È È dy œ 3 6 3 2 0 1 17. fax, y, zb œ 2x y z œ 2 Ê ™ f œ 2i j k Ê k ™ f k œ È6 and p œ k Ê k ™ f † pk œ 1 Ê d5 œ 2 c 2x 2 c 2x Ê ' ' G d5 œ '0 '1 c x x ya2 2x ybÈ6 dy dx œ È6'0 '1 c x a2x y 2x2 y x y2 b dy dx 1 1 È6 1 dy dx S È œ È6' ˆ 2 x 2x2 2x3 2 x4 ‰ dx œ 6 1 3 0 3 30 18. fax, y, zb œ x y œ " Ê ™ f œ i j Ê k ™ f k œ È2 and p œ j Ê k ™ f † pk œ 1 È2 1 dz dx Ê Ê d5 œ œ È2' ' ' G d5 œ ' ' ax a1 xb zb È2 dz dx œ È2' ' a2x z 1b dz dx S 1 1 1 1 0 0 0 0 È ˆ2x 32 ‰dx œ 22 0 1 19. Let the parametrization be r(xß y) œ xi yj a4 y# b k , 0 Ÿ x Ÿ 1, 2 Ÿ y Ÿ 2; z œ 0 Ê 0 œ 4 y# â â k â âi j â â 0 â œ 2yj k Ê F † n d5 Ê y œ „ 2; rx œ i and ry œ j 2yk Ê rx ‚ ry œ â 1 0 â â â 0 1 2y â œ F † krxx ‚ryyk krx ‚ ry k dy dx œ (2xy 3z) dy dx œ c2xy 3a4 y# bd dy dx Ê ' ' F † n d5 r ‚r œ '0 'c2 a2xy 3y 12b dy dx œ '0 cxy y 1 2 1 # # S $ # 12yd # dx œ '0 32 dx œ 32 1 20. Let the parametrization be r(xß y) œ xi x# j zk , 1 Ÿ x Ÿ 1, 0 Ÿ z Ÿ 2 Ê rx œ i 2xj and rz œ k â â â i j kâ â â rz # Ê rx ‚ rz œ â 1 2x 0 â œ 2xi j Ê F † n d5 œ F † krrxx ‚ ‚rz k krx ‚ rz k dz dx œ x dz dx â â â0 0 1â Ê ' ' F † n d5 œ 'c1 '0 x# dz dx œ 43 1 2 S 21. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i (a sin 9 sin ))j (a cos 9)k (spherical coordinates with 3 œ a, a 0, on the sphere), 0 Ÿ 9 Ÿ 1# (for the first octant) ß 0 Ÿ ) Ÿ 1# (for the first octant) Ê r9 œ (a cos 9 cos ))i (a cos 9 sin ))j (a sin 9)k and r) œ (a sin 9 sin ))i (a sin 9 cos ))j â â i j k â â â â Ê r9 ‚ r) œ â a cos 9 cos ) a cos 9 sin ) a sin 9 â â â 0 â a sin 9 sin ) a sin 9 cos ) â r ‚r œ aa# sin# 9 cos )b i aa# sin# 9 sin )b j aa# sin 9 cos 9b k Ê F † n d5 œ F † kr99 ‚r)) k kr9 ‚ r) k d) d9 œ a$ cos# 9 sin 9 d) d9 since F œ zk œ (a cos 9)k Ê ' ' F † n d5 œ '0 1Î2 S '01Î2 a$ cos# 9 sin 9 d9 d) œ 1a6 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. $ 976 Chapter 16 Integration in Vector Fields 22. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i (a sin 9 sin ))j (a cos 9)k (spherical coordinates with 3 œ a, a 0, on the sphere), 0 Ÿ 9 Ÿ 1, 0 Ÿ ) Ÿ 21 Ê r9 œ (a cos 9 cos ))i (a cos 9 sin ))j (a sin 9)k and r) œ (a sin 9 sin ))i (a sin 9 cos ))j â â i j k â â â â Ê r9 ‚ r) œ â a cos 9 cos ) a cos 9 sin ) a sin 9 â â â 0 â a sin 9 sin ) a sin 9 cos ) â r ‚r œ aa# sin# 9 cos )b i aa# sin# 9 sin )b j aa# sin 9 cos 9b k Ê F † n d5 œ F † kr99 ‚r)) k kr9 ‚ r) k d) d9 œ aa$ sin$ 9 cos# 9 a$ sin$ 9 sin# ) a$ sin 9 cos# 9b d) d9 œ a$ sin 9 d) d9 since F œ xi yj zk œ (a sin 9 cos ))i (a sin 9 sin ))j (a cos 9)k Ê ' ' F † n d5 œ '0 '0 a$ sin 9 d9 d) œ 41a$ 21 1 S 23. Let the parametrization be r(xß y) œ xi yj (2a x y)k , 0 Ÿ x Ÿ a, 0 Ÿ y Ÿ a Ê rx œ i k and ry œ j k â â âi j k â â â r ‚r Ê rx ‚ ry œ â 1 0 1 â œ i j k Ê F † n d5 œ F † krxx ‚ryyk krx ‚ ry k dy dx â â â 0 1 1 â œ [2xy 2y(2a x y) 2x(2a x y)] dy dx since F œ 2xyi 2yzj 2xzk œ 2xyi 2y(2a x y)j 2x(2a x y)k Ê ' ' F † n d5 S a a a a œ '0 '0 [2xy 2y(2a x y) 2x(2a x y)] dy dx œ '0 '0 a4ay 2y# 4ax 2x# 2xyb dy dx a œ '0 ˆ 43 a$ 3a# x 2ax# ‰ dx œ ˆ 43 3# 23 ‰ a% œ 13a 6 % 24. Let the parametrization be r()ß z) œ (cos ))i (sin ))j zk , 0 Ÿ z Ÿ a, 0 Ÿ ) Ÿ 21 (where r œ Èx# y# œ 1 on â â j kâ â i â â the cylinder) Ê r) œ ( sin ))i (cos ))j and rz œ k Ê r) ‚ rz œ â sin ) cos ) 0 â œ (cos ))i (sin ))j â â 0 1â â 0 rz # # Ê F † n d5 œ F † krr)) ‚ ‚rz k kr) ‚ rz k dz d) œ acos ) sin )b dz d) œ dz d), since F œ (cos ))i (sin ))j zk Ê ' ' F † n d5 œ '0 21 S '0a 1 dz d) œ 21a 25. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j rk , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â Ê rr œ (cos ))i (sin ))j k and r) œ (r sin ))i (r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â â â sin ) 1 â â cos ) rr $ # # œ (r cos ))i (r sin ))j rk Ê F † n d5 œ F † krr)) ‚ ‚rr k kr) ‚ rr k d) dr œ ar sin ) cos ) r b d) dr since F œ ar# sin ) cos )b i rk Ê ' ' F † n d5 œ '0 '0 ar$ sin ) cos# ) r# b dr d) œ '0 ˆ "4 sin ) cos# ) "3 ‰ d) 21 #1 " œ 12 cos$ ) 3) ‘ ! œ 231 21 1 S 26. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j 2rk , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 2) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â Ê rr œ (cos ))i (sin ))j 2k and r) œ (r sin ))i (r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â â â sin ) 2 â â cos ) rr œ (2r cos ))i (2r sin ))j rk Ê F † n d5 œ F † krr)) ‚ ‚rr k kr) ‚ rr k d) dr œ a2r$ sin# ) cos ) 4r$ cos ) sin ) rb d) dr since F œ ar# sin# )b i a2r# cos )b j k Ê ' ' F † n d5 œ '0 '0 a2r$ sin# ) cos ) 4r$ cos ) sin ) rb dr d) 21 1 S 21 #1 œ '0 ˆ "2 sin# ) cos ) cos ) sin ) "2 ‰ d) œ "6 sin$ ) 12 sin# ) "# )‘ ! œ 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.6 Surface Integrals 27. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j rk , 1 Ÿ r Ÿ 2 (since 1 Ÿ z Ÿ 2) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â Ê rr œ (cos ))i (sin ))j k and r) œ (r sin ))i (r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â â â sin ) 1 â â cos ) rr # # # # $ œ (r cos ))i (r sin ))j rk Ê F † n d5 œ F † krr)) ‚ ‚rr k kr) ‚ rr k d) dr œ ar cos ) r sin ) r b d) dr œ ar# r$ b d) dr since F œ (r cos ))i (r sin ))j r# k Ê ' ' F † n d5 œ '0 '1 ar# r$ b dr d) œ 7361 21 2 S 28. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j r# k , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â Ê rr œ (cos ))i (sin ))j 2rk and r) œ (r sin ))i (r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â â â sin ) 2r â â cos ) rr $ # $ # œ a2r# cos )b i a2r# sin )b j rk Ê F † n d5 œ F † krr)) ‚ ‚rr k kr) ‚ rr k d) dr œ a8r cos ) 8r sin ) 2rb d) dr œ a8r$ 2rb d) dr since F œ (4r cos ))i (4r sin ))j 2k Ê ' ' F † n d5 œ '0 '0 a8r$ 2rb dr d) œ 21 21 1 S 29. g(xß yß z) œ z, p œ k Ê ™ g œ k Ê k ™ gk œ 1 and k ™ g † pk œ 1 Ê Flux œ ' ' F † n d5 œ ' ' (F † k) dA R S œ '0 '0 3 dy dx œ 18 2 3 30. g(xß yß z) œ y, p œ j Ê ™ g œ j Ê k ™ gk œ 1 and k ™ g † pk œ 1 Ê Flux œ ' ' F † n d5 œ ' ' (F † j) dA R S 2 7 2 œ 'c1 '2 2 dz dx œ 'c1 2(7 2) dx œ 10(2 1) œ 30 2yj 2zk 31. ™ g œ 2xi 2yj 2zk Ê k ™ gk œ È4x# 4y# 4z# œ 2a; n œ #2xÈi œ xi yaj zk Ê F † n œ za ; x # y# z# # ' ' Š za ‹ ˆ az ‰ dA œ ' ' z dA œ ' ' Èa# ax# y# b dx dy k ™ g † kk œ 2z Ê d5 œ 2a 2z dA Ê Flux œ # œ '0 1Î2 '0 Èa# r# r dr d) œ 1a6 a R R R $ 2yj 2zk 32. ™ g œ 2xi 2yj 2zk Ê k ™ gk œ È4x# 4y# 4z# œ 2a; n œ #2xÈi œ xi yaj zk Ê F † n œ axy xy a x # y# z# ' ' F † n d5 œ ' ' 0 d 5 œ 0 œ 0; k ™ g † kk œ 2z Ê d5 œ 2a 2z dA Ê Flux œ S S xy z z ' ' ˆ az ‰ ˆ za ‰ dA 33. From Exercise 31, n œ xi yaj zk and d5 œ za dA Ê F † n œ xy a a a œ a Ê Flux œ œ ' ' 1 dA œ 1a4 R # R # # # # 34. From Exercise 31, n œ xi yaj zk and d5 œ az dA Ê F † n œ zxa zya za œ z Š x ya z ‹ œ az # $ Ê Flux œ ' ' (za) ˆ az ‰ dx dy œ ' ' a# dx dy œ a# (Area of R) œ 4" 1a% R R # 35. From Exercise 31, n œ xi yaj zk and d5 œ za dA Ê F † n œ xa ya za œ a Ê Flux # œ ' ' a ˆ az ‰ dA œ ' ' R R a# z dA œ ' ' È # a# # œ '0 a# ’Èa# r# “ d) œ 1a# 1Î2 a R a ax y# b dA œ '0 1Î2 '0a È a # a # r# # r dr d) $ ! Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 977 978 Chapter 16 Integration in Vector Fields y# # 36. From Exercise 31, n œ xi yaj zk and d5 œ az dA Ê F†nœ Ê Flux œ ' ' az dx dy œ ' ' Èa# aax# y# b dx dy œ '0 1Î2 R R # Š xa ‹ Œ a Š za ‹ È x # y # z# '0 È a a a # r# # Š aa ‹ œ a r dr d) œ 1#a œ1 # k 37. g(xß yß z) œ y# z œ 4 Ê ™ g œ 2yj k Ê k ™ gk œ È4y# 1 Ê n œ È2y4yj #1 2xy 3z Ê F†nœ È ; p œ k Ê k ™ g † pk œ 1 Ê d5 œ È4y# 1 dA Ê Flux 4y# 1 2xy 3z œ ' ' ŠÈ ‹ È4y# 1 dA œ ' ' (2xy 3z) dA; z œ 0 and z œ 4 y# Ê y# œ 4 4y# 1 R R Ê Flux œ ' ' c2xy 3 a4 y bd dA œ '0 'c2 a2xy 12 3y# b dy dx œ '0 cxy# 12y y$ d # dx 1 # 2 1 # R œ '0 32 dx œ 32 1 38. g(xß yß z) œ x# y# z œ 0 Ê ™ g œ 2xi 2yj k Ê k ™ gk œ È4x# 4y# 1 œ È4 ax# y# b 1 # # i 2yj k 8y 2 Ê n œ È2x Ê F † n œ È8x ; p œ k Ê k ™ g † pk œ 1 Ê d5 œ È4 ax# y# b 1 dA 4 ax# y# b 1 4 ax# y# b 1 8y 2 Ê Flux œ ' ' Š È8x ‹ È4 ax# y# b 1 dA œ ' ' a8x# 8y# 2b dA; z œ 1 and x# y# œ z 4 ax # y # b 1 # # R Ê x y œ 1 Ê Flux œ '0 '0 a8r 2b r dr d) œ 21 # 21 # 1 R # 39. g(xß yß z) œ y ex œ 0 Ê ™ g œ ex i j Ê k ™ gk œ Èe2x 1 Ê n œ Èe 2xi j x e 1 Ê k ™ g † pk œ ex Ê d5 œ Èe2x 1 dA ex Ê Flux œ ' ' Š È2e2x 2y ‹ Š x R œ ' ' 4 dA œ ' ' 4 dy dz œ 4 R 1 2 0 1 e 1 Èe2x 1 ‹ dA œ ex 40. g(xß yß z) œ y ln x œ 0 Ê ™ g œ "x i j Ê k ™ gk œ É x"# 1 œ Ê nœ Š "x i j‹ Œ Ê Flux œ ' ' Š È2xy ‹Š # 1 R x È 1 x# ‹ dA œ x x e 1 ' ' 2exex 2ex dA R È 1 x# since 1 Ÿ x Ÿ e x 2xy i x j È1 x# œ È1 x# Ê F † n œ È1 x# ; p œ j Ê k ™ g † pk œ 1 Ê d5 œ x Ê F † n œ È2e2x 2y ; p œ i È 1 x# dA x '01 '1e 2y dx dz œ '1e '01 2 ln x dz dx œ '1e 2 ln x dx œ 2 cx ln x xd " œ 2(e e) 2(0 1) œ 2 e 41. On the face z œ a: g(xß yß z) œ z Ê ™ g œ k Ê k ™ gk œ 1; n œ k Ê F † n œ 2xz œ 2ax since z œ a; d5 œ dx dy Ê Flux œ ' ' 2ax dx dy œ '0 '0 2ax dx dy œ a% . a a R On the face z œ 0: g(xß yß z) œ z Ê ™ g œ k Ê k ™ gk œ 1; n œ k Ê F † n œ 2xz œ 0 since z œ 0; d5 œ dx dy Ê Flux œ ' ' 0 dx dy œ 0. R On the face x œ a: g(xß yß z) œ x Ê ™ g œ i Ê k ™ gk œ 1; n œ i Ê F † n œ 2xy œ 2ay since x œ a; d5 œ dy dz Ê Flux œ '0 '0 2ay dy dz œ a% . a a On the face x œ 0: g(xß yß z) œ x Ê ™ g œ i Ê k ™ gk œ 1; n œ i Ê F † n œ 2xy œ 0 since x œ 0 Ê Flux œ 0. On the face y œ a: g(xß yß z) œ y Ê ™ g œ j Ê k ™ gk œ 1; n œ j Ê F † n œ 2yz œ 2az since y œ a; d5 œ dz dx Ê Flux œ '0 '0 2az dz dx œ a% . a a On the face y œ 0: g(xß yß z) œ y Ê ™ g œ j Ê k ™ gk œ 1; n œ j Ê F † n œ 2yz œ 0 since y œ 0 Ê Flux œ 0. Therefore, Total Flux œ 3a% . Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.6 Surface Integrals 979 42. Across the cap: g(xß yß z) œ x# y# z# œ 25 Ê ™ g œ 2x i 2yj 2zk Ê k ™ gk œ È4x# 4y# 4z# œ 10 # g x i y j zk Ê n œ k™ Ê F † n œ x5z y5z 5z ; p œ k Ê k ™ g † pk œ 2z since z ™gk œ 5 # 0 Ê d5 œ 10 2z dA Ê Fluxcap œ ' ' F † n d5 œ ' ' Š x5z y5z 5z ‹ ˆ 5z ‰ dA œ ' ' ax# y# 1b dx dy œ '0 '0 ar# 1b r dr d) cap 21 # # R 21 œ '0 72 d) œ 1441. 4 R Across the bottom: g(xß yß z) œ z œ 3 Ê ™ g œ k Ê k ™ gk œ 1 Ê n œ k Ê F † n œ 1; p œ k Ê k ™ g † pk œ 1 Ê d5 œ dA Ê Fluxbottom œ ' ' F † n d5 œ ' ' 1 dA œ 1(Area of the circular region) œ 161. Therefore, R bottom Flux œ Fluxcap Fluxbottom œ 1281 43. ™ f œ 2x i 2yj 2zk Ê k ™ f k œ È4x# 4y# 4z# œ 2a; p œ k Ê k ™ f † pk œ 2z since z 0 Ê d5 œ 2a 2z dA œ az dA; M œ ' ' $ d5 œ 8$ (surface area of sphere) œ $1#a ; Mxy œ ' ' z$ d5 œ $ ' ' z ˆ za ‰ dA œ a$ ' ' dA # œ a$ '0 1Î2 ' S $ $ M r dr d) œ $14a Ê z œ Mxy œ Š $14a ‹ ˆ $12a# ‰ œ #a . 0 R R S a Because of symmetry, x œ y œ 2a Ê the centroid is ˆ #a ß #a ß #a ‰ . 44. ™ f œ 2y j 2zk Ê k ™ f k œ È4y# 4z# œ È4 ay# z# b œ 6; p œ k Ê k ™ f † kk œ 2z since z 6 0 Ê d5 œ 2z dA œ 3z dA; M œ ' ' 1 d5 œ 'c3 '0 3z dx dy œ 'c3 '0 È93 y# dx dy œ 91; Mxy œ ' ' z d5 œ 'c3 '0 z ˆ 3z ‰ dx dy œ 54; 3 3 3 3 3 S Mxz œ ' ' y d5 œ ' ' y ˆ 3z ‰ dx dy œ ' ' S Therefore, x œ 3 3 3 c3 0 c3 Š 27 # 1‹ 91 3 S ' ' x d5 œ ' ' 3 3y dx dy œ 0; Myz œ 0 È 9 y# S 3 3 c3 0 3x 27 È9 y# dx dy œ # 1. œ 3# , y œ 0, and z œ 9541 œ 16 45. Because of symmetry, x œ y œ 0; M œ ' ' $ d5 œ $ ' ' d5 œ (Area of S)$ œ 31È2 $ ; ™ f œ 2x i 2yj 2zk S S 2È x y z Ê k ™ f k œ È4x# 4y# 4z# œ 2Èx# y# z# ; p œ k Ê k ™ f † pk œ 2z Ê d5 œ dA # œ È 2 È x# y# Èx# y# ax# y# b dA œ dA Ê Mxy œ $ z z È œ $ '0 '1 È2 r# dr d) œ 1413 2 $ Ê z œ 21 œ'' R 2 È 2 È x# y# ax# y# b Š ‹ $ dA œ $ È2 z È # ' ' z Š È2 Èzx# y# ‹ dA œ $ ' ' È2 Èx# y# dA 141 2 Œ 3 $ 31 È 2 $ # #z R R 14 ‰ ' ' ax# y# b $ d5 ˆ œ 14 9 Ê axß yß zb œ 0ß 0ß 9 . Next, Iz œ S ' ' ax# y# b dA œ $ È2 ' ' 21 0 R È r$ dr d) œ 151# 2 $ 1 2 Iz Ê Rz œ É M œ È10 # 46. f(xß yß z) œ 4x# 4y# z# œ 0 Ê ™ f œ 8xi 8yj 2zk Ê k ™ f k œ È64x# 64y# 4z# È œ 2È16x# 16y# z# œ 2È4z# z# œ 2È5 z since z 0; p œ k Ê k ™ f † pk œ 2z Ê d5 œ 2 2z5 z dA œ È5 dA Ê Iz œ ' ' ax# y# b $ d5 œ $ È5 ' ' ax# y# b dx dy œ $ È5 'c1Î2 '0 1Î2 2 cos ) R S È r$ dr d) œ 3 #51$ 47. (a) Let the diameter lie on the z-axis and let f(xß yß z) œ x# y# z# œ a# , z 0 be the upper hemisphere Ê ™ f œ 2xi 2yj 2zk Ê k ™ f k œ È4x# 4y# 4z# œ 2a, a 0; p œ k Ê k ™ f † pk œ 2z since z y Ê d5 œ az dA Ê Iz œ ' ' $ ax# y# b ˆ az ‰ d5 œ a$ ' ' Èa# x dA œ a$ '0 '0 È #r ax # y # b # S œ a$ '0 ’r# Èa# r# 23 aa# r# b 21 “ d) œ a$ '0 $Î# a ! # R 21 2 $ 41 % 3 a d) œ 3 a $ 21 a # a r# r dr d) Ê the moment of inertia is 831 a% $ for the whole sphere Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 0 980 Chapter 16 Integration in Vector Fields (b) IL œ Ic.m. mh# , where m is the mass of the body and h is the distance between the parallel lines; now, Ic.m. œ 831 a% $ (from part a) and m# œ ' ' $ d5 œ $ ' ' ˆ az ‰ dA œ a$ ' ' È # " # # dy dx œ a$ '0 '0 È "# 21 a r# R S a ax y b R r dr d) œ a$ '0 ’Èa# r# “ d) œ a$ '0 a d) œ 21a# $ and h œ a 21 a 21 a ! Ê IL œ 831 a% $ 41a# $ a# œ 2031 a% $ 48. Let z œ ha Èx# y# be the cone from z œ 0 to z œ h, h 0. Because of symmetry, x œ 0 and y œ 0; # # # 2yh z œ ha Èx# y# Ê f(xß yß z) œ ha# ax# y# b z# œ 0 Ê ™ f œ 2xh a# i a# j 2zk # % # % % # # # Ê k ™ f k œ É 4xa%h 4ya%h 4z# œ 2É ha% ax# y# b ha# ax# y# b œ 2Éˆ ha# ‰ ax# y# b ˆ ha# 1‰ # # œ 2Éz# ˆ h a# a ‰ œ ˆ 2za ‰ Èh# a# since z œ È h # a# dA; M œ a 0; p œ k Ê k ™ f † pk œ 2z Ê d5 œ ˆ 2z ‰ È h # a# a dA 2z ' ' d5 œ ' ' Èh#a a# dA œ Èh#a a# a1a# b œ 1aÈh# a# ; R S # # # È # È # È # Mxy œ ' ' z d5 œ ' ' z Š h a a ‹ dA œ h a a ' ' ha Èx# y# dx dy œ h ha# a '0 '0 r# dr d) 21 R S # È # œ 21ah 3h a R M z œ Mxy œ 2h 3 Ê a ‰ Ê the centroid is ˆ0ß 0ß 2h 3 16.7 STOKES' THEOREM â i â â 1. curl F œ ™ ‚ F œ â ``x â # âx j ` `y 2x k ââ ` â ` z â œ 0i 0j (2 0)k œ 2k and n œ k Ê curl F † n œ 2 Ê d5 œ dx dy â z# â Ê )C F † dr œ ' ' 2 dA œ 2(Area of the ellipse) œ 41 R â i â â 2. curl F œ ™ ‚ F œ â ``x â â 2y j ` `y 3x k ââ ` â ` z â œ 0i 0j (3 2)k œ k and n œ k Ê curl F † n œ 1 Ê d5 œ dx dy â z# â Ê )C F † dr œ ' ' dx dy œ Area of circle œ 91 R â i â â 3. curl F œ ™ ‚ F œ â ``x â â y j ` `y xz k ââ ` â ijk Ê curl F † n ` z â œ xi 2xj (z 1)k and n œ È3 â #â x œ È"3 (x 2x z 1) Ê d5 œ 1cx œ '0 '0 1 È3 1 dA Ê )C F † dr œ ' ' 1cx [3x (1 x y) 1] dy dx œ '0 '0 1 œ '0 ˆ "# 3x #7 x# ‰ dx œ 56 R " È3 dA È3 (3x z 1) (4x y) dy dx œ '0 4x(1 x) "# (1 x)# ‘ dx 1 1 â i â â 4. curl F œ ™ ‚ F œ â ``x â # â y z# j ` `y x# z# â â â â œ (2y 2z)i (2z 2x)j (2x 2y)k and n œ i Èj 3 k â x# y# â k ` `z Ê curl F † n œ È"3 (2y 2z 2z 2x 2x 2y) œ 0 Ê )C F † dr œ ' ' 0 d5 œ 0 S Copyright © 2010 Pearson Education Inc. 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Section 16.7 Stoke's Theorem â i â â 5. curl F œ ™ ‚ F œ â ``x â # â y z# j 981 â â â â œ 2yi (2z 2x)j (2x 2y)k and n œ k â # #â x y k ` `y x# y# ` `z Ê curl F † n œ 2x 2y Ê d5 œ dx dy Ê )C F † dr œ 'c1 'c1 (2x 2y) dx dy œ 'c1 cx# 2xyd " dy 1 1 1 " œ 'c1 4y dy œ 0 1 â i â â 6. curl F œ ™ ‚ F œ â ``x â # $ âx y k ââ ` â 2xi 2yj 2zk xi yj zk # # ` z â œ 0i 0j 3x y k and n œ 2Èx# y# z# œ 4 â z â j ` `y 1 Ê curl F † n œ 34 x# y# z; d5 œ 4z dA (Section 16.6, Example 6, with a œ 4) œ 3 '0 '0 ar cos )b ar sin )b r dr d) œ 3' 21 2 # # # Ê )C F † dr œ ' ' ˆ 34 x# y# z‰ ˆ 4z ‰ dA ' # ’ r6 “ (cos ) sin ))# d) œ 32 0 ! # %1 21 '0 21 R " # 4 sin 2) d) œ 4 '041 sin# u du œ 4 u2 sin42u ‘ ! œ 81 7. x œ 3 cos t and y œ 2 sin t Ê F œ (2 sin t)i a9 cos# tb j a9 cos# t 16 sin% tb sin eÈÐ6 sin t cos tÑÐ0Ñ k at the base of the shell; r œ (3 cos t)i (2 sin t)j Ê dr œ (3 sin t)i (2 cos t)j Ê F † ddtr œ 6 sin# t 18 cos$ t Ê ' ' ™ ‚ F † n d5 œ '0 a6 sin# t 18 cos$ tb dt œ 3t 32 sin 2t 6(sin t) acos# t 2b‘ ! œ 61 21 #1 S â i â â ` 8. curl F œ ™ ‚ F œ ââ `x â z " â #x â â â ` ` â œ 2j ; f(xß yß z) œ 4x# y z# Ê ™ f œ 8xi j 2zk `y `z â tan" y x 4 " z ââ j k k™f k f " 2 Ê n œ k™ ™f k and p œ j Ê k ™ f † pk œ 1 Ê d5 œ k™f†pk dA œ k ™ f k dA; ™ ‚ F † n œ k™f k (2j † ™ f) œ k™f k Ê ™ ‚ F † n d5 œ 2 dA Ê ' ' ™ ‚ F † n d5 œ ' ' 2 dA œ 2(Area of R) œ 2(1 † 1 † 2) œ 41, where R R S is the elliptic region in the xz-plane enclosed by 4x# z# œ 4. 9. Flux of ™ ‚ F œ ' ' ™ ‚ F † n d5 œ )C F † dr, so let C be parametrized by r œ (a cos t)i (a sin t)j , S 0 Ÿ t Ÿ 21 Ê ddtr œ (a sin t)i (a cos t)j Ê F † ddtr œ ay sin t ax cos t œ a# sin# t a# cos# t œ a# Ê Flux of ™ ‚ F œ )C F † dr œ '0 a# dt œ 21a# 21 â i â â 10. ™ ‚ (yi) œ â ``x â â y j ` `y 0 k ââ ™f 2xi 2yj 2zk ` â ` z â œ k ; n œ k™f k œ 2Èx# y# z# œ xi yj zk â 0 â Ê ™ ‚ (yi) † n œ z; d5 œ "z dA (Section 16.6, Example 6, with a œ 1) Ê ' ' ™ ‚ (yi) † n d5 S œ ' ' (z) ˆ "z dA‰ œ ' ' dA œ 1, where R is the disk x# y# Ÿ 1 in the xy-plane. R R 11. Let S" and S# be oriented surfaces that span C and that induce the same positive direction on C. Then ' ' ™ ‚ F † n" d5" œ ) F † dr œ ' ' ™ ‚ F † n# d5# S" C S# Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 982 Chapter 16 Integration in Vector Fields 12. ' ' ™ ‚ F † n d5 œ ' ' ™ ‚ F † n d5 ' ' ™ ‚ F † n d5, and since S" and S# are joined by the simple S S" S# closed curve C, each of the above integrals will be equal to a circulation integral on C. But for one surface the circulation will be counterclockwise, and for the other surface the circulation will be clockwise. Since the integrands are the same, the sum will be 0 Ê ' ' ™ ‚ F † n d5 œ 0. S â i j k ââ â â ` â ` 13. ™ ‚ F œ â ` x ` y ``z â œ 5i 2j 3k ; rr œ (cos ))i (sin ))j 2rk and r) œ (r sin ))i (r cos ))j â â â 2z 3x 5y â â â i j k â â â â r) sin ) 2r â œ a2r# cos )b i a2r# sin )b j rk ; n œ krrrr ‚ Ê rr ‚ r) œ â cos ) ‚r) k and d5 œ krr ‚ r) k dr d) â â 0 â â r sin ) r cos ) Ê ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) œ a10r# cos ) 4r# sin ) 3rb dr d) Ê ' ' ™ ‚ F † n d5 œ '0 '0 a10r cos ) 4r sin ) 3rb dr d) œ ' 21 2 # S 21 "30 r$ cos ) 43 r$ sin ) 3# r# ‘ # d) ! 0 # 32 ‰ œ '0 ˆ 80 3 cos ) 3 sin ) 6 d) œ 6(21) œ 121 21 â i j â â ` 14. ™ ‚ F œ â ``x `y â ây z z x k ââ â ` # # ` z â œ i 2j 2k ; rr ‚ r) œ a2r cos )b i a2r sin )b j rk and â x zâ ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) (see Exercise 13 above) Ê ' ' ™ ‚ F † n d5 œ '0 '0 a2r cos ) 4r sin ) 2rb dr d) œ ' 21 3 # S 21 23 r$ cos ) 43 r$ sin ) r# ‘ $ d) ! 0 # œ '0 a18 cos ) 36 sin ) 9b d) œ 9(21) œ 181 21 â i â â j k ââ i j kâ â â â ` â â ` ` â $ # sin ) 1 â 15. ™ ‚ F œ â ` x `y ` z â œ 2y i 0j x k ; rr ‚ r) œ â cos ) â # â â â â x y 2y$ z 3z â â r sin ) r cos ) 0 â œ (r cos ))i (r sin ))j rk and ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) (see Exercise 13 above) Ê ' ' ™ ‚ F † n d5 œ ' ' a2ry$ cos ) rx# b dr d) œ '0 '0 a2r% sin3 ) cos ) r$ cos# )b dr d) 21 œ' 1 R S 21 #1 " ˆ 25 sin3 ) cos ) 4" cos# )‰ d) œ 10 sin4 ) 4" ˆ #) sin42) ‰‘ ! œ 14 0 â i â â 16. ™ ‚ F œ â ``x â âx y â â k ââ i j k â â â â â ` sin ) 1 â ` z â œ i j k ; rr ‚ r) œ â cos ) â â â â r sin ) r cos ) 0 â y z z xâ œ (r cos ))i (r sin ))j rk and ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) (see Exercise 13 above) j ` `y ‰ Ê ' ' ™ ‚ F † n d5 œ '0 '0 (r cos ) r sin ) r) dr d) œ '0 ’(cos ) sin ) 1) r# “ d) œ ˆ 25 # (21) œ 251 21 21 5 # & ! S â i â â i j k j k ââ â â â â ` â âÈ â ` ` È È 3 cos cos 3 cos sin 3 sin 9 ) 9 ) 9 i j k r r œ ‚ œ 17. ™ ‚ F œ â ` x 0 0 5 ; â â â 9 ) `y `z â â â â # È È â 3y 5 2x z 2 â â 3 sin 9 sin ) â 3 sin 9 cos ) 0 # # œ a3 sin 9 cos )b i a3 sin 9 sin )b j (3 sin 9 cos 9)k ; ™ ‚ F † n d5 œ ( ™ ‚ F) † (r9 ‚ r) ) d9 d) (see Exercise # ‘ 13 above) Ê ' ' ™ ‚ F † n d5 œ '0 '0 15 cos 9 sin 9 d9 d) œ '0 15 2 cos 9 ! 21 S 1/2 21 1Î# d) œ '0 15 # d) œ 151 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 21 Section 16.7 Stoke's Theorem 983 â â â j k ââ i j k â i â â â ` â â ` ` â 18. ™ ‚ F œ â ` x ` y ` z â œ 2zi j 2yk ; r9 ‚ r) œ â 2 cos 9 cos ) 2 cos 9 sin ) 2 sin 9 â â # â â â 0 ây â 2 sin 9 sin ) 2 sin 9 cos ) â z# x â # # œ a4 sin 9 cos )b i a4 sin 9 sin )b j (4 sin 9 cos 9)k ; ™ ‚ F † n d5 œ ( ™ ‚ F) † (r9 ‚ r) ) d9 d) (see Exercise 13 above) Ê ' ' ™ ‚ F † n d5 œ ' ' a8z sin# 9 cos ) 4 sin# 9 sin ) 8y sin 9 cos )b d9 d) R S 21 1/2 œ '0 '0 a16 sin# 9 cos 9 cos ) 4 sin# 9 sin ) 16 sin# 9 sin ) cos )b d9 d) 21 1Î# œ '0 "36 sin$ 9 cos ) 4 ˆ 9# sin429 ‰ (sin )) 16 ˆ 9# sin429 ‰ (sin ) cos ))‘ ! d) 21 # ‘ #1 ‰ "6 œ '0 ˆ 16 3 cos ) 1 sin ) 41 sin ) cos ) d) œ 3 sin ) 1 cos ) 21 sin ) ! œ 0 19. (a) F œ 2xi 2yj 2zk Ê curl F œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 œ 0 S S (b) Let f(xß yß z) œ x y z Ê ™ ‚ F œ ™ ‚ ™ f œ 0 Ê curl F œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 # # $ S S œ0 (c) F œ ™ ‚ (xi yj zk) œ 0 Ê ™ ‚ F œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 œ 0 S S (d) F œ ™ f Ê ™ ‚ F œ ™ ‚ ™ f œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 œ 0 S 20. F œ ™ f œ "# ax# y# z# b $Î# # $Î# (2x)i "# ax# y# z# b $Î# # $Î# S (2y)j "# ax# y# z# b $Î# (2z)k # $Î# i y ax# y# z b j z ax# y# z b k œ x ax# y# z b dr (a) r œ (a cos t)i (a sin t)j , 0 Ÿ t Ÿ 21 Ê dt œ (a sin t)i (a cos t)j Ê F † ddtr œ x ax# y# z# b $Î# (a sin t) y ax# y# z# b $Î# (a cos t) t‰ t‰ œ ˆ a cos (a sin t) ˆ a sin (a cos t) œ 0 Ê )C F † dr œ 0 a$ a$ (b) )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' ™ ‚ ™ f † n d5 œ ' ' 0 † n d5 œ ' ' 0 d5 œ 0 S S S â i â â 21. Let F œ 2yi 3zj xk Ê ™ ‚ F œ â ``x â â 2y j ` `y 3z S k ââ 2i 2j k ` â ` z â œ 3i j 2k ; n œ 3 â x â Ê ™ ‚ F † n œ 2 Ê )C 2y dx 3z dy x dz œ )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 2 d5 S S œ 2 ' ' d5, where ' ' d5 is the area of the region enclosed by C on the plane S: 2x 2y z œ 2 S S â i â â 22. ™ ‚ F œ â ``x â â x j ` `y y k ââ ` â `z â œ 0 â z â 23. Suppose F œ Mi Nj Pk exists such that ™ ‚ F œ Š `` Py ``Nz ‹ i ˆ ``Mz `` Px ‰ j Š ``Nx ``My ‹ k # # œ xi yj zk . Then ``x Š `` Py ``Nz ‹ œ ``x (x) Ê ``x`Py ``x`Nz œ 1. Likewise, ``y ˆ ``Mz `` Px ‰ œ ``y (y) # # # # Ê ``y`Mz ``y`Px œ 1 and ``z Š ``Nx ``My ‹ œ ``z (z) Ê ``z`Nx ``z`My œ 1. Summing the calculated equations # # # # # # Ê Š ``x`Py ``y`Px ‹ Š ``z`Nx ``x`Nz ‹ Š ``y`Mz ``z`My ‹ œ 3 or 0 œ 3 (assuming the second mixed partials are equal). This result is a contradiction, so there is no field F such that curl F œ xi yj zk . Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 984 Chapter 16 Integration in Vector Fields 24. Yes: If ™ ‚ F œ 0 , then the circulation of F around the boundary C of any oriented surface S in the domain of F is zero. The reason is this: By Stokes's theorem, circulation œ )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 † n d5 œ 0. S S # 25. r œ Èx# y# Ê r% œ ax# y# b Ê F œ ™ ar% b œ 4x ax# y# b i 4y ax# y# b j œ Mi Nj Ê )C ™ ar% b † n ds œ )C F † n ds œ )C M dy N dx œ ' ' Š ``Mx ``Ny ‹ dx dy R œ ' ' c4 ax# y# b 8x# 4 ax# y# b 8y# d dA œ ' ' 16 ax# y# b dA œ 16 ' ' x# dA 16 ' ' y# dA R R R R œ 16Iy 16Ix . # # # # # # # # 26. `` Py œ 0, ``Nz œ 0, ``Mz œ 0, `` Px œ 0, ``Nx œ axy# yx# b# , ``My œ axy# yx# b# Ê curl F œ ’ axy# yx# b# axy# yx# b# “ k œ 0 . However, x# y# œ 1 Ê r œ (cos t)i (sin t)j Ê ddtr œ (sin t)i (cos t)j #1 Ê F œ a sin tb i acos tb j Ê F † ddtr œ sin# t cos# t œ 1 Ê )C F † dr œ ) 1 dt œ 21 which is not zero. ! 16.8 THE DIVERGENCE THEOREM AND A UNIFIED THEORY y i x j 1. F œ È Ê div F œ x# y# xy xy œ0 ax# y# b$Î# 2. F œ xi yj Ê div F œ 1 1 œ 2 # # $Î# # # # # "Î# # ax y z b i yj zk) 3. F œ GM(x Ê div F œ GM ” ax y z abx# y3x • # z # b$ ax# y# z# b$Î# # # $Î# # # # # "Î# # # # $Î# # # # # # "Î# ax y z b ax y z b 3z ax y z b GM ” ax y z abx# y3y • GM ” • # z # b$ ax# y# z# b$ # # # # # # # # # # z b ax y z b œ GM ’ 3 ax y z b # 3 ax# y #(Î# “œ0 ax y z b 4. z œ a# r# in cylindrical coordinates Ê z œ a# ax# y# b Ê v œ aa# x# y# b k Ê div v œ 0 5. ` ` ` ` x (y x) œ 1, ` y (z y) œ 1, ` z (y x) œ 0 Ê 6. ` ` ` # # # ` x ax b œ 2x, ` y ay b œ 2y, ` x az b œ 2z ™ † F œ 2 Ê Flux œ 'c1 'c1 'c1 2 dx dy dz œ 2 a2$ b œ 16 1 1 1 Ê ™ † F œ 2x 2y 2z (a) Flux œ '0 '0 '0 (2x 2y 2z) dx dy dz œ '0 '0 cx# 2x(y z)d ! dy dz œ '0 '0 (1 2y 2z) dy dz 1 1 1 1 1 1 " 1 œ '0 cy(1 2z) y# d ! dz œ '0 (2 2z) dz œ c2z z# d ! œ 3 1 1 " " (b) Flux œ 'c1 'c1 'c1 (2x 2y 2z) dx dy dz œ 'c1 'c1 cx# 2x(y z)d " dy dz œ 'c1 'c1 (4y 4z) dy dz 1 1 1 1 1 1 " 1 œ 'c1 c2y# 4yzd " dz œ 'c1 8z dz œ c4z# d " œ 0 1 " 1 " (c) In cylindrical coordinates, Flux œ ' ' ' (2x 2y 2z) dx dy dz D # œ '0 '0 '0 (2r cos ) 2r sin ) 2z) r dr d) dz œ '0 '0 23 r$ cos ) 23 r$ sin ) zr# ‘ ! d) dz 1 21 2 1 21 16 ' "36 sin ) 163 cos ) 4z)‘ #!1 dz œ '0 81z dz œ c41z# d "! œ 41 ‰ œ '0 '0 ˆ 16 3 cos ) 3 sin ) 4z d) dz œ 0 1 7. 21 1 ` ` ` ` x (y) œ 0, ` y (xy) œ x, ` z (z) œ 1 1 Ê ™ † F œ x 1; z œ x# y# Ê z œ r# in cylindrical coordinates Ê Flux œ ' ' ' (x 1) dz dy dx œ '0 '0 '0 (r cos ) 1) dz r dr d) œ '0 '0 ar$ cos ) r# b r dr d) 21 œ' D # r& r% cos ) d) œ ’ “ 5 4 0 ! 21 2 r# 21 2 '021 ˆ 325 cos ) 4‰ d) œ 325 sin ) 4)‘ #!1 œ 81 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.8 The Divergence Theorem and a Unified Theory 8. 985 Ê ™ † F œ 2x 3 Ê Flux œ ' ' ' (2x 3) dV ` ` ` # ` x ax b œ 2x, ` y (xz) œ 0, ` z (3z) œ 3 D # 21 2 21 1 1 œ '0 '0 '0 (23 sin 9 cos ) 3) a3# sin 9b d3 d9 d) œ '0 '0 ’ 32 sin 9 cos ) 3$ “ sin 9 d9 d) % ! 21 21 21 1 1 œ '0 '0 (8 sin 9 cos ) 8) sin 9 d9 d) œ '0 8 ˆ 92 sin429 ‰ cos ) 8 cos 9‘ ! d) œ '0 (41 cos ) 16) d) œ 321 9. Ê Flux œ ' ' ' 3x dx dy dz ` ` ` # ` x ax b œ 2x, ` y (2xy) œ 2x, ` z (3xz) œ 3x D 2 1Î2 1 Î2 1 Î2 1 Î2 1 Î2 œ '0 '0 '0 (33 sin 9 cos )) a3# sin 9b d3 d9 d) œ '0 '0 12 sin# 9 cos ) d9 d) œ '0 31 cos ) d) œ 31 10. ``x a6x# 2xyb œ 12x 2y, ``y a2y x# zb œ 2, ``z a4x# y$ b œ 0 Ê ™ † F œ 12x 2y 2 Ê Flux œ ' ' ' (12x 2y 2) dV œ '0 '0 3 D 1Î2 '02 (12r cos ) 2r sin ) 2) r dr d) dz 16 ‰ ' ˆ ‰ œ '0 '0 ˆ32 cos ) 16 3 sin ) 4 d) dz œ 0 32 21 3 dz œ 112 61 3 1Î2 3 11. ``x (2xz) œ 2z, ``y (xy) œ x, ``z az# b œ 2z Ê ™ † F œ x Ê Flux œ ' ' ' x dV È16 c 4x œ '0 '0 2 # 4cy '0 È16 c 4x x dz dy dx œ '0 '0 2 D # (xy 4x) dy dx œ ' 2 ’ 12 x a16 4x# b 4xÈ16 4x# “ dx 0 $Î# # œ ’4x# "2 x% "3 a16 4x# b “ œ 40 3 ! 12. ``x ax$ b œ 3x# , ``y ay$ b œ 3y# , ``z az$ b œ 3z# Ê ™ † F œ 3x# 3y# 3z# Ê Flux œ ' ' ' 3 ax# y# z# b dV 21 a 21 21 1 1 œ 3 '0 '0 '0 3# a3# sin 9b d3 d9 d) œ 3 '0 '0 a5 sin 9 d9 d) œ 3 '0 2a5 d) œ 1251a & & D & 13. Let 3 œ Èx# y# z# . Then `` 3x œ 3x , `` 3y œ 3y , `` 3z œ 3z Ê ``x (3x) œ Š `` 3x ‹ x 3 œ x3 3 , ``y (3y) œ Š `` 3y ‹ y 3 # # # # # œ y3 3, ``z (3z) œ Š `` 3z ‹ z 3 œ z3 3 Ê ™ † F œ x y3 z 33 œ 43, since 3 œ Èx# y# z# # È2 Ê Flux œ ' ' ' 43 dV œ '0 '0 '1 21 1 D (43) a3# sin 9b d3 d9 d) œ '0 '0 3 sin 9 d9 d) œ '0 6 d) œ 121 21 1 21 14. Let 3 œ Èx# y# z# . Then `` 3x œ 3x , `` 3y œ 3y , `` 3z œ 3z Ê ``x Š 3x ‹ œ 3" Š 3x# ‹ `` 3x œ 31 3x$ . Similarly, # y y# ` " ` z " z# ` y Š 3 ‹ œ 3 3$ and ` z Š 3 ‹ œ 3 3$ Ê Flux œ ' ' ' D 2 3 dV œ # # # Ê ™ † F œ 33 x 3y$ z œ 32 '021 '01 '12 Š 32 ‹ a3# sin 9b d3 d9 d) œ '021 '01 3 sin 9 d9 d) œ '021 6 d) œ 121 15. ``x a5x$ 12xy# b œ 15x# 12y# , ``y ay$ ey sin zb œ 3y# ey sin z, ``z a5z$ ey cos zb œ 15z# ey sin z È2 Ê ™ † F œ 15x# 15y# 15z# œ 153# Ê Flux œ ' ' ' 153# dV œ '0 '0 '1 21 1 a153# b a3# sin 9b d3 d9 d) D œ '0 '0 Š12È2 3‹ sin 9 d9 d) œ '0 Š24È2 6‹ d) œ Š48È2 12‹ 1 21 1 21 ` ˆ 2z " y ‰ ˆ 2z ‰ 16. ``x cln ax# y# bd œ x# 2x y# , ` y x tan x œ x – Š "x ‹ 1 ˆx‰ — y # ` ˆ È # # ‰ œ Èx# y# œ x# 2z y# , ` z z x y 2z Èx# y# Ê Flux œ ' ' ' Š # 2x # # 2z # Èx# y# ‹ dz dy dx Ê ™ † F œ x# 2x y# x# y# x y x y D Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 986 Chapter 16 Integration in Vector Fields È2 œ '0 '1 21 È ) 'c21 ˆ 2r cos ' 21 '1 2 ˆ6 cos ) 3r 3r# ‰ dr d) ‰ 2z r r r dz r dr d) œ 0 # # œ '0 ’6 ŠÈ2 1‹ cos ) 3 ln È2 2È2 1“ d) œ 21 Š 3# ln 2 2È2 1‹ 21 17. (a) G œ Mi Nj Pk Ê ™ ‚ G œ curl G œ Š `` Py ``Nz ‹ i ˆ ``Mz `` Px ‰ k Š ``Nx ``My ‹ k Ê ™ † ™ ‚ G œ div(curl G) œ ``x Š `` Py ``Nz ‹ ``y ˆ ``Mz `` Px ‰ ``z Š ``Nx ``My ‹ # # # # # # œ ``x`Py ``x`Nz ``y`Mz ``y`Px ``z`Nx ``z`My œ 0 if all first and second partial derivatives are continuous (b) By the Divergence Theorem, the outward flux of ™ ‚ G across a closed surface is zero because outward flux of ™ ‚ G œ ' ' ( ™ ‚ G) † n d5 œ ' ' ' ™ † ™ ‚ G dV S [Divergence Theorem with F œ ™ ‚ G] D œ ' ' ' (0) dV œ 0 [by part (a)] D 18. (a) Let F" œ M" i N" j P" k and F# œ M# i N# j P# k Ê aF" bF# œ (aM" bM# )i (aN" bN# )j (aP" bP# )k Ê ™ † (aF" bF# ) œ â ``Mx" b ``Mx# ‰ Ša ``Ny" b ``Ny# ‹ â ``Pz" b ``Pz# ‰ œ a Š ``Mx" ``Ny" ``Pz" ‹ b Š ``Mx# ``Ny# ``Pz# ‹ œ a( ™ † F" ) b( ™ † F# ) (b) Define F" and F# as in part a Ê ™ ‚ (aF" bF# ) œ ’Ša ``Py" b ``Py# ‹ â ``Nz" b ``Nz# ‰“ i â ``Mz" b ``Mz# ‰ â ``Px" b ``Px# ‰‘ j ’â ``Nx" b ``Nx# ‰ Ša ``My" b ``My# ‹“ k œ a ’Š ``Py" ``Nz" ‹ i ˆ ``Mz" ``Px" ‰ j Š ``Nx" ``My" ‹ k“ b ’Š ``Py# ``Nz# ‹ i ˆ ``Mz# ``Px# ‰ j Š ``Nx# ``My# ‹ k“ œ a ™ ‚ F" b ™ ‚ F# â â j kâ â i â â (c) F" ‚ F# œ â M" N" P" â œ (N" P# P" N# )i (M" P# P" M# )j (M" N# N" M# )k Ê ™ † (F" ‚ F# ) â â â M# N# P# â œ ™ † [(N" P# P" N# )i (M" P# P" M# )j (M" N# N" M# )k] œ ``x (N" P# P" N# ) ``y (M" P# P" M# ) ``z (M" N# N" M# ) œ ˆP# ``Nx" N" ``Px# N# ``Px" P" ``Nx# ‰ ŠM" ``Py# P# ``My" P" ``My# M# ``Py" ‹ ˆM" ``Nz# N# ``Mz" N" ``Mz# M# ``Nz" ‰ œ M# Š ``Py" ``Nz" ‹ N# ˆ ``Mz" ``Px" ‰ P# Š ``Nx" ``My" ‹ M" Š ``Nz# ``Py# ‹ N" ˆ ``Px# ``Mz# ‰ P" Š ``My# ``Nx# ‹ œ F# † ™ ‚ F" F" † ™ ‚ F# 19. (a) div(gF) œ ™ † gF œ ``x (gM) ``y (gN) ``z (gP) œ Šg ``Mx M `` xg ‹ Šg ``Ny N `` yg ‹ Šg ``Pz P `` gz ‹ œ ŠM `` gx N `` gy P `` gz ‹ g Š ``Mx ``Ny ``Pz ‹ œ g ™ † F ™ g † F (b) ™ ‚ (gF) œ ’ ``y (gP) ``z (gN)“ i ``z (gM) ``x (gP)‘ j ’ ``x (gN) ``y (gM)“ k œ ŠP `` gy g `` Py N `` gz g ``Nz ‹ i ŠM `` gz g ``Mz P `` xg g `` Px ‹ j ŠN `` xg g ``Nx M `` yg g ``My ‹ k œ ŠP `` yg N `` gz ‹ i Šg `` Py g ``Nz ‹ i ŠM `` gz P `` xg ‹ j ˆg ``Mz g `` Px ‰ j ŠN `` xg M `` yg ‹ k Šg ``Nx g ``My ‹ k œ g ™ ‚ F ™ g ‚ F Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Section 16.8 The Divergence Theorem and a Unified Theory 987 20. Let F" œ M" i N" j P" k and F# œ M# i N# j P# k . (a) F" ‚ F# œ (N" P# P" N# )i (P" M# M" P# )j (M" N# N" M# )k Ê ™ ‚ (F" ‚ F# ) œ ’ ``y (M" N# N" M# ) ``z (P" M# M" P# )“ i ``z (N" P# P" N# ) ``x (M" N# N" M# )‘ j ’ ``x (P" M# M" P# ) ``y (N" P# P" N# )“ k and consider the i-component only: ``y (M" N# N" M# ) ``z (P" M# M" P# ) œ N# ``My" M" ``Ny# M# ``Ny" N" ``My# M# ``Pz" P" ``Mz# P# ``Mz" M" ``Pz# œ ŠN# ``My" P# ``Mz" ‹ ŠN" ``My# P" ``Mz# ‹ Š ``Ny# ``Pz# ‹ M" Š ``Ny" ``Pz" ‹ M# œ ŠM# ``Mx" N# ``My" P# ``Mz" ‹ ŠM" ``Mx# N" ``My# P" ``Mz# ‹ Š ``Mx# ``Ny# ``Pz# ‹ M" Š ``Mx" ``Ny" ``Pz" ‹ M# . Now, i-comp of (F# † ™ )F" œ ŠM# ``x N# ``y P# ``z ‹ M" œ ŠM# ``Mx" N# ``My" P# ``Mz" ‹ ; likewise, i-comp of (F" † ™ )F# œ ŠM" ``Mx# N" ``My# P" ``Mz# ‹ ; i-comp of ( ™ † F# )F" œ Š ``Mx# ``Ny# ``Pz# ‹ M" and i-comp of ( ™ † F" )F# œ Š ``Mx" ``Ny" ``Pz" ‹ M# . Similar results hold for the j and k components of ™ ‚ (F" ‚ F# ). In summary, since the corresponding components are equal, we have the result ™ ‚ ( F " ‚ F # ) œ ( F # † ™ ) F " ( F " † ™ ) F # ( ™ † F # ) F " ( ™ † F" ) F# (b) Here again we consider only the i-component of each expression. Thus, the i-comp of ™ (F" † F# ) œ ``x (M" M# N" N# P" P# ) œ ˆM" ``Mx# M# ``Mx" N" ``Nx# N# ``Nx" P" ``Px# P# ``Px" ‰ i-comp of (F" † ™ )F# œ ŠM" ``Mx# N" ``My# P" ``Mz# ‹ , i-comp of (F# † ™ )F" œ ŠM# ``Mx" N# ``My" P# ``Mz" ‹ , i-comp of F" ‚ ( ™ ‚ F# ) œ N" Š ``Nx# ``My# ‹ P" ˆ ``Mz# ``Px# ‰ , and i-comp of F# ‚ ( ™ ‚ F" ) œ N# Š ``Nx" ``My" ‹ P# ˆ ``Mz" ``Px" ‰ . Since corresponding components are equal, we see that ™ (F" † F# ) œ (F" † ™ )F# (F# † ™ )F" F" ‚ ( ™ ‚ F# ) F# ‚ ( ™ ‚ F" ), as claimed. 21. The integral's value never exceeds the surface area of S. Since kFk Ÿ 1, we have kF † nk œ kFk knk Ÿ (1)(1) œ 1 and ' ' ' ™ † F d 5 œ ' ' F † n d5 D [Divergence Theorem] S Ÿ ' ' kF † nk d5 [A property of integrals] S Ÿ ' ' (1) d5 ckF † nk Ÿ 1d S œ Area of S. 22. Yes, the outward flux through the top is 5. The reason is this: Since ™ † F œ ™ † (xi 2yj (z 3)k œ 1 2 1 œ 0, the outward flux across the closed cubelike surface is 0 by the Divergence Theorem. The flux across the top is therefore the negative of the flux across the sides and base. Routine calculations show that the sum of these latter fluxes is 5. (The flux across the sides that lie in the xz-plane and the yz-plane are 0, while the flux across the xy-plane is 3.) Therefore the flux across the top is 5. 23. (a) ` ` ` ` x (x) œ 1, ` y (y) œ 1, ` z (z) œ 1 Ê ™ † F œ 3 Ê Flux œ ' ' ' 3 dV œ 3 ' ' ' dV œ 3(Volume of the solid) D D (b) If F is orthogonal to n at every point of S, then F † n œ 0 everywhere Ê Flux œ ' ' F † n d5 œ 0. But the flux is S 3 (Volume of the solid) Á 0, so F is not orthogonal to n at every point. Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 988 Chapter 16 Integration in Vector Fields 24. ™ † F œ 2x 4y 6z 12 Ê Flux œ '0 '0 '0 (2x 4y 6z 12) dz dy dx œ '0 '0 (2x 4y 9) dy dx a b 1 a b œ '0 a2xb 2b# 9bb dx œ a# b 2ab# 9ab œ ab(a 2b 9) œ f(aß b); `` af œ 2ab 2b# 9b and a `f `f `f # ` b œ a 4ab 9a so that ` a œ 0 and ` b œ 0 Ê b(2a 2b 9) œ 0 and a(a 4b 9) œ 0 Ê b œ 0 or 2a 2b 9 œ 0, and a œ 0 or a 4b 9 œ 0. Now b œ 0 or a œ 0 Ê Flux œ 0; 2a 2b 9 œ 0 and a 4b 9 œ 0 Ê 3a 9 œ 0 Ê a œ 3 Ê b œ 3# so that f ˆ3ß 3# ‰ œ 27 # is the maximum flux. 25. ' ' F † n d5 œ ' ' ' ™ † F dV œ ' ' ' 3 dV Ê "3 ' ' F † n d5 œ ' ' ' dV œ Volume of D D S D D S 26. F œ C Ê ™ † F œ 0 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV œ ' ' ' 0 dV œ 0 D S D 27. (a) From the Divergence Theorem, ' ' ™ f † n d5 œ ' ' ' ™ † ™ f dV œ ' ' ' ™ # f dV œ ' ' ' 0 dV œ 0 S D S D D D (b) From the Divergence Theorem, ' ' f ™ f † n d5 œ ' ' ' ™ † f ™ f dV. Now, f ™ f œ ˆf `` xf ‰ i Šf `` yf ‹ j ˆf `` zf ‰ k # # # # # # Ê ™ † f ™ f œ ’f `` xf# ˆ `` xf ‰ “ ”f `` yf# Š `` yf ‹ • ’f `` zf# ˆ `` zf ‰ “ œ f ™ # f k ™ f k# œ 0 k ™ f k# since f is harmonic Ê ' ' f ™ f † n d5 œ ' ' ' k ™ f k# dV, as claimed. D S 28. From the Divergence Theorem, ' ' ™ f † n d5 œ ' ' ' ™ † ™ f dV œ ' ' ' Š `` xf# `` yf# `` zf# ‹ dV. Now, # D S # # D f(xß yß z) œ ln Èx# y# z# œ "# ln ax# y# z# b Ê `` xf œ x# yx# z# , `` yf œ x# yy# z# , `` zf œ x# yz# z# # # # # # # # # # z z Ê `` xf# œ ax#xy#yz#zb# , `` yf# œ ax#xyy# , ` f œ ax#xyy# , Ê `` xf# `` yf# `` zf# z# b# ` z# z# b# # # # z œ ax#xyy# œ x# y"# z# Ê ' ' ™ f † n d5 œ ' ' ' z # b# # œ '0 1Î2 # # D S # dV x # y # z# # œ '0 1Î2 # '01Î2 '0a 3 3sin 9 d3 d9 d) # # '01Î2 a sin 9 d9 d) œ '01Î2 ca cos 9d 1! Î# d) œ '01Î2 a d) œ 1#a 29. ' ' f ™ g † n d5 œ ' ' ' ™ † f ™ g dV œ ' ' ' ™ † Šf `` gx i f `` gy j f `` gz k‹ dV D S D # # # œ ' ' ' Šf `` xg# `` xf `` gx f `` yg# `` yf `` gy f `` zg# `` zf `` gz ‹ dV D # # # œ ' ' ' ’f Š `` xg# `` yg# `` zg# ‹ Š `` xf `` gx `` yf `` gy `` zf `` gz ‹“ dV œ ' ' ' af ™ # g ™ f † ™ gb dV D D 30. By Exercise 29, ' ' f ™ g † n d5 œ ' ' ' af ™ # g ™ f † ™ gb dV and by interchanging the roles of f and g, D S ' ' g ™ f † n d5 œ ' ' ' ag ™ # f ™ g † ™ f b dV. Subtracting the second equation from the first yields: D S ' ' a f ™ g g ™ f b † n d5 œ ' ' ' af ™ # g g ™ # f b dV since ™ f † ™ g œ ™ g † ™ fÞ D S 31. (a) The integral ' ' ' p(tß xß yß z) dV represents the mass of the fluid at any time t. The equation says that D the instantaneous rate of change of mass is flux of the fluid through the surface S enclosing the region D: the mass decreases if the flux is outward (so the fluid flows out of D), and increases if the flow is inward (interpreting n as the outward pointing unit normal to the surface). (b) ' ' ' D `p d ` t dV œ dt ' ' ' p dV œ ' ' pv † n d5 œ ' ' ' ™ † pv dV Ê ``3t œ ™ † pv D S D Since the law is to hold for all regions D, ™ † pv ``pt œ 0, as claimed Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 16 Practice Exercises 32. (a) ™ T points in the direction of maximum change of the temperature, so if the solid is heating up at the point the temperature is greater in a region surrounding the point Ê ™ T points away from the point Ê ™ T points toward the point Ê ™ T points in the direction the heat flows. (b) Assuming the Law of Conservation of Mass (Exercise 31) with k ™ T œ pv and c3T œ p, we have d dt ' ' ' c3T dV œ ' ' k ™ T † n d5 Ê the continuity equation, ™ † (k ™ T) ``t (c3T) œ 0 D S Ê c3 ``Tt œ ™ † (k ™ T) œ k ™ # T Ê ``Tt œ ck3 ™ # T œ K ™ # T, as claimed CHAPTER 16 PRACTICE EXERCISES dy 1. Path 1: r œ ti tj tk Ê x œ t, y œ t, z œ t, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ 3 3t# and dx dt œ 1, dt œ 1, dz dt œ 1 ‰ Š dy ˆ dz ‰ dt œ È3 dt Ê ' f(xß yß z) ds œ ' È3 a3 3t# b dt œ 2È3 Ê Êˆ dx dt dt ‹ dt C 0 # # # 1 dy Path 2: r" œ ti tj , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ 2t 3t# 3 and dx dt œ 1, dt œ 1, dz dt œ 0 ‰ Š dy ˆ dz ‰ dt œ È2 dt Ê ' f(xß yß z) ds œ ' È2 a2t 3t# 3b dt œ 3È2 ; Ê Êˆ dx dt dt ‹ dt 0 # # # 1 C" dy dz r# œ i j tk Ê x œ 1, y œ 1, z œ t Ê f(g(t)ß h(t)ß k(t)) œ 2 2t and dx dt œ 0, dt œ 0, dt œ 1 ‰ Š dy ˆ dz ‰ dt œ dt Ê ' f(xß yß z) ds œ ' (2 2t) dt œ 1 Ê Êˆ dx dt dt ‹ dt C 0 # # # 1 # Ê 'C f(xß yß z) ds œ 'C f(xß yß z) ds 'C f(xß yß z) œ 3È2 1 " # dy dz 2. Path 1: r" œ ti Ê x œ t, y œ 0, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ t# and dx dt œ 1, dt œ 0, dt œ 0 ‰# Š dy ˆ dz ‰# dt œ dt Ê ' f(xß yß z) ds œ ' t# dt œ "3 ; Ê Êˆ dx dt dt ‹ dt C 0 # 1 " dy dz r# œ i tj Ê x œ 1, y œ t, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ 1 t and dx dt œ 0, dt œ 1, dt œ 0 ‰# Š dy ˆ dz ‰# dt œ dt Ê ' f(xß yß z) ds œ ' (1 t) dt œ 32 ; Ê Êˆ dx dt dt ‹ dt C 0 # 1 # dy dz r$ œ i j tk Ê x œ 1, y œ 1, z œ t Ê f(g(t)ß h(t)ß k(t)) œ 2 t and dx dt œ 0, dt œ 0, dt œ 1 ‰# Š dy ˆ dz ‰# dt œ dt Ê ' f(xß yß z) ds œ ' (2 t) dt œ 32 Ê Êˆ dx dt dt ‹ dt C 0 # 1 $ Ê 'Path 1 f(xß yß z) ds œ 'C f(xß yß z) ds 'C f(xß yß z) ds 'C f(xß yß z) ds œ 10 3 " # $ dy dz Path 2: r% œ ti tj Ê x œ t, y œ t, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ t# t and dx dt œ 1, dt œ 1, dt œ 0 ‰# Š dy ˆ dz ‰# dt œ È2 dt Ê ' f(xß yß z) ds œ ' È2 at# tb dt œ 56 È2; Ê Êˆ dx dt dt ‹ dt C 0 # 1 % r$ œ i j tk (see above) Ê 'C f(xß yß z) ds œ 32 $ È Ê 'Path 2 f(xß yß z) ds œ 'C f(xß yß z) ds 'C f(xß yß z) ds œ 56 È2 3# œ 5 26 9 $ % dy dz Path 3: r& œ tk Ê x œ 0, y œ 0, z œ t, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ t and dx dt œ 0, dt œ 0, dt œ 1 ‰ Š dy ˆ dz ‰ dt œ dt Ê ' f(xß yß z) ds œ ' t dt œ "2 ; Ê Êˆ dx dt dt ‹ dt C 0 # # # 1 & dy dz r' œ tj k Ê x œ 0, y œ t, z œ 1, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ t 1 and dx dt œ 0, dt œ 1, dt œ 0 ‰ Š dy ˆ dz ‰ dt œ dt Ê ' f(xß yß z) ds œ ' (t 1) dt œ 12 ; Ê Êˆ dx dt dt ‹ dt C 0 # # # 1 ' dy dz r( œ ti j k Ê x œ t, y œ 1, z œ 1, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ t# and dx dt œ 1, dt œ 0, dt œ 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 989 990 Chapter 16 Integration in Vector Fields ‰# Š dy ˆ dz ‰# dt œ dt Ê ' f(xß yß z) ds œ ' t# dt œ 13 Ê Êˆ dx dt dt ‹ dt C 0 # 1 ( Ê 'Path 3 f(xß yß z) ds œ 'C f(xß yß z) ds 'C f(xß yß z) ds 'C f(xß yß z) ds œ "# "# "3 œ 32 & ' ( 3. r œ (a cos t)j (a sin t)k Ê x œ 0, y œ a cos t, z œ a sin t Ê fag(t)ß h(t)ß k(t)b œ Èa# sin# t œ a ksin tk and dy dx dz dt œ 0, dt œ a sin t, dt œ a cos t # # # ‰ Š dy ˆ dz ‰ dt œ a dt Ê Êˆ dx dt dt ‹ dt Ê 'C f(xß yß z) ds œ '0 a# ksin tk dt œ '0 a# sin t dt '1 a# sin t dt œ 4a# 21 1 21 4. r œ (cos t t sin t)i (sin t t cos t)j Ê x œ cos t t sin t, y œ sin t t cos t, z œ 0 Ê fag(t)ß h(t)ß k(t)b œ È(cos t t sin t)# (sin t t cos t)# œ È1 t# and dx œ sin t sin t t cos t dt dz œ t cos t, dy dt œ cos t cos t t sin t œ t sin t, dt œ 0 Ê # ‰# Š dy ˆ dz ‰# dt Êˆ dx dt dt ‹ dt È3 œ Èt# cos# t t# sin# t dt œ ktk dt œ t dt since 0 Ÿ t Ÿ È3 Ê 'C f(xß yß z) ds œ '0 5. tÈ1 t# dt œ 73 `P " $Î# œ ``Nz , ``Mz œ "# (x y z)$Î# œ `` Px , ``Nx œ "# (x y z)$Î# œ ``My ` y œ # (x y z) Ê M dx N dy P dz is exact; `` xf œ Èx "y z Ê f(xß yß z) œ 2Èx y z g(yß z) Ê `` yf œ Èx "y z `` gy œ Èx "y z Ê `` gy œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ 2Èx y z h(z) Ê `` zf œ Èx "y z hw (z) Ð4ß 3ß0Ñ dx dy dz œ Èx "y z Ê hw (x) œ 0 Ê h(z) œ C Ê f(xß yß z) œ 2Èx y z C Ê 'Ð 1ß1ß1Ñ Èx y z œ f(4ß 3ß 0) f(1ß 1ß 1) œ 2È1 2È1 œ 0 6. `P " `N `M `P `N `M ` y œ #Èyz œ ` z , ` z œ 0 œ ` x , ` x œ 0 œ ` y Ê M dx N dy T dz is exact; `` xf œ 1 Ê f(xß yß z) œ x g(yß z) Ê `` yf œ `` yg œ É yz Ê g(yß z) œ 2Èyz h(z) Ê f(xß yß z) œ x 2Èyz h(z) Ê `` zf œ É yz hw (z) œ É yz Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x 2Èyz C Ê 'Ð1ß1ß1Ñ dx É yz dy É yz dz œ f(10ß 3ß 3) f(1ß 1ß 1) œ (10 2 † 3) (1 2 † 1) œ 4 1 œ 5 Ð10ß3ß3Ñ 7. `M `P ` z œ y cos z Á y cos z œ ` x Ê F is not conservative; r œ a2 cos tbi a2 sin tbj k, 0 Ÿ t Ÿ 21 Ê dr œ a2 sin tbi a2 cos tbj Ê 'C F † dr œ '0 c a2 sin tbasina1bba2 sin tb a2 cos tbasina1bba2 cos tbd dt 21 œ 4 sina1b'0 asin2 t cos2 tbdt œ 81 sina1b 21 8. `P `N `M `P `N `M # ` y œ 0 œ ` z , ` z œ 0 œ ` x , ` x œ 3x œ ` y Ê F is conservative Ê 'C F † dr œ 0 9. Let M œ 8x sin y and N œ 8y cos x Ê ``My œ 8x cos y and ``Nx œ 8y sin x Ê 'C 8x sin y dx 8y cos x dy œ ' ' (8y sin x 8x cos y) dy dx œ '0 1/2 R '01/2 (8y sin x 8x cos y) dy dx œ '01/2 a1# sin x 8xb dx œ 1# 1# œ 0 10. Let M œ y# and N œ x# Ê ``My œ 2y and ``Nx œ 2x Ê 'C y# dx x# dy œ ' ' (2x 2y) dx dy 21 2 21 œ '0 '0 (2r cos ) 2r sin )) r dr d) œ '0 16 3 (cos ) sin )) d) œ 0 R Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 16 Practice Exercises 991 11. Let z œ 1 x y Ê fx (xß y) œ 1 and fy (xß y) œ 1 Ê Éfx# fy# 1 œ È3 Ê Surface Area œ ' ' È3 dx dy R œ È3(Area of the circular region in the xy-plane) œ 1È3 12. ™ f œ 3i 2yj 2zk , p œ i Ê k ™ f k œ È9 4y# 4z# and k ™ f † pk œ 3 Ê Surface Area œ ' ' R È9 4y# 4z# dy dz œ 3 È '021 '0 3 È9 3 4r r dr d) œ "3 '021 Š 74 È21 94 ‹ d) œ 16 Š7È21 9‹ # 13. ™ f œ 2xi 2yj 2zk , p œ k Ê k ™ f k œ È4x# 4y# 4z# œ 2Èx# y# z# œ 2 and k ™ f † pk œ k2zk œ 2z since z 0 Ê Surface Area œ ' ' R œ '0 ’È1 r# “ 21 "ÎÈ# ! 2 2z dA œ ' ' "z dA œ ' ' È R R " dx dy œ 1 x# y# È '021 '01Î 2 È " 1 r# r dr d) d) œ '0 Š1 È"2 ‹ d) œ 21 Š1 È"2 ‹ 21 14. (a) ™ f œ 2xi 2yj 2zk , p œ k Ê k ™ f k œ È4x# 4y# 4z# œ 2Èx# y# z# œ 4 and k ™ f † pk œ 2z since 4 0 Ê Surface Area œ ' ' 2z dA œ ' ' z R 2 z dA œ 2 R '01/2 '02 cos ) 2 È4 r# r dr d) œ 41 8 (b) r œ 2 cos ) Ê dr œ 2 sin ) d); ds# œ r# d)# dr# (Arc length in polar coordinates) Ê ds# œ (2 cos ))# d)# dr# œ 4 cos# ) d)# 4 sin# ) d)# œ 4 d)# Ê ds œ 2 d); the height of the cylinder is z œ È4 r# œ È4 4 cos# ) œ 2 ksin )k œ 2 sin ) if 0 Ÿ ) Ÿ 1# Ê Surface Area œ 'c1Î2 h ds 1/2 œ 2 '0 (2 sin ))(2 d)) œ 8 1/2 15. f(xß yß z) œ xa yb cz œ 1 Ê ™ f œ ˆ "a ‰ i ˆ b" ‰ j ˆ "c ‰ k Ê k ™ f k œ É a"# b"# c"# and p œ k Ê k ™ f † pk œ "c since c 0 Ê Surface Area œ ' ' É "# "# "# R a b c Š "c ‹ dA œ cÉ a"# b"# c"# ' ' dA œ #" abcÉ a"# b"# c"# , R since the area of the triangular region R is "# ab. To check this result, let v œ ai ck and w œ ai bj; the area can be found by computing "# kv ‚ wk. 16. (a) ™ f œ 2yj k , p œ k Ê k ™ f k œ È4y# 1 and k ™ f † pk œ 1 Ê d5 œ È4y# 1 dx dy Ê ' ' g(xß yß z) d5 œ ' ' È4yyz# 1 È4y# 1 dx dy œ ' ' y ay# 1b dx dy œ 'c1 '0 ay$ yb dx dy 1 S œ 'c1 3 ay 1 $ R 3 R % # " yb dy œ 3 ’ y4 y# “ œ 0 " (b) ' ' g(xß yß z) d5 œ ' ' È4yz# 1 È4y# 1 dx dy œ 'c1 '0 ay# 1b dx dy œ 'c1 3 ay# 1b dy 1 3 1 R S " $ œ 3 ’ y3 y“ œ 4 " 17. ™ f œ 2yj 2zk , p œ k Ê k ™ f k œ È4y# 4z# œ 2Èy# z# œ 10 and k ™ f † pk œ 2z since z 5 ' ' g(xß yß z) d5 œ ' ' ax% yb ay# z# b ˆ 5z ‰ dx dy Ê d5 œ 10 2z dx dy œ z dx dy œ S œ'' R % ax yb (25) Š È255 y# ‹ dx dy œ '0 ' 4 R 1 125y x% dx dy œ 0 È25 y# 0 '04 È2525y y dy œ 50 # 18. Define the coordinate system so that the origin is at the center of the earth, the z-axis is the earth's axis (north is the positive z direction), and the xz-plane contains the earth's prime meridian. Let S denote the surface which is Wyoming so "Î# then S is part of the surface z œ aR# x# y# b . Let Rxy be the projection of S onto the xy-plane. The surface area of Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 992 Chapter 16 Integration in Vector Fields Wyoming is ' ' 1 d5 œ ' ' Ê1 ˆ `` xz ‰ Š `` yz ‹ dA œ ' ' É R# xx# y# R# yx# y# 1 dA œ ' ' # # S Rxy œ ') 'R sin 45° R aR# r# b )# R sin 49° Rxy "Î# " œ ') R aR# r# b )# "Î# R sin 49° " R sin 45° ¹ # # Rxy R dA aR# x# y# b"Î# r dr d) (where )" and )# are the radian equivalent to 104°3w and 111°3w , respectively) œ ') R aR# R# sin# 45°b )# "Î# " R aR# R# sin# 49°b "Î# d) 71 71 œ ()# )" )R# (cos 45° cos 49°) œ 180 R# (cos 45° cos 49°) œ 180 (3959)# (cos 45° cos 49°) ¸ 97,751 sq. mi. 19. A possible parametrization is r(9ß )) œ (6 sin 9 cos ))i (6 sin 9 sin ))j (6 cos 9)k (spherical coordinates); now 3 œ 6 and z œ 3 Ê 3 œ 6 cos 9 Ê cos 9 œ " Ê 9 œ 21 and z œ 3È3 Ê 3È3 œ 6 cos 9 Ê È cos 9 œ #3 # Ê 9 œ 16 Ê 3 1 21 6 Ÿ 9 Ÿ 3 ; also 0 Ÿ ) Ÿ 21 # 20. A possible parametrization is r(rß )) œ (r cos ))i (r sin ))j Š r# ‹ k (cylindrical coordinates); # # now r œ Èx# y# Ê z œ r# and 2 Ÿ z Ÿ 0 Ê 2 Ÿ r# Ÿ 0 Ê 4 r# 0 Ê 0 Ÿ r Ÿ 2 since r 0; also 0 Ÿ ) Ÿ 21 21. A possible parametrization is r(rß )) œ (r cos ))i (r sin ))j (1 r)k (cylindrical coordinates); now r œ Èx# y# Ê z œ 1 r and 1 Ÿ z Ÿ 3 Ê 1 Ÿ 1 r Ÿ 3 Ê 0 Ÿ r Ÿ 2; also 0 Ÿ ) Ÿ 21 22. A possible parametrization is r(xß y) œ xi yj ˆ3 x y# ‰ k for 0 Ÿ x Ÿ 2 and 0 Ÿ y Ÿ 2 23. Let x œ u cos v and z œ u sin v, where u œ Èx# z# and v is the angle in the xz-plane with the x-axis Ê r(uß v) œ (u cos v)i 2u# j (u sin v)k is a possible parametrization; 0 Ÿ y Ÿ 2 Ê 2u# Ÿ 2 Ê u# Ÿ 1 Ê 0 Ÿ u Ÿ 1 since u 0; also, for just the upper half of the paraboloid, 0 Ÿ v Ÿ 1 24. A possible parametrization is ŠÈ10 sin 9 cos )‹ i ŠÈ10 sin 9 sin )‹ j ŠÈ10 cos 9) k , 0 Ÿ 9 Ÿ 12 and 0 Ÿ ) Ÿ 1# â âi â 25. ru œ i j , rv œ i j k Ê ru ‚ rv œ â " â â" â j kâ â " 0 â œ i j 2k Ê kru ‚ rv k œ È6 â " " â Ê Surface Area œ ' ' kru ‚ rv k du dv œ '0 '0 È6 du dv œ È6 1 1 Ruv 26. ' ' axy z# b d5 œ '0 '0 c(u v)(u v) v# d È6 du dv œ È6'0 '0 au# 2v# b du dv 1 S 1 1 1 œ È6 '0 ’ u3 2uv# “ dv œ È6 '0 ˆ "3 2v# ‰ dv œ È6 "3 v 23 v$ ‘ ! œ 36 œ É 23 1 $ " 1 " È ! â â i j kâ â â â sin ) 0 â 27. rr œ (cos ))i (sin ))j , r) œ (r sin ))i (r cos ))j k Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) " â œ (sin ))i (cos ))j rk Ê krr ‚ r) k œ Èsin# ) cos# ) r# œ È1 r# Ê Surface Area œ ' ' krr ‚ r) k dr d) œ '0 21 Rr) '0 È1 r# dr d) œ '0 ’ 2r È1 r# "# ln Šr È1 r# ‹“ d) œ '0 ’ "# È2 "# ln Š1 È2‹“ d) 21 1 " 21 ! œ 1 ’È2 ln Š1 È2‹“ Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 16 Practice Exercises 993 28. ' ' Èx# y# 1 d5 œ '0 '0 Èr# cos# ) r# sin# ) 1 È1 r# dr d) œ '0 '0 a1 r# b dr d) 21 S œ '0 ’r r3 “ d) œ '0 21 $ " 21 ! 21 1 1 4 8 3 d) œ 3 1 29. `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 0 œ ``My Ê Conservative 30. `` Py œ 3zy œ ``Nz , ``Mz œ # # 3xz # &Î# œ `` Px , ``Nx œ # #3xy # &Î# œ ``My ax# y# z# b &Î# ax y z b ax y z b Ê Conservative 31. `` Py œ 0 Á yez œ ``Nz Ê Not Conservative y `P `N z `M 32. `` Py œ (x xyz)# œ ``Nz , ``Mz œ (x yz) # œ ` x , ` x œ (x yz)# œ ` y Ê Conservative 33. `` xf œ 2 Ê f(xß yß z) œ 2x g(yß z) Ê `` yf œ `` yg œ 2y z Ê g(yß z) œ y# zy h(z) Ê f(xß yß z) œ 2x y# zy h(z) Ê `` zf œ y hw (z) œ y 1 Ê hw (z) œ 1 Ê h(z) œ z C Ê f(xß yß z) œ 2x y# zy z C 34. `` xf œ z cos xz Ê f(xß yß z) œ sin xz g(yß z) Ê `` yf œ `` yg œ ey Ê g(yß z) œ ey h(z) Ê f(xß yß z) œ sin xz ey h(z) Ê `` zf œ x cos xz hw (z) œ x cos xz Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ sin xz ey C 35. Over Path 1: r œ ti tj tk , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ t and dr œ (i j k) dt Ê F œ 2t# i j t# k Ê F † dr œ a3t# 1b dt Ê Work œ '0 a3t# 1b dt œ 2; 1 Over Path 2: r" œ ti tj , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ 0 and dr" œ (i j) dt Ê F" œ 2t# i j t# k Ê F" † dr" œ a2t# 1b dt Ê Work" œ '0 a2t# 1b dt œ 53 ; r# œ i j tk , 0 Ÿ t Ÿ 1 Ê x œ 1, y œ 1, z œ t and 1 dr# œ k dt Ê F# œ 2i j k Ê F# † dr# œ dt Ê Work# œ '0 dt œ 1 Ê Work œ Work" Work# œ 53 1 œ 83 1 36. Over Path 1: r œ ti tj tk , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ t and dr œ (i j k) dt Ê F œ 2t# i t# j k Ê F † dr œ a3t# 1b dt Ê Work œ '0 a3t# 1b dt œ 2; 1 Over Path 2: Since f is conservative, )C F † dr œ 0 around any simple closed curve C. Thus consider 'curve F † dr œ 'C F † dr 'C F † dr , where C" is the path from (0ß 0ß 0) to (1ß 1ß 0) to ("ß "ß ") and C# is the path from (1ß 1ß 1) to (!ß !ß !). Now, from Path 1 above, 'C F † dr œ 2 Ê 0 œ 'curve F † dr œ 'C F † dr (2) Ê 'C F † dr œ 2 " # # " " 37. (a) r œ aet cos tb i aet sin tb j Ê x œ et cos t, y œ et sin t from (1ß 0) to ae21 ß 0b Ê 0 Ÿ t Ÿ 21 Ê ddtr œ aet cos t et sin tb i aet sin t et cos tb j and F œ # t t xi yj œ ae2t cos#tb i 2tae sin# tb$Î#j ax# y# b$Î# ae cos t e sin tb # t‰ t‰ dr cos t sin t cos t t œ ˆ cos i ˆ sin sinet t sin tecos ‹ œ et t e2t e2t j Ê F † dt œ Š et et Ê Work œ '0 et dt œ 1 e21 21 (b) F œ œ xi yj ax# y# b$Î# y ax# y# b$Î# 21 Ê `` xf œ x ax# y# b$Î# Ê f(xß yß z) œ ax# y# b Ê g(yß z) œ C Ê f(xß yß z) œ ax# y# b "Î# "Î# g(yß z) Ê `` yf œ y `` gy ax# y# b$Î# is a potential function for F Ê 'C F † dr œ f ae ß 0b f(1ß 0) œ 1 e21 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 994 Chapter 16 Integration in Vector Fields 38. (a) F œ ™ ax# zey b Ê F is conservative Ê )C F † dr œ 0 for any closed path C (b) 'C F † dr œ 'Ð1ß0ß0Ñ Ð1ß0ß21Ñ â i â â 39. ™ ‚ F œ â ``x â # ây ™ ax# zey b † dr œ ax# zey bk Ð"ß!ß#1Ñ ax# zey bk Ð"ß!ß!Ñ œ 21 0 œ 21 k ââ ` â 2i 6 j 3k 2 6 3 ` z â œ 2yk; unit normal to the plane is n œ È4 36 9 œ 7 i 7 j 7 k â y 3z# â j ` `y k™f k 7 Ê ™ ‚ F † n œ 67 y; p œ k and f(xß yß z) œ 2x 6y 3z Ê k ™ f † pk œ 3 Ê d5 œ k™ f†pk dA œ 3 dA Ê )C F † dr œ ' ' 76 y d5 œ ' ' ˆ 76 y‰ ˆ 37 dA‰ œ ' ' 2y dA œ '0 '0 2r sin ) r dr d) œ '0 21 R â i â â 40. ™ ‚ F œ â ``x â # âx y n œ È"2 j È"2 k R 21 1 R 2 3 sin ) d) œ 0 â â â â œ 8yi ; the circle lies in the plane f(xß yß z) œ y z œ 0 with unit normal â 4y z â j k ` `y ` `z # xy Ê ™ ‚ F † n œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 œ 0 R R È2, dy œ È2, dz œ 2t 41. (a) r œ È2ti È2tj a4 t# b k , 0 Ÿ t Ÿ 1 Ê x œ È2t, y œ È2t, z œ 4 t# Ê dx dt œ dt dt ‰ Š dy ˆ dz ‰ dt œ È4 4t# dt Ê M œ ' $ (xß yß z) ds œ ' 3tÈ4 4t# dt œ "4 (4 4t)$Î# ‘ Ê Êˆ dx dt dt ‹ dt ! C 0 # # # " 1 œ 4È2 2 (b) M œ 'C $ (xß yß z) ds œ '0 È4 4t# dt œ ’tÈ1 t# ln Št È1 t# ‹“ œ È2 ln Š1 È2‹ " 1 ! dy dz "Î# 42. r œ ti 2tj 32 t$Î# k , 0 Ÿ t Ÿ 2 Ê x œ t, y œ 2t, z œ 32 t$Î# Ê dx dt œ 1, dt œ 2, dt œ t ‰# Š dy ˆ dz ‰# dt œ Èt 5 dt Ê M œ ' $ (xß yß z) ds œ ' 3È5 t Èt 5 dt Ê Êˆ dx dt dt ‹ dt C 0 # 2 œ '0 3(t 5) dt œ 36; Myz œ 'C x$ ds œ '0 3t(t 5) dt œ 38; Mxz œ 'C y$ ds œ '0 6t(t 5) dt œ 76; 2 2 2 È2 Ê x œ yz œ 38 œ 19 , y œ Mxz œ 76 œ 19 , z œ xy œ Mxy œ 'C z$ ds œ '0 2t$Î# (t 5) dt œ 144 7 M 36 18 M 36 9 M 2 M M È 2‹ Š 144 7 36 œ 47 È2 È È # # dy È2 t"Î# , dz œ t 43. r œ ti Š 2 3 2 t$Î# ‹ j Š t# ‹ k , 0 Ÿ t Ÿ 2 Ê x œ t, y œ 2 3 2 t$Î# , z œ t# Ê dx dt œ 1, dt œ dt # # # ‰ Š dy ˆ dz ‰ dt œ È1 2t t# dt œ È(t 1)# dt œ kt 1k dt œ (t 1) dt on the domain given. Ê Êˆ dx dt dt ‹ dt Then M œ 'C $ ds œ '0 ˆ t " 1 ‰ (t 1) dt œ '0 dt œ 2; Myz œ 'C x$ ds œ '0 t ˆ t " 1 ‰ (t 1) dt œ '0 t dt œ 2; 2 2 2 2 32 Mxz œ 'C y$ ds œ '0 Š 2 3 2 t$Î# ‹ ˆ t " 1 ‰ (t 1) dt œ '0 2 3 2 t$Î# dt œ 15 ; Mxy œ 'C z$ ds 2 È 2 È œ '0 Š t# ‹ ˆ t"1 ‰ (t 1) dt œ '0 t# dt œ 43 Ê x œ Myz œ #2 œ 1; y œ MMxz œ 2 œ 2 # # ˆ 43 ‰ 2 # œ 3 ; Ix œ M ˆ 32 ‰ Mxy 16 15 # œ 15 ; z œ M 64 'C ay# z# b $ ds œ '02 Š 89 t$ t4 ‹ dt œ 232 ' # # '2 # t 45 ; Iy œ C ax z b $ ds œ 0 Št 4 ‹ dt œ 15 ; % % Iz œ 'C ay# x# b $ ds œ '0 ˆt# 89 t$ ‰ dt œ 56 9 2 44. z œ 0 because the arch is in the xy-plane, and x œ 0 because the mass is distributed symmetrically with respect # # # ‰ Š dy ˆ dz ‰ dt to the y-axis; r(t) œ (a cos t)i (a sin t)j , 0 Ÿ t Ÿ 1 Ê ds œ Êˆ dx dt dt ‹ dt œ È(a sin t)# (a cos t)# dt œ a dt, since a 0; M œ 'C $ ds œ 'C (2a y) ds œ '0 (2a a sin t) a dt 1 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 16 Practice Exercises 995 œ 2a2 1 2a# ; Mxz œ 'C y$ dt œ 'C y(2a y) ds œ '0 (a sin t)(2a a sin t) dt œ '0 a2a# sin t a# sin# tb dt 1 1 # Š4a# a 1 ‹ 1 # 81 ˆ 81 ‰ œ 2a# cos t a# ˆ 2t sin4 2t ‰ ‘ ! œ 4a# a#1 Ê y œ 2a# 1 2a # œ 41 4 Ê axß yß zb œ 0ß 41 4 ß 0 # t t 45. r(t) œ aet cos tb i aet sin tb j et k , 0 Ÿ t Ÿ ln 2 Ê x œ et cos t, y œ et sin t, z œ et Ê dx dt œ ae cos t e sin tb , dy dz t t t dt œ ae sin t e cos tb, dt œ e # # # ‰ Š dy ˆ dz ‰ dt Ê Êˆ dx dt dt ‹ dt œ Éaet cos t et sin tb# aet sin t et cos tb# aet b# dt œ È3e2t dt œ È3 et dt; M œ 'C $ ds œ '0 È3 et dt ln 2 œ È3; Mxy œ 'C z$ ds œ '0 ln 2 È M ŠÈ3 et ‹ aet b dt œ '0 È3 e2t dt œ 3 # 3 Ê z œ Mxy œ ln 2 Œ 3 È3 # È3 œ 3# ; È Iz œ 'C ax# y# b $ ds œ '0 ae2t cos# t e2t sin# tb ŠÈ3 et ‹ dt œ '0 È3 e3t dt œ 7 3 3 ln 2 ln 2 dy 46. r(t) œ (2 sin t)i (2 cos t)j 3tk , 0 Ÿ t Ÿ 21 Ê x œ 2 sin t, y œ 2 cos t, z œ 3t Ê dx dt œ 2 cos t, dt œ 2 sin t, dz dt œ 3 ‰# Š dy ˆ dz ‰# dt œ È4 9 dt œ È13 dt; M œ ' $ ds œ ' $ È13 dt œ 21$ È13; Ê Êˆ dx dt dt ‹ dt C 0 # 21 Mxy œ 'C z$ ds œ '0 (3t) Š$ È13‹ dt œ 6$1# È13; Myz œ 'C x$ ds œ '0 (2 sin t) Š$ È13‹ dt œ 0; 21 21 13 Mxz œ 'C y$ ds œ '0 (2 cos t) Š$ È13‹ dt œ 0 Ê x œ y œ 0 and z œ Mxy œ 62$1 œ 31 Ê (!ß !ß 31) is the $1È13 21 #È M center of mass 47. Because of symmetry x œ y œ 0. Let f(xß yß z) œ x# y# z# œ 25 Ê ™ f œ 2xi 2yj 2zk Ê k ™ f k œ È4x# 4y# 4z# œ 10 and p œ k Ê k ™ f † pk œ 2z, since z 0 Ê M œ ' ' $ (xß yß z) d5 R ' ' 5 dA œ 5(Area of the circular region) œ 801; Mxy œ ' ' z$ d5 œ ' ' 5z dA ‰ œ ' ' z ˆ 10 #z dA œ R R R œ ' ' 5È25 x# y# dx dy œ '0 '0 Š5È25 r# ‹ r dr d) œ '0 21 21 4 R R 490 980 3 d) œ 3 1 Ê zœ Š 980 3 1‹ 801 œ 49 12 ' ' ax# y# b $ d5 œ ' ' 5 ax# y# b dx dy œ ' ' 5r$ dr d) œ ' 320 d) œ 6401 ‰ Ê axß yß zb œ ˆ0ß !ß 49 12 ; Iz œ 0 0 0 21 R 4 21 R 48. On the face z œ 1: g(xß yß z) œ z œ 1 and p œ k Ê ™ g œ k Ê k ™ gk œ 1 and k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' ax# y# b dA œ 2 '0 1/4 R '0sec r$ dr d) œ 23 ; On the face z œ 0: g(xß yß z) œ z œ 0 Ê ™ g œ k and p œ k ) Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' ax# y# b dA œ 23 ; On the face y œ 0: g(xß yß z) œ y œ 0 R Ê ™ g œ j and p œ j Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' ax# 0b dA œ '0 '0 x# dx dz œ "3 ; 1 1 R On the face y œ 1: g(xß yß z) œ y œ 1 Ê ™ g œ j and p œ j Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' ax# 1# b dA œ '0 '0 ax# 1b dx dz œ 43 ; On the face x œ 1: g(xß yß z) œ x œ 1 Ê ™ g œ i and p œ i 1 1 R Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' a1# y# b dA œ '0 '0 a1 y# b dy dz œ 43 ; On the face 1 1 R x œ 0: g(xß yß z) œ x œ 0 Ê ™ g œ i and p œ i Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' a0# y# b dA œ '0 '0 y# dy dz œ 13 Ê Iz œ 23 23 "3 43 43 3" œ 14 3 1 1 R Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 996 Chapter 16 Integration in Vector Fields 49. M œ 2xy x and N œ xy y Ê ``Mx œ 2y 1, ``My œ 2x, ``Nx œ y, ``Ny œ x 1 Ê Flux œ ' ' Š ``Mx ``Ny ‹ dx dy 1 1 œ ' ' (2y 1 x 1) dy dx œ '0 '0 (2y x) dy dx œ 3# ; Circ œ ' ' Š ``Nx ``My ‹ dx dy R œ ' ' (y 2x) dy dx œ '0 ' 1 R R R 1 (y 2x) dy dx œ "# 0 50. M œ y 6x# and N œ x y# Ê ``Mx œ 12x, ``My œ 1, ``Nx œ 1, ``Ny œ 2y Ê Flux œ ' ' Š ``Mx ``Ny ‹ dx dy R œ ' ' (12x 2y) dx dy œ '0 'y (12x 2y) dx dy œ '0 a4y# 2y 6b dy œ 11 3 ; 1 1 1 R Circ œ ' ' Š ``Nx ``My ‹ dx dy œ ' ' (1 1) dx dy œ 0 R R 51. M œ cosx y and N œ ln x sin y Ê ``My œ sinx y and ``Nx œ sinx y Ê )C ln x sin y dy cosx y dx œ ' ' Š ``Nx ``My ‹ dx dy œ ' ' Š sinx y sinx y ‹ dx dy œ 0 R R 52. (a) Let M œ x and N œ y Ê ``Mx œ 1, ``My œ 0, ``Nx œ 0, ``Ny œ 1 Ê Flux œ ' ' Š ``Mx ``Ny ‹ dx dy R œ ' ' (1 1) dx dy œ 2 ' ' dx dy œ 2(Area of the region) R R (b) Let C be a closed curve to which Green's Theorem applies and let n be the unit normal vector to C. Let F œ xi yj and assume F is orthogonal to n at every point of C. Then the flux density of F at every point of C is 0 since F † n œ 0 at every point of C Ê ``Mx ``Ny œ 0 at every point of C Ê Flux œ ' ' Š ``Mx ``Ny ‹ dx dy œ ' ' 0 dx dy œ 0. But part (a) above states that the flux is R R 2(Area of the region) Ê the area of the region would be 0 Ê contradiction. Therefore, F cannot be orthogonal to n at every point of C. 53. ``x (2xy) œ 2y, ``y (2yz) œ 2z, ``z (2xz) œ 2x Ê ™ † F œ 2y 2z 2x Ê Flux œ ' ' ' (2x 2y 2z) dV D 1 1 1 1 1 1 œ '0 '0 '0 (2x 2y 2z) dx dy dz œ '0 '0 (1 2y 2z) dy dz œ '0 (2 2z) dz œ 3 54. ``x (xz) œ z, ``y (yz) œ z, ``z (1) œ 0 Ê ™ † F œ 2z Ê Flux œ ' ' ' 2z r dr d) dz È25 c r œ '0 '0 '3 21 4 D # 2z dz r dr d) œ '0 21 '0 ra16 r# b dr d) œ '0 64 d) œ 1281 21 4 55. ``x (2x) œ 2, ``y (3y) œ 3, ``z (z) œ 1 Ê ™ † F œ 4; x# y# z# œ 2 and x# y# œ z Ê z œ 1 È2cr Ê x# y# œ 1 Ê Flux œ ' ' ' 4 dV œ 4 '0 '0 'r# 21 D 1 # dz r dr d) œ 4 '0 '0 ŠrÈ2 r# r$ ‹ dr d) 21 1 7 œ 4'0 Š 12 32 È2‹ d) œ 23 1 Š7 8È2‹ 21 56. ``x (6x y) œ 6, ``y (x z) œ 0, ``z (4yz) œ 4y Ê ™ † F œ 6 4y; z œ Èx# y# œ r Ê Flux œ ' ' ' (6 4y) dV œ '0 1/2 D '01 '0r (6 4r sin )) dz r dr d) œ '01/2 '01 a6r# 4r$ sin )b dr d) œ '0 (2 sin )) d) œ 1 1 1/2 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 16 Additional and Advanced Exercises 997 57. F œ yi zj xk Ê ™ † F œ 0 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV œ 0 D S 58. F œ 3xz# i yj z$ k Ê ™ † F œ 3z# 1 3z# œ 1 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV È16cx Î2 œ '0 '0 4 # D S '0 yÎ2 1 dz dy dx œ ' % 16x# x$ dx œ x œ 83 Š ‹ ’ “ 16 48 0 ! 4 59. F œ xy# i x# yj yk Ê ™ † F œ y# x# 0 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV S œ ' ' ' ax# y# b dV œ '0 '0 'c1 r# dz r dr d) œ '0 '0 2r$ dr d) œ '0 21 1 21 1 21 1 D D " # d) œ 1 60. (a) F œ (3z 1)k Ê ™ † F œ 3 Ê Flux across the hemisphere œ ' ' F † n d5 œ ' ' ' ™ † F dV œ ' ' ' 3 dV D S œ 3 ˆ "# ‰ ˆ 43 1a$ ‰ œ 21a$ D (b) f(xß yß z) œ x# y# z# a# œ 0 Ê ™ f œ 2xi 2yj 2zk Ê k ™ f k œ È4x# 4y# 4z# œ È4a# œ 2a since a 0 Ê n œ 2xi 2y#aj 2zk œ xi yaj zk Ê F † n œ (3z 1) ˆ za ‰ ; p œ k Ê ™ f † p œ ™ f † k œ 2z Ê k ™ f † pk œ 2z since z k™f k 2a a ' ' F † n d5 œ ' ' (3z 1) ˆ za ‰ ˆ az ‰ dA 0 Ê d5 œ k™ f†pk œ 2z dA œ z dA Ê S Rxy œ ' ' (3z 1) dx dy œ ' ' ˆ3Èa# x# y# 1‰ dx dy œ '0 '0 Š3Èa# r# 1‹ r dr d) 21 Rxy a Rxy œ '0 Š a# a$ ‹ d) œ 1a# 21a$ , which is the flux across the hemisphere. Across the base we find 21 # F œ [3(0) 1]k œ k since z œ 0 in the xy-plane Ê n œ k (outward normal) Ê F † n œ 1 Ê Flux across the base œ ' ' F † n d5 œ ' ' 1 dx dy œ 1a# . Therefore, the total flux across the closed surface is Rxy S # $ # a1a 21a b 1a œ 21a$ . CHAPTER 16 ADDITIONAL AND ADVANCED EXERCISES 1. dx œ (2 sin t 2 sin 2t) dt and dy œ (2 cos t 2 cos 2t) dt; Area œ "# )C x dy y dx œ "# '0 [(2 cos t cos 2t)(2 cos t 2 cos 2t) (2 sin t sin 2t)(2 sin t 2 sin 2t)] dt 21 œ "# '0 [6 (6 cos t cos 2t 6 sin t sin 2t)] dt œ "# '0 (6 6 cos t) dt œ 61 21 21 2. dx œ (2 sin t 2 sin 2t) dt and dy œ (2 cos t 2 cos 2t) dt; Area œ "# )C x dy y dx œ "# '0 [(2 cos t cos 2t)(2 cos t 2 cos 2t) (2 sin t sin 2t)(2 sin t 2 sin 2t)] dt 21 œ "# '0 [2 2(cos t cos 2t sin t sin 2t)] dt œ "# '0 (2 2 cos 3t) dt œ "# 2t 23 sin 3t‘ ! œ 21 21 21 #1 3. dx œ cos 2t dt and dy œ cos t dt; Area œ "# )C x dy y dx œ "# '0 ˆ "# sin 2t cos t sin t cos 2t‰ dt 1 œ #" '0 csin t cos# t (sin t) a2 cos# t 1bd dt œ "# '0 a sin t cos# t sin tb dt œ "# "3 cos$ t cos t‘ ! œ 3" 1 œ 23 1 1 4. dx œ (2a sin t 2a cos 2t) dt and dy œ (b cos t) dt; Area œ "# )C x dy y dx œ "# '0 ca2ab cos# t ab cos t sin 2tb a2ab sin# t 2ab sin t cos 2tbd dt 21 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 1 998 Chapter 16 Integration in Vector Fields œ "# '0 c2ab 2ab cos# t sin t 2ab(sin t) a2 cos# t 1bd dt œ "# '0 a2ab 2ab cos# t sin t 2ab sin tb dt 21 21 #1 œ "# 2abt 23 ab cos$ t 2ab cos t‘ ! œ 21ab 5. (a) F(xß yß z) œ zi xj yk is 0 only at the point (0ß 0ß 0), and curl F(xß yß z) œ i j k is never 0 . (b) F(xß yß z) œ zi yk is 0 only on the line x œ t, y œ 0, z œ 0 and curl F(xß yß z) œ i j is never 0 . (c) F(xß yß z) œ zi is 0 only when z œ 0 (the xy-plane) and curl F(xß yß z) œ j is never 0 . y j zk x i y j zk cy # 6. F œ yz# i xz# j 2xyzk and n œ Èxxi , so F is parallel to n when yz# œ cx # y # z# œ R R , xz œ R , # # yz xz c # # # # È2x. Also, and 2xyz œ cz R Ê x œ y œ 2xy Ê y œ x Ê y œ „ x and z œ „ R œ 2x Ê z œ „ x# y# z# œ R# Ê x# x# 2x# œ R# Ê 4x# œ R# Ê x œ „ R# . Thus the points are: Š R# ß R# ß Š R# ß R# ß È2R È2R È2R R R È2R R R R R È2R R R # ‹ , Š # ß # ß # ‹ , Š # ß # ß # ‹ , Š # ß # ß # ‹ , Š # ß # ß # ‹ , Š R# ß R# ß È2R È2R R R # ‹ , Š # ß # ß # ‹ È2R # ‹, 7. Set up the coordinate system so that (aß bß c) œ (0ß Rß 0) Ê $ (xß yß z) œ Èx# (y R)# z# œ Èx# y# z# 2Ry R# œ È2R# 2Ry ; let f(xß yß z) œ x# y# z# R# and p œ i ™f k 2R Ê ™ f œ 2xi 2yj 2zk Ê k ™ f k œ 2Èx# y# z# œ 2R Ê d5 œ kk™ f†ik dz dy œ 2x dz dy Ê Mass œ ' ' $ (xß yß z) d5 œ ' ' È2R# 2Ry ˆ Rx ‰ dz dy œ R ' ' Ryz S È c R œ 4R 'cR '0 R# y# Ryz È2R# 2Ry ÈR# y# z# dz dy œ 4R È2R# 2Ry ÈR# y# z# dz dy 'cRR É2R2 2Ry sin1 Š ÈR z y ‹» 2 ÈR cy # # dy 2 0 3/2 1‰ 2 œ 21R'cR É2R2 2Ry dy œ 21Rˆ 3R a2R 2Ryb » R R cR œ 1613R $ â â i j kâ â â â sin ) 0 â 8. r(rß )) œ (r cos ))i (r sin ))j )k , 0 Ÿ r Ÿ 1, 0 Ÿ ) Ÿ 21 Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) " â œ (sin ))i (cos ))j rk Ê krr ‚ r) k œ È1 r# ; $ œ 2Èx# y# œ 2Èr# cos# ) r# sin# ) œ 2r Ê Mass œ ' ' $ (xß yß z) d5 œ '0 21 S '01 2rÈ1 r# dr d) œ '021 ’ 23 a1 r# b$Î# “ " d) œ '021 23 Š2È2 1‹ d) ! œ 431 Š2È2 1‹ 9. M œ x# 4xy and N œ 6y Ê ``Mx œ 2x 4y and ``Nx œ 6 Ê Flux œ '0 '0 (2x 4y 6) dx dy b a œ '0 aa# 4ay 6ab dy œ a# b 2ab# 6ab. We want to minimize f(aß b) œ a# b 2ab# 6ab œ ab(a 2b 6). b Thus, fa (aß b) œ 2ab 2b# 6b œ 0 and fb (aß b) œ a# 4ab 6a œ 0 Ê b(2a 2b 6) œ 0 Ê b œ 0 or b œ a 3. Now b œ 0 Ê a# 6a œ 0 Ê a œ 0 or a œ 6 Ê (0ß 0) and (6ß 0) are critical points. On the other hand, b œ a 3 Ê a# 4a(a 3) 6a œ 0 Ê 3a# 6a œ 0 Ê a œ 0 or a œ 2 Ê (0ß 3) and (#ß ") are also critical points. The flux at (0ß 0) œ 0, the flux at (6ß 0) œ 0, the flux at (0ß 3) œ 0 and the flux at (2ß 1) œ 4. Therefore, the flux is minimized at (2ß 1) with value 4. 10. A plane through the origin has equation ax by cz œ 0. Consider first the case when c Á 0. Assume the plane is given by z œ ax by and let f(xß yß z) œ x# y# z# œ 4. Let C denote the circle of intersection of the plane with the sphere. By Stokes's Theorem, )C F † dr œ ' ' ™ ‚ F † n d5 , where n is a unit normal to the plane. Let S Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 16 Additional and Advanced Exercises 999 â â â i j kâ â â r(xß y) œ xi yj (ax by)k be a parametrization of the surface. Then rx ‚ ry œ â " 0 a â œ ai bj k â â â0 " bâ â i j k ââ â â â k ` ` Ê d5 œ krx ‚ ry k dx dy œ Èa# b# 1 dx dy. Also, ™ ‚ F œ â ` x ` y ``z â œ i j k and n œ Èaai #bjb #1 â â â z x y â Ê ' ' ™ ‚ F † n d5 œ ' ' a b 1 Èa# b# 1 dx dy œ ' ' (a b 1) dx dy œ (a b 1) ' ' dx dy. Now Rxy S x# y# (ax by)# œ 4 Ê È a# b# 1 Rxy # # ‰ Š a 4 1 ‹ x# Š b 4 1 ‹ y# ˆ ab # xy œ 1 Rxy Ê the region Rxy is the interior of the ellipse # # b 1 Ax# Bxy Cy# œ 1 in the xy-plane, where A œ a 4 1 , B œ ab # , and C œ 4 . The area of the ellipse is 21 41 È4AC B# œ Èa# b# 1 41(a b 1) 1) Ê )C F † dr œ h(aß b) œ È . Thus we optimize H(aß b) œ (aa#bb #1 : a# b # 1 # # 2(a b 1) ab# 1 a abb `H œ 0 and ``Hb œ 2(a b aa#1)aab# 11b2 b abb œ 0 à œ aa# b# 1b2 Ê a b 1 œ 0, or b# 1 a ab œ 0 and a# 1 b ab œ 0 Ê a b 1 œ 0, or a# b# (b a) œ 0 Ê a b 1 œ 0, or (a b)(a b 1) œ 0 Ê a b 1 œ 0 or a œ b. The critical values a b 1 œ 0 give a saddle. If a œ b, then 0 œ b# 1 a ab Ê a# 1 a a# œ 0 Ê a œ 1 Ê b œ 1. Thus, the point (aß b) œ (1ß 1) gives a local extremum for )C F † dr Ê z œ x y Ê x y z œ 0 is the desired plane, if c Á 0. Note: Since h(1ß 1) is negative, the circulation about n is clockwise, so n is the correct pointing normal for the counterclockwise circulation. Thus ' ' ™ ‚ F † (n) d5 actually gives the maximum circulation. S If c œ 0, one can see that the corresponding problem is equivalent to the calculation above when b œ 0, which does not lead to a local extreme. 11. (a) Partition the string into small pieces. Let ?i s be the length of the ith piece. Let (xi ß yi ) be a point in the ith piece. The work done by gravity in moving the ith piece to the x-axis is approximately Wi œ (gxi yi ?i s)yi where xi yi ?i s is approximately the mass of the ith piece. The total work done by gravity in moving the string to the x-axis is D Wi œ D gxi y#i ?i s Ê Work œ 'C gxy# ds i i 1/2 1/2 (b) Work œ 'C gxy# ds œ '0 g(2 cos t) a4 sin# tbÈ4 sin# t 4 cos# t dt œ 16g '0 cos t sin# t dt $ œ ’16g Š sin3 t ‹“ 1Î# ! œ 16 3 g ' x(xy) ds ' y(xy) ds (c) x œ 'C and y œ 'C ; the mass of the string is 'C xy ds and the weight of the string is xy ds xy ds C C g 'C xy ds. Therefore, the work done in moving the point mass at axß yb to the x-axis is W œ Šg 'C xy ds‹ y œ g 'C xy# ds œ 16 3 g. 12. (a) Partition the sheet into small pieces. Let ?i 5 be the area of the ith piece and select a point (xi ß yi ß zi ) in the ith piece. The mass of the ith piece is approximately xi yi ?i 5 . The work done by gravity in moving the ith piece to the xy-plane is approximately (gxi yi ?i 5 )zi œ gxi yi zi ?i 5 Ê Work œ ' ' gxyz d5. (b) ' ' gxyz d5 œ g ' ' xy(1 x y)È1 (1)# (1)# dA œ È3g Rxy S ' 1 1cx '0 '0 1 S axy x# y xy# b dy dx ' 1 "x œ È3g 0 "2 xy# "# x# y# "3 xy$ ‘ ! dx œ È3g 0 "6 x #" x# "# x$ "6 x% ‘ dx È " # " &‘ " " ‰ œ È3g 12 x 6" x$ 6" x% 30 x ! œ È3g ˆ 1"# 30 œ 203g Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. 1000 Chapter 16 Integration in Vector Fields (c) The center of mass of the sheet is the point axß yß zb where z œ Mxy with Mxy œ ' ' xyz d5 and M S M œ ' ' xy d5. The work done by gravity in moving the point mass at axß yß zb to the xy-plane is S gMz œ gM Š MMxy ‹ œ gMxy œ ' ' gxyz d5 œ S È3g 20 . 13. (a) Partition the sphere x# y# (z 2)# œ 1 into small pieces. Let ?i 5 be the surface area of the ith piece and let (xi ß yi ß zi ) be a point on the ith piece. The force due to pressure on the ith piece is approximately w(4 zi )?i 5 . The total force on S is approximately D w(4 zi )?i 5. This gives the actual force to be ' ' w(4 z) d5. i S (b) The upward buoyant force is a result of the k-component of the force on the ball due to liquid pressure. The force on the ball at (xß yß z) is w(4 z)(n) œ w(z 4)n , where n is the outer unit normal at (xß yß z). Hence the k-component of this force is w(z 4)n † k œ w(z 4)k † n . The (magnitude of the) buoyant force on the ball is obtained by adding up all these k-components to obtain ' ' w(z 4)k † n d5. S (c) The Divergence Theorem says ' ' w(z 4)k † n d5 œ ' ' ' div(w(z 4)k) dV œ ' ' ' w dV, where D D S D is x# y# (z 2)# Ÿ 1 Ê ' ' w(z 4)k † n d5 œ w ' ' ' 1 dV œ 43 1w, the weight of the fluid if it D S were to occupy the region D. 14. The surface S is z œ Èx# y# from z œ 1 to z œ 2. Partition S into small pieces and let ?i 5 be the area of the ith piece. Let (xi ß yi ß zi ) be a point on the ith piece. Then the magnitude of the force on the ith piece due to liquid pressure is approximately Fi œ w(2 zi )?i 5 Ê the total force on S is approximately # D Fi œ D w(2 zi )?i 5 Ê the actual force is ' ' w(2 z) d5 œ ' ' w ˆ2 Èx# y# ‰ É1 x# x y# x# y y# dA # i S Rxy # œ ' ' È2 w ˆ2 Èx# y# ‰ dA œ '0 '1 È2w(2 r) r dr d) œ '0 È2w r# 3" r$ ‘ " d) œ '0 21 2 21 21 Rxy È 2È2w d) œ 4 321w 3 15. Assume that S is a surface to which Stokes's Theorem applies. Then )C E † dr œ ' ' ( ™ ‚ E) † n d5 S œ ' ' ˆ ``Bt ‰ † n d5 œ ``t ' ' B † n d5. Thus the voltage around a loop equals the negative of the rate of S S change of magnetic flux through the loop. 16. According to Gauss's Law, ' ' F † n d5 œ 41GmM for any surface enclosing the origin. But if F œ ™ ‚ H S then the integral over such a closed surface would have to be 0 by the Divergence Theorem since div F œ 0. 17. )C f ™ g † dr œ ' ' ™ ‚ (f ™ g) † n d5 (Stokes's Theorem) S œ ' ' (f ™ ‚ ™ g ™ f ‚ ™ g) † n d5 (Section 16.8, Exercise 19b) S œ ' ' [(f)(0) ™ f ‚ ™ g] † n d5 (Section 16.7, Equation 8) S œ ' ' ( ™ f ‚ ™ g) † n d5 S 18. ™ ‚ F" œ ™ ‚ F# Ê ™ ‚ (F# F" ) œ 0 Ê F# F" is conservative Ê F# F" œ ™ f; also, ™ † F" œ ™ † F# Ê ™ † (F# F" ) œ 0 Ê ™ # f œ 0 (so f is harmonic). Finally, on the surface S, ™ f † n œ (F# F" ) † n œ F# † n F" † n œ 0. Now, ™ † (f ™ f) œ ™ f † ™ f f ™ # f so the Divergence Theorem gives Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. Chapter 16 Additional and Advanced Exercises 1001 ' ' ' k ™ f k# dV ' ' ' f ™ # f dV œ ' ' ' ™ † (f ™ f) dV œ ' ' f ™ f † n d5 œ 0, and since ™ # f œ 0 we have D D D S ' ' ' k ™ f k# dV 0 œ 0 Ê ' ' ' kF# F" k # dV œ 0 Ê F# F" œ 0 Ê F# œ F" , as claimed. D D 19. False; let F œ yi xj Á 0 Ê â i â â ™ † F œ ``x (y) ``y (x) œ 0 and ™ ‚ F œ â ``x â â x j ` `y y k ââ ` â ` z â œ 0i 0j 0k œ 0 â 0 â 20. kru ‚ rv k# œ kru k# krv k# sin# ) œ kru k# krv k# a1 cos# )b œ kru k# krv k# kru k# krv k# cos# ) œ kru k# krv k# (ru † rv )# Ê kru ‚ rv k# œ EG F# Ê d5 œ kru ‚ rv k du dv œ ÈEG F# du dv 21. r œ xi yj zk Ê ™ † r œ 1 1 1 œ 3 Ê ' ' ' ™ † r dV œ 3 ' ' ' dV œ 3V Ê V œ "3 ' ' ' ™ † r dV œ "3 ' ' r † n d5, by the Divergence Theorem D D S Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley. D

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