Thomas Calculus 11E Giordano

  

to accompany

T HOMAS

  

’ C

ALCULUS E LEVENTH E DITION

B ASED ON THE O RIGINAL W ORK BY

George B. Thomas, Jr.

  

Massachusetts Institute of Technology

AS R EVISED BY

  

Maurice D. Weir

Naval Postgraduate School

  

J oel Hass

University of California, Davis

  

Frank R. Giordano

Naval Postgraduate School

  I NSTRUCTOR ’ S S OLUTIONS M ANUAL P ART O NE

  

A RDIS • B ORZELLINO • B UCHANAN • M OGILL • N ELSON

  Reproduced by Pearson Addison-Wesley from electronic files supplied by the authors. Copyright © 2005 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 0-321-22653-4 1 2 3 4 5 6 BB 07 06 05 04

PREFACE TO THE INSTRUCTOR

  This Instructor's Solutions Manual contains the solutions to every exercise in the 11th Edition of THOMAS' CALCULUS by Maurice Weir, Joel Hass and Frank Giordano, 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 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 ì

  MATHEMATICA

  Every CAS exercise is solved in both the MAPLE and 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).

  Acknowledgments

  Solutions Writers William Ardis, Collin County Community College-Preston Ridge Campus Joseph Borzellino, California Polytechnic State University Linda Buchanana, Howard College

  Tim Mogill Patricia Nelson, University of Wisconsin-La Crosse

  Accuracy Checkers Karl Kattchee, University of Wisconsin-La Crosse Marie Vanisko, California State University, Stanislaus Tom Weigleitner, VISTA Information Technologies

  Thanks to Rachel Reeve, Christine O'Brien, Sheila Spinney, Elka Block, and Joe Vetere for all their guidance and help at every step.

  CHAPTER 1 PRELIMINARIES

1.1 REAL NUMBERS AND THE REAL LINE

  "

  2

  3

  8

  9

  1. Executing long division, 0.1, 0.2, 0.3, 0.8, 0.9 œ œ œ œ œ

  9

  9

  9

  9

  9 "

  2

  3

  9

  11

  2. Executing long division, 0.09, 0.18, 0.27, 0.81, 0.99 œ œ œ œ œ

  11

  11

  11

  11

  11 3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6.

  a) NNT. 5 is a counter example.

  b) NT. 2 < x < 6 Ê 2 2 < x 2 < 6 2 Ê 0 < x 2 < 2.

  c) NT. 2 < x < 6 2/2 < x/2 < 6/2 1 < x < 3.

  Ê Ê d) NT. 2 < x < 6 1/2 > 1/x > 1/6 1/6 < 1/x < 1/2. Ê Ê e) NT. 2 < x < 6 1/2 > 1/x > 1/6 1/6 < 1/x < 1/2 6(1/6) < 6(1/x) < 6(1/2) 1 < 6/x < 3. Ê Ê Ê Ê f) NT. 2 < x < 6 Ê x < 6 Ê (x 4) < 2 and 2 < x < 6 Ê x > 2 Ê x < 2 Ê x + 4 < 2 Ê (x 4) < 2.

  The pair of inequalities (x 4) < 2 and (x 4) < 2 Ê | x 4 | < 2.

  g) NT. 2 < x < 6 Ê 2 > x > 6 Ê 6 < x < 2. But 2 < 2. So 6 < x < 2 < 2 or 6 < x < 2.

  h) NT. 2 < x < 6 Ê 1(2) > 1(x) < 1(6) Ê 6 < x < 2 4. NT = necessarily true, NNT = Not necessarily true. Given: 1 < y 5 < 1.

  a) NT. 1 < y 5 < 1 1 + 5 < y 5 + 5 < 1 + 5 4 < y < 6.

  Ê Ê

  b) NNT. y = 5 is a counter example. (Actually, never true given that 4 y 6) c) NT. From a), 1 < y 5 < 1, 4 < y < 6 y > 4.

  Ê Ê d) NT. From a), 1 < y 5 < 1, Ê 4 < y < 6 Ê y < 6.

  e) NT. 1 < y 5 < 1 Ê 1 + 1 < y 5 + 1 < 1 + 1 Ê 0 < y 4 < 2.

  f) NT. 1 < y 5 < 1 Ê (1/2)( 1 + 5) < (1/2)(y 5 + 5) < (1/2)(1 + 5) Ê 2 < y/2 < 3.

  g) NT. From a), 4 < y < 6 Ê 1/4 > 1/y > 1/6 Ê 1/6 < 1/y < 1/4.

  h) NT. 1 < y 5 < 1 Ê y 5 > 1 Ê y > 4 Ê y < 4 Ê y + 5 < 1 Ê (y 5) < 1.

  Also, 1 < y 5 < 1 Ê y 5 < 1. The pair of inequalities (y 5) < 1 and (y 5) < 1 Ê | y 5 | < 1. 5. 2x 4 Ê x

  2 6. 8 3x   5 Ê 3x   3 Ê x Ÿ 1 ïïïïïïïïïñqqqqqqqqp x

  1

  5 7. 5x $ Ÿ ( 3x Ê 8x Ÿ 10 Ê x Ÿ

  4 8. 3(2 x) 2(3 x) Ê 6 3x 6 2x

  Ê 5x Ê x ïïïïïïïïïðqqqqqqqqp x " 7 "

  7 9. 2x   7x Ê   5x

  # 6 #

  6 " 10 "

  Ê ˆ ‰   x or   x

  5

  6

  3 6 x 3x 4 10. Ê 12 2x 12x

  16

  4

  2 Ê 28 14x Ê 2 x qqqqqqqqqđỉỉỉỉỉỉỉỉĩ x

  2

  2

Chapter 1 Preliminaries

  4 "

  11. (x 2) (x 6) Ê 12(x 2) 5(x 6)

  5

  3

  6 Ê 12x 24 5x 30 Ê 7x 6 or x

  7 x 5 12 3x 12. (4x 20) 24 6x

  Ÿ Ê Ÿ

  2

  4

  22 Ê 44 Ÿ 10x Ê Ÿ x qqqqqqqqqñïïïïïïïïî x

  5 22/5

  13. y œ 3 or y œ

  3 14. y 3 7 or y

  3 7 y 10 or y

  4 œ œ Ê œ œ

  "

  9 15. 2t 5 4 or 2t 4 2t 1 or 2t 9 t or t

  œ & œ Ê œ œ Ê œ œ # #

  16. 1 œ t 1 or 1 œ t 1 Ê œ ! t or t œ 2 Ê t œ 0 or t œ

  2

  9

  9

  7

  25

  7

  25 17. 8 3s œ or 8 3s œ Ê 3s œ or 3s œ Ê s œ or s œ

  2 # # #

  6

  6 s s s s 18. œ 1 1 or œ

  1 1 Ê œ 2 or œ ! Ê s œ 4 or s œ # # # # 19. 2 x 2; solution interval ( 2 2) ß 20. 2 x 2; solution interval [ 2 2] x

  Ÿ Ÿ ß qqqqñïïïïïïïïñqqqqp

  2

  2 21. Ÿ Ÿ 3 t

  1 3 Ê Ÿ Ÿ 2 t 4; solution interval [ 2 4] ß 22. 1 t 2 1 3 t 1;

  Ê solution interval ( 3 1) t

  ß qqqqđỉỉỉỉỉỉỉỉđqqqqp

  3

  1

  11 23. % 3y

  7 4 Ê 3 3y 11 Ê 1 y ;

  3

  11 solution interval ˆ 1 ß ‰

  3 24. 1 2y " Ê

  5 6 2y 4 Ê 3 y 2; solution interval ( 3 2) y

  ß qqqqđỉỉỉỉỉỉỉỉđqqqqp

  3

  2 z z

  25. 1 Ÿ Ÿ

  1 1 Ê Ÿ Ÿ 2 Ê Ÿ Ÿ z 10;

  5

  5 solution interval [0 10] ß

  3z 3z

  2 26. 2 Ÿ Ÿ

  1 2 Ê Ÿ 1 Ÿ 3 Ê Ÿ Ÿ z 2; # #

  3

  2 solution interval 2 ‘ z

  ß qqqqñïïïïïïïïñqqqqp

  3 2/3

  2 " " " 7 "

  5 7 "

  5 27.

  3 Ê Ê

  # x # # x # # x #

  2 2 2 2 x ; solution interval ˆ ‰

  Ê ß

  7 5 7 5

  2 2 x " 28. 3

  4 3 Ê 1 ( Ê

  1 x x

  7 #

Section 1.1 Real Numbers and the Real Line

  3

  29. 2s   4 or 2s   4 Ê s   2 or s Ÿ 2; solution intervals ( _ß 2] [2 ß _ )

  5

  7 " "

  30. s   3 or (s 3)   Ê s   or s   # # # #

  5

  7 s or s ;

  Ê   Ÿ # #

  7

  5 solution intervals ˆ _ß ‘ ß _ ‰ ïïïïïïñqqqqqqñïïïïïïî s

  # # 7/2 5/2

  31. 1 x 1 or ( " x) 1 Ê x 0 or x

  2 Ê x 0 or x 2; solution intervals ( _ß ! ) (2 ß _ )

  32. 2 3x 5 or (2 3x) 5 3x 3 or 3x

  7 Ê

  7 x 1 or x ;

  Ê

  3

  7 solution intervals ( _ß 1) ˆ ß _ ‰ ỉỉỉỉỉỉđqqqqqqđỉỉỉỉỉỉĩ x

  3 1 7/3 r r 1

  " 33.   1 or ˆ ‰   1 Ê r   1 2 or r Ÿ

  1

  2 # #

  Ê r   1 or r Ÿ 3; solution intervals ( _ß 3] [1 ß _ ) 3r 2 3r

  2 34. " or ˆ " ‰

  5

  5

  5

  5 3r 7 3r

  3

  7 or r or r

  1 Ê Ê

  5

  5

  5

  5

  3

  7 solution intervals ( _ß " ) ˆ ß _ ‰ ỉỉỉỉỉỉđqqqqqqđỉỉỉỉỉỉĩ r

  3 1 7/3

  # È È

  35. x # Ê x 2 2 Ê x 2 ; k k È È È solution interval

  2 ß 2 qqqqqqđỉỉỉỉỉỉđqqqqqqp x Š ‹

  È È # #

  # 36. 4 x 2 x x 2 or x 2;

  Ÿ Ê Ÿ Ê   Ÿ k k solution interval ( _ß 2] [2 ß _ ) ïïïïïïñqqqqqqñïïïïïïî r

  2

  2 #

  37. 4 x 9 Ê 2 x 3 Ê 2 x 3 or 2 x

  3 k k 2 x 3 or 3 x 2;

  Ê solution intervals ( 3 ß 2) (2 3) ß qqqqđỉỉỉỉđqqqqđỉỉỉỉđqqqp x

  3

  2

  2

  3 " # " " " " " " "

  9

  4 3 # 3 # 3 # " " " " x or x ;

  38. x x Ê x Ê or x k k

  Ê

  3

  3 # #

  " " " " solution intervals ˆ ß ‰ ˆ ß ‰ qqqqđỉỉỉỉđqqqqđỉỉỉỉđqqqp x

  #

  3 3 # 1/2 1/3 1/3 1/2

  # 39. (x 1) 4 Ê x

  1 2 Ê 2 x

  1

  2 k k Ê 1 x 3; solution interval ( "ß $ ) qqqqqqđỉỉỉỉỉỉỉỉđqqqqp x

  1

  3 #

  40. (x 3) # Ê x

  3

  2 k k È È È È È

  Ê 2 x 3 2 or 3 2 x 3 2 ; È È solution interval

  3 2 ß

  3 2 qqqqqqđỉỉỉỉỉỉỉỉđqqqqp x Š ‹

  È È 3 # 3 #

  4

Chapter 1 Preliminaries

  b) or a b (a b); k k k k œ œ both squared equal (a b)

  # # Ÿ Ÿ Ÿ

  (4) x y implies x y or x y for all nonnegative real numbers x and y. Let x a b and # #

  # # # #

  a, so a a ; likewise, b b k k k k k k k k œ œ œ œ

  (3) a a or a

  # (2) ab ab a b Ÿ œ k k k k k k

  45. (1) a b (a

  È k k y a b so that a b a b a b a b . œ Ÿ Ê Ÿ k k k k k k a b k k k k k k k k k k

  1 k k k k œ Í œ Í   Í Ÿ

  1 1 x (x 1) 1 x 1 x x

  43. True if a 0; False if a 0.   44. x

  ˆ ‰ The solution interval is ( 1] [2 ) _ß ß _

  2 ¹ ¹

  2

  œ È

  # #

  41. x x x x + < x < x < < x < 0 < x < 1.

  % % | 2x + 3 3 | < . But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) f(0) | < . %

  a. So again œ Ê œ œ œ | a | | a|. œ

  b. By definition, | b | b and thus | a|

  a. œ œ œ œ ii) For a < 0, | a |

  | a | ( a) a and | a | | a |

  b. Since b = a, œ Ê œ

  % 51. Consider: i) a > 0; ii) a < 0; iii) a = 0.

  50. Let > 0 be any positive number and f(x) = 2x + 3. Suppose that | x 0 | < /2. Then 2| x 0 | < and %

  46. If a 0 and b 0, then ab 0 and ab ab a b .       œ œ k k k k k k If a 0 and b 0, then ab 0 and ab ab ( a)( b) a b . œ œ œ k k k k k k If a 0 and b 0, then ab 0 and ab (ab) (a)( b) a b .   Ÿ œ œ œ k k k k k k If a 0 and b 0, then ab 0 and ab (ab) ( a)(b) a b .

  | 2x | < 2 | (2x + 1) 3 | < 2 | f(x) f(1) | < 2 # Ê Ê $ $ $

  $ $ $ Ê Ê

  49. Let be a real number > 0 and f(x) = 2x + 1. Suppose that | x 1 | < . Then | x 1 | < 2| x 1 | < 2 $

  48. Graph of x y 1 is the interior k k k k Ÿ of “diamond-shaped" region.

  " " # #

    Ÿ œ œ œ k k k k k k 47. 3 x 3 and x x 3. Ÿ Ÿ Ê Ÿ

  2

  2

  2

  1

  4

  2

  4

  4

  1

  1

  1

  2

  1

  1

  1

  1

  1

  # # Ê Ê Ê Ê Ê

  2

  2

  2

  3

  4

  4

  3

  1

  3

  1

  1

  2

  9

  1

  # #   Ê   Ê   Ê     Ê   Ÿ

  42. x x 2 x x + x x or x x 2 or x 1.

  ¹ ¹ So the solution is the interval (0 1) ß

  2 ˆ ‰

  2

i) For a > 0, | a | a by definition. Now, a > 0 a < 0. Let a = b. By definition, | b |

a. Now, a < 0 a > 0. Let a

Section 1.2 Lines, Circles and Parabolas

  5

54. Prove S a a for any real number a and any positive integer n. n

1.2 LINES, CIRCLES, AND PARABOLAS

  È

  œ $ œ œ œ œ œ È È

  # # 3. x 8.1 ( 3.2) 4.9, y 2 ( 2) 0; d ( 4.9)

  4.9 ? ?

  œ œ œ œ œ œ È

  # # 4. x 2 2, y

  1.5 4 2.5; d 2 ( 2.5)

  8.25 ? ?

  œ œ œ œ œ œ È È È

  ÊŠ ‹ È

  # # 5. Circle with center ( ) and radius 1.

  6. Circle with center ( ) and radius 2. !ß ! !ß !

  ? ?

  !ß ! È 8. The origin (a single point).

  9. m 3 10. m œ œ œ œ œ œ

  2

  ? ? y y x

  2 ( 1) x 2 ( 2)

  4

  1

  2 3 # " perpendicular slope perpendicular slope

  œ œ

  "

  3

  3

  4

  5 ? ?

  4

  52. i) Prove | x | &gt; 0 x &gt; a or x &lt; a for any positive number, a. Ê For x 0, | x | x. | x | &gt; a x &gt; a.

  ¸ ¸ " "

    œ Ê For x &lt; 0, | x | x. | x | &gt; a x &gt; a x &lt; a.

  œ Ê Ê ii) Prove x &gt; a or x &lt; a | x | &gt; 0 for any positive number, a. Ê a &gt; 0 and x &gt; a | x | x. So x &gt; a | x | &gt; a. Ê œ Ê For a &gt; 0, a &lt; 0 and x &lt; a x &lt; 0 | x | x. So x &lt; a x &gt; a | x | &gt; a. Ê Ê œ Ê Ê

  53. a) 1 = 1 | 1 | = 1 b b Ê Ê œ Ê œ Ê œ Ê œ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹

  † †

  " " " l l l l l l l l b b b b b b b b b b b b b

  ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹

  † † " b

  "

  b) a a a l l l l l l l l l l

  " " a a b b b b b

  œ œ œ œ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹

  † † † "

  n n œ œ k k k k a a a, so S is true. Now, assume that S a a is true form some positive integer . k k k k k k

  " œ œ œ œ

  # # 2. x ( 1) 2, y 2 ( 2) 4; d ( 2)

  5 k k k

  Since a a and a a , we have a a a a a a a a . Thus, k k k k k k k k k k k k k k ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸

  " " " " " " " œ œ œ † œ œ œ k k k k k k k+

  S a a is also true. Thus by the Principle of Mathematical Induction, S a a k n k k+ n n

  " " "

  œ œ œ l l œ l l ¸ ¸ k k is true for all n positive integers.

  1. x 1 ( 3) 2, y

  2 2 4; d ( x) ( y)

  4

  16

  2

  5 ? ? ? ?

  œ œ œ œ œ œ œ È È È

7. Disk (i.e., circle together with its interior points) with center ( ) and radius 3.

  6

Chapter 1 Preliminaries

  ? ? y

  3

  3 y #

  11. m œ œ œ 12. m œ œ ; no slope ?

  1

  2 x # # ( ) perpendicular slope does not exist perpendicular slope

  ? x

  œ

  1 È

  13. (a) x 1 14. (a) x 2 15. (a) x 16. (a) x œ œ œ œ

  4 È

  (b) y œ (b) y œ 1.3 (b) y œ 2 (b) y œ

  3

  17. P( 1 1), m ß œ 1 Ê y œ 1 1 x ( 1) Ê y œ x a b " " "

  18. P(2 ß 3), m œ Ê y ( 3) œ (x 2) Ê y œ x

  4 # # #

  ? y 5 4 " " "

  23

  19. P(3 4), Q( 2 5) m y 4 (x 3) y x ß ß Ê œ œ œ Ê œ Ê œ

  2

  3

  5

  5

  5

  5 ? y

  ? x

  3

  3

  3

  3

  24

  20. P( 8 0), Q( 1 3) ß ß Ê m œ œ œ Ê y œ x ( 8) Ê y œ x ? a b x

  1 ( 8)

  7

  7

  7

  7

  5 5 " " 21. m œ , b œ 6 Ê y œ x 6 22. m œ , b œ 3 Ê y œ x

  3

  4

  4 # #

  " " 23. m œ 0, P( 12 ß 9) Ê y œ

  9

24. No slope, P ˆ ß % Ê ‰ x œ

  3

  3 ? y

  4 25. a 1, b 4 (0 4) and ( 0) are on the line m 4 y 4x

  4 œ œ Ê ß "ß Ê œ œ œ Ê œ

  1 ? y

  ? x

  6 26. a œ 2, b œ 6 Ê (2 0) and ( ß !ß 6) are on the line Ê m œ œ œ 3 Ê y œ 3x

  6 ? x

  2

  2

  2

  2

  27. P(5 1), L: 2x 5y 15 m parallel line is y ( 1) (x 5) y x

  1 ß œ Ê L œ Ê œ Ê œ

  5

  5

  5 È 2 È 2 È

  2

  8 È È È

  È

  28. P 2 2 , L: ß 2x 5y œ 3 Ê m L œ Ê parallel line is y œ 2 x 2 Ê y œ x Š ‹

  Š Š ‹ ‹

  5

  5

  5

  5 " " "

  29. P(4 10), L: 6x ß 3y œ 5 Ê m œ 2 Ê m œ Ê perpendicular line is y 10 œ (x 4) Ê y œ x

  12 L ¼ # # #

  8

  13

  13

  30. P( 1), L: 8x 13y 13 m m perpendicular line is y x

  1 !ß œ Ê L œ Ê ¼ œ Ê œ

  13

  8

  8

Section 1.2 Lines, Circles and Parabolas

  7

  31. x-intercept œ 4, y-intercept œ 3 32. x-intercept œ 4, y-intercept œ

  2 È È

  33. x-intercept œ 3, y-intercept œ 2 34. x-intercept œ 2, y-intercept œ

  3 A C " B C # A B

  35. Ax By C y x and Bx Ay C y x . Since ˆ ‰ ˆ ‰ 1 is the œ " Í œ œ # Í œ œ

  B B A A B A product of the slopes, the lines are perpendicular. A C A C

  " #

  36. Ax By œ C Í y œ x and Ax By œ C Í y œ x . Since the lines have the same " #

  B B B B A slope , they are parallel. B

  ? ?

  37. New position œ x old x y ß old y œ # &amp;ß ( 3 ( 6)) œ $ß ( 3). a b

  ? ?

  38. New position œ x x y ß y œ (6 ß ( 6) 0 0) œ (0 0). ß a old old b ? ? ?

  39. x œ 5, y œ 6, B(3 ß 3). Let A œ (x y). Then x ß œ x x Ê 5 œ 3 x Ê x œ 2 and # "

  ? y œ y y Ê 6 œ 3 y Ê y œ 9. Therefore, A œ #ß ( 9).

  # " ? ?

  40. x œ " " œ ! , y œ ! ! œ !

  8

Chapter 1 Preliminaries

  # # # #

  41. C( 2), a !ß œ 2 Ê x (y 2) œ

  4

  42. C( $ß 0), a œ 3 Ê (x 3) y œ

  9 # #

  È

  43. C( 1 5), a 10 (x 1) (y 5)

  10 ß œ Ê œ

  # # È

  44. C( "ß " ), a œ 2 Ê (x 1) (y 1) œ

  2 # # # x œ Ê (0 1) (y 1) œ 2 Ê (y 1) œ

  1 Ê y œ „

  1 1 Ê y œ 0 or y œ 2. Similarly, y œ Ê x œ 0 or x œ

  2 #

  # È È

  45. C 3 ß 2 , a œ 2 Ê x 3 (y 2) œ 4, Š ‹ Š ‹

  # # #

  È x œ Ê 3 (y 2) œ 4 Ê (y 2) œ

  1 Š ‹ y

  2 1 y 1 or y

3. Also, y

  Ê œ „ Ê œ œ œ # #

  # È È x

  3 (0 2) 4 x

  3 Ê œ Ê œ

  Š ‹ Š ‹ Ê x œ È

  3 #

  " # "

  46. C 3 ˆ ‰ , a 5 (x 3) ˆ y ‰ 25, so ß œ Ê œ

  # # #

  # " x œ Ê (0 3) ˆ y ‰ œ

  25 #

  # " "

  9 ˆ y ‰ 16 y 4 y

  Ê œ Ê œ „ Ê œ # # # 7 " #

  # or y œ . Also, y œ Ê (x 3) ˆ ‰ œ

  25 # #

  È

  3

  11 #

  99 Ê (x 3) œ Ê x œ „

  3 4 # 3 È

  11 Ê x œ „

  3 #

Section 1.2 Lines, Circles and Parabolas

  9

  # # 47. x y 4x 4y % œ

  # # Ê x %B y 4y œ

  4 # #

  Ê x 4x 4 y 4y œ

  4

  4 # #

  Ê (x 2) (y 2) œ 4 Ê C œ ß ( 2 2), a œ 2.

  # # 48. x y 8x 4y 16 œ

  # # Ê x 8x y 4y œ

  16 # # x 8x 16 y 4y

  4

  4 Ê œ

  # # (x 4) (y 2)

  4 Ê œ

  C ( 2), a 2. Ê œ %ß œ

  # # # # 49. x y 3y œ 4 Ê x y 3y œ

  4 # #

  9

  25 x y 3y

  Ê œ

  4

  4 #

  3

  25

  3 #

  Ê x ˆ y ‰ œ Ê C œ ˆ ß ‰ , # 4 #

  5 a .

  œ #

  # #

  9 50. x y 4x

  œ

  4

  9 # #

  Ê x 4x y œ

  4 # #

  25 x 4x 4 y

  Ê œ

  4

  25 # #

  Ê (x 2) y œ

  4

5 C (2 0), a .

  Ê œ ß œ #

  # # 51. x y 4x 4y

  œ # #

  Ê x 4x y 4y œ # #

  Ê x 4x 4 y 4y œ

  4

  8 # #

  (x Ê 2) (y 2) œ

8 C(2 2), a 8.

  Ê ß œ È

  10

Chapter 1 Preliminaries

  3 # #

  1 Ê œ œ

  4 # a 2( 1) y (2) 4(2) œ 4 Ê œ

  # V (2 4). If x 0 then y 0. Ê œ ß œ œ

  Also, y x 4x œ Ê œ # x(x 4) x 4 or x 0. Ê œ Ê œ œ

  Axis of parabola is x 2. œ 56. x œ 2 œ œ b

  4 a 2( 1) # y (2) 4(2)

  5

  # V (2 1). If x 0 then y 5. Ê œ ß œ œ

  3. Axis of parabola is x 2. œ œ 55. x

  Also, y x 4x Ê 5 œ œ # x 4x Ê 5 x œ Ê œ

  # „

  #

  4

  4 È no x intercepts. Axis of parabola is x 2.

  Ê œ

  2 œ œ œ b

  Ê œ Ê œ x

  œ x 2x 1 y

  2 a 2(1) # y (1) 2(1)

  4 Ê œ

  # # (x 1) y

  4 Ê œ

  # # C ( 1 0), a 2. Ê œ ß œ

  53. x

  1 œ œ œ b

  3 œ 4 Ê œ #

  52. x y 2x

  V ( 4). If x 0 then y 3. Ê œ "ß œ œ Also, y x 2x Ê 3 œ œ

  # (x 3)(x 1) x 3 or Ê œ Ê œ x

  1. Axis of parabola is x 1. œ œ 54. x œ 2 œ œ b

  4 # a 2(1) y ( 2) 4( 2)

  3 œ 1 Ê œ #

  V ( 2 1). If x 0 then y 3. Ê œ ß œ œ Also, y x 4x

  3 œ Ê œ

  # (x 1)(x 3) x 1 or

Section 1.2 Lines, Circles and Parabolas

  11

1. Axis of parabola is x 3.

23 V . If x 0 then y 3.

2 V . If x 0 then y 4.

  2 # a 2( 1/4) y (4) 2(4)

  Ê œ ß œ œ Also, y 0 x 2x Ê 4 œ œ

  " #

  8 Ê œ œ

  4

  Axis of parabola is x 1.

  4 œ œ œ b

  œ 60. x

  4 x

  1 È

  7

  " #

  2

  4

  4 2. Ê œ œ „ „

  Ê œ Ê „

  8 1/2

  È È

  Axis of parabola is x 4. œ

  61. The points that lie outside the circle with center ( 0) and radius 7.

  !ß È

  62. The points that lie inside the circle with center ( 0) and radius 5. !ß È

  63. The points that lie on or inside the circle with center ( 0) and radius 2. "ß

  64. The points lying on or outside the circle with center ( 2) and radius 2. !ß

  65. The points lying outside the circle with center ( 0) and radius 1, but inside the circle with center ( 0), !ß !ß and radius 2 (i.e., a washer).

  1

  # x no x intercepts.

  57. x œ 3 œ œ b

  4

  6 # a 2( 1) y ( 3) 6( 3)

  5 œ 4 Ê œ #

  V ( 3 ). If x 0 then y 5. Ê œ ß % œ œ Also, y x 6x Ê 5 œ œ

  # (x 5)(x 1) x 5 or Ê œ Ê œ x

  œ œ 58. x œ œ œ b

  1 a 2(2) 4 # " y

  2 œ 3 Ê œ ˆ ‰ " "

  #

  4

  4

  8

  Ê œ ß œ œ ˆ ‰

  "

  8

  2 Also, y x x Ê 4 œ œ " #

  23 Also, y 2x x Ê 3 œ œ # x no x intercepts. Ê œ Ê

  1

  23

  4 „

  È Axis of parabola is x . œ

  "

  4 59. x œ 1 œ œ b

  1 # a 2(1/2) y ( 1) ( 1) œ 4 Ê œ

  " #

  #

  7

  Ê œ "ß œ œ ˆ ‰

  7

4 V (4 8) . If x 0 then y 4.

  12

Chapter 1 Preliminaries

  2

  67. x y 6y x (y 3) 9.

  # # # # Ê

  The interior points of the circle centered at ( 3) with radius 3, but above the line !ß y

  3. œ 68. x y 4x 2y 4 (x 2) (y 1) 9.

  # # # # Ê

  The points exterior to the circle centered at (2 1) with radius 3 and to the right of the ß line x

  2. œ 69. (x 2) (y 1) 6 70. (x 4) (y 2)

  16 71. x y 2, x 1 72. x y 4, (x 1) (y 3)

  10 # #

  # # # # Ÿ  

  73. x y 1 and y 2x 1 x 4x 5x # # # # #

  œ œ Ê œ œ Ê x and y or x and y . œ œ œ œ

  66. The points on or inside the circle centered at ( ) with radius 2 and on or inside the !ß ! circle centered at ( 2 0) with radius 2. ß

2 Thus, A , B are the

  5

  5

  5

  5

  Š ‹ Š ‹ " "

  È È È È

  5

  5

  5

  5

  2

  2 ß ß points of intersection.

  Š ‹ Š ‹ " "

  È È È È

Section 1.2 Lines, Circles, and Parabolas

  13

  # # 74. x œ y 1 and (x 1) y œ

  1 # # #

  1 ( y) y 2y Ê œ œ

  " " Ê y œ and x œ " or

  Š ‹ È È

  2

  2 " " y œ and x œ 1 . Thus,

  Š ‹ È 2 È

  2 " " " "

  A and B 1 " ß ß

  Š ‹ Š ‹ È È È È

  2

  2

  2

  2 are intersection points.

  # # 75. y x 1 and y x x x

  1 œ œ Ê œ

  È 1 „

  5 # x x 1 x . Ê œ Ê œ

  # 1 È 5 3 È

  5 If x œ , then y œ œ x 1 . # #

  È È

  1

  5

  3

  5 If x , then y x 1 .

  œ œ œ # # 1 È

  5 3 È

  5 1 È

  5 3 È

  5 Thus, A ß and B ß Š ‹ Š ‹

  # # # # are the intersection points.

  # # 76. y x and (x 1) (x 1) x

  œ C œ Ê œ 3 „ È

  5 # x 3x x . If

  Ê " œ Ê œ #

  È È

  3 5 5 3 x œ , then y œ œ x . If

  # # È È

  3

  5

  3

  5 x , then y x .

  œ œ œ # # 3 È

  5 È 5 3 3 È

  5 3 È

5 Thus, A ß and B ß

  Š ‹ Š ‹ # # # # are the intersection points.

  # # # 77. y 2x 1 x 3x

  1 œ œ Ê œ

  " " " " x and y or x and y .

  Ê œ œ œ œ

  3

  3 È È

  3

  3 " " " "

  Thus, A ß and B ß are the Š ‹ Š ‹

  È 3 È

  3

  3

  3 intersection points.

  14

Chapter 1 Preliminaries

  # #

  x 3x #

  78. y œ œ (x 1) Ê œ 2x

  1

  4

  4 #

  Ê œ 3x 8x œ 4 (3x 2)(x 2)

  #

  x

  2 x 2 and y 1, or x and

  Ê œ œ œ œ

  4

  3

  #

  x " 2 " y œ œ . Thus, A(2 1) and B ß ˆ ß ‰

  4

  9

  3

  9 are the intersection points.

  # # # # 79. x y œ 1 œ (x 1) y

  # # # x (x 1) x 2x

  1 Ê œ œ

  " Ê œ 2x 1 Ê x œ . Hence

  # È

  3

  3 # # y œ " x œ or y œ „ . Thus,

  4 # È 3 È

  3 " "

  A and B are the ß ß

  Š ‹ Š ‹ # # # # intersection points.

  # # # # 80. x y œ 1 œ x y Ê y œ y

  Ê y(y 1) œ Ê y œ 0 or y œ 1.

  # # If y œ 1, then x œ " y œ 0 or x œ 0.

  # # If y œ 0, then x œ 1 y œ 1 or x œ „ 1.

  Thus, A(0 1), B( 0), and C( 1 0) are the ß "ß ß intersection points.

  68° 69° 81. (a) A (69° 0 in), B (68° .4 in) m 2.5°/in.

  ¸ ß ¸ ß Ê œ ¸ .4

  10° 68° (b) A ¸ (68° .4 in), B ß ¸ (10° 4 in) ß Ê m œ ¸ 16.1°/in.

  4 .4 5° 10° (c) A ¸ (10° 4 in), B ß ¸ (5° 4.6 in) ß Ê m œ ¸ 8.3°/in.

  4.6

  4 ?

  U ?

  &gt; ?

  X to the temperature gradient across the material, (the slopes from the previous problem), and to a constant characteristic

  ? B

  ? U

  ? ? ? Î A ?

  U U X ? &gt;

  X of the material. œ -kA Ê k = . Note that and are of opposite sign because heat flow is toward lower

  ? ? ?

  X ? ?

  &gt; B &gt; B

  ? B

  temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are ?

  "

  X not changing), we may define another constant, K, characteristics of the material: K œ Þ Using the values of from

  ? X ?

  B

  ? B

  the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the poorest insulator, with K œ

  0.4.

  10.94 "

  83. p œ kd 1 and p œ 10.94 at d œ 100 Ê k œ œ 0.0994. Then p œ 0.0994d 1 is the diver's 100 pressure equation so that d œ 50 Ê p œ (0.0994)(50) œ 1 5.97 atmospheres.

84. The line of incidence passes through ( 1) and ( 0) !ß "ß Ê The line of reflection passes through ( 0) and ( "ß #ß " )

  1 0 m 1 y 1(x 1) y x 1 is the line of reflection. Ê œ œ Ê œ Ê œ

  1 #

Section 1.2 Lines, Circles, and Parabolas

  15

  4 " increments from the fourth vertex D(x y) to A must equal the increments from C to B 2 x x 4 and

  " "ß D ( 2) be located vertically downward from A so

  Since BC is vertical and has length BC 3, let k k œ D ( 4) be located vertically upward from A and

  œ ß œ ß œ ß

  5 9 x 3(2 ( 5)) x 12. œ Ê œ Ê œ Ê œ Ê œ Therefore, A ( 12 2), C (9 5), and B ( 12 5).

  90. Let A (x 2) and C (9 y) B (x y). Then 9 x AD and 2 y DC 2(9 x) 2(2 y) ß 56 œ œ ß Ê œ ß œ œ Ê œ k k k k and 9 x 3(2 y) 2(3(2 y)) 2(2 y) 56 y

  ?

  ß Ê œ œ ? 1 y y x 2 and y

  1

  3. Denote the point k k k k k k œ œ œ

  4

  # # # # # # # # Also, slope AB and slope BC , so AB BC. Thus, the points are vertices of a square. The coordinate œ œ ¼

  È ? ? ? ?

  È È È

  È È

  1 17. œ œ œ œ œ œ

  4

  # "ß that BC AD AD

  " # D (x y). Since the slope of AB equals the slope of

  85. C (F 32) and C F F F F or F 40° gives the same numerical reading. œ œ Ê œ Ê œ œ

  2 œ Ê œ œ Solving the system of equations we find x 5 and y 2 yielding the vertex D (5 ). x 3y

  " œ œ m y x will be y, x or y, x , the first of these corresponds to a counter-clockwise rotation, the latter to a clockwise a b a b rotation.

  ‰ rotation gives a segment with slope m . If this segment has length equal to the original segment, its endpoint w

  Á ! Á ! !ß ! y x

  92. Let x, y , x and/or y be a point on the coordinate plane. The slope, m, of the segment to x, y is . A 90 a b a b a b

  œ œ ß # $

  œ "" œ

  2x 3y

  3

  $ ß

  2

  $ y x

  4.

  11. Likewise, the slope of AC equals the slope œ of BD so that 3y 2x 4 or 2x 3y

  3 œ Ê œ x 3y

  " y 3 x 2

  CD we have 3y 9 x 2 or $

  1 4 17 and length BC ( x) ( y)

  89. Length AB ( x) ( y)

  #

  14

  9 œ 5 œ œ È È

  16

  87. length AB (5 1) (5 2)

  # #

  ? É ˆ ‰

  .371 ?

  14 14 100 x .371

  37.1

  È È #

  ¸

  9 86. m x . Therefore, distance between first and last rows is (14) 40.25 ft. œ œ Ê œ

  9

  9

  9

  9

  5 5 160 4 160

  # # length AC (4 1) ( )

  9

  16 # # 5 œ œ œ È È

  # # length BC (4 5) ( 5)