lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR'S SOLUTIONS MANUAL for Telegram: @uni_k lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL Robert A. Adams University of British Columbia Calculus Ninth Edition Robert A. Adams University of British Columbia Christopher Essex University of Western Ontario dumperina ISBN: 978-0-13-452876-2 Copyright © 2018 Pearson Canada Inc., Toronto, Ontario. All rights reserved. This work is protected by Canadian copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the Internet) will destroy the integrity of the work and is not permitted. The copyright holder grants permission to instructors who have adopted Calculus: A Complete Course, Single-Variable Calculus, or Calculus of Several Variables by Adams and Essex, to post this material online only if the use of the website is restricted by access codes to students in the instructor’s class that is using the textbook and provided the reproduced material bears this copyright notice. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in FOREWORD These solutions are provided for the benefit of instructors using the textbooks: Calculus: A Complete Course (9th Edition), Single-Variable Calculus (9th Edition), and Calculus of Several Variables (9th Edition) by R. A. Adams and Chris Essex, published by Pearson Canada. For the most part, the solutions are detailed, especially in exercises on core material and techniques. Occasionally some details are omitted—for example, in exercises on applications of integration, the evaluation of the integrals encountered is not always given with the same degree of detail as the evaluation of integrals found in those exercises dealing specifically with techniques of integration. Instructors may wish to make these solutions available to their students. However, students should use such solutions with caution. It is always more beneficial for them to attempt exercises and problems on their own, before they look at solutions done by others. If they examine solutions as “study material” prior to attempting the exercises, they can lose much of the benefit that follows from diligent attempts to develop their own analytical powers. When they have tried unsuccessfully to solve a problem, then looking at a solution can give them a “hint” for a second attempt. Separate Student Solutions Manuals for the books are available for students. They contain the solutions to the even-numbered exercises only. November, 2016. R. A. Adams adms@math.ubc.ca Telegram: @uni_k Chris Essex essex@uwo.ca Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in CONTENTS Solutions for Chapter P 1 Solutions for Chapter 1 23 Solutions for Chapter 2 39 Solutions for Chapter 3 81 Solutions for Chapter 4 108 Solutions for Chapter 5 177 Solutions for Chapter 6 213 Solutions for Chapter 7 267 Solutions for Chapter 8 316 Solutions for Chapter 9 351 Solutions for Chapter 10 392 Solutions for Chapter 11 420 Solutions for Chapter 12 448 Solutions for Chapter 13 491 Solutions for Chapter 14 538 Solutions for Chapter 15 579 Solutions for Chapter 16 610 Solutions for Chapter 17 637 Solutions for Chapter 18 644 Solutions for Chapter 18-cosv9 671 Solutions for Appendices 683 NOTE: “Solutions for Chapter 18-cosv9” is only needed by users of Calculus of Several Variables (9th Edition), which includes extra material in Sections 18.2 and 18.5 that is found in Calculus: a Complete Course and in Single-Variable Calculus in Sections 7.9 and 3.7 respectively. Solutions for Chapter 18-cosv9 contains only the solutions for the two Sections 18.2 and 18.9 in the Several Variables book. All other Sections are in “Solutions for Chapter 18.” It should also be noted that some of the material in Chapter 18 is beyond the scope of most students in single-variable calculus courses as it requires the use of multivariable functions and partial derivatives. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHAPTER P. SECTION P.1 (PAGE 10) PRELIMINARIES 19. Given: 1=.2 x/ < 3. CASE I. If x < 2, then 1 < 3.2 x/ D 6 3x, so 3x < 5 and x < 5=3. This case has solutions x < 5=3. CASE II. If x > 2, then 1 > 3.2 x/ D 6 3x, so 3x > 5 and x > 5=3. This case has solutions x > 2. Solution: . 1; 5=3/ [ .2; 1/. 20. Given: .x C 1/=x 2. CASE I. If x > 0, then x C 1 2x, so x 1. CASE II. If x < 0, then x C 1 2x, so x 1. (not possible) Solution: .0; 1. 21. Given: x 2 2x 0. Then x.x 2/ 0. This is only possible if x 0 and x 2. Solution: Œ0; 2. Section P.1 Real Numbers and the Real Line (page 10) 1. 2 D 0:22222222 D 0:2 9 2. 1 D 0:09090909 D 0:09 11 3. If x D 0:121212 , then 100x D 12:121212 D 12 C x. Thus 99x D 12 and x D 12=99 D 4=33. 4. If x D 3:277777 , then 10x 32 D 0:77777 and 100x 320 D 7 C .10x 32/, or 90x D 295. Thus x D 295=90 D 59=18. 5. 1=7 D 0:142857142857 D 0:142857 22. Given 6x 2 5x 1, then .2x 1/.3x 1/ 0, so either x 1=2 and x 1=3, or x 1=3 and x 1=2. The latter combination is not possible. The solution set is Œ1=3; 1=2. 23. Given x 3 > 4x, we have x.x 2 4/ > 0. This is possible if x < 0 and x 2 < 4, or if x > 0 and x 2 > 4. The possibilities are, therefore, 2 < x < 0 or 2 < x < 1. Solution: . 2; 0/ [ .2; 1/. 2=7 D 0:285714285714 D 0:285714 3=7 D 0:428571428571 D 0:428571 4=7 D 0:571428571428 D 0:571428 note the same cyclic order of the repeating digits 5=7 D 0:714285714285 D 0:714285 24. 6=7 D 0:857142857142 D 0:857142 6. Two different decimal expansions can represent the same number. For instance, both 0:999999 D 0:9 and 1:000000 D 1:0 represent the number 1. 7. 25. x 0 and x 5 define the interval Œ0; 5. 8. x < 2 and x 3 define the interval Œ 3; 2/. 6 defines the union . 1; 6/ [ . 5; 1/. 9. x > 5 or x < 10. x 1 defines the interval . 1; 1. x> 2 defines the interval . 2; 1/. 11. 2x > 4, then x < If 3x C 5 8, then 3x 8 . 1; 1 15. If 5x 3 7 . 1; 5=4 6 x 3x 5 27. If jxj D 3 then x D ˙3. 28. If jx 29. If j2t C 5j D 4, then 2t C 5 D ˙4, so t D t D 1=2. 30. Ifj1 31. If j8 3sj D 9, then 8 3s D ˙9, so 3s D s D 1=3 or s D 17=3. 3 and x 1. Solution: 3x, then 8x 10 and x 5=4. Solution: 4 , then 6 x 6x 4 2 and x 2. Solution: . 1; 2 16. If 17. 8. Thus 14 7x If 3.2 x/ < 2.3 C x/, then 0 < 5x and x > 0. Solution: .0; 1/ 2 18. If x < 9, then jxj < 3 and . 3; 3/ 3 < x < 3. Solution: 2 3 < . x 1 xC1 CASE I. If x > 1 then .x 1/.x C 1/ > 0, so that 3.x C 1/ < 2.x 1/. Thus x < 5. There are no solutions in this case. CASE II. If 1 < x < 1, then .x 1/.x C 1/ < 0, so 3.x C 1/ > 2.x 1/. Thus x > 5. In this case all numbers in . 1; 1/ are solutions. CASE III. If x < 1, then .x 1/.x C 1/ > 0, so that 3.x C 1/ < 2.x 1/. Thus x < 5. All numbers x < 5 are solutions. Solutions: . 1; 5/ [ . 1; 1/. Given: 2. Solution: . 1; 2/ 14. 4 x 1C . 2 x CASE I. If x > 0, then x 2 2x C 8, so that x 2 2x 8 0, or .x 4/.x C 2/ 0. This is possible for x > 0 only if x 4. CASE II. If x < 0, then we must have .x 4/.x C 2/ 0, which is possible for x < 0 only if x 2. Solution: Œ 2; 0/ [ Œ4; 1/. Given: 26. 12. x < 4 or x 2 defines the interval . 1; 1/, that is, the whole real line. 13. If Given x 2 x 2, then x 2 x 2 0 so .x 2/.x C1/ 0. This is possible if x 2 and x 1 or if x 2 and x 1. The latter situation is not possible. The solution set is Œ 1; 2. 3j D 7, then x t j D 1, then 1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 3 D ˙7, so x D 4 or x D 10. 9=2 or t D ˙1, so t D 0 or t D 2. 1 or 17, and 1 lOMoARcPSD|6566483 www.konkur.in SECTION P.1 (PAGE 10) ADAMS and ESSEX: CALCULUS 9 ˇs ˇ s ˇ ˇ 32. If ˇ 1ˇ D 1, then 1 D ˙1, so s D 0 or s D 4. 2 2 33. If jxj < 2, then x is in . 2; 2/. 4. From A.0:5; 3/ to B.2; 3/, x D 2 y D 3 3 D 0. jABj D 1:5. 5. Starting point: . 2; 3/. Increments x D 4, y D New position is . 2 C 4; 3 C . 7//, that is, .2; 4/. 6. Arrival point: . 2; 2/. Increments x D 5, y D 1. Starting point was . 2 . 5/; 2 1/, that is, .3; 3/. 7. x 2 C y 2 D 1 represents a circle of radius 1 centred at the origin. p x 2 C y 2 D 2 represents a circle of radius 2 centred at the origin. 34. If jxj 2, then x is in Œ 2; 2. 35. If js 1j 2, then 1 36. If jt C 2j < 1, then . 3; 1/. 2 s 1 C 2, so s is in Œ 1; 3. 2 2 C 1, so t is in 1 < t < 37. If j3x 7j < 2, then 7 2 < 3x < 7C2, so x is in .5=3; 3/. 38. If j2x C 5j < 1, then 5 1 < 2x < 5 C 1, so x is in . 3; 2/. ˇ ˇx x ˇ ˇ 1ˇ 1, then 1 1 1 C 1, so x is in Œ0; 4. 39. If ˇ 2 2 ˇ ˇ x 1 ˇ ˇ 40. If ˇ2 ˇ < , then x=2 lies between 2 .1=2/ and 2 2 2 C .1=2/. Thus x is in .3; 5/. 41. The inequality jx C 1j > jx 3j says that the distance from x to 1 is greater than the distance from x to 3, so x must be to the right of the point half-way between 1 and 3. Thus x > 1. jxj jx C yj Apply this inequality with x D a ja bj jaj jyj: b and y D b to get jbj: ˇ ˇ ˇ ˇ Similarly, ja bj D jb aj jbj jaj. Since ˇjaj jbjˇ is equal to either jaj jbj or jbj jaj, depending on the sizes of a and b, we have ja ˇ ˇ bj ˇjaj ˇ ˇ jbjˇ: Section P.2 Cartesian Coordinates in the Plane (page 16) 1. From A.0; 3/ to B.4; 0/, xp D 4 0 D 4 and y D 0 3 D 3. jABj D 42 C . 3/2 D 5. 2. From A. 1; 2/ to B.4; 10/, xpD 4 . 1/ D 5 and y D 10 2 D 12. jABj D 52 C . 12/2 D 13. 3. From A.3; 2/ to B. 1; 2/, x and pD 1 3 D 4p y D 2 2 D 4. jABj D . 4/2 C . 4/2 D 4 2. 2 Telegram: @uni_k 11. x 2 C y 2 1 represents points inside and on the circle of radius 1 centred at the origin. y x 2 represents all points lying on or above the parabola y D x 2 . 12. y < x 2 represents all points lying below the parabola y D x2. The vertical line through . 2; 5=3/ is x D 2; the horizontal line through that point is y D 5=3. p p The vertical line through . 2; 1:3/ is x D 2; the horizontal line through that point is y D 1:3. 15. Line through . 1; 1/ with slope m D 1 is y D 1C1.xC1/, or y D x C 2. 16. Line through . 2; 2/ with slope m D 1=2 is y D 2 C .1=2/.x C 2/, or x 2y D 6. 17. Line through .0; b/ with slope m D 2 is y D b C 2x. 1/, 45. The triangle inequality jx C yj jxj C jyj implies that 7. 10. x 2 C y 2 D 0 represents the origin. 14. 43. jaj D a if and only if a 0. It is false if a < 0. .x 9. 13. 42. jx 3j < 2jxj , x 2 6x C 9 D .x 3/2 < 4x 2 , 3x 2 C 6x 9 > 0 , 3.x C 3/.x 1/ > 0. This inequality holds if x < 3 or x > 1. 44. The equation jx 1j D 1 x holds if jx 1j D that is, if x 1 0, or, equivalently, if x 1. 8. 0:5 D 1:5 and 18. Line through .a; 0/ with slope m D or y D 2a 2x. 2 is y D 0 2.x a/, 19. At x D 2, the height of the line 2x C 3y D 6 is y D .6 4/=3 D 2=3. Thus .2; 1/ lies above the line. 20. At x D 3, the height of the line x 4y D 7 is y D .3 7/=4 D 1. Thus .3; 1/ lies on the line. 21. The line through .0; 0/ and .2; 3/ has slope m D .3 0/=.2 0/ D 3=2 and equation y D .3=2/x or 3x 2y D 0. 22. The line through . 2; 1/ and .2; 2/ has slope m D . 2 1/=.2 C 2/ D 3=4 and equation y D 1 .3=4/.x C 2/ or 3x C 4y D 2. 23. The line through .4; 1/ and . 2; 3/ has slope m D .3 1/=. 2 4/ D or x C 3y D 7. 24. 25. 1=3 and equation y D 1 The line through . 2; 0/ and .0; 2/ has slope m D .2 0/=.0 C 2/ D 1 and equation y D 2 C x. p If m D 2 and p b D 2, then the line has equation y D 2x C 2. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 .x 4/ 3 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.2 (PAGE 16) y 26. If m D 1=2 and b D 3, then the line has equation y D .1=2/x 3, or x C 2y D 6. 1:5x 27. 3x C 4y D 12 has x-intercept a D 12=3 D 4 and yintercept b D 12=4 D 3. Its slope is b=a D 3=4. y 2y D 3 x Fig. P.2-30 3x C 4y D 12 31. line through .2; 1/ parallel to y D x C 2 is y D x perpendicular to y D x C 2 is y D x C 3. 32. line through . 2; 2/ parallel to 2xCy D 4 is 2xCy D line perpendicular to 2x C y D 4 is x 2y D 6. 33. We have x Fig. P.2-27 3x C 4y D 6 2x 3y D 13 28. x C 2y D 4 has x-intercept a D 4 and y-intercept b D 4=2 D 2. Its slope is b=a D 2=. 4/ D 1=2. y 2; 6x C 8y D 12 6x 9y D 39: Subtracting these equations gives 17y D 51, so y D 3 and x D .13 9/=2 D 2. The intersection point is .2; 3/. 34. x x C 2y D ÷ 1; line We have 2x C y D 8 5x 7y D 1 4 ÷ 14x C 7y D 56 5x 7y D 1: Adding these equations gives 19x D 57, so x D 3 and y D 8 2x D 2. The intersection point is .3; 2/. 35. If a ¤ 0 and b ¤ 0, then .x=a/ C .y=b/ D 1 represents a straight line that is neither horizontal nor vertical, and does not pass through the origin. Putting y D 0 we get x=a D 1, so the x-intercept of this line is x D a; putting x D 0 gives y=b D 1, so the y-intercept is y D b. 36. The line .x=2/ .y=3/ D 1 has x-intercept a D 2, and y-intercept b D 3. y Fig. P.2-28 29. p p p p 2x 3y D 2 has x-intercept a D 2= 2 D 2 p and y-intercept 2= 3. Its slope is p b pD b=a D 2= 6 D 2=3. y 2 p 2x p x x 2 3y D 2 x y D1 3 3 Fig. P.2-29 Fig. P.2-36 30. 1:5x 2y D 3 has x-intercept a D 3=1:5 D 2 and y-intercept b D 3=. 2/ D 3=2. Its slope is b=a D 3=4. 37. The line through .2; 1/ and .3; 1/ has slope m D . 1 1/=.3 2/ D 2 and equation y D 1 2.x 2/ D 5 2x. Its y-intercept is 5. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 3 lOMoARcPSD|6566483 www.konkur.in SECTION P.2 (PAGE 16) ADAMS and ESSEX: CALCULUS 9 38. The line through . 2; 5/ and .k; 1/ has x-intercept 3, so also passes through .3; 0/. Its slope m satisfies 1 k Thus k 3D 0 5 0 DmD D 3 3C2 43. B D .1; 3/; C D . 3; 2/ p p jABj D .1 2/2 C .3 C 1/2 D 17 p p p p jAC j D . 3 2/2 C .2 C 1/2 D 34 D 2 17 p p jBC j D . 3 1/2 C .2 3/2 D 17: p Since jABj D jBC j and jAC j D 2jABj, triangle ABC is an isosceles right-angled triangle with right angle at B. Thus ABCD is a square if D is displaced from C by the same amount A is from B, that is, by increments x D 2 1 D 1 and y D 1 3 D 4. Thus D D . 3 C 1; 2 C . 4// D . 2; 2/. 44. If M D .xm ; ym / is the midpoint of P1 P2 , then the displacement of M from P1 equals the displacement of P2 from M : 1: 1, and so k D 2. 39. C D Ax C B. If C D 5; 000 when x D 10; 000 and C D 6; 000 when x D 15; 000, then 10; 000A C B D 5; 000 15; 000A C B D 6; 000 Subtracting these equations gives 5; 000A D 1; 000, so A D 1=5. From the first equation, 2; 000 C B D 5; 000, so B D 3; 000. The cost of printing 100,000 pamphlets is $100; 000=5 C 3; 000 D $23; 000. xm 40ı and 40ı is the same temperature on both the Fahrenheit and Celsius scales. C 40 xq C DF -20 -30 46. 10 20 30 40 50 60 70 80 F C D 5 .F 9 47. A D .2; 1/; B D .6; 4/; C D .5; 3/ p p jABj D .6 2/2 C .4 1/2 D 25 D 5 p p jAC j D .5 2/2 C . 3 1/2 D 25 D 5 p p p jBC j D .6 5/2 C .4 C 3/2 D 50 D 5 2: Since jABj D jAC j, triangle ABC is isosceles. p A D .0; 0/; B D .1; 3/; C D .2; 0/ q p p jABj D .1 0/2 C . 3 0/2 D 4 D 2 p p jAC j D .2 0/2 C .0 0/2 D 4 D 2 q p p 3/2 D 4 D 2: jBC j D .2 1/2 C .0 Since jABj D jAC j D jBC j, triangle ABC is equilateral. 4 Telegram: @uni_k x1 D 2.x2 xq /; .x C X/=2 D 2; Fig. P.2-40 42. y1 D y2 ym : yq y1 D 2.y2 yq /: Let the coordinates of P be .x; 0/ and those of Q be .X; 2X/. If the midpoint of PQ is .2; 1/, then 32/ -40 . 40; 40/ -50 41. ym Thus xq D .x1 C 2x2 /=3 and yq D .y1 C 2y2 /=3. 10 -50 -40 -30 -20 -10 -10 xm ; If Q D .xq ; yq / is the point on P1 P2 that is two thirds of the way from P1 to P2 , then the displacement of Q from P1 equals twice the displacement of P2 from Q: 30 20 x1 D x2 Thus xm D .x1 C x2 /=2 and ym D .y1 C y2 /=2. 45. 40. A D .2; 1/; 48. .0 2X/=2 D 1: The second equation implies that X D 1, and the second then implies that x D 5. Thus P is .5; 0/. p .x 2/2 C y 2 D 4 says that the distance of .x; y/ from .2; 0/ is 4, so the equation represents a circle of radius 4 centred at .2; 0/. p p .x 2/2 C y 2 D x 2 C .y 2/2 says that .x; y/ is equidistant from .2; 0/ and .0; 2/. Thus .x; y/ must lie on the line that is the right bisector of the line from .2; 0/ to .0; 2/. A simpler equation for this line is x D y. 49. The line 2x C ky D 3 has slope m D 2=k. This line is perpendicular to 4x C y D 1, which has slope 4, provided m D 1=4, that is, provided k D 8. The line is parallel to 4x C y D 1 if m D 4, that is, if k D 1=2. 50. For any value of k, the coordinates of the point of intersection of x C 2y D 3 and 2x 3y D 1 will also satisfy the equation .x C 2y Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 3/ C k.2x 3y C 1/ D 0 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.3 (PAGE 22) because they cause both expressions in parentheses to be 0. The equation above is linear in x and y, and so represents a straight line for any choice of k. This line will pass through .1; 2/ provided 1 C 4 3 C k.2 6 C 1/ D 0, that is, if k D 2=3. Therefore, the line through the point of intersection of the two given lines and through the point .1; 2/ has equation x C 2y 2 3 C .2x 3 3y C 1/ D 0; 13. Together, x 2 C y 2 > 1 and x 2 C y 2 < 4 represent annulus (washer-shaped region) consisting of all points that are outside the circle of radius 1 centred at the origin and inside the circle of radius 2 centred at the origin. 14. Together, x 2 C y 2 4 and .x C 2/2 C y 2 4 represent the region consisting of all points that are inside or on both the circle of radius 2 centred at the origin and the circle of radius 2 centred at . 2; 0/. 15. Together, x 2 Cy 2 < 2x and x 2 Cy 2 < 2y (or, equivalently, .x 1/2 C y 2 < 1 and x 2 C .y 1/2 < 1) represent the region consisting of all points that are inside both the circle of radius 1 centred at .1; 0/ and the circle of radius 1 centred at .0; 1/. 16. x2 C y2 4x C 2y > 4 can be rewritten .x 2/2 C .y C 1/2 > 9. This equation, taken together with x C y > 1, represents all points that lie both outside the circle of radius 3 centred at .2; 1/ and above the line x C y D 1. 17. The interior of the circle with centre . 1; 2/ and radius p 6 is given by .x C 1/2 C .y 2/2 < 6, or 2 x C y 2 C 2x 4y < 1. or, on simplification, x D 1. Section P.3 Graphs of Quadratic Equations (page 22) 1. x 2 C y 2 D 16 2. x 2 C .y 2/2 D 4, or x 2 C y 2 4y D 0 3. .x C 2/2 C y 2 D 9, or x 2 C y 2 C 4y D 5 4. .x 5. x2 C y2 x 6. 3/2 C .y C 4/2 D 25, or x 2 C y 2 2 6x C 8y D 0. 2x D 3 2x C 1 C y 2 D 4 .x 1/2 C y 2 D 4 centre: .1; 0/; radius 2. 18. The exterior of the circle with centre .2; 3/ and radius 4 is given by .x 2/2 C .y C 3/2 > 16, or x 2 C y 2 4x C 6y > 3. x 2 C y 2 C 4y D 0 19. 2 2 x C y C 4y C 4 D 4 2 x1 1/2 C .y 3/2 < 10 x C y > 4; 21. 2x C 4y D 4 The parabola with focus .0; 4/ and directrix y D equation x 2 D 16y. 22. .x 1/2 C .y C 2/2 D 9 centre: .1; 2/; radius 3. The parabola with focus .0; 1=2/ and directrix y D 1=2 has equation x 2 D 2y. 23. x2 C y2 The parabola with focus .2; 0/ and directrix x D equation y 2 D 8x. 24. The parabola with focus . 1; 0/ and directrix x D 1 has equation y 2 D 4x. 25. y D x 2 =2 has focus .0; 1=2/ and directrix y D 2 x C .y C 2/ D 4 centre: .0; 2/; radius 2. x2 C y2 x 8. 2 20. 2 7. x 2 C y 2 < 2; 2 x2 2x C 1 C y 2 C 4y C 4 D 9 2x yC1D0 2x C 1 C y 2 y C 41 D 14 2 .x 1/2 C y 12 D 41 centre: .1; 1=2/; radius 1=2. .x yDx 2 =2 1=2. .0;1=2/ 10. x 2 C y 2 < 4 represents the open disk consisting of all points lying inside the circle of radius 2 centred at the origin. x yD 1=2 .x C 1/2 C y 2 4 represents the closed disk consisting of all points lying inside or on the circle of radius 2 centred at the point . 1; 0/. 12. x 2 C .y 2/2 4 represents the closed disk consisting of all points lying inside or on the circle of radius 2 centred at the point .0; 2/. Fig. P.3-25 26. yD x 2 has focus .0; 1=4/ and directrix y D 1=4. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 has y 9. x 2 C y 2 > 1 represents all points lying outside the circle of radius 1 centred at the origin. 11. 4 has Downloaded by ted cage (sxnbyln180@questza.com) 5 lOMoARcPSD|6566483 www.konkur.in SECTION P.3 (PAGE 22) ADAMS and ESSEX: CALCULUS 9 y y Version (c) yD1=4 y D x2 x .3; 3/ Version (b) .0; 1=4/ yD x 2 Fig. P.3-26 x 4 Version (d) .4; 2/ 27. xD Version (a) 3 y 2 =4 has focus . 1; 0/ and directrix x D 1. y Fig. P.3-29 a) has equation y D x 2 xD1 . 1;0/ b) has equation y D .x x c) has equation y D .x d) has equation y D .x xD y 2 =4 30. Fig. P.3-27 28. x D y 2 =16 has focus .4; 0/ and directrix x D 4. y 31. .4;0/ 32. x 33. xD 4 xDy 2 =16 34. Fig. P.3-28 6 Telegram: @uni_k 8x C 16. 3/2 C 3 or y D x 2 4/2 2, or y D x 2 6x C 12. 8x C 14. b) If y D mx is shifted vertically by amount y1 , the equation y D mx C y1 results. If .a; b/ satisfies this equation, then b D ma C y1 , and so y1 D b ma. Thus the shifted equation is y D mx C b ma D m.x a/ C b, the same equation obtained in part (a). p y D .x=3/ C 1 p 4y D x C 1 p y D .3x=2/ C 1 p .y=2/ D 4x C 1 x 2 shifted down 1, left 1 gives y D 35. yD1 36. x2 C y2 D .x C 4/2 C .y 38. 4/2 or y D x 2 a) If y D mx is shifted to the right by amount x1 , the equation y D m.x x1 / results. If .a; b/ satisfies this equation, then b D m.a x1 /, and so x1 D a .b=m/. Thus the shifted equation is y D m.x a C .b=m// D m.x a/ C b. 37. y D .x y D .x 29. 3. .x C 1/2 . 5 shifted up 2, left 4 gives 2/2 D 5. 1/2 2/2 1 shifted down 1, right 1 gives 2. p p y D x shifted down 2, left 4 gives y D x C 4 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2. lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.3 (PAGE 22) y 39. y D x 2 C 3, y D 3x C 1. Subtracting these equations gives x 2 3x C 2 D 0, or .x 1/.x 2/ D 0. Thus x D 1 or x D 2. The corresponding values of y are 4 and 7. The intersection points are .1; 4/ and .2; 7/. 9x 2 C16y 2 D144 x 40. y D x 2 6, y D 4x x 2 . Subtracting these equations gives 2x 2 4x 6 D 0, or 2.x 3/.x C 1/ D 0. Thus x D 3 or x D 1. The corresponding values of y are 3 and 5. The intersection points are .3; 3/ and . 1; 5/. 41. x 2 C y 2 D 25, 3x C 4y D 0. The second equation says that y D 3x=4. Substituting this into the first equation gives 25x 2 =16 D 25, so x D ˙4. If x D 4, then the second equation gives y D 3; if x D 4, then y D 3. The intersection points are .4; 3/ and . 4; 3/. Note that having found values for x, we substituted them into the linear equation rather than the quadratic equation to find the corresponding values of y. Had we substituted into the quadratic equation we would have got more solutions (four points in all), but two of them would have failed to satisfy 3x C 4y D 12. When solving systems of nonlinear equations you should always verify that the solutions you find do satisfy the given equations. 42. 2x 2 C 2y 2 D 5, xy D 1. The second equation says that y D 1=x. Substituting this into the first equation gives 2x 2 C .2=x 2 / D 5, or 2x 4 5x 2 C 2 D 0. This equation factors to p .2x 2 1/.x 2 p2/ D 0, so its solutions are x D ˙1= 2 and x D ˙ 2. The corresponding values of y are givenpby p y D 1=x. p Therefore, p the p intersection p points are .1= 2; 2/, . 1= 2; 2/, . 2; 1= 2/, and p p 2; 1= 2/. . Fig. P.3-44 45. 3/2 .y C 2/2 D 1 is an ellipse with centre at 9 4 .3; 2/, major axis between .0; 2/ and .6; 2/ and minor axis between .3; 4/ and .3; 0/. .x C y x .3; 2/ .x 3/2 .yC2/2 C D1 9 4 Fig. P.3-45 46. 43. .x 2 =4/ C y 2 D 1 is an ellipse with major axis between . 2; 0/ and .2; 0/ and minor axis between .0; 1/ and .0; 1/. .y C 1/2 D 4 is an ellipse with centre at 4 .1; 1/, major axis between .1; 5/ and .1; 3/ and minor axis between . 1; 1/ and .3; 1/. .x 1/2 C y .x 1/2 C .yC1/2 D4 4 y x2 2 4 Cy D1 x .1; 1/ x Fig. P.3-46 Fig. P.3-43 44. 9x 2 C 16y 2 D 144 is an ellipse with major axis between . 4; 0/ and .4; 0/ and minor axis between .0; 3/ and .0; 3/. 47. .x 2 =4/ y 2 D 1 is a hyperbola with centre at the origin and passing through .˙2; 0/. Its asymptotes are y D ˙x=2. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 7 lOMoARcPSD|6566483 www.konkur.in SECTION P.3 (PAGE 22) ADAMS and ESSEX: CALCULUS 9 y y x2 4 y 2 D1 yD x=2 .x x 1/.y C 2/ D 1 yD x yDx=2 2 xD1 Fig. P.3-47 Fig. P.3-50 51. 48. x 2 y 2 D 1 is a rectangular hyperbola with centre at the origin and passing through .0; ˙1/. Its asymptotes are y D ˙x. y x2 y2 D 1 yDx a) Replacing x with x replaces a graph with its reflection across the y-axis. b) Replacing y with y replaces a graph with its reflection across the x-axis. 52. Replacing x with x and y with y reflects the graph in both axes. This is equivalent to rotating the graph 180ı about the origin. 53. jxj C jyj D 1. In the first quadrant the equation is x C y D 1. In the second quadrant the equation is x C y D 1. In the third quadrant the equation is x y D 1. In the fourth quadrant the equation is x y D 1. x yD x y Fig. P.3-48 1 jxj C jyj D 1 1 49. xy D 4 is a rectangular hyperbola with centre at the origin and passing through .2; 2/ and . 2; 2/. Its asymptotes are the coordinate axes. x 1 y 1 Fig. P.3-53 x Section P.4 Functions and Their Graphs (page 32) xyD 4 Fig. P.3-49 50. .x 1/.y C 2/ D 1 is a rectangular hyperbola with centre at .1; 2/ and passing through .2; 1/ and .0; 3/. Its asymptotes are x D 1 and y D 2. 8 Telegram: @uni_k 1. f .x/ D 1 C x 2 ; domain R, range Œ1; 1/ 2. f .x/ D 1 3. G.x/ D 4. F .x/ D 1=.x 1/; domain . 1; 1/ [ .1; 1/, range . 1; 0/ [ .0; 1/ p p 8 x; domain Œ0; 1/, range . 1; 1 2x; domain . 1; 4, range Œ0; 1/ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.4 (PAGE 32) t ; domain . 1; 2/, range R. (The equation 2 t y D h.t / can be squared and rewritten as t 2 C y 2 t 2y 2 D 0, a quadratic equation in t having real solutions for every real value of y. Thus the range of h contains all real numbers.) 5. h.t / D p 1 p ; domain Œ2; 3/ [ .3; 1/, range 1 x 2 . 1; 0/ [ .0; 1/. The equation y D g.x/ can be solved for x D 2 .1 .1=y//2 so has a real solution provided y ¤ 0. y b) is the graph of x 2 if x < 1. x 3 D x 2 .1 x/, which is positive d) is the graph of x 3 x 4 , which is positive if 0 < x < 1 and behaves like x 3 near 0. 9. x 0 ˙0:5 ˙1 ˙1:5 ˙2 y graph (i) x/2 , which is positive for x > 0. c) is the graph of x x 4 , which is positive if 0 < x < 1 and behaves like x near 0. 6. g.x/ D 7. a) is the graph of x.1 graph (ii) f .x/ D x 4 0 0:0625 1 5:0625 16 y x y y D x4 x y graph (iii) graph (iv) x x x Fig. P.4-7 Graph (ii) is the graph of a function because vertical lines can meet the graph only once. Graphs (i), (iii), and (iv) do not have this property, so are not graphs of functions. Fig. P.4-9 10. 8. y y graph (a) graph (b) x f .x/ D x 2=3 0 ˙0:5 ˙1 ˙1:5 ˙2 0 0:62996 1 1:3104 1:5874 y x y y graph (c) yDx x 2=3 graph (d) x Fig. P.4-10 11. x Fig. P.4-8 x f .x/ D x 2 C 1 is even: f . x/ D f .x/ 12. f .x/ D x 3 C x is odd: f . x/ D f .x/ x is odd: f . x/ D f .x/ 13. f .x/ D 2 x 1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 9 lOMoARcPSD|6566483 www.konkur.in SECTION P.4 (PAGE 32) 1 14. f .x/ D x2 15. f .x/ D x 16. f .x/ D f. 4 17. 1 1 2 ADAMS and ESSEX: CALCULUS 9 26. is even: f . x/ D f .x/ is odd about .2; 0/: f .2 y x/ D f .2 C x/ 1 is odd about . 4; 0/: xC4 x/ D f . 4 C x/ f .x/ D x 2 x 18. f .x/ D x 3 2 is odd about .0; 2/: f . x/ C 2 D .f .x/ C 2/ 3 yD.x 1/2 C1 6x is even about x D 3: f .3 x/ D f .3 C x/ 27. y 3 19. f .x/ D jx j D jxj is even: f . x/ D f .x/ 20. f .x/ D jx C 1j is even about x D 1: f . 1 x/ D f . 1 C x/ p 21. f .x/ D 2x has no symmetry. p 22. f .x/ D .x 1/2 is even about x D 1: f .1 x/ D f .1 C x/ 23. yD1 x 3 x 28. y y yD x 2 x x yD.xC2/3 24. 29. y y yD1 x 2 p yD xC1 x x 25. 30. y y yD.x 1/2 p yD xC1 x x 10 Telegram: @uni_k Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.4 (PAGE 32) 31. 36. y y x xD2 x yD jxj yD 32. 1 2 x 37. y y x yD xC1 yDjxj 1 yD1 x x xD 1 33. 38. y y xD1 yDjx 2j x yD 1 x yD 1 x x 34. y 39. yD1Cjx 2j y y .1;3/ yDf .x/C2 2 .2;2/ .1;1/ yDf .x/ x 2 x x Fig. P.4.39(a) 35. Fig. P.4.39(b) y 40. y y .1;3/ yDf .x/C2 xD 2 2 .2;2/ x 1 x 1 yD 2 xC2 Fig. P.4.40(a) Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k yDf .x/ 1 x .2; 1/ Downloaded by ted cage (sxnbyln180@questza.com) Fig. P.4.40(b) 11 lOMoARcPSD|6566483 www.konkur.in SECTION P.4 (PAGE 32) ADAMS and ESSEX: CALCULUS 9 y 0.8 0.6 41. y y D 0:68 x C 2 yD 2 x C 2x C 3 0.4 0.2 . 1;1/ yDf .xC2/ x 2 -5 -4 -3 -2 -1 -0.2 1y D2 0:18 3 4 x -0.4 -0.6 42. y .2;1/ 1 -0.8 -1.0 Fig. P.4-47 yDf .x 1/ 3 x 48. Range is approximately . 1; 0:1 [ Œ2:9; 1/. y 43. 4 y 3 2 2 yD f .x/ .1; 1/ 1 x -4 -3 -2 -1 1 -1 44. y -2 2 3 4 5 x 2 yD x.x C 2/ 6 x -3 . 1;1/ yDf . x/ -4 x 2 -5 Fig. P.4-48 45. y 49. y yDf .4 x/ 5 .3;1/ 4 4 x 2 3 2 46. y 1 -5 . 1;1/ -4 -3 .1;1/ Telegram: @uni_k 1 Range is approximately Œ 0:18; 0:68. 2 y D x4 Fig. P.4-49 x 12 -1 -1 yD1 f .1 x/ 47. -2 3 4 x 6x 3 C 9x 2 1 Apparent symmetry about x D 1:5. This can be confirmed by calculating f .3 x/, which turns out to be equal to f .x/. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.5 (PAGE 38) Apparent symmetry about . 2; 2/. This can be confirmed by calculating shifting the graph right by 2 (replace x with x 2) and then down 2 (subtract 2). The result is 5x=.1 C x 2 /, which is odd. 50. y yD 3 2 2 2 2x C x 2x C x 2 53. 1 -5 -4 -3 -2 -1 1 2 3 4 Section P.5 Combining Functions to Make New Functions (page 38) Apparent symmetry about x D 1. This can be confirmed by calculating f .2 x/, which turns out to be equal to f .x/. 1. 51. y 4 yDx 1 x 1 x 2 3 yD 2 1 -2 -1 1 2 3 4 -1 5 xC3 2. Fig. P.4-51 Apparent symmetry about .2; 1/, and about the lines y D x 1 and y D 3 x. These can be confirmed by noting that f .x/ D 1 C 1 x 2 so the graph is that of 1=x shifted right 2 units and up one. yD 2x 2 C 3x y x 2 C 4x C 5 p f .x/ D x, g.x/ D x 1. D.f / D R, D.g/ D Œ1; 1/. D.f C g/ D D.f g/ D D.fg/ D D.g=f / D Œ1; 1/, D.f =g/ D .1; 1/. p .f C g/.x/ D x C x 1 p .f g/.x/ D x x 1 p .fg/.x/ D x x 1 p .f =g/.x/ D x= x 1 p .g=f /.x/ D . 1 x/=x x 6 yD -2 52. f .x/, x -1 Fig. P.4-50 -3 If f is both even and odd the f .x/ D f . x/ D so f .x/ D 0 identically. , p p f .x/ D 1 x, g.x/ D 1 C x. D.f / D . 1; 1, D.g/ D Œ 1; 1/. D.f C g/ D D.f g/ D D.fg/ D Œ 1; 1, D.f =g/ D . 1;p1, D.g=fp/ D Œ 1; 1/. .f C g/.x/ D 1 x C 1 C x p p .f g/.x/ D 1 x 1Cx p 2 .fg/.x/ D 1 x p .f =g/.x/ D .1 x/=.1 C x/ p .g=f /.x/ D .1 C x/=.1 x/ 3. y 5 4 yDx 3 x2 x 2 yDx 1 -7 -6 -5 -4 -3 -2 -1 1 2 x -1 -2 yD x2 Fig. P.4-52 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 13 lOMoARcPSD|6566483 www.konkur.in SECTION P.5 (PAGE 38) ADAMS and ESSEX: CALCULUS 9 4. 8. y yD x 1 -2 -1 x 1 y D x3 x f .x/ D 2=x, g.x/ D x=.1 x/. f ı f .x/ D 2=.2=x/ D xI D.f ı f / D fx W x ¤ 0g f ı g.x/ D 2=.x=.1 x// D 2.1 x/=xI D.f ı g/ D fx W x ¤ 0; 1g g ı f .x/ D .2=x/=.1 .2=x// D 2=.x 2/I D.g ı f / D fx W x ¤ 0; 2g g ı g.x/ D .x=.1 x//=.1 .x=.1 x/// D x=.1 2x/I D.g ı g/ D fx W x ¤ 1=2; 1g -1 y D x3 -2 9. 5. y y D x C jxj y D jxj p f .x/ D 1=.1 x/, g.x/ D x 1. f ı f .x/ D 1=.1 .1=.1 x/// D .x 1/=xI D.f ı f / D fx W x ¤ 0; 1g p x 1/I f ı g.x/ D 1=.1 D.f ı g/ D fx W x 1; x ¤ 2g p p g ı f .x/ D .1=.1 x// 1 D x=.1 x/I D.g ı f / D Œ0; 1/ q p g ı g.x/ D x 1 1I D.g ı g/ D Œ2; 1/ y D x D jxj x yDx 6. y 4 y D jxj C jx 3 2j y D jxj 2 y D jx 1 -2 -1 1 2 3 4 10. f .x/ D .x C 1/=.x 1/ D 1 C 2=.x 1/, g.x/ D sgn .x/. f ı f .x/ D 1 C 2=.1 C .2=.x 1/ 1// D xI D.f ı f / D fx W x ¤ 1g sgn x C 1 f ı g.x/ D D 0I D.f ı g/ D . 1; 0/ sgn x 1 n xC1 1 if x < 1 or x > 1 g ı f .x/ D sgn D I 1 if 1 < x < 1 x 1 D.g ı f / D fx W x ¤ 1; 1g g ı g.x/ D sgn .sgn .x// D sgn .x/I D.g ı g/ D fx W x ¤ 0g 2j 5 11. 12. 13. 14. 15. 16. x -1 7. g.x/ f ı g.x/ 2 xC1 xC4 x2 x 1=3 1=.x 1/ x 1 .x C 1/2 x jxj 2x C 3 x 1=x 2 x xp 4 x 2x 3 C 3 .x C 1/=x 1=.x C 1/2 f .x/ D x C 5, g.x/ D x 2 3. f ı g.0/ D f . 3/ D 2; g.f .0// D g.5/ D 22 f .g.x// D f .x 2 3/ D x 2 C 2 g ı f .x/ D g.f .x// D g.x C 5/ D .x C 5/2 f ı f . 5/ D f .0/ D 5; g.g.2// D g.1/ D f .f .x// D f .x C 5/ D x C 10 g ı g.x/ D g.g.x// D .x 2 14 Telegram: @uni_k f .x/ 3/2 3 3 2 17. p y D x. p y D 2 C px: previous graph is raised 2 units. y D 2 C 3p C x: previous graph is shiftend left 3 units. y D 1=.2 C 3 C x/: previous graph turned upside down and shrunk vertically. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.5 (PAGE 38) y 21. y y D2C y D2C p xC3 y D 1=.2 C p p .1=2;1/ yDf .2x/ x yD x C 3/ x 1 p x 22. y x yDf .x=3/ Fig. P.5-17 18. yD p 1 23. y 2x y y D 2x yD p y D 2x 1 1 6 x 3 . 2;2/ 1 yD1Cf . x=2/ 2x x x yD p 24. 1 1 2x y 1 yD2f ..x 1/=2/ yD1 2x Fig. P.5-18 1 5 x 19. y 25. y .1;2/ y D f .x/ .1; 1/ yD2f .x/ 82 x 2 x 26. 20. -2y y y D g.x/ .1; 1/ 2 x yD .1=2/f .x/ x Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 15 lOMoARcPSD|6566483 www.konkur.in SECTION P.5 (PAGE 38) 27. ADAMS and ESSEX: CALCULUS 9 F .x/ D Ax C B (a) F ı F .x/ D F .x/ ) A.Ax C B/ C B D Ax C B ) AŒ.A 1/x C B D 0 Thus, either A D 0 or A D 1 and B D 0. (b) F ı F .x/ D x ) A.Ax C B/ C B D x ) .A2 1/x C .A C 1/B D 0 Thus, either A D 1 or A D 1 and B D 0 28. bxc D 0 for 0 x < 1; dxe D 0 for 35. a) Let E.x/ D 12 Œf .x/ C f . x/. Then E. x/ D 21 Œf . x/ C f .x/ D E.x/. Hence, E.x/ is even. Let O.x/ D 12 Œf .x/ f . x/. Then O. x/ D 12 Œf . x/ f .x/ D O.x/ and O.x/ is odd. E.x/ C O.x/ D 12 Œf .x/ C f . x/ C 12 Œf .x/ 1 x < 0. f . x/ D f .x/: 29. bxc D dxe for all integers x. Hence, f .x/ is the sum of an even function and an odd function. 30. d xe D bxc is true for all real x; if x D n C y where n is an integer and 0 y < 1, then x D n y, so that d xe D n and bxc D n. 31. y y D x bxc b) If f .x/ D E1 .x/ C O1 .x/ where E1 is even and O1 is odd, then E1 .x/ C O1 .x/ D f .x/ D E.x/ C O.x/: Thus E1 .x/ E.x/ D O.x/ O1 .x/. The left side of this equation is an even function and the right side is an odd function. Hence both sides are both even and odd, and are therefore identically 0 by Exercise 36. Hence E1 D E and O1 D O. This shows that f can be written in only one way as the sum of an even function and an odd function. x 32. f .x/ is called the integer part of x because jf .x/j is the largest integer that does not exceed x; i.e. jxj D jf .x/j C y, where 0 y < 1. y Section P.6 Polynomials and Rational Functions (page 45) x y D f .x/ 1. x 2 7x C 10 D .x C 5/.x C 2/ The roots are 5 and 2. 2. x 2 3x 10 D .x 5/.x C 2/ The roots are 5 and 2. 34. f even , f . x/ D f .x/ f odd , f . x/ D f .x/ f even and odd ) f .x/ D ) f .x/ D 0 16 Telegram: @uni_k f .x/ ) 2f .x/ D 0 4 8 D 1 ˙ i. If x C 2x C 2 D 0, then x D 2 The roots are 1 C i and 1 i . x 2 C 2x C 2 D .x C 1 i /.x C 1 C i /. 4. Rather than use the quadratic formula this time, let us complete the square. x2 6x C 13 D x 2 D .x D .x f ı g. x/ D f .g. x// D f . g.x// D f .g.x// D f ı g.x/ .fg/. x/ D f . x/g. x/ D f .x/Œ g.x/ D f .x/g.x/ D .fg/.x/: The others are similar. p 3. Fig. P.5-32 33. If f is even and g is odd, then: f 2 , g 2 , f ı g, g ı f , and f ı f are all even. fg, f =g, g=f , and g ı g are odd, and f C g is neither even nor odd. Here are two typical verifications: 2˙ 2 The roots are 3 C 2i and 3 6x C 9 C 4 3/2 C 22 3 2i /.x 3 C 2i /: 2i . 5. 16x 4 8x 2 C 1 D .4x 2 1/2 D .2x 1/2 .2x C 1/2 . There are two double roots: 1=2 and 1=2. 6. x 4 C 6x 3 C 9x 2 D x 2 .x 2 C 6x C 9/ D x 2 .x C 3/2 . There are two double roots, 0 and 3. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 7. SECTION P.6 (PAGE 45) x 3 C 1 D .x C 1/.x 2 x C 1/. One root is 1. The other two are the solutions of x 2 x C 1 D 0, namely xD 1˙ p 1 4 p 3 1 ˙ i: 2 2 D 2 15. The denominator is x 3 C x 2 D x 2 .x C 1/ which is zero only if x D 0 or x D 1. Thus the rational function is defined for all real numbers except 0 and 1. 16. The denominator is x 2 Cx 1, which is a quadratic polynomial whose roots can p be found with the quadratic formula. They are x D . 1 ˙ 1 C 4/=2. Hence the given rational p 5/=2 function is p defined for all real numbers except . 1 and . 1 C 5/=2. We have p 3 i 2 1 2 x 3 C 1 D .x C 1/ x 8. x 4 1 D .x 2 1/.x 2 C 1/ D .x The roots are 1, 1, i , and i . ! p ! 1 3 C i : 2 2 x 1/.x C 1/.x i /.x C i /. 9. x 6 3x 4 C 3x 2 1 D .x 2 1/3 D .x 1/3 .x C 1/3 . The roots are 1 and 1, each with multiplicity 3. 10. x 5 x 4 16x C 16 D .x The roots are 1, 2, 11. 1/.x 4 2, 2i , and 2i . x 5 C x 3 C 8x 2 C 8 D .x 2 C 1/.x 3 C 8/ i /.x C i /.x 2 D .x C 2/.x 18. 16/ 1/.x 2 4/.x 4 C 4/ 1/.x 2/.x C 2/.x D .x D .x 17. 2i /.x C 2i /: 19. p p 20. x Cx C8x C8 D .xC2/.x i /.xCi /.x aC 3 i /.x a 3 i /: 12. 3 2 x9 4x 7 x 6 C 4x 4 D x 4 .x 5 4 D x .x x2 3 1/.x 4 D x .x xD 1˙ 2 1 4 D The required factorization of x 9 x 4 .x 1/.x 2/.xC2/ x 1 2 x.x 3 C x 2 C 1/ x 3 x C x 2 x3 C x2 C 1 3 .x C x 2 C 1/ C x 2 C 1 x C x 2 DxC x3 C x2 C 1 2 2x xC1 : D x 1C 3 x C x2 C 1 D 22. Following the method of Example 6, we calculate .x 2 bxCa2 /.x 2 CbxCa2 / D x 4 Ca4 C.2a2 b 2 /x 2 D x 4 Cx 2 C1 provided a D 1 and b 2 D 1 C 2a2 D 1, so b D ˙1. Thus P .x/ D .x 2 x C 1/.x 2 C x C 1/. x 6 C 4x 4 is p ! 1 3 C i x 2 2 x 3 C 2x 2 C 3x 2x 2 3x x 2 C 2x C 3 x.x 2 C 2x C 3/ 2x 2 3x D x 2 C 2x C 3 2.x 2 C 2x C 3/ 4x 6 C 3x Dx x 2 C 2x C 3 xC6 Dx 2C 2 : x C 2x C 3 D 4 2 As in Example p 6, we wantpa D 4, so a D 2 and a D 2, b D ˙ 2a D ˙2. Thus P .x/ D .x 2 2x C 2/.x 2 C 2x C 2/. 3 1 ˙ i: 2 2 ! 3 i : 2 p 13. The denominator is x 2 C 2x C 2 D .x C 1/2 C 1 which is never 0. Thus the rational function is defined for all real numbers. 14. x4 C x2 x3 C x2 C 1 The denominator is x 3 x D x.x 1/.x C 1/ which is zero if x D 0, 1, or 1. Thus the rational function is defined for all real numbers except 0, 1, and 1. 23. 24. Let P .x/ D an x n C an 1 x n 1 C C a1 x C a0 , where n 1. By the Factor Theorem, x 1 is a factor of P .x/ if and only if P .1/ D 0, that is, if and only if an C an 1 C C a1 C a0 D 0. Let P .x/ D an x n C an 1 x n 1 C C a1 x C a0 , where n 1. By the Factor Theorem, x C 1 is a factor of P .x/ if and only if P . 1/ D 0, that is, if and only if a0 a1 C a2 a3 C C . 1/n an D 0. This condition says that the sum of the coefficients of even powers is equal to the sum of coefficients of odd powers. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 3 21. p 4x 7 x3 x 2 C 2x C 3 4/ Seven of the nine roots are: 0 (with multiplicity 4), 1, 2, and 2. The other two roots are solutions of x 2 C x C 1 D 0, namely p x 2 C 5x C 3 5x x2 D x 2 C 5x C 3 x 2 C 5x C 3 5x 3 : D1C 2 x C 5x C 3 4x 3 C 4/ 2 2/.x C 2/.x 2 C x C 1/: 1/.x 1 x 3 2x C 2x 1 D 2 x2 2 x.x 2 2/ C 2x 1 D x2 2 2x 1 DxC 2 : x 2 2x C 4/ Three of the five roots are 2, i and i . The remain2 ing two are 2x C 4 D 0, namely p solutions of x p 2 ˙ 4 16 xD D 1 ˙ 3 i . We have 2 5 x3 x2 Downloaded by ted cage (sxnbyln180@questza.com) 17 lOMoARcPSD|6566483 www.konkur.in SECTION P.6 (PAGE 45) ADAMS and ESSEX: CALCULUS 9 25. Let P .x/ D an x n C an 1 x n 1 C C a1 x C a0 , where the coefficients ak , 0 k n are all real numbers, so that ak D ak . Using the facts about conjugates of sums and products mentioned in the statement of the problem, we see that if z D x C iy, where x and y are real, then 4. sin 5. cos 6. sin P .z/ D an z n C an 1 z n 1 C C a1 z C a0 D an z n C an 1 z n 1 C C a1 z C a0 D P .z/: If z is a root of P , then P .z/ D P .z/ D 0 D 0, and z is also a root of P . 26. By the previous exercise, z D u iv is also a root of P . Therefore P .x/ has two linear factors x u iv and x u C iv. The product of these factors is the real quadratic factor .x u/2 i 2 v 2 D x 2 2ux C u2 C v 2 , which must also be a factor of P .x/. 27. By the previous exercise P .x/ D 2ux C u2 C v 2 .x x2 u P .x/ iv/.x u C iv/ D Q1 .x/; where Q1 , being a quotient of two polynomials with real coefficients, must also have real coefficients. If z D u C iv is a root of P having multiplicity m > 1, then it must also be a root of Q1 (of multiplicity m 1), and so, therefore, z must be a root of Q1 , as must be the real quadratic x 2 2ux C u2 C v 2 . Thus .x 2 P .x/ D 2 2ux C u2 C v 2 /2 x Q1 .x/ D Q2 .x/; 2ux C u2 C v 2 where Q2 is a polynomial with real coefficients. We can continue in this way until we get .x 2 7. P .x/ D Qm .x/; 2ux C u2 C v 2 /m 7 12 1. cos 3 4 D cos 2. tan 3 D 4 3. sin 2 D sin 3 18 Telegram: @uni_k tan D 4 3 D 4 cos D 4 11 D sin 12 12 D sin 3 4 cos sin D sin cos 3 4 3 4 p ! 3 1 1 1 D p p 2 2 2 2 p 3 1 D p 2 2 cos. C x/ D cos 2 . x/ D cos . x/ D cos. 8. sin.2 x/ D 9. sin x 10. 3 Cx cos 2 3 2 ! x/ D cos x sin x D sin x 2 D sin x 2 D sin x 2 D cos x D cos 3 cos x 2 sin 3 sin x 2 D . 1/. sin x/ D sin x 1 p 2 11. 1 C 2 5 D cos 12 3 4 2 2 D cos cos C sin sin 3 4 3 4 p ! 1 3 1 1 C D p p 2 2 2 2 p 3 1 D p 2 2 where Qm no longer has z (or z) as a root. Thus z and z must have the same multiplicity as roots of P . Section P.7 The Trigonometric Functions (page 57) 4 3 D sin cos C cos sin 4 3 p 4 3 p 1 1 3 1 1C 3 D p p Cp D 22 2 2 2 2 D sin cos x sin x C cos x sin x sin2 x C cos2 x D cos x sin x 1 D cos x sin x tan x C cot x D p 3 D sin D 3 3 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 12. cos x sin x tan x cot x cos x sin x D cos x sin x tan x C cot x C cos x sin x ! 2 sin x cos2 x cos x sin x ! D sin2 x C cos2 x cos x sin x D sin2 x 13. cos4 x sin4 x D .cos2 x 16. 17. .cos x sin x/2 cos x sin x D cos x C sin x .cos x C sin x/.cos x sin x/ cos2 x 2 sin x cos x C sin2 x D cos2 x sin2 x 1 sin.2x/ D cos.2x/ D sec.2x/ tan.2x/ D 2 sin x cos x C sin x.1 D 3 sin x sin x has period 4. 2 y sin2 x D cos.2x/ sin 3x D sin.2x C x/ D sin 2x cos x C cos 2x sin x 1 Fig. P.7-20 21. sin x has period 2. y y D sin.x/ 1 1 D 2 cos3 x D 4 cos3 x sin x/ C sin x Fig. P.7-21 22. cos x has period 4. 2 y 1 1 x 1 Fig. P.7-22 2 2 sin3 x 23. y 4 sin x 2 2.1 5 3 3 cos x 4 x 3 1 y D 2 cos x 3 1 2 sin2 x cos x cos2 x/ cos x 3 cos x x -1 -2 -3 19. cos 2x has period . Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 2 sin x/ 2 1/ cos x 2 x 1 cos 3x D cos.2x C x/ D cos 2x cos x sin 2x sin x D .2 cos2 x 2 x Fig. P.7-19 sin2 x/.cos2 x C sin2 x/ 2 18. =2 .1 cos x/.1 C cos x/ D 1 cos2 x D sin2 x implies 1 cos x sin x D . Now sin x 1 C cos x x 1 cos 2 1 cos x x 2 D sin x sin 2 2 x 1 1 2 sin2 2 D x x 2 sin cos 2 2 x sin 2 D tan x D x 2 cos 2 x 2 sin2 x 1 cos x 2x D tan2 D 1 C cos x 2 2 cos2 2 D 2 sin x.1 y D cos.2x/ 1 20. D cos x 15. y cos2 x 2 14. SECTION P.7 (PAGE 57) Downloaded by ted cage (sxnbyln180@questza.com) 19 lOMoARcPSD|6566483 www.konkur.in SECTION P.7 (PAGE 57) ADAMS and ESSEX: CALCULUS 9 24. y D 1 C sin x C 4 y 2 29. sin x D cos x D 1 ; 2p <x< 3 2 3 2 1 tan x D p 3 1 p 1 3 ; <x< 5 2 4 3 cos x D ; tan x D 5 4 Fig. P.7-29 5 3 x 4 Fig. P.7-25 . Then 2 2 2 sec x D 1 C tan x D 1 C 4 D 5. Hence, p 1 1 sec x D 5 and cos x D D p , sec x 5 2 sin x D tan x cos x D p . 5 26. tan x D 2 where x is in Œ0; cos x D sin x D tan x D <x<0 2 8 2p D 2 3 3 p p 8 D 2 2 1 1 ; 3p 1 x 3 p 8 Fig. P.7-27 h i 5 where x is in ; . Hence, 13 2 r p 12 25 D , sin x D 1 cos2 x D 1 169 13 12 tan x D . 5 28. cos x D 20 Telegram: @uni_k 2 sin x D 3 1 where x is in Œ; . Then, 2 2 1 5 sec2 x D 1 C D . Hence, 4 p 4 5 2 ; cos x D p ; sec x D 2 5 1 sin x D tan x cos x D p . 5 31. c D 2; B D 3 1 a D c cos B D 2 D 1 2 p p 3 b D c sin B D 2 D 3 2 32. b D 2; B D 3 p 2 2 D tan B D 3 ) a D p c a 3 B p 4 2 3 D sin B D )cD p a c 2 3 33. a D 5; B D C 6 1 5 b D a tan B D 5 p D p 3 3 r p 10 25 D p c D a2 C b 2 D 25 C 3 3 a 34. sin A D ) a D c sin A c a D tan A ) a D b tan A 35. b a 36. cos B D ) a D c cos B c b D tan B ) a D b cot B 37. a a a 38. sin A D ) c D c sin A b D cos A ) c D b sec A 39. c 30. 27. x x -1 25. 3 tan x D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) A b lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 40. sin A D a c 41. sin A D a D c 42. sin A D a a D p c a2 C b 2 SECTION P.7 (PAGE 57) 50. p c 2 b2 c 43. a D 4, b D 3, A D 4 sin A 3 1 3 sin B D b D p D p a 4 2 4 2 c 2 D a2 C b 2 2ab cos C p D 1 C 2 2 2 cos C p p D 3 .1 3/ or 3 .1 C 3/ p p D 2 C 3 or 2 3: A c b B 45. a D 2, b D 3, c D 4 b 2 D a2 C c 2 2ac cos B 4 C 16 9 11 Thus cos B D D 2 2 p 4 16 r p 112 256 121 135 D D sin B D 1 2 16 16 16 a C 46. Given that a D 2; b D 3; C D : 4 12 2 2 2 c D a Cb 2ab cos C D 4C9 2.2/.3/ cos D 13 p . 4 2 s 12 Hence, c D 13 p 2:12479. 2 5 implies C D c D 3, A D , B D 4 3 12 c 1 a 3 D )aD p 5 sin A sin C 2 sin 12 1 3 aD p 7 2 sin 12 p 3 2 2 D p p (by #5) 21C 3 6 D p 1C 3 Hence, c D p 2C p 3 or p 2 p 3. C p 2 1 1 =6 A B0 B 00 Fig. P.7-50 51. Let h be the height of the pole and x be the distance from C to the base of the pole. Then h D x tan 50ı and h D .x C 10/ tan 35ı Thus x tan 50ı D x tan 35ı C 10 tan 35ı so 10 tan 35ı tan 50ı tan 35ı 10 tan 50ı tan 35ı hD 16:98 tan 50ı tan 35ı xD The pole is about 16.98 metres high. 52. ı 48. Given that a D 2; b D 3; C D 35 . Then c 2 D 4 C 9 2.2/.3/ cos 35ı , hence c 1:78050. See the following diagram. Since tan 40ı D h=a, therefore a D h= tan 40ı . Similarly, b D h= tan 70ı . Since a C b D 2 km, therefore, 49. a D 4, B D 40ı , C D 70ı Thus A D 70ı . b 4 sin 40ı D so b D 4 D 2:736 ı ı sin 40 sin 70 sin 70ı h h C D2 tan 40ı tan 70ı ı 2.tan 40 tan 70ı / 1:286 km: hD tan 70ı C tan 40ı Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2; A D 30ı , then Hence, 44. Given that a D 2; b D 2; c D 3: Since a2 D b 2 C c 2 2bc cos A, a2 b 2 c 2 cos A D 2bc 3 4 4 9 D . D 2.2/.3/ 4 47. p sin B sin A 1 D D . b a 2 p 2 1 3 Thus sin B D D p ,BD or , and 2 4 2 4 3 7 C D C or C D C . D D 4 6 12 4 6 p 12 3 7 1 p or Thus, cos C D cos D cos C D 12 4 3 p 2 2 1C 3 cos C D cos D cos D p . 12 3 4 2 2 If a D 1; b D Downloaded by ted cage (sxnbyln180@questza.com) 21 lOMoARcPSD|6566483 www.konkur.in SECTION P.7 (PAGE 57) ADAMS and ESSEX: CALCULUS 9 54. Balloon h sin B D 40ı 70ı b a A D B 1 ah ac sin B ab sin C 53. Area 4ABC D jBC jh D D D 2 2 2 2 1 By symmetry, area 4ABC also D bc sin A 2 A h s.s a/.s Thus p s.s b P Fig. P.7-53 22 Telegram: @uni_k 1 p 2 2 a C c 2 b2 2ac a4 b4 c 4 C 2a2 b 2 C 2b 2 c 2 C 2a2 c 2 : 2ac p a4 b4 square units. Since, C B s Hence, Area D Fig. P.7-52 c From Exercise 53, area D 12 ac sin B. By Cosine Law, a2 C c 2 b 2 . Thus, cos B D 2ac c4 C 2a2 b2 C 2b2 c2 C 2a2 c2 4 b/.s c/ bCcCa bCc a a bCc aCb c D 2 2 2 2 1 2 2 2 2 D .b C c/ a a .b c/ 16 1 a4 .b 2 c 2 /2 a2 .b C c/2 C .b c/2 D 16 1 2 2 D 2a b C 2a2 c 2 a4 b 4 c 4 C 2b 2 c 2 16 a/.s Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) b/.s c/ = Area of triangle. lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHAPTER 1. SECTION 1.1 (PAGE 63) LIMITS AND CONTINUITY 8. 3.t C k/2 Section 1.1 Examples of Velocity, Growth Rate, and Area (page 63) 1. Average velocity = .t C h/2 x D t h t2 D m/s. D 2. Avg. vel. over Œ2; 2 C h h 1 0:1 0:01 0:001 0:0001 k; t C k is 12.t C k/ C 1 Œ3.t k/2 .t C k/ .t k/ 1 2 3t C 6t k C 3k 2 12t 2k C 12t 12k C 1 12t k 24k D 6t 2k 12.t 12k C 1 k/ C 1 3t 2 C 6t k 3k 2 12 m/s; which is the velocity at time t from Exercise 7. 5.0000 4.1000 4.0100 4.0010 4.0001 9. y 2 3. Guess velocity is v D 4 m/s at t D 2 s. 4 D y D2C 1 4. Average velocity on Œ2; 2 C h is 4 C 4h C h2 .2 C h/2 4 D .2 C h/ 2 h Average velocity over Œt 4h C h2 D 4 C h: h 1 2 1 sin. t / 3 Fig. 1.1-9 4 5 t As h approaches 0 this average velocity approaches 4 m/s 5. 6. x D 3t 2 12t C 1 m at time t s. Average velocity over interval Œ1; 2 is .3 22 12 2 C 1/ .3 12 12 1 C 1/ D 3 m/s. 2 1 Average velocity over interval Œ2; 3 is .3 32 12 3 C 1/ .3 22 12 2 C 1/ D 3 m/s. 3 2 Average velocity over interval Œ1; 3 is .3 32 12 3 C 1/ .3 12 12 1 C 1/ D 0 m/s. 3 1 At t D 1 the height is y D 2 ft and the weight is moving downward. 10. 1 1 sin .1 C h/ 2 C sin h sin cos.h/ C cos sin.h/ sin. C h/ D D h h sin.h/ D : h 2C Average velocity over Œt; t C h is 3.t C h/2 12.t C h/ C 1 .3t 2 12t C 1/ .t C h/ t 6t h C 3h2 12h D 6t C 3h 12 m/s: D h This average velocity approaches 6t 12 m/s as h approaches 0. At t D 1 the velocity is 6 1 12 D 6 m/s. At t D 2 the velocity is 6 2 12 D 0 m/s. At t D 3 the velocity is 6 3 12 D 6 m/s. 7. Average velocity over Œ1; 1 C h is At t D 1 the velocity is v D 6 < 0 so the particle is moving to the left. At t D 2 the velocity is v D 0 so the particle is stationary. At t D 3 the velocity is v D 6 > 0 so the particle is moving to the right. h 1:0000 0:1000 0:0100 0:0010 11. Avg. vel. on Œ1; 1 C h The velocity at t D 1 is about v D 1 ft/s. The “ ” indicates that the weight is moving downward. 12. We sketched a tangent line to the graph on page 55 in the text at t D 20. The line appeared to pass through the points .10; 0/ and .50; 1/. On day 20 the biomass is growing at about .1 0/=.50 10/ D 0:025 mm2 /d. 13. The curve is steepest, and therefore the biomass is growing most rapidly, at about day 45. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 0 -0.983631643 -0.999835515 -0.999998355 Downloaded by ted cage (sxnbyln180@questza.com) 23 lOMoARcPSD|6566483 www.konkur.in SECTION 1.1 (PAGE 63) 14. a) ADAMS and ESSEX: CALCULUS 9 profit 175 150 125 100 75 50 25 3. 4. lim g.x/ D 0 x!1C 5. lim g.x/ D 0 x!3C 6. 2011 2012 2013 2014 2015 7. Fig. 1.1-14 8. b) Average rate of increase in profits between 2010 and 2012 is 174 62 112 D D 56 (thousand$/yr). 2012 2010 2 c) Drawing a tangent line to the graph in (a) at t D 2010 and measuring its slope, we find that the rate of increase of profits in 2010 is about 43 thousand$/year. Section 1.2 Limits of Functions (page 71) y D f .x/ 9. 10. 11. 12. 1 x 1 we see that lim f .x/ D 0; x!0 lim f .x/ D 1: x!1 2. From inspecting the graph y y D g.x/ 1 4x C 1/ D 42 lim 3.1 x/.2 lim xC3 x!3 x C 6 t2 lim t! 4 4 1 2 3 x Fig. 1.2-2 we see that t x!2 24 Telegram: @uni_k x!3 . 4/2 D2 4C4 x2 1 D lim .x x! 1 x! 1 x C 1 lim 1/ D 2 .x 3/2 6x C 9 D lim x!3 9 .x 3/.x C 3/ 0 x 3 D D0 D lim x!3 x C 3 6 lim x2 x2 x 2 C 2x x D lim D x! 2 x 2 x! 2 x 4 2 15. limh!2 lim 2 1 D 4 2 1 does not exist; denominator approaches 0 4 h2 but numerator does not approach 0. 3 C 4h 3h C 4h2 D lim does not exist; denomih2 h3 h2 h!0 h nator approaches 0 but numerator does not approach 0. p p p x 3 . x 3/. x C 3/ D lim p 17. lim x!9 .x x!9 x 9 9/. x C 3/ 1 1 x 9 p D lim p D D lim x!9 x!9 .x 6 9/. x C 3/ xC3 p 4Ch 2 18. lim h h!0 4Ch 4 D lim p h!0 h. 4 C h C 2/ 1 1 D lim p D 4 h!0 4ChC2 16. limh!0 19. lim .x 02 /2 D 2 D0 x lim jx 2j D j x! 2 (left limit is 1, right limit is 0) lim g.x/ D 1; lim g.x/ D 0: 2/ D 0 3C3 2 D 3C6 3 14. 20. x!1 D x/ D 3. 1/.2 12 1 0 x 1 D D D0 x!1 x C 1 1C1 2 x! lim g.x/ does not exist D 4.4/ C 1 D 1 lim x!3 Fig. 1.2-1 lim f .x/ D 1; lim .x 2 2 13. 1 x! 1 x!4 x!2 From inspecting the graph y lim g.x/ D 0 x!3 year 1. lim g.x/ D 1 x!1 21. lim x!0 jx x 4j D 4 2j j 2j D D 2 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 22. SECTION 1.2 (PAGE 71) 2j 1; if x > 2 D lim 1; if x < 2. x!2 2 jx 2j Hence, lim does not exist. x!2 x 2 jx x!2 x lim 33. lim t2 lim t!1 t 2 34. x 1 1 2 x2 4 xC2 1 D lim x!2 .x 2/.x C 2/ xC1 does not exist. D lim x!2 .x 2/.x C 2/ x!2 35. 4 1 2 x2 4 xC2 4 1 1 D lim D lim D x!2 .x x!2 x C 2 2/.x C 2/ 4 x!2 1 2t C 1 t C1 .t 1/.t C 1/ D lim does not exist lim 2 t!1 t!1 .t 1/ t 1 (denominator ! 0, numerator ! 2.) p 4 4x C x 2 24. lim x!2 x 2 jx 2j does not exist. D lim x!2 x 2 p p t t. 4 C t C 4 t/ 25. lim p p D lim t!0 t!0 .4 C t / .4 t / 4Ct 4 t p p 4Ct C 4 t D lim D2 t!0 2 p x2 1 .x 1/.x C 1/. x C 3 C 2/ 26. lim p D lim x!1 x C 3 x!1 .x C 3/ 4 2 p p D lim .x C 1/. x C 3 C 2/ D .2/. 4 C 2/ D 8 23. lim lim x p x!0 2 C x2 p 2 x2 x2 .2 C x 2 / .2 x 2 / p p 2 C x2 C 2 x2/ 2x 2 p D lim p x!0 2 x 2 C x2/ C 2 x2 1 2 p D p D p 2C 2 2 D lim x!0 x 2 . x!1 27. 28. t 2 C 3t t!0 .t C 2/2 .t 2/2 t .t C 3/ D lim 2 t!0 t C 4t C 4 .t 2 t C3 3 D lim D t!0 8 8 36. lim s!0 .s C 1/2 .s s 1/2 4t C 4/ D lim s!0 4s D4 s 31. lim 32. 38. lim x 4 16 x!2 x 3 8 .x 2/.x C 2/.x 2 C 4/ D lim x!2 .x 2/.x 2 C 2x C 4/ 8 .4/.8/ D D 4C4C4 3 1j f .x C h/ h h!0 p 4 yC3 29. lim y!1 y 2 p1 p . y 1/. y 3/ 2 1 D lim p D D p y!1 . y 1/. y C 1/.y C 1/ 4 2 x3 C 1 x! 1 x C 1 .x C 1/.x 2 x C 1/ D3 D lim x! 1 xC1 j3x 37. y 30. j3x C 1j x .3x 1/2 .3x C 1/2 D lim x!0 x .j3x 1j C j3x C 1j/ 12 12x D D D lim x!0 x .j3x 1j C j3x C 1j/ 1C1 lim x!0 lim f .x C h/ h h!0 lim f .x/ x 2=3 4 x!8 x 1=3 2 .x 1=3 2/.x 1=3 C 2/ D lim x!8 .x 1=3 2/ D lim .x 1=3 C 2/ D 4 .x C h/2 x 2 h h!0 2hx C h2 D lim 2x C h D 2x D lim h h!0 h!0 D lim f .x/ D x 3 f .x/ .x C h/3 x 3 h h!0 3x 2 h C 3xh2 C h3 D lim h h!0 D lim 3x 2 C 3xh C h2 D 3x 2 D lim h!0 f .x/ D 1=x 39. f .x C h/ lim h h!0 x!8 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k f .x/ D x 2 lim lim 6 Downloaded by ted cage (sxnbyln180@questza.com) f .x/ 1 1 x C h x D lim h h!0 x .x C h/ D lim h!0 h.x C h/x 1 D lim D h!0 .x C h/x 1 x2 25 lOMoARcPSD|6566483 www.konkur.in SECTION 1.2 (PAGE 71) f .x/ D 1=x 2 40. lim h!0 f .x C h/ h 41. f .x C h/ lim h h!0 48. 1 1 f .x/ .x C h/2 x 2 D lim h h!0 x 2 .x 2 C 2xh C h2 / D lim h.x C h/2 x 2 h!0 2x C h 2x D lim D D x4 h!0 .x C h/2 x 2 f .x/ D f .x/ p lim h!0 f .x C h/ h f .x/ x p p lim sin x D sin =2 D 1 50. 51. 52. 53. lim p 2 lim p p x3 lim x!0 3 1 1 p x 54. x does not exist. 2 2 55. 56. p x3 x does not exist. (See # 9.) lim p x2 x4 D 0 x!0C 57. jx x2 aj a2 jx aj D D lim x!a .x a/.x C a/ lim x!a 58. lim jx lim 0 x2 4 D D0 jx C 2j 4 x!aC x 2 x!2 3=2 sin x D 1. x!0 x It appears that lim 26 Telegram: @uni_k aj x D lim 2 x!aC x a2 63. 1 a D a2 2a x! 1 lim f .x/ D lim x 2 C 1 D 1 C 1 D 2 x! 1C lim f .x/ D lim .x C /2 D 2 x!0C lim f .x/ D lim x 2 C 1 D 1 x!0 65. .a ¤ 0/ x2 4 0 D D0 x!2C jx C 2j 4 8 if x 1 <x 1 if 1 < x 0 f .x/ D x 2 C 1 : .x C /2 if x > 0 lim f .x/ D lim x 1 D 1 1 D x!0C 64. 1 2a lim x! 1 47. 0:84147098 0:99833417 0:99998333 0:99999983 1:00000000 x does not exist. lim x! 1C ˙1:0 ˙0:1 ˙0:01 ˙0:001 0:0001 xD2 x < 0 if 0 < x < 1/ p x3 x D 0 lim x!0C 62. .sin x/=x xD2 x!0 61. x 1 cos x D . x2 2 .x lim cos x D cos =3 D 1=2 x!2=3 1 xD0 2 x! 2C 59. p lim p x! 2 60. sin x D sin 2=3 D p x!2C p lim cos x D cos =4 D 1= 2 lim lim x!2 h!0 x!=3 46. 0:45969769 0:49958347 0:49999583 0:49999996 0:50000000 It appears that lim 49. p xCh h p p x xCh D lim p p h!0 h x x C h x .x C h/ D lim p p p p h!0 h x x C h. x C x C h/ 1 D lim p p p p h!0 x x C h. x C x C h/ 1 D 3=2 2x D lim x!=4 45. cos x/=x 2 .1 ˙1:0 ˙0:1 ˙0:01 ˙0:001 0:0001 2 x3 xCh x D lim h h!0 xCh x D lim p p h!0 h. x C h C x/ 1 1 D lim p p D p 2 x h!0 xChC x x!=2 44. x x!0 p f .x/ D 1= x 42. 43. ADAMS and ESSEX: CALCULUS 9 x!0 If lim f .x/ D 2 and lim g.x/ D 3, then x!4 x!4 a) lim g.x/ C 3 D 3 C 3 D 0 x!4 b) lim xf .x/ D 4 2 D 8 x!4 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 1.2 (PAGE 71) 2 c) lim g.x/ D . 3/2 D 9 70. y x!4 d) lim g.x/ x!4 f .x/ 1 D 3 D 3 1 2 0.8 0.6 yD 0.4 66. If lim x ! af .x/ D 4 and lim g.x/ D x!a 2, then 0.2 a) lim f .x/ C g.x/ D 4 C . 2/ D 2 -0.08 x!a b) lim f .x/ g.x/ D 4 . 2/ D c) lim 4g.x/ D 4. 2/ D x!a 4 f .x/ D D d) lim x!a g.x/ 2 67. -0.4 8 Fig. 1.2-70 2 limx!0 sin.2x/= sin.3x/ D 2=3 71. f .x/ 5 D 3, then x!2 x 2 x!2 y f .x/ 5 .x 5 D lim x!2 x 2 2/ D 3.2 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 2/ D 0: Thus limx!2 f .x/ D 5. 68. If lim x!0 f .x/ D x2 2 then f .x/ D 0 . 2/ D 0, x2 f .x/ f .x/ and similarly, limx!0 D lim x 2 D 0. 2/ D 0. x!0 x x limx!0 f .x/ D limx!0 x 2 0.2 0.4 0.6 0.8 1.0 x 0.8 x Fig. 1.2-71 x!1 69. p sin 1 x yD p 1 x2 -0.1 lim p sin 1 x p 0:7071 1 x2 72. y 1.2 1.0 0.8 y yD sin x x 0.2 -0.6 x yD p -0.8 -2 -1 -0.2 -0.4 1 2 sin x D1 x!0 x p x sin x x -1.0 -1.2 Fig. 1.2-69 lim 0.6 -0.4 0.2 -3 0.4 -0.2 0.6 0.4 Fig. 1.2-72 lim x!0C x p p x sin x Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 0.08 x 0.04 -0.2 If lim lim f .x/ -0.04 8 x!a sin.2x/ sin.3x/ Downloaded by ted cage (sxnbyln180@questza.com) D 1 27 lOMoARcPSD|6566483 www.konkur.in SECTION 1.2 (PAGE 71) ADAMS and ESSEX: CALCULUS 9 73. yD -0.2 y x f .x/ D s sin 79. jf .x/j g.x/ ) g.x/ f .x/ g.x/ Since lim g.x/ D 0, therefore 0 lim f .x/ 0. yDx y D x sin.1=x/ 0.1 -0.1 x 0.1 1 is defined for all x ¤ 0; its domain is x . 1; 0/ [ .0; 1/. Since j sin t j 1 for all t , we have jf .x/j jxj and jxj f .x/ jxj for all x ¤ 0. Since limx!0 D . jxj/ D 0 D limx!0 jxj, we have limx!0 f .x/ D 0 by the squeeze theorem. 78. x!a x!a Hence, lim f .x/ D 0. x!a -0.1 If lim g.x/ D 3, then either 3 lim f .x/ 3 or x!a x!a limx!a f .x/ does not exist. -0.2 Fig. 1.2-73 f .x/ D x sin.1=x/ oscillates infinitely often as x approaches 0, but the amplitude of the oscillations decreases and, in fact, limx!0 f .x/ D 0. This is predictable because jx sin.1=x/j jxj. (See Exercise 95 below.) p p 74. Since p 5 2x 2 f .x/ 5p x 2 for 1p x 1, and 2 D lim 5 x 2 D 5, we have limx!0 5 2xp x!0 limx!0 f .x/ D 5 by the squeeze theorem. 75. Since 2 x 2 g.x/ 2 cos x for all x, and since limx!0 .2 x 2 / D limx!0 2 cos x D 2, we have limx!0 g.x/ D 2 by the squeeze theorem. 76. Section 1.3 Limits at Infinity and Infinite Limits (page 78) 1. 2. 3. a) y 3 yDx 4. 4 2 . 1; 1/ 1 y D x2 .1; 1/ 5. -2 -1 x 1 Fig. 1.2-76 2 b) Since the graph of f lies between those of x and x 4 , and since these latter graphs come together at .˙1; 1/ and at .0; 0/, we have limx!˙1 f .x/ D 1 and limx!0 f .x/ D 0 by the squeeze theorem. 77. x 1=3 < x 3 on . 1; 0/ and .1; 1/. x 1=3 > x 3 on . 1; 1/ and .0; 1/. The graphs of x 1=3 and x 3 intersect at . 1; 1/, .0; 0/, and .1; 1/. If the graph of h.x/ lies between those of x 1=3 and x 3 , then we can determine limx!a h.x/ for a D 1, a D 0, and a D 1 by the squeeze theorem. In fact lim h.x/ D x! 1 28 Telegram: @uni_k 1; lim h.x/ D 0; x!0 lim h.x/ D 1: x!1 6. 7. lim lim 3 D lim 1 1 D .3=x/ 2 4 D lim 1=x 0 D D0 .4=x 2 / 1 x x!1 2x x x!1 x 2 x!1 2 x!1 1 3x 3 5x 2 C 7 x!1 8 C 2x 5x 3 5 7 3 C 3 x x D D lim 2 x!1 8 C 5 x3 x2 lim x2 2 x! 1 x x2 2 1 2 1 x D D lim D x! 1 1 1 1 x 3 5 lim 1 1 3 C 3 x2 C 3 x x D lim D0 lim 2 x! 1 x! 1 x 3 C 2 1C 3 x sin x 1C 2 1 x 2 C sin x x lim D lim cos x D 1 D 1 x!1 x 2 C cos x x!1 1C 2 x sin x We have used the fact that limx!1 2 D 0 (and simx ilarly for cosine) because the numerator is bounded while the denominator grows large. p 3x C 2 x lim x!1 1 x 2 3C p x D lim D x!1 1 1 x Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 3 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 8. 2x lim p SECTION 1.3 (PAGE 78) 1 3x 2 Cx C 1 1 x 2 x r D lim (but jxj D x as x ! 1/ x!1 1 1 jxj 3 C C 2 x x 1 2 2 x D lim r D p x!1 1 1 3 3C C 2 x x 2x 1 9. lim p x! 1 3x 2 C x C 1 1 2 2 x D lim D p , r x! 1 1 1 3 3C C 2 x x p because x ! 1 implies that x < 0 and so x 2 D x. x!1 10. 11. 2x lim 5 x! 1 j3x C 2j 1 lim D lim x! 1 2x 5 D .3x C 2/ 14. 15. 16. 17. 1 x D1 x!3C 3 x D lim 2x C 5 lim x!3 3 1 lim x! 5=2 5x C 2 19. 20. 28. 29. 1 D 0 D0 25 C2 2 30. 2x C 5 does not exist. x! 2=5 5x C 2 2x C 5 D 5x C 2 lim 2x C 5 lim x! 2=5C 5x C 2 x lim x/3 x!2C .2 22. x 2 C 2x p x!1 x 2 C 2x C x 2 2x 4x D lim r r x!1 2 2 x 1C Cx 1 x x 4 4 D D2 r D lim r x!1 2 2 2 1C C 1 x x 1 1 D1 1 x2 1 D1 lim x!1C jx 1j lim x!1 23. 24. lim 1 jx 1j x 31. lim x!1C x 2 C 2x 1 x2 2x p x 2 2x C x p p x!1 . x 2 2x C x/. x 2 2x x/ p x 2 2x C x D lim 2 x!1 x 2 p 2x x x. 1 .2=x/ C 1/ 2 D D 1 D lim x!1 2x 2 D1 x2 x 1 D lim p D 2 x!1C x x2 x x 1 32. lim p x! 1 1 x 2 C 2x Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x D lim 1 p lim p x!1 3 x 3 D D lim x!2 .x 4x C 4 2/2 x!2 x 2 .x 2 C 2x/ .x 2 2x/ p D lim p x! 1 x 2 C 2x C x 2 2x 4x ! D lim r r x! 1 2 2 . x/ 1C C 1 x x 4 D 2 D 1C1 p p 2 2 lim x C 2x x 2x D lim p x lim p 3 xC 2 x3 C 3 x lim D1 D lim 2 x!1 x 2 C 2 x!1 1C 2 x p p 2x C 3 x xC1 1 lim 2 x!1 7 r6x C 4x ! ! r 1 1 3 2 1C 2C x p x x x D lim x!1 6 7 x2 C4 x2 x p 1. 2/ 1p 2 D D 4 4 2 x x2 2x 2 lim D 2 D lim 2 x!1 x C 1 x!1 x x 1 1 p p lim x 2 C 2x x 2 2x x!1 D1 D x C x3 C x5 x! 1 x!1 21. 27. lim x! .2=5/ 18. 26. does not exist. x!3 3 lim x!1 1 C x 2 C x 3 1 C 1 C x2 2 x D1 D lim 1 x!1 1 C C1 3 x x 2 3 x 1 D1 12. lim x!3 .3 x/2 13. 25. Downloaded by ted cage (sxnbyln180@questza.com) x D lim p x! 1 jxj. 1 1 C .2=x/ C 1 D0 29 lOMoARcPSD|6566483 www.konkur.in SECTION 1.3 (PAGE 78) ADAMS and ESSEX: CALCULUS 9 33. By Exercise 35, y D 1 is a horizontal asymptote (at the 1 . Since right) of y D p 2 x 2x x lim p x! 1 1 x2 2x x p D lim x! 1 jxj. 1 1 .2=x/ C 1 D 0; 46. 47. horizontal: y D 1; vertical: x D 1, x D 3. lim bxc D 3 x!3C lim bxc D 2 x!3 49. lim bxc does not exist x!3 50. lim bxc D 2 x!2:5 51. lim b2 x!0C 52. xc D lim bxc D 1 x!2 lim bxc D x! 3 53. 4 lim C.t / D C.t0 / except at integers t0 t!t0 lim C.t / D C.t0 / everywhere t!t0 lim f .x/ D 1 lim C.t / D C.t0 / if t0 ¤ an integer x!0C 36. lim f .x/ D 1 x!1 48. y D 0pis also a horizontal asymptote (at the left). Now x 2 2x x D 0 if and only if x 2 2x D x 2 , that is, if and only if x D 0. The given function is undefined at x D 0, and where x 2 2x < 0, that is, on the interval Œ0; 2. Its only vertical asymptote is at x D 0, where 1 D 1. limx!0 p x 2 2x x 2 2x 5 2 2x 5 D and lim D , 34. Since lim x! 1 j3x C 2j x!1 j3x C 2j 3 3 y D ˙.2=3/ are horizontal asymptotes of y D .2x 5/=j3x C 2j. The only vertical asymptote is x D 2=3, which makes the denominator zero. 35. 45. t!t0 C lim f .x/ D 1 lim C.t / D C.t0 / C 1:5 if t0 is an integer x!1 t!t0 C 37. y 6:00 y 3 y D f .x/ 4:50 2 y D C.t / 3:00 1:50 1 1 1 2 3 4 -1 5 6 x 2 3 4 x Fig. 1.3-53 54. lim f .x/ D L x!0C (a) If f is even, then f . x/ D f .x/. Hence, lim f .x/ D L. Fig. 1.3-37 x!0 (b) If f is odd, then f . x/ D Therefore, lim f .x/ D L. limx!2C f .x/ D 1 38. lim f .x/ D 2 x!2 39. f .x/. x!0 55. lim f .x/ D A; lim f .x/ D 1 x!0C lim f .x/ D 1 a) lim f .x/ D 2 b) lim f .x 3 lim f .x/ D B x!0 x!3 40. x!3C 41. x!0 x!4C 42. lim f .x/ D 0 x!4 43. lim f .x/ D 1 x!5 44. lim f .x/ D 0 x!5C 30 Telegram: @uni_k lim f .x 3 x!0C 1 < x < 0) c) lim f .x 2 x!0 d) lim f .x 2 x!0C 0 < jxj < 1) Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x/ D B (since x 3 x/ D A (because x 3 x < 0 if 0 < x < 1) x > 0 if x4/ D A x 4 / D A (since x 2 x 4 > 0 for lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL Section 1.4 Continuity 1. SECTION 1.4 (PAGE 87) (page 87) 8. g is continuous at x D 2, discontinuous at x D 1; 0; 1, and 2. It is left continuous at x D 0 and right continuous at x D 1. y .1; 2/ 2 y D g.x/ . 1; 1/ -2 1 lim f .x/ D x! 1 1 2 x Fig. 1.4-1 2. g has removable discontinuities at x D 1 and x D 2. Redefine g. 1/ D 1 and g.2/ D 0 to make g continuous at those points. D f . 1/ D 4. Function f is discontinuous at x D 1; 2; 3; 4, and 5. f is left continuous at x D 4 and right continuous at x D 2 and x D 5. y 13. 15. 4 5 6 x 16. -1 Fig. 1.4-4 5. f cannot be redefined at x D 1 to become continuous there because limx!1 f .x/ .D 1/ does not exist. (1 is not a real number.) 6. sgn x is not defined at x D 0, so cannot be either continuous or discontinuous there. (Functions can be continuous or discontinuous only at points in their domains!) x if x < 0 is continuous everywhere on the 7. f .x/ D x 2 if x 0 real line, even at x D 0 where its left and right limits are both 0, which is f .0/. The least integer function dxe is continuous everywhere on 12. C.t / is discontinuous only at the integers. It is continuous on the left at the integers, but not on the right. 1 17. x2 4 Since D x C 2 for x ¤ 2, we can define the x 2 function to be 2 C 2 D 4 at x D 2 to make it continuous there. The continuous extension is x C 2. 1 C t3 .1 C t /.1 t C t 2 / 1 t C t2 D D for 2 1 t .1 C t /.1 t / 1 t t ¤ 1, we can define the function to be 3=2 at t D 1 to make it continuous there. The continuous extension is 1 t C t2 . 1 t Since .t 2/.t 3/ t 2 t 2 5t C 6 D D for t ¤ 3, 2 t t 6 .t C 2/.t 3/ t C2 we can define the function to be 1=5 at t D 3 to make it t 2 continuous there. The continuous extension is . t C2 Since p p p xC 2 x2 2 .x 2/.x C 2/ p p D p D x4 4 2/.x C 2/.x 2 C 2/ .x C 2/.x 2 C 2/ p.x for x p ¤ 2, we can define the function to be 1=4 at there. The continuous x D 2 to make it continuous p xC 2 extension is p . (Note: cancelling the .x C 2/.x 2 C 2/ p 2 factors provides a further continuous extension to xC p 2. xD Since limx!2C f .x/ D k 4 and limx!2 f .x/ D 4 D f .2/. Thus f will be continuous at x D 2 if k 4 D 4, that is, if k D 8. 18. limx!3 g.x/ D 3 m and limx!3C g.x/ D 1 3m D g.3/. Thus g will be continuous at x D 3 if 3 m D 1 3m, that is, if m D 1. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k lim f .x/: x! 1C right continuous. 2 3 lim x 2 D x! 1C R except at the integers, where it is left continuous but not 14. y D f .x/ 2 1¤1 2 if x ¤ 0 is continuous everywhere except f .x/ D 1=x 0 if x D 0 at x D 0, where it is neither left nor right continuous since it does not have a real limit there. 2 if x 1 is continuous everywhere ex10. f .x/ D x 0:987 if x > 1 cept at x D 1, where it is left continuous but not right continuous because 0:987 ¤ 1. Close, as they say, but no cigar. 11. 3. g has no absolute maximum value on Œ 2; 2. It takes on every positive real value less than 2, but does not take the value 2. It has absolute minimum value 0 on that interval, assuming this value at the three points x D 2, x D 1, and x D 1. 1 lim x D x! 1 9. -1 3 x if x < 1 is continuous everywhere on the x 2 if x 1 real line except at x D 1 where it is right continuous, but not left continuous. f .x/ D Downloaded by ted cage (sxnbyln180@questza.com) 31 lOMoARcPSD|6566483 www.konkur.in SECTION 1.4 (PAGE 87) ADAMS and ESSEX: CALCULUS 9 19. x 2 has no maximum value on 1 < x < 1; it takes all positive real values less than 1, but it does not take the value 1. It does have a minimum value, namely 0 taken on at x D 0. 20. The Max-Min Theorem says that a continuous function defined on a closed, finite interval must have maximum and minimum values. It does not say that other functions cannot have such values. The Heaviside function is not continuous on Œ 1; 1 (because it is discontinuous at x D 0), but it still has maximum and minimum values. Do not confuse a theorem with its converse. 21. Let the numbers be x and y, where x 0, y 0, and x C y D 8. If P is the product of the numbers, then P D xy D x.8 x/ D 8x x 2 D 16 .x f .x/ D 29. f .x/ D x 3 C x 1, f .0/ D 1, f .1/ D 1. Since f is continuous and changes sign between 0 and 1, it must be zero at some point between 0 and 1 by IVT. 30. f .x/ D x 3 15x C 1 is continuous everywhere. f . 4/ D 3; f . 3/ D 19; f .1/ D 13; f .4/ D 5: Because of the sign changes f has a zero between 4 and 3, another zero between 3 and 1, and another between 1 and 4. 31. F .x/ D .x a/2 .x b/2 C x. Without loss of generality, we can assume that a < b. Being a polynomial, F is continuous on Œa; b. Also F .a/ D a and F .b/ D b. Since a < 21 .a C b/ < b, the Intermediate-Value Theorem guarantees that there is an x in .a; b/ such that F .x/ D .a C b/=2. 32. Let g.x/ D f .x/ x. Since 0 f .x/ 1 if 0 x 1, therefore, g.0/ 0 and g.1/ 0. If g.0/ D 0 let c D 0, or if g.1/ D 0 let c D 1. (In either case f .c/ D c.) Otherwise, g.0/ > 0 and g.1/ < 0, and, by IVT, there exists c in .0; 1/ such that g.c/ D 0, i.e., f .c/ D c. 33. The domain of an even function is symmetric about the y-axis. Since f is continuous on the right at x D 0, therefore it must be defined on an interval Œ0; h for some h > 0. Being even, f must therefore be defined on Œ h; h. If x D y, then 4/2 : Therefore P 16, so P is bounded. Clearly P D 16 if x D y D 4, so the largest value of P is 16. 22. Let the numbers be x and y, where x 0, y 0, and x C y D 8. If S is the sum of their squares then S D x 2 C y 2 D x 2 C .8 D 2x 2 x/2 4/2 C 32: 16x C 64 D 2.x Since 0 x 8, the maximum value of S occurs at x D 0 or x D 8, and is 64. The minimum value occurs at x D 4 and is 32. 23. Since T D 100 30x C 3x 2 D 3.x 5/2 C 25, T will be minimum when x D 5. Five programmers should be assigned, and the project will be completed in 25 days. 24. lim f .x/ D lim f . y/ D lim f .y/ D f .0/: x!0 If x desks are shipped, the shipping cost per desk is C D 245x 30x 2 C x 3 D x2 x D .x 34. This cost is minimized if x D 15. The manufacturer should send 15 desks in each shipment, and the shipping cost will then be $20 per desk. x2 Telegram: @uni_k x!0C x!0 x, we obtain t!0C D t!0C f .t / f .0/ D f . 0/ D f .0/: Therefore f is continuous at 0 and f .0/ D 0. .x 1/.x C 1/ x2 1 D f .x/ D 2 x 4 .x 2/.x C 2/ f D 0 at x D ˙1. f is not defined at x D ˙2. f .x/ > 0 on . 1; 2/, . 1; 1/, and .2; 1/. f .x/ < 0 on . 2; 1/ and .1; 2/. 32 f odd , f . x/ D f .x/ f continuous on the right , lim f .x/ D f .0/ Therefore, letting t D .x 26. f .x/ D x 2 C 4x C 3 D .x C 1/.x C 3/ f .x/ > 0 on . 1; 3/ and . 1; 1/ f .x/ < 0 on . 3; 1/. 27. y!0C lim f .x/ D lim f . t / D lim 1/.x C 1/ 25. f .x/ D D x x f D 0 at x D ˙1. f is not defined at 0. f .x/ > 0 on . 1; 0/ and .1; 1/. f .x/ < 0 on . 1; 1/ and .0; 1/. 1 y!0C Thus, f is continuous on the left at x D 0. Being continuous on both sides, it is therefore continuous. 30x C 245 15/2 C 20: .x C 2/.x 1/ x2 C x 2 D 3 x x3 f .x/ > 0 on . 2; 0/ and .1; 1/ f .x/ < 0 on . 1; 2/ and .0; 1/. 28. 35. max 1:593 at 0:831, min 0:756 at 0:629 36. max 0:133 at x D 1:437; min 0:232 at x D 1:805 37. max 10:333 at x D 3; min 4:762 at x D 1:260 38. max 1:510 at x D 0:465; min 0 at x D 0 and x D 1 39. root x D 0:682 40. root x D 0:739 41. roots x D 0:637 and x D 1:410 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 42. roots x D SECTION 1.5 (PAGE 92) 0:7244919590 and x D 1:220744085 43. fsolve gives an approximation to the single real root to 10 significant figures; solve gives the three roots (including a complex conjugate pair) in exact form involving the quan p 1=3 ; evalf(solve) gives approximations tity 108 C 12 69 to the three roots using 10 significant figures for the real and imaginary parts. Section 1.5 The Formal Definition of Limit (page 92) 1. We require 39:9 L 40:1. Thus 10. We need 1 0:05 1=.x C 1/ 1 C 0:05, or 1:0526 x C 1 0:9524. This will occur if 0:0476 x 0:0526. In this case we can take ı D 0:0476. 11. x!1 Proof: Let > 0 be given. Then j.3x C 1/ 4j < holds if 3jx 1j < , and so if jx 1j < ı D =3. This confirms the limit. x!2 Proof: Let > 0 be given. Then j.5 2x/ 1j < holds if j2x 4j < , and so if jx 2j < ı D =2. This confirms the limit. Let > 0 be given. p Then jx 2 jx 0j D jxj < ı D . 14. ı The temperature should be kept between 12 C and 20 C. 4. 5. 6. 0:02 2x 1 3 C 0:02 3:98 2x 4:02 1:99 x 2:01 16. 0:01 3 Here ı D 0:02995 will do. 9. We need 8 0:2 x 3 8:2, or 1:9832 x 2:0165. Thus, we need 0:0168 x 2 0:0165. Here ı D 0:0165 will do. ˇ ˇ 2ˇˇ D j.1 C 2x/ 2j D j2x 1 2 j < ı D =2. ˇ ˇ 1j D 2 ˇˇx ˇ 1 ˇˇ < 2ˇ x 2 C 2x D 2. x! 2 x C 2 Proof: Let > 0 be given. For x ¤ 2 we have To be proved: lim ˇ ˇ . 2/ˇˇ D jx C 2j < provided jx C 2j < ı D . This completes the proof. 17. 1 1 . 2 Proof: Let > 0 be given. We have ˇ ˇ ˇ ˇ ˇ 1 1 ˇˇ ˇˇ 1 x ˇˇ jx 1j ˇ D ˇ x C 1 2 ˇ ˇ 2.x C 1/ ˇ D 2jx C 1j : To be proved: lim x!1 x C 1 D If jx 1j < 1, then 0 < x < 2 and 1 < x C 1 < 3, so that jx C 1j > 1. Let ı D min.1; 2/. If jx 1j < ı, then ˇ ˇ ˇ 1 1 ˇˇ 2 jx 1j ˇ ˇ x C 1 2 ˇ D 2jx C 1j < 2 D : This establishes the required limit. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2j < lim ˇ 2 ˇ x C 2x ˇ ˇ xC2 7. p 2:99 2x C 3 3:01 8:9401 2x C 3 9:0601 2:97005 x 3:03005 0:02995 x 3 0:03005: D 0. 1 4x 2 D 2. 2x x!1=2 1 Proof: Let > 0 be given. Then if x ¤ 1=2 we have provided jx We need 0:03 .3x C 1/ 7 0:03, which is equivalent to 0:01 x 2 0:01 Thus ı D 0:01 will do. p 8. We need 0:01 2x C 3 3 0:01. Thus 2 2j < ı D . To be proved: ˇ ˇ 1 4x 2 ˇ ˇ 1 2x 1 2 C 0:01 x 1 1 x 2:01 1:99 0:5025 x 0:4975 x x!2 1 C x 2 provided jx 15. 4 0:1 x 2 4 C 0:1 1:9749 x 2:0024 p 1 0:1 x 1:1 0:81 x 1:21 2 To be proved: lim 0j < holds if Proof: Let > 0 be given. Then ˇ ˇ ˇ ˇ x 2 ˇ D jx 2j jx ˇ 0 ˇ ˇ 1 C x2 1 C x2 2. Since 1.2% of 8,000 is 96, we require the edge length x of the cube to satisfy 7904 x 3 8096. It is sufficient that 19:920 x 20:079. The edge of the cube must be within 0.079 cm of 20 cm. 3 To be proved: lim x 2 D 0. x!0 39:9 39:6 C 0:025T 40:1 0:3 0:025T 0:5 12 T 20: 3. 2x/ D 1. 12. To be proved: lim .5 13. ı To be proved: lim .3x C 1/ D 4. Downloaded by ted cage (sxnbyln180@questza.com) 33 lOMoARcPSD|6566483 www.konkur.in SECTION 1.5 (PAGE 92) ADAMS and ESSEX: CALCULUS 9 1 xC1 D . 1 2 Proof: Let > 0 be given. If x ¤ 1, we have ˇ ˇ ˇ ˇ ˇ xC1 jx C 1j 1 ˇˇ ˇˇ 1 1 ˇˇ ˇ D D : ˇ x2 1 2 ˇ ˇx 1 2 ˇ 2jx 1j 18. To be proved: lim If jx C 1j < 1, then 2 < x < 0, so 3 < x 1 < 1 and jx 1j > 1. Ler ı D min.1; 2/. If 0 < jx . 1/j < ı then jx 1j > 1 and jx C 1j < 2. Thus ˇ ˇ ˇ xC1 1 ˇˇ jx C 1j 2 ˇ D < D : ˇ x2 1 2 ˇ 2jx 1j 2 This completes the required proof. p 19. To be proved: lim x D 1. Proof: Let > 0 be given. We have ˇ ˇ ˇ x 1 ˇ p ˇ jx j x 1j D ˇˇ p x C 1ˇ 26. 28. 3 20. To be proved: lim x D 8. x!2 Proof: Let > 0 be given. We have jx 3 8j D jx 2jjx 2 C2xC4j. If jx 2j < 1, then 1 < x < 3 and x 2 < 9. Therefore jx 2 C 2x C 4j 9 C 2 3 C 4 D 19. If jx 2j < ı D min.1; =19/, then jx 3 19 D : 2jjx 2 C 2x C 4j < 19 8j D jx 29. implies jf .x/ ı<x<a x< R implies jf .x/ 1 To be proved: limx!1 D 1. Proof: Let B > 0 x 1 1 < B if 0 > x 1 > 1=B, be given. We have x 1 that is, if 1 ı < x < 1, where ı D 1=B:. This completes the proof. To be proved: limx!1 p 23. We say that limx!a f .x/ D 1 if the following condition holds: for every number B > 0 there exists a number ı > 0, depending on B, such that 0 < jx 24. aj < ı B: We say that limx!1 f .x/ D 1 if the following condition holds: for every number B > 0 there exists a number R > 0, depending on B, such that x>R 34 Telegram: @uni_k implies f .x/ < implies f .x/ > B: 1 x2 C 1 D 0. Proof: Let > 0 provided x > R, where R D 1=. This completes the proof. 30. Lj < : implies f .x/ > B: ˇ ˇ ˇ ˇ 1 ˇp 1 ˇD p 1 < < ˇ ˇ 2 2 x x C1 x C1 Lj < : 22. We say that limx! 1 f .x/ D L if the following condition holds: for every number > 0 there exists a number R > 0, depending on , such that ı<x<a be given. We have We say that limx!a f .x/ D L if the following condition holds: for every number > 0 there exists a number ı > 0, depending on , such that a B: 1 To be proved: limx!1C D 1. Proof: Let B > 0 x 1 1 > B if 0 < x 1 < 1=B, that be given. We have x 1 is, if 1 < x < 1 C ı, where ı D 1=B. This completes the proof. This completes the proof. 21. implies f .x/ < We say that limx!a f .x/ D 1 if the following condition holds: for every number B > 0 there exists a number ı > 0, depending on B, such that a 1j < 1j < ı D . This completes the proof. We say that limx!aC f .x/ D 1 if the following condition holds: for every number B > 0 there exists a number ı > 0, depending on R, such that a <x <aCı 27. x!1 provided jx 25. x! 1 x 2 31. p To be proved: lim px!1 x D 1. Proof: Let B >2 0 be given. We have x > B if x > R where R D B . This completes the proof. To be proved: if lim f .x/ D L and lim f .x/ D M , then x!a x!a L D M. Proof: Suppose L ¤ M . Let D jL M j=3. Then > 0. Since lim f .x/ D L, there exists ı1 > 0 such that x!a jf .x/ Lj < if jx aj < ı1 . Since lim f .x/ D M , there x!a exists ı2 > 0 such that jf .x/ M j < if jx Let ı D min.ı1 ; ı2 /. If jx aj < ı, then 3 D jL aj < ı2 . M j D j.f .x/ M / C .L f .x/j jf .x/ M j C jf .x/ Lj < C D 2: This implies that 3 < 2, a contradiction. Thus the original assumption that L ¤ M must be incorrect. Therefore L D M. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES 1 (PAGE 93) 32. To be proved: if lim g.x/ D M , then there exists ı > 0 This completes the proof. x!a such that if 0 < jx aj < ı, then jg.x/j < 1 C jM j. Proof: Taking D 1 in the definition of limit, we obtain a number ı > 0 such that if 0 < jx aj < ı, then jg.x/ M j < 1. It follows from this latter inequality that 36. To be proved: if lim f .x/ D L and lim f .x/ D M ¤ 0, x!a x!a L f .x/ D . x!a g.x/ M Proof: By Exercises 33 and 35 we have then lim jg.x/j D j.g.x/ M /CM j jG.x/ M jCjM j < 1CjM j: lim f .x/ x!a g.x/ D lim f .x/ x!a 1 1 L DL D : g.x/ M M 33. To be proved: if lim f .x/ D L and lim g.x/ D M , then x!a x!a lim f .x/g.x/ D LM . x!a 37. To be proved: if f is continuous at L and lim g.x/ D L, Proof: Let > 0 be given. Since lim f .x/ D L, there x!c x!a then lim f .g.x// D f .L/. exists ı1 > 0 such that jf .x/ Lj < =.2.1 C jM j// if 0 < jx aj < ı1 . Since lim g.x/ D M , there ex- x!c Proof: Let > 0 be given. Since f is continuous at L, there exists a number > 0 such that if jy Lj < , then jf .y/ f .L/j < . Since limx!c g.x/ D L, there exists ı > 0 such that if 0 < jx cj < ı, then jg.x/ Lj < . Taking y D g.x/, it follows that if 0 < jx cj < ı, then jf .g.x// f .L/j < , so that limx!c f .g.x// D f .L/. x!a ists ı2 > 0 such that jg.x/ M j < =.2.1 C jLj// if 0 < jx aj < ı2 . By Exercise 32, there exists ı3 > 0 such that jg.x/j < 1 C jM j if 0 < jx aj < ı3 . Let ı D min.ı1 ; ı2 ; ı3 /. If jx aj < ı, then jf .x/g.x/ LM D jf .x/g.x/ Lg.x/ C Lg.x/ LM j D j.f .x/ L/g.x/ C L.g.x/ M /j j.f .x/ L/g.x/j C jL.g.x/ M /j D jf .x/ Ljjg.x/j C jLjjg.x/ M j < .1 C jM j/ C jLj 2.1 C jM j/ 2.1 C jLj/ C D : 2 2 38. Thus lim f .x/g.x/ D LM . x!a 34. To be proved: if lim g.x/ D M where M ¤ 0, then jg.x/ x!a there exists ı > 0 such that if 0 < jx aj < ı, then jg.x/j > jM j=2. Proof: By the definition of limit, there exists ı > 0 such that if 0 < jx aj < ı, then jg.x/ M j < jM j=2 (since jM j=2 is a positive number). This latter inequality implies that jM j D jg.x/C.M g.x//j jg.x/jCjg.x/ M j < jg.x/jC It follows that jg.x/j > jM j required. jM j : 2 Lj D jg.x/ f .x/ C f .x/ Lj jg.x/ f .x/j C jf .x/ Lj jh.x/ f .x/j C jf .x/ Lj D jh.x/ L C L f .x/j C jf .x/ Lj jh.x/ Lj C jf .x/ Lj C jf .x/ Lj < C C D : 3 3 3 Thus limx!a g.x/ D L. .jM j=2/ D jM j=2, as 35. To be proved: if lim g.x/ D M where M ¤ 0, then x!a 1 1 D . lim x!a g.x/ M Proof: Let > 0 be given. Since lim g.x/ D M ¤ 0, To be proved: if f .x/ g.x/ h.x/ in an open interval containing x D a (say, for a ı1 < x < a C ı1 , where ı1 > 0), and if limx!a f .x/ D limx!a h.x/ D L, then also limx!a g.x/ D L. Proof: Let > 0 be given. Since limx!a f .x/ D L, there exists ı2 > 0 such that if 0 < jx aj < ı2 , then jf .x/ Lj < =3. Since limx!a h.x/ D L, there exists ı3 > 0 such that if 0 < jx aj < ı3 , then jh.x/ Lj < =3. Let ı D min.ı1 ; ı2 ; ı3 /. If 0 < jx aj < ı, then Review Exercises 1 (page 93) 1. The average rate of change of x 3 over Œ1; 3 is 33 3 x!a 2 there exists ı1 > 0 such that jg.x/ M j < jM j =2 if 0 < jx aj < ı1 . By Exercise 34, there exists ı2 > 0 such that jg.x/j > jM j=2 if 0 < jx aj < ı3 . Let ı D min.ı1 ; ı2 /. If 0 < jx aj < ı, then ˇ ˇ ˇ 1 1 ˇˇ jM g.x/j jM j2 2 ˇ D < D : ˇ g.x/ M ˇ jM jjg.x/j 2 jM j2 2. The average rate of change of 1=x over Œ 2; 1 is .1=. 1// .1=. 2// D 1 . 2/ Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 13 26 D D 13: 1 2 Downloaded by ted cage (sxnbyln180@questza.com) 1=2 D 1 1 : 2 35 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 1 (PAGE 93) ADAMS and ESSEX: CALCULUS 9 3. The rate of change of x 3 at x D 2 is .2 C h/3 h h!0 23 lim 8 C 12h C 6h2 C h3 8 h h!0 D lim .12 C 6h C h2 / D 12: 18. h!0 19. lim h!0 1 3=2 5. lim .x 4x C 7/ D 1 x!1 3=2 is 20. 2 2 C 3 3 h h!0 2.3 C 2h 3/ D lim 3/h h!0 3.2h 4 4 D lim D : 3/ 9 h!0 3.2h D lim 2h lim 9. 10. x2 4 xC2 .x 2/.x C 2/ D lim D lim x!2 x 2 x!2 x x!2 4x C 4 .x 2/2 2 does not exist. The denominator approaches 0 (from both sides) while the numerator does not. lim x2 lim x2 x!2 11. 12. xC2 4 D lim D x!2 x 4x C 4 2 x2 lim 4 x! 2C x 2 C 4x C 4 D p lim x x lim p x!3 x 1 2 x! 2C x C 2 4 x x D lim p x!4 .2 C 4 x/.x 2 x!4 x lim 2 13. 2/.x C 2/ xC2 D lim D x!2 x 2/.x 3/ 3 4/ 26. 4 27. lim p h!0 15. x C 3h lim p lim p x!0C 16. x!0 x x 28. Telegram: @uni_k x4 x!1 x 2 lim p 4 D lim x!1 1 1 x lim p x D 1 D 1 4 29. D1 x2 1 D p D2 1=4 lim sin x does not exist; sin x takes the values 1 and 1 in any interval .R; 1/, and limits, if they exist, must be unique. cos x D 0 by the squeeze theorem, since lim x!1 x 1 cos x 1 for all x > 0 x x x and limx!1 . 1=x/ D limx!1 .1=x/ D 0. 1 D 0 by the squeeze theorem, since x 1 jxj x sin jxj for all x ¤ 0 x and limx!0 . jxj/ D limx!0 jxj D 0. lim x sin 1 does not exist; sin.1=x 2 / takes the values 1 x2 and 1 in any interval . ı; ı/, where ı > 0, and limits, if they exist, must be unique. p lim Œx C x 2 4x C 1 lim sin x! 1 D lim .x 2 p x2 4x C 1/ 4x C 1 4x 1 p jxj 1 .4=x/ C .1=x 2 / xŒ4 .1=x/ D lim p x! 1 x C x 1 .4=x/ C .1=x 2 / 4 .1=x/ D lim p D 2: x! 1 1 C 1 .4=x/ C .1=x 2 / Note how we have used jxj D x (in the second last line), because x ! 1. p lim Œx C x 2 4x C 1 D 1 C 1 D 1 D lim x! 1 x p p h. x C 3h C x/ .x C 3h/ x h!0 x p p p 2 x x C 3h C x D D lim 3 3 h!0 30. p x2 x! 1 x D lim x 2 does not exist because x2 D1 .4=x 2 / x2 1 1 x!0 x2 D 0 fined for x < 0. 36 p lim x!0 p p 9 .x 3/.x C 3/. x C 3/ p D lim x!3 x 3 3 p p p D lim .x C 3/. x C 3/ D 12 3 h 1 3 2x C 100 .2=x/ C .100=x 2 / D lim D0 2 x! 1 x C 3 x! 1 1 C .3=x 2 / lim x!3 14. x 2 is not de- lim x!1 8. x .1=x 2 / 1 D .1=x/ .1=x 2 / 22. 4 3 4 .x D lim x!2 .x 5x C 6 x!1 3x 2 x!0C lim x2 x2 D lim x!1 3 x 1 1 lim x .1=x 2 / x3 1 D lim D 2 x! 1 1 C .4=x 2 / x! 1 x C 4 25. x2 does not exist. The denominator approaches 0 x!1 1 x2 (from both sides) while the numerator does not. p fined for x > 1. p lim x x2 D 0 x!1=2 7. x!2 x 2 x 2 does not exist because x 21. 23. 4C7D4 22 x2 D D 6. lim 2 x!2 1 x 1 22 p x!1 24. 2 lim x!1 D lim 4. The rate of change of 1=x at x D 1 .3=2/ C h h 17. x!1 x x 2 is not de- 31. f .x/ D x 3 4x 2 C 1 is continuous on the whole real line and so is discontinuous nowhere. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 1 (PAGE 94) x is continuous everywhere on its domain, xC1 which consists of all real numbers except x D 1. It is discontinuous nowhere. x 2 if x > 2 is defined everywhere and disf .x/ D x if x 2 continuous at x D 2, where it is, however, left continuous since limx!2 f .x/ D 2 D f .2/. x 2 if x > 1 is defined and continuous evf .x/ D x if x 1 erywhere, and so discontinuous nowhere. Observe that limx!1 f .x/ D 1 D limx!1C f .x/. n 1 if x 1 is defined everywhere f .x/ D H.x 1/ D 0 if x < 1 and discontinuous at x D 1 where it is, however, right continuous. n 1 if 3 x 3 is defined f .x/ D H.9 x 2 / D 0 if x < 3 or x > 3 everywhere and discontinuous at x D ˙3. It is right continuous at 3 and left continuous at 3. 32. f .x/ D 33. 34. 35. 36. 37. f .x/ D jxj C jx C 1j is defined and continuous everywhere. It is discontinuous nowhere. n jxj=jx C 1j if x ¤ 1 38. f .x/ D is defined everywhere 1 if x D 1 and discontinuous at x D 1 where it is neither left nor right continuous since limx! 1 f .x/ D 1, while f . 1/ D 1. Challenging Problems 1 1. 2. 3. .c C h/3 h h!0 lim p c3 4. 3c 2 h C 3ch2 C h3 D 3c 2 : h h!0 .a2 C ab C b 2 /=3, then 3c 2 D a2 C ab C b 2 , so If c D the average rate ofpchange over Œa; b is the instantaneous rate of change at .a2 C ab C b 2 /=3. p 2 Claim: .a C ab C b 2 /=3 > .a C b/=2. Proof: Since a2 2ab C b 2 D .a b/2 > 0, we have 2 2 2 4a C 4ab C 4b > 3a C 6ab C 3b s 2 a2 C ab C b 2 a2 C 2ab C b 2 > D 3 4 a2 C ab C b 2 aCb > : 3 2 aCb 2 2 jx C 1j j5 2xj jx 5j j3x x D lim x!0 .1 x/ .x C 1/ D 1 : 2 2xj D 2x 5, jx 2j D x 7j D 3x 7. Thus 2x 5 .x 2j D lim x!3 5 7j x .3x x 3 D lim D x!3 4.3 x/ x 1=3 x!64 x 1=2 lim D lim y C2 y!2 y 2 C 2y C 4 5. x and jx C 1j D x C 1. 2, 2/ 7/ 1 : 4 Let y D x 1=6 . Then we have Use a b D have a3 D y2 4 D lim 3 8 y!2 y 4 1 D : 12 3 b3 a2 C ab C b 2 4 8 .y 2/.y C 2/ D lim y!2 .y 2/.y 2 C 2y C 4/ to handle the denominator. We p 3Cx 2 lim p x!1 3 7 C x 2 .7 C x/2=3 C 2.7 C x/1=3 C 4 3Cx 4 D lim p x!1 .7 C x/ 8 3CxC2 .7 C x/2=3 C 2.7 C x/1=3 C 4 4C4C4 D 3: D lim p D x!1 2C2 3CxC2 1C p 1Ca p 1Ca , r .a/ D . a a a) lima!0 r .a/ does not exist. Observe that the right limit is 1 and the left limit is 1. rC .a/ D 1 b) From the following table it appears that lima!0 rC .a/ D 1=2, the solution of the linear equation 2x 1 D 0 which results from setting a D 0 in the quadratic equation ax 2 C 2x 1 D 0. a rC .a/ 1 0:1 0:1 0:01 0:01 0:001 0:001 0:41421 0:48810 0:51317 0:49876 0:50126 0:49988 0:50013 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1j D 1 For x near 3 we have j5 jx 5j D 5 x, and j3x lim 6. D lim 1j x!3 jx Let 0 < a < b. The average rate of change of x 3 over Œa; b is b 3 a3 D b 2 C ab C a2 : b a The instantaneous rate of change of x at x D c is x lim x!0 jx (page 94) 3 For x near 0 we have jx Thus Downloaded by ted cage (sxnbyln180@questza.com) 37 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 1 (PAGE 94) p 1Ca 1 a!0 a .1 C a/ 1 D lim p a!0 a. 1 C a C 1/ 1 1 D : D lim p a!0 2 1CaC1 8. c) lim rC .a/ D lim a!0 7. ADAMS and ESSEX: CALCULUS 9 TRUE or FALSE a) If limx!a f .x/ exists and limx!a g.x/ does not exist, then limx!a f .x/ C g.x/ does not exist. TRUE, because if limx!a f .x/ C g.x/ were to exist then lim g.x/ D lim f .x/ C g.x/ f .x/ x!a x!a lim f .x/ D lim f .x/ C g.x/ 9. x!a x!a would also exist. b) If neither limx!a f .x/ nor limx!a g.x/ exists, then limx!a f .x/ C g.x/ does not exist. FALSE. Neither limx!0 1=x nor limx!0 . 1=x/ exist, but limx!0 .1=x/ C . 1=x/ D limx!0 0 D 0 exists. c) If f is continuous at a, then so is jf j. TRUE. For any two real numbers u and v we have ˇ ˇ ˇjuj This follows from juj D ju jvj D jv ˇ ˇ jvjˇ ju v C vj ju u C uj jv Now we have ˇ ˇ ˇjf .x/j vj: vj C jvj; and uj C juj D ju vj C juj: ˇ ˇ jf .a/jˇ jf .x/ f .a/j d) If jf j is continuous at a, then so is f . 1 if x < 0 is FALSE. The function f .x/ D 1 if x 0 discontinuous at x D 0, but jf .x/j D 1 everywhere, and so is continuous at x D 0. e) If f .x/ < g.x/ in an interval around a and if limx!a f .x/ D L and limx!a g.x/ D M both exist, then L < M . 2 if x ¤ 0 and let FALSE. Let g.x/ D x 1 if x D 0 f .x/ D g.x/. Then f .x/ < g.x/ for all x, but limx!0 f .x/ D 0 D limx!0 g.x/. (Note: under the given conditions, it is TRUE that L M , but not necessarily true that L < M .) Telegram: @uni_k b) If the domain of the continuous function f is an open interval, the range of f can be any interval (open, closed, half open, finite, or infinite). n x2 1 1 if 1 < x < 1 D . f .x/ D 2 1 if x < 1 or x > 1 jx 1j f is continuous wherever it is defined, that is at all points except x D ˙1. f has left and right limits 1 and 1, respectively, at x D 1, and has left and right limits 1 and 1, respectively, at x D 1. It is not, however, discontinuous at any point, since 1 and 1 are not in its domain. 1 1 1 D D 1 2 . 1 2 1 x x2 x C x x 12 4 4 4 Observe that f .x/ f .1=2/ D 4 for all x in .0; 1/. 10. f .x/ D 11. Suppose f is continuous on Œ0; 1 and f .0/ D f .1/. a) To be proved: f .a/ D f .a C 12 / for some a in Œ0; 21 . Proof: If f .1=2/ D f .0/ we can take a D 0 and be done. If not, let g.x/ D f .x C 21 / so the left side approaches zero whenever the right side does. This happens when x ! a by the continuity of f at a. 38 a) To be proved: if f is a continuous function defined on a closed interval Œa; b, then the range of f is a closed interval. Proof: By the Max-Min Theorem there exist numbers u and v in Œa; b such that f .u/ f .x/ f .v/ for all x in Œa; b. By the Intermediate-Value Theorem, f .x/ takes on all values between f .u/ and f .v/ at values of x between u and v, and hence at points of Œa; b. Thus the range of f is Œf .u/; f .v/, a closed interval. f .x/: Then g.0/ ¤ 0 and g.1=2/ D f .1/ f .1=2/ D f .0/ f .1=2/ D g.0/: Since g is continuous and has opposite signs at x D 0 and x D 1=2, the Intermediate-Value Theorem assures us that there exists a between 0 and 1/2 such that g.a/ D 0, that is, f .a/ D f .a C 12 /. b) To be proved: if n > 2 is an integer, then f .a/ D f .a C n1 / for some a in Œ0; 1 n1 . Proof: Let g.x/ D f .x C n1 / f .x/. Consider the numbers x D 0, x D 1=n, x D 2=n, : : : , x D .n 1/=n. If g.x/ D 0 for any of these numbers, then we can let a be that number. Otherwise, g.x/ ¤ 0 at any of these numbers. Suppose that the values of g at all these numbers has the same sign (say positive). Then we have f .1/ > f . n n 1 / > > f . n2 / > n1 > f .0/; which is a contradiction, since f .0/ D f .1/. Therefore there exists j in the set f0; 1; 2; : : : ; n 1g such that g.j=n/ and g..j C 1/=n/ have opposite sign. By the Intermediate-Value Theorem, g.a/ D 0 for some a between j=n and .j C 1/=n, which we had to prove. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHAPTER 2. SECTION 2.1 (PAGE 100) DIFFERENTIATION 7. Slope of y D x C 1 at x D 3 is p p 4Ch 2 4ChC2 p m D lim h h!0 4ChC2 4Ch 4 p D lim h!0 h hChC2 1 1 D : D lim p 4 h!0 4ChC2 Section 2.1 Tangent Lines and Their Slopes (page 100) 1. Slope of y D 3x m D lim h!0 1 at .1; 2/ is 3.1 C h/ 1 .3 1 h 1/ D lim h!0 3h D 3: h The tangent line is y 2 D 3.x 1/, or y D 3x 1. (The tangent to a straight line at any point on it is the same straight line.) Tangent line is y 8. 2.2 C h/2 5 .2.22 / h h!0 8 C 8h C 2h2 8 D lim h h!0 D lim .8 C 2h/ D 8 5/ 3 D 8.x 4. The slope of y D 6 x 6 m D lim h!0 . 2 C h/ h2 3h D lim h!0 2/ or y D 8x x 2 at x D h h 13. 9. Slope of y D . 2 C h/2 4 h/ D 3: h!0 The tangent line at . 2; 4/ is y D 3x C 10. 5. Slope of y D x 3 C 8 at x D . 2 C h/3 C 8 . 8 C 8/ h h!0 8 C 12h 6h2 C h3 C 8 D lim h h!0 D lim 12 6h C h2 D 12 6. The slope of y D 0 m D lim 1 h!0 h 0 D 12.x C 2/ or y D 12x C 24. 1 h2 C 1 1 D lim The tangent line at .0; 1/ is y D 1. h h!0 h2 C 1 D 0: 1 .x 4 9/, or 2/, p 2 h .1 C h/2 C 2 The tangent line at .1; 2/ is y D 2 y D 52 12 x. 11. 1 2 .x D 1 2 1/, or Slope of y D x 2 at x D x0 is .x0 C h/2 h h!0 m D lim Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1D D lim p h!0 5 1 at .0; 1/ is x2 C 1 ! 1 .x 54 5 x 2 at x D 1 is p 5 .1 C h/2 2 m D lim h h!0 5 .1 C h/2 4 D lim p h!0 h 5 .1 C h/2 C 2 10. The slope of y D h!0 Tangent line is y 5. 2x at x D 2 is xC2 Tangent line is y or x 4y D 2. 2 is m D lim 4y D 2.2 C h/ 1 2ChC2 m D lim h h!0 4 C 2h 2 h 2 D lim h.2 C h C 2/ h!0 1 h D : D lim 4 h!0 h.4 C h/ 2 is D lim .3 3/, or x The tangent line at .9; 13 / is y D 31 1 y D 12 54 x. h!0 Tangent line is y 1 .x 4 ! 1 1 1 m D lim p h!0 h 9Ch 3 p p 3 9Ch 3C 9Ch D lim p p h!0 3h 9 C h 3C 9Ch 9 9 h D lim p p h!0 3h 9 C h.3 C 9 C h/ 1 1 D : D 3.3/.6/ 54 5 at .2; 3/ is m D lim 2D 1 The slope of y D p at x D 9 is x 2. Since y D x=2 is a straight line, its tangent at any point .a; a=2/ on it is the same line y D x=2. 3. Slope of y D 2x 2 p Downloaded by ted cage (sxnbyln180@questza.com) x02 2x0 h C h2 D 2x0 : h h!0 D lim 39 lOMoARcPSD|6566483 www.konkur.in SECTION 2.1 (PAGE 100) ADAMS and ESSEX: CALCULUS 9 Tangent line is y x02 D 2x0 .x or y D 2x0 x x02 . x0 /, 19. 1 at .a; a1 / is 12. The slope of y D x ! 1 1 1 a a h 1 . C D D lim m D lim a a2 h!0 h.a C h/.a/ h!0 h a C h 1 1 1 The tangent line at .a; / is y D .x a/, or a a a2 x 2 . yD a a2 p j0 C hj 0 1 D lim does not 13. Since limh!0 h h!0 jhjsgn .h/ p exist (and is not 1 or 1), the graph of f .x/ D jxj has no tangent at x D 0. 14. The slope of f .x/ D .x m D lim h!0 .1 C h 0 1=3 D lim h h!0 Slope of y D x 3 at x D a is .a C h/3 a3 h h!0 a3 C 3a2 h C 3ah2 C h3 D lim h h!0 D lim .3a2 C 3ah C h2 / D 3a2 m D lim a3 h!0 b) We have m D 3 if 3a2 D 3, i.e., if a D ˙1. Lines of slope 3 tangent to y D x 3 are y D 1 C 3.x 1/ and y D 1 C 3.x C 1/, or y D 3x 2 and y D 3x C 2. 20. The slope of y D x 3 3x at x D a is i 1h .a C h/3 3.a C h/ .a3 3a/ h!0 h 1h 3 a C 3a2 h C 3ah2 C h3 3a 3h D lim h!0 h D lim Œ3a2 C 3ah C h2 3 D 3a2 3: m D lim 1/4=3 at x D 1 is 1/4=3 h a) D 0: a3 C 3a h!0 The graph of f has a tangent line with slope 0 at x D 1. Since f .1/ D 0, the tangent has equation y D 0 15. The slope of f .x/ D .x C 2/3=5 at x D . 2 C h C 2/3=5 h h!0 0 m D lim At points where the tangent line is parallel to the x-axis, the slope is zero, so such points must satisfy 3a2 3 D 0. Thus, a D ˙1. Hence, the tangent line is parallel to the x-axis at the points .1; 2/ and . 1; 2/. 2 is D lim h 2=5 D 1: 21. h!0 The graph of f has vertical tangent x D 2 at x D f .0 C h/ h h!0C f .0 C h/ lim h h!0 f .0/ p h D1 h p h D1 h D lim h!0C f .0/ D lim h!0 m D lim The tangent at x D a is parallel to the line y D 2x C 5 if 3a2 1 D 2, that is, if a D ˙1. The corresponding points on the curve are . 1; 1/ and .1; 1/. 22. h!0 1 h .x02 1/ 2x0 h C h2 D 2x0 : h h!0 1 : a2 The tangent at x D a is perpendicular to the line y D 4x 3 if 1=a2 D 1=4, that is, if a D ˙2. The corresponding points on the curve are . 2; 1=2/ and .2; 1=2/. 23. The slope of the curve y D x 2 at x D a is m D lim If m D 3, then x0 D 32 . The tangent line with slope m D 3 at . 32 ; 54 / is y D 45 3.x C 23 /, that is, 13 y D 3x 4 . Telegram: @uni_k 1 a D lim a .a C h/ D h!0 ah.a C h/ .a C h/2 h h!0 D lim 40 The slope of the curve y D 1=x at x D a is 1 a C h m D lim h h!0 1 at x D x0 is Œ.x0 C h/2 .a C h/3 h!0 Thus the graph of f has a vertical tangent x D 0. 18. The slope of y D x 2 x C 1 at x D a is .a C h/ C 1 .a3 a C 1/ h h!0 3a2 h C 3ah2 C a3 h D lim h h!0 D lim .3a2 C 3ah C h2 1/ D 3a2 1: m D lim 2. 16. The slope of f .x/ D jx 2 1j at x D 1 is j.1 C h/2 1j j1 1j j2h C h2 j m D limh!0 D lim , h h h!0 which does not exist, and is not 1 or 1. The graph of f has no tangent at x D 1. p x if x 0 p 17. If f .x/ D , then x if x < 0 lim The slope of the curve y D x 3 a2 D lim .2a C h/ D 2a: h!0 The normal at x D a has slope y a2 D 1 .x 2a Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) a/; or 1=.2a/, and has equation x 1 C y D C a2 : 2a 2 i lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 2.1 (PAGE 100) y This is the line x C y D k if 2a D 1, and so k D .1=2/ C .1=2/2 D 3=4. 24. The curves y D kx 2 and y D k.x The slope of y D kx 2 at x D 1 is k.1 C h/2 m1 D lim h h!0 The slope of y D k.x m2 D lim k.2 h!0 k 2 2/2 intersect at .1; k/. 1 -3 -2 -1 D lim .2 C h/k D 2k: -2 2/2 at x D 1 is k D lim . 2 C h/k D h!0 x 2 -1 h!0 .1 C h//2 h 1 1j x -3 Fig. 2.1-27 2k: The two curves intersect at right angles if 2k D 1=. 2k/, that is, if 4k 2 D 1, which is satisfied if k D ˙1=2. y D jx 2 28. 25. Horizontal tangents at .0; 0/, .3; 108/, and .5; 0/. y .3; 108/ Horizontal tangent at .a; 2/ and . a; 2/ for all a > 1. No tangents at .1; 2/ and . 1; 2/. y y D jx C 1j jx 1j 2 1 100 80 -3 -2 -1 60 1 2 x -1 40 y D x 3 .5 20 -1 1 2 -2 x/2 3 4 5 -3 Fig. 2.1-28 x -20 Fig. 2.1-25 29. 26. Horizontal tangent at . 1; 8/ and .2; 19/. y Horizontal tangent at .0; 1/. The tangents at .˙1; 0/ are vertical. y y D .x 2 20 -2 . 1; 8/ 10 y D 2x 3 3x 2 12x C 1 -1 1 2 3 1/1=3 2 1 -3 x -2 -1 1 2 x -1 -10 -2 -20 .2; 19/ -3 Fig. 2.1-29 -30 Fig. 2.1-26 27. Horizontal tangent at . 1=2; 5=4/. No tangents at . 1; 1/ and .1; 1/. 30. Horizontal tangent at .0; 1/. No tangents at . 1; 0/ and .1; 0/. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 41 lOMoARcPSD|6566483 www.konkur.in SECTION 2.1 (PAGE 100) ADAMS and ESSEX: CALCULUS 9 y 2. y y D ..x 2 1/2 /1=3 2 1 y D g 0 .x/ -1 -2 1 2 x x Fig. 2.1-30 31. The graph of the function f .x/ D x 2=3 (see Figure 2.1.7 in the text) has a cusp at the origin O, so does not have a tangent line there. However, the angle between OP and the positive y-axis does ! 0 as P approaches 0 along the graph. Thus the answer is NO. 3. y y D h0 .x/ x 32. The slope of P .x/ at x D a is P .a C h/ h h!0 m D lim P .a/ : Since P .a C h/ D a0 C a1 h C a2 h2 C C an hn and P .a/ D a0 , the slope is a0 C a1 h C a2 h2 C C an hn a0 h h!0 D lim a1 C a2 h C C an hn 1 D a1 : m D lim 4. y h!0 Thus the line y D `.x/ D m.x a/ C b is tangent to y D P .x/ at x D a if and only if m D a1 and b D a0 , that is, if and only if x P .x/ `.x/ D a2 .x a/2 C a3 .x a/3 C C an .x a/n h i D .x a/2 a2 C a3 .x a/ C C an .x a/n 2 D .x y D k 0 .x/ a/2 Q.x/ where Q is a polynomial. Section 2.2 The Derivative (page 107) 5. Assuming the tick marks are spaced 1 unit apart, the function f is differentiable on the intervals . 2; 1/, . 1; 1/, and .1; 2/. 6. Assuming the tick marks are spaced 1 unit apart, the function g is differentiable on the intervals . 2; 1/, . 1; 0/, .0; 1/, and .1; 2/. 7. y D f .x/ has its minimum at x D 3=2 where f 0 .x/ D 0 1. y 0 y D f .x/ x 42 Telegram: @uni_k Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 2.2 (PAGE 107) y y D f .x/ D 3x x 2 y D f .x/ D jx 3 y 1j 1 x x y y D f 0 .x/ y y D f 0 .x/ x x Fig. 2.2-7 Fig. 2.2-9 8. y D f .x/ has horizontal tangents at the points near 1=2 and 3=2 where f 0 .x/ D 0 y 10. y D f .x/ is constant on the intervals . 1; 2/, . 1; 1/, and .2; 1/. It is not differentiable at x D ˙2 and x D ˙1. y y D f .x/ D jx 2 1j jx 2 4j x x y D f .x/ D x 3 3x 2 C 2x C 1 y y D f 0 .x/ y x x y D f 0 .x/ Fig. 2.2-10 Fig. 2.2-8 11. y D x2 y 0 D lim h!0 9. y D f .x/ fails to be differentiable at x D 1, x D 0, and x D 1. It has horizontal tangents at two points, one between 1 and 0 and the other between 0 and 1. 3x .x C h/2 2xh C h2 h h!0 dy D .2x 3/ dx D lim Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 3.x C h/ .x 2 h 3h D 2x 3 3x/ 43 lOMoARcPSD|6566483 www.konkur.in SECTION 2.2 (PAGE 107) 12. f .x/ D 1 C 4x f 0 .x/ D lim h!0 5x 2 13. 17. 1 C 4.x C h/ 10xh h h!0 df .x/ D .4 10x/ dx D lim ADAMS and ESSEX: CALCULUS 9 4h 5.x C h/2 h 5h2 D4 .1 C 4x 5x 2 / 10x 2 D lim p p h!0 2.t C h/ C 1 C 2t C 1 1 D p 2t C 1 1 dF .t / D p dt 2t C 1 f .x/ D x 3 .x C h/3 x 3 h h!0 3x 2 h C 3xh2 C h3 D lim D 3x 2 h h!0 df .x/ D 3x 2 dx f 0 .x/ D lim 18. f .x/ D 34 0 14. 15. 16. 1 3 C 4t ds 1 1 1 D lim dt 3 C 4t h!0 h 3 C 4.t C h/ 4 3 C 4t 3 4t 4h D D lim .3 C 4t /2 h!0 h.3 C 4t /Œ3 C .4t C h/ 4 dt ds D .3 C 4t /2 2 x g.x/ D 2Cx 2 .x C h/ 2 x 2CxCh 2Cx 0 g .x/ D lim h h!0 .2 x h/.2 C x/ .2 C x C h/.2 D lim h.2 C x C h/.2 C x/ h!0 4 D .2 C x/2 4 dx dg.x/ D .2 C x/2 y 0 D lim x 1h h!0 h 3 1 3 .x C h/ .x C h/ . 13 x 3 h 1 2 x h C xh2 C 31 h3 h!0 h D lim .x 2 C xh C 31 h2 1/ D x 2 D lim h!0 2 dy D .x 44 Telegram: @uni_k 1/ dx 1 i x/ p f .x/ D lim 2 3 4 h!0 3 D lim h!0 4 sD y D 31 x 3 p 2t C 1 p p 2.t C h/ C 1 2t C 1 F 0 .t / D lim h h!0 2t C 2h C 1 2t 1 D lim p p h!0 h 2.t C h/ C 1 C 2t C 1 F .t / D " 3 3 4 .x C h/ h 2 p h. 2 df .x/ D p 8 2 x 3 p dx 8 2 x yDxC 1 x D 19. x p 2 p 2 x 2Cx p .x C h/ C 2 x h x/ # 1 1 x x C h x y D lim h h!0 x x h D lim 1 C h.x C h/x h!0 1 1 D 1 C lim D1 x2 h!0 .x C h/x 1 dx dy D 1 x2 0 x/ 20. xChC s 1Cs dz 1 sCh s D lim ds 1Cs h!0 h 1 C s C h .s C h/.1 C s/ s.1 C s C h/ 1 D lim D h.1 C s/.1 C s C h/ .1 C s/2 h!0 1 dz D ds .1 C s/2 zD Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL F .x/ D p 21. SECTION 2.2 (PAGE 107) 1 25. 1 C x2 1 1 p p 2 1 C x2 1 C .x C h/ F 0 .x/ D lim h h!0 p 26. p 1 C x2 1 C .x C h/2 D lim p p h!0 h 1 C .x C h/2 1 C x 2 1 C x 2 1 x 2 2hx h2 p 27. D lim p p p h!0 h 1 C .x C h/2 1 C x 2 1 C x 2 C 1 C .x C h/2 2x x D 2 3=2 2.1 C x / .1 C x 2 /3=2 x dF .x/ D dx .1 C x 2 /3=2 D 22. yD 1 x2 28. 23. yD p p 2x f .x/ 1 ˇ ˇ ˇ 2x/ˇ ˇ 1 p 1Cx t2 3 t2 C 3 t2 3 1 .t C h/2 3 f 0 .t / D lim .t C h/2 C 3 t 2 C 3 h!0 h 2 Œ.t C h/ 3.t 2 C 3/ .t 2 3/Œ.t C h/2 C 3 D lim h.t 2 C 3/Œ.t C h/2 C 3 h!0 12t h C 6h2 12t D lim D 2 2 2 .t C 3/2 h!0 h.t C 3/Œ.t C h/ C 3 12t df .t / D 2 dt .t C 3/2 f .1/ x 1 0:71000 0:97010 0:99700 0:99970 d 3 .x dx 1CxCh y 0 .x/ D lim h h!0 p p 1Cx 1CxCh D lim p p h!0 h 1 C x C h 1 C x 1Cx 1 x h p D lim p p p h!0 h 1 C x C h 1 C x 1CxC 1CxCh 1 p D lim p p p h!0 1CxCh 1Cx 1CxC 1CxCh 1 D 2.1 C x/3=2 1 dy D dx 2.1 C x/3=2 24. y D x3 0:9 0:99 0:999 0:9999 1 1Cx h.x/ D jx 2 C 3x C 2j fails to be differentiable where x 2 C 3x C 2 D 0, that is, at x D 2 and x D 1. Note: both of these are single zeros of x 2 C 3x C 2. If they were higher order zeros (i.e. if .x C 2/n or .x C 1/n were a factor of x 2 C 3x C 2 for some integer n 2) then h would be differentiable at the corresponding point. x 1 1 1 2 x2 h!0 h .x C h/ 2 2 x .x C h/ 2 D lim D x3 h!0 hx 2 .x C h/2 2 dx dy D x3 y 0 D lim Since f .x/ D x sgn x D jxj, for x ¤ 0, f will become continuous at x D 0 if we define f .0/ D 0. However, f will still not be differentiable at x D 0 since jxj is not differentiable at x D 0. 2 Since g.x/ D x 2 sgn x D xjxj D x 2 if x > 0 , g will x if x < 0 become continuous and differentiable at x D 0 if we define g.0/ D 0. xD1 f .x/ x 1 1:31000 1:03010 1:00300 1:00030 1:1 1:01 1:001 1:0001 D lim h!0 f .1/ .1 C h/3 2.1 C h/ h . 1/ h C 3h2 C h3 h h!0 D lim 1 C 3h C h2 D 1 D lim h!0 29. f .x/ D 1=x f .x/ x f .2/ x 2 0:26316 0:25126 0:25013 0:25001 1:9 1:99 1:999 1:9999 x 2:1 2:01 2:001 2:0001 f .x/ f .2/ x 2 0:23810 0:24876 0:24988 0:24999 1 2 2 .2 C h/ 2Ch f .2/ D lim D lim h h!0 h!0 h.2 C h/2 1 1 D lim D .2 C h/2 4 h!0 0 f .t / D 30. The slope of y D 5 C 4x x 2 at x D 2 is ˇ 5 C 4.2 C h/ .2 C h/2 dy ˇˇ D lim ˇ ˇ dx h h!0 9 xD2 D lim h!0 h2 D 0: h Thus, the tangent line at x D 2 has the equation y D 9. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x Downloaded by ted cage (sxnbyln180@questza.com) 45 lOMoARcPSD|6566483 www.konkur.in SECTION 2.2 (PAGE 107) 31. y D ADAMS and ESSEX: CALCULUS 9 p x C 6. Slope at .3; 3/ is p 1 9Ch 3 9Ch 9 p D lim m D lim D : h 6 h!0 h h!0 9ChC3 Tangent line is y 32. The slope of y D ˇ dy ˇˇ ˇ dt ˇ tD 2 3D 1 .x 6 t t2 2 3/, or x at t D 6y D 2 and y D 2Ch 1 2 h!0 h . 2 C h/2 D lim D lim h!0 44. 3 : 2 45. yD 2 t2 C t 37. 38. 39. 40. 41. 42. 43. 46. 46 Telegram: @uni_k 1 . a2 1 D a2 .x a a/, tD4 The intersection points of y D x 2 and x C 4y D 18 satisfy 4x 2 C x .4x C 9/.x Therefore x D 18 D 0 2/ D 0: 9 4 or x D 2. dy The slope of y D x 2 is m1 D D 2x. dx 9 9 , m1 D . At x D 2, m1 D 4. At x D 4 2 The slope of x C 4y D 18, i.e. y D 14 x C 18 4 , is m2 D 41 . Thus, at x D 2, the product of these slopes is .4/. 41 / D 1. So, the curve and line intersect at right angles at that point. 17x 18 for x ¤ 0 1 dy D x 2=3 for x ¤ 0 dx 3 dy 1 4=3 D x for x ¤ 0 dx 3 d 2:25 t D 2:25t 3:25 for t > 0 dt d 119=4 119 115=4 s D s for s > 0 ds 4 ˇ ˇ 1 d p ˇˇ 1 ˇˇ D : sˇ D p ˇ ˇ ds 6 2 sˇ sD9 sD9 1 1 0 1 ; F D 16 F .x/ D ; F 0 .x/ D x x2 4 ˇ 2 5=3 ˇˇ 1 0 f .8/ D x D ˇ ˇ 3 48 xD8 ˇ ˇ ˇ ˇ dy ˇ 1 1 ˇ D t 3=4 ˇ D p ˇ ˇ dt ˇ 4 8 2 tD4 xDa D Slope at t D a is 35. g 0 .t / D 22t 21 for all t 36. ˇ 1 ˇˇ ˇ x2 ˇ Normal has slope a , and equation y 1 or y D a2 x a3 C a 2 2 .a C h/2 C .a C h/ a2 C a m D lim h h!0 2.a2 C a a2 2ah h2 a h/ D lim hŒ.a C h/2 C a C h.a2 C a/ h!0 4a 2h 2 D lim 2 h!0 Œ.a C h/ C a C h.a2 C a/ 4a C 2 D .a2 C a/2 2.2a C 1/ 2 .t a/ Tangent line is y D 2 a C a .a2 C a/2 34. f 0 .x/ D 1 at x D a is Slope of y D x 2 Thus, the tangent line has the equation y D 1 23 .t C 2/, that is, y D 32 t 4. 33. 1 D p : 2 x0 Thus, the equation of the tangent line is x C x0 1 p y D x0 C p .x x0 /, that is, y D p . 2 x0 2 x0 1 is D x at x D x0 is xDx0 15. 2 p ˇ dy ˇˇ ˇ dx ˇ . 1/ 2 C h C Œ. 2 C h/2 hŒ. 2 C h/2 2 The slope of y D 47. Let theˇ point of tangency be .a; a2 /. Slope of tangent is d 2 ˇˇ x ˇ D 2a dx ˇ xDa This is the slope from .a; a2 / to .1; 3/, so a2 C 3 D 2a, and a 1 a2 C 3 D 2a2 2a a2 2a 3 D 0 a D 3 or 1 The two tangent lines are (for a D 3): y 9 D 6.x 3/ or 6x 9 (for a D 1): y 1 D 2.x C 1/ or y D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2x 1 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 2.2 (PAGE 107) p p 4a2 4b D a ˙ a2 b. 2 p If b < a2 , i.e. a2 b > 0, then t D a ˙ a2 b has two real solutions. Therefore, there will be two distinct tangent lines passing through .a; b/ with equations p y D b C 2 a ˙ a2 b .x a/. If b D a2 , then t D a. There will be only one tangent line with slope 2a and equation y D b C 2a.x a/. If b > a2 , then a2 b < 0. There will be no real solution for t . Thus, there will be no tangent line. y Hence t D .a;a2 / y D x2 x 2a ˙ .1; 3/ Fig. 2.2-47 48. The slope of y D 1 at x D a is x ˇ dy ˇˇ ˇ dx ˇ xDa D 51. Suppose f is odd: f . x/ D f .x/. Then f . x C h/ f . x/ f 0 . x/ D lim h h!0 f .x h/ f .x/ D lim h h!0 .let h D k/ f .x C k/ f .x/ D lim D f 0 .x/ k k!0 Thus f 0 is even. Now suppose f is even: f . x/ D f .x/. Then f . x C h/ f . x/ f 0 . x/ D lim h h!0 f .x h/ f .x/ D lim h h!0 f .x C k/ f .x/ D lim k k!0 D f 0 .x/ so f 0 is odd. 52. Let f .x/ D x n . Then 1 : a2 1 1 D 2, or a D ˙ p . Therea2 2 fore, the equations of the are two straight lines p p 1 1 and y D 2 2 xCp , yD 2 2 x p 2 p 2 or y D 2x ˙ 2 2. If the slope is 2, then p 49. Let the point of tangency be ˇ .a; a/ d p ˇˇ 1 Slope of tangent is xˇ D p ˇ dx 2 a xDa p a 0 1 , so a C 2 D 2a, and a D 2. Thus p D aC2 2 a 1 The required slope is p . 2 2 y p .a; a/ p yD x x 2 Fig. 2.2-49 50. If aˇ line is tangent to y D x 2 at .t; t 2 /, then its slope is dy ˇˇ D 2t . If this line also passes through .a; b/, then ˇ dx ˇ xDt its slope satisfies t2 b D 2t; t a that is t 2 2at C b D 0: .x C h/ n x n h h!0 1 1 1 D lim .x C h/n x n h!0 h n x .x C h/n D lim n h!0 hx .x C h/n x .x C h/ D lim h!0 hx n ..x C h/n x n 1 C x n 2 .x C h/ C C .x C h/n 1 f 0 .x/ D lim D 1 nx n 1 D x 2n Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) nx .nC1/ : 47 lOMoARcPSD|6566483 www.konkur.in SECTION 2.2 (PAGE 107) 53. ADAMS and ESSEX: CALCULUS 9 f .x/ D x 1=3 If f 0 .aC/ is finite, call the half-line with equation y D f .a/ C f 0 .aC/.x a/, (x a), the right tangent line to the graph of f at x D a. Similarly, if f 0 .a / is finite, call the half-line y D f .a/ C f 0 .a /.x a/, (x a), the left tangent line. If f 0 .aC/ D 1 (or 1), the right tangent line is the half-line x D a, y f .a/ (or x D a, y f .a/). If f 0 .a / D 1 (or 1), the right tangent line is the half-line x D a, y f .a/ (or x D a, y f .a/). The graph has a tangent line at x D a if and only if f 0 .aC/ D f 0 .a /. (This includes the possibility that both quantities may be C1 or both may be 1.) In this case the right and left tangents are two opposite halves of the same straight line. For f .x/ D x 2=3 , f 0 .x/ D 32 x 1=3 : At .0; 0/, we have f 0 .0C/ D C1 and f 0 .0 / D 1. In this case both left and right tangents are the positive y-axis, and the curve does not have a tangent line at the origin. For f .x/ D jxj, we have .x C h/1=3 x 1=3 h h!0 .x C h/1=3 x 1=3 D lim h h!0 .x C h/2=3 C .x C h/1=3 x 1=3 C x 2=3 .x C h/2=3 C .x C h/1=3 x 1=3 C x 2=3 xCh x D lim h!0 hŒ.x C h/2=3 C .x C h/1=3 x 1=3 C x 2=3 1 D lim h!0 .x C h/2=3 C .x C h/1=3 x 1=3 C x 2=3 1 1 D D x 2=3 3 3x 2=3 f 0 .x/ D lim 54. Let f .x/ D x 1=n . Then .x C h/1=n f 0 .x/ D lim h h!0 a b D lim n bn a!b a x 1=n f 0 .x/ D sgn .x/ D (let x C h D an , x D b n ) a!b an 1 C an 2 b C an 3 b 2 C C b n 1 D 55. 1 1 D x .1=n/ 1 : nb n 1 n d n .x C h/n x n x D lim dx h h!0 n.n 1/ n 2 2 1 n n n 1 hC x h x C x D lim 1 12 h!0 h n.n 1/.n 2/ n 3 3 C x h C C hn x n 123 n.n 1/ n 2 x h D lim nx n 1 C h 12 h!0 ! n.n 1/.n 2/ n 3 2 n 1 C x h C C h 123 D nx n 1 1. y D 3x 2 2. y D 4x 1=2 3. f .x/ D Ax 2 C Bx C C; 4. f .x/ D 5. zD 6. y D x 45 7. 8. 9. f .a C h/ h f .a C h/ 0 f .a / D lim h h!0 f .a/ h!0C 48 Telegram: @uni_k f .a/ 1 if x > 0 if x < 0. Section 2.3 Differentiation Rules (page 115) 56. Let f 0 .aC/ D lim 1 At .0; 0/, f 0 .0C/ D 1, and f 0 .0 / D 1. In this case the right tangent is y D x, .x 0/, and the left tangent is y D x, .x 0/. There is no tangent line. 1 D lim n 10. 5x 7; 5 ; x s3 15 2; f 0 .x/ D 2Ax C B: f 0 .x/ D dz 1 D s4 dx 3 ; 5: y 0 D 2x 1=2 C 5x 2 2 6 C 2 x3 x s5 y 0 D 6x x 45 18 x4 4 x3 1 2 s : 5 y 0 D 45x 44 C 45x 46 g.t / D t 1=3 C 2t 1=4 C 3t 1=5 1 3 1 g 0 .t / D t 2=3 C t 3=4 C t 4=5 3 2 5 p 2 3 2=3 2 p D 3t yD3 t 2t 3=2 t3 dy D 2t 1=3 C 3t 5=2 dt 3 5 3=5 u D x 5=3 x 5 3 du D x 2=3 C x 8=5 dx F .x/ D .3x F 0 .x/ D 3.1 2/.1 5x/ 5x/ C .3x 2/. 5/ D 13 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 30x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 11. yD p x 5 3p x 2 5 y0 D p 2 x 1 12. g.t / D 13. 2t 3 4 3 x 15. f .t / D 2 16. g.y/ D 20. 21. 22. x 3=2 1 5=2 x 3 23. 5 3=2 x 6 g 0 .t / D ; y0 D ; 2 .2t 3/2 24. 2x C 5 .x 2 C 5x/2 4 .3 x/2 25. t 2 . / D 2 .2 t / .2 t /2 f 0 .t / D 19. p D5 x 1 x 2 C 5x 1 y0 D .2x C 5/ D .x 2 C 5x/2 yD 18. yD 14. 17. x2 3 x SECTION 2.3 (PAGE 115) 2 y2 1 1 ; 4x 2 g 0 .y/ D .1 4y y 2 /2 4 f .x/ D Dx x3 x 4x 2 3 f 0 .x/ D 3x 4 C 4x 2 D x4 p u u 3 D u 1=2 3u 2 g.u/ D u2 p 1 3=2 12 u u g 0 .u/ D u C 6u 3 D 2 2u3 26. 3 27. p 2 C t C t2 D 2t 1=2 C t C t 3=2 p t 1 3p dy 3t 2 C t 2 D t 3=2 C p C p tD dt 2 2 t 2t t yD x 1 z D 2=3 D x 1=3 x 2=3 x dz 1 2=3 2 5=3 xC2 D x C x D dx 3 3 3x 5=3 3 4x 3 C 4x .3 C 4x/. 4/ .3 f 0 .x/ D .3 C 4x/2 24 D .3 C 4x/2 x3 4 xC1 .x C 1/.3x 2 / .x 3 0 f .x/ D .x C 1/2 3 2x C 3x 2 C 4 D .x C 1/2 f .x/ D 4/.1/ ax C b cx C d .cx C d /a .ax C b/c f 0 .x/ D .cx C d /2 ad bc D .cx C d /2 f .x/ D t 2 C 7t 8 t2 t C 1 2 .t t C 1/.2t C 7/ .t 2 C 7t F 0 .t / D .t 2 t C 1/2 2 8t C 18t 1 D .t 2 t C 1/2 F .t / D 8/.2t 1/ f .x/ D .1 C x/.1 C 2x/.1 C 3x/.1 C 4x/ f 0 .x/ D .1 C 2x/.1 C 3x/.1 C 4x/ C 2.1 C x/.1 C 3x/.1 C 4x/ C 3.1 C x/.1 C 2x/.1 C 4x/ C 4.1 C x/.1 C 2x/.1 C 3x/ OR f .x/ D Œ.1 C x/.1 C 4x/ Œ.1 C 2x/.1 C 3x/ D 1 C 10x C 25x 2 C 10x 2 .1 C 5x/ C 24x 4 0 D 1 C 10x C 35x 2 C 50x 3 C 24x 4 f .x/ D 10 C 70x C 150x 2 C 96x 3 28. f .r/ D .r 2 C r 3 f 0 .r/ D . 2r 3 4x/.4/ or f .r/ D 0 f .r/ D t 2 C 2t t2 1 2 .t 1/.2t C 2/ .t 2 C 2t /.2t / z0 D .t 2 1/2 2.t 2 C t C 1/ D .t 2 1/2 29. 4/.r 2 C r 3 C 1/ 3r 4 /.r 2 C r 3 C 1/ C .r 2 C r 3 4/.2r C 3r 2 / 2Cr 1 Cr 2Cr 3Cr 2 3 4 2r 3r C 1 8r p y D .x 2 C 4/. x C 1/.5x 2=3 2/ p y 0 D 2x. x C 1/.5x 2=3 2/ 1 C p .x 2 C 4/.5x 2=3 2/ 2 x p 10 C x 1=3 .x 2 C 4/. x C 1/ 3 r Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k p 1 .1 C t /. p / 2 t p 2 t/ D .1 C 5x C 4x 2 /.1 C 5x C 6x 2 / f .x/ D zD p 1C t p 1 t p 1 .1 t/ p ds 2 t D dt .1 1 p D p t .1 t /2 sD Downloaded by ted cage (sxnbyln180@questza.com) 4r 2 4r 3 12r 2 49 lOMoARcPSD|6566483 www.konkur.in SECTION 2.3 (PAGE 115) 30. 31. 32. ADAMS and ESSEX: CALCULUS 9 .x 2 C 1/.x 3 C 2/ .x 2 C 2/.x 3 C 1/ x 5 C x 3 C 2x 2 C 2 D 5 x C 2x 3 C x 2 C 2 .x 5 C 2x 3 C x 2 C 2/.5x 4 C 3x 2 C 4x/ y0 D .x 5 C 2x 3 C x 2 C 2/2 5 .x C x 3 C 2x 2 C 2/.5x 4 C 6x 2 C 2x/ .x 5 C 2x 3 C x 2 C 2/2 2x 7 3x 6 3x 4 6x 2 C 4x D .x 5 C 2x 3 C x 2 C 2/2 7 2x 3x 6 3x 4 6x 2 C 4x D .x 2 C 2/2 .x 3 C 1/2 yD yD 36. 37. 1 3x C 1 .6x 2 C 2x C 1/.6x C 1/ .3x 2 C x/.12x C 2/ y0 D .6x 2 C 2x C 1/2 6x C 1 D .6x 2 C 2x C 1/2 2x C p . x 33. 34. 35. d dx d dx 39. 1 p x x2 f .x/ ˇˇ ˇ ˇ ˇ ˇ f .x/ ˇˇ ˇ x2 ˇ xD2 ˇ f .x/.2x/ x 2 f 0 .x/ ˇˇ D ˇ ˇ Œf .x/2 D xD2 4f .2/ 4f 0 .2/ D Œf .2/2 ˇ x 2 f 0 .x/ 2xf .x/ ˇˇ D ˇ ˇ x4 ! 40. xD2 18 14 1 .4 C f .2//f 0 .2/ f .2/.4 C f 0 .2// D D D .4 C f .2//2 62 9 ˇ 2 ˇ x 4 d d 8 ˇ j D 1 ˇ xD 2 dx x 2 C 4 dx x2 C 4 ˇ xD 2 ˇ ˇ 8 ˇ D 2 .2x/ˇ ˇ .x C 4/2 D p #ˇ t .1 C t / ˇˇ ˇ ˇ 5 t ˇ " #tD4 d t C t 3=2 ˇˇ D ˇ ˇ dt 5 t d dt " 32 D 64 1 2 tD4 ˇ t /.1 C 23 t 1=2 / .t C t 3=2 /. 1/ ˇˇ ˇ ˇ .5 t /2 .5 tD4 .12/. 1/ D D 16 .1/2 p x f .x/ D xC1 p 1 .x C 1/ p x.1/ 2 x f 0 .x/ D .x C 1/2 p 3 p 2 1 2 2 f 0 .2/ D p D 9 18 2 ˇ ˇ d Œ.1 C t /.1 C 2t /.1 C 3t /.1 C 4t /ˇˇ dt tD0 D .1/.1 C 2t /.1 C 3t /.1 C 4t / C .1 C t /.2/.1 C 3t /.1 C 4t /C ˇ ˇ .1 C t /.1 C 2t /.3/.1 C 4t / C .1 C t /.1 C 2t /.1 C 3t /.4/ˇˇ tD0 D 1 C 2 C 3 C 4 D 10 xD2 4 D 4 1 41. y D 2 p , y0 D 4 x 3 2 p 2 4 x 3 4 p 2 x 8 D4 . 1/2 2 Tangent line has the equation y D 2 C 4.x 1/ or y D 4x 6 Slope of tangent at .1; 2/ is m D xD2 4 1 4f 0 .2/ 4f .2/ D D D 16 16 4 xD2 ˇˇ ˇ ˇ ˇ xD2 ˇ .x 2 C f .x//f 0 .x/ f .x/.2x C f 0 .x// ˇˇ D ˇ ˇ .x 2 C f .x//2 D .3 C 2x/. 1 ˇˇ d 2 x f .x/ ˇˇ dx 50 Telegram: @uni_k f .x/ x 2 C f .x/ .1/.4/ 4x C 3x 2 / .2 x 2x 2 C x 3 /.2/ .3 C 2x/2 2 .2 x/.1 x / D 2x 3=2 .3 C 2x/ ! 1 4x 3 C 5x 2 12x 7 C 1 p .3 C 2x/2 x 38. 2 1/.2 x/.1 x / p x.3 C 2x/ ! 1 2 x 2x 2 C x 3 D 1 p 3 C 2x x ! 1 3=2 2 x 2x 2 C x 3 x C 1 f 0 .x/ D 2 3 C 2x f .x/ D xD 2 3x 2 C x D 2 6x C 2x C 1 x d dx ˇˇ D 2xf .x/ C x 2 f 0 .x/ ˇˇ D 4f .2/ C 4f 0 .2/ D 20 42. xD2 For y D xC1 we calculate x 1 y0 D .x Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1/.1/ .x .x C 1/.1/ D 1/2 2 .x 1/2 : lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 2.3 (PAGE 115) y At x D 2 we have y D 3 and y 0 D 2. Thus, the equation of the tangent line is y D 3 2.x 2/, or y D 2x C 7. The normal line is y D 3 C 12 .x 2/, or y D 12 x C 2. b a; 1 1 0 , y D1 x x2 1 For horizontal tangent: 0 D y 0 D 1 so x 2 D 1 and x2 x D ˙1 The tangent is horizontal at .1; 2/ and at . 1; 2/ 43. y D x C 44. If y D x 2 .4 0 y D 2x.4 Fig. 2.3-47 2 2 3 x / C x . 2x/ D 8x 4x D 4x.2 2 x /: 48. Thus 2x C 1 D 0 and x D 1 2 The tangent is horizontal only at 2x C 1 . .x 2 C x C 1/2 or .x C 2/ 50. .x 1/ at .a; a2 =.a ˇ 1/2x x 2 .1/ ˇˇ a2 D ˇ 2 ˇ .x 1/ .a xDa 1// has slope 2a : 1/2 The equation of the tangent is 5 . 2 At x D 3 , 2 Hence, the is parallel to y D 4x at the points tangent 5 3 ; 1 and ; 3 . 2 2 Let the point of tangency be .a; a1 /. The slope of the tanb a1 1 2 gent is D . Thus b a1 D a1 and a D . a2 0 a b b2 b2 so has equation y D b x. Tangent has slope 4 4 y a2 a 1 D a2 .a 2a .x 1/2 a/: This line passes through .2; 0/ provided 0 a2 a 1 D a2 .a 2a .2 1/2 a/; or, upon simplification, 3a2 4a D 0. Thus we can have either a D 0 or a D 4=3. There are two tangents through .2; 0/. Their equations are y D 0 and y D 8x C 16. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2/2 .a C 1/: The tangent to y D x 2 =.x mD D 41 : 3a2 C 4 D .a The two tangent lines to y D x 3 passing through .2; 8/ correspond to a D 2 and a D 1, so their equations are y D 12x 16 and y D 3x C 2. 1 .x C 2/.1/ .x C 1/.1/ D : .x C 2/2 .x C 2/2 2 1. Hence, the The tangent to y D x 3 at .a; a3 / has equation y D a3 C 3a2 .x a/, or y D 3a2 x 2a3 . This line passes through .2; 8/ if 8 D 6a2 2a3 or, equivalently, if a3 3a2 C 4 D 0. Since .2; 8/ lies on y D x 3 , a D 2 must be a solution of this equation. In fact it must be a double root; .a 2/2 must be a factor of a3 3a2 C 4. Dividing by this factor, we find that the other factor is a C 1, that is, a3 Hence x C 2 D ˙ 12 , and x D 32 or y D 1, and at x D 52 , y D 3. 47. 49. 1 4 ; . 2 3 In order to be parallel to y D 4x, the tangent line must have slope equal to 4, i.e., xD1 The product of the slopes is .2/ 12 D two curves intersect at right angles. xC1 , then xC2 1 D 4; .x C 2/2 1 Since p D y D x 2 ) x 5=2 D 1, therefore x D 1 at x theˇ intersection point. The slope of y D x 2 at x D 1 is ˇ 1 2x ˇˇ D 2. The slope of y D p at x D 1 is x xD1 ˇ ˇ dy ˇˇ 1 3=2 ˇˇ 1 x : D D ˇ ˇ ˇ dx ˇ 2 2 xD1 2x C 1 2 .x C x C 1/2 For horizontal tangent we want 0 D y 0 D y0 D x 2 /, then 1 , y0 D 2 x CxC1 46. If y D 1 a x The slope of a horizontal line must be zero, so p 4x.2 x 2 / D 0, which impliesp that x D 0 or x D ˙ 2. At x D 0; y D 0 and at x D ˙ 2; y D 4. Hence, there are two horizontal lines that are tangent to the curve. Their equations are y D 0 and y D 4. 45. y D 1 x yD Downloaded by ted cage (sxnbyln180@questza.com) 51 lOMoARcPSD|6566483 www.konkur.in SECTION 2.3 (PAGE 115) 51. ADAMS and ESSEX: CALCULUS 9 p p f .x C h/ f .x/ h f .x C h/ f .x/ 1 p D lim p h h!0 f .x C h/ C f .x/ d p f .x/ D lim dx h!0 Proof: The case n D 2 is just the Product Rule. Assume the formula holds for n D k for some integer k > 2. Using the Product Rule and this hypothesis we calculate .f1 f2 fk fkC1 /0 D Œ.f1 f2 fk /fkC1 0 0 D .f1 f2 fk /0 fkC1 C .f1 f2 fk /fkC1 f 0 .x/ D p 2 f .x/ 2x x d p 2 x C1D p D p dx 2 x2 C 1 x2 C 1 3 52. f .x/ D jx 3 j D x 3 if x 0 . Therefore f is differenx if x < 0 tiable everywhere except possibly at x D 0, However, f .0 C h/ h h!0C f .0 C h/ lim h h!0 lim f .0/ D .f10 f2 fk C f1 f20 fk C C f1 f2 fk0 /fkC1 0 C .f1 f2 fk /fkC1 D f10 f2 fk fkC1 C f1 f20 fk fkC1 C 0 C f1 f2 fk0 fkC1 C f1 f2 fk fkC1 D lim h2 D 0 so the formula is also true for n D k C 1. The formula is therefore for all integers n 2 by induction. D lim . h2 / D 0: Section 2.4 The Chain Rule h!0C f .0/ h!0 Thus f 0 .0/ exists and equals 0. We have 2 f 0 .x/ D 3x 2 3x 1. 2. if x 0 if x < 0. n d n=2 D x .n=2/ 1 for n D 1, 2, 3, : : : . x 53. To be proved: dx 2 Proof: It is already known that the case n D 1 is true: the derivative of x 1=2 is .1=2/x 1=2 . Assume that the formula is valid for n D k for some positive integer k: 3. 4. 5. d k=2 k x D x .k=2/ 1 : dx 2 Then, by the Product Rule and this hypothesis, 6. d .kC1/=2 d 1=2 k=2 x D x x dx dx 1 k k C 1 .kC1/=2 1 D x 1=2 x k=2 C x 1=2 x .k=2/ 1 D x : 2 2 2 Thus the formula is also true for n D k C 1. Therefore it is true for all positive integers n by induction. For negative n D m (where m > 0) we have d 1 d n=2 x D dx dx x m=2 1 m D m x .m=2/ 1 x 2 n m .m=2/ 1 x D x .n=2/ 1 : D 2 2 .f1 f2 fn /0 D f10 f2 fn C f1 f20 fn C C f1 f2 fn0 52 Telegram: @uni_k y D .2x C 3/6 ; y 0 D 6.2x C 3/5 2 D 12.2x C 3/5 x 99 yD 1 3 x 98 1 x 98 D 33 1 y 0 D 99 1 3 3 3 f .x/ D .4 x 2 /10 f 0 .x/ D 10.4 x 2 /9 . 2x/ D 20x.4 x 2 /9 d p 3x dy 6x D D p 1 3x 2 D p 2 dx dx 2 1 3x 1 3x 2 10 3 F .t / D 2 C t 30 3 11 3 11 3 D 2 C F 0 .t / D 10 2 C t t2 t2 t z D .1 C x 2=3 /3=2 z 0 D 32 .1 C x 2=3 /1=2 . 23 x 1=3 / D x 1=3 .1 C x 2=3 /1=2 7. yD 3 5 0 y D 4x 3 12 . 4/ D .5 4x/2 .5 4x/2 y D .1 2t 2 / 3=2 9. y D j1 x 2 j; 10. f .t / D j2 C t 3 j 8. y0 D 3 2 .1 2t 2 / 5=2 . 4t / D 6t .1 y0 D 2xsgn .1 f 0 .t / D Œsgn .2 C t 3 /.3t 2 / D 11. 54. To be proved: (page 120) y D 4x C j4x 1j y 0 D 4 C 4.sgn .4x 1// 8 if x > 14 D 0 if x < 14 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2t 2 / 5=2 x2/ D 2x 3 2x j1 x 2 j 3t 2 .2 C t 3 / j2 C t 3 j lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 12. SECTION 2.4 (PAGE 120) 18. y D .2 C jxj3 /1=3 y y 0 D 31 .2 C jxj3 / 2=3 .3jxj2 /sgn .x/ x D jxj2 .2 C jxj3 / 2=3 D xjxj.2 C jxj3 / 2=3 jxj 13. r f .x/ D 4 1 C D 16. 17. 2 3 r z D uC D 3 2 p 2 3x C 4 2 C 3x C 4 f .x/ D 1 C dz D du slope 0 p 0 15. yD4xCj4x 1j 1 yD p 2 C 3x C 4 3 1 p y0 D 2 p 2 3x C 4 2 C 3x C 4 D 14. slope 8 3 !4 x 2 x r 2 3 3 x 2 1 !3 1C r 1 2 x r 2 3 3 x !3 ! 1 2 3 5=3 u 1 8=3 5 1 1 uC 1 3 u 1 .u 1/2 8=3 5 1 1 1 u C 3 .u 1/2 u 1 p x5 3 C x6 yD .4 C x 2 /3 p 3x 5 1 2 3 4 6 C x5 p 3 C x y0 D .4 C x / 5x .4 C x 2 /6 3 C x6 ! h i p x 5 3 C x 6 3.4 C x 2 /2 .2x/ h i .4 C x 2 / 5x 4 .3 C x 6 / C 3x 10 x 5 .3 C x 6 /.6x/ p D .4 C x 2 /4 3 C x 6 60x 4 3x 6 C 32x 10 C 2x 12 D p .4 C x 2 /4 3 C x 6 21=3 d 1=4 d x D dx dx q 20. d 3=4 d x D dx dx q 21. 3 1 d 3=2 d p 3 x D p .3x 2 / D x 1=2 x D dx dx 2 2 x3 22. d f .2t C 3/ D 2f 0 .2t C 3/ dt 23. d f .5x dx 24. 25. 26. 27. 28. 29. t p 1 1 1 x D pp p D x 3=4 4 2 x 2 x p 1 x xD p p 2 x x x 2 / D .5 2x/f 0 .5x p x xC p 2 x D 3 1=4 x 4 x2/ 2 3 2 2 2 2 d D3 f f0 f dx x x x x2 2 6 0 2 2 D f f 2 x x x 2f 0 .x/ d p f 0 .x/ 3 C 2f .x/ D p D p dx 2 3 C 2f .x/ 3 C 2f .x/ p p d 2 f . 3 C 2t / D f 0 . 3 C 2t/ p dt 2 3 C 2t p 1 0 D p f . 3 C 2t / 3 C 2t p p d 1 f .3 C 2 x/ D p f 0 .3 C 2 x/ dx x d f 2f 3f .x/ dt 0 D f 2f 3f .x/ 2f 0 3f .x/ 3f 0 .x/ D 6f 0 .x/f 0 3f .x/ f 0 2f 3f .x/ d f 2 3f .4 5t / dx D f 0 2 3f .4 5t / 3f 0 .4 5t / . 5/ D 15f 0 .4 5t /f 0 2 3f .4 5t / Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x 19. y yDj2Ct 3 j 1 4 ;1 Downloaded by ted cage (sxnbyln180@questza.com) 53 lOMoARcPSD|6566483 www.konkur.in SECTION 2.4 (PAGE 120) 30. 31. 32. !ˇ x 2 1 ˇˇ ˇ x2 C 1 ˇ xD 2 p x 2 x 2 1.2x/ ˇˇ .x C 1/ p 2 ˇ x 1 D ˇ ˇ .x 2 C 1/2 xD 2 p 2 p 3. 4/ .5/ 2 3 D p D 25 25 3 p d dx ˇ ˇ ˇ 7ˇ ˇ d p 3t dt 3 D p 2 3t tD3 1 f .x/ D p 2x C 1 0 f .4/ D 33. ADAMS and ESSEX: CALCULUS 9 ˇ ˇ 1 ˇ ˇ .2x C 1/3=2 ˇ ˇ ˇ ˇ ˇ 7ˇ xD4 xD 1 The tangent p line at . 1; 23=2 / has equation 3=2 2.x C 1/. yD2 38. b The slope of y D .ax C b/8 at x D is a ˇ ˇ ˇ dy ˇˇ 7ˇ D 8a.ax C b/ ˇ D 1024ab 7 : ˇ ˇ dx ˇ xDb=a 3 D p 2 2 tD3 y D .2b/8 D 256b 8 is D 256b 8 C 1024ab 7 x y D 210 ab 7 x 1 27 y D .x 3 C 9/17=2 ˇ ˇ ˇ ˇ 17 3 0ˇ 15=2 2ˇ yˇ D .x C 9/ 3x ˇ ˇ ˇ 2 xD 2 D 40. Given that f .x/ D .x f 0 .x/ D m.x 2.1 C x/.2 C x/.3 C x/3 .4 C x/4 C D .x 3.1 C x/.2 C x/2 .3 C x/2 .4 C x/4 C F 0 .0/ D .22 /.33 /.44 / C 2.1/.2/.33 /.44 /C D 4.22 33 44 / D 110; 592 y D D 36. The slope of y D ˇ dy ˇˇ ˇ dx ˇ 1=2 6 ˇ ˇ ˇ D p ˇ 2 1 C 2x 2 ˇ 4x xD2 D Telegram: @uni_k .x b/n C n.x n 1 b/ .mx a/m .x b/n 1 mb C nx na/: mb C nx na D 0; n m aC b: mCn mCn This point lies lies between a and b. 41. x.x 4 C 2x 2 2/=.x 2 C 1/5=2 42. 4.7x 4 43. 857; 592 44. 5=8 45. The Chain Rule does not enable you to calculate the derivatives of jxj2 and jx 2 j at x D 0 directly as a composition of two functions, one of which is jxj, because jxj is not differentiable at x D 0. However, jxj2 D x 2 and jx 2 j D x 2 , so both functions are differentiable at x D 0 and have derivative 0 there. 46. It may happen that k D g.x C h/ g.x/ D 0 for values of h arbitrarily close to 0 so that the division by k in the “proof” is not justified. 4 : 3 Thus, the equation of the tangent line at .2; 3/ is y D 3 C 43 .x 2/, or y D 43 x C 31 : 54 a/ xD 1 C 2x 2 at x D 2 is xD2 m 1 b/n then which is equivalent to 1=2 7 5 6 x C .3x/ 2 3=2 1 .3x/5 2 5.3x/4 3 1 2 3=2 15 6 1 .3x/4 .3x/5 2 2 1=2 7 x C .3x/5 2 p a/m 1 .x mx 3.1/.22 /.32 /.44 / C 4.1/.22 /.33 /.43 / 0 a/m .x If x ¤ a and x ¤ b, then f 0 .x/ D 0 if and only if 4.1 C x/.2 C x/2 .3 C x/3 .4 C x/3 2 b , or a Slope of y D 1=.x 2 x Cˇ3/3=2 at x D 2 is ˇ 3 5 3 2 D .x xC3/ 5=2 .2x 1/ˇˇ .9 5=2 /. 5/ D 2 2 162 xD 2 1 The tangent line at . 2; / has equation 27 5 1 C .x C 2/. yD 27 162 F 0 .x/ D .2 C x/2 .3 C x/3 .4 C x/4 C 35. 3 28 b 8 . b and a 39. 17 .12/ D 102 2 F .x/ D .1 C x/.2 C x/2 .3 C x/3 .4 C x/4 y D x C .3x/5 xDb=a The equation of the tangent line at x D y D xD 2 34. 37. Slope of y D .1 C x 2=3 /3=2ˇ at x D 1 is p 2 1=3 ˇˇ 3 .1 C x 2=3 /1=2 x D 2 ˇ ˇ 2 3 49x 2 C 54/=x 7 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 2.5 (PAGE 126) Section 2.5 Derivatives of Trigonometric Functions (page 126) 24. 1. d d 1 csc x D D dx dx sin x cos x D sin2 x 2. d d cos x cot x D D dx dx sin x cos2 x sin2 x D sin2 x 3. y D cos 3x; y0 D 4. y D sin y0 D x ; 5 5. y D tan x; 7. y D cot.4 csc x cot x 25. csc 2 x 26. 3 sin 3x 1 x cos : 5 5 27. y 0 D sec2 x 28. y 0 D a sec ax tan ax: 6. y D sec ax; 23. y 0 D 3 csc2 .4 3x/; 29. 3x/ x 1 x d sin D cos dx 3 3 3 9. f .x/ D cos.s rx/; f 0 .x/ D r sin.s 8. rx/ 30. 0 10. y D sin.Ax C B/; y D A cos.Ax C B/ d sin.x 2 / D 2x cos.x 2 / dx p p d 1 12. cos. x/ D p sin. x/ dx 2 x p sin x 13. y D 1 C cos x; y 0 D p 2 1 C cos x 11. 14. 15. 16. 31. 32. d sin.2 cos x/ D cos.2 cos x/. 2 sin x/ dx D 2 sin x cos.2 cos x/ f .x/ D cos.x C sin x/ f 0 .x/ D .1 C cos x/ sin.x C sin x/ 33. g. / D tan. sin / 34. g 0 . / D .sin C cos / sec2 . sin / 17. u D sin3 .x=2/; 18. y D sec.1=x/; 19. u0 D y0 D F .t / D sin at cos at 0 F .t / D a cos at cos at . D a cos 2at / 20. 21. 22. 3 cos.x=2/ sin2 .x=2/ 2 .1=x 2 / sec.1=x/ tan.1=x/ 1 sin 2at / 2 a sin at sin at .D 35. 36. sin a cos b a cos b cos a C b sin a sin b G 0 . / D : cos2 b d sin.2x/ cos.2x/ D 2 cos.2x/ C 2 sin.2x/ dx d d .cos2 x sin2 x/ D cos.2x/ dx dx D 2 sin.2x/ D 4 sin x cos x G. / D d .tan x C cot x/ D sec2 x csc2 x dx d .sec x csc x/ D sec x tan x C csc x cot x dx d .tan x x/ D sec2 x 1 D tan2 x dx d d tan.3x/ cot.3x/ D .1/ D 0 dx dx d .t cos t sin t / D cos t t sin t cos t D t sin t dt d .t sin t C cos t / D sin t C t cos t sin t D t cos t dt d .1 C cos x/.cos x/ sin.x/. sin x/ sin x D dx 1 C cos x .1 C cos x/2 1 cos x C 1 D D 2 .1 C cos x/ 1 C cos x d cos x .1 C sin x/. sin x/ cos.x/.cos x/ D dx 1 C sin x .1 C sin x/2 sin x 1 1 D D .1 C sin x/2 1 C sin x d 2 x cos.3x/ D 2x cos.3x/ 3x 2 sin.3x/ dx p g.t / D .sin t /=t t cos t sin t 1 g 0 .t / D p t2 2 .sin t /=t t cos t sin t D p 2t 3=2 sin t v D sec.x 2 / tan.x 2 / v 0 D 2x sec.x 2 / tan2 .x 2 / C 2x sec3 .x 2 / p sin x zD p 1 C cos x p p p p p p .1 C cos x/.cos x=2 x/ .sin x/. sin x=2 x/ p 2 z0 D .1 C cos x/ p 1 1 C cos x D p p p D p 2 x.1 C cos x/2 2 x.1 C cos x/ d sin.cos.tan t // D dt f .s/ D cos.s C cos.s C cos s// f 0 .s/ D Œsin.s C cos.s C cos s// Œ1 .sin.s C cos s//.1 sin s/ 37. Differentiate both sides of sin.2x/ D 2 sin x cos x and divide by 2 to get cos.2x/ D cos2 x sin2 x. sin2 x and 38. Differentiate both sides of cos.2x/ D cos2 x divide by 2 to get sin.2x/ D 2 sin x cos x. 39. Slope of y D sin x at .; 0/ is cos D 1. Therefore the tangent and normal lines to y D sin x at .; 0/ have equations y D .x / and y D x , respectively. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k .sec2 t /.sin.tan t // cos.cos.tan t // Downloaded by ted cage (sxnbyln180@questza.com) 55 lOMoARcPSD|6566483 www.konkur.in SECTION 2.5 (PAGE 126) ADAMS and ESSEX: CALCULUS 9 40. The slope of y D tan.2x/ at .0; 0/ is 2 sec2 .0/ D 2. Therefore the tangent and normal lines to y D tan.2x/ at .0; 0/ have equations y D 2x and y D x=2, respectively. p 41. Thepslope of y D 2 cos.x=4/ at .; 1/ is . 2=4/ sin.=4/ Dp 1=4. Therefore the tangent and normal lines to y D 2 cos.x=4/ at .; 1/ have equations y D 1 .x /=4 and y D 1 C 4.x /, respectively. 42. The slope of y Dpcos2 x at .=3; 1=4/ is sin.2=3/ D 3=2. Therefore the tangent and normal lines to y D tan.2x/ at .0; 0/ have equations p y D .1=4/ . 3=2/.x .=3// and p y D .1=4/ C .2= 3/.x .=3//, respectively. x x is y 0 D cos . 43. Slope of y D sin.x ı / D sin 180 180 180 At x D 45 the tangent line has equation 1 yD p C p .x 45/. 2 180 2 x 44. For y D sec .x ı / D sec we have 180 x x dy D sec tan : dx 180 180 180 p p 3 .2 3/ D : 180 90 90 Thus, the normal line has slope p and has equation 3 90 yD2 p .x 60/. 3 49. y D x C sin x has a horizontal tangent at x D because dy=dx D 1 C cos x D 0 there. 50. y D 2x C sin x has no horizontal tangents because dy=dx D 2 C cos x 1 everywhere. 51. y D x C 2 sin x has horizontal tangents at x D 2=3 and x D 4=3 because dy=dx D 1 C 2 cos x D 0 at those points. 52. y D x C 2 cos x has horizontal tangents at x D =6 and x D 5=6 because dy=dx D 1 2 sin x D 0 at those points. 53. 54. 2 sin.2x/ t an.2x/ D lim D12D2 x!0 x 2x cos.2x/ lim sec.1 C cos x/ D sec.1 x! 55. 56. 57. 58. At x D 60 the slope is 45. The slope of y D tan x at x D a is sec2 a. The tangent there is parallel to y D 2x if sec2 a D 2, or p cos a D ˙1= 2. The only solutions in . =2; =2/ are a D ˙=4. The corresponding points on the graph are .=4; 1/ and . =4; 1/. lim x!0 x 2 cos x D 12 1 D 1 x!0 x!0 sin x cos2 x sin x 2 lim cos D cos D 1 D lim cos x!0 x!0 x2 x 1 2 sin2 .h=2/ 1 sin.h=2/ 2 1 cos h D D lim D lim lim h2 h2 h=2 2 h!0 h!0 2 h!0 lim x 2 csc x cot x D lim f will be differentiable at x D 0 if 2 sin 0 C 3 cos 0 D b; and ˇ ˇ d D a: .2 sin x C 3 cos x/ˇˇ dx xD0 Thus we need b D 3 and a D 2. 59. There are infinitely many lines through the origin that are tangent to y D cos x. The two with largest slope are shown in the figure. y 46. The slope of y D tan.2x/ at x D a is 2 sec2 .2a/. The tangent there is normal to y D x=8 if 2 sec2 .2a/ D 8, or cos.2a/ D ˙1=2. The only solutions in . =4; =4/ are a D points on the graph are p ˙=6. The corresponding p .=6; 3/ and . =6; 3/. 47. 48. 56 Telegram: @uni_k 2 x y D cos x d sin x D cos x D 0 at odd multiples of =2. dx d cos x D sin x D 0 at multiples of . dx d sec x D sec x tan x D 0 at multiples of . dx d csc x D csc x cot x D 0 at odd multiples of =2. dx Thus each of these functions has horizontal tangents at infinitely many points on its graph. d tan x D sec2 x D 0 nowhere. dx d cot x D csc2 x D 0 nowhere. dx Thus neither of these functions has a horizontal tangent. 1/ D sec 0 D 1 Fig. 2.5-59 The tangent to y D cos x at x D a has equation y D cos a .sin a/.x a/. This line passes through the origin if cos a D a sin a. We use a calculator with a “solve” function to find solutions of this equation near a D and a D 2 as suggested in the figure. The solutions are a 2:798386 and a 6:121250. The slopes of the corresponding tangents are given by sin a, so they are 0:336508 and 0:161228 to six decimal places. 60. 61. 1 p 2 C 3.2 3=2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 4 C 3/= lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 62. SECTION 2.6 (PAGE 131) a) As suggested by the figure in the problem, the square of the length of chord AP is .1 cos /2 C.0 sin /2 , and the square of the length of arc AP is 2 . Hence 5. y D x 1=3 x 1=3 1 1 y 0 D x 2=3 C x 4=3 3 3 2 5=3 4 7=3 00 y D x x 9 9 10 8=3 28 10=3 x C x y 000 D 27 27 6. y D x 10 C 2x 8 .1 C cos /2 C sin2 < 2 ; and, since squares cannot be negative, each term in the sum on the left is less than 2 . Therefore 0 j1 cos j < j j; 0 y D 10x C 16x 0 j sin j < j j: 7. Since lim!0 j j D 0, the squeeze theorem implies that lim 1 cos D 0; lim sin D 0: !0 !0 From the first of these, lim!0 cos D 1. b) Using the result of (a) and the addition formulas for cosine and sine we obtain 8. lim cos.0 C h/ D lim .cos 0 cos h h!0 h!0 sin 0 sin h/ D cos 0 lim sin.0 C h/ D lim .sin 0 cos h C cos 0 sin h/ D sin 0 : h!0 h!0 9. This says that cosine and sine are continuous at any point 0 . y D .3 1. 0 2. 3. 14.3 2x/6 y 00 D 168.3 2x/5 000 1680.3 D 12. 2x/4 1 x 1 y 0 D 2x C 2 x y 00 D 2 y D x2 y 000 D 6 D 6.x .x 1/2 0 y D 12.x 1/ 3 yD y 00 D 36.x y 000 D 4. 2x/ y D y p ax C b a y0 D p 2 ax C b yD 2 x3 6 x4 13. 1/ 2 4 .x C 1/3 12 y 000 D .x C 1/4 y 00 D y 00 D 2 sec2 x tan x y D tan x 2 y 000 D 2 sec4 x C 4 sec2 x tan2 x y D sec x y 0 D sec x tan x y 000 D sec x tan3 x C 5 sec3 x tan x y D cos.x 2 / 2x sin.x 2 / y 00 D y 000 D 2 sin.x 2 / 4x 2 cos.x 2 / 12x cos.x 2 / C 8x 3 sin.x 2 / sin x x cos x sin x 0 y D x x2 2 .2 x / sin x 2 cos x y 00 D x3 x2 2 2 3.x 2/ sin x .6 x / cos x C y 000 D 3 4 x x yD 1 Dx 1 x f 0 .x/ D x 2 f .x/ D f 000 .x/ D 3Šx 4 .4/ 1/ 5 a2 4.ax C b/3=2 3a3 y 000 D 8.ax C b/5=2 y 00 D f .x/ D 4Šx 5 Guess: f .n/ .x/ D . 1/n nŠx .nC1/ ./ Proof: (*) is valid for n D 1 (and 2, 3, 4). .kC1/ Assume f .k/ .x/ D . 1/k kŠx for some k 1 .kC1/ k Then f .x/ D . 1/ kŠ .k C 1/ x .kC1/ 1 D . 1/kC1 .k C 1/Šx ..kC1/C1/ which is (*) for n D k C 1. Therefore, (*) holds for n D 1; 2; 3; : : : by induction. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k y 00 D sec x tan2 x C sec3 x f 00 .x/ D 2x 3 1/ 4 144.x x 1 xC1 2 0 y D .x C 1/2 y0 D 7 y 000 D 720x 7 C 672x 5 yD y D sec x 11. y 00 D 90x 8 C 112x 6 7 p y D .x 2 C 3/ x D x 5=2 C 3x 1=2 5 3 y 0 D x 3=2 C x 1=2 2 2 15 1=2 3 3=2 y 00 D x x 4 4 15 1=2 9 5=2 x C x y 000 D 8 8 0 10. Section 2.6 Higher-Order Derivatives (page 131) 9 Downloaded by ted cage (sxnbyln180@questza.com) 57 lOMoARcPSD|6566483 www.konkur.in SECTION 2.6 (PAGE 131) 14. ADAMS and ESSEX: CALCULUS 9 1 Dx 2 x2 f 0 .x/ D 2x 3 f 00 .x/ D 2. 3/x 4 D 3Šx 4 f .3/ .x/ D 2. 3/. 4/x 5 D 4Šx 5 Conjecture: Thus, the formula is also true for n D k C 1. Hence, it is true for n 2 by induction. f .x/ D f .n/ .x/ D . 1/n .n C 1/Šx .nC2/ 17. f 00 .x/ D 2b 2 .a C bx/ 3 for n D 1; 2; 3; : : : f 000 .x/ D 3Šb 3 .a C bx/ 4 Guess: f .n/ .x/ D . 1/n nŠb n .a C bx/ .nC1/ ./ Proof: (*) holds for n D 1; 2; 3 Assume (*) holds for n D k: f .k/ .x/ D . 1/k kŠb k .a C bx/ .kC1/ Then f .kC1/ .x/ D . 1/k kŠb k .k C 1/ .a C bx/ .kC1/ 1 .b/ Proof: Evidently, the above formula holds for n D 1; 2 and 3. Assume it holds for n D k, i.e., f .k/ .x/ D . 1/k .k C 1/Šx .kC2/ : Then d .k/ f .x/ dx D . 1/k .k C 1/ŠŒ. 1/.k C 2/x .kC2/ 1 f .kC1/ .x/ D D . 1/kC1 .k C 1/Šb kC1 .a C bx/..kC1/C1/ So (*) holds for n D k C 1 if it holds for n D k. Therefore, (*) holds for n D 1; 2; 3; 4; : : : by induction. D . 1/kC1 .k C 2/Šx Œ.kC1/C2 : Thus, the formula is also true for n D k C 1. Hence it is true for n D 1; 2; 3; : : : by induction. 15. 1 D .2 2 x f 0 .x/ D C.2 x/ 2 f .x/ D f 00 .x/ D 2.2 x/ 1 x/ 3 f 000 .x/ D C3Š.2 x/ 4 Guess: f .n/ .x/ D nŠ.2 x/ .nC1/ ./ Proof: (*) holds for n D 1; 2; 3. Assume f .k/ .x/ D kŠ.2 x/ .kC1/ (i.e., (*) holds for n D k) Then f .kC1/ .x/ D kŠ .k C 1/.2 x/ .kC1/ 1 . 1/ 18. f .x/ D x 2=3 f 0 .x/ D 32 x 1=3 f 00 .x/ D 32 . 13 /x 4=3 f 000 .x/ D 32 . 13 /. 34 /x 7=3 Conjecture: 1 4 7 .3n 5/ .3n 2/=3 f .n/ .x/ D 2. 1/n 1 x for 3n n 2. Proof: Evidently, the above formula holds for n D 2 and 3. Assume that it holds for n D k, i.e. f .k/ .x/ D 2. 1/k 1 ..kC1/C1/ D .k C 1/Š.2 x/ : Thus (*) holds for n D k C 1 if it holds for k. Therefore, (*) holds for n D 1; 2; 3; : : : by induction. p 16. f .x/ D x D x 1=2 f 0 .x/ D 21 x 1=2 f 00 .x/ D 12 . 12 /x 3=2 f 000 .x/ D 21 . 12 /. 32 /x 5=2 f .4/ .x/ D 21 . 12 /. 32 /. 25 /x 7=2 Conjecture: f .n/ .x/ D . 1/n 1 1 3 5 .2n 2n 3/ x .2n 1/=2 f .k/ .x/ D . 1/k 1 1 3 5 .2k 2k 3/ x .2k 1/=2 : Then d .k/ f .kC1/ .x/ D f .x/ dx .2k 1/ 1 3 5 .2k 3/ D . 1/k 1 x Œ.2k 1/=2 1 2 2k 1 3 5 .2k 3/Œ2.k C 1/ 3 Œ2.kC1/ 1=2 x : D . 1/.kC1/ 1 2kC1 58 Telegram: @uni_k 1 4 7 .3k 3k 5/ x .3k 2/=3 : Then, d .k/ f .x/ dx .3k 2/ 1 4 7 .3k 5/ D 2. 1/k 1 x Œ.3k 2/=3 1 3 3k 1 4 7 .3k 5/Œ3.k C 1/ 5 Œ3.kC1/ 2=3 x : D 2. 1/.kC1/ 1 3. k C 1/ f .kC1/ .x/ D Thus, the formula is also true for n D k C 1. Hence, it is true for n 2 by induction. .n 2/: Proof: Evidently, the above formula holds for n D 2; 3 and 4. Assume that it holds for n D k, i.e. 1 D .a C bx/ 1 a C bx f 0 .x/ D b.a C bx/ 2 f .x/ D 19. f 00 .x/ D f .x/ D cos.ax/ f 0 .x/ D a sin.ax/ a2 cos.ax/ f 000 .x/ D a3 sin.ax/ f .4/ .x/ D a4 cos.ax/ D a4 f .x/ It follows that f .n/ .x/ D a4 f .n 4/ .x/ for n 4, and 8̂ n a cos.ax/ < n a sin.ax/ .n/ f .x/ D n :̂ a cos.ax/ n a sin.ax/ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) if n D 4k if n D 4k C 1 .k D 0; 1; 2; : : :/ if n D 4k C 2 if n D 4k C 3 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 2.6 (PAGE 131) Differentiating any of these four formulas produces the one for the next higher value of n, so induction confirms the overall formula. 20. f .x/ D x cos x f 0 .x/ D cos x x sin x f 00 .x/ D 2 sin x x cos x f 000 .x/ D 3 cos x C x sin x The pattern suggests that nŠjxj .nC1/ sgn x f .n/ .x/ D nŠjxj .nC1/ 23. f .4/ .x/ D 4 sin x C x cos x This suggests the formula (for k D 0; 1; 2; : : :) 8̂ n sin x C x cos x < n cos x x sin x .n/ f .x/ D :̂ n sin x x cos x n cos x C x sin x if n D 4k if n D 4k C 1 if n D 4k C 2 if n D 4k C 3 Differentiating any of these four formulas produces the one for the next higher value of n, so induction confirms the overall formula. 21. f .x/ D x sin.ax/ f 0 .x/ D sin.ax/ C ax cos.ax/ f 00 .x/ D 2a cos.ax/ 000 f .x/ D 2 3a sin.ax/ a2 x sin.ax/ a3 x cos.ax/ f 4/ .x/ D 4a3 cos.ax/ C a4 x sin.ax/ This suggests the formula 8̂ nan 1 cos.ax/ C an x sin.ax/ ˆ < n 1 na sin.ax/ C an x cos.ax/ f .n/ .x/ D ˆ nan 1 cos.ax/ an x sin.ax/ :̂ nan 1 sin.ax/ an x cos.ax/ if n D 4k if n D 4k C 1 if n D 4k C 2 if n D 4k C 3 for k D 0; 1; 2; : : :. Differentiating any of these four formulas produces the one for the next higher value of n, so induction confirms the overall formula. 22. f .x/ D 24. jxj 2 sgn x: 25. .sgn x/ D 1: f 00 .x/ D 2jxj 3 .sgn x/2 D 2jxj 3 3Šjxj 4 sgn x .x/ D 4Šjxj 5 : D k 2 y.2 sec2 .kx/ 2 Thus we can calculate successive derivatives of f using the product rule where necessary, but will get only one nonzero term in each case: f If y D sec.kx/, then y 0 D k sec.kx/ tan.kx/ and y 00 D k 2 .sec2 .kx/ tan2 .kx/ C sec3 .kx// d sgn x D 0 and dx .4/ If y D tan.kx/, then y 0 D k sec2 .kx/ and D 2k 2 .1 C tan2 .kx// tan.kx/ D 2k 2 y.1 C y 2 /: If x ¤ 0 we have f .3/ .x/ D Differentiating this formula leads to the same formula with n replaced by n C 1 so the formula is valid for all n 1 by induction. p f .x/ D 1 3x D .1 3x/1=2 1 f 0 .x/ D . 3/.1 3x/ 1=2 2 1 1 f 00 .x/ D . 3/2 .1 3x/ 3=2 2 2 1 3 1 . 3/3 .1 3x/ 5=2 f 000 .x/ D 2 2 2 1 1 3 5 f .4/ .x/ D . 3/4 .1 3x/7=2 2 2 2 2 1 3 5 .2n 3/ n 3 Guess: f .n/ .x/ D 2n .2n 1/=2 .1 3x/ ./ Proof: (*) is valid for n D 2; 3; 4; (but not n D 1) Assume (*) holds for n D k for some integer k 2 1 3 5 : : : .2k 3/ k i.e., f .k/ .x/ D 3 2k .1 3x/ .2k 1/=2 1 3 5 .2k 3/ k .kC1/ Then f .x/ D 3 2k 2.k 1/ .1 3x/ .2k 1/=2 1 . 3/ 2 1 3 5 2.k C 1/ 1 D 3kC1 2kC1 .1 3x/ .2.kC1/ 1/=2 Thus (*) holds for n D k C 1 if it holds for n D k. Therefore, (*) holds for n D 2; 3; 4; : : : by induction. y 00 D 2k 2 sec 2 .kx/t an.kx/ d 1 D jxj 1 . Recall that jxj D sgn x, so jxj dx f 0 .x/ D 26. 1/ D k 2 y.2y 2 1/: To be proved: if f .x/ D sin.ax C b/, then . 1/k an sin.ax C b/ if n D 2k f .n/ .x/ D . 1/k an cos.ax C b/ if n D 2k C 1 for k D 0; 1; 2; : : : Proof: The formula works for k D 0 (n D 2 0 D 0 and n D 2 0 C 1 D 1): f .0/ .x/ D f .x/ D . 1/0 a0 sin.ax C b/ D sin.ax C b/ f .1/ .x/ D f 0 .x/ D . 1/0 a1 cos.ax C b/ D a cos.ax C b/ Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k if n is odd if n is even Downloaded by ted cage (sxnbyln180@questza.com) 59 lOMoARcPSD|6566483 www.konkur.in SECTION 2.6 (PAGE 131) ADAMS and ESSEX: CALCULUS 9 Section 2.7 Using Differentials and Derivatives (page 137) Now assume the formula holds for some k 0. If n D 2.k C 1/, then f .n/ .x/ D d .2kC1/ d .n 1/ f .x/ D f .x/ dx dx d D . 1/k a2kC1 cos.ax C b/ dx D . 1/kC1 a2kC2 sin.ax C b/ and if n D 2.k C 1/ C 1 D 2k C 3, then d . 1/kC1 a2kC2 sin.ax C b/ f .n/ .x/ D dx 1. 2. 3. D . 1/kC1 a2kC3 cos.ax C b/: 27. 3 dx 3 f .x/ df .x/ D p D .0:08/ D 0:06 4 2 3x C 1 f .1:08/ f .1/ C 0:06 D 2:06. t 1 1 h.t / dh.t / D sin dt .1/ D . 4 4 4 10 40 1 1 1 D : h.2/ h 2C 10 40 40 s 1 1 u du D sec2 ds D .2/. 0:04/ D 0:04. 4 4 4 If s D 0:06, then u 1 0:04 0:96. Thus the formula also holds for k C 1. Therefore it holds for all positive integers k by induction. 4. If y D tan x, then 5. If y D x 2 , then y dy D 2x dx. If dx D .2=100/x, then y .4=100/x 2 D .4=100/y, so y increases by about 4%. 6. If y D 1=x, then y dy D . 1=x 2 / dx. If dx D .2=100/x, then y . 2=100/=x D . 2=100/y, so y decreases by about 2%. 7. If y D 1=x 2 , then y dy D . 2=x 3 / dx. If dx D .2=100/x, then y . 4=100/=x 2 D . 4=100/y, so y decreases by about 4%. 8. If y D x 3 , then y dy D 3x 2 dx. If dx D .2=100/x, then y .6=100/x 3 D .6=100/y, so y increases by about 6%. p p x, then dy dy D .1=2 If y D p x/ dx. If x D .2=100/x, then y .1=100/ x D .1=100/y, so y increases by about 1%. y 0 D sec2 x D 1 C tan2 x D 1 C y 2 D P2 .y/; where P2 is a polynomial of degree 2. Assume that y .n/ D PnC1 .y/ where PnC1 is a polynomial of degree n C 1. The derivative of any polynomial is a polynomial of one lower degree, so y .nC1/ D dy d PnC1 .y/ D Pn .y/ D Pn .y/.1Cy 2 / D PnC2 .y/; dx dx a polynomial of degree n C 2. By induction, .d=dx/n tan x D PnC1 .tan x/, a polynomial of degree n C 1 in tan x. 28. .fg/00 D .f 0 g C fg 0 / D f 00 g C f 0 g 0 C f 0 g 0 C fg 00 D f 00 g C 2f 0 g 0 C fg 00 29. .fg/.3/ D d .fg/00 dx d D Œf 00 g C 2f 0 g 0 C fg 00 dx D f .3/ g C f 00 g 0 C 2f 00 g 0 C 2f 0 g 00 C f 0 g 00 C fg .3/ .3/ .4/ .fg/ 00 0 0 00 D f g C 3f g C 3f g C fg : d .fg/.3/ D dx d D Œf .3/ g C 3f 00 g 0 C 3f 0 g 00 C fg .3/ dx D f .4/ g C f .3/ g 0 C 3f .3/ g 0 C 3f 00 g 00 C 3f 00 g 00 D f .4/ g C 4f .3/ g 0 C 6f 00 g 00 C 4f 0 g .3/ C fg .4/ : nŠ f .n 2/ g 00 .fg/.n/ D f .n/ g C nf .n 1/g 0 C 2Š.n 2/Š nŠ f .n 3/ g .3/ C C nf 0 g .n 1/ C fg .n/ C 3Š.n 3/Š n X nŠ f .n k/ g .k/ : D kŠ.n k/Š kD0 60 9. 10. If y D x 2=3 , then y dy D . 2=3/x 5=3 dx. If dx D .2=100/x, then y . 4=300/x 2=3 D . 4=300/y, so y decreases by about 1.33%. 11. .3/ C 3f 0 g .3/ C f 0 g .3/ C fg .4/ Telegram: @uni_k 0:01 1 dx D D 0:0025. y dy D x2 22 If x D 2:01, then y 0:5 0:0025 D 0:4975. If V D 34 r 3 , then V d V D 4 r 2 dr. If r increases by 2%, then dr D 2r=100 and V 8 r 3 =100. Therefore V =V 6=100. The volume increases by about 6%. 12. If V is the volume and x is the edge length of the cube then V D x 3 . Thus V d V D 3x 2 x. If V D .6=100/V , then 6x 3 =100 3x 2 dx, so dx .2=100/x. The edge of the cube decreases by about 2%. 13. 14. Rate change of Area A with respect to side s, where dA A D s 2 , is D 2s: When s D 4 ft, the area is changing ds 2 at rate 8 ft /ft. p p If A D s 2 , then s D A and ds=dA D 1=.2 A/. If A D 16 m2 , then the side is changing at rate ds=dA D 1=8 m/m2 . Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 15. SECTION 2.7 (PAGE 137) The diameter D and area A of a circle are related by p D D 2 A=. The rate of change of diameter with respect p to area is dD=dA D 1=.A/ units per square unit. The flow rate will increase by 10% if the radius is increased by about 2.5%. 23. F D k=r 2 implies that dF=dr D 2k=r 3 . Since dF=dr D 1 pound/mi when r D 4; 000 mi, we have 2k D 4; 0003 . If r D 8; 000, we have dF=dr D .4; 000=8; 000/3 D 1=8. At r D 8; 000 mi F decreases with respect to r at a rate of 1/8 pounds/mi. 24. If price = $p, then revenue is $R D 4; 000p 16. Since A D D 2 =4, the rate of change of area with respect to diameter is dA=dD D D=2 square units per unit. 17. Rate of change of V D 4 3 r with respect to radius r is 3 dV D 4 r 2 . When r D 2 m, this rate of change is 16 dr 3 m /m. a) Sensitivity of R to p is dR=dp D 4; 000 20p. If p D 100, 200, and 300, this sensitivity is 2,000 $/$, 0 $/$, and 2; 000 $/$ respectively. 18. Let A be the area of a square, s be its side length and L be its diagonal. Then, L2 D s 2 C s 2 D 2s 2 and dA D L. Thus, the rate of change of A D s 2 D 12 L2 , so dL the area of a square with respect to its diagonal L is L. b) The distributor should charge $200. This maximizes the revenue. 25. 19. If the radius of the circle is r then C D 2 r and A D r 2. r p p A Thus C D 2 D 2 A. Rate of p change of C with respect to A is dC 1 D p D . dA r A 20. Let s be the side length and V be the volume of a cube. ds D 31 V 2=3 . Hence, Then V D s 3 ) s D V 1=3 and dV the rate of change of the side length of a cube with respect to its volume V is 31 V 2=3 . 21. tD5 D 700.20 ˇ ˇ ˇ t /ˇ ˇ tD5 D 26. tD15 D 700.20 ˇ ˇ ˇ t /ˇ ˇ tD15 b) To maximize daily profit, production should be 800 sheets/day. 10; 500: D 27. 3; 500: b) Average rate of change between t D 5 and t D 15 is 350 .25 V .5/ D 5 10 225/ D 80; 000 n2 C 4n C n 100 n 80; 000 dC C4C : D dn n2 50 dC (a) n D 100; D 2. Thus, the marginal cost of dn production is $2. C D 82 dC D 9:11. Thus, the marginal cost dn 9 of production is approximately $9.11. (b) n D 300; Water is draining out at 3,500 L/min at that time. V .15/ 15 Daily profit if production is x sheets per day is $P .x/ where P .x/ D 8x 0:005x 2 1; 000: a) Marginal profit P 0 .x/ D 8 0:01x. This is positive if x < 800 and negative if x > 800. Water is draining out at 10,500 L/min at that time. At t D 15, water volume is changing at rate ˇ d V ˇˇ ˇ dt ˇ 7; 000: The average rate of draining is 7,000 L/min over that interval. 22. Flow rate F D kr 4 , so F 4kr 3 r. If F D F=10, then kr 4 F D D 0:025r: r 3 40kr 40kr 3 28. Daily profit P D 13x C x D 13x 10x 20 x2 1000 x2 D 3x 20 1000 Graph of P is a parabola opening downward. P will be maximum where the slope is zero: 0D dP D3 dx 2x so x D 1500 1000 Should extract 1500 tonnes of ore per day to maximize profit. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 0:5x 2 if x units are b) C.101/ C.100/ D 43; 299:50 43; 000 D $299:50 which is approximately C 0 .100/. a) At t D 5, water volume is changing at rate ˇ d V ˇˇ ˇ dt ˇ Cost is $C.x/ D 8; 000 C 400x manufactured. a) Marginal cost if x D 100 is C 0 .100/ D 400 100 D $300. t /2 L at t min. Volume in tank is V .t / D 350.20 10p 2 . Downloaded by ted cage (sxnbyln180@questza.com) 61 lOMoARcPSD|6566483 www.konkur.in SECTION 2.7 (PAGE 137) ADAMS and ESSEX: CALCULUS 9 29. One of the components comprising C.x/ is usually a fixed cost, $S, for setting up the manufacturing operation. On a per item basis, this fixed cost $S=x, decreases as the number x of items produced increases, especially when x is small. However, for large x other components of the total cost may increase on a per unit basis, for instance labour costs when overtime is required or maintenance costs for machinery when it is over used. C.x/ Let the average cost be A.x/ D . The minimal avx erage cost occurs at point where the graph of A.x/ has a horizontal tangent: 0D If f .x/ D cos x C .x 2 =2/, then f 0 .x/ D x sin x > 0 for x > 0. By the MVT, if x > 0, then f .x/ f .0/ D f 0 .c/.x 0/ for some c > 0, so f .x/ > f .0/ D 1. Thus cos x C .x 2 =2/ > 1 and cos x > 1 .x 2 =2/ for x > 0. Since both sides of the inequality are even functions, it must hold for x < 0 as well. 5. Let f .x/ D tan x. If 0 < x < =2, then by the MVT f .x/ f .0/ D f 0 .c/.x 0/ for some c in .0; =2/. Thus tan x D x sec2 c > x, since secc > 1. 6. dA xC 0 .x/ C.x/ : D dx x2 C.x/ D A.x/. x Thus the marginal cost C 0 .x/ equals the average cost at the minimizing value of x. Hence, xC 0 .x/ 4. C.x/ D 0 ) C 0 .x/ D 7. Let f .x/ D .1 C x/r 1 rx where r > 1. Then f 0 .x/ D r.1 C x/r 1 r. If 1 x < 0 then f 0 .x/ < 0; if x > 0, then f 0 .x/ > 0. Thus f .x/ > f .0/ D 0 if 1 x < 0 or x > 0. Thus .1 C x/r > 1 C rx if 1 x < 0 or x > 0. Let f .x/ D .1 C x/r where 0 < r < 1. Thus, f 0 .x/ D r.1 C x/r 1 . By the Mean-Value Theorem, for x 1, and x ¤ 0, f .x/ f .0/ D f 0 .c/ x 0 .1 x/r 1 D r.1 C c/r 1 ) x 30. If y D Cp r , then the elasticity of y is p dy D y dp p . r/Cp r 1 D r: Cp r for some c between 0 and x. Thus, .1 C x/r D 1 C rx.1 C c/r 1 . If 1 x < 0, then c < 0 and 0 < 1 C c < 1. Hence Section 2.8 The Mean-Value Theorem (page 144) 1. f .x/ D x 2 ; bCa D b2 b f .2/ 2 where c D f .a/ a bCa )cD 2 p f .1/ 1 D 1 2 c>0 1Cc >1 .1 C c/r 1 < 1 rx.1 C c/r 1 < rx: 1 then x2 1D 1 D 2 1 D f 0 .c/ c2 Telegram: @uni_k Hence, .1 C x/r < 1 C rx in this case also. Hence, .1 C x/r < 1 C rx for either 1 x < 0 or x > 0. 8. If f .x/ D x 3 12x C 1, then f 0 .x/ D 3.x 2 4/. The critical points of f are x D ˙2. f is increasing on . 1; 2/ and .2; 1/ where f 0 .x/ > 0, and is decreasing on . 2; 2/ where f 0 .x/ < 0. 9. If f .x/ D x 2 4, then f 0 .x/ D 2x. The critical point of f is x D 0. f is increasing on .0; 1/ and decreasing on . 1; 0/. 2 lies between 1 and 2. 3. f .x/ D x 3 3x C 1, f 0 .x/ D 3x 2 3, a D 2, b D 2 f .b/ f .a/ f .2/ f . 2/ D b a 4 8 6 C 1 . 8 C 6 C 1/ D 4 4 D D1 4 0 2 f .c/ D 3c 3 2 3c 2 3 D 1 ) 3c 2 D 4 ) c D ˙ p 3 (Both points will be in . 2; 2/.) 62 1 < 0/; .since x < 0/: Hence, .1 C x/r < 1 C rx. If x > 0, then f .b/ a2 D a b 1 , and f 0 .x/ D x .since r rx.1 C c/r 1 < rx f 0 .x/ D 2x D f 0 .c/ D 2c 2. If f .x/ D .1 C c/r 1 > 1 10. If y D 1 x x 5 , then y 0 D 1 5x 4 < 0 for all x. Thus y has no critical points and is decreasing on the whole real line. 11. If y D x 3 C 6x 2 , then y 0 D 3x 2 C 12x D 3x.x C 4/. The critical points of y are x D 0 and x D 4. y is increasing on . 1; 4/ and .0; 1/ where y 0 > 0, and is decreasing on . 4; 0/ where y 0 < 0. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 2.8 (PAGE 144) 12. If f .x/ D x 2 C 2x C 2 then f 0 .x/ D 2x C 2 D 2.x C 1/. Evidently, f 0 .x/ > 0 if x > 1 and f 0 .x/ < 0 if x < 1. Therefore, f is increasing on . 1; 1/ and decreasing on . 1; 1/. 22. There is no guarantee that the MVT applications for f and g yield the same c. 23. CPs x D 0:535898 and x D 7:464102 13. f .x/ D x 3 4x C 1 f 0 .x/ D 3x 2 4 24. CPs x D 1:366025 and x D 0:366025 25. CPs x D 0:518784 and x D 0 26. CP x D 0:521350 27. If x1 < x2 < : : : < xn belong to I , and f .xi / D 0, .1 i n/, then there exists yi in .xi ; xiC1 / such that f 0 .yi / D 0, .1 i n 1/ by MVT. 28. For x ¤ 0, we have f 0 .x/ D 2x sin.1=x/ cos.1=x/ which has no limit as x ! 0. However, f 0 .0/ D limh!0 f .h/= h D limh!0 h sin.1= h/ D 0 does exist even though f 0 cannot be continuous at 0. 29. If f 0 exists on Œa; b and f 0 .a/ ¤ f 0 .b/, let us assume, without loss of generality, that f 0 .a/ > k > f 0 .b/. If g.x/ D f .x/ kx on Œa; b, then g is continuous on Œa; b because f , having a derivative, must be continuous there. By the Max-Min Theorem, g must have a maximum value (and a minimum value) on that interval. Suppose the maximum value occurs at c. Since g 0 .a/ > 0 we must have c > a; since g 0 .b/ < 0 we must have c < b. By Theorem 14, we must have g 0 .c/ D 0 and so f 0 .c/ D k. Thus f 0 takes on the (arbitrary) intermediate value k. 2 f .x/ D x C 2x sin.1=x/ if x ¤ 0 0 if x D 0. 2 f 0 .x/ > 0 if jxj > p 3 2 0 f .x/ < 0 if jxj < p 3 2 2 p / and . p ; 1/: 3 3 2 2 f is decreasing on . p ; p /: 3 3 f is increasing on . 1; 14. If f .x/ D x 3 C 4x C 1, then f 0 .x/ D 3x 2 C 4. Since f 0 .x/ > 0 for all real x, hence f .x/ is increasing on the whole real line, i.e., on . 1; 1/. 15. f .x/ D .x 2 4/2 f 0 .x/ D 2x2.x 2 4/ D 4x.x 2/.x C 2/ f 0 .x/ > 0 if x > 2 or 2 < x < 0 f 0 .x/ < 0 if x < 2 or 0 < x < 2 f is increasing on . 2; 0/ and .2; 1/. f is decreasing on . 1; 2/ and .0; 2/. 2x 1 then f 0 .x/ D . Evidently, x2 C 1 .x 2 C 1/2 f 0 .x/ > 0 if x < 0 and f 0 .x/ < 0 if x > 0. Therefore, f is increasing on . 1; 0/ and decreasing on .0; 1/. 16. If f .x/ D 17. f .x/ D x 3 .5 0 2 f .x/ D 3x .5 2 D x .5 x/2 30. 2 3 x/.15 5x/ x/ C 2x .5 x/. 1/ f .0 C h/ f .0/ h h!0 h C 2h2 sin.1= h/ D lim h h!0 D lim .1 C 2h sin.1= h/ D 1; a) f 0 .0/ D lim D 5x 2 .5 x/.3 x/ f .x/ > 0 if x < 0, 0 < x < 3, or x > 5 f 0 .x/ < 0 if 3 < x < 5 f is increasing on . 1; 3/ and .5; 1/. f is decreasing on .3; 5/. 0 h!0 because j2h sin.1= h/j 2jhj ! 0 as h ! 0. b) For x ¤ 0, we have 18. If f .x/ D x 2 sin x, then f 0 .x/ D 1 2 cos x D 0 at x D ˙=3 C 2nfor n D 0; ˙1; ˙2; : : :. f is decreasing on . =3 C 2n; C 2n/. f is increasing on .=3 C 2n; =3 C 2.n C 1// for integers n. f 0 .x/ D 1 C 4x sin.1=x/ There are numbers x arbitrarily close to 0 where f 0 .x/ D 1; namely, the numbers x D ˙1=.2n/, where n D 1, 2, 3, : : : . Since f 0 .x/ is continuous at every x ¤ 0, it is negative in a small interval about every such number. Thus f cannot be increasing on any interval containing x D 0. 19. If f .x/ D x C sin x, then f 0 .x/ D 1 C cos x 0 f 0 .x/ D 0 only at isolated points x D ˙; ˙3; :::. Hence f is increasing everywhere. 20. If f .x/ D x C 2 sin x, then f 0 .x/ D 1 C 2 cos x > 0 if cos x > 1=2. Thus f is increasing on the intervals . .4=3/ C 2n; .4=3/ C 2n/ where n is any integer. 21. f .x/ D x 3 is increasing on . 1; 0/ and .0; 1/ because f 0 .x/ D 3x 2 > 0 there. But f .x1 / < f .0/ D 0 < f .x2 / whenever x1 < 0 < x2 , so f is also increasing on intervals containing the origin. 31. Let a, b, and c be three points in I where f vanishes; that is, f .a/ D f .b/ D f .c/ D 0. Suppose a < b < c. By the Mean-Value Theorem, there exist points r in .a; b/ and s in .b; c/ such that f 0 .r/ D f 0 .s/ D 0. By the MeanValue Theorem applied to f 0 on Œr; s, there is some point t in .r; s/ (and therefore in I ) such that f 00 .t / D 0. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 cos.1=x/: Downloaded by ted cage (sxnbyln180@questza.com) 63 lOMoARcPSD|6566483 www.konkur.in SECTION 2.8 (PAGE 144) ADAMS and ESSEX: CALCULUS 9 32. If f .n/ exists on interval I and f vanishes at n C 1 distinct points of I , then f .n/ vanishes at at least one point of I . Proof: True for n D 2 by Exercise 8. Assume true for n D k. (Induction hypothesis) Suppose n D k C 1, i.e., f vanishes at k C 2 points of I and f .kC1/ exists. By Exercise 7, f 0 vanishes at k C 1 points of I . By the induction hypothesis, f .kC1/ D .f 0 /.k/ vanishes at a point of I so the statement is true for n D k C 1. Therefore the statement is true for all n 2 by induction. (case n D 1 is just MVT.) 3. x 2 C xy D y 3 Differentiate with respect to x: 2x C y C xy 0 D 3y 2 y 0 2x C y y0 D 2 3y x 4. x 3 y C xy 5 D 2 3x 2 y C x 3 y 0 C y 5 C 5xy 4 y 0 D 0 3x 2 y y 5 y0 D 3 x C 5xy 4 5. x 2 y 3 D 2x y 2xy 3 C 3x 2 y 2 y 0 D 2 2 2xy 3 y0 D 3x 2 y 2 C 1 33. Given that f .0/ D f .1/ D 0 and f .2/ D 1: a) By MVT, f 0 .a/ D 1 f .0/ D 0 2 f .2/ 2 0 1 D 0 2 6. for some a in .0; 2/. b) By MVT, for some r in .0; 1/, f 0 .r/ D f .1/ 1 7. f .0/ 0 D 0 1 0 D 0: 0 Also, for some s in .1; 2/, f 0 .s/ D f .2/ 2 f .1/ 1 D 1 2 0 D 1: 1 8. Then, by MVT applied to f 0 on the interval Œr; s, for some b in .r; s/, 1 f 0 .s/ f 0 .r/ D s r s 1 1 D > s r 2 f 00 .b/ D since s 0 r r < 2. c) Since f 00 .x/ exists on Œ0; 2, therefore f 0 .x/ is continuous there. Since f 0 .r/ D 0 and f 0 .s/ D 1, and since 0 < 71 < 1, the Intermediate-Value Theorem assures us that f 0 .c/ D 71 for some c between r and s. Section 2.9 Implicit Differentiation (page 149) 1. xy x C 2y D 1 Differentiate with respect to x: y C xy 0 1 C 2y 0 D 0 1 y Thus y 0 D 2Cx 2. x 3 C y 3 D 1 3x 2 C 3y 2 y 0 D 0, so y 0 D 64 Telegram: @uni_k 9. x2 . y2 x 2 C 4.y 2x C 8.y y0 1/2 D 4 1/y 0 D 0, so y 0 D x 4.1 y/ x2 x2 C y x y D C1D xCy y y Thus xy y 2 D x 3 Cx 2 y Cxy Cy 2 , or x 3 Cx 2 y C2y 2 D 0 Differentiate with respect to x: 3x 2 C 2xy C x 2 y 0 C 4yy 0 D 0 3x 2 C 2xy y0 D x 2 C 4y p x x C y D 8 xy 1 p xCyCx p .1 C y 0 / D y xy 0 2 xCy p 2.x C y/ C x.1 C y 0 /p D 2 x C y.y C xy 0 / 3x C 2y C 2y x C y y0 D p x C 2x x C y 2x 2 C 3y 2 D 5 4x C 6yy 0 D 0 At .1; 1/: 4 C 6y 0 D 0, y 0 D 2 Tangent line: y 1 D .x 3 2 3 1/ or 2x C 3y D 5 10. x 2 y 3 x 3 y 2 D 12 2xy 3 C 3x 2 y 2 y 0 3x 2 y 2 2x 3 yy 0 D 0 At . 1; 2/: 16 C 12y 0 12 C 4y 0 D 0, so the slope is 28 7 12 C 16 D D : y0 D 12 C 4 16 4 Thus, the equation of the tangent line is y D 2 C 74 .x C 1/, or 7x 4y C 15 D 0: 11. x y 3 C D2 y x x 4 C y 4 D 2x 3 y 4x 3 C 4y 3 y 0 D 6x 2 y C 2x 3 y 0 at . 1; 1/: 4 4y 0 D 6 2y 0 2y 0 D 2, y 0 D 1 Tangent line: y C 1 D 1.x C 1/ or y D x. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 12. x C 2y C 1 D SECTION 2.9 (PAGE 149) y2 19. 1 1/2yy 0 y 2 .1/ 1 C 2y D .x 1/2 At .2; 1/ we have 1 C 2y 0 D 2y 0 1 so y 0 D Thus, the equation of the tangent is y D 1 21 .x 2/, or x C 2y D 0: p 13. 2x C y p 2 sin.xy/ D =2 2 C y0 2 cos.xy/.y C xy 0 / D 0 At .=4; 1/: 2 C y 0 .1 C .=4/y 0 / D 0, so 0 y D 4=.4 /. The tangent has equation 4 yD1 14. 3x 2 1 2. 20. : 4 x 2 1 C .x C /: 2 4. 1/ 17. 21. y 00 D D 22. 0 y 00 D 8.y 0 /2 D 8y 1 4y x2 D 16y 3 4y 2 x 2 D 16y 3 1 : 4y 3 Ax By 2A C 2B.y 0 /2 C 2Byy 00 D 0. Thus, A B.y 0 /2 D By D 2 C 8.y 0 /2 C 8yy 00 D 0. x y x2 y2 Ax 2 C By 2 D C 23. Maple gives 0 for the value. 24. Maple gives the slope as 25. Maple gives the value 26. Maple gives the value Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1C 2Ax C 2Byy 0 D 0 ) y 0 D 0 18. x 2 C 4y 2 D 4; 2x C 8yy 0 D 0; x Thus, y 0 D and 4y 0 2 1 C .y / D y y a2 y2 C x2 D y3 y3 y 00 D y 1 y C xy D 1 C y ) y D 1 x y 0 C y 0 C xy 00 D y 00 2.y 1/ 2y 0 D Therefore, y 00 D 1 x .1 x/2 2 x 2 C y 2 D a2 2x C 2yy 0 D 0 so x C yy 0 D 0 and y 0 D 1 C y 0 y 0 C yy 00 D 0 so xy D x C y 0 x 3 3xy C y 3 D 1 3x 2 3y 3xy 0 C 3y 2 y 0 D 0 6x 3y 0 3y 0 3xy 00 C 6y.y 0 /2 C 3y 2 y 00 D 0 Thus 2 17 x D x y 2 y i .xy 0 y/ 2xy x 2 y 0 sin D : 2 x p x y2 3 .3y 0 1/ D 6 9y 0 , At .3; 1/: 2 p 9 p so y 0 D .108 3/=.162 3 3/. The tangent has equation p 3 108 p .x 3/: y D1C 162 3 3 16. cos h 6x y x2 y2 x 2x C 2y 0 2y.y 0 /2 y 00 D y2 x " 2 # 2 y x2 y x2 D 2 xC y y x y2 x y2 x 2 4xy 2xy D 2 D : y x .y 2 x/2 .x y 2 /3 x sin.xy y 2 / D x 2 1 sin.xy y 2 / C x.cos.xy y 2 //.y C xy 0 2yy 0 / D 2x. At .1; 1/: 0 C .1/.1/.1 y 0 / D 2, so y 0 D 1. The tangent has equation y D 1 .x 1/, or y D 2 x. y 2yy 0 C 3y 2 y 0 D 1 ) y 0 D y0 D tan.xy 2 / D .2=/xy .sec2 .xy 2 //.y 2 C 2xyy 0 / D .2=/.y C xy 0 /. At . ; 1=2/: 2..1=4/ y 0 / D .1=/ 2y 0 , so y 0 D . 2/=.4. 1//. The tangent has equation yD 15. 4 y2 C y3 D x 1 3x 2 3y 2 2y 6x 2.y 0 /2 2yy 00 C 6y.y 0 /2 C 3y 2 y 00 D 0 .1 3x 2 /2 .2 6y/ 0 2 .2 6y/.y / 6x .3y 2 2y/2 y 00 D D 3y 2 2y 3y 2 2y .2 6y/.1 3x 2 /2 6x D 2 3 2 .3y 2y/ 3y 2y x .x 0 x3 Downloaded by ted cage (sxnbyln180@questza.com) A B Ax By 2 By A.By 2 C Ax 2 / D B 2y3 AC : B 2y3 206 . 55 26. 855; 000 . 371; 293 65 lOMoARcPSD|6566483 www.konkur.in SECTION 2.9 (PAGE 149) 27. ADAMS and ESSEX: CALCULUS 9 Ellipse: x 2 C 2y 2 D 2 2x C 4yy 0 D 0 30. x Slope of ellipse: 2y Hyperbola: 2x 2 2y 2 D 1 4x 4yy 0 D 0 x 0 Slope of hyperbola: yH D 2 y 2 x C 2y D 2 At intersection points 2x 2 2y 2 D 1 1 3x 2 D 3 so x 2 D 1, y 2 D 2 x2 x x 0 0 D 1 D yH D Thus yE 2y y 2y 2 Therefore the curves intersect at right angles. 0 yE D 2x C 4yy 0 C y C xy 0 D 0 Similarly, the slope of the hyperbola .x; y/ satisfies 2x A2 2y 0 y D 0; B2 y2 D 1 at B2 x2 A2 or y 0 D B 2x : A2 y 2 2 2 2 Section 2.10 Antiderivatives and Initial-Value Problems (page 155) 1. b CB A a x D 2 . y2 B 2 b2 a A2 Since a2 b 2 D A2 C B 2 , therefore B 2 C b 2 D a2 A2 , x2 A2 a2 and 2 D 2 2 . Thus, the product of the slope of the y B b two curves at .x; y/ is 3. 4. 5. Thus, b2 x B 2 x D a2 y A2 y b 2 B 2 A2 a2 D a2 A2 B 2 b 2 1: Therefore, the curves intersect at right angles. 6. 7. 8. 9. 29. If z D tan.x=2/, then 1 C tan2 .x=2/ dx 1 C z 2 dx 1 dx D D : 1 D sec2 .x=2/ 2 dz 2 dz 2 dz 10. Thus dx=dz D 2=.1 C z 2 /. Also 11. cos x D 2 cos2 .x=2/ D 2 1 C z2 1D 1D 2 sec2 .x=2/ 1 z2 1 C z2 sin x D 2 sin.x=2/ cos.x=2/ D 66 Telegram: @uni_k 2x C y : 4y C x the only solution is x D 0, y D 0, and these values do not satisfy the original equation. There are no points on the given curve. 2. If the point .x; y/ is an intersection of the two curves, then y2 x2 y2 x2 C D 2 2 b A2 B 2 a 1 1 1 1 2 2 D y : C x A2 a2 B2 b2 2 y0 D 1 7 y 7 x C xy C y 2 C y 2 D 0; or .x C /2 C y 2 D 0; 4 4 2 4 b2 x : a2 y i.e. y 0 D ) However, since x 2 C 2y 2 C xy D 0 can be written x2 y2 28. The slope of the ellipse 2 C 2 D 1 is found from a b 2x 2y C 2 y 0 D 0; a2 b x x y D C 1 , xy y 2 D x 2 C xy C xy C y 2 xCy y , x 2 C 2y 2 C xy D 0 Differentiate with respect to x: 1 12. 2 tan.x=2/ 2z : D 2 1 C tan .x=2/ 1 C z2 13. Z Z Z Z Z 5 dx D 5x C C x 2 dx D 31 x 3 C C p x dx D 2 3=2 x CC 3 1 13 x 12 dx D 13 x CC x 3 dx D 1 4 x CC 4 Z .x C cos x/ dx D Z 1 C cos3 x dx D cos2 x x2 C sin x C C 2 Z Z tan x cos x dx D sin x dx D cos x C C Z Z Z .a2 Z .sec2 xCcos x/ dx D tan xCsin xCC x 2 / dx D a2 x 1 3 x CC 3 .A C Bx C C x 2 / dx D Ax C .2x 1=2 C 3x 1=3 dx D B 2 C 3 x C x CK 2 3 4 3=2 9 4=3 x C x CC 3 4 Z Z 6.x 1/ dx D .6x 1=3 6x 4=3 / dx x 4=3 D 9x 2=3 C 18x 1=3 C C Z 3 x2 1 4 1 3 1 2 x C x 1 dx D x x C x 3 2 12 6 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x CC lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 14. 105 Z SECTION 2.10 .1 C t 2 C t 4 C t 6 / dt 29. 30. Given that D 105.t C 13 t 3 C 15 t 5 C 17 t 7 / C C D 105t C 35t 3 C 21t 5 C 15t 7 C C Z 1 15. cos.2x/ dx D sin.2x/ C C 2 Z x x 16. sin dx D 2 cos CC 2 2 Z 1 dx D CC 17. .1 C x/2 1Cx Z 18. sec.1 x/ tan.1 x/ dx D sec.1 19. Z p 20. Since 21. Z p 23. 24. 25. 26. 27. Z Z Z Z ( p tan2 x dx D Z 2x x2 C 1 .sec2 x sin x cos x dx D cos2 x dx D sin2 x dx D 0 y Dx 2 Thus y D 1 2 x 2 y.0/ D 3 Z 32. Given that then y D then y D Z 1/ dx D tan x 1 sin.2x/ dx D 2 yD .x 2 and 0 D y. 1/ D x 3 / dx D 34. For . 1/ 1 C 12 . 1/ 2 C C so C D 32 . 1 1 3 C which is valid on the 2 x 2x 2 Z cos x dx D sin x C C Z C D 3 2 1 cos.2x/ C C 2 1 1 1 1D cos C C D C C ÷ C D 2 2 2 1 yD 1 cos.2x/ (for all x): 2 sin.2x/ dx D y 0 D sec2 x ; we have y.0/ D 1 Z y D sec2 x dx D tan x C C For 1 D tan 0 C C D C ÷ C D 1 y D tan x C 1 (for =2 < x < =2/. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2=7 7 C C. 2x y 0 D sin.2x/ , we have y.=2/ D 1 yD 35. x 1 C 12 x 2 C C x 9=7 dx D 1 CC D CC ÷ 6 2 3 (for all x): y D sin x C 2 2x C 3 for all x. y0 D x 2 x 3 y. 1/ D 0; y 0 D x 9=7 y.1/ D 4; 2 D sin 1 cos.2x/ C C 4 cos.2x/ x sin.2x/ dx D CC 2 2 4 1 ) y D x 2 2x C C 2 ) 3 D 0 C C therefore C D 3 interval . 1; 0/. 33. Z Also, 4 D y.1/ D 72 C C , so C D 12 . Hence, y D 72 x 2=7 21 , which is valid in the interval .0; 1/. 0 y D cos x , we have For y.=6/ D 2 xCC 1 Hence, y.x/ D p dx D 2 x 2 C 1 C C: x sin.2x/ 1 C cos.2x/ dx D C CC 2 2 4 28. Given that Z Z x 1=3 dx D 43 x 4=3 C C and 5 D y.0/ D C . Since y 0 D Ax 2 C Bx C C we have B A y D x 3 C x 2 C C x C D. Since y.1/ D 1, therefore 3 2 A B A B C, 1 D y.1/ D C C C C D. Thus D D 1 3 2 3 2 and B A y D .x 3 1/ C .x 2 1/ C C.x 1/ C 1 for all x 3 2 cos.x 2 / C C d p 2 x 22. Since x C1D p , therefore 2 dx x C1 Z y 0 D x 1=3 y.0/ D 5; 31. x/ C C p 4 dx D 8 x C 1 C C: xC1 2x sin.x 2 / dx D 15 Hence, y.x/ D 43 x 4=3 C 5 which is valid on the whole real line. d p 1 xC1D p , therefore dx 2 xC1 Z Z then y D 1 .2x C 3/3=2 C C 3 2x C 3 dx D p y 0 D 3 x ) y D 2x 3=2 C C y.4/ D 1 ) 1 D 16 C C so C D Thus y D 2x 3=2 15 for x > 0. (PAGE 155) Downloaded by ted cage (sxnbyln180@questza.com) 67 lOMoARcPSD|6566483 www.konkur.in SECTION 2.10 (PAGE 155) 36. For ADAMS and ESSEX: CALCULUS 9 8 00 < y D cos x 41. For y.0/ D 0 we have : 0 y .0/ D 1 y 0 D sec2 x , we have y./ D 1 yD Z sec2 x dx D tan x C C y0 D 1 D tan C C D C ÷ C D 1 y D tan x C 1 (for =2 < x < 3=2/: then y 0 D Z x 4 dx D 0 D cos 0 C 0 C C2 y D 1 C x cos x: 42. 8 < y 00 D x 4 y 0 .1/ D 2 : y.1/ D 1; Z 1 x 3 C C. 3 1 x 3 C 73 3 dx D 61 x 2 C 73 x C D; and 1 D y.1/ D 16 C 73 C D, so that D D 23 . Hence, y.x/ D 16 x 2 C 37 x 32 , which is valid in the interval .0; 1/. 1 4 x x C C1 . 4 0 Since y .0/ D 0, therefore 0 D 0 0 C C1 , and 1 y 0 D x 4 x. 4 1 5 1 2 x x C C2 . Thus y D 20 2 Since y.0/ D 8, we have 8 D 0 0 C C2 . 1 5 1 2 Hence y D x x C 8 for all x. 20 2 40. Given that 8 < y 00 D 5x 2 3x 1=2 y 0 .1/ D 2 : y.1/ D 0; Z we have y 0 D 5x 2 3x 1=2 dx D 53 x 3 6x 1=2 C C . 39. Since y 00 D x 3 y 0 D 35 x 3 yD Z 6 C C so that C D 6x 1=2 C 19 3 , and 5 3 x 3 6x 1=2 C 19 3 19 . Thus, 3 Telegram: @uni_k y0 D Z yD x3 6 .x C sin x/ dx D sin x C x C 2: B . Then y 0 D A x Thus, for all x ¤ 0, Let y D Ax C x 2 y 00 C xy 0 yD 4x 3=2 C 19 x C D: 3 11 4 . 2B C Ax x B 2B , and y 00 D 3 . x2 x B x Ax B D 0: x We will also have y.1/ D 2 and y 0 .1/ D 4 provided A C B D 2; and A B D 4: These equations have solution A D 3, B D 1, so the initial value problem has solution y D 3x .1=x/. 44. Let r1 and r2 be distinct rational roots of the equation ar.r 1/ C br C c D 0 Let y D Ax r1 C Bx r2 .x > 0/ Then y 0 D Ar1 x r1 1 C Br2 x r2 1 , and y 00 D Ar1 .r1 1/x r1 2 C Br2 .r2 1/x r2 2 . Thus ax 2 y 00 C bxy 0 C cy 1/x r1 2 C Br2 .r2 r1 1 5 4 x dx D 12 C2 D 1 8 00 < y D x C sin x we have For y.0/ D 2 : 0 y .0/ D 0 D ax 2 .Ar1 .r1 5 4 C 19 Finally, 0 D y.1/ D 12 3 C D so that D D 5 4 11 3=2 Hence, y.x/ D 12 x 4x C 19 3 x 4 . 68 43. 1, therefore y 0 D Also, 2 D y 0 .1/ D 53 ÷ x2 cos x C C1 2 0 D 0 cos 0 C C1 ÷ C1 D 1 Z 2 x x3 yD cos x C 1 dx D sin x C x C C2 2 6 2 D 0 sin 0 C 0 C C2 ÷ C2 D 2 Since 2 D y 0 .1/ D 31 C C , therefore C D 37 , and y 0 D 13 x 3 C 73 . Thus yD cos x dx D sin x C C1 1 D sin 0 C C1 ÷ C1 D 1 Z y D .sin x C 1/ dx D cos x C x C C2 37. Since y 00 D 2, therefore y 0 D 2x C C1 . Since y 0 .0/ D 5, therefore 5 D 0 C C1 , and y 0 D 2x C 5. Thus y D x 2 C 5x C C2 . Since y.0/ D 3, therefore 3 D 0C0CC2 , and C2 D 3. Finally, y D x 2 C 5x 3, for all x. 38. Given that Z r2 1 1/x r2 2 C bx.Ar1 x C Br2 x / C c.Ax r1 C Bx r2 / D A ar1 .r1 1/ C br1 C c x r1 C B.ar2 .r2 1/ C br2 C c x r2 D 0x r1 C 0x r2 0 .x > 0/ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 8 < 4x 2 y 00 C 4xy 0 45. y.4/ D 2 : 0 y .4/ D 2 yD0 Auxilary Equation: 4r.r 4r 2 ./ SECTION 2.11 ) a D 4; b D 4; c D 1/ C 4r 1 d) never accelerating to the left e) particle is speeding up for t > 2 1D0 1D0 1 r D˙ 2 By #31, y D Ax 1=2 C Bx 1=2 solves ./ for x > 0. A B 3=2 Now y 0 D x 1=2 x 2 2 Substitute the initial conditions: 2 D 2A C 2D A 4 B 2 B 16 )1 D A C ) f) slowing down for t < 2 g) the acceleration is 2 at all times h) average velocity over 0 t 4 is B : 4 2. 2t , a D 2. e) The particle is speeding up if v and a have the same sign, i.e., for t > 52 . f) The particle is slowing down if v and a have opposite sign, i.e., for t < 52 . g) Since a D 6D0 r2 r 6 D 0 .r 3/.r C 2/ D 0: There are two roots: r1 D 2, and r2 D 3. Thus the differential equation has solutions of the form y D Ax 2 C Bx 3 . Then y 0 D 2Ax 3 C 3Bx 2 . Since 1 D y.1/ D A C B and 1 D y 0 .1/ D 2A C 3B, therefore A D 52 and B D 35 . Hence, y D 25 x 2 C 53 x 3 . Section 2.11 Velocity and Acceleration (page 162) dx dv xDt 4t C 3, v D D 2t 4, a D D2 dt dt a) particle is moving: to the right for t > 2 2 b) to the left for t < 2 c) particle is always accelerating to the right 3. 2 at all t , a D 2 at t D 52 when v D 0. h) The average velocity over Œ0; 4 is 8 4 x.4/ x.0/ D D 1. 4 4 dx dv x D t 3 4t C 1, v D D 3t 2 4, a D D 6t dt dt p a) particlepmoving: to the right for t < 2= 3 or t > 2= 3, b) to the left for p p 2= 3 < t < 2= 3 c) particle is accelerating: to the right for t > 0 d) to the left for t < 0 p e) particle p is speeding up for t > 2= 3 or for 2= 3 < t < 0 p f) particle is slowing down for t < 2= 3 or for p 0 < t < 2= 3 p g) velocity is zeropat t D ˙2= 3. Acceleration at these times is ˙12= 3. h) average velocity on Œ0; 4 is Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k D0 d) The point is accelerating to the left if a < 0, i.e., for all t . Since this equation must hold for all x > 0, we must have 1. t 2, v D 5 3 c) The point is accelerating to the right if a > 0, but a D 2 at all t ; hence, the point never accelerates to the right. 1/x r 2 6x r D 0 Œr.r 1/ 6x r D 0: 1/ x D 4 C 5t 16 C 3 4 b) The point is moving to the left if v < 0, i.e., when t > 25 . Let y D x r ; y 0 D rx r 1 ; y 00 D r.r 1/x r 2 . Substituting these expressions into the differential equation we obtain r.r 16 x.0/ D 0 a) The point is moving to the right if v > 0, i.e., when t < 25 . B 7 , so B D 18, A D . 2 2 7 1=2 x C 18x 1=2 (for x > 0). Thus y D 2 46. Consider 8 < x 2 y 00 6y D 0 y.1/ D 1 : 0 y .1/ D 1: Hence 9 D x 2 Œr.r x.4/ 4 B 4 8DA (PAGE 162) Downloaded by ted cage (sxnbyln180@questza.com) 43 44C1 4 0 1 D 12 69 lOMoARcPSD|6566483 www.konkur.in SECTION 2.11 (PAGE 162) 4. ADAMS and ESSEX: CALCULUS 9 .t 2 C 1/.1/ .t /.2t / 1 t2 D 2 ; 2 2 .t C 1/ .t C 1/2 2t .t 2 3/ .t 2 C 1/2 . 2t / .1 t 2 /.2/.t 2 C 1/.2t / aD D 2 : 2 4 .t C 1/ .t C 1/3 xD t t2 C 1 ; vD 7. D D t 2 , D in metres, t in seconds dD D 2t velocity v D dt Aircraft becomes airborne if 500 200; 000 D m/s: v D 200 km/h D 3600 9 250 Time for aircraft to become airborne is t D s, that is, 9 about 27:8 s. Distance travelled during takeoff run is t 2 771:6 metres. 8. Let y.t / be the height of the projectile t seconds after it is fired upward from ground level with initial speed v0 . Then a) The point is moving to the right if v > 0, i.e., when 1 t 2 > 0, or 1 < t < 1. b) The point is moving to the left if v < 0, i.e., when t < 1 or t > 1. c) The point is accelerating to the right if a > 0, i.e., whenp2t .t 2 p3/ > 0, that is, when t > 3 or 3 < t < 0. y 00 .t / D d) The point p to the left if a < 0, i.e., for p is accelerating t< 3 or 0 < t < 3. Two antidifferentiations give 2. 2/ D At t D 1, a D .2/3 yD 2. 2/ 1 D . .2/3 2 1 . 2 9. tD1 Ball strikes the ground when y D 0, .t > 0/, i.e., 0 D t .9:8 4:9t / so t D 2. Velocity at t D 2 is 9:8 9:8.2/ D 9:8 m/s. Ball strikes the ground travelling at 9.8 m/s (downward). hD 6. Given that y D 100 2t 4:9t , the time t at which the ball reaches the ground is the positive root of the equation y D 0, i.e., 100 2t 4:9t 2 D 0, namely, tD 2C p 4 C 4.4:9/.100/ 4:318 s: 9:8 100 The average velocity of the ball is D 23:16 m=s. 4:318 Since 23:159 D v D 2 9:8t , then t ' 2:159 s. 70 Telegram: @uni_k v2 v0 g v02 2 C v0 D 0: 2 g g 2g An initial speed of 2v0 means the maximum height will be 4v02 =2g D 4h. To get a maximum height of 2h an p initial speed of 2v0 is required. 10. To get to 3h metres above Mars, the ball would have to be thrown upward with speed q p p vM D 6gM h D 6gM v02 =.2g/ D v0 3gM =g: Since gM D 3:72 and g D 9:80, we have vM 1:067v0 m/s. 11. 2 1:86t /: The height of the ball after t seconds is y.t / D .g=2/t 2 C v0 t m if its initial speed was v0 m/s. Maximum height h occurs when dy=dt D 0, that is, at t D v0 =g. Hence 2 4:9t metres (t in seconds) dy D 9:8 9:8t velocity v D dt dv acceleration a D D 9:8 dt The acceleration is 9:8 m/s2 downward at all times. Ball is at maximum height ˇwhen v D 0, i.e., at t D 1. ˇ Thus maximum height is y ˇ D 9:8 4:9 D 4:9 metres. 1:86t 2 C v0 t D t .49 The time taken to fall back to ground level on Mars would be t D 49=1:86 26:3 s. h) The average velocity over Œ0; 4 is 4 0 x.4/ x.0/ 1 D 17 D . 4 4 17 5. y D 9:8t 4:9t /: Since the projectile returns to the ground at t D 10 s, we have y.10/ D 0, so v0 D 49 m/s. On Mars, the acceleration of gravity is 3.72 m/s2 rather than 9.8 m/s2 , so the height of the projectile would be f) The particle is slowing down if v and a have opposite p 3 < t < 1, or 0 < t < 1 or sign,pi.e., for t > 3. 1, a D 4:9t 2 C v0 t D t .v0 yD e) The particle is speeding p up if v and a have the same sign, i.e.,pfor t < 3, or 1 < t < 0 or 1 < t < 3. g) v D 0 at t D ˙1. At t D 9:8; y 0 .0/ D v0 ; y.0/ D 0: If the cliff is h ft high, then the height of the rock t seconds after it falls is y D h p 16t 2 ft. The rock hits the ground (y D 0) at time p t D h=16 s. Its speedpat that time is v D 32t D 8 h D 160 ft/s. Thus h D 20, and the cliff is h D 400 ft high. 12. If the cliff is h ft high, then the height of the rock t seconds after it is thrown down is y D h 32t 16t 2 ft. The rock hits the ground (y D 0) at time p 1p 32 C 322 C 64h D 1C tD 16 C h s: 32 4 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 2.11 v Its speed at that time is vD 32 32t D p 8 16 C h D (PAGE 162) .4; 96/ 160 ft/s: Solving this equation for h gives the height of the cliff as 384 ft. t 13. Let x.t / be the distance travelled by the train in the t seconds after the brakes are applied. Since d 2 x=dt 2 D 1=6 m/s2 and since the initial speed is v0 D 60 km/h D 100=6 m/s, we have 1 2 100 x.t / D t C t: 12 6 The speed of the train at time t is v.t / D .t =6/ C .100=6/ m/s, so it takes the train 100 s to come to a stop. In that time it travels x.100/ D 1002 =12 C 1002 =6 D 1002 =12 833 metres. 14. 15. x D At 2 C Bt C C; v D 2At C B. The average velocity over Œt1 ; t2 is x.t2 / x.t1 / t2 t1 At22 C Bt1 C C At12 Bt1 C D t2 t1 A.t22 t12 / C B.t2 t1 / D .t2 t1 / A.t2 C t1 /.t2 t1 / C B.t2 t1 / D .t2 t1 / D A.t2 C t1 / C B. The velocity atthe midpoint of Œt1 ; t2 is instantaneous t2 C t1 t 2 C t1 D 2A C B D A.t2 C t1 / C B. v 2 2 Hence, the average velocity over the interval is equal to the instantaneous velocity at the midpoint. 8 < t2 0t 2 s D 4t 4 2<t <8 : 68 C 20t t 2 8 t 10 Note: s is continuous at 2 and 8 since 22 D 4.2/ 4 and 4.8/ 4 D 68 C 160 64 ( 2t if 0 < t < 2 ds D 4 velocity v D if 2 < t < 8 dt 20 2t if 8 < t < 10 Since 2t ! 4 as t ! 2 , therefore, v is continuous at 2 (.v.2/ D 4). Since 20 2t ! 4 as t ! 8C, therefore v is continuous at 8 .v.8/ D 4/. Hence the velocity is continuous for 0 < t < 10 ( 2 if 0 < t < 2 dv acceleration a D D 0 if 2 < t < 8 dt 2 if 8 < t < 10 is discontinuous at t D 2 and t D 8 Maximum velocity is 4 and is attained on the interval 2 t 8. .14; 224/ Fig. 2.11-16 The rocket’s acceleration while its fuel lasted is the slope of the first part of the graph, namely 96=4 D 24 ft/s. 17. The rocket was rising until the velocity became zero, that is, for the first 7 seconds. 18. As suggested in Example 1 on page 154 of the text, the distance travelled by the rocket while it was falling from its maximum height to the ground is the area between the velocity graph and the part of the t -axis where v < 0. The area of this triangle is .1=2/.14 7/.224/ D 784 ft. This is the maximum height the rocket achieved. 19. The distance travelled upward by the rocket while it was rising is the area between the velocity graph and the part of the t -axis where v > 0, namely .1=2/.7/.96/ D 336 ft. Thus the height of the tower from which the rocket was fired is 784 336 D 448 ft. 20. Let s.t / be the distance the car travels in the t seconds after the brakes are applied. Then s 00 .t / D t and the velocity at time t is given by s 0 .t / D . t / dt D t2 C C1 ; 2 where C1 D 20 m/s (that is, 72km/h) as determined in Example 6. Thus Z s.t / D 20 t2 2 dt D 20t t3 C C2 ; 6 where C2 D 0 because s.0/ D 0. The time taken p to come to a stop is given by s 0 .t / D 0, so it is t D 40 s. The distance travelled is 16. This exercise and the next three refer to the following figure depicting the velocity of a rocket fired from a tower as a function of time since firing. p s D 20 40 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Z Downloaded by ted cage (sxnbyln180@questza.com) 1 3=2 40 84:3 m. 6 71 lOMoARcPSD|6566483 www.konkur.in SECTION 2.11 (PAGE 162) ADAMS and ESSEX: CALCULUS 9 Review Exercises 2 (page 163) 1. 8. y D .3x C 1/2 .3x C 3h C 1/2 .3x C 1/2 dy D lim dx h h!0 9x 2 C 18xh C 9h2 C 6x C 6h C 1 D lim h h!0 D lim .18x C 9h C 6/ D 18x C 6 d d 1 C x C x2 C x3 D .x 4 C x 3 C x 2 C x 1 / 4 dx x dx D 4x 5 3x 4 2x 3 x 2 .9x 2 C 6x C 1/ 9. d .4 dx x 2=5 / 5=2 D D x 3=5 .4 h!0 2. 3. d p 1 dx p p .x C h/2 1 x2 h h!0 1 .x C h/2 .1 x 2 / p D lim p h!0 h. 1 .x C h/2 C 1 x 2 / 2x h x D lim p p D p h!0 1 .x C h/2 C 1 x 2 1 x2 x 2 D lim 1 g.t / D t 5 p 1C t 4Ch p 1 1C 9Ch g 0 .9/ D lim h h!0 p p .3 C h 9 C h/.3 C h C 9 C h/ D lim p p h!0 h.1 C 9 C h/.3 C h C 9 C h/ 9 C 6h C h2 .9 C h/ D lim p p h!0 h.1 C 9 C h/.3 C h C 9 C h/ 5Ch p p D lim h!0 .1 C 9 C h/.3 C h C 9 C h/ 5 D 24 5. The tangent to y D cos.x/ at x D 1=6 has slope Its equation is ˇ dy ˇˇ ˇ dx ˇ D sin 3 2 2 xD1=6 yD p x D 6 : 2 1 : 6 d dx x 72 Telegram: @uni_k 1 D sin x 1 .x cos x sin x/2 d p 2 cos x sin x sin x cos x 2 C cos2 x D p D p dx 2 2 C cos2 x 2 C cos2 x 11. d .tan d sec2 / D sec2 12. 1 13. 14. sec2 2 sec2 tan p d 1 C t2 1 p dt 1 C t 2 C 1 p p t . 1 C t 2 C 1/ p . 1 C t2 2 1Ct p D . 1 C t 2 C 1/2 2t p D p 1 C t 2 . 1 C t 2 C 1/2 .x C h/20 h h!0 p 4x C 1 lim x!2 x 2 lim x 20 2 sec2 tan 1/ p t 1 C t2 d 20 x D 20x 19 dx p 3 9 C 4h 3 D lim 4 h!0 ˇ 4h d p ˇˇ 2 4 D D p D 4 xˇ ˇ dx 3 2 9 D xD9 15. lim x!=6 16. cos..=3/ C 2h/ cos.2x/ .1=2/ D lim 2 x =6 2h h!0 ˇ ˇ d ˇ D2 cos x ˇ ˇ dx xD=3 p D 2 sin.=3/ D 3 cos.=3/ 1 1 .1=x 2 / .1=a2 / . a C h/2 . a/2 lim D lim x! a xCa h h!0 ˇ ˇ d 1 ˇ 2 D D 3 ˇ dx x 2 ˇ a xD a 6. At x D the curve y D tan.x=4/ has slope .sec2 .=4//=4 D 1=2. The normal to the curve there has equation y D 1 2.x /. 7. x 2=5 / 7=2 10. D f .x/ D 4=x 2 4 1 .2 C h/2 f 0 .2/ D lim h h!0 4 h 4 .4 C 4h C h2 / D lim D D lim 2 h.2 C h/ h!0 .2 C h/2 h!0 4. 4 C 3x C 2x 2 C x 3 x5 5 2 3=5 .4 x 2=5 / 7=2 x 2 5 D 17. d f .3 dx 18. p p p p p 1 f . x/f 0 . x/ d p Œf . x/2 D 2f . x/f 0 . x/ p D dx 2 x x 19. x2/ D 2xf 0 .3 x2/ p p f .2x/g 0 .x=2/ d f .2x/ g.x=2/ D 2f 0 .2x/ g.x=2/ C p dx 4 g.x=2/ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 20. d f .x/ g.x/ dx f .x/ C g.x/ 1 D f .x/ C g.x//.f 0 .x/ .f .x/ C g.x//2 .f .x/ g.x//.f 0 .x/ C g 0 .x/ D 21. 22. 23. 24. REVIEW EXERCISES 2 (PAGE 163) 32. g.x/ D g 0 .x// 33. d f .x C .g.x//2 / D .1 C 2g.x/g 0 .x//f 0 .x C .g.x//2 / dx g.x 2 / 2x 2 g 0 .x 2 / g.x 2 / 0 g.x 2 / d f f D dx x x2 x .sin x/f .sin x/g 0 .cos x/ 34. x sin x dx D 3 3 4 31. If f .x/ D 12x C 12x , then f .x/ D 4x C 3x C C . If f .1/ D 0, then 4 C 3 C C D 0, so C D 7 and f .x/ D 4x 3 C 3x 4 7. Conjecture: g .n/ .x/ D .n C x/f .x/ for n D 1, 2, 3, : : : Proof: The formula is true for n D 1, 2, and 3 as shown above. Suppose it is true for n D k; that is, suppose g .k/ .x/ D .k C x/f .x/. Then d .k C x/f .x/ dx D f .x/ C .k C x/f 0 .x/ D ..k C 1/ C x/f .x/: g .kC1/ .x/ D Thus the formula is also true for n D k C 1. It is therefore true for all positive integers n by induction. 35. 36. The tangent to y D x 3 C 2 at x D a has equation y D a3 C 2 C 3a2 .x a/, or y D 3a2 x 2a3 C 2. This line passes through the origin if 0 D 2a3 C 2, that is, if a D 1. The line then has equation y D 3x. p 2 C x 2 at x D a has slope The ptangent to y D 2 a= 2 C a and equation yD p a 2 C a2 C p .x 2 C a2 a/: This line passes through .0; 1/ provided p a2 1 D 2 C a2 p 2 C a2 p 2 2 2Ca D2Ca a2 D 2 2 C a2 D 4 p The possibilities are a D ˙ 2, and the equations of the p corrresponding tangent lines are y D 1 ˙ .x= 2/. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x cos x C sin x C C If f 0 .x/ D f .x/ and g.x/ D x f .x/, then 16x 5 C 8x 4 C 8x 3 C 4x 2 C x C C 5 or, equivalently, Z .2x C 1/5 .2x C 1/4 dx D CC 10 2 x cos x dx D x sin x C cos x C C g 0 .x/ D f .x/ C xf 0 .x/ D .1 C x/f .x/ g 00 .x/ D f .x/ C .1 C x/f 0 .x/ D .2 C x/f .x/ g 000 .x/ D f .x/ C .2 C x/f 0 .x/ D .3 C x/f .x/ 25. If x 3 yC2xy 3 D 12, then 3x 2 yCx 3 y 0 C2y 3 C6xy 2 y 0 D 0. At .2; 1/: 12 C 8y 0 C 2 C 12y 0 D 0, so the slope there is y 0 D 7=10. The tangent line has equation 7 y D 1 10 .x 2/ or 7x C 10y D 24. p 26. 3p2x sin.y/ C 8ypcos.x/ D 2 3 2 sin.y/ C 3 2x cos.y/y 0 C 8y 0 cos.x/ 8y sin.x/ D 0 p At .1=3; 1=4/: 3pC y 0 C 4y 0 3 D 0, so the slope 3 3 . there is y 0 D C4 Z Z 1 C x4 x3 1 1 2 27. dx D C x C CC dx D x2 x2 x 3 Z Z p 2 1Cx 28. p dx D .x 1=2 C x 1=2 / dx D 2 x C x 3=2 C C 3 x Z Z 2 C 3 sin x 29. dx D .2 sec2 x C 3 sec x tan x/ dx cos2 x D 2 tan x C 3 sec x C C Z Z 30. .2x C 1/4 dx D .16x 4 C 32x 3 C 24x 2 C 8x C 1/ dx 0 d .x sin x C cos x/ D sin x C x cos x sin x D x cos x dx d .x cos x sin x/ D cos x x sin x cos x D x sin x dx Z Z f 0 .x/ sin f .x/ sin g.x/ g 0 .x/ cos f .x/ cos g.x/ .sin g.x//2 D 3 cos.x=3/ C 6 sin.x=6/ C C: If .; 2/ lies on y D g.x/, then .3=2/ C 3 C C D 2, so C D 1=2 and g.x/ D 3 cos.x=3/ C 6 sin.x=6/ C .1=2/: 2.f 0 .x/g.x/ f .x/g 0 .x// .f .x/ C g.x//2 d f .sin x/g.cos x/ dx D .cos x/f 0 .sin x/g.cos x/ s cos f .x/ d dx sin g.x/ s 1 sin g.x/ D 2 cos f .x/ If g 0 .x/ D sin.x=3/ C cos.x=6/, then Downloaded by ted cage (sxnbyln180@questza.com) 73 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 2 (PAGE 163) 37. ADAMS and ESSEX: CALCULUS 9 d n sin x sin.nx/ dx D n sinn 1 x cos x sin.nx/ C n sinn x cos.nx/ Observe that this rate is half the rate at which F decreases when r increases from R. 42. D n sinn 1 xŒcos x sin.nx/ C sin x cos.nx/ D n sinn 1 x sin..n C 1/x/ y D sinn x sin.nx/ has a horizontal tangent at x D m=.n C 1/, for any integer m. d n sin x cos.nx/ 38. dx D n sinn 1 x cos x cos.nx/ n sinn x sin.nx/ D n sinn 1 xŒcos x cos.nx/ Thus the isothermal compressibility of the gas is 1 1 dV D V dP V D n sin x cos..n C 1/x/ d n cos x sin.nx/ dx D n cosn 1 x sin x sin.nx/ C n cosn x cos.nx/ D n cosn 1 xŒcos x cos.nx/ 43. sin x sin.nx/ D n cos x cos..n C 1/x/ d n cos x cos.nx/ dx D n cosn 1 x sin x cos.nx/ 40. The average profit per tonne if x tonnes are exported is P .x/=x, that is the slope of the line joining .x; P .x// to the origin. This slope is maximum if the line is tangent to the graph of P .x/. In this case the slope of the line is P 0 .x/, the marginal profit. ( mgR2 if r R 41. F .r/ D r2 mkr if 0 r < R b) As r increases from R, F changes at rate D 2mgR2 D R3 y2 D ˇ d ˇ D .mkr/ˇ rDR dr mk D mg : R 100 9:8 100 4:9 100 DhC : .9:8/2 19:6 400 9:8 400 4:9 400 D : .9:8/2 19:6 These two maximum heights are equal, so hC 100 400 D ; 19:6 19:6 which gives h D 300=19:6 15:3 m as the height of the building. 44. The first ball has initial height 60 m and initial velocity 0, so its height at time t is y1 D 60 4:9t 2 m: The second ball has initial height 0 and initial velocity v0 , so its height at time t is 2mg : R As r decreases from R, F changes at rate 4:9t 2 : It reaches maximum height when dy2 =dt D 20 9:8t D 0, that is, at t D 20=9:8 s. The maximum height of the second ball is a) For continuity of F .r/ at r D R we require mg D mkR, so k D g=R. Telegram: @uni_k 1 : P The height of the second ball at time t during its motion is y2 D 20t 4:9t 2 : 39. Q D .0; 1/. If P D .a; a2 / on the curve y D x 2 , then the slope of y D x 2 at P is 2a, and the slope of PQ is .a2 1/=a. PQ is normal to y D x 2 if a D 0 or Œ.a2 1/=a.2a/ D 1, that is, ifpa D 0 or a2 D 1=2. The points P are .0; 0/ and .˙1= p2; 1=2/. The distances from these points to Q are 1 and 3=2, respectively. The distance from Q to the curve y D x 2 is the shortest of p these distances, namely 3=2 units. 74 D Let the building be h m high. The height of the first ball at time t during its motion is y1 D h C n cosn x sin.nx/ n cosn 1 x sin..n C 1/x/ rDR It reaches maximum height when dy1 =dt D 10 9:8t D 0, that is, at t D 10=9:8 s. The maximum height of the first ball is n cosn 1 xŒsin x cos.nx/ C cos x sin.nx/ ˇ d mgR2 ˇˇ ˇ dr r 2 ˇ V P y1 D h C 10t n 1 D sin x sin.nx/ n 1 D P V D kT . Differentiate with respect to P holding T constant to get dV D0 V CP dP y2 D v0 t 4:9t 2 m: The two balls collide at a height of 30 m (at time T , say). Thus 30 D 60 4:9T 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 30 D v0 T 4:9T 2 : lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 2 (PAGE 164) Thus v0 T D 60 and T 2 D 30=4:9. The initial upward speed of the second ball is v0 D 60 D 60 T r 4:9 24:25 m/s: 30 Challenging Problems 2 (page 164) 1. At time T , the velocity of the first ball is ˇ dy1 ˇˇ ˇ dt ˇ tDT D 9:8T x2 24:25 m/s: tDT D v0 9:8T D 0 m/s: 45. Let the car’s initial speed be v0 . The car decelerates at 20 ft/s2 starting at t D 0, and travels distance s in time t , where d 2 s=dt 2 D 20. Thus 2. f 0 .x/ D 1=x, f .2/ D 9. f .9 C 4h C h2 / f .x 2 C 5/ f .9/ D lim x!2 x 2 h h!0 f .9 C 4h C h2 / f .9/ 4h C h2 D lim 4h C h2 h h!0 f .9 C k/ f .9/ D lim lim .4 C h/ k k!0 h!0 4 0 D f .9/ 4 D 9 p p f .x/ 3 f .2 C h/ 3 D lim b) lim x!2 x 2 h h!0 f .2 C h/ 9 1 D lim p h h!0 f .2 C h/ C 3 1 1 D f 0 .2/ D : 6 12 The car stops at time t D v0 =20. The stopping distance is s D 160 ft, so v02 20 a2 D 0 a) lim ds D v0 20t dt x D v0 t 10t 2 : 160 D mx C ma In order that this quadratic have only one solution x D a, the left side must be .x a/2 , so that m D 2a. The tangent has slope 2a. This won’t work for more general curves whose tangents can intersect them at more than one point. At time T , the velocity of the second ball is ˇ dy2 ˇˇ ˇ dt ˇ The line through .a; a2 / with slope m has equation y D a2 C m.x a/. It intersects y D x 2 at points x that satisfy x 2 D a2 C mx ma; or v02 v2 D 0: 40 40 The car’s p initial speed cannot exceed v0 D 160 40 D 80 ft/s. p 46. P D 2 L=g D 2L1=2 g 1=2 . f .9/ a) If L remains constant, then 3. dP g D L1=2 g 3=2 g dg 1 g L1=2 g 3=2 P : g D 1=2 1=2 P 2 g 2L g P If g increases by 1%, then g=g D 1=100, and P =P D 1=200. Thus P decreases by 0:5%. f 0 .4/ D 3, g 0 .4/ D 7, g.4/ D 4, g.x/ ¤ 4 if x ¤ 4. f .x/ f .4/ .x 4/ a) lim f .x/ f .4/ D lim x!4 x!4 x 4 D f 0 .4/.4 4/ D 0 b) lim x!4 f .x/ x2 b) If g remains constant, then dP L D L 1=2 g 1=2 L P dL 1 L P L 1=2 g 1=2 L D : P 2 L 2L1=2 g 1=2 If L increases by 2%, then L=L D 2=100, and P =P D 1=100. Thus P increases by 1%. p f .x/ f .4/ f .x/ f .4/ p . x C 2/ D lim x!4 x!4 x 4 x 2 D f 0 .4/ 4 D 12 c) lim f .x/ 1 x!4 x d) lim Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 f .4/ f .x/ f .4/ D lim x!4 16 x 4 xC4 3 1 D f 0 .4/ D 8 8 Downloaded by ted cage (sxnbyln180@questza.com) x 4 f .x/ f .4/ f .4/ D lim 1 x!4 x 4 .4 x/=4x 4 D f 0 .4/ . 16/ D 48 75 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 2 (PAGE 164) f .x/ f .x/ f .4/ x e) lim D lim x!4 g.x/ x!4 g.x/ 4 x 3 f 0 .4/ D D 0 g .4/ 7 f) lim x!4 n ADAMS and ESSEX: CALCULUS 9 f .4/ 4 g.4/ 4 7. a) g.0/ D g.0 C 0/ D g.0/ C g.0/. Thus g.0/ D 0. g.x C h/ g.x/ h g.x/ C g.h/ g.x/ g.h/ g.0/ D lim D lim h h h!0 h!0 D g 0 .0/ D k: b) g 0 .x/ D lim h!0 f .g.x// f .4/ x 4 f .g.x// f .4/ g.x/ g.4/ D lim x!4 g.x/ 4 x 4 D f 0 .g.4// g 0 .4/ D f 0 .4/ g 0 .4/ D 3 7 D 21 x if x D 1; 1=2; 1=3; : : : . x 2 otherwise a) f is continuous except at 1=2, 1=3, 1=4, : : : . It is continuous at x D 1 and x D 0 (and everywhere else). Note that 4. f .x/ D c) If h.x/ D g.x/ kx, then h0 .x/ D g 0 .x/ k D 0 for all x. Thus h.x/ is constant for all x. Since h.0/ D g.0/ 0 D 0, we have h.x/ D 0 for all x, and g.x/ D kx. 8. lim x 2 D 1 D f .1/; x!1 lim x 2 D lim x D 0 D f .0/ x!0 x!0 b) If a D 1=2 and b D 1=3, then f .a/ C f .b/ 1 D 2 2 1 1 C 2 3 D 5 : 12 If 1=3 < x < 1=2, then f .x/ D x 2 < 1=4 < 5=12. Thus the statement is FALSE. 2 lim h 0h D 1 ¤ 0 D lim h!0 h 0 h f .x C k/ f .x/ (let k D h) k k!0 f .x/ f .x f .x h/ f .x/ D lim D lim h h h!0 h!0 1 0 0 0 f .x/ C f .x/ f .x/ D 2 1 f .x C h/ f .x/ D lim 2 h!0 h f .x/ f .x h/ C lim h h!0 f .x C h/ f .x h/ D lim : 2h h!0 b) The change of variables used in the first part of (a) shows that a) f 0 .x/ D lim f .x C h/ h h!0 lim h!0 9. p ˇ ˇ ˇ f .h/ f .0/ ˇ ˇ D jf .h/j > jhj ! 1 ˇ ˇ ˇ h jhj jhj f .0 C h/ 76 Telegram: @uni_k h/ D lim h!0 27a3 27 C 2 8 4a 27 27a3 C 2 8 4a f .0/ D 1: Thus f .0/ D 1. jhj h jhj D lim 0 h!0 h 3a : 2 x a 3a 2 D 0: If a ¤ 0, the x-axis is another tangent to y D x 3 that passes through .a; 0/. The number of tangents to y D x 3 that pass through .x0 ; y0 / is f .x/ f .x/ h!0 h/ This line passes through .a; 0/ because 6. Given that f .0/ D k, f .0/ ¤ 0, and f .x C y/ D f .x/f .y/, we have f .x C h/ f .x/ D lim h h!0 f .x/f .h/ D lim h h!0 f .x h The tangent to y D x 3 at x D 3a=2 has equation 0 0 f .0 2h yD as h ! 0. Therefore f 0 .0/ does not exist. f .0/ D 0 or f .x/ : c) If f .x/ D jxj, then f 0 .0/ does not exist, but : 5. If h ¤ 0, then ÷ lim and h/ are always equal if either exists. f is differentiable elsewhere, including at x D 1 where its derivative is 2. f .0/ D f .0C0/ D f .0/f .0/ f .x/ lim c) By (a) f cannot be differentiable at x D 1=2, 1=2, : : :. It is not differentiable at x D 0 either, since h!0 Given that g 0 .0/ D k and g.x C y/ D g.x/ C g.y/, then D f .x/f 0 .0/ D kf .x/: three, if x0 ¤ 0 and y0 is between 0 and x03 ; two, if x0 ¤ 0 and either y0 D 0 or y0 D x03 ; Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) D 0: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 2 (PAGE 164) The curve y D Ax 2 C Bx C C has slope m D 2Aa C B at .a; Aa2 C Ba C C /. Thus a D .m B/=.2A/, and the tangent has equation one, otherwise. This is the number of distinct real solutions b of the cubic equation 2b 3 3b 2 x0 C y0 D 0, which states that the tangent to y D x 3 at .b; b 3 / passes through .x0 ; y0 /. y D Aa2 C Ba C C C m.x 10. By symmetry, any line tangent to both curves must pass through the origin. y y D x 2 C 4x C 1 where f .m/ D C 14. x a/ B/2 B.m B/ D mx C C CC 4A 2A 2 .m B/ .m B/2 D mx C C C 4A 2A D mx C f .m/; .m B/2 =.4A/. .m Parabola y D x 2 has tangent y D 2ax a2 at .a; a2 /. Parabola y D Ax 2 C Bx C C has tangent Ab 2 C C y D .2Ab C B/x x 2 C 4x yD 1 at .b; Ab 2 C Bb C C /. These two tangents coincide if 2Ab C B D 2a Fig. C-2-10 Ab 2 The tangent to y D x C 4x C 1 at x D a has equation y D a2 C 4a C 1 C .2a C 4/.x D .2a C 4/x .a2 a/ 11. The slope of y D x 2 at x D a is 2a. The slope of the line from .0; b/ to .a; a2 / is .a2 b/=a. This line is normal to y D x 2 if either a D 0 or 2a..a2 b/=a/ D 1, that is, if a D 0 or 2a2 D 2b 1. There are three real solutions for a if b > 1=2 and only one (a D 0) if b 1=2. 2 2 12. The point Q D .a; a / on y D x that is closest to P D .3; 0/ is such that PQ is normal to y D x 2 at Q. Since PQ has slope a2 =.a 3/ and y D x 2 has slope 2a at Q, we require a2 1 D ; a 3 2a which simplifies to 2a3 C a 3 D 0. Observe that a D 1 is a solution of this cubic equation. Since the slope of y D 2x 3 C x 3 is 6x 2 C 1, which is always positive, the cubic equation can have only one real solution. Thus Q D .1; 1/ is the point on y D x 2 that is closest p to P . The distance from P to the curve is jPQj D 5 units. 13. The curve y D x 2 has slope m D 2a at .a; a2 /. The tangent there has equation y D a2 C m.x a/ D mx m2 : 4 ./ 2 C Da : .2Ab C B/2 D 4Ab 2 4C; or, on simplification, 4A.A 1/b 2 C 4ABb C .B 2 C 4C / D 0: This quadratic equation in b has discriminant D D 16A2 B 2 16A.A 1/.B 2 C4C / D 16A.B 2 4.A 1/C /: There are five cases to consider: CASE I. If A D 1, B ¤ 0, then ./ gives bD B 2 C 4C ; 4B aD B2 4C : 4B There is a single common tangent in this case. CASE II. If A D 1, B D 0, then ./ forces C D 0, which is not allowed. There is no common tangent in this case. CASE III. If A ¤ 1 but B 2 D 4.A bD B 2.A 1/ 1/C , then D a: There is a single common tangent, and since the points of tangency on the two curves coincide, the two curves are tangent to each other. CASE IV. If A ¤ 1 and B 2 4.A 1/C < 0, there are no real solutions for b, so there can be no common tangents. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 The two curves have one (or more) common tangents if ./ has real solutions for a and b. Eliminating a between the two equations leads to 1/; which passes through the origin if a D ˙1. The two common tangents are y D 6x and y D 2x. m.m B/ 2A Downloaded by ted cage (sxnbyln180@questza.com) 77 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 2 (PAGE 164) ADAMS and ESSEX: CALCULUS 9 b) The tangent to y D x 4 CASE V. If A ¤ 1 and B 2 4.A 1/C > 0, there are two distinct real solutions for b, and hence two common tangent lines. y y y D a4 D 4a.a 2x 2 at x D a has equation 2a2 C .4a3 2 4a/.x a/ 3a4 C 2a2 : 1/x Similarly, the tangent at x D b has equation x y D 4b.b 2 x one common two common tangent tangents tangent curves y 1/x These tangents are the same line (and hence a double tangent) if 4a.a2 y 1/ D 4b.b 2 3a4 C 2a2 D x x Fig. C-2-14 a) The tangent to y D x 3 at .a; a3 / has equation y D 3a2 x 0 D a2 C b 2 2a3 : x .x 2 3a x C 2a D 0 a/2 .x C 2a/ D 0: b) The slope of y D x 3 at x D 2a is 3. 2a/2 D 12a2 , which is four times the slope at x D a. 17. a) The tangent to y D f .x/ D ax 4 C bx 3 C cx 2 C dx C e 3 c) If the tangent to y D x at x D a were also tangent at x D b, then the slope at b would be four times that at a and the slope at a would be four times that at b. This is clearly impossible. d) No line can be tangent to the graph of a cubic polynomial P .x/ at two distinct points a and b, because if there was such a double tangent y D L.x/, then .x a/2 .x b/2 would be a factor of the cubic polynomial P .x/ L.x/, and cubic polynomials do not have factors that are 4th degree polynomials. 16. a) y D x 4 2x 2 has horizontal tangents at points x satisfying 4x 3 4x D 0, that is, at x D 0 and x D ˙1. The horizontal tangents are y D 0 and y D 1. Note that y D 1 is a double tangent; it is tangent at the two points .˙1; 1/. 78 Telegram: @uni_k b/2 > 0: c) If y D Ax C B is a double tangent to y D x 4 2x 2 C x, then y D .A 1/x C B is a double tangent to y D x 4 2x 2 . By (b) we must have A 1 D 0 and B D 1. Thus the only double tangent to y D x 4 2x 2 C x is y D x 1. 3 The tangent also intersects y D x 3 at .b; b 3 /, where b D 2a. 2ab D .a Thus y D 1 is the only double tangent to y D x 4 2x 2 . For intersections of this line with y D x 3 we solve 3 1/ 3b 4 C 2b 2 : The second equation says that either a2 D b 2 or 3.a2 C b 2 / D 2; the first equation says that a3 b 3 D a b, or, equivalently, a2 C ab C b 2 D 1. If a2 D b 2 , then a D b (a D b is not allowed). Thus a2 D b 2 D 1 and the two points are .˙1; 1/ as discovered in part (a). If a2 Cb 2 D 2=3, then ab D 1=3. This is not possible since it implies that no common tangent 15. 3b 4 C 2b 2 : at x D p has equation y D .4ap 3 C3bp 2 C2cpCd /x 3ap 4 2bp 3 cp 2 Ce: This line meets y D f .x/ at x D p (a double root), and xD 2ap b˙ p b2 4ac 2a 4abp 8a2 p 2 : These two latter roots are equal (and hence correspond to a double tangent) if the expression under the square root is 0, that is, if 8a2 p 2 C 4abp C 4ac Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) b 2 D 0: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 2 (PAGE 164) This quadratic has two real solutions for p provided its discriminant is positive, that is, provided 16a2 b 2 4.8a2 /.4ac b 2 / > 0: This condition simplifies to 3b 2 > 8ac: For example, for y D x 4 2x 2 C x 1, we have a D 1, b D 0, and c D 2, so 3b 2 D 0 > 16 D 8ac, and the curve has a double tangent. b) From the discussion above, the second point of tangency is so the formula above is true for n D 1. Assume it is true for n D k, where k is a positive integer. Then d kC1 d k k cos.ax/ D a cos ax C dx 2 dx kC1 k k Da a sin ax C 2 .k C 1/ kC1 Da cos ax C : 2 Thus the formula holds for n D 1; 2; 3; : : : by induction. b) Claim: qD 2ap b D 2a p b : 2a The slope of PQ is f .q/ q b3 f .p/ D p 4abc C 8a2 d : 8a2 Calculating f 0 ..p C q/=2/ leads to the same expression, so the double tangent PQ is parallel to the tangent at the point horizontally midway between P and Q. c) The inflection points are the real zeros of f 00 .x/ D 2.6ax 2 C 3bx C c/: This equation has distinct real roots provided 9b 2 > 24ac, that is, 3b 2 > 8ac. The roots are rD sD p 9b 2 p12a 3b C 9b 2 12a 3b b3 f .r/ D r so the formula above is true for n D 1. Assume it is true for n D k, where k is a positive integer. Then d kC1 k d k a sin ax C sin.ax/ D dx 2 dx kC1 k k D a a cos ax C 2 .k C 1/ kC1 Da sin ax C : 2 c) Note that 4abc C 8a2 d ; 8a2 so this line is also parallel to the double tangent. 18. d sin.ax/ D a cos.ax/ D a sin ax C ; dx 2 : The slope of the line joining these inflection points is f .s/ s Proof: For n D 1 we have Thus the formula holds for n D 1; 2; 3; : : : by induction. 24ac 24ac n dn sin.ax/ D an sin ax C . n dx 2 dn n n cos.ax/ D a cos ax C . dx n 2 Proof: For n D 1 we have d .cos4 x C sin4 x/ D 4 cos3 x sin x C 4 sin3 x cos x dx D 4 sin x cos x.cos2 sin2 x/ D 2 sin.2x/ cos.2x/ D sin.4x/ D cos 4x C : 2 It now follows from part (a) that a) Claim: d cos.ax/ D dx a sin.ax/ D a cos ax C ; 2 n dn .cos4 x C sin4 x/ D 4n 1 cos 4x C : n dx 2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 79 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 2 (PAGE 164) 19. ADAMS and ESSEX: CALCULUS 9 v (m/s) d) The upward acceleration in Œ0; 3 was 39:2=3 13:07 m/s2 . .3; 39:2/ 40 e) The maximum height achieved by the rocket is the distance it fell from t D 7 to t D 15. This is the area under the t -axis and above the graph of v on that interval, that is, 30 20 10 t (s) -10 2 4 6 8 10 12 14 12 .15; 1/ 7 2 .49/ C 49 C 1 .15 2 12/ D 197:5 m: -20 -30 f) During the time interval Œ0; 7, the rocket rose a distance equal to the area under the velocity graph and above the t -axis, that is, -40 .12; 49/ Fig. C-2-19 1 .7 2 a) The fuel lasted for 3 seconds. b) Maximum height was reached at t D 7 s. c) The parachute was deployed at t D 12 s. 80 Telegram: @uni_k 0/.39:2/ D 137:2 m: Therefore the height of the tower was 197:5 137:2 D 60:3 m. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHAPTER 3. FUNCTIONS TRANSCENDENTAL Section 3.1 Inverse Functions 1. SECTION 3.1 (PAGE 171) f .x1 / D f .x2 / , x12 D x22 ; .x1 0; x2 0/ , x1 D x2 Thus f is one-to-one. Let y D f 1 .x/. Then x D f .y/ p D y 2 .y 0/. p x and f 1 .x/ D x. therefore y D D.f / D . 1; 0 D R.f 1 /, D.f 1 / D Œ0; 1/ D R.f /. (page 171) f .x/ D x 1 f .x1 / D f .x2 / ) x1 1 D x2 1 ) x1 D x2 . Thus f is one-to-one. Let y D f 1 .x/. Then x D f .y/ D y 1 and y D x C 1. Thus f 1 .x/ D x C 1. D.f / D D.f 1 / D R D R.f / D R.f 1 /. 2. f .x/ D 2x 1. If f .x1 / D f .x2 /, then 2x1 1 D 2x2 1. Thus 2.x1 x2 / D 0 and x1 D x2 . Hence, f is one-toone. Let y D f 1 .x/. Thus x D f .y/ D 2y 1, so y D 21 .x C 1/. Thus f 1 .x/ D 12 .x C 1/. D.f / D R.f 1 / D . 1; 1/. R.f / D D.f 1 / D . 1; 1/. 8. f .x/ D .1 2x/3 . If f .x1 / D f .x2 /, then .1 2x1 /3 D .1 2x2 /3 and x1 D x2 . Thus, f is one-toone. 1 3 Let y D f p .x/. Then x D f .y/ D .1 2y/ p so y D 21 .1 3 x/. Thus, f 1 .x/ D 12 .1 3 x/. D.f / D R.f 1 / D . 1; 1/. R.f / D D.f 1 / D . 1; 1/. 9. f .x/ D p 3. f .x/ D x 1 p p f .x1 / D f .x2 / , x1 1 D x2 1; .x1 ; x2 1/ , x1 1 D x2 1 D 0 , x1 D x2 Thus f is one-to-one.pLet y D f 1 .x/. Then x D f .y/ D y 1, and y D 1 C x 2 . Thus f 1 .x/ D 1 C x 2 , .x 0/. D.f / D R.f 1 / D Œ1; 1/, R.f / D D.f 1 / D Œ0; 1/. p x 1 for x 1. 4. f .x/ D p p If f .x1 / D f .x2 /, then x1 1 D x2 1 and x1 1 D x2 1. Thus x1 D x2 and f is one-to-one. p y 1 so Let y D f 1 .x/. Then x D f .y/ D 2 2 1 x D y 1 and y D x C 1. Thus, f .x/ D x 2 C 1. D.f / D R.f 1 / D Œ1; 1/. R.f / D D.f 1 / D . 1; 0. 5. f .x/ D x 3 f .x1 / D f .x2 / , x13 D x23 ) .x1 x2 /.x12 C x1 x2 C x22 / D 0 ) x1 D x2 Thus f is one-to-one. Let y D f 1 .x/. Then x D f .y/ D y 3 so y D x 1=3 . Thus f 1 .x/ D x 1=3 . D.f / D D.f 1 / D R D R.f / D R.f 1 /. p 6. f .x/ pD 1 C 3 x. p If f .x1 / D f .x2 /, then 1 C 3 x 1 D 1 C 3 x 2 so x1 D x2 . Thus, f is oneto-one. p Let y D f 1 .x/ so that x D f .y/ D 1 C 3 y. Thus 3 1 3 y D .x 1/ and f .x/ D .x 1/ . D.f / D R.f 1 / D . 1; 1/. R.f / D D.f 1 / D . 1; 1/. f .x/ D x 2 ; .x 0/ 7. 1 : D.f / D fx W x ¤ 1g D R.f xC1 1 1 D f .x1 / D f .x2 / , x1 C 1 x2 C 1 , x2 C 1 D x1 C 1 , x2 D x1 Thus f is one-to-one; Let y D f 1 .x/. 1 Then x D f .y/ D y C1 1 1 and y D f 1 .x/ D 1. so y C 1 D x x 1 D.f / D fx W x ¤ 0g D R.f /. /. x . If f .x1 / D f .x2 /, then 10. f .x/ D 1Cx x1 x2 D . Hence x1 .1 C x2 / D x2 .1 C x1 / 1 C x1 1 C x2 and, on simplification, x1 D x2 . Thus, f is one-to-one. y Let y D f 1 .x/. Then x D f .y/ D and 1Cy x D f 1 .x/. x.1 C y/ D y. Thus y D 1 x D.f / D R.f 1 / D . 1; 1/ [ . 1; 1/. R.f / D D.f 1 / D . 1; 1/ [ .1; 1/. 11. 1 2x : D.f / D fx W x ¤ 1g D R.f 1 / 1Cx 1 2x2 1 2x1 D f .x1 / D f .x2 / , 1 C x1 1 C x2 , 1 C x2 2x1 2x1 x2 D 1 C x1 2x2 2x1 x2 , 3x2 D 3x1 , x1 D x2 Thus f is one-to-one. Let y D f 1 .x/. 1 2y Then x D f .y/ D 1Cy so x C xy D 1 2y 1 x . and f 1 .x/ D y D 2Cx 1 D.f / D fx W x ¤ 2g D R.f /. f .x/ D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 Downloaded by ted cage (sxnbyln180@questza.com) 81 lOMoARcPSD|6566483 www.konkur.in SECTION 3.1 (PAGE 171) ADAMS and ESSEX: CALCULUS 9 x . If f .x1 / D f .x2 /, then x2 C 1 x1 x2 q D q : ./ 2 x1 C 1 x22 C 1 12. f .x/ D p 20. Thus x12 .x22 C 1/ D x22 .x12 C 1/ and x12 D x22 . From (*), x1 and x2 must have the same sign. Hence, x1 D x2 and f is one-to-one. y , and Let y D f 1 .x/. Then x D f .y/ D p 2 y C1 x2 x 2 .y 2 C 1/ D y 2 . Hence y 2 D . Since f .y/ and y 1 x2 x , so have the same sign, we must have y D p 1 x2 x . f 1 .x/ D p 1 x2 1 D.f / D R.f / D . 1; 1/. R.f / D D.f 1 / D . 1; 1/. 21. 13. g.x/ D f .x/ 2 Let y D g 1 .x/. Then x D g.y/ D f .y/ 2, so f .y/ D x C 2 and g 1 .x/ D y D f 1 .x C 2/. 14. 15. y 1 y D f .x/ k.x/ D 3f .x/. Let y D k 1 .x/. Then x and x D k.y/ D 3f .y/, so f .y/ D 3 x 1 1 . k .x/ D y D f 3 1 . Let y D p 1 .x/. p.x/ D 1 C f .x/ 1 1 so f .y/ D Then x D p.y/ D 1 C f .y/ x 1 1 . and p 1 .x/ D y D f 1 x x Fig. 3.1-21 22. g.x/ D x 3 if x 0, and g.x/ D x 1=3 if x < 0. Suppose f .x1 / D f .x2 /. If x1 0 and x2 0 then x13 D x23 so x1 D x2 . Similarly, x1 D x2 if both are negative. If x1 and x2 have opposite sign, then so do g.x1 / and g.x2 /. Therefore g is Let y D g 1 .x/. Then one-to-one. 3 y if y 0 x D g.y/ D y 1=3 if y < 0. 1=3 if x 0 Thus g 1 .x/ D y D x 3 x if x < 0. 23. If x1 and x2 are both positive or both negative, and h.x1 / D h.x2 /, then x12 D x22 so x1 D x2 . If x1 and x2 have opposite sign, then h.x1 / and h.x2 / are on opposite sides of 1, so cannot be equal. Hence h is one-to-one. y2 C 1 if y 0 . If If y D h 1 .x/, then x D h.y/ D 2 y C 1 if y < 0 p p y 0, then y D px 1. If y < 0, then y D 1 x. Thus h 1 .x/ D px 1 if x 1 1 x if x < 1 24. y D f 1 .x/ , x D f .y/ D y 3 C y. To find y D f 1 .2/ we solve y 3 C y D 2 for y. Evidently y D 1 is the only solution, so f 1 .2/ D 1. 1, f .x/ 3 Let y D q 1 .x/. Then 2 f .y/ 3 and f .y/ D 2x C 3. Hence x D q.y/ D 2 1 1 q .x/ D y D f .2x C 3/. 18. q.x/ D 19. r.x/ D 1 2f .3 4x/ Let y D r 1 .x/. Then x D r.y/ D 1 f .3 3 4y/ D 82 1 f x 2 1 4y D f 1 and r 1 .x/ D y D 3 4 Telegram: @uni_k f .x/ D x 2 C 1 if x 0, and f .x/ D x C 1 if x < 0. If f .x1 / D f .x2 / then if x1 0 and x2 0 then x12 C 1 D x22 C 1 so x1 D x2 ; if x1 0 and x2 < 0 then x12 C 1 D x2 C 1 so x2 D x12 (not possible); if x1 < 0 and x2 0 then x1 D x22 (not possible); if x1 < 0 and x2 < 0 then x1 C 1 D x2 C 1 so x1 D x2 . Therefore f is Let y D f 1 .x/. Then one-to-one. y 2 C 1 if y 0 x D f .y/ D y C 1 if y < 0. p 1 x 1 if x 1 Thus f .x/ D y D x 1 if x < 1. h.x/ D f .2x/. Let y D h 1 .x/. Then x D h.y/ D f .2y/ and 2y D f 1 .x/. Thus h 1 .x/ D y D 21 f 1 .x/. 16. m.x/ D f .x 2/. Let y D m 1 .x/. Then x D m.y/ D f .y 2/, and y 2 D f 1 .x/. Hence m 1 .x/ D y D f 1 .x/ C 2. 17. 1 C f .x/ . Let y D s 1 .x/. 1 f .x/ 1 C f .y/ Then x D s.y/ D . Solving for f .y/ we obtain 1 f .y/ x 1 x 1 . Hence s 1 .x/ D y D f 1 . f .y/ D xC1 xC1 s.x/ D 1 2f .3 1 x 2 1 x 2 . 4y/. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 3.1 (PAGE 171) 25. g.x/ D 1 if x 3 C x D 10, that is, if x D 2. Thus g 1 .1/ D 2. 26. h.x/ D 3 if xjxj D h 1 . 3/ D 2. 27. If y D f 4, that is, if x D 31. 2. Thus 32. 1 .x/ then x D f .y/. dy dy 1 1 Thus 1 D f 0 .y/ so D 0 D Dy 1 dx dx f .y/ y (since f 0 .x/ D 1=x). dy dx D .2 C cos y/ dx dx ˇ dy ˇˇ 1 1 0 .g / .2/ D 0:36036: D ˇ dx ˇ 2 C cos y 1D xD2 4x 3 , then x2 C 1 4x 2 .x 2 C 3/ .x 2 C 1/.12x 2 / 4x 3 .2x/ D : f .x/ D 2 2 .x C 1/ .x 2 C 1/2 33. 0 Since f 0 .x/ > 0 for all x, except x D 0, f must be oneto-one and so it has an inverse. 4y 3 , and If y D f 1 .x/, then x D f .y/ D 2 y C1 1 D f 0 .y/ D .y 2 C 1/2 . Since f .1/ D 2, therefore 4y 4 C 12y 2 1 f .2/ D 1 and f ˇ .y 2 C 1/2 ˇˇ .2/ D 4 ˇ 4y C 12y 2 ˇ yD1 p 30. If f .x/ D x 3pC x 2 and y D f x D f .y/ D y 3 C y 2 , so, p 2yy 0 1 D y 3 C y2 C y p 2 3 C y2 0 Since f . 1/ D 2 implies that f f 1 0 If f .x/ D x sec x, then f 0 .x/ D sec x C x sec x tan x 1 for x in . =2; =2/. Thus f is increasing, and so oneto-one on that interval. Moreover, limx! .=2/C f .x/ D 1 and limx!.=2/C f .x/ D 1, so, being continuous, f has range . 1; 1/, and so f 1 has domain . 1; 1/. Since f .0/ D 0, we have f 1 .0/ D 0, and .f .y 2 C 1/.12y 2 y 0 / 4y 3 .2yy 0 / : .y 2 C 1/2 Thus y 0 D 1 0 g.x/ D 2x C sin x ) g 0 .x/ D 2 C cos x 1 for all x. Therefore g is increasing, and so one-to-one and invertible on the whole real line. y D g 1 .x/ , x D g.y/ D 2y C sin y. For y D g 1 .2/, we need to solve 2y C sin y 2 D 0. The root is between 0 and 1; to five decimal places g 1 .2/ D y 0:68404. Also 28. f .x/ D 1 C 2x 3 Let y D f 1 .x/. Thus x D f .y/ D 1 C 2y 3 . 1 dy 1 dy D so .f 1 /0 .x/ D D 1 D 6y 2 dx dx 6y 2 6Œf 1 .x/2 29. If f .x/ D p y D f 1 .2/ , 2 D f .y/ D y 2 =.1 C y/. We must solve p 2 2 C 2 y D y for y. There is a root between 2 and 3: f 1 .2/ 2:23362 to 5 decimal places. 1 D 1 : 4 35. .x/, then ) 1 0 y D . 2/ D ˇ p 3 C y 2 ˇˇ . 2/ D ˇ 3 C 2y 2 ˇ yD 1 D 34. p 3 C y2 : 3 C 2y 2 1, we have 2 : 5 p Note: f .x/ D x 3 C x 2 D 2 ) x 2 .3 C x 2 / D 4 4 2 ) x C 3x 4 D 0 ) .x 2 C 4/.x 2 1/ D 0. 2 Since .x C 4/ D 0 has no real solution, therefore x 2 1 D 0 and x D 1 or 1. Since it is given that f .x/ D 2, therefore x must be 1. 1 0 / .0/ D 1 .0/ D 1 D 1: f 0 .0/ If y D .f ı g/ 1 .x/, then x D f ı g.y/ D f .g.y//. Thus g.y/ D f 1 .x/ and y D g 1 .f 1 .x// D g 1 ı f 1 .x/. That is, .f ı g/ 1 D g 1 ı f 1 . x a bx c y a and Let y D f 1 .x/. Then x D f .y/ D by c cx a bxy cx D y a so y D . We have bx 1 cx a x a D , which simplifies to f 1 .x/ D f .x/ if bx c bx 1 f .x/ D b.1 c/x 2 C .c 2 1/x C a..1 c/ 0: This equation is satisfied for (almost) all values of x provided that c D 1 a and b arbitrary, or c D 1 and a D b D 0. In the latter case f .x/ x, which is certainly self-inverse. For the former case, f will be self-inverse provided f is one-to-one and so has an inverse. Since, for c D 1, ab 1 f 0 .x/ D .bx 1/2 is positive (or negative) for all x ¤ Ê1=b provided ab > 1 (or ab < 1), we have that f is self-inverse in c D 1, and a and b are arbitrary except that ab ¤ 1. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 f 0 .f Downloaded by ted cage (sxnbyln180@questza.com) 83 lOMoARcPSD|6566483 www.konkur.in SECTION 3.1 (PAGE 171) ADAMS and ESSEX: CALCULUS 9 36. Let f .x/ be an even function. Then f .x/ D f . x/. Hence, f is not one-to-one and it is not invertible. Therefore, it cannot be self-inverse. An odd function g.x/ may be self-inverse if its graph is symmetric about the line x D y. Examples are g.x/ D x and g.x/ D 1=x. 37. No. A function that is one-to-one on a single interval need not be either increasing or decreasing. For example, consider the function defined on Œ0; 2 by f .x/ D x x if 0 x 1 if 1 < x 2. 7. log1=3 32x D log1=3 8. 43=2 D 8 9. 10 log10 .1=x/ D 38. First we consider the case where the domain of f is a closed interval. Suppose that f is one-to-one and continuous on Œa; b, and that f .a/ < f .b/. We show that f must be increasing on Œa; b. Suppose not. Then there are numbers x1 and x2 with a x1 < x2 b and f .x1 / > f .x2 /. If f .x1 / > f .a/, let u be a number such that u < f .x1 /, f .x2 / < u, and f .a/ < u. By the Intermediate-Value Theorem there exist numbers c1 in .a; x1 / and c2 in .x1 ; x2 / such that f .c1 / D u D f .c2 /, contradicting the one-to-oneness of f . A similar contradiction arises if f .x1 / f .a/ because, in this case, f .x2 / < f .b/ and we can find c1 in .x1 ; x2 / and c2 in .x2 ; b/ such that f .c1 / D f .c2 /. Thus f must be increasing on Œa; b. ) p 2log4 8 D 23=2 D 2 2 1 Dx 1=x 10. Since loga x 1=.loga x/ D x 1=.loga x/ D a1 D a. 1 loga x D 1, therefore loga x .loga b/.logb a/ D loga a D 1 12. logx x.logy y 2 / D logx .2x/ D logx x C logx 2 D 1 C logx 2 D 1 C 1 log2 x 1 D1 2 13. .log4 16/.log4 2/ D 2 14. log15 75 C log15 3 D log15 225 D 2 .since 152 D 225/ 15. 16. 17. A similar argument shows that if f .a/ > f .b/, then f must be decreasing on Œa; b. log6 9 C log6 4 D log6 36 D 2 2 2 4 3 2 log3 12 4 log3 6 D log3 24 34 D log3 .3 2 / D 2 loga .x 4 C 3x 2 C 2/ C loga .x 4 C 5x 2 C 6/ p 4 loga x 2 C 2 D loga .x 2 C 2/.x 2 C 1/ C loga .x 2 C 2/.x 2 C 3/ 2 log1 .x 2 C 2/ Finally, if the interval I where f is defined is not necessarily closed, the same argument shows that if Œa; b is a subinterval of I on which f is increasing (or decreasing), then f must also be increasing (or decreasing) on any intervals of either of the forms Œx1 ; b or Œa; x2 , where x1 and x2 are in I and x1 a < b x2 . So f must be increasing (or decreasing) on the whole of I . 18. Section 3.2 Exponential and Logarithmic Functions (page 175) 19. p 33 p D 33 5=2 D 31=2 D 3 5 3 20. log3 5 D .log10 5/=.log10 3 1:46497 21. 22x D 5xC1 , 2x log10 2 D .x C 1/ log10 5, x D .log10 5/=.2 log10 2 log10 5/ 7:21257 p p x D log10 3, x 2 D 3, 2 log p 10 x D 10.log10 3/= 2 2:17458 2. 2 1=2 1=2 3. .x 8 3 / 2 D2 Dx 1=2 3=2 2 6 D loga .x 2 C 1/ C loga .x 2 C 3/ D loga .x 4 C 4x 2 C 3/ 2 D2 D4 22. 2x 4. . 21 /x 4x=2 D x D 1 2 6. If log4 . 18 / D y then 4y D 18 , or 22y D 2 3 . Thus 2y D 3 and log4 . 18 / D y D 32 . 84 log .1 cos x/ C log .1 C cos x/ 2 log sin x sin2 x .1 cos x/.1 C cos x/ D log D log sin2 x sin2 x D log 1 D 0 p p 2 yD3 p , log10 y D 2 log10 3, y D 10 2 log10 3 4:72880 23. logx 3 D 5, .log10 3/=.log10 x/ D 5, log10 x D .log10 3/=5, x D 10.log10 3/=5 1:24573 24. log3 x D 5, .log10 x/=.log10 3/ D 5, log10 x D 5 log10 3, x D 105 log10 3 D 35 D 243 5. log5 125 D log5 53 D 3 Telegram: @uni_k log4 8 D 32 ) 2x 11. It is one-to-one but neither increasing nor decreasing on all of Œ0; 2. 1. 2x 1 D 3 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 3.3 (PAGE 183) 1 1 then au D D x 1 . Hence, a u D x x x and u D logax. 1 D loga x. Thus, loga x 25. Let u D loga 26. Let loga x D u, loga y D v. Then x D au , y D av . au x D v D au v Thus y a x and loga D u v D loga x y 27. 35. h!0 D k. Thus axCh ax h h!0 ax ah ax D lim h h!0 h a 1 D ax f 0 .0/ D ax k D kf .x/: D ax lim h h!0 loga y. 36. Let u D loga .x y /, then au D x y and au=y D x. u D loga x, or u D y loga x. Therefore y y Thus, loga .x / D y loga x. 1 yDf Thus .f .x/ ) x D f .y/ D ay dx dy )1D D kay dx dx 1 1 dy D D : ) dx kay kx 1 0 / .x/ D 1=.kx/. Section 3.3 The Natural Logarithm and Exponential Functions (page 183) 1. p e3 p D e 3 5=2 D e 1=2 D e e5 2. ln.e 1=2 e 2=3 / D 21 C 32 D 76 3. e 5 ln x D x 5 4. e .3 ln 9/=2 D 93=2 D 27 5. ln log3 x 2 C log3 x 1=2 D 10 6. x 5=2 D 310 ; so x D .310 /2=5 D 34 D 81 2 e 2 ln cos x C ln e sin x D cos2 x C sin2 x D 1 7. 3 ln 4 8. 4 ln 9. 2 ln x C 5 ln.x log4 .x C 4/ 2 log4 .x C 1/ D 1 2 xC4 1 D 2 .x C 1/ 2 xC4 1=2 D4 D2 .x C 1/2 2x 2 C 3x 2 D 0 but we need x C 1 > 0, so x D 1=2. log4 30. First observe that log9 x D log3 x= log3 9 D 21 log3 x. Now 2 log3 x C log9 x D 10 log3 x 5=2 D 10 31. Note that logx 2 D 1= log2 x. Since limx!1 log2 x D 1, therefore limx!1 logx 2 D 0. 32. Note that logx .1=2/ D logx 2 D 1= log2 x. Since limx!0C log2 x D 1, therefore limx!0C logx .1=2/ D 0. 33. Note that logx 2 D 1= log2 x. Since limx!1C log2 x D 0C, therefore limx!1C logx 2 D 1. 34. Note that logx 2 D 1= log2 x. Since limx!1 log2 x D 0 , therefore limx!1 logx 2 D 1. 1 D ln e 3x D e 3x p 4 ln 3 D ln 3x 64 43 D ln 34 81 x C 6 ln.x 1=3 / D 2 ln x C 2 ln x D 4 ln x 2/ D ln x 2 .x 2/5 10. ln.x 2 C 6x C 9/ D lnŒ.x C 3/2 D 2 ln.x C 3/ 11. 2xC1 D 3x .x C 1/ ln 2 D x ln 3 ln 2 ln 2 xD D ln 3 ln 2 ln.3=2/ 12. 3x D 91 x ) 3x D 32.1 x/ ) x D 2.1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 h f 0 .x/ D lim 28. Let logb x D u, logb a D v. Thus b u D x and b v D a. Therefore x D b u D b v.u=v/ D au=v u logb x and loga x D D . v logb a 29. ah f .x/ D ax and f 0 .0/ D lim Downloaded by ted cage (sxnbyln180@questza.com) x/ ) x D 23 85 lOMoARcPSD|6566483 www.konkur.in SECTION 3.3 (PAGE 183) 13. 14. ADAMS and ESSEX: CALCULUS 9 5 1 D xC3 2x 8 x ln 2 D ln 5 .x C 3/ ln 8 D ln 5 .3x C 9/ ln 2 2x ln 2 D ln 5 9 ln 2 ln 5 9 ln 2 xD 2 ln 2 2x 2 2 3 D 4x D 22x ) x 2 33. y D ln ln x 34. y D x ln x x 1 y D ln x C x x 35. 3/.x C 1/ D 0 x2 2 x2 0 y D 2x ln x C x 36. y D ln j sin xj; 37. y D 52xC1 16. ln.x 2 x 2/ D lnŒ.x 2/.x C 1/ is defined if .x 2/.x C 1/ > 0, that is, if x < 1 or x > 2. The domain is the union . 1; 1/ [ .2; 1/. 38. 17. ln.2x 5/ > ln.7 2x/ holds if 2x 5 > 0, 7 2x > 0, and 2x 5 > 7 2x, that is, if x > 5=2, x < 7=2, and 4x > 12 (i.e., x > 3). The solution set is the interval .3; 7=2/. 2 2 18. ln.x 2 2/ ln x holds 2 x. p if x > 2, x > 0, and x Thus we need x > 2 and x 2 x 2 0. This latter inequality says that .x 2/.x C 1/ 0, so it holds for 1 x 2. The solution set of the given inequality is p . 2; 2. 39. D .1 2x/e 22. y D x 2 e x=2 ; 23. y D ln.3x 24. y D ln j3x y0 D 2j; 0 y D f 0 .s/ D 42. 25. y D ln.1 C e x / 2 26. f .x/ D e x ; 27. yD f 0 .x/ D .2x/e x ex C e x ; 2 y0 D x 30. y D 32. y D e x cos x; 86 Telegram: @uni_k 2 e x 2 x y 0 D e x e .e / D e xCe ex D1 1 C ex 31. y D e x sin x; ex 44. 45. dx 1 D 3e 3t ln t C e 3t dt t 28. x D e 3t ln t; 29. y D e .e / ; 2 3x 2 ex y0 D 1 C ex ex .1 C e x /2 y0 D y0 D e x cos x y 0 D e x .sin x C cos x/ xt ; e x sin x y 0 D .2x 46. 3/.ln 2/2.x 2 h0 .t / D xt x 1 x t ln x ln.bs C c/ ln a b .bs C c/ ln a ln.2x C 3/ g.x/ D logx .2x C 3/ D ln x 2 1 ln x Œln.2x C 3/ 2x C 3 x 0 g .x/ D .ln x/2 2x ln x .2x C 3/ ln.2x C 3/ D x.2x C 3/.ln x/2 p 3xC8/ g 0 .x/ D t x x t ln t C t xC1 x t 1 p D e x ln x p p ln x x y 0 D e x ln x p C x 2 x p 1 1 ln x C 1 Dx x p x 2 ln x 1 Given that y D , let u D ln x. Then x D e u and x u 1 2 yD D .e u /u D e u . Hence, eu 1 dy du 2 ln x 1 ln x dy 2 : D D . 2ue u / D dx du dx x x x yDx x y D ln j sec x C tan xj sec x tan x C sec2 x sec x C tan x D sec x p y D ln jx C x 2 a2 j 2x 1C p 2p x 2 a2 D p 1 y0 D x C x 2 a2 x 2 a2 y0 D x 1 ; 1 C ex g.x/ D t x x t ; ; f .s/ D loga .bs C c/ D 43. 3 3xC8/ 41. 3 3x 2 h.t / D t x y 0 D 2xe x=2 C 12 x 2 e x=2 2/ y D 2.x 40. 1 2x 2x D 2x ln x 2 cos x D cot x y0 D sin x y 0 D 2.52xC1 / ln 5 D .2 ln 5/52xC1 19. y D e 5x ; y 0 D 5e 5x 20. y D xe x x; y 0 D e x C xe x x 21. y D 2x D xe 2x e y 0 D e 2x 2xe 2x 1 D ln x y D x 2 ln x ln.x=.2 x// is defined if x=.2 x/ > 0, that is, if 0 < x < 2. The domain is the interval .0; 2/. 15. 1 x ln x 0 3 D 2x x 2x 3 D 0 ) .x Hence; x D 1 or 3: y0 D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 47. 48. p y D ln. x 2 C a2 x/ x 1 p x 2 C a2 0 y D p x 2 C a2 x 1 D p x 2 C a2 53. x 2 f .x/ D .x x /x D x .x / 2 x tan x/ 1 sin x ln x C cos x x f 0 D x x C1 .2 ln x C 1/ x g.x/ D x x ln g D x x ln x xx 1 0 g D x x .1 C ln x/ ln x C 0 g x 1 0 xx x g Dx x C ln x C .ln x/2 x Evidently g grows more rapidly than does f as x grows large. b) " cos x a) ln f .x/ D x 2 ln x 1 0 f D 2x ln x C x f x cos x D e x ln cos x e .cos x/.ln x/ # 1 0 x ln cos x . sin x/ ln cos x C x y De cos x " # 1 .cos x/.ln x/ sin x ln x C cos x e x y D .cos x/x D .cos x/x .ln cos x 49. SECTION 3.3 (PAGE 183) ! f .x/ D xe ax f 0 .x/ D e ax .1 C ax/ 54. f 00 .x/ D e ax .2a C a2 x/ Given that x x f 000 .x/ D e ax .3a2 C a3 x/ :: : :: x: D a where a > 0, then ln a D x x f .n/ .x/ D e ax .nan 1 C an x/ Thus ln x D 50. Since :: x: ln x D a ln x: 1 ln a D ln a1=a , so x D a1=a . a d .ax 2 C bx C c/e x D .2ax C b/e x C .ax 2 C bx C c/e x dx D Œax 2 C .2a C b/x C .b C c/e x D ŒAx 2 C Bx C C e x : 2 55. x Thus, differentiating .ax C bx C c/e produces another function of the same type with different constants. Any number of differentiations will do likewise. 51. y D ex 2 y 0 D 2xe x 2 2 2 y 00 D 2e x C 4x 2 e x D 2.1 C 2x 2 /e x 2 2 2 y .4/ D 4.3 C 6x 2 /e x C 4.3x C 2x 3 /2xe x 52. f .x/ D ln.2x C 1/ 00 2 f .x/ D . 1/2 .2x C 1/ f .4/ .x/ D 2 2 2 2 .3Š/24 .2x C 1/ 4 56. f 0 .x/ D 2.2x C 1/ 1 f 000 .x/ D .2/23 .2x C 1/ 3 Thus, if n D 1; 2; 3; : : : we have f .n/ .x/ D . 1/n 1 .n 1/Š2n .2x C 1/ n . p 1 C x.1 x/1=3 .1 C 5x/4=5 1 ln F .x/ D 2 ln.1 C x/ C 13 ln.1 x/ 54 ln.1 C 5x/ 1 1 4 F 0 .x/ D F .x/ 2.1 C x/ 3.1 x/ .1 C 5x/ # " 1 1 4 23 1 1 0 4 D F .0/ D F .0/ D .1/ 2 3 1 2 3 6 F .x/ D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 4/ 2 y 000 D 2.4x/e x C 2.1 C 2x 2 /2xe x D 4.3x C 2x 3 /e x D 4.3 C 12x 2 C 4x 4 /e x f .x/ D .x 1/.x 2/.x 3/.x 4/ ln f .x/ D ln.x 1/ C ln.x 2/ C ln.x 3/ C ln.x 1 1 1 1 1 f 0 .x/ D C C C f .x/ x 1 x 2 x 3 x 4 1 1 1 1 C C C f 0 .x/ D f .x/ x 1 x 2 x 3 x 4 Downloaded by ted cage (sxnbyln180@questza.com) 87 lOMoARcPSD|6566483 www.konkur.in SECTION 3.3 (PAGE 183) 57. ADAMS and ESSEX: CALCULUS 9 .x 2 1/.x 2 2/.x 2 3/ .x 2 C 1/.x 2 C 2/.x 2 C 3/ 321 1 f .2/ D D ; f .1/ D 0 567 35 2 2 ln f .x/ D ln.x 1/ C ln.x 2/ C ln.x 2 61. f .x/ D 2 2 3/ xDa 2 ln.x C 1/ ln.x C 2/ ln.x C 3/ 1 2x 2x 2x 0 f .x/ D 2 C 2 C 2 f .x/ x 1 x 2 x 3 2x 2x 2x x2 C 1 x2 C 2 x2 C 3 1 1 1 C 2 C 2 f 0 .x/ D 2xf .x/ 2 x 1 x 2 x 3 1 1 1 x2 C 1 x2 C 2 x2 C 3 4 1 1 1 1 1 1 f 0 .2/ D C C 35 3 2 1 5 6 7 139 556 4 D D 35 105 3675 2 Since f .x/ D .x 1/g.x/ where g.1/ ¤ 0, then f 0 .x/ D 2xg.x/ C .x 2 1/g 0 .x/ and 1 . 1/. 2/ D . f 0 .1/ D 2g.1/ C 0 D 2 234 6 Therefore, a D 1 and line has slope e. The line has equation y D ex. y .a;e a / y D ex x Fig. 3.3-61 62. 2 58. Since y D x 2 e x , then y 0 D 2xe x 2 Let the point of tangency be .a; e a /. Tangent line has slope ˇ d x ˇˇ ea 0 D e ˇ D ea : a 0 dx ˇ ˇ ˇ 1 1 ˇ The slope of y D ln x at x D a is y 0 D ˇ D : The xˇ a xDa line from .0; 0/ to .a; ln a/ is tangent to y D ln x if ln a 0 1 D a 0 a 2 2 2x 3 e x D 2x.1 x/.1 C x/e x : The tangent is horizontal at .0; 0/ and i.e., if ln a D 1, or a D e. Thus, the line is y D 1 . ˙1; e y x . e .a; ln a/ 59. f .x/ D xe x f 0 .x/ D e x .1 x/, C.P. x D 1, f .1/ D f 0 .x/ > 0 if x < 1 (f increasing) f 0 .x/ < 0 if x > 1 (f decreasing) x 1 e y D ln x Fig. 3.3-62 y .1;1=e/ y D x e x 63. x Let the point of tangency be .a; 2a /. Slope of the tangent is ˇ d x ˇˇ 2a 0 D 2 ˇ D 2a ln 2: a 1 dx ˇ xDa 1 1 ; a D1C . Thus a 1 D ln 2 ln 2 So the slope is 2a ln 2 D 21C.1= ln 2/ ln 2 D 2e ln 2. 1 (Note: ln 21= ln 2 D ln 2 D 1 ) 21= ln 2 D e) ln 2 The tangent line has equation y D 2e ln 2.x 1/. Fig. 3.3-59 64. 1 D 4 then x D 14 and x y D ln 41 D ln 4. The tangent line of slope 4 is y D ln 4 C 4.x 14 /, i.e., y D 4x 1 ln 4. 60. Since y D ln x and y 0 D 88 Telegram: @uni_k The tangent line to y D ax which passes through the origin is tangent at the point .b; ab / where ˇ d x ˇˇ ab 0 D a ˇ D ab ln a: b 0 dx ˇ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) xDb lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 3.3 (PAGE 183) 1 D ln a, so ab D a1= ln a D e. The line y D x b will intersect y D ax provided the slope of this tangent e line does not exceed 1, i.e., provided 1, or e ln a 1. b Thus we need a e 1=e . Thus 70. y (a) If Aa CBb D 1 and Ba Ab D 0, then A D .b; ab / and B D Z y D ax x 1 x DxC y y y y xy 0 x D1 e xy .y C xy 0 / ln C e xy y x y2 1 At e; we have e 1 1 C ey 0 2 C e 2 .e e 3 y 0 / D 1 e 2 y 0 e e e 2 C 2e 2 y 0 C 1 e 2 y 0 D 1 e 2 y 0 . 1 Thus the slope is y 0 D . e2 b (b) If Aa CBb D 0 and Ba Ab D 1, then A D 2 a C b2 a and B D 2 . Thus a C b2 65. e xy ln 1 0 y y2 Z e ax sin bx dx 1 ae ax sin bx D 2 2 a Cb 66. xe y C y 2x D ln 2 ) e y C xe y y 0 C y 0 2 D 0: At .1; ln 2/, 2 C 2y 0 C y 0 2 D 0 ) y 0 D 0. Therefore, the tangent line is y D ln 2. 68. b . Thus a2 C b 2 f .x/ D Ax cos ln x C Bx sin ln x f 0 .x/ D A cos ln x A sin ln x C B sin ln x C B cos ln x D .A C B/ cos ln x C .B A/ sin ln x 1 If A D B D then f 0 .x/ D cos ln x. Z 2 1 1 Therefore cos ln x dx D x cos ln x C x sin ln x C C . 2 2 1 1 If B D , A D then f 0 .x/ D sin ln x. 2 Z 2 1 1 x cos ln x C C . Therefore sin ln x dx D x sin ln x 2 2 FA;B .x/ D Ae x cos x C Be x sin x d FA;B .x/ dx D Ae x cos x Ae x sin x C Be x sin x C Be x cos x D .A C B/e x cos x C .B A/e x sin x D FACB;B A .x/ 71. be ax cos bx C C: 1 1 d 1 1 C D ln C ln x D dx x 1=x x 2 x 1 1 C D 0: x x 1 C ln x D C (constant). Taking x D 1, we x 1 get C D ln 1 C ln 1 D 0. Thus ln D ln x. x x 1 1 ln D ln x D ln x C ln D ln x ln y: y y y Therefore ln 72. 73. r rx r 1 r r d Œln.x r / r ln x D D D 0. dx xr x x x Therefore ln.x r / r ln x D C (constant). Taking x D 1, we get C D ln 1 r ln 1 D 0 0 D 0. Thus ln.x r / D r ln x. 74. Let x > 0, and F .x/ be the area bounded by y D t 2 , the t -axis, t D 0 and t D x. For h > 0, F .x C h/ F .x/ is the shaded area in the following figure. y d FA;B .x/ D FACB;B A .x/ we have 69. Since dx 2 d d FA;B .x/ D FACB;B A .x/ D F2B; 2A .x/ a) dx 2 dx d3 x d3 d b) e cos x D F1;0 .x/ D F0; 2 .x/ 3 3 dx dx dx x x D F 2; 2 .x/ D 2e cos x 2e sin x Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k a a2 C b 2 e ax cos bx dx 1 ae ax cos bx C be ax sin bx C C: D 2 2 a Cb Fig. 3.3-64 67. d .Ae ax cos bx C Be ax sin bx/ dx D Aae ax cos bx Abe ax sin bx C Bae ax sin bx C Bbe ax cos bx D .Aa C Bb/e ax cos bx C .Ba Ab/e ax sin bx: Downloaded by ted cage (sxnbyln180@questza.com) y D t2 x xCh t Fig. 3.3-74 89 lOMoARcPSD|6566483 www.konkur.in SECTION 3.3 (PAGE 183) ADAMS and ESSEX: CALCULUS 9 c) The tangent to y D 1=t at t D 2 has slope equation is Comparing this area with that of the two rectangles, we see that hx 2 < F .x C h/ F .x/ < h.x C h/2 : Hence, the Newton quotient for F .x/ satisfies F .x C h/ x < h 2 F .x/ F .x C h/ h h!0C < .x C h/ : F .x/ yD 1 3 F .x C h/ h F .x/ h!0 F .x C h/ h F .x/ < x2; A2 D D x2: 1 1 t A2 2 0 3 1 C 4 2 1 2 4 1 C 9 3 x : 9 D 5 : 8 D 7 : 18 2. (page 191) x3 D 0 (exponential wins) x!1 e x ex D1 x!1 x 3 lim x 3 e x D lim x!1 3. 2e x 3 2 3e x 2 0 D D lim D2 x x!1 e C 5 x!1 1 C 5e x 1C0 4. 1 2=.xe x / 1 0 x 2e x D D1 D lim x x!1 x C 3e x!1 1 C 3=.xe x / 1C0 lim lim lim x ln x D 0 .power wins/ x!0C t 6. lim x!0C 2 b) If f .t / D 1=t , then f .t / D 1=t and f 00 .t / D 2=t 3 > 0 for t > 0. Thus f 0 .t / is an increasing function of t for t > 0, and so the graph of f .t / bends upward away from any of its tangent lines. (This kind of argument will be explored further in Chapter 5.) Telegram: @uni_k 2 3 lim x 3 e x D lim x!1 5. 3 Fig. 3.3-75 90 1. yD1=t 2 1 2 Section 3.4 Growth and Decay a) The shaded area A in part (i) of the figure is less than the area of the rectangle (actually a square) with base from t D 1 to t D 2 and height 1=1 D 1. Since ln 2 D A < 1, we have 2 < e 1 D e; i.e., e > 2. (i) (ii) y y A1 3/ or y D 1=9. Its 7 73 5 C D > 1. 8 18 72 1 Thus 3 > e D e. Combining this with the result of (a) we conclude that 2 < e < 3. F .0/ D C D 0, therefore F .x/ D 31 x 3 . For x D 2, the area of the region is F .2/ D 83 square units. A x : 4 e) ln 3 > A1 C A2 D F .x C h/ F .x/ d F .x/ D lim D x2: dx h h!0 Z Therefore F .x/ D x 2 dx D 31 x 3 C C . Since yD1=t or y D 1 The trapezoid bounded by x D 2, x D 3, y D 0, and y D .2=3/ .x=9/ has area Combining these two limits, we obtain 75. 1 .x 9 A1 D so similarly, lim 2/ d) The trapezoid bounded by x D 1, x D 2, y D 0, and y D 1 .x=4/ has area D x2: If h < 0 and 0 < x C h < x, then .x C h/2 < 1 .x 4 The tangent to y D 1=t at t D 3 has slope equation is 2 Letting h approach 0 from the right (by the Squeeze Theorem applied to one-sided limits) lim 1 2 yD 1=4. Its 7. ln x D x 1 lim x.ln jxj/2 D 0 x!0 8. .ln x/3 p D0 x!1 x lim Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) .power wins/ lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 3.4 (PAGE 191) 9. Let N.t / be the number of bacteria present after t hours. Then N.0/ D 100; N.1/ D 200: dN D kN we have N.t / D N.0/e k t D 100e k t . Since dt k Thus 200 D and k D ln 2. 100e 5 Finally, N D 100e .5=2/ ln 2 565:685. 2 There will be approximately 566 bacteria present after another 1 21 hours. 10. Let y.t / be the number of kg undissolved after t hours. Thus, y.0/ D 50 and y.5/ D 20. Since y 0 .t / D ky.t /, therefore y.t / D y.0/e k t D 50e k t . Then 20 D y.5/ D 50e 5k 107 D N.7/ D N.0/e 7k ) N.0/ D 107 e .7=3/ ln 3 770400: There were approximately 770,000 bacteria in the culture initially. (Note that we are approximating a discrete quantity (number of bacteria) by a continuous quantity N.t / in this exercise.) 15. 16. 17. $P invested at 4% compounded continuously grows to $P .e 0:04 /7 D $P e 0:28 in 7 years. This will be $10,000 if $P D $10; 000e 0:28 D $7; 557:84. 18. Let y.t / be the value of the investment after t years. Thus y.0/ D 1000 and y.5/ D 1500. Since y.t / D 1000e k t and 1500 D y.5/ D 1000e 5k , therefore, k D 51 ln 32 . a) Let t be the time such that y.t / D 2000, i.e., 1000e k t D 2000 5 ln 2 1 D 8:55: ) t D ln 2 D k ln. 32 / Hence, the doubling time for the investment is about 8.55 years. 1 1 ln 0:0004101: 1690 2 b) Let r% be the effective annual rate of interest; then r 1000.1 C / D y.1/ D 1000e k 100 )r D 100.e k 1/ D 100Œexp . 51 ln 32 / 1 a) P .100/ D 100e 100k 95:98, i.e., about 95.98% remains after 100 years. D 8:447: b) P .1000/ D 100e 1000k 66:36, i.e., about 66.36% remains after 1000 years. The effective annual rate of interest is about 8.45%. 19. 13. Let P .t / be the percentage of the initial amount remaining after t years. Then P .t / D 100e k t and 99:57 D P .1/ D 100e k . Thus k D ln.0:9957/: The half-life T satisfies 50 D P .T / D 100e kT , 1 ln.0:5/ so T D ln.0:5/ D 160:85. k ln.0:995/ The half-life is about 160.85 years. Since I 0 .t / D kI.t / ) I.t / D I.0/e k t D 40e k t ; 15 3 1 ln D 100 ln ; 15 D I.0:01/ D 40e 0:01k ) k D 0:01 40 8 thus, 100t 3 3 I.t / D 40 exp 100t ln : D 40 8 8 12. Let P .t / be the percentage remaining after t years. Thus P 0 .t / D kP .t / and P .t / D P .0/e k t D 100e k t . Then, 50 D P .1690/ D 100e 1690k ) k D Let the purchasing power of the dollar be P .t / cents after t years. Then P .0/ D 100 and P .t / D 100e k t . Now 91 D P .1/ D 100e k so k D ln.0:91/. If 25 D P .t / D 100k t then ln.0:25/ 1 14:7. t D ln.0:25/ D k ln.0:91/ The purchasing power will decrease to $0.25 in about 14.7 years. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 ln.1:1/. 14 Five days after birth, the baby weighs W .5/ D 4000e .5=14/ ln.1:1/ 4138:50 4139 grams. 1 1 5 ln.0:1/ ln D 12:56: k 10 ln.0:4/ Let P .t / be the percentage undecayed after t years. Thus P .0/ D 100; P .15/ D 70. dP D kP , we have P .t / D P .0/e k t D 100e k t : Since dt 1 Thus 70 D P .15/ D 100e 15k so k D ln.0:7/. 15 The half-life T satisfies if 50 D P .T / D 100e kT , so 15 ln.0:5/ 1 29:15. T D ln.0:5/ D k ln.0:7/ The half-life is about 29.15 years. Let W .t / be the weight t days after birth. Thus W .0/ D 4000 and W .t / D 4000e k t : Also 4400 D W .14/ D 4000e 14k , is k D Hence, 90% of the sugar will dissolved in about 12.56 hours. 11. Let N.t / be the number of bacteria in the culture t days after the culture was set up. Thus N.3/ D 3N.0/ and N.7/ D 10 106 . Since N.t / D N.0/e k t , we have 3N.0/ D N.3/ D N.0/e 3k ) k D 31 ln 3: ) k D 15 ln 25 : If 90% of the sugar is dissolved at time T then 5 D y.T / D 50e kT , so T D 14. Downloaded by ted cage (sxnbyln180@questza.com) 91 lOMoARcPSD|6566483 www.konkur.in SECTION 3.4 (PAGE 191) ADAMS and ESSEX: CALCULUS 9 20. Let i % be the effective rate, then ! an original investment of i in one year. Let r% be $A will grow to $A 1 C 100 the nominal rate per annum compounded n times per year, then an original investment of $A will grow to r $A 1 C 100n a) If u.x/ D a C bf .x/, then u0 .x/ D bf 0 .x/ D bŒa C bf .x/ D bu.x/. This equation for u is the equation of exponential growth/decay. Thus u.x/ D C1 e bx ; 1 C1 e bx f .x/ D b !n ! !12 r 9:5 D $A 1 C $A 1 C 100 1200 p 12 1:095 1 D 9:1098: )r D 1200 dy D a C by and y.0/ D y0 , then, from part (a), dx a a ; y0 D C e 0 : y D C e bx b b Thus C D y0 C .a=b/, and a bx a y D y0 C : e b b 24. The nominal rate of interest is about 9.1098%. Let x.t / be the number of rabbits on the island t years after they were introduced. Thus x.0/ D 1;000, x.3/ D 3;500, and x.7/ D 3;000. For t < 5 we have dx=dt D k1 x, so ÷ e 2k1 D 3:5 5=2 x.5/ D 1;000e 5k1 D 1;000 e 2k1 D 1;000.3:5/5=2 x.2/ D 1;000e 2k1 D 3;500 22;918: c) We will have x.t / D 21 .a=b/ if 1 e bt D 12 , that is, if e bt D 12 , or bt D ln.1=2/ D ln 2. The time required to attain half the limiting concentration is t D .ln 2/=b. For t > 5 we have dx=dt D k2 x, so that x.t / D x.5/e k2 .t 5/ 3;000 22;918 5=2 3;000 5=2 x.10/ D x.5/35k2 D x.5/ e 2k2 22;918 22;918 142: ÷ 25. e 2k2 so there are approximately 142 rabbits left after 10 years. 22. Let N.t / be the number of rats on the island t months after the initial population was released and before the first cull. Thus N.0/ D R and N.3/ D 2R. Since N.t / D Re k t , we have e 3k D 2, so e k D 21=3 . Hence N.5/ D Re 5k D 25=3 R. After the first 1,000 rats are killed the number remaining is 25=3 R 1;000. If this number is less than R, the number at the end of succeeding 5-year periods will decline. The minimum value of R for which this won’t happen must satisfy 25=3 R 1;000 D R, that is, R D 1;000=.25=3 1/ 459:8. Thus R D 460 rats should be brought to the island initially. dx D a bx.t /. dt This says that x.t / is increasing if it is less than a=b and decreasing if it is greater than a=b. Thus, the limiting concentration is a=b. a) The concentration x.t / satisfies b) The differential equation for x.t / resembles that of Exercise 21(b), except that y.x/ is replaced by x.t /, and b is replaced by b. Using the result of Exercise 21(b), we obtain, since x.0/ D 0, a a bt e C x.t / D x.0/ b b a D 1 e bt : b x.t / D x.0/e k1 t D 1;000e k1 t x.7/ D x.5/e 2k2 D 3;000 26. Let T .t / be the reading t minutes after the Thermometer is moved outdoors. Thus T .0/ D 72, T .1/ D 48. dT D k.T 20/. By Newton’s law of cooling, dt dV If V .t / D T .t / 20, then D kV , so dt kt kt V .t / D V .0/e D 52e . Also 28 D V .1/ D 52e k , so k D ln.7=13/. Thus V .5/ D 52e 5 ln.7=13/ 2:354. At t D 5 the thermometer reads about T .5/ D 20 C 2:354 D 22:35 ı C. Let T .t / be the temperature of the object t minutes after its temperature was 45 ı C. Thus T .0/ D 45 and dT D k.T C 5/. Let T .40/ D 20. Also dt u.t / D T .t / C 5, so u.0/ D 50, u.40/ D 25, and dT du D D k.T C 5/ D ku. Thus, dt dt u.t / D 50e k t ; 23. f 0 .x/ D a C bf .x/. 92 Telegram: @uni_k a : b b) If in one year, if compounding is performed n times per year. For i D 9:5 and n D 12, we have 21. a D C e bx 25 D u.40/ D 50e 40k ; 1 25 1 1 )k D ln D ln : 40 50 40 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 3.5 (PAGE 199) We wish to know t such that T .t / D 0, i.e., u.t / D 5, hence 5 D u.t / D 50e k t 5 40 ln 50 D 132:88 min: tD 1 ln 2 Hence, it will take about .132:88 more to cool to 0 ı C. 27. 30. 40/ D 92:88 minutes Let T .t / be the temperature of the body t minutes after it was 5 ı . Thus T .0/ D 5, T .4/ D 10. Room temperature = 20ı . dT By Newton’s law of cooling (warming) D k.T 20/. dt dV If V .t / D T .t / 20 then D kV , dt so V .t / D V .0/e k t D 15e k t . 2 1 . Also 10 D V .4/ D 15e 4k , so k D ln 4 3 kt If T .t / D 15ı , then 5 D V.t / D 15e 1 ln 1 1 3 D 4 10:838. so t D ln 2 k 3 ln 3 It will take a further 6.84 minutes to warm to 15 ı C. 31. Ly0 ; y0 C .L y0 /e k y2 D .L y1 /y0 y1 .L y0 / 2 D Assuming k and L are positive, but y0 is negative, we have t > 0. The solution is therefore valid on . 1; t /. The solution approaches 1 as t ! t . 32. Ly0 y0 C .L y0 /e 2k .L y2 /y0 y2 .L y0 / t!1 33. dy D ky 1 dt Since dy D dt k y L L 2 y : L 2 C kL ; 4 L 10; 000 D 7671 cases 1 C 49.9=49/3 1 C Me 3k LkMe 3k 3; 028 cases/week: y 0 .3/ D .1 C Me 3k /2 y.3/ D Section 3.5 The Inverse Trigonometric Functions (page 199) Assuming L ¤ 0, L D 29. The rate of growth of y in the logistic equation is L 1 C Me k t L 200 D y.0/ D 1CM L 1; 000 D y.1/ D 1 C Me k 10; 000 D lim y.t / D L y.t / D Thus 200.1 C M / D L D 10; 000, so M D 49. Also 1; 000.1 C 49e k / D L D 10; 000, so e k D 9=49 and k D ln.49=9/ 1:695. Now simplify: y0 y2 .L y1 /2 D y12 .L y0 /.L y2 / y0 y2 L2 2y1 y0 y2 LCy0 y12 y2 D y12 L2 y12 .y0 Cy2 /LCy0 y12 y2 y12 .y0 C y2 / 2y0 y1 y2 . y12 y0 y2 If y0 D 3, y1 D 5, y2 D 6, then 25.9/ 180 45 LD D 6:429. 25 18 7 1. 2. p 3 D 2 3 2 1 D cos 1 2 3 sin 1 4 3. tan 1 . 1/ D 4. sec 1 5. sin.sin 1 0:7/ D 0:7 p 2D 4 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Ly0 y0 C .L y0 /e k t of the logistic equation is valid on any interval containing t D 0 and not containing any point where the denominator is zero. The denominator is zero if y0 D .y0 L/e k t , that is, if y0 1 ln : t Dt D k y0 L Thus y1 .L y0 /e k D .L y1 /y0 , and y2 .L y0 /e 2k D .L y2 /y0 . Square the first equation and thus eliminate e k : The solution yD 28. By the solution given for the logistic equation, we have y1 D L dy is greatest when y D . dt 2 Ly0 The solution y D is valid on the y0 C .L y0 /e k t largest interval containing t D 0 on which the denominator does not vanish. If y0 > L then y0 C .L y0 /e k t D 0 if y0 1 ln . t D t D k y0 L Then the solution is valid on .t ; 1/. limt!t C y.t / D 1. thus Downloaded by ted cage (sxnbyln180@questza.com) 93 lOMoARcPSD|6566483 www.konkur.in SECTION 3.5 (PAGE 199) 6. q cos.sin 1 0:7/ D p D 7. sin2 . arcsin 0:7/ p 0:49 D 0:51 1 1 p 2 tan 1 tan 3/ D D tan 1 . 3 8. sin 1 .cos 40ı / D 90ı 9. 10. 11. ADAMS and ESSEX: CALCULUS 9 19. y0 D s 3 D p cos 1 .cos 40ı / D 50ı sin 1 sin. 0:2/ cos 1 sin. 0:2/ D 2 D C 0:2 2 sin cos 1 . 1 3/ q 1 q D 1 D p cos2 . arccos . 13 / p p 8 2 2 1 D D 9 3 3 D 1 D cos tan 1 2 14. 15. 16. sin.cos cos.sin 1 q x/ D 1 cos.tan 1 x/ D 24. sin x D p tan. arctan x/ D x ) sec. arctan x/ D p ) sin. arctan x/ D p 18. cos.sec 1 x/ D 94 Telegram: @uni_k 1 25. 1 C x2 b a 1 1 .x 2 a b/ a2 1 .x 1 C x2 x 26. f .x/ D x sin 1 x f .t / D t tan 1 t x 1 x2 : t 1 C t2 u D z 2 sec 1 .1 C z 2 / F .x/ D .1 C x 2 / tan 1 x y D sin 1 p 1 x2 1 1 D 2 x jxj p ) tan.sec 1 x/ D x 2 1 sgn x p x2 1 if x 1 p D x 2 1 if x 1 a x h 1 1 a 2 G.x/ D sin 1 x sin 1 .2x/ y0 D r 1 C x2 27. r (assuming) a > 0/: b/2 F 0 .x/ D 2x tan 1 x C 1 1 1 1 ) sin.sec 1 x/ D x a2 a : 1 C .ax C b/2 2z 2 sgn .z/ p D 2z sec 1 .1 C z 2 / C .1 C z 2 / z 2 C 2 sin.cos x/ cos.cos 1 x p 1 x2 .by # 13/ D x tan.cos 1 x/ D D p 1 x y0 D z 2 .2z/ du D 2z sec 1 .1 C z 2 / C p 2 dz .1 C z / .1 C z 2 /2 x2 1 1 D p sec.tan 1 x/ 1 C x2 ) cos. arctan x/ D p 17. x2 2Cx r 2 2 3 4x C 1/ f 0 .t / D tan 1 t C x2 sin 1 3 2 .4x 2 1 9 f 0 .x/ D sin 1 x C p cos2 .cos 1 x/ 1 2x y D cos 1 23. 2 3 1 1 1 21. 22. 12. tan.tan 1 200/ D 200 13. 2x y D tan 1 .ax C b/; 1 p x/ D 1 p D 1 20. y0 D 1 sec tan 1 2 2 1 D s D p5 1 1 C tan2 tan 1 2 1 y D sin 1 x .jxj > jaj/ a i a D p x2 jxj x 2 sin 1 .2x/ p 1 sin 1 x p 2 1 1 4x 2 2 sin 1 .2x/ p p 1 4x 2 sin 1 .2x/ 2 1 x 2 sin 1 x D 2 p p 1 x 2 1 4x 2 sin 1 .2x/ G 0 .x/ D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x2 a2 1 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 28. H.t / D sin 1 t sin t sin t 0 H .t / D D 29. 30. p SECTION 3.5 (PAGE 199) 35. 1 1 1 p .sin t / 1 t2 sin2 t t2 sin 1 t cos t csc t cot t sin 1 t f .x/ D .sin 1 x 2 /1=2 2x 1 f 0 .x/ D .sin 1 x 2 / 1=2 p 2 1 x4 x D p p 1 x 4 sin 1 x 2 a 1 y D cos p a2 C x 2 ! 1=2 " # a 2 a2 2 3=2 0 .a C x / .2x/ y D 1 a2 C x 2 2 asgn .x/ a2 C x 2 p x 31. y D a2 x 2 C a sin 1 a x a 1 0 y D p Cr 2 2 2 a a x x 1 a2 r a x a x D D p .a > 0/ aCx a2 x 2 p x 32. y D a cos 1 1 2ax x 2 .a > 0/ a " # 1=2 1 2a 2x x 2 0 p y D a 1 1 a a 2 2ax x 2 x D p 2ax x 2 2x x 33. tan 1 D 2 y y 1 2y 2xy 0 y 2 2xyy 0 D 2 2 y y4 4x 1C 2 y 4 4y 0 1 4 2y 0 D At .1; 2/ 2 4 16 2 8 4y 0 D 4 4y 0 ) y 0 D 1 2 At .1; 2/ the slope is 1 36. 37. D 34. If y D sin 1 x, then y 0 D p then p 1 x2 1 1 D 2 so that x D ˙ 1 of the two tangentplines are 3 and y D y D C2 x 3 2 x2 p . If the slope is 2 3 : Thus the equations 2 p 3 C2 xC . 3 2 1 d sin 1 x D p > 0 on . 1; 1/. dx 1 x2 1 Therefore, sin is increasing. 1 d tan 1 x D > 0 on . 1; 1/: dx 1 C x2 1 Therefore tan is increasing. d 1 < 0 on . 1; 1/. cos 1 x D p dx 1 x2 1 Therefore cos is decreasing. Since the domain of sec 1 consists of two disjoint intervals . 1; 1 and Œ1; 1/, the fact that the derivative of sec 1 is positive wherever defined does not imply that sec 1 is increasing over its whole domain, only that it is increasing on each of those intervals taken independently. In fact, sec 1 . 1/ D > 0 D sec 1 .1/ even though 1 < 1. d d 1 csc 1 x D sin 1 dx dx x 1 1 D r x2 1 1 x2 1 p D jxj x 2 1 y .1;=2/ x y D csc 1 x . 1; =2/ Fig. 3.5-37 38. cot 1 x D arctan .1=x/I d 1 1 cot 1 x D D 1 x2 dx 1C 2 x y =2 y D cot 1 x x =2 Fig. 3.5-38 Remark: the domain of cot 1 can be extended to include 0 by defining, say, cot 1 0 D =2. This will make cot 1 right-continuous (but not continuous) at x D 0. It is also possible to define cot 1 in such a way that it is continuous on the whole real line, but we would then lose the identity cot 1 x D tan 1 .1=x/, which we prefer to maintain for calculation purposes. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 1 C x2 Downloaded by ted cage (sxnbyln180@questza.com) 95 lOMoARcPSD|6566483 www.konkur.in SECTION 3.5 (PAGE 199) 39. ADAMS and ESSEX: CALCULUS 9 d d 1 .tan 1 x C cot 1 x/ D tan 1 x C tan 1 dx dx x 1 1 1 D 0 if x ¤ 0 D C 1 1 C x2 x2 1C 2 x Thus tan 1 x C cot 1 x D C1 (const. for x > 0) At x D 1 we have C D C1 4 4 for x > 0. Thus tan 1 x C cot 1 x D 2 1 1 Also tan x C cot x D C2 for .x < 0/. D C2 . At x D 1, we get 4 4 Thus tan 1 x C cot 1 x D for x < 0. 2 42. 1 d sin 1 .cos x/ D p . sin x/ dx 1 cos2 x n 1 if sin x > 0 D 1 if sin x < 0 sin 1 .cos x/ is continuous everywhere and differentiable everywhere except at x D n for integers n. y y D sin 1 .cos x/ =2 x 40. If g.x/ D tan.tan 1 x/ then 2 Fig. 3.5-42 1 sec .tan x/ 1 C x2 1 C x2 1 C Œtan.tan 1 x/2 D D 1: D 2 1Cx 1 C x2 g 0 .x/ D 43. If h.x/ D tan 1 .tan x/ then h is periodic with period , and sec2 x h0 .x/ D D1 1 C tan2 x tan 1 .tan x/ is continuous and differentiable everywhere except at x D .2n C 1/=2 for integers n. It is not defined at those points. y y D tan 1 .tan x/ =2 provided that x ¤ .k C 21 / where k is an integer. h.x/ is not defined at odd multiples of . 2 y d 1 tan 1 .tan x/ D .sec2 x/ D 1 except at odd dx 1 C tan2 x multiples of =2. y .=2;=2/ yDtan.tan yDtan 41. x Fig. 3.5-43 1 .tan x/ Fig. 3.5.40(b) 1 d cos 1 .cos x/ D p . sin x/ 2x dx 1 cos n 1 if sin x > 0 D 1 if sin x < 0 44. 1 cos .cos x/ is continuous everywhere and differentiable everywhere except at x D n for integers n. y y D cos 1 .cos x/ 1 d tan 1 .cot x/ D . csc2 x/ D dx 1 C cot2 x integer multiples of . x Fig. 3.5-41 Telegram: @uni_k 1 except at tan 1 .cot x/ is continuous and differentiable everywhere except at x D n for integers n. It is not defined at those points. y y D tan 1 .cot x/ =2 96 x 1 x/ x Fig. 3.5.40(a) Fig. 3.5-44 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 45. If jxj < 1 and y D tan 1 p and tan y D p x x2 1 SECTION 3.5 (PAGE 199) because both x and sec y are negative. Thus y D sec 1 x in this case also. , then y > 0 , x > 0 x x2 1 x2 D sec2 y D 1 C 1 x2 1 x2 sin2 y D 1 cos2 y D 1 .1 x 2 / D x 2 sin y D x: x Thus y D sin 1 x and sin 1 x D tan 1 p . 1 x2 An alternative method of proof involves showing that the derivative of the left side minus the right side is 0, and both sides are 0 at x D 0. p p 46. If x 1 and y D tan 1 x 2 1, then tan y D x 2 1 and sec y D x, so that y D secp1 x. If x 1 and y D tan 1 x 2 1, then 2 < y < 3 2 , so sec y < 0. Therefore p p tan y D tan. tan 1 x 2 1/ D x2 1 1 sec2 y D 1 C .x 2 sec y D x; 49. f 0 .x/ 0 on . 1; 1/ x 1 tan 1 x D C on . 1; 1/: Thus f .x/ D tan 1 xC1 Evaluate the limit as x ! 1: x! 1 Thus tan 1 50. x 1 xC1 Since f .x/ D x 1/ D x 2 cos2 y D 1 x x2 1 D 3 on . 1; 1/: 4 tan 1 .tan x/ then sec2 x D1 1 C tan2 x 1D0 .k C 21 / where k is an integer. Thus, f is constant on intervals not containing odd multiples of . 2 f .0/ D 0 but f ./ D 0 D . There is no contradiction here because f 0 is not defined, so f is not constant 2 on the interval containing 0 and . 51. sin 1 .sin x/ . x / 1 f 0 .x/ D 1 p cos x 1 sin2 x cos x D1 j cos xj 8 < 0 if <x< 2 2 D : 2 if < x < or < x < 2 2 Note: f is not differentiable at ˙ . 2 f .x/ D x 1 x2 y .;/ =2 =2 x y D f .x/ . ; / sec2 y D x 2 sec y D x; Fig. 3.5-51 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k tan 1 x D if x ¤ Thus sec y D x and y D sec 1p x. x2 1 If x 1 and y D C sin 1 , then 2 y < 3 2 x and sec y < 0. Therefore ! p p x2 1 x2 1 1 sin y D sin C sin D x x 2 f 0 .x/ D 1 because both x and sec y are negative. Thus y D sec 1 x in this case also. x , then y > 0 , x > 0 and 47. If y D sin 1 p 1 C x2 x sin y D p 1 C x2 1 x2 D cos2 y D 1 sin2 y D 1 2 1Cx 1 C x2 2 2 2 tan y D sec y 1 D 1 C x 1 D x2 tan y D x: x Thus y D tan 1 x and tan 1 x D sin 1 p . 1 C x2 p x2 1 48. If x 1 and y D sin 1 , then 0 y < 2 and x p x2 1 sin y D x 1 x2 1 D 2 cos2 y D 1 x2 x sec2 y D x 2 : 3 D 2 4 lim f .x/ D tan 1 1 Downloaded by ted cage (sxnbyln180@questza.com) 97 lOMoARcPSD|6566483 www.konkur.in SECTION 3.5 (PAGE 199) 52. ADAMS and ESSEX: CALCULUS 9 1 ) y D tan 1 x C C 1 C x2 y.0/ D C D 1 y0 D Thus; y D tan 1 x C 1: 8̂ 1 1 x < y0 D ) y D tan 1 C C 9 C x2 3 3 53. 1 :̂ y.3/ D 2 2 D tan 1 1 C C C D2 3 1 x Thus y D tan 1 C 2 . 3 3 12 1 54. y 0 D p ) y D sin 1 x C C 1 x2 y. 12 / D sin 1 . 12 / C C D 1 : ) CC D1)C D1 6 6 Thus; y D sin 1 x C 1 : 6 8 x 4 < 0 ) y D 4sin 1 C C y D p 55. 2 5 25 x : y.0/ D 0 0D0CC )C D0 x Thus y D 4sin 1 . 5 sinh.x ˙ y/ cosh.x ˙ y/ sinh x cosh y ˙ cosh x sinh y D cosh x cosh y ˙ sinh x sinh y tanh x ˙ tanh y D 1 ˙ tanh x tanh y 3. tanh.x ˙ y/ D 4. y D coth x D 12 ex C e x ex e x y D sech x D y 1 y y D coth x 1 y D sech x x 1 Fig. 3.6.4(a) y D csch x D 2 ex C e x x Fig. 3.6.4(b) 2 ex e x y Section 3.6 Hyperbolic Functions (page 205) 1. 2. y D csch x d 1 d sech x D dx dx cosh x 1 D sinh x D sech x tanh x cosh2 x d d 1 csch x D dx dx sinh x 1 D cosh x D csch x coth x sinh2 x d cosh d coth x D dx dx sinh x 1 sinh2 x cosh2 x D D csch 2 x D 2 sinh x sinh2 x cosh x cosh y C sinh x sinh y D 41 Œ.e x C e x /.e y C e y / C .e x e x /.e y e y / D 14 .2e xCy C 2e x y / D 12 .e xCy C e .xCy/ / D cosh.x C y/: sinh x cosh y C cosh x sinh y D 41 Œ.e x D 12 .e xCy e x /.e y C e y / C .e x C e x /.e y e .xCy/ / D sinh.x C y/: cosh.x y/ D coshŒx C . y/ D cosh x cosh. y/ C sinh x sinh. y/ D cosh x cosh y sinh x sinh y: sinh.x y/ D sinhŒx C . y/ D sinh x cosh. y/ C cosh x sinh. y/ D sinh x cosh y cosh x sinh y: 98 Telegram: @uni_k e y / x Fig. 3.6-4 5. x 1C p p 2 d d x C1 p sinh 1 x D ln.x C x 2 C 1/ D dx dx x C x2 C 1 1 D p x2 C 1 x 1C p p 2 d d 1 cosh 1 x D ln.x C x 2 1/ D px dx dx x C x2 1 1 D p x2 1 1Cx d 1 d tanh 1 x D ln dx dx 2 1 x 1 1 x 1 x .1 C x/. 1/ 1 D D 21Cx .1 x/2 1 x2 R dx D sinh 1 x C C p x2 C 1 R dx p D cosh 1 x C C .x > 1/ x2 1 R dx D tanh 1 x C C . 1 < x < 1/ 1 x2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 3.6 (PAGE 205) y 6. Let y D sinh 1 Thus, x a dy , x D a sinh y ) 1 D a.cosh y/ . dx y D coth 1 x x d 1 sinh 1 D dx a a cosh y 1 1 D q D p 2 C x2 2 a a 1 C sinh y Z x dx p D sinh 1 C C: .a > 0/ 2 2 a a Cx x , x D a Cosh y D a cosh y a dy for y 0, x a. We have 1 D a.sinh y/ . Thus, dx Let y D cosh 1 1 Fig. 3.6-8 9. x 1 d cosh 1 D dx a a sinh y 1 1 D q D p 2 2 x a2 a cosh y 1 Z x dx .a > 0; x a/ D cosh 1 C C: p a x 2 a2 Since sech 1 x D cosh 1 .1=x/ is defined in terms of the restricted function Cosh, its domain consists of the reciprocals of numbers in Œ1; 1/, and is therefore the interval .0; 1. The range of sech 1 is the domain of Cosh, that is, Œ0; 1/. Also, d sech dx x dy Let y D tanh 1 , x D a tanh y ) 1 D a.sech2 y/ . a dx Thus, x 1 d tanh 1 D dx a a sech2 y a a D D 2 2 2 tanh2 x a x2 a a Z 1 x dx D tanh 1 C C: a2 x 2 a a 1 1 1 x2 1 7. a) sinh ln x D .e ln x e ln x / D x D 2 2 x 2x 2 1 1 x C1 1 ln x xC D b) cosh ln x D .e C e ln x / D 2 2 x 2x sinh ln x x2 1 c) tanh ln x D D 2 cosh ln x x C1 x 2 C 1 C .x 2 1/ cosh ln x C sinh ln x D 2 D x2 d) cosh ln x sinh ln x .x C 1/ .x 2 1/ 1 xC1 1 8. The domain of coth x D ln consists of all x 2 x 1 xC1 > 0. Since satisfying jxj > 1. For such x, we have x 1 this fraction takes very large values for x close to 1 and values close to 0 for x close to 1, the range of coth 1 x consists of all real numbers except 0. 1 d 1 cosh 1 dx x 1 1 1 D p : D s 2 2 x x 1 x2 1 1 x xD y y D Sech 1 x 1 x Fig. 3.6-9 10. csch 1 has domain and range consisting of all real numbers x except x D 0. We have d 1 d csch 1 x D sinh 1 dx dx x 1 1 1 D s 2 x 2 D jxjpx 2 C 1 : 1 1C x d 1 d coth 1 x D tanh 1 dx dx x 1 1 1 D D 2 : 2 2 1 .1=x/ x x 1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x 1 Downloaded by ted cage (sxnbyln180@questza.com) y y D csch 1 x x Fig. 3.6-10 99 lOMoARcPSD|6566483 www.konkur.in SECTION 3.6 (PAGE 205) 11. ADAMS and ESSEX: CALCULUS 9 fA;B .x/ D Ae kx C Be kx 2. 0 fA;B .x/ D kAe kx kBe kx 00 fA;B .x/ D k 2 Ae kx C k 2 Be kx 00 Thus fA;B k 2 fA;B D 0 auxiliary eqn a/ C M sinh k.x Telegram: @uni_k 3 D 0 ) .2r C 1/.2r 1 2; ) rD 5; 2 2 3/ D 0 and y D Ae .1=2/t C Be .3=2/t : r 2 C 8r C 16 D 0 auxiliary eqn ) rD 4; 4 y D Ae 4t C Bt e 4t y 00 2y 0 C y D 0 y 00 auxiliary eqn a/ r 6y 0 C 10y D 0 2 6r C 10 D 0 y D Ae 8. 3t ) r D3˙i cos t C Be 3t sin t 9y 00 C 6y 0 C y D 0 9r 2 C 6r C 1 D 0 ) .3r C 1/2 D 0 Thus; r D auxiliary eqn y 00 C 7y 0 C 10y D 0 r2 D 23 ; y 00 C 8y 0 C 16y D 0 5. M /. Section 3.7 Second-Order Linear DEs with Constant Coefficients (page 212) ) r D 0; 1; r D 3 2t 3y D 0 9. k 2 y D 0 ) y D hL;M .x/ D L cosh k.x a/ C M sinh k.x y.a/ D y0 ) y0 D L C 0 ) L D y0 , v0 y 0 .a/ D v0 ) v0 D 0 C M k ) M D k Therefore y D hy0 ;v0 =k .x/ D y0 cosh k.x a/ C .v0 =k/ sinh k.x a/. 100 4r Thus; r1 D k 2 y D 0 and y 00 y D Ae 5t C Be 2t 4y 0 2 7. where A D 12 e ka .L C M / and B D 12 e ka .L auxiliary eqn r 2 C 7r C 10 D 0 .r C 5/.r C 2/ D 0 4y 00 4r a/ D Ae kx C Be kx D fA;B .x/ 1. r 2 C 2r D 0 ) rD r 2 2r C 1 D 0 ) .r 1/2 D 0 Thus; r D 1; 1; and y D Ae t C Bt e t : hL;M .x/ M L kx ka e C e kxCka C e kx ka e kxCka D 2 2 ! ! L ka M ka kx L ka M ka e C e e e D e C e kx 2 2 2 2 13. 4. a/ a/ C M k 2 sinh k.x hence, hL;M .x/ is a solution of y 00 3D0 y D A C Be 6. D k hL;M .x/ 2r 3y D 0 y 00 C 2y 0 D 0 auxiliary eqn 12. Since 2 2 r 3. 00 gC;D .x/ D k 2 C cosh kx C k 2 D sinh kx 00 k 2 gC;D D 0 Thus gC;D cosh kx C sinh kx D e kx cosh kx sinh kx D e kx Thus fA;B .x/ D .A C B/ cosh kx C .A B/ sinh kx, that is, fA;B .x/ D gACB;A B .x/, and c D gC;D .x/ D .e kx C e kx / C .e kx e kx /, 2 2 that is gC;D .x/ D f.C CD/=2;.C D/=2 .x/. 00 hL;M .x/ D Lk 2 cosh k.x 2y 0 y D Ae t C Be 3t gC;D .x/ D C cosh kx C D sinh kx 0 gC;D .x/ D kC cosh kx C kD sinh kx hL;M .x/ D L cosh k.x y 00 1 3; 1 3; and y D Ae .1=3/t C Bt e .1=3/t : y 00 C 2y 0 C 5y D 0 r 2 C 2r C 5 D 0 ) r D 1 ˙ 2i y D Ae t cos 2t C Be t sin 2t 10. For y 00 4y 0 C 5y D 0 the auxiliary equation is r 2 4r C 5 D 0, which has roots r D 2 ˙ i . Thus, the general solution of the DE is y D Ae 2t cos t C Be 2t sin t . 11. For y 00 C 2y 0 C 3y D 0 the auxiliary equation isp r 2 C 2r C 3 D 0, which has solutions r D 1 ˙ 2i . Thus the general solution of the givenpequation is p y D Ae t cos. 2t / C Be t sin. 2t /. 12. Given that y 00 C y 0 C y D 0, hence r 2 C r C 1 D 0. Since a D 1, b D 1 and c D 1, the discriminant is D D b 2 4ac D 3 < 0 and .b=2a/ D 21 and p ! D 3=2. Thus, the general solution is p p 3 3 t C Be .1=2/t sin t . y D Ae .1=2/t cos 2 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 3.7 (PAGE 212) 8 00 < 2y C 5y 0 3y D 0 13. y.0/ D 1 : 0 y .0/ D 0 The DE has auxiliary equation 2r 2 C 5y 3 D 0, with roots r D 12 and r D 3. Thus y D Ae t=2 C Be 3t . A Now 1 D y.0/ D A C B, and 0 D y 0 .0/ D 3B. 2 Thus B D 1=7 and A D 6=7. The solution is 6 1 y D e t=2 C e 3t . 7 7 14. 00 0 Given that y C 10y C 25y D 0, hence r 2 C 10r C 25 D 0 ) .r C 5/2 D 0 ) r D 5. Thus, 5e 5t .A C Bt / C Be 5t : Since 0 D y.1/ D Ae 5 C Be 5 2 D y 0 .1/ D 5e 5 .A C B/ C Be 5 ; 5 1/e 5.t 1/ . 8 00 < y C 4y 0 C 5y D 0 y.0/ D 2 : 0 y .0/ D 0 The auxiliary equation for the DE is r 2 C 4r C 5 D 0, which has roots r D 2 ˙ i . Thus y D Ae 2t cos t C Be 2t sin t y 0 D . 2Ae 2t C Be 2t / cos t Since b2 p ˙ b2 p b ˙ b2 .Ae 2t C 2Be 2t / sin t: Now 2 D y.0/ D A ) A D 2, and 2 D y 0 .0/ D 2A C B ) B D 6. Therefore y D e 2t .2 cos t C 6 sin t /. 4ac < b 4ac < 0 If t ! 1, then e r1 t ! 0 and e r2 t ! 0. Thus, lim y.t / D 0. t!1 Case 2: If D D b 2 4ac D 0 then the two equal roots r1 D r2 D b=.2a/ are negative. The general solution is y.t / D Ae r1 t C Bt e r2 t : If t ! 1, then e r1 t ! 0 and e r2 t ! 0 at a faster rate than Bt ! 1. Thus, lim y.t / D 0. t!1 Case 3: If D D b 2 e .1C/t e t lim y .t / D lim !0 !0 e tCh e t D t lim h h!0 d t Dt e D t et dt which is, along with e t , a solution of the CASE II DE y 00 2y 0 C y D 0. 4ac < 0 then the general solution is y D Ae .b=2a/t cos.!t / C Be .b=2a/t sin.!t / p 4ac b 2 . If t ! 1, then the amplitude of 2a .b=2a/t both terms Ae ! 0 and Be .b=2a/t ! 0. Thus, lim y.t / D 0. where ! D t!1 18. The auxiliary equation ar 2 C br C c D 0 has roots r1 D 16. The auxiliary equation r 2 .2 C /r C .1 C / factors to .r 1 /.r 1/ D 0 and so has roots r D 1 C and r D 1. Thus the DE y 00 .2 C /y 0 C .1 C /y D 0 has general solution y D Ae .1C/t C Be t . The function e .1C/t e t is of this form with A D B D 1=. y .t / D We have, substituting D h=t , b p 2a D ; r2 D p bC D ; 2a where D D p b 2 4ac. Note that a.r2 r1 / D D D .2ar1 C b/. If y D e r1 t u, then y 0 D e r1 t .u0 Cr1 u/, and y 00 D e r1 t .u00 C2r1 u0 Cr12 u/. Substituting these expressions into the DE ay 00 C by 0 C cy D 0, and simplifying, we obtain e r1 t .au00 C 2ar1 u0 C bu0 / D 0; or, more simply, u00 .r2 r1 /u0 D 0. Putting v D u0 reduces this equation to first order: v 0 D .r2 r1 /v; which has general solution v D C e .r2 r1 /t : Hence Z u D C e .r2 r1 /t dt D Be .r2 r1 /t C A; and y D e r1 t u D Ae r1 t C Be r2 t . Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 4ac < b 2 therefore r1 and r2 are negative. The general solution is 5 we have A D 2e and B D 2e . Thus, y D 2e 5 e 5t C 2t e 5 e 5t D 2.t 15. Given that a > 0, b > 0 and c > 0: Case 1: If D D b 2 4ac > 0 then the two roots are p b ˙ b 2 4ac : r1;2 D 2a y.t / D Ae r1 t C Be r2 t : y D Ae 5t C Bt e 5t y0 D 17. Downloaded by ted cage (sxnbyln180@questza.com) 101 lOMoARcPSD|6566483 www.konkur.in SECTION 3.7 (PAGE 212) ADAMS and ESSEX: CALCULUS 9 by the previous problem. Therefore ay 00 C by 0 C cy D 0 has general solution 19. If y D A cos !t C B sin !t then y 00 C ! 2 y D A! 2 cos !t B! 2 sin !t y D Ae k t cos.!t / C Be k t sin.!t /: 2 C ! .A cos !t C B sin !t / D 0 for all t . So y is a solution of (†). 20. If f .t / is any solution of .†/ then f 00 .t / D all t . Thus, ! 2 f .t / for 24. 2 2 i d h 2 ! f .t / C f 0 .t / dt D 2! 2 f .t /f 0 .t / C 2f 0 .t /f 00 .t / D 2! 2 f .t /f 0 .t / y 0 .0/ D If g.t / satisfies .†/ and also g.0/ D g 0 .0/ D 0, then by Exercise 20, 2 2 ! 2 g.t / C g 0 .t / 2 2 D ! 2 g.0/ C g 0 .0/ D 0: 25. Since a sum of squares cannot vanish unless each term vanishes, g.t / D 0 for all t . 22. If f .t / is any solution of .†/, let g.t / D f .t / A cos !t B sin !t where A D f .0/ and B! D f 0 .0/. Then g is also solution of .†/. Also g.0/ D f .0/ A D 0 and g 0 .0/ D f 0 .0/ B! D 0. Thus, g.t / D 0 for all t by Exercise 24, and therefore f .x/ D A cos !t C B sin !t . Thus, it is proved that every solution of .†/ is of this form. 26. 4ac b 2 b and ! 2 D which is 2a 4a2 kt positive for Case III. If y D e u, then 8 00 < y C 100y D 0 y.0/ D 0 : 0 y .0/ D 3 y D A cos.10t / C B sin.10t / A D y.0/ D 0; 10B D y 0 .0/ D 3 3 yD sin.10t / 10 y D A cos !.t c/ C B sin !.t c/ D A cos !t C B sin !t where A D A cos.!c/ B sin.!c/ and B D A sin.!c/ C B cos.!c/ 27. For y 00 C y D 0, we have y D A sin t C B cos t . Since, y.2/ D 3 D A sin 2 C B cos 2 y 0 .2/ D 4 D A cos 2 B sin 2; Substituting into ay 00 C by 0 C cy D 0 leads to 0 D e k t au00 C .2ka C b/u0 C .ak 2 C bk C c/u D e k t au00 C 0 C ..b 2 =.4a/ .b 2 =.2a/ C c/u D a e k t u00 C ! 2 u : 00 u D A cos.!t / C B sin.!t / Telegram: @uni_k therefore A D 3 sin 2 4 cos 2 B D 4 sin 2 C 3 cos 2: Thus, 2 Thus u satisfies u C ! u D 0, which has general solution 102 5 2: (easy tocalculate y 00 C ! 2 y D 0) y D A cos.!t / cos.!c/ C sin.!t / sin.!c/ C B sin.!t / cos.!c/ cos.!t / sin.!c/ D A cos.!c/ B sin.!c/ cos !t C A sin.!c/ C B cos.!c/ sin !t 23. We are given that k D y 0 D e k t u0 C ku y 00 D e k t u00 C 2ku0 C k 2 u : 5)B D Thus, y D 2 cos 2t 25 sin 2t . circular frequency = ! D 2, frequency = 1 ! D 0:318 2 2 period = D 3:14 !q amplitude = .2/2 C . 52 /2 ' 3:20 2! 2 f .t /f 0 .t / D 0 2 2 for all t . Thus, ! 2 f .t / C f 0 .t / is constant. (This can be interpreted as a conservation of energy statement.) 21. Because y 00 C 4y D 0, therefore y D A cos 2t C B sin 2t . Now y.0/ D 2 ) A D 2; y D .3 sin 2 D 3 cos.t Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 4 cos 2/ sin t C .4 sin 2 C 3 cos 2/ cos t 2/ 4 sin.t 2/: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 3.7 (PAGE 212) 8 < y 00 C ! 2 y D 0 28. y.a/ D A : 0 y .a/ D B B y D A cos !.t a/ C sin !.t a/ ! 29. From Example 9, the spring constant is k D 9 104 gm=sec2 . For a frequency of 10 Hz (i.e., a circular frequency ! D 20 rad=sec.), a mass m satisfying p k=m D 20 should be used. So, 32. y D e k t ŒA cosh !.t t0 /B sinh !.t A !.t t0 / A !.t t0 / e C e D ek t 2 2 B !.t t0 / B !.t t0 / e C e 2 2 where A1 D .A=2/e !t0 C .B=2/e !t0 and B1 D .A=2/e !t0 .B=2/e !t0 . Under the conditions of this problem we know that Rr D k ˙ ! are the two real roots of the auxiliary equation ar 2 C br C c D 0, so e .k˙!/t are independent solutions of ay 00 C by 0 C cy D 0, and our function y must also be a solution. Since it involves two arbitrary constants, it is a general solution. The motion is determined by 8 < y 00 C 400 2 y D 0 y.0/ D 1 : 0 y .0/ D 2 therefore, y D A cos 20 t C B sin 20 t and 1)AD 1 2 1 y 0 .0/ D 2 ) B D D : 20 10 33. t0 /B sin !.t t0 / D e k t ŒA cos !t cos !t0 C A sin !t sin !t0 C B sin !t cos !t0 B cos !t sin !t0 D e k t ŒA1 cos !t C B1 sin !t ; where A1 D A cos !t0 B sin !t0 and B1 D A sin !t0 C B cos !t0 . Under the conditions of this problem we know that e k t cos !t and e k t sin !t are independent solutions of ay 00 C by 0 C cy D 0, so our function y must also be a solution, and, since it involves two arbitrary constants, it is a general solution. 8 00 < y C 2y 0 C 5y D 0 y.3/ D 2 : 0 y .3/ D 0 The DE has auxiliary equation r 2 C 2r C 5 D 0 with roots r D 1 ˙ 2i . By the second previous problem, a general solution can be expressed in the form y D e t ŒA cos 2.t 3/ C B sin 2.t 3/ for which 1 Thus, y D cos 20 t C sin 20 t , with y in cm and t 10 in r second, gives the displacement at time t . The amplitude 1 2 is . 1/2 C . / 1:0005 cm. 10 k ! , !2 D (k = spring const, m = mass) 30. Frequency D 2 m Since the spring does not change, ! 2 m D k (constant) For m D 400 gm, ! D 2.24/ (frequency = 24 Hz) 4 2 .24/2 .400/ If m D 900 gm, then ! 2 D 900 2 24 2 D 32. so ! D 3 32 Thus frequency = D 16 Hz 2 4 2 .24/2 400 For m D 100 gm, ! D 100 ! so ! D 96 and frequency = D 48 Hz. 2 31. Using the addition identities for cosine and sine, y D e k t ŒA cos !.t y0 D e t ŒA cos 2.t 3/ C B sin 2.t 3/ C e t Œ 2A sin 2.t 3/ C 2B cos 2.t 3/: The initial conditions give 2 D y.3/ D e 3 A 0 D y 0 .3/ D Thus A D 2e 3 and B D solution e 3 .A C 2B/ A=2 D y D e 3 t Œ2 cos 2.t 34. 3/ e 3 . The IVP has sin 2.t 3/: 8 00 < y C 4y 0 C 3y D 0 y.3/ D 1 : 0 y .3/ D 0 The DE has auxiliary equation r 2 C 4r C 3 D 0 with roots r D 2 C 1 D 1 and r D 2 1 D 3 (i.e. k ˙ !, where k D 2 and ! D 1). By the second previous problem, a general solution can be expressed in the form y D e 2t ŒA cosh.t 3/ C B sinh.t 3/ for which y0 D 2e 2t ŒA cosh.t C e 2t ŒA sinh.t Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k t0 / D A1 e .kC!/t C B1 e .k !/t 9 104 k D D 22:8 gm: mD 400 2 400 2 y.0/ D Expanding the hyperbolic functions in terms of exponentials, Downloaded by ted cage (sxnbyln180@questza.com) 3/ C B sinh.t 3/ C B cosh.t 3/ 3/: 103 lOMoARcPSD|6566483 www.konkur.in SECTION 3.7 (PAGE 212) ADAMS and ESSEX: CALCULUS 9 The initial conditions give The conditions for stopping the motion are met at t D 2; the mass remains at rest thereafter. Thus 84 1 < 5 cos t C 5 if 0 t 2 1 x.t / D 5 cos t 5 if < t 2 :1 if t > 2 5 1 D y.3/ D e 6 A 0 D y 0 .3/ D e 6 . 2A C B/ Thus A D e 6 and B D 2A D 2e 6 . The IVP has solution y D e 6 2t Œcosh.t 3/ C 2 sinh.t 3/: Review Exercises 3 (page 213) 1. 35. Let u.x/ D c Also u0 .x/ D u00 .x/ D k 2 y.x/. Then u.0/ D c k 2 a. k 2 y 0 .x/, so u0 .0/ D k 2 b. We have k 2 y 00 .x/ D k2 c k 2 y.x/ D k 2 u.x/ This IVP for the equation of simple harmonic motion has solution u.x/ D .c k 2 a/ cos.kx/ xD0 2. kb sin.kx/ so that 1 y.x/ D 2 c u.x/ k c D 2 c .c k 2 a/ cos.kx/ C kb sin.kx/ k b c D 2 .1 cos.kx/ C a cos.kx/ C sin.kx/: k k 36. Since x 0 .0/ D 0 and x.0/ D 1 > 1=5, the motion will be governed by x 00 D x C .1=5/ until such time t > 0 when x 0 .t / D 0 again. Let u D x .1=5/. Then u00 D x 00 D .x 1=5/ D u, u.0/ D 4=5, and u0 .0/ D x 0 .0/ D 0. This simple harmonic motion initial-value problem has solution u.t / D .4=5/ cos t . Thus x.t / D .4=5/ cos t C .1=4/ and x 0 .t / D u0 .t / D .4=5/ sin t . These formulas remain valid until t D when x 0 .t / becomes 0 again. Note that x./ D .4=5/ C .1=5/ D .3=5/. 104 Telegram: @uni_k f .x/ D sec2 x tan x ) f 0 .x/ D 2 sec2 x tan2 x C sec4 x > 0 for x in . =2; =2/, so f is increasing and therefore one-to-one and invertible there. The domain of f 1 is . 1; 1/, the range of f . Since f .=4/ D 2, therefore f 1 .2/ D =4, and .f 3. 1 0 / .2/ D lim f .x/ D x!˙1 4. 5. 1 f 0 .f x lim x!˙1 e x 2 1 .2// D 1 1 D : f 0 .=4/ 8 D 0: 2 Observe f 0 .x/ D e x .1 2x 2 / is positive if x 2 < 1=2 2 and ispnegative p if x > 1=2. Thus f is increasing p on 2; 1= 2/ and is decreasing on . 1; 1= 2/ and . 1= p on .1= 2; 1/. p p The maxpand min values p of f are 1= 2e (at x D 1= 2) and 1= 2e (at x D 1= 2). 6. y D e x sin x, .0 x 2/ has a horizontal tangent where dy 0D D e x .cos x sin x/: dx This occurs if tan x D 1, p The pso x D =4 or x D 5=4. points are .=4; e =4 = 2/ and .5=4; e 5=4 = 2/. 7. If f 0 .x/ D x for all x, then Since x./ < .1=5/, the motion for t > will be governed by x 00 D x .1=5/ until such time t > when x 0 .t / D 0 again. Let v D x C .1=5/. Then v 00 D x 00 D .x C 1=5/ D v, v./ D .3=5/ C .1=5/ D .2=5/, and v 0 ./ D x 0 ./ D 0. Thius initial-value problem has solution v.t / D .2=5/ cos.t / D .2=5/ cos t , so that x.t / D .2=5/ cos t .1=5/ and x 0 .t / D .2=5/ sin t . These formulas remain valid for t until t D 2 when x 0 becomes 0 again. We have x.2/ D .2=5/ .1=5/ D 1=5 and x 0 .2/ D 0. f .x/ D 3x C x 3 ) f 0 .x/ D 3.1 C x 2 / > 0 for all x, so f is increasing and therefore one-to-one and invertible. Since f .0/ D 0, therefore f 1 .0/ D 0, and ˇ ˇ 1 1 1 d ˇ 1 .f /.x/ˇ D 0 D : D 0 ˇ dx f .f 1 .0// f .0/ 3 d f .x/ f 0 .x/ xf .x/ D D 0: 2 =2 2 x dx e e x =2 2 Thus f .x/=e x =2 D C (constant) for all x. Since f .2/ D 3, we have C D 3=e 2 and 2 2 f .x/ D .3=e 2 /e x =2 D 3e .x =2/ 2 . 8. Let the length, radius, and volume of the clay cylinder at time t be `, r, and V , respectively. Then V D r 2 `, and dr d` dV D 2 r` C r2 : dt dt dt Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES 3 (PAGE 213) Since d V =dt D 0 and d `=dt D k` for some constant k > 0, we have 2 r` dr D dt k r 2 `; ) dr D dt 13. kr : 2 y D aa C aa .1 C ln a/.x a) An investment of $P at r% compounded continuously grows to $P e rT =100 in T years. This will be $2P provided e rT =100 D 2, that is, rT D 100 ln 2. If T D 5, then r D 20 ln 2 13:86%. 14. b) Since the doubling time is T D 100 ln 2=r, we have If r D 13:863% and r D T 0:5%, then 100 ln 2 . 0:5/ 0:1803 years: 13:8632 The doubling time will increase by about 66 days. ˇ ah 1 a0Ch a0 d x ˇˇ 10. a) lim D lim D a ˇ D ln a. h h dx ˇ h!0 h!0 xD0 Putting h D 1=n, we get lim n a1=n 1 D ln a. n!1 11. ln 2 ln x D is satisfied if x D 2 or x D 4 (because x 2 ln 4 D 2 ln 2). 1 ln b DmD : b b Thus ln b D 1, and b D e. 15. b) Using the technique described in the exercise, we calculate 10 210 21=2 1 0:69338183 11 211 21=2 1 0:69326449 Thus ln 2 0:693. 2 2 d f .x/ D f 0 .x/ dx 2 ) 2f .x/f 0 .x/ D f 0 .x/ a) b) The line y D mx through the origin intersects the curve y D ln x at .b; ln b/ if m D .ln b/=b. The same line intersects y D ln x at a different point .x; ln x/ if .ln x/=x D m D .ln b/=b. This equation will have only one solution x D b if the line y D mx intersects the curve y D ln x only once, at x D b, that is, if the line is tangent to the curve at x D b. In this case m is the slope of y D ln x at x D b, so 100 ln 2 r: r2 dT T r D dr Let the rate be r%. The interest paid by account A is 1; 000.r=100/ D 10r. The interest paid by account B is 1; 000.e r=100 1/. This is $10 more than account A pays, so 1; 000.e r=100 If y D cos 1 x, then x D cos y and 0 y . Thus tan y D sgn x 2 D 2 1 cos x D 2 D 1 x2 1D p x2 1 x : x 2 /=x/. 1 D x p 2 cos 1 sin 1 x. If y D cot 1 x, then x D cot y and 0 < y < =2. Thus p p csc y D sgn x 1 C cot2 y D sgn x 1 C x 2 sgn x : sin y D p 1 C x2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 D sgn x r 1 x 1 .1=x/2 tan 1 1=x p 1 x 2 1: sgn xtan csc 1 x D sin 1 17. and e < e follows because ln is increasing. sec2 y Since cot x D 1= tan x, cot 1 x D tan 1 .1=x/. 12. If f .x/ D .ln x/=x, then f 0 .x/ D .1 ln x/=x 2 . Thus f 0 .x/ > 0 if ln x < 1 (i.e., x < e) and f 0 .x/ < 0 if ln x > 1 (i.e., x > e). Since f is increasing to the left of e and decreasing to the right, it has a maximum value f .e/ D 1=e at x D e. Thus, if x > 0 and x ¤ e, then ln. e / D e ln < D ln e D ln e ; p p Thus cos 1 x D tan 1 .. 1 ) f 0 .x/ D 0 or f 0 .x/ D 2f .x/: Since f .x/ is given to be nonconstant, we have f 0 .x/ D 2f .x/. Thus f .x/ D f .0/e 2x D e 2x . Putting x D we obtain .ln /= < 1=e. Thus 1/ D 10r C 10: A TI-85 solve routine gives r 13:8165%. 16. ln x 1 < : x e a/: This line passes through the origin if 0 D aa Œ1 a.1 C ln a/, that is, if .1 C ln a/a D 1. Observe that a D 1 solves this equation. Therefore the slope of the line is 11 .1 C ln 1/ D 1, and the line is y D x. That is, r is decreasing at a rate proportional to itself. 9. y D x x D e x ln x ) y 0 D x x .1 C ln x/. The tangent to y D x x at x D a has equation Downloaded by ted cage (sxnbyln180@questza.com) 105 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 3 (PAGE 213) ADAMS and ESSEX: CALCULUS 9 1 sgn x D sgn xsin 1 p : Thus cot 1 x D sin 1 p 1 C x2 1 C x2 1 csc 1 x D sin 1 . x 18. Let T .t / be the temperature of the milk t minutes after it is removed from the refrigerator. Let U.t / D T .t / 20. By Newton’s law, U 0 .t / D kU.t / 21. ex > 1 C x C g.t / D e t Now T .0/ D 5 ) U.0/ D 15 and T .12/ D 12 ) U.12/ D 8. Thus sD 2D 15e sk . Thus 1. U.t / D U.0/e k t : T .0/ D 96 ) U.0/ D 96 R T .20/ D 40 ) U.20/ D 40 R ) 40 R ) 60 ln y D x ln x ln.ln y/ D ln x C ln.ln x/ g.y/ ln.ln y/ x.ln x C ln.ln x// D lim lim x!1 y!1 ln y x ln x ln.ln x/ : D lim 1 C x!1 ln x R D .96 R D .96 R/e 10k R/e 20k : p Now ln x <p x for sufficiently large x, so ln.ln x/ < ln x for sufficiently large x. 1 ln.ln x/ < p Therefore, 0 < ! 0 as x ! 1, and ln x ln x so g.y/ ln.ln y/ D 1 C 0 D 1: lim y!1 ln y Thus 40 R 60 R 2 D e 20k D 96 R 96 R 2 .60 R/ D .96 R/.40 R/ 3600 120R C R2 D 3840 16R D 240 R D 15: 136R C R2 2. ı Room temperature is 15 . 20. Let f .x/ D e x 1 x. Then f .0/ D 0 and by the MVT, f .x/ f .x/ D x x f .0/ D f 0 .c/ D e c 0 1 for some c between 0 and x. If x > 0, then c > 0, and f 0 .c/ > 0. If x < 0, then c < 0, and f 0 .c/ < 0. In either case f .x/ D xf 0 .c/ > 0, which is what we were asked to show. 106 Telegram: @uni_k g.0/ D g 0 .c/ 0 a) .d=dx/x x D x x .1 C ln x/ > 0 if ln x > 1, that is, if x > e 1 . Thus x x is increasing on Œe 1 ; 1/. We have T .10/ D 60 ) U.10/ D 60 t kC1 .k C 1/Š b) Being increasing on Œe 1 ; 1/, f .x/ D x x is invertible on that interval. Let g D f 1 . If y D x x , then x D g.y/. Note that y ! 1 if and only if x ! 1. We have 19. Let R be the temperature of the room, Let T .t / be the temperature of the water t minutes after it is brought into the room. Let U.t / D T .t / R. Then ) Challenging Problems 3 (page 214) 12 D 26:46 min for the milk to U 0 .t / D kU.t / t2 2Š for some c in .0; x/. Since x and g 0 .c/ are both positive, so is g.x/. This completes the induction and shows the desired inequality holds for x > 0 for all positive integers k. ln.2=15/ ln.2=15/ D 12 38:46: k ln.8=15/ It will take another 38:46 warm up to 18ı . t g.x/ g.x/ D x x 15e 12k 1 ln.8=15/: k D 12 2, so 1 on the interval .0; x/ (where x > 0) to obtain 8 D U.12/ D U.0/e 12k D If T .s/ D 18, then U.s/ D sk D ln.2=15/, and x2 xk C C 2Š kŠ holds for all x > 0. This is certainly true for k D 1, as shown in the previous exercise. Apply the MVT to U.t / D U.0/e k t : ) e 12k D 8=15; Suppose that for some positive integer k, the inequality dv D dt g kv. a) Let u.t / D g kv.t /. Then and du D dt k dv D dt ku, u.t / D u.0/e k t D .g C kv0 /e k t 1 1 g C u.t / D g .g C kv0 /e k t : v.t / D k k b) lim t!1 v.t / D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) g=k lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL c) 3. CHALLENGING PROBLEMS 3 (PAGE 214) dy g g C kv0 k t D v.t / D C e ; y.0/ D y0 dt k k g C kv0 k t gt e CC y.t / D k k2 g C kv0 g C kv0 y0 D 0 C C ) C D y0 C k2 k2 gt g C kv0 kt y.t / D y0 1 e C k k2 dv D dt g C kv 2 (k > 0) a) Let u D 2t p gk. If v.t / D kv g 1 eu , then k 1 C eu g .1 C e u /. e u / .1 k .1 C e u /2 u 4ge D .1 C e u /2 .1 e u /2 gDg 1 .1 C e u /2 4ge u dv D D : .1 C e u /2 dt dv D dt 2 r r e u /e u p 2 gk 4. dy dp D e bt by . If p D e bt y, then dt dt dp p therefore transforms to The DE D kp 1 dt e bt M dy p D by C kpe bt 1 dt e bt M 2 ky y D .b C k/y D Ky 1 ; M L bCk M . This is a standard k Logistic equation with solution (as obtained in Section 3.4) given by Ly0 yD ; y0 C .L y0 /e Kt where K D b C k and L D where y0 D y.0/ D p.0/ D p0 . Converting this solution back in terms of the function p.t /, we obtain Lp0 e bt p0 C .L p0 /e .bCk/t .b C k/Mp0 : D bt p0 ke C .b C k/M kp0 e k t p.t / D p g 1 e 2t gk p . Thus v.t / D k 1 C e 2t gk p r r g e 2t gk 1 g b) lim v.t / D lim p D t!1 t!1 k e 2t gk C 1 k p r g 1 1 C e 2t gk c) If y.t / D y0 C t ln , then k k 2 y.0/ D y0 and p p r 1 2 gke 2t gk dy g D p dt k k 1 C e 2t gk p r g 1 e 2t gk p D v.t /: D k 1 C e 2t gk r Thus y.t / gives the height of the object at time t during its fall. Since p represents a percentage, we must have .b C k/M=k < 100. If k D 10, b D 1, M D 90, and p0 D 1, then bCk M D 99 < 100. The numerator of the final exk pression for p.t / given above is a constant. Therefore p.t / will be largest when the derivative of the denominator, f .t / D p0 ke bt C .b Ck/M is zero. Since f 0 .t / D 10e t 9; 800e 10t , this will happen at t D ln.980/=11. The value of p at this t is approximately 48.1. Thus the maximum percentage of potential clients who will adopt the technology is about 48.1%. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k kp0 e k t D 10e t C980e 10t Downloaded by ted cage (sxnbyln180@questza.com) 107 lOMoARcPSD|6566483 www.konkur.in SECTION 4.1 (PAGE 220) ADAMS and ESSEX: CALCULUS 9 CHAPTER 4. MORE APPLICATIONS OF DIFFERENTIATION 7. V D dV dr 4 3 r , so D 4 r 2 . 3 dt dt When r D 30 cm and d V =dt D 20 cm3 /s, we have Section 4.1 Related Rates 1. (page 220) 20 D 4.30/2 If the side and area of the square at time t are x and A, respectively, then A D x 2 , so dA dx D 2x : dt dt 20 1 dr D D : dt 3600 180 The radius is increasing at 1=.180/ cm/s. 8. The volume V of the ball is given by If x D 8 cm and dx=dt D 2 cm/min, then the area is increasing at rate dA=dt D 32 cm2 /min. 4 4 V D r3 D 3 3 2. As in Exercise 1, dA=dt D 2x dx=dt . If dA=dt D 2 ft2 /s and x D 8 ft, then dx=dt D 2=.16/. The side length is decreasing at 1/8 ft/s. D D When D D 6 cm, dD=dt D 4 cm/s, then A D r ) r D 9. ) 1 p 2 A dA : dt dr 1 p . The D dt 10 1 radius is decreasing at the rate p cm/min when the 10 area is 100 cm2 . When dA D dt dr D dt 2, and A D 100, 5. For A D r 2 , we have dA=dt D 2 r dr=dt . If dA=dt D 1=3 km2 /h, p then (a) dr=dt p D 1=.6 r/ km/h, or (b) dr=dt D 1=.6 A=/ D 1=.6 A/ km/h 6. Let the length, width, and area be l, w, and A at time t . Thus A D lw. dw dl dA Dl Cw dt dt dt When l D 16, w D 12, dl dl ) D dt dt The length is decreasing at 4 m/s. 108 Telegram: @uni_k :5 cm/h. At that time 9 28:3: 48 D 12 dS dx D 12x : dt dt If V D 64 cm3 and d V =dt D 2 cm3 /s, then x D 4 cm and dx=dt D 2=.3 16/ D 1=24 cm/s. Therefore, dS=dt D 12.4/.1=24/ D 2. The surface area is increasing at 2 cm2 /s. 10. Let V , r and h denote the volume, radius and height of the cylinder at time t . Thus, V D r 2 h and dV dr dh D 2 rh C r2 : dt dt dt If V D 60, dV dr D 2, r D 5, D 1, then dt dt 60 12 V D D r 2 25 5 dh 1 dr dV D 2 rh dt r 2 dt dt 1 12 22 D : 2 10 D 25 5 25 hD dA dw D 3, D 0, we have dt dt 0 D 16 3 C 12 3 D ; 6 The volume V , surface area S, and edge length x of a cube are related by V D x 3 and S D 6x 2 , so that dx dV D 3x 2 ; dt dt A D The volume is decreasing at about 28.3 cm3 /h. 4. Let A and r denote the area and radius of the circle. Then r 3 dV D .36/. 0:5/ D dt 2 dA D 40.4/ D 160. dt The area is increasing at 160 cm2 /s. 2 D 2 dV dD D D2 : dt 2 dt dr dA D 2 r : dt dt If r where D D 2r is the diameter of the ball. We have 3. Let the radius and area of the ripple t seconds after impact be r and A respectively. Then A D r 2 . We have dr 20 cm and dt dr dt 4 The height is decreasing at the rate Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 22 cm/min. 25 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 11. SECTION 4.1 (PAGE 220) Let the length, width, depth, and volume at time t be l, w, h and V respectively. Thus V D lwh, and dl dw dh dV D wh C lh C lw : dt dt dt dt If l D 6 cm, w =5cm, h= 4cm, dw D dt The pdistance from the origin is increasing at a rate of 2= 5. 16. From the figure, x 2 C k 2 D s 2 . Thus dl dh D D 1m/s, and dt dt x 2cm/s, then dV D 20 dt p When angle P CA D 45ı , x D k and s D 2k. The radar gun indicates p that ds=dt D 100 km/h. Thus dx=dt D 100 2k=k 141. The car is travelling at about 141 km/h. 48 C 30 D 2: The volume is increasing at a rate of 2 cm3 /s. A 12. Let the length, width and area at time t be x, y and A respectively. Thus A D xy and k dA dx D 5, D 10, x D 20, y D 16, then dt dt dy dy 5 D 20 C 16.10/ ) D dt dt Thus, the width is decreasing at 13. y D x 2 . Thus then 14. dy D dt P Fig. 4.1-16 2 and dx D dt 3, 4. 3/ D 12. y is increasing at rate 12. Since x 2 y 3 D 72, then 2xy 3 dx dy dy C 3x 2 y 2 D0) D dt dt dt 2y dx : 3x dt dx dy 8 D 2, then D . Hence, the dt dt 9 8 vertical velocity is units/s. 9 We have dx dy Cy D1 xy D t ) x dt dt dy dx y D tx 2 ) D x 2 C 2xt dt dt If x D 3, y D 2, 15. 17. We continue the notation of Exercise 16. If dx=dt p D 90 km/h, and p angle P CA D 30ı , p then s D 2k, x D 3k, and ds=dt D . 3k=2k/.90/ D 45 3 D 77:94. The radar gun will read about 78 km/h. 18. Let the distances x and y be as shown at time t . Thus dx dy x 2 C y 2 D 25 and 2x C 2y D 0. dt dt dx 1 4 dy If D and y D 3, then x D 4 and C 3 D 0 so dt 3 3 dt dy 4 D . dt 9 4 m/s. The top of the ladder is slipping down at a rate of 9 At t D 2 we have xy D 2, y D 2x 2 ) 2x 3 D 2 ) x D 1, y D 2. dy dx dx dy Thus C2 D 1, and 1 C 4 D . dt dt dt dt dx dy dx D1) D0) D 1 ). So 1 C 6 dt dt dt p Distance D from origin satisfies D D x 2 C y 2 . So 1 dy dx dD D p C 2y 2x dt dt dt 2 x2 C y2 1 2 D p 1.0/ C 2.1/ D p : 5 5 5 m y Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k s 31 : 4 31 m/s. 4 dy dx D 2x . If x D dt dt C x dA dy dx Dx Cy : dt dt dt If ds dx Ds : dt dt Downloaded by ted cage (sxnbyln180@questza.com) x 1/3 m/s Fig. 4.1-18 109 lOMoARcPSD|6566483 www.konkur.in SECTION 4.1 (PAGE 220) ADAMS and ESSEX: CALCULUS 9 We are given that dx=dt D 2 ft/s, so dy=dt D 24=13 ft/s when x D 12 ft. Now the similar triangles in the figure show that s sCy D ; 6 15 so that s D 2y=3. Hence ds=dt D 48=39. The woman’s shadow is changing at rate 48/39 ft/s when she is 12 ft from the point on the path nearest the lamppost. 19. Let x and y be the distances shown in the following figure. From similar triangles: x xCy 2y dx 2 dy D )xD ) D : 2 5 3 dt 3 dt Since dy D dt 1 , then 2 dx D dt 1 d and .x C y/ D 3 dt 1 2 1 D 3 5 : 6 21. Hence, the man’s shadow is decreasing at 31 m/s and the shadow of his head is moving towards the lamppost at a rate of 65 m/s. If dC =dt D 600 when x D 12; 000, then dx=dt D 100. The production is increasing at a rate of 100 tonnes per day. 22. 5 m 2 m y x2 C D 10; 000 C 3x C 8; 000 x dx dC D 3C : dt 4; 000 dt Let x, y be distances travelled by A and B from their positions at 1:00 pm in t hours. dx dy Thus D 16 km/h, D 20 km/h. dt dt Let s be the distance between A and B at time t . Thus s 2 D x 2 C .25 C y/2 x 2s Fig. 4.1-19 ds dx dy D 2x C 2.25 C y/ dt dt dt At 1:30 t D 21 we have x D 8, y D 10, p p s D 82 C 352 D 1289 so 20. p 1289 ds D 8 16 C 35 20 D 828 dt ds 828 D p 23:06. At 1:30, the ships are sepadt 1289 rating at about 23.06 km/h. 15 and 6 y 5 s x A pos. of A at 1:00 p.m. 16 km/h x 25 km s pos. of B at 1:00 p.m. Fig. 4.1-20 y Refer to the figure. s, y, and x are, respectively, the length of the woman’s shadow, the distances from the woman to the lamppost, and the distances from the woman to the point on the path nearest the lamppost. From one of triangles in the figure we have If x D 12, then y D 13. Moreover, 2y 110 Telegram: @uni_k B Fig. 4.1-22 23. y 2 D x 2 C 25: dy dx D 2x : dt dt 20 km/h Let and ! be the angles that the minute hand and hour hand made with the vertical t minutes after 3 o’clock. Then d D rad=min dt 30 d! D rad=min: dt 360 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL Since D 0 and ! D D t 30 SECTION 4.1 (PAGE 220) at t D 0, therefore 2 and ! D 25. tC : 360 2 V D 13 r 2 h D 31 h3 dV dh dh 1 dV D h2 ) D : dt dt dt h2 dt At the first time after 3 o’clock when the hands of the clock are together, i.e., D !, 1 dV 1 dV D and h D 3, then D . Hence, the dt 2 dt 18 1 m/min. height of the pile is increasing at 18 Let r, h, and V be the top radius, depth, and volume of 10 r and the water in the tank at time t . Then D h 8 1 2 25 3 V D r h D h . We have 3 3 16 If 180 tD tC )t D : ) 30 360 2 11 4 Thus, the hands will be together at 16 11 minutes after 3 o’clock. Let V , r and h be the volume, radius and height of the cone. Since h D r, therefore 26. 12 ! 9 25 2 dh dh 16 1 : D 3h ) D 10 3 16 dt dt 250h2 3 1 dh D . dt 250 1 m/min when The water level is rising at a rate of 250 depth is 4 m. When h D 4 m, we have 6 Fig. 4.1-23 24. Let y be the height of balloon t seconds after release. Then y D 5t m. Let be angle of elevation at B of balloon at time t . Then tan D y=100. Thus 10 m r 1 dy 5 1 d D D D sec dt 100 dt 100 20 d 1 1 C tan2 D dt 20 y 2 d 1 D : 1C 100 dt 20 2 d 1 d 1 . D so D dt 20 dt 100 The angle of elevation of balloon at B is increasing at a 1 rad/s. rate of 100 When y D 200 we have 5 8 m h Fig. 4.1-26 27. Let r and h be the radius and height of the water in the tank at time t . By similar triangles, 10 5 r D ) r D h: h 8 4 The volume of water in the tank at time t is V D 1 2 25 3 r h D h : 3 48 Thus, y dV 25 2 dh dh 16 d V D h ) D : dt 16 dt dt 25h2 dt If B 100 m Fig. 4.1-24 A h3 and h D 4, then 1000 16 43 9 1 dh D : D dt .25/.4/2 10 1000 6250 dV 1 D dt 10 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 111 lOMoARcPSD|6566483 www.konkur.in SECTION 4.1 (PAGE 220) ADAMS and ESSEX: CALCULUS 9 9 m/min Hence, the depth of water is increasing at 6250 when the water is 4 m deep. The maximum depth occurs dh when D 0, i.e., dt 16 25h2 1 10 h3 1000 x 10 m/min 30 m h3 D0 1000 p 3 ) h D 100: D0) 1 10 Thus, the maximum depth the water in the tank can get is p 3 100 4:64 m. Fig. 4.1-29 30. 28. Let r, h, and V be the top radius, depth, and volume of the water in the tank at time t . Then Let P , x, and y be your position, height above centre, and horizontal distance from centre at time t . Let be the angle shown. Then y D 10 sin , and x D 10 cos . We have dy d D 10 cos ; dt dt 3 1 r D D h 9 3 1 3 V D r 2h D h 3 27 dh dV D h2 : dt 9 dt If s d D 1 rpm D 2 rad/min: dt 6 dy 6 When x D 6, then cos D , so D 10 12. 10 dt 10 You are rising or falling at a rate of 12 m/min at the time in question. 2 dh D 20 cm/h D m/h when h D 6 m, then dt 10 P dV 2 4 D 36 D 2:51 m3 /h. dt 9 10 5 10 m y Since water is coming in at a rate of 10 m3 /h, it must be leaking out at a rate of 10 2:51 7:49 m3 /h. C x 3 m r Fig. 4.1-30 9 m 31. h Let x and y denote the distances of the two aircraft east and north of the airport respectively at time t as shown in the following diagram. Also let the distance between the two aircraft be s, then s 2 D x 2 C y 2 . Thus, 2s Fig. 4.1-28 29. Let x and s be the distance as shown. Then s 2 D x 2 C302 and dx ds x dx ds D 2x ) D : 2s dt dt dt s dt When x D 40, p dx D 10, s D 402 C 302 D 50, then dt ds 40 D .10/ D 8. Hence, one must let out line at 8 dt 50 m/min. 112 Telegram: @uni_k dx dy ds D 2x C 2y : dt dt dt dx dy D 200 and D 150 when x D 144 and dt p dt 2 y D 60, we have s D 144 C 602 D 156, and Since 1 ds D Œ144. 200/ C 60.150/ dt 156 126:9: Thus, the distance between the aircraft is decreasing at about 126.9 km/h. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.1 (PAGE 220) 150 km/h 34. s Let x and y be the distances travelled from the intersection point by the boat and car respectively in t minutes. Then y 1000 1000 dx D 20 D m/min dt 60 3 dy 1000 4000 D 80 D m/min dt 60 3 The distance s between the boat and car satisfy x airport 200 km/h Fig. 4.1-31 s 2 D x 2 C y 2 C 202 ; 32. 1 0:6 0:4 x y 3 dP 0:6 0:4 0:4 dx 0:4 0:6 0:6 dy D x y C x y : dt 3 dt 3 dt P D After one minute, x D Thus 1374:5 If dP =dt D 0, x D 40, dx=dt D 1, and y D 10; 000, then dy D dt 6y 0:4 y 0:6 dx D x 0:4 4x 0:6 dt 6y dx D 4x dt Hence 375: s ds dx dy Dx Cy : dt dt dt 4000 1000 ,yD so s 1374: m. 3 3 1000 1000 4000 4000 ds D C 1; 888; 889: dt 3 3 3 3 ds 1374:2 m/min 82:45 km/h after 1 minute. dt The daily expenses are decreasing at $375 per day. y 20 m 33. Let the position of the ant be .x; y/ and the position of its shadow be .0; s/. By similar triangles, Car x s s y x D y )sD 3 x 3.3 ds D dt x/ 3y 3 : x Boat Then, If the ant is at .1; 2/ and 3.2/. ds D dt dy dx C 3y dt dt : .3 x/2 1 dy dx D , D dt 3 dt 1 , then 4 1 1 4 / C 3.2/. 3 / 1 : 8 4 D Hence, the ant’s shadow is moving at 18 units/s upwards along the y-axis. y S Fig. 4.1-34 35. Let h and b (measured in metres) be the depth and the surface width of the water in the trough at time t . We have p 2 h D tan 60ı D 3 ) b D p h: 1 . 2 b/ 3 Thus, the volume of the water is 10 1 hb .10/ D p h2 ; V D 2 3 and y ant p dV 20 dh dh 3 dV D p h ) D : dt dt dt 20h dt 3 dV 1 D and h D 0:2 metres, then dt 4 p p 3 3 1 dh D : D dt 20.0:2/ 4 16 p 3 m/min. Hence, the water level is rising at 16 If x Fig. 4.1-33 3 lamp x Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 113 lOMoARcPSD|6566483 www.konkur.in SECTION 4.1 (PAGE 220) ADAMS and ESSEX: CALCULUS 9 ı 60 b=2 10 m b=2 30 cm p y h ı 30 3 m 32 Cx 2 x s Fig. 4.1-37 Fig. 4.1-35 a) By similar triangles, Thus, 36. Let V and h be the volume and depth of water in the pool at time t . If h 2, then x 20 D D 10; h 2 so V D 1D m/min. If x D 4 and 2/. 1 D 5 24 : 125 Hence, the free top end of the ladder is moving vertically downward at 24/125 m/s. 1 160 b) By similar triangles, p Then, dh dh dV D 80h D 80 . dt dt dt 1 m/min. So surface of water is dropping at a rate of 80 b) If h D 1m, then 1 dx D , then dt 5 dy 30.4/ D dt .9 C 16/3=2 dV dh D 160 . dt dt So surface of water is dropping at a rate of a) If h D 2:5m, then dy dx dy dx 30x D D : dt dx dt .9 C x 2 /3=2 dt 1 xh8 D 40h2 : 2 If 2 h 3, then V D 160 C 160.h y 3 30 D p )yD p : 2 2 10 3 Cx 9 C x2 1D D s 10x : )sD p 10 9 C x2 ds ds dx D dt dx dt p 2x 2 . 9 C x /.10/ .10x/ p 2 9 C x 2 dx D .9 C x 2 / dt 90 dx D : .9 C x 2 /3=2 dt 20 8 3 1 x 32 C x 2 If x D 4 and x h 1 dx D , then dt 5 ds 90 D dt .9 C 16/3=2 1 18 : D 5 125 This is the rate of change of the length of the horizontal projection of the ladder. The free top end of the ladder is moving horizontally to the right at rate dx dt Fig. 4.1-36 38. 37. Let the various distances be as shown in the figure. 114 Telegram: @uni_k ds 1 D dt 5 18 7 D m/s: 125 125 Let x, y, and s be distances shown at time t . Then s 2 D x 2 C 16; ds dx s Dx ; dt dt Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) s/2 D y 2 C 16 ds dy .15 s/ Dy : dt dt .15 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.1 (PAGE 220) 1 dx D , then s D 5 and When x D 3 and dt 2 p p 2 D y D 102 4 84. 3 1 3 ds D D so Also dt 5 2 10 dy D dt 10 3 p D 84 10 40. Let y be height of ball t seconds after it drops. dy d 2y D 9:8, j tD0 D 0, yj tD0 D 20, and Thus dt 2 dt 3 p 0:327: 84 Crate B is moving toward Q at a rate of 0.327 m/s. dy D dt 4:9t 2 C 20; yD 9:8t: Let s be distance of shadow of ball from base of pole. s 10 s By similar triangles, D . y 20 200 20s 200 D sy, s D 20 y ds ds dy 20 Dy Cs . dt dt dt P 4 15 s B y a) At t D 1, we have s x A Fig. 4.1-38 39. Let be the angle of elevation, and x and y the horizontal and vertical distances from the launch site. We have dx dy y x d sec2 D dt 2 dt : dt x ) At the instant in question b) As the r ball hits the ground, y Dr0, s D 10, 20 20 dy t D , and D 9:8 , so 4:9 dt 4:9 dy ds D 0 C 10 . 20 dt dt r 20 Now y D 0 implies that t D . Thus 4:9 100 D 2, sec2 D 1 C tan2 D 5, and 50 p p d 1 2 3 1 50.2/ 100.2 3/ D D dt 5 .50/2 125 r 1 20 .9:8/ 2 4:9 ds D dt 9:90: The shadow is moving at about 9.90 m/s when the ball hits the ground. 10 m p dx dy D 4 cos 30ı D 2 3; D 4 sin 30ı D 2; dt dt x D 50 km; y D 100 km: Thus tan D 9:8, y D 15:1, ds 200 4:9 D . 9:8/. dt 4:9 The shadow is moving at a rate of 81.63 m/s after one second. Q y tan D x dy D dt 20 y 20 m y 0:0197: 10 s 10 s Therefore, the angle of elevation is decreasing at about 0.0197 rad/s. Fig. 4.1-40 4 km/s ı 30 y 41. Let y.t / be the height of the rocket t seconds after it blasts off. We have d 2y D 10; dt 2 x Fig. 4.1-39 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) dy DyD0 dt 115 lOMoARcPSD|6566483 www.konkur.in SECTION 4.1 (PAGE 220) ADAMS and ESSEX: CALCULUS 9 at t D 0. Hence y D 5t 2 , (y in metres, t in seconds). Now y d dy=dt tan D , so sec2 D , and 2000 dt 2000 y 2 d 10t t 1C D D 2000 dt 2000 200 t 1 d D dt 200 25t 4 1C 20002 t 800t 1 D D : 200 4002 C t 4 t4 1C 4002 At t D 10, we have d 8000 0:047 rad/s. D dt 4002 C 1002 Starting with x0 D 1:5, get x4 D x5 D 1:73205080757. 9. g.x/ D x 10. f .x/ D x 3 C 2x 2 2, f 0 .x/ D 3x 2 C 4x. Newton’s formula xnC1 D g.xn /, where Starting with x0 D 1:5, get x5 D x6 D 0:839286755214. 11. f .x/ D x 4 8x 2 x C 16, f 0 .x/ D 4x 3 Newton’s formula xnC1 D g.xn /, where g.x/ D x Fig. 4.1-41 Section 4.2 Finding Roots of Equations (page 230) x 4 8x 2 x C 16 3x 4 8x 2 16 D : 3 4x 16x 1 4x 3 16x 1 Because f . 3/ D 1, f . 2/ D 3, f . 1/ D 1, f .0/ D 1, f .1/ D 3, there are roots between 3 and 2, between 1 and 0, and between 0 and 1. Starting with x0 D 2:5, get x5 D x6 D 2:87938524157. Starting with x0 D 0:5, get x4 D x5 D 0:652703644666. Starting with x0 D 0:5, get x4 D x5 D 0:532088886328. 2. To solve 1 C 14 sin x D x, start with x0 D 1 and iterate xnC1 D 1 C 41 sin xn . x5 and x6 round to 1.23613. 3. To solve cos.x=3/ D x, start with x0 D 0:9 and iterate xnC1 D cos.xn =3/. x4 and x5 round to 0.95025. 4. To solve .x C 9/1=3 D x, start with x0 D 2 and iterate xnC1 D .xn C 9/1=3 . x4 and x5 round to 2.24004. 13. f .x/ D sin x 1 C x, f 0 .x/ D cos x C 1. Newton’s formula is xnC1 D g.xn /, where 5. To solve 1=.2 C x 2 / D x, start with x0 D 0:5 and iterate xnC1 D 1=.2 C xn2 /. x6 and x7 round to 0.45340. 6. To solve x 3 C 10x 10 D 0, start with x0 D 1 and iterate 1 3 xn . x7 and x8 round to 0.92170. xnC1 D 1 10 f .x/ D x 2 2, f 0 .x/ D 2x. Newton’s formula xnC1 D g.xn /, where 2x 3 C 3x 2 C 1 x 3 C 3x 2 1 D : 2 3x C 6x 3x 2 C 6x g.x/ D x 1 xn e starting with x C 0 D 0:3. Both x10 2 and x11 round to 0.35173. g.x/ D x x2 2 x2 C 2 D : 2x 2x 8. f .x/ D x 2 3, f 0 .x/ D 2x. Newton’s formula xnC1 D g.xn /, where 116 Telegram: @uni_k x 2x 3 sin x 1 C x : cos x C 1 The graphs of sin x and 1 x suggest a root near x D 0:5. Starting with x0 D 0:5, get x3 D x4 D 0:510973429389. y yD1 x y D sin x Starting with x0 D 1:5, get x3 D x4 D 1:41421356237. g.x/ D x 0.5 1.0 2 D 1. 12. f .x/ D x 3 C 3x 2 1, f 0 .x/ D 3x 2 C 6x. Newton’s formula xnC1 D g.xn /, where Iterate xnC1 D 2 16x Starting with x0 D 1:5, get x4 D x5 D 1:64809536561. Starting with x0 D 2:5, get x5 D x6 D 2:35239264766. 2 km g.x/ D x x 3 C 2x 2 2 2x 3 C 2x 2 C 2 D : 2 3x C 4x 3x 2 C 4x g.x/ D x 7. 2x 3 C 1 x 3 C 2x 1 D : 2 3x C 2 3x 2 C 2 Starting with x0 D 0:5, get x3 D x4 D 0:45339765152. y 1. f .x/ D x 3 C 2x 1, f 0 .x/ D 3x 2 C 2. Newton’s formula xnC1 D g.xn /, where x C3 : 2x Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) Fig. 4.2-13 1.5x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 14. SECTION 4.2 (PAGE 230) f .x/ D x 2 cos x, f 0 .x/ D 2x C sin x. Newton’s formula is xnC1 D g.xn /, where g.x/ D x has a root near x D 0:6. Use of a solve routine or Newton’s Method gives x D 0:56984029099806. x 2 cos x : 2x C sin x sin x . Since jf .x/j 1=.1 C x 2 / ! 0 1 C x2 as x ! ˙1 and f .0/ D 0, the maximum and minimum values of f will occur at the two critical points of f that are closest to the origin on the right and left, respectively. For CP: 18. Let f .x/ D The graphs of cos x and x 2 , suggest a root near x D ˙0:8. Starting with x0 D 0:8, get x3 D x4 D 0:824132312303. The other root is the negative of this one, because cos x and x 2 are both even functions. y .1 C x 2 / cos x 2x sin x .1 C x 2 /2 2 0 D .1 C x / cos x 2x sin x 0 D f 0 .x/ D y D x2 y D cos x 19. -1.5 -1.0 -0.5 0.5 1.0 1.5x Fig. 4.2-14 15. Since tan x takes all real values between any two consecutive odd multiples of =2, its graph intersects y D x infinitely often. Thus, tan x D x has infinitely many solutions. The one between =2 and 3=2 is close to 3=2, so start with x0 D 4:5. Newton’s formula here is xnC1 D xn with 0 < x < for the maximum and < x < 0 for the minimum. Solving this equation using a solve routine or Newton’s Method starting, say, with x0 D 1:5, we get x D ˙0:79801699184239: The corresponding max and min values of f are ˙0:437414158279. cos x Let f .x/ D . Note that f is an even function, and 1 C x2 that f has maximum value 1 at x D 0. (Clearly f .0/ D 1 and jf .x/j < 1 if x ¤ 0.) The minimum value will occur at the critical points closest to but not equal to 0. For CP: .1 C x 2 /. sin x/ 2x cos x .1 C x 2 /2 2 0 D .1 C x / sin x C 2x cos x: 0 D f 0 .x/ D The first CP to the right of zero is between =2 and 3=2, so start with x D 2:5, say, and get x D 2:5437321475261. The minimum value is f .x/ D 0:110639672192. tan xn xn : sec2 xn 1 We get x3 D x4 D 4:49340945791. y 20. For x 2 D 0 we have xnC1 D xn .xn2 =.2xn // D xn =2. If x0 D 1, then x1 D 1=2, x2 D 1=4, x3 D 1=8. a) xn D 1=2n , by induction. b) xn approximates the root x D 0 to within 0.0001 provided 2n > 10; 000. We need n 14 to ensure this. yDx c) To ensure that xn2 is within 0.0001 of 0 we need .1=2n /2 < 0:0001, that is, 22n > 10; 000. We need n 7. x y D tan x 21. Fig. 4.2-15 d) Convergence of Newton approximations to the root x D 0 of x 2 D 0 is slower than usual because the derivative 2x of x 2 is zero at the root. p x if x 0 , f .x/ D p x if x < 0 p x/ if x > 0 1=.2 p f 0 .x/ D . 1=.2 x/ if x < 0 The Newton’s Method formula says that 16. A graphing calculator shows that the equation p .1 C x 2 / x 1D0 xnC1 D xn Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) f .xn / D xn f 0 .xn / 2xn D xn : 117 lOMoARcPSD|6566483 www.konkur.in SECTION 4.2 (PAGE 230) ADAMS and ESSEX: CALCULUS 9 If x0 D a, then x1 D a, x2 D a, and, in general, xn D . 1/n a. The approximations oscillate back and forth between two numbers. If one observed that successive approximations were oscillating back and forth between two values a and b, one should try their average, .aCb/=2, as a new starting guess. It may even turn out to be the root! 22. Newton’s Method formula for f .x/ D x 1=3 is Hence Since the values of xn are assumed to neither converge nor diverge, the exponential factor 2n will dominate for large n 26. 1=3 xn xnC1 D xn .1=3/xn 2=3 D xn 3xn D 2xn : If x0 D 1, then x1 D 2, x2 D 4, x3 D 8, x4 D 16, and, in general, xn D . 2/n . The successive “approximations” oscillate ever more widely, diverging from the root at x D 0. 23. Newton’s Method formula for f .x/ D x 2=3 is xnC1 D xn .2=3/xn 1=3 3 2 xn D D xn 1 2 xn : 27. If x0 D 1, then x1 D 1=2, x2 D 1=4, x3 D 1=8, x4 D 1=16, and, in general, xn D . 1=2/n . The successive approximations oscillate around the root x D 0, but still converge to it (though more slowly than is usual for Newton’s Method). 24. Since xnC1 D xn2 1 2xn 2 D xn2 C 1 2xn 2 jx1 jx2 : vj rj D jf .x0 / rj D jf .x1 / f .r/j Kjx0 f .r/j Kjx1 rj rj K 2 jx0 rj; and, in general, by induction 2 2xn ynC1 D xn2 C 1 1 yn D 4yn2 D 4yn .1 yn jxn a) Since sin2 .unC1 / D 4 sin2 .un /.1 sin2 .un // D 4 sin2 .un / cos2 .un / D sin2 .2un /; we have unC1 D 2un . Thus unC1 D 2n u0 . It follows that 1 , we have 1 C xn2 dyn D 1. 2. 2xn dxn : .1 C xn2 /2 rj: Section 4.3 Indeterminate Forms (page 235) dyn D 2 sin.un / cos.un / 2n du0 : b) Since yn D rj K n jx0 Since K < 1, limn!1 K n D 0, so limn!1 xn D r. The iterates converge to the fixed point as claimed in Theorem 6. yn /: 25. Let yj D sin2 .uj /. Telegram: @uni_k f .v/j Kju holds whenever u and v are in Œa; b. Pick any x0 in Œa; b, and let x1 D f .x0 /, x2 D f .x1 /, and, in general, xnC1 D f .xn /. Let r be the fixed point of f in Œa; b found in Exercise 24. Thus f .r/ D r. We have It follows that 118 We are given that there is a constant K satisfying 0 < K < 1, such that jf .u/ xn2 1 , we have 2xn 2 1 C xnC1 D1C Let g.x/ D f .x/ x for a x b. g is continuous (because f is), and since a f .x/ b whenever a x b (by condition (i)), we know that g.a/ 0 and g.b/ 0. By the Intermediate-Value Theorem there exists r in Œa; b such that g.r/ D 0, that is, such that f .r/ D r. The fixed point r is unique because if there were two such fixed points, say r1 and r2 , then condition (ii) would imply that jr1 r2 j D jf .r1 / f .r2 /j Kjr1 r2 j; which is impossible if r1 ¤ r2 and K < 1. 2=3 xn .1 C xn2 /2 2 sin.un / cos.un / 2n du0 : 2xn dxn D 0 0 3 3 D D lim x!0 4 sec2 4x 4 0 ln.2x 3/ lim x!2 x 2 4 0 2 1 2x 3 D : D 2x 2 3x lim x!0 tan 4x Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 0 sin ax x!0 sin bx 0 a cos ax a D D lim x!0 b cos bx b 1 cos ax 0 4. lim x!0 1 cos bx 0 0 a sin ax D lim x! 0 b sin bx 0 a2 cos ax a2 D lim 2 D 2: x!0 b cos bx b sin 1 x 0 5. lim x!0 tan 1 x 0 1 C x2 D lim p D1 x!0 1 x2 x 1=3 1 0 lim 2=3 6. x!1 x 0 1 . 13 /x 2=3 1 D lim 2 D : x!1 . /x 1=3 2 3 3. 7. lim lim x cot x Œ0 1 x cos x D lim x!0 sin x 0 x D 1 lim x!0 sin x 0 1 D1 D lim x!0 cos x 0 1 cos x lim x!0 ln.1 C x 2 / 0 sin x D lim 2x x!0 1 C x2 sin x D lim .1 C x 2 / lim x!0 x!0 2x 1 cos x D : D lim x!0 2 2 sin2 t 0 lim t! t 0 2 sin t cos t D0 D lim t! 1 0 10x e x lim x!0 x 0 10x ln 10 e x D lim D ln 10 1: x!0 1 0 cos 3x lim 2x 0 x!=2 3 sin 3x 3 3 D lim D . 1/ D 2 2 2 x!=2 SECTION 4.3 (PAGE 235) 12. 13. 9. 10. 11. lim x sin x!1 1 x sin D lim D lim x!1 14. 15. 16. 17. Œ1 0 1 x 0 0 1 x 1 1 cos 2 x x D lim cos 1 D 1: 1 x!1 x x2 x!1 x!0 8. 0 ln.ex/ 1 x!1 sin x 0 1 1 x D lim D : x!1 cos.x/ lim 0 sin x x!0 x3 0 0 1 cos x D lim x!0 3x 2 0 sin x 0 D lim x!0 6x 0 1 cos x D : D lim x!0 6 6 lim x sin x 0 tan x 0 0 1 cos x D lim 2 x!0 1 sec x 0 1 cos x 2 D lim .cos x/ 2 x!0 cos x 1 cos x 1 D 1 lim x!0 .cos x 1/.cos x C 1/ 1 D 2 x x!0 x lim 0 x!0 x4 0 0 2x C 2 sin x D lim 3 x!0 4x 0 1 x sin x D lim 3 2 x!0 x 1 1 1 D (by Exercise 14): D 2 6 12 lim 2 x2 2 cos x 0 sin2 x lim x!0C tan x x 0 2 sin x cos x 0 D lim x!0C sec2 x 1 0 cos x D1 D 2 1 lim x!0C 2 sec2 x tan x Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 119 lOMoARcPSD|6566483 www.konkur.in SECTION 4.3 (PAGE 235) 18. 19. 20. 21. 0 ln sin r 0 r!=2 cos r cos r sin r D 0: D lim sin r r!=2 25. lim 2 sin t D t!=2 t lim 0 cos 1 x lim x!1 x 1 0 1 p 1 x2 D D lim x!1 1 lim x.2 tan 1 x / x!1 2 tan 1 x 1 x!1 x 1 2 . D lim x!1 1 C x 2 x2 2x 2 D 2 D lim x!1 1 C x 2 D lim lim 22. t!.=2/ 23. 24. tan t / .sec t 1 D t!.=2/ D t!.=2/ lim lim Since p x x!0C x!0C p x ln x 1 Œ1 1 x!1C x 1 ln x x ln x x C 1 0 D lim x!1C .x 1/.ln x/ 0 0 ln x D lim 1 x!1C 0 ln x C 1 x 1 x D lim 1 x!1C 1 C 2 x x x 1 D lim D D : x!1C xC1 2 lim x 1/ p lim x 26. 1 ln x x ln x D lim x!0C x 1=2 x!0C 1 x D 0; D lim 1 x!0C x 3=2 2 lim D lim e 120 Œ0 1 0 0 sin t cos t cos t D 0: sin t 1 t 2 Let y D .csc x/sin x . Then ln y D sin2 x ln.csc x/ h1i ln.csc x/ lim ln y D lim x!0C x!0C csc2 x 1 csc x cot x csc x D lim x!0C 2 csc2 x cot x 1 D lim D 0: x!0C 2 csc2 x 2 Thus limx!0C .csc x/sin x D e 0 D 1. 1: Œ1 0 0 1 .1 t!0 t e at e at 1 0 D lim t!0 t e at 0 ae at D lim at Da t!0 e C at e at lim hence Telegram: @uni_k ADAMS and ESSEX: CALCULUS 9 D e 0 D 1: 0 0 27. 0 3 sin t sin 3t t!0 3 tan t tan 3t 0 3.cos t cos 3t / 0 D lim 2 t!0 3.sec2 t sec 3t / 0 cos t cos 3t D lim t!0 cos2 3t cos2 t 2 cos t cos2 3t cos 3t cos t D lim t!0 cos2 3t cos2 t 1 1 D D lim t!0 cos 3t C cos t 2 lim Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 28. Let y D sin x x 1=x 2 SECTION 4.3 (PAGE 235) 32. : sin x 0 x lim ln y D lim x!0 x!0 x2 0 x x cos x sin x sin x x2 D lim x!0 2x 0 x cos x sin x D lim x!0 2x 2 sin x 0 x sin x D lim x!0 4x sin x C 2x 2 cos x 0 sin x D lim x!0 4 sin x C 2x cos x 0 cos x 1 D lim D : x!0 6 cos x 2x sin x 6 2 sin x 1=x Thus; lim D e 1=6 : x!0 x ln x!0 33. 34. 2 29. Let y D .cos 2t /1=t . ln.cos 2t / . We have Then ln y D t2 ln.cos 2t / 0 t!0 t2 0 0 2 tan 2t D lim t!0 2t 0 2 2 sec 2t D lim D 2: t!0 1 lim ln y D lim t!0 2 Therefore lim t!0 .cos 2t /1=t D e 2 . 30. 31. h 1i 1 csc x cot x h 1 i D lim 1 x!0C 1 x x cos x 0 D lim x!0C sin2 x 0 1 D lim cos x lim x!0C x!0C 2 sin x cos x D 1: csc x lim x!0C ln x h1i ln sin x lim x!1 csc x 1 cos x sin x D lim x!1 csc x cot x D lim tan x D 0 x!1 Let y D .1 C tan x/1=x : 0 ln.1 C tan x/ lim ln y D lim x!0 x!0 x 0 sec2 x D lim D 1: x!0 1 C tan x Thus; lim .1 C tan x/1=x D e: 35. 0 2f .x/ C f .x h/ 2 h 0 h!0 f 0 .x C h/ f 0 .x h/ 0 D lim 2h 0 h!0 f 00 .x C h/ C f 00 .x h/ D lim 2 h!0 2f .x/ 00 D f .x/ D 2 lim f .x C h/ 3f .x C h/ C 3f .x h/ f .x 3h/ h3 h!0 3f 0 .x C 3h/ 3f 0 .x C h/ 3f 0 .x h/ C 3f 0 .x 3h/ D lim 3h2 h!0 3f 00 .x C 3h/ f 00 .x C h/ C f 00 .x h/ 3f 00 .x 3h/ D lim 2h h!0 9f 000 .x C 3h/ f 000 .x C h/ f 000 .x h/ C 9f 000 .x 3h/ D lim 2 h!0 D8f 000 .x/: lim f .x C 3h/ Suppose that f and g are continuous on Œa; b and differentiable on .a; b/ and g.x/ 6D 0 there. Let a < x < t < b, and apply the Generalized Mean-Value Theorem; there exists c in .x; t / such that ) ) ) ) ) f .x/ f .t / f 0 .c/ D 0 g.x/ g.t / g .c/ f .x/ f .t / g.x/ f 0 .c/ D 0 g.x/ g.x/ g.t / g .c/ f .x/ f .t / f 0 .c/ g.x/ g.t / D 0 g.x/ g.x/ g .c/ g.x/ f 0 .c/ g.t / f 0 .c/ f .t / f .x/ D 0 C g.x/ g .c/ g.x/ g 0 .c/ g.x/ " # f 0 .c/ 1 f .x/ f 0 .c/ D 0 C f .t / g.t / 0 g.x/ g .c/ g.x/ g .c/ " # f .x/ f 0 .c/ 1 f 0 .c/ LD 0 LC f .t / g.t / 0 : g.x/ g .c/ g.x/ g .c/ Since jm C nj jmj C jnj, therefore, " ˇ ˇ 0 ˇ ˇ 0 ˇ# ˇ ˇ ˇ f .c/ ˇ ˇ f .c/ ˇ ˇ f .x/ 1 ˇ ˇ ˇ ˇ ˇ ˇ ˇ g.x/ Lˇ ˇ g 0 .c/ Lˇ C jg.x/j jf .t /jCjg.t /jˇ g 0 .c/ ˇ : Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 121 lOMoARcPSD|6566483 www.konkur.in SECTION 4.3 (PAGE 235) ADAMS and ESSEX: CALCULUS 9 Now suppose that is an arbitrary small positive number. Since limc!aC f 0 .c/=g 0 .c/ D L, and since a < x < c < t , we can choose t sufficiently close to a to ensure that ˇ 0 ˇ f .c/ ˇ ˇ g 0 .c/ In particular, ˇ ˇ Lˇˇ < : 2 ˇ 0 ˇ ˇ f .c/ ˇ ˇ ˇ ˇ g 0 .c/ ˇ < jLj C 2 : f .x/ D x 5 C x 3 C 2x on .a; b f 0 .x/ D 5x 4 C 3x 2 C 2 > 0 for all x. f has no min value, but has abs max value b 5 C b 3 C 2b at x D b. 1 1 10. f .x/ D . Since f 0 .x/ D < 0 for all x in x 1 .x 1/2 the domain of f , therefore f has no max or min values. 9. x 12. f .x/ D x Since limx!aC jg.x/j D 1, we can choose x between a and t sufficiently close to a to ensure that 1 h i jf .t /j C jg.t /j jLj C < : jg.x/j 2 2 It follows that ˇ ˇ f .x/ ˇ ˇ g.x/ ˇ ˇ Lˇˇ < C D : 2 2 1. (page 242) f .x/ D x C 2 on Œ 1; 1 f 0 .x/ D 1 so f is increasing. f has absolute minimum 1 at x D mum 3 at x D 1. 1 and absolute maxi- 3. f .x/ D x C 2 on Œ 1; 1/ f has absolute minimum 1 at x D maximum. 4. f .x/ D x 2 1 no max, abs min 1 on Œ2; 3 1 at x D 3, abs max 1 at x D 2. Let f .x/ D jx 1j on Œ 2; 2: f . 2/ D 3, f .2/ D 1. f 0 .x/ D sgn .x 1/. No CP; SP x D 1, f .1/ D 0. Max value of f is 3 at x D 2; min value is 0 at x D 1. 14. Let f .x/ D jx 2 x 2j D j.x 2/.x C 1/j on Œ 3; 3: f . 3/ D 10, f .3/ D 4. f 0 .x/ D .2x 1/sgn .x 2 x 2/. CP x D 1=2; SP x D 1, and x D 2. f .1=2/ D 9=4, f . 1/ D 0, f .2/ D 0. Max value of f is 10 at x D 3; min value is 0 at x D 1 or x D 2. 1 2x f .x/ D 2 , f 0 .x/ D x C1 .x 2 C 1/2 f has abs max value 1 at x D 0; f has no min values. 16. 17. 1 and has no absolute 1 13. 15. 2. f .x/ D x C 2 on . 1; 0 abs max 2 at x D 0, no min. on .0; 1/ 1 < 0 on .0; 1/ f 0 .x/ D .x 1/2 f has no max or min values. abs min 21 f .x/ D L: Thus limx!aC g.x/ Section 4.4 Extreme Values 1 f .x/ D 11. f .x/ D .x C 2/.2=3/ no max, abs min 0 at x D f .x/ D .x 2/1=3 , f 0 .x/ D f has no max or min values. 2 y D .x f .x/ D x 3 C x 4 on Œa; b f 0 .x/ D 3x 2 C 1 > 0 for all x. Therefore f has abs min a3 C a b 3 C b 4 at x D b. x 2/1=3 Fig. 4.4-17 4 at x D a and abs max 18. f .x/ D x 2 C 2x, f 0 .x/ D 2x C 2 D 2.x C 1/ Critical point: x D 1. f .x/ ! 1 as x ! ˙1. 8. f .x/ D x 3 C x 4 on .a; b/ Since f 0 .x/ D 3x 2 C 1 > 0 for all x, therefore f is increasing. Since .a; b/ is open, f has no max or min values. Telegram: @uni_k 2/ 2=3 > 0 1 at x D 0. 6. f .x/ D x 2 1 on .2; 3/ no max or min values. 122 1 .x 3 y 5. f .x/ D x 2 1 on Œ 2; 3 f has abs min 1 at x D 0, abs max 8 at x D 3, and local max 3 at x D 2. 7. 2. f0 f Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) & CP 1 j abs min C % !x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.4 (PAGE 242) Hence, f .x/ has no max value, and the abs min is x D 1. y 1 at y D .x 2 y 4/2 16 y D x 2 C 2x x 2 2 x Fig. 4.4-20 . 1; 1/ 21. Fig. 4.4-18 19. f .x/ D x 3 3x 2 f 0 .x/ D 3x 2 3 D 3.x f0 C % f D x 2 .x 1/.5x 3 CP x D 0; ; 1 5 1/.x C 1/ CP 1 j loc max CP 1 j loc min & f .x/ D x 3 .x 1/2 f 0 .x/ D 3x 2 .x 1/2 C 2x 3 .x C % f0 !x f f has no absolute extrema. C % CP 0 j 1/ 3/ CP CP 1 j loc min 3 5 C j loc max & % C % !x f has no absolute extrema. y y 1 x 3 108 5 ; 55 x 1 y D x3 3x 2 y D x 3 .x .1; 4/ Fig. 4.4-19 Fig. 4.4-21 22. 20. f .x/ D .x 2 4/2 , f 0 .x/ D 4x.x 2 Critical points: x D 0; ˙2. f .x/ ! 1 as x ! ˙1. f0 f & CP 2 j abs min C % CP 0 j loc max 4/ D 4x.x C 2/.x & CP C2 j abs min C % 2/ !x Hence, f .x/ has abs min 0 at x D ˙2 and loc max 16 at x D 0. f .x/ D x 2 .x 1/2 , f 0 .x/ D 2x.x 1/2 C 2x 2 .x 1/ D 2x.2x Critical points: x D 0; 21 and 1. f .x/ ! 1 as x ! ˙1. f0 f & CP 0 j abs min CP C % 1 2 j loc max & CP 1 j abs min 1/.x C % 1/ !x 1 at x D 21 and abs min 0 at Hence, f .x/ has loc max 16 x D 0 and x D 1. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1/2 Downloaded by ted cage (sxnbyln180@questza.com) 123 lOMoARcPSD|6566483 www.konkur.in SECTION 4.4 (PAGE 242) ADAMS and ESSEX: CALCULUS 9 y y .1;0:5/ yD . 1; 0:5/ y D x 2 .x 1/2 f .x/ D f 0 .x/ D x2 2x .x 2 C 1/2 f .x/ D x.x 2 0 f .x/ D .x D .x D .x 2 D .x 1/2 C 2x.x 2 2 1/.x 2 1/2x 1/ p 1/.x C 1/. 5x CP 1 j loc & f % max CP p 1/. 5x C 1/ yD CP 1 C j j j !x loc % loc & loc % max min min C x Fig. 4.4-25 26. 1= 5 x 1 Telegram: @uni_k % y C % 27. CP C1 j abs max & Hence, f has abs max 21 at x D 1 and abs min x D 1. 124 C CP C1 j abs max & !x Hence, f has abs max p1 at x D 1 and abs min 2 x D 1. 1/2 1 x2 x , f 0 .x/ D 2 f .x/ D 2 x C1 .x C 1/2 Critical point: x D ˙1. f .x/ ! 0 as x ! ˙1. & CP 1 j abs min & f Fig. 4.4-23 f x4 .x 4 C 1/3=2 1 , f 0 .x/ D f0 p p 1= 5 f0 x f .x/ D p x4 C 1 Critical points: x D ˙1. f .x/ ! 0 as x ! ˙1. y CP 1 j abs min x2 x2 C 1 p1 5 p f .˙1/ D 0, f .˙1= 5/ D ˙16=25 5 24. !x CP p1 5 y D x.x 2 % yD1 p 1 C y 2 1 C 4x / 1/.5x 2 f0 C & f 1/2 2 CP 0 j abs min f0 Fig. 4.4-22 23. 1 <1 x2 C 1 D1 x2 C 1 x 1 x Fig. 4.4-24 25. 1 1 2 ; 16 x x2 C 1 !x p f .x/ D x 2 p f 0 .x/ D 2 x 2 f0 1 2 x2 at f SP p 2 j loc & max Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 1; p 2 p1 at 2 1 1; p 2 yD p x x x4 C 1 Fig. 4.4-26 p .jxj 2/ x2 2.1 x 2 / p D p 2 x2 2 x2 CP 1 C j abs min % CP 1 j abs & max SP p 2 j !x loc min lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL y SECTION 4.4 (PAGE 242) .1;1/ 30. f .x/ D x x2 1 2 D 2 2 1Cx x C1 2 tan 1 x, f 0 .x/ D 1 Critical points: x D ˙1. f .x/ ! ˙1 as x ! ˙1. p 2 p p yDx 2 2 x f0 x2 % f . 1; 1/ CP 1 j loc max C CP C1 j loc min & C % !x Fig. 4.4-27 28. f .x/ D x C sin x, f 0 .x/ D 1 C cos x 0 f 0 .x/ D 0 at x D ˙; ˙3; ::: f .x/ ! ˙1 as x ! ˙1. Hence, f has no max or min values. Hence, f has loc max 1 at x D 1. 2 1C at x D 2 1 and loc min y y .2;2/ 1; 1C 2 .;/ x x y D x C sin x 1;1 2 yDx 2 tan 1 x Fig. 4.4-30 Fig. 4.4-28 29. f .x/ D x f 0 .x/ D 1 2 sin x 2 cos x CP: x D ˙ C 2n 3 n D 0; ˙1; ˙2; alternating local maxima and minima 31. y yDx 3 yDx 2 sin x x sin 1 x . 1 x 1/ 1 f 0 .x/ D 2 p 2 p 1 x 2 2 1 x 1 D p 2 1 x 3 4x 2 D p p 2 1 x .2 1 x 2 C 1/ p 3 ; SP: (EP:) x D ˙1 CP: x D ˙ 2 ! p p 3 3 f ˙ D˙ 2 3 f .x/ D 2x f Fig. 4.4-29 0 f SP CP CP SP 1 j loc max & C j abs min % j abs & max 1 j !x loc min Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) p 3 2 p 3 2 125 lOMoARcPSD|6566483 www.konkur.in SECTION 4.4 (PAGE 242) ADAMS and ESSEX: CALCULUS 9 y 1 34. p 3 2 2 f0 y D 2x p 1 3 2 1 sin C x f x 2 f .x/ D x 2 e x , f 0 .x/ D 2xe x .1 Critical points: x D 0; ˙1. f .x/ ! 0 as x ! ˙1. % CP 1 j abs max & CP 0 j abs min x2/ CP 1 j abs max C % & !x Hence, f has abs max 1=e at x D ˙1 and abs min 0 at x D 0. y . 1;1=e/ Fig. 4.4-31 2 32. f .x/ D e x =2 , f 0 .x/ D Critical point: x D 0. f .x/ ! 0 as x ! ˙1. f0 C % f y D x2 e x 2 xe x =2 CP 0 j abs max & 35. !x y 1 2 y D e x =2 x Fig. 4.4-32 f .x/ D x2 f 0 .x/ D 2 x C x. 2 x ln 2/ D 2 x .1 x ln 2/ f C f y % ln x .x > 0/ x x ln x 1 ln x f 0 .x/ D x 2 D x x2 f .x/ ! 1 as x ! 0C (vertical asymptote), f .x/ ! 0 as x ! 1 (horizontal asymptote). ASY CP f0 0 C e j j !x abs & f % max 1 e; e CP 1= ln 2 j abs max x yD & ln x x !x Fig. 4.4-35 1 1 ln 2 ; e ln 2 36. Since f .x/ D jx C 1j, x y Dx2 x Fig. 4.4-33 126 Telegram: @uni_k x f .x/ D y x 0 2 Fig. 4.4-34 Hence,f has abs max 1 at x D 0 and no min value. 33. .1;1=e/ f 0 .x/ D sgn .x C 1/ D 1; 1; if x > 1; if x < 1. 1 is a singular point; f has no max but has abs min 0 at x D 1. f .x/ ! 1 as x ! ˙1. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.4 (PAGE 242) y y 1 x y D jx C 1j y D sin jxj 1 Fig. 4.4-38 x 39. Fig. 4.4-36 f .x/ D jx 2 37. 1j f .x/ D 2xsgn .x 2 CP: x D 0 SP: x D ˙1 f f 0 0 & SP 1 C j abs min % 1/ y D j sin xj y CP 0 j loc & max SP 1 C j !x abs % min 40. f .x/ D .x 1/2=3 .x C 1/2=3 f 0 .x/ D 32 .x 1/ 1=3 32 .x C 1/ 1=3 Singular point at x D ˙1. For critical points: .x 1/ 1=3 D .x C 1/ 1=3 ) x 1 D x C 1 ) 2 D 0, so there are no critical points. 1j 1 f0 C f % SP 1 j abs max SP C1 j abs min & Hence, f has abs max 22=3 at x D at x D 1. . 1;22=3 / 1 C % !x 1 and abs min 22=3 y y D .x x 1 x 2 Fig. 4.4-39 y y D jx 2 f .x/ D j sin xj .2n C 1/ CP: x D ˙ , SP D ˙n 2 f has abs max 1 at all CP. f has abs min 0 at all SP. 1/2=3 .x C 1/2=3 x .1; 22=3 / Fig. 4.4-37 Fig. 4.4-40 38. f .x/ D sin jxj 3 5 f 0 .x/ D sgn .x/ cos jxj D 0 at x D ˙ ; ˙ ; ˙ ; ::: 2 2 2 0 is a singular point. Since f .x/ is an even function, its graph is symmetric about the origin. f 0 f & CP 3 C 2 j abs min % CP 2 j abs max & SP 0 C j loc min % CP 2 j abs & max CP 3 C 2 j !x abs % min Hence, f has abs max 1 at x D ˙.4k C 1/ and abs min 2 1 at x D ˙.4k C 3/ where k D 0, 1, 2, : : : and loc 2 min 0 at x D 0. p 41. f .x/ D x= x 2 C 1. Since p 0 f .x/ D 42. 2x x p 1 2 x2 C 1 D 2 > 0; x2 C 1 .x C 1/3=2 x2 C 1 for all x, f cannot have any maximum or minimum value. p f .x/ D x= x 4 C 1. f is continuous on R, and limx!˙1 f .x/ D 0. Since f .1/ > 0 and f . 1/ < 0, f must have both maximum and minimum values. p f 0 .x/ D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 4x 3 x p 1 x4 2 x4 C 1 D : x4 C 1 .x 4 C 1/3=2 x4 C 1 Downloaded by ted cage (sxnbyln180@questza.com) 127 lOMoARcPSD|6566483 www.konkur.in SECTION 4.4 (PAGE 242) ADAMS and ESSEX: CALCULUS 9 p p CP x D ˙1. f .˙1/ p D ˙1= 2. f has max value 1= 2 and min value 1= 2. y 1 1; p 2 p 43. f .x/ D x 4 f .˙2/ D 0. f 0 .x/ D 1 1; p 2 yD p x 48. 49. x x4 C 1 Fig. 4.4-42 x 2 is continuous on Œ 2; 2, and p 4 2.2 x 2 / 2x D p : x2 C x p 2 4 x2 4 x2 p p maximum value 2 CP x Dp˙ 2. f .˙ 2/ D ˙2. f hasp at x D 2 and min value 2 at x D 2. p 2 2 44. f .x/ D x = 4 x is continuous on . 2; 2/, and limx! 2C f .x/ D limx!2 f .x/ D 1. Thus f can have no maximum value, but will have a minimum value. p 2x 4 f 0 .x/ D x2 4 x2 p 2 4 x2 x2 D 8x x 3 : .4 x 2 /3=2 p p CP x D 0, x D ˙ 8. f .0/ D 0, and ˙ 8 is not in the domain of f . f has minimum value 0 at x D 0. 45. f .x/ D 1=Œx sin x is continuous on .0; /, and limx!0C f .x/ D 1 D limx! f .x/. Thus f can have no maximum value, but will have a minimum value. Since f is differentiable on .0; /, the minimum value must occur at a CP in that interval. 46. f .x/ D .sin x/=x is continuous and differentiable on R except at x D 0 where it is undefined. Since limx!0 f .x/ D 1 (Theorem 8 of Section 2.5), and jf .x/j < 1 for all x ¤ 0 (because j sin xj < jxj), f cannot have a maximum value. Since limx!˙1 f .x/ D 0 and since f .x/ < 0 at some points, f must have a minimum value occurring at a critical point. In fact, since jf .x/j 1=jxj for x ¤ 0 and f is even, the minimum value will occur at the two critical points closest to x D 0. (See Figure 2.20 In Section 2.5 of the text.) 47. If it exists, an absolute max value is the maximum of the set of all the local max values. Hence, if a function has an absolute max value, it must have one or more local max values. On the other hand, if a function has a local max value, it may or may not have an absolute max value. Since a local max value, say f .x0 / at the point x0 , is defined such that it is the max within some interval jx x0 j < h where h > 0, the function may have greater values, and may even approach 1 outside this interval. There is no absolute max value in this latter case. 128 Telegram: @uni_k Section 4.5 Concavity and Inflections (page 246) 1. 2x No. f .x/ D x 2 has abs max value 0, but g.x/ D jf .x/j D x 2 has no abs max value. ( 1 if x > 0 f .x/ D x sin x 0 if x < 0 jf .x/j jxj if x > 0 so limx!0C f .x/ D 0 D f .0/. 1 Therefore f is continuous at x D 0. Clearly x sin is x continuous at x > 0. Therefore f is continuous on Œ0; 1/. Given any h > 0 there exists x1 in .0; h/ and x2 in .0; h/ such that f .x1 / > 0 D f .0/ and f .x2 / < 0 D f .0/. Therefore f cannot be a local max or min value at 0. 1 Specifically, let positive integer n satisfy 2n > h 1 1 . and let x1 D , x2 D 3 2n C 2n C 2 2 Then f .x1 / D x1 > 0 and f .x2 / < 0. 2. 3. 4. 5. p 1 1 3=2 x x, f 0 .x/ D p , f 00 .x/ D 4 2 x 00 f .x/ < 0 for all x > 0. f is concave down on .0; 1/. f .x/ D f .x/ D 2x x 2 , f 0 .x/ D 2 2x, f 00 .x/ D Thus, f is concave down on . 1; 1/. f .x/ D x 2 C 2x C 3, f 0 .x/ D 2x C 2, f 00 .x/ D 2 > 0. f is concave up on . 1; 1/. f .x/ D x x 3 , f 0 .x/ D 1 f 00 .x/ D 6x. 3x 2 , f 00 C f ^ f .x/ D 10x 3 3x 5 ; 00 x 3 / D 60x.1 f 0 .x/ D 30x 2 f 00 C f ^ 0 j infl _ !x 15x 4 ; f .x/ D 60.x 6. 2 < 0. 1 j infl _ 0 j infl x/.1 C x/: C ^ 1 j infl f .x/ D 10x 3 C 3x 5 , f 0 .x/ D 30x 2 C 15x 4 , f 00 .x/ D 60x C 60x 3 D 60x.1 C x 2 /. f 00 f Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) _ 0 j infl C ^ !x _ !x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 7. x 2 /2 ; f .x/ D .3 0 f .x/ D 13. 2 3 x /D 4x.3 f 00 .x/ D SECTION 4.5 (PAGE 246) 12x C 4x ; 12 C 12x 2 D 12.x f 00 C f ^ 1 j infl 1/.x C 1/: _ C 1 j infl .2n 1/ ; n , and conf is concave up on intervals 2 .2n C 1/ n cave down on intervals n; are . Points 2 2 inflection points. !x ^ 14. 8. f .x/ D .2 C 2x 00 x 2 /2 ; f 0 .x/ D 2.2 C 2x 2 f .x/ D 2.2 2x/ C 2.2 C 2x D 12x.x 2/: f 00 C f 9. ^ f .x/ D .x 2 f 00 .x/ D 6.x 2 D 6.x C f ^ _ 2x/; 2 x /. 2/ 15. C 2 j infl ^ f .x/ D x 2 sin x, f 0 .x/ D 1 2 cos x, f 00 .x/ D 2 sin x. Inflection points: x D n for n D 0; ˙1; ˙2; :::. f is concave down on .2nC1/; .2nC2/ and concave up on .2n/; .2n C 1/ . f .x/ D tan 1 x; f 0 .x/ D f 00 .x/ D !x 2x . .1 C x 2 /2 4/3 ; f 0 .x/ D 6x.x 2 f 00 0 j infl x 2 /.2 f .x/ D x C sin 2x; f 0 .x/ D 1 C 2 cos 2x; f 00 .x/ D 4 sin 2x: 2 f 00 C f ^ 4/2 ; 4/2 C 24x 2 .x 2 4/.5x 2 j infl _ 2 4/ 16. 4/: p2 5 j infl C ^ p2 5 j infl _ f 00 f 17. x 3 x , , f 0 .x/ D 2 x2 C 3 .x C 3/2 2x.x 2 9/ . f 00 .x/ D .x 2 C 3/3 10. f .x/ D f 11. _ 3 j infl C ^ 0 j infl _ C ^ !x f .x/ D sin x; f 0 .x/ D cos x; f 00 .x/ D sin x. f is concave down on intervals .2n; .2n C 1// and concave up on intervals ..2n 1/; 2n/, where n ranges over the integers. Points x D n are inflection points. 12. f .x/ D cos 3x, f 0 .x/ D 3 sin 3x, f 00 .x/ D 9 cos 3x. Inflection points: x D n C 12 for n D 0; ˙1; ˙2; :::. 3 4n C 1 4n C 3 f is concave up on ; and concave 6 6 4n C 3 4n C 5 ; . down on 6 6 2 !x ^ !x 2 C ^ p1 2 j infl p1 2 _ j infl C ^ !x 2 ln.x 2 / ln.x 2 / , f 0 .x/ D , x x2 2 6 C 2 ln.x / f 00 .x/ D . x3 f has inflection point at x D ˙e 3=2 and f is undefined at x D 0. f is concave up on . e 3=2 ; 0/ and .e 3=2 ; 1/; and concave down on . 1; e 3=2 / and .0; e 3=2 /. 18. f .x/ D 19. 2x ; 1 C x2 2.1 x 2 / .1 C x 2 /.2/ 2x.2x/ D : f 00 .x/ D 2 2 .1 C x / .1 C x 2 /2 f .x/ D ln.1 C x 2 /; f 00 f Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k C 2 j infl _ f 00 3 j infl _ f .x/ D e x ; f 0 .x/ D 2xe x ; 2 f 00 .x/ D e x .4x 2 2/. f f 00 0 j infl f .x/ D xe x , f 0 .x/ D e x .1 C x/, f 00 .x/ D e x .2 C x/. 2 C j !x infl ^ 2 1 ; 1 C x2 Downloaded by ted cage (sxnbyln180@questza.com) _ f 0 .x/ D 1 j infl C ^ 1 j infl _ !x 129 lOMoARcPSD|6566483 www.konkur.in SECTION 4.5 (PAGE 246) 20. f .x/ D .ln x/2 , f 0 .x/ D f 00 .x/ D 2.1 2 ln x, x 2 f 00 .x/ D 6x C 3 : x 1 1 00 > 0; f < 0: p f 00 p 4 4 3 3 1 1 Therefore f has a loc min at p and a loc max at p . 4 4 3 3 ln x/ for all x > 0. x2 f 00 x3 3 f 0 .x/ D x 2 f 00 .x/ D 2x C 0 j ^ e j infl 4x 2 C 12x 25 ; 3 f 21. ADAMS and ESSEX: CALCULUS 9 f .x/ D !x _ 8x C 12; 8 D 2.x 4/: f 00 f _ 29. C 4 j infl ^ !x 22. f .x/ D .x 1/1=3 C .x C 1/1=3 , f 0 .x/ D 31 Œ.x 1/ 2=3 C .x C 1/ 2=3 , f 00 .x/ D 29 Œ.x 1/ 5=3 C .x C 1/ 5=3 . f .x/ D 0 , x 1 D .x C 1/ , x D 0. Thus, f has inflection point at x D 0. f 00 .x/ is undefined at x D ˙1. f is defined at ˙1 and x D ˙1 are also inflection points. f is concave up on . 1; 1/ and .0; 1/; and down on . 1; 0/ and .1; 1/. 23. According to Definition 4.3.1 and the subsequent discussion, f .x/ D ax C b has no concavity and therefore no inflections. 24. f .x/ D 3x 3 36x 3, f 0 .x/ D 9.x 2 4/, f 00 .x/ D 18x. The critical points are x D 2; f 00 .2/ > 0 ) local min; x D 2; f 00 . 2/ < 0 ) local max: 25. f .x/ D x.x f 0 .x/ D 3x 2 2/2 C 1 D x 3 8x C 4 D .x 2 CP: x D 2, x D 3 4x 2 C 4x C 1 2/.3x 27. f 00 .2/ D 4 > 0; f .x/ D x 3 C 1 x f 0 .x/ D 3x 2 1 3x 4 1 D ; x2 x2 130 Telegram: @uni_k 8; 30. 31. f .x/ D xe x , f 0 .x/ D e x .1 C x/, f 00 .x/ D e x .2 C x/. The critical point is x D 1. f 00 . 1/ > 0; ) local min: f .x/ D x ln x; CP: x D 1 e 1 1 ; f 00 . / D e > 0: x e 1 f has a loc min at . e f 00 .x/ D 32. f .x/ D .x 2 4/2 , f 0 .x/ D 4x 3 16x, f 00 .x/ D 12x 2 16. The critical points are x D 0; f 00 .0/ < 0 ) local max; x D 2; f 00 .2/ > 0 ) local min; x D 2; f 00 . 2/ > 0 ) local min: 33. f .x/ D .x 2 f 00 1 CP: x D ˙ p : 4 3 x 1 x ln 2 , f 0 .x/ D , 2x 2x ln 2.x ln 2 2/ f 00 .x/ D . 2x The critical point is 1 1 ; f 00 < 0 ) local max: xD ln 2 ln 2 x f .x/ D 1 C x2 .1 C x 2 / x2x 1 x2 f 0 .x/ D D .1 C x 2 /2 .1 C x 2 /2 CP: x D ˙1 .1 C x/2 . 2x/ .1 x 2 /2.1 C x 2 /2x f 00 .x/ D .1 C x 2 /4 3 6x C 2x 3 2x 2x 4x C 4x 3 D D 2 3 .1 C x / .1 C x 2 /3 1 1 , f 00 . 1/ D . f 00 .1/ D 2 2 f has a loc max at 1 and a loc min at 1. f .x/ D f 0 .x/ D 1 C ln x; 2/ 2 D 4 < 0. 3 Therefore, f has a loc min at x D 2 and a loc max at 2 xD . 3 4 4 , f 00 .x/ D 8x 3 . 26. f .x/ D x C , f 0 .x/ D 1 x x2 The critical points are x D 2; f 00 .2/ > 0 ) local min; x D 2; f 00 . 2/ < 0 ) local max: f 00 .x/ D 6x 28. 0 4/3 2 f .x/ D 6x.x 4/2 CP: x D 0, x D ˙2 f 00 .x/ D 6.x 2 4/2 C 24x 2 .x 2 2 2 4/ D 6.x 4/.5x 4/ f 00 .0/ > 0; f 00 .˙2/ D 0: f has a loc min at x D 0. Second derivative test yields no direct information about ˙2. However, since f 00 has opposite signs on opposite sides of the points 2 and 2, each of these points is an inflection point of f , and therefore f cannot have a local maximum or minimum value at either. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 34. f .x/ D .x 2 0 SECTION 4.5 (PAGE 246) 3/e x ; 2 f .x/ D .x C 2x 38. x 3/e D .x C 3/.x x 1/e ; f 00 .x/ D .x 2 C 4x 1/e x : The critical points are x D 3; f 00 . 3/ < 0 ) local max; x D 1; f 00 .1/ > 0 ) local min. 35. 2 f .x/ D x e Suppose that f has an inflection point at x0 . To be specific, suppose that f 00 .x/ < 0 on .a; x0 / and f 00 .x/ > 0 on .x0 ; b/ for some numbers a and b satisfying a < x0 < b. If the graph of f has a non-vertical tangent line at x0 , then f 0 .x0 / exists. Let F .x/ D f .x/ 2x 2 2 f 0 .x/ D e 2x .2x 4x 3 / D 2.x 2x 3 /e 2x f .x/ D x2 x2 if x 0 if x < 0, we have n 2x if x 0 D 2jxj 2x if x < 0 n 2 if x > 0 f 00 .x/ D D 2sgn x: 2 if x < 0 f 0 .x/ D f 0 .x/ D 0 if x D 0. Thus, x D 0 is a critical point of f . It is also an inflection point since the conditions of Definition 3 are satisfied. f 00 .0/ does not exist. If a the graph of a function has a tangent line, vertical or not, at x0 , and has opposite concavity on opposite sides of x0 , the x0 is an inflection point of f , whether or not f 00 .x0 / even exists. 00 37. Suppose f is concave up (i.e., f .x/ > 0) on an open interval containing x0 . Let h.x/ D f .x/ f .x0 / f 0 .x0 /.x x0 /. Since h0 .x/ D f 0 .x/ f 0 .x0 / D 0 at x D x0 , x D x0 is a CP of h. Now h00 .x/ D f 00 .x/. Since h00 .x0 / > 0, therefore h has a min value at x0 , so h.x/ h.x0 / D 0 for x near x0 . Since h.x/ measures the distance y D f .x/ lies above the tangent line y D f .x0 / C f 0 .x0 /.x x0 / at x, therefore y D f .x/ lies above that tangent line near x0 . Note: we must have h.x/ > 0 for x near x0 , x ¤ x0 , for otherwise there would exist x1 ¤ x0 , x1 near x0 , such that h.x1 / D 0 D h.x0 /. If x1 > x0 , there would therefore exist x2 such that x0 < x2 < x1 and f 0 .x2 / D f 0 .x0 /. Therefore there would exist x3 such that x0 < x3 < x2 and f 0 .x3 / D 0, a contradiction. The same contradiction can be obtained if x1 < x0 . 39. f .x/ D x n g.x/ D x n D x0 /: f .x/; n D 2; 3; 4; : : : fn0 .x/ D nx n 1 D 0 at x D 0 If n is even, fn has a loc min, gn has a loc max at x D 0. If n is odd, fn has an inflection at x D 0, and so does gn . 40. Let there be a function f such that f 0 .x0 / D f 00 .x0 / D ::: D f .k 1/ .x0 / D 0; f .k/ .x0 / ¤ 0 for some k 2: If k is even, then f has a local min value at x D x0 when f .k/ .x0 / > 0, and f has a local max value at x D x0 when f .k/ .x0 / < 0. If k is odd, then f has an inflection point at x D x0 . 1=x 2 if x ¤ 0 41. f .x/ D e 0 if x D 0 a) e 1=x x!0C xn lim x n f .x/ D lim x!0C 2 .put y D 1=x/ 2 D lim y n e y D 0 by Theorem 5 of Sec. 4.4 y!1 Similarly, limx!0 x n f .x/ D 0, and limx!0 x n f .x/ D 0. P b) If P .x/ D jnD0 aj x j then by (a) lim P x!0 n X 1 f .x/ D aj lim x j f .x/ D 0: x!0 x Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k f 0 .x0 /.x F .x/ represents the signed vertical distance between the graph of f and its tangent line at x0 . To show that the graph of f crosses its tangent line at x0 , it is sufficient to show that F .x/ has opposite signs on opposite sides of x0 . Observe that F .x0 / D 0, and F 0 .x/ D f 0 .x/ f 0 .x0 /, so that F 0 .x0 / D 0 also. Since F 00 .x/ D f 00 .x/, the assumptions above show that F 0 has a local minimum value at x0 (by the First Derivative Test). Hence F .x/ > 0 if a < x < x0 or x0 < x < b. It follows (by Theorem 6) that F .x/ < 0 if a < x < x0 , and F .x/ > 0 if x0 < x < b. This completes the proof for the case of a nonvertical tangent. If f has a vertical tangent at x0 , then its graph necessarily crosses the tangent (the line x D x0 ) at x0 , since the graph of a function must cross any vertical line through a point of its domain that is not an endpoint. 2 1 CP: x D 0, x D ˙ p 2 2 f 00 .x/ D e 2x .2 20x 2 C 16x 4 / 1 4 < 0: f 00 .0/ > 0; f 00 ˙ p D e 2 Therefore, f has a loc (and abs) min value at 0, and loc 1 (and abs) max values at ˙ p . 2 36. Since f .x0 / Downloaded by ted cage (sxnbyln180@questza.com) j D0 131 lOMoARcPSD|6566483 www.konkur.in SECTION 4.5 (PAGE 246) ADAMS and ESSEX: CALCULUS 9 c) If x ¤ 0 and P1 .t / D 2t 3 , then f 0 .x/ D 2 1=x 2 e D P1 x3 If x D 0, then 1 f .x/: x f .x/ D lim 1 Assume that f .k/ .x/ D Pk f .x/ for some x k 1, where Pk is a polynomial. Then 1 1 1 0 1 P f .x/ C P P1 f .x/ k x2 k x x x 1 D PkC1 f .x/; x where PkC1 .t / D t 2 Pk0 .t / C P1 .t /Pk .t / is a polynomial. 1 f .x/ for n ¤ 0, where By induction, f .n/ D Pn n Pn is a polynomial. f .h/ f .0/ D lim h 1 f .h/ D 0 by h h!0 (a). Suppose that f .k/ .0/ D 0 for some k 1. Then f .kC1/ .0/ D lim f .k/ .h/ f .k/ .0/ h D lim h 1 f .k/ .h/ h!0 1 f .h/ D 0 D lim h 1 Pk h h!0 1. by (b). Thus f .n/ .0/ D 0 for n D 1; 2; : : : by induction. e) Since f 0 .x/ < 0 if x < 0 and f 0 .x/ > 0 if x > 0, therefore f has a local min value at 0 and f has a loc max value there. Function (d) appears to be the derivative of function (c), and function (b) appears to be the derivative of function (d). Thus graph (c) is the graph of f , (d) is the graph of f 0 , (b) is the graph of f 00 , and (a) must be the graph of the other function g. y y (a) (b) 5 4 3 2 1 1 2 3 4 5 1 x 2 sin ; if x ¤ 0; x 0; if x D 0. 5 4 If x ¤ 0, then 1 1 f 0 .x/ D 2x sin cos x x 2 1 1 cos f 00 .x/ D 2 sin x x x 132 Telegram: @uni_k 1 2 3 4x y (c) 3 2 11 2 3 4 5 5 4 3 2 (d) 4 3 2 1 42. We are given that f .x/ D 4 3 2 1 4 3 2 1 0 f) If g.x/ D xf .x/ then g .x/ D f .x/ C xf .x/, g 00 .x/ D 2f 0 .x/ C xf 00 .x/. In general, g .n/ .x/ D nf .n 1/ .x/ C xf .n/ .x/ (by induction). Then g .n/ .0/ D 0 for all n (by (d)). Since g.x/ < 0 if x < 0 and g.x/ > 0 if x > 0, g cannot have a max or min value at 0. It must have an inflection point there. ( 1 2 3 4x 5 4 3 2 Fig. 4.6-1 1 1 sin : x2 x D 0: Section 4.6 Sketching the Graph of a Function (page 255) h!0 0 h!0 0 Thus 0 is a critical point of f . There are points x arbitrarily close to 0 where f .x/ > 0, for example 2 x D , and other such points where f .x/ < 0, .4n C 1/ 2 for example x D . Therefore f does not have .4n C 3/ a local max or min at x D 0. Also, there are points arbitrarily close to 0 where f 00 .x/ > 0, for example 1 , and other such points where f 00 .x/ < 0, xD .2n C 1/ 1 for instance x D . Therefore f does not have con2n stant concavity on any interval .0; a/ where a > 0, so 0 is not an inflection point of f either. f .kC1/ .x/ D d) f 0 .0/ D limh!0 1 h h h2 sin 0 2. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 1 2 3 4 5 1 2 3 4x 1 2 3 4x y 4 3 2 1 11 2 3 4 5 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL (a) y 5 4 3 2 1 1 2 3 4 5 (c) 5 4 4 3 2 1 1 2 3 4x y 3 2 11 2 3 4 5 (b) y 5 1 1 2 3 4 5 1 2 3 4x 5 has inflections at 0 and ˙1:7 (approximately). 4 3 2 1 4 3 2 3. 1 2 3 4x 1 2 3 4x y (d) 4 3 2 1 SECTION 4.6 (PAGE 255) 4 3 2 1 4 3 2 11 2 3 4 5 f .x/ D x=.1 x 2 / has slope 1 at the origin, so its graph must be (c). g.x/ D x 3 =.1 x 4 / has slope 0 at the origin, but has the same sign at all points as does f .x/, so its graph must be (b). p h.x/ D .x 3 x/= 1 C x 6 has no vertical asymptotes, so its graph must p be (d). k.x/ D x 3 = jx 4 1j is positive for all positive x ¤ 1, so its graph must be (a). 4. (a) Fig. 4.6-2 The function graphed in Fig. 4.2(a): is odd, is asymptotic to y D 0 at ˙1, is increasing on . 1; 1/ and .1; 1/, is decreasing on . 1; 1/, has CPs at x D 1 (max) and 1 (min), is concave up on . 1; 2/ and .0; 2/ (approximately), is concave down on . 2; 0/ and .2; 1/ (approximately), has inflections at x D ˙2 (approximately). The function graphed in Fig. 4.2(b): is even, is asymptotic to y D 0 at ˙1, is increasing on . 1:7; 0/ and .1:7; 1/ (approximately), is decreasing on . 1; 1:7/ and .0; 1:7/ (approximately), has CPs at x D 0 (max) and ˙1:7 (min) (approximately), is concave up on . 2:5; 1/ and .1; 2:5/ (approximately), is concave down on . 1; 2:5/, . 1; 1/, and .2:5; 1/ (approximately), has inflections at ˙2:5 and ˙1 (approximately). The function graphed in Fig. 4.2(c): is even, is asymptotic to y D 2 at ˙1, is increasing on .0; 1/, is decreasing on . 1; 0/, has a CP at x D 0 (min), is concave up on . 1; 1/ (approximately), is concave down on . 1; 1/ and .1; 1/ (approximately), has inflections at x D ˙1 (approximately). The function graphed in Fig. 4.2(d): is odd, is asymptotic to y D 0 at ˙1, is increasing on . 1; 1/, is decreasing on . 1; 1/ and .1; 1/, has CPs at x D 1 (min) and 1 (max), is concave down on . 1; 1:7/ and .0; 1:7/ (approximately), is concave up on . 1:7; 0/ and .1:7; 1/ (approximately), 5 4 (b) y 3 2 2 1 1 3 2 1 1 1 2 3 5 4 3 2 5 4 1 1 2 2 3 3 (d) 3 3 4x 1 2 3 4x y 3 2 2 1 1 3 2 1 1 1 2 4 y 1 2 3 4x 5 4 3 2 1 1 2 2 3 3 4 4 Fig. 4.6-4 The function graphed in Fig. 4.4(a): is odd, is asymptotic to x D ˙1 and y D x, is increasing on . 1; 1:5/, . 1; 1/, and .1:5; 1/ (approximately), is decreasing on . 1:5; 1/ and .1; 1:5/ (approximately), has CPs at x D 1:5, x D 0, and x D 1:5, is concave up on .0; 1/ and .1; 1/, is concave down on . 1; 1/ and . 1; 0/, has an inflection at x D 0. The function graphed in Fig. 4.4(b): is odd, is asymptotic to x D ˙1 and y D 0, is increasing on . 1; 1/, . 1; 1/, and .1; 1/, has a CP at x D 0, is concave up on . 1; 1/ and .0; 1/, is concave down on . 1; 0/ and .1; 1/, has an inflection at x D 0. The function graphed in Fig. 4.4(c): is odd, is asymptotic to x D ˙1 and y D 0, is increasing on . 1; 1/, . 1; 1/, and .1; 1/, has no CP, is concave up on . 1; 1/ and .0; 1/, is concave down on . 1; 0/ and .1; 1/, has an inflection at x D 0. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 4x 4 (c) y 3 Downloaded by ted cage (sxnbyln180@questza.com) 133 lOMoARcPSD|6566483 www.konkur.in SECTION 4.6 (PAGE 255) ADAMS and ESSEX: CALCULUS 9 y The function graphed in Fig. 4.4(d): is odd, is asymptotic to y D ˙2, is increasing on . 1; 0:7/ and .0:7; 1/ (approximately), is decreasing on . 0:7; 0:7/ (approximately), has CPs at x D ˙0:7 (approximately), is concave up on . 1; 1/ and .0; 1/ (approximately), is concave down on . 1; 0/ and .1; 1/ (approximately), has an inflection at x D 0 and x D ˙1 (approximately). 5. f .0/ D 1 f .˙1/ D 0 f .2/ D 1 limx!1 f .x/ D 2, limx! 1 f .x/ D f0 SP 0 j loc max C % f CP 1 j loc min & y D f .x/ 2 .1;1/ .3;1/ 1 2 x 1 C % yDx 1 !x Fig. 4.6-6 00 f f C C 0 j ^ 2 j infl ^ _ !x 0 must be a SP because f 00 > 0 on both sides and it is a loc max. 1 must be a CP because f 00 is defined there so f 0 must be too. y yD2 y D f .x/ 1 .2;1/ 1 7. x 1 y D .x 2 y D 6x.x yD 1 00 f % SP 1 j C % CP 0 j loc max CP 2 j loc min & C % C f ^ Since lim x!˙1 1 j infl f .x/ C 1 oblique asymptote. 134 Telegram: @uni_k _ 1 j infl 2 1/2 .x C 1/2 2 2 1/2 C 4x 2 .x 2 y0 !x Inflection points: x D 1; 1; 3. f0 1/2 1/ D 6.x 1/.5x 1/ p p D 6.x 1/.x C 1/. 5x 1/. 5x C 1/ From y: Asymptotes: none. Symmetry: even. Intercepts: x D ˙1. From y 0 : CP: x D 0, x D ˙1. SP: none. 6. According to the given properties: Oblique asymptote: y D x 1. Critical points: x D 0; 2. Singular point: x D 1. Local max 2 at x D 0; local min 0 at x D 2. C 2 D 6x.x y D 6Œ.x Fig. 4.6-5 f0 1/3 0 y & CP 1 j CP CP 0 C 1 C j j !x abs % & min % 1 From y 00 : y 00 D 0 at x D ˙1, x D ˙ p . 5 C ^ 3 j infl _ x D 0, the line y D x !x 1 is an y 00 C 1 j y ^ infl _ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) p1 5 j infl C p1 5 1 C j j !x ^ infl _ infl ^ lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.6 (PAGE 255) y 9. y D .x 2 1/3 4 2 2 2 x ; y 00 D 3 : D 1; y 0 D yD x x x2 x From y: Asymptotes: x D 0, y D 1. Symmetry: none obvious. Intercept: .2; 0/. Points: . 1; 3/. From y 0 : CP: none. SP: none. 1 x 1 & y Fig. 4.6-7 y 00 2 0 2 2 00 2 8. y D x.x 1/ , y D .x 1/.5x 1/, y D 4x.5x 3/. From y: Intercepts: .0; 0/, .1; 0/. Symmetry: odd (i.e., about the origin). 1 From y 0 : Critical point: x D ˙1; ˙ p . 5 CP y0 C CP 1 C p 5 j loc % min 1 j loc & max y % CP 1 p 5 j loc max & y y 00 y _ 3 5 j infl C ^ yD !x 2 x x C 1 .2;0/ x j !x loc % min 1 . 1; 3/ 0 j infl ^ C CP 3 . 5 q _ ASY 0 j y From y 00 : q Inflection points at x D 0; ˙ !x & From y 00 : y 00 D 0 nowhere. 1 2 ASY 0 j y0 p 1= 5 p 1= 5 _ q 3 5 j infl C ^ Fig. 4.6-9 !x 4 2 2 x 1 , y 00 D . D1 , y0 D xC1 xC1 .x C 1/2 .x C 1/3 From y: Intercepts: .0; 1/, .1; 0/. Asymptotes: y D 1 (horizontal), x D 1 (vertical). No obvious symmetry. Other points: . 2; 3/. From y 0 : No critical point. 10. y D y q p1 3 1 5 5 y0 C y % x ASY 1 j C % !x From y 00 : No inflection point. y D x.x 2 2 1/ Fig. 4.6-8 y 00 C y ^ Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) ASY 1 j _ !x 135 lOMoARcPSD|6566483 www.konkur.in SECTION 4.6 (PAGE 255) ADAMS and ESSEX: CALCULUS 9 y yD y x3 1Cx yD x 1 xC1 . 2;3/ yD1 xD 1 x 1 xD 1 3 27 2; 4 x 1 Fig. 4.6-11 1 2x 6x 2 8 , y0 D , y 00 D . 2 2 2 4Cx .4 C x / .4 C x 2 /3 From y: Intercept: .0; 14 /. Asymptotes: y D 0 (horizontal). Symmetry: even (about y-axis). From y 0 : Critical point: x D 0. 12. y D Fig. 4.6-10 CP 0 j abs y % max 2 00 00 From y : y D 0 at x D ˙ p . 3 y0 3 11. x 1Cx 3x 2 C 2x 3 .1 C x/3x 2 x 3 D y0 D 2 .1 C x/ .1 C x/2 2 2 .1 C x/ .6x C 6x / .3x 2 C 2x 3 /2.1 C x/ y 00 D .1 C x/4 2 2 6x.1 C x/ 6x 4x 3 6x C 6x 2 C 2x 3 D D .1 C x/3 .1 C x/3 2x.3 C 3x C x 2 / D .1 C x/3 From y: Asymptotes: x D 1. Symmetry: none. Intercepts .0; 0/. Points . 3=2; 27=4/. 3 . From y 0 CP: x D 0, x D 2 yD CP 3 2 y0 y j loc min & C % ASY 1 j C % CP 0 j % !x From y 00 : y 00 D 0 only at x D 0. 00 C y ^ y 136 Telegram: @uni_k ASY 1 j _ 0 j infl C ^ y 00 C y ^ _ 2 p 3 j infl C ^ 1=4 yD p2 3 !x 1 4 C x2 p2 3 x Fig. 4.6-12 2x ; y0 D x2 .2 x 2 /2 2 8x 2 4 C 6x 2 y 00 D C D 2 2 2 3 .2 x / .2 x / .2 px 2 /3 From y: Asymptotes: y D 0, x D ˙ 2. Symmetry: even. Intercepts .0; 12 /. Points .˙2; 21 /. From y 0 : CP x D 0. yD 1 2 y 00 !x 2 p 3 j infl & !x y 13. C C y & ASY p 2 j Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) & CP 0 C j loc min % ASY p 2 j C !x % lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.6 (PAGE 255) y y 00 W y 00 D 0 nowhere. ASY p 2 j y 00 y _ C ^ _ p x2 xD 1 !x 2 2; 3 y xD x yD ASY p 2 j 2 2; 3 1 x xD1 p xD 2 2 1=2 x 1 2; 2 1 2; 2 yD Fig. 4.6-14 15. 1 x2 2 Fig. 4.6-13 x2 1 D1C 2 1 x 1 2x 0 y D 2 .x 1/2 2 2.3x 2 C 1/ .x 1/2 x2.x 2 1/2x D y 00 D 2 2 4 .x 1/ .x 2 1/3 From y: Asymptotes: y D 1, x D ˙1. Symmetry: even. 4 Intercepts .0; 0/. Points ˙2; . 3 0 From y : CP x D 0. yD x2 y0 14. 2x.x 2 C 3/ x2 C 1 , y 00 D . 2 2 1 .x 1/ .x 2 1/3 From y: Intercept: .0; 0/. Asymptotes: y D 0 (horizontal), x D ˙1 (vertical). Symmetry: odd. Other points: .2; 23 /, . 2; 23 /. From y 0 : No critical or singular points. yD x x2 , y0 D C y % 00 00 ASY 1 j CP 0 j loc & max C % y0 y & & ASY 1 j & & !x From y : y D 0 nowhere. y 00 C y ^ ASY 1 j _ ASY 1 j y ASY 1 j ASY 1 j !x 4 2; 3 C ^ !x x2 yD x2 4 2; 3 1 yD1 From y 00 : y 00 D 0 at x D 0. y 00 y _ ASY 1 j C ^ x xD 1 0 j infl _ ASY 1 j C ^ !x Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) xD1 Fig. 4.6-15 137 lOMoARcPSD|6566483 www.konkur.in SECTION 4.6 (PAGE 255) x3 ADAMS and ESSEX: CALCULUS 9 x 2 .x 2 3/ 00 2x.x 2 C 3/ ,y D . 2 2 1 .x 1/ .x 2 1/3 From y: Intercept: .0; 0/. Asymptotes: x D ˙1 (vertical), y D p x (oblique). Symmetry: odd. Other points: p 3 3 . ˙ 3; ˙ 2 p From y 0 : Critical point: x D 0; ˙ 3. 16. y D , y0 D x2 CP ASY CP ASY CP p p 3 1 0 1 3 C j j j j j !x loc loc y % max & & & & min % From y 0 : CP: x D 0. C y % CP 0 j y 00 p C y j infl ^ _ From y : y D 0 at x D 0. ASY 1 j y _ C ^ !x % 0 C j infl ^ 3 00 y 00 C p From y 00 : y 00 D 0 at x D 0, x D ˙ 3. y0 C 00 y0 p 3 j !x infl _ y 0 j infl ASY 1 j _ C ^ !x x3 yD p 3; y p yDx 3; p 3 3 4 p 3 3 4 x2 C 1 x xD 1 yDx p 3 p 3 x Fig. 4.6-17 xD1 yD x2 18. y D 1 Fig. 4.6-16 17. x3 C x x x Dx 2 2 x C1 x C1 x 4 C 3x 2 x 2 .x 2 C 3/ .x 2 C 1/3x 2 x 3 2x 0 D 2 D y D 2 2 2 .x C 1/ .x C 1/ .x 2 C 1/2 2 2 3 4 2 2 .x C 1/ .4x C 6x/ .x C 3x /2.x C 1/2x y 00 D .x 2 C 1/4 5 3 4x C 10x C 6x 4x 5 12x 3 D .x 2 C 1/3 2x.3 x 2 / D .x 2 C 1/3 From y: Asymptotes: y D x (oblique). Symmetry: odd. Intercepts p .0; 0/. p Points .˙ 3; ˙ 34 3/. yD 138 Telegram: @uni_k x3 x2 C 1 x2 2.1 3x 2 / . .x 2 C 1/3 From y: Intercept: .0; 0/. Asymptotes: y D 1 (horizontal). Symmetry: even. From y 0 : Critical point: x D 0. x3 x2 C 1 D , y0 D 2x .x 2 C 1/2 CP 0 j abs min y0 & y , y 00 D C % !x 1 From y 00 : y 00 D 0 at x D ˙ p . 3 y 00 y _ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 p 3 j infl C ^ 1 p 3 j infl _ !x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.6 (PAGE 255) y 20. yD1 x2 yD p1 3 x2 C 1 p1 3 x x2 2 0 2x 2.3x 2 C 1/ ,y D 2 , y 00 D . 2 2 x 1 .x 1/ p .x 2 1/3 From y: Intercept: .0; 2/, .˙ 2; 0/. Asymptotes: y D 1 (horizontal), x D ˙1 (vertical). Symmetry: even. From y 0 : Critical point: x D 0. yD Fig. 4.6-18 f0 & f x 4 3 Dx 1 xC1 xC1 .x C 1/2 C 3 3 D y0 D 1 C 2 .x C 1/ .x C 1/2 6 y 00 D .x C 1/3 From y: Asymptotes: y D x 1 (oblique), x D Symmetry: none. Intercepts .0; 4/, .˙2; 0/. From y 0 : CP: none. & CP 0 C j loc min % yD y0 y _ ASY 1 j C ^ C y % ASY 1 j ASY 1 j % !x p C y ^ _ !x x 2 yD ASY 1 j _ yD1 p 2 From y 00 : y 00 D 0 nowhere. y 00 % !x xD1 2 C C y 1. xD 1 y0 ASY 1 j From y 00 : y 00 D 0 nowhere. 2 19. ASY 1 j x2 x2 2 1 !x Fig. 4.6-20 y 21. 2 1 yDx 1 Fig. 4.6-19 2 4 x2 4 yD xC1 x x 3 4x x.x 2/.x C 2/ D x2 1 x2 1 2 2 .x 1/.3x 4/ .x 3 4x/2x y0 D 2 2 .x 1/ 3x 4 7x 2 C 4 2x 4 C 8x 2 D .x 2 1/2 x4 C x2 C 4 D .x 2 1/2 .x 2 1/2 .4x 3 C 2x/ .x 4 C x 2 C 4/2.x 2 y 00 D .x 2 1/4 5 3 4x 2x 2x 4x 5 4x 3 16x D .x 2 1/3 6x 3 18x x2 C 3 D D 6x 2 2 3 .x 1/ .x 1/3 From y: Asymptotes: y D x (oblique), x D ˙1. Symmetry: odd. Intercepts .0; 0/, .˙2; 0/. yD Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 1/2x 139 lOMoARcPSD|6566483 www.konkur.in SECTION 4.6 (PAGE 255) ADAMS and ESSEX: CALCULUS 9 y From y 0 : CP: none. y0 C y % yD1 ASY 1 j C % ASY 1 j C % 1 y ^ ASY 1 j 0 C j _ infl ^ ASY 1 j x !x yD From y 00 : y 00 D 0 at x D 0. y 00 C 1 x2 1 x2 !x _ Fig. 4.6-22 y xD1 xD 1 2 x 2 23. yDx yD x 3 4x x2 1 Fig. 4.6-21 1 2 6 x2 1 D1 , y 0 D 3 , y 00 D . x2 x2 x x4 From y: Intercepts: .˙1; 0/. Asymptotes: y D 1 (horizontal), x D 0 (vertical). Symmetry: even. From y 0 : No critical points. 22. y D y0 C y % y0 y & ASY 0 j From y 00 : y 00 is negative for all x. 140 Telegram: @uni_k C % x5 2x 3 x .x 2 1/2 2 2 4 5 .x 1/ 5x x 2.x 2 1/2x y0 D 2 .x 1/4 6 4 6 5x 5x 4x x 4 .x 2 5/ D D .x 2 1/3 .x 2 1/3 2 3 5 3 .x 1/ .6x 20x / .x 6 5x 4 /3.x 2 1/2 2x y 00 D .x 2 1/6 7 5 3 6x 26x C 20x 6x 7 C 30x 5 D 2 4 .x 1/ 4x 3 .x 2 C 5/ D .x 2 1/4 From y: Asymptotes: y D x, x D ˙1. Symmetry: odd. p 25 p 5 . Intercepts .0; 0/. Points ˙ 5; ˙ 16 p From y 0 : CP x D 0, x D ˙ 5. yD .x 2 DxC 1/2 CP ASY CP ASY CP p p 5 1 C 0 C 1 5 C j j j j j loc loc % % & max & min From y 00 : y 00 D 0 if x D 0. !x y 00 y _ ASY 1 j Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) _ 0 C j infl ^ ASY 1 j C ^ !x !x % lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.6 (PAGE 255) y yDx p 5 1 p 5 1 x x5 yD .x 2 1/2 Fig. 4.6-23 24. yD 00 x/2 .2 x3 2 y D 2.x 2/.x 6/ , x4 p 2.x 6 C 2 3/.x 12x C 24/ D x5 x5 , y0 D .x 6 p 2 3/ From y: Intercept: .2; 0/. Asymptotes: y D 0 (horizontal), x D 0 (vertical). Symmetry: none obvious. Other points: . 2; 2/, . 10; 0:144/. From y 0 : Critical points: x D 2; 6. ASY 0 j CP 2 C j loc % y & & min p From y 00 : y 00 D 0 at x D 6 ˙ 2 3. y0 y 00 y _ 0 j C ^ p 6C2 3 j infl _ CP 6 j loc max 6 . 1 1 D 4x x.x 2/.x C 2/ 3x 2 4 3x 2 4 D y0 D .x 3 4x/2 x 2 .x 2 4/2 3 2 .x 4x/ .6x/ .3x 2 4/2.x 3 4x/.3x 2 4/ y 00 D .x 3 4x/4 4 2 4 6x 24x 18x C 48x 2 32 D 3 .x 4x/3 2 2 12.x 1/ C 20 D x 3 .x 2 4/3 From y: Asymptotes: y D 0, x D 0; 2; 2. Symmetry: odd. No intercepts. 16 1 2 Points: ˙ p ; ˙ p , ˙3; ˙ 15 3 3 3 2 0 From y : CP: x D ˙ p . 3 ASY CP ASY CP ASY p2 y0 2 C 0 C p2 2 3 3 j j j j j !x loc % loc & y & & min % max & From y 00 : y 00 D 0 nowhere. 25. yD x3 ASY ASY ASY 2 C 0 2 C j j j !x y _ ^ _ ^ y 00 y & !x yD p 2 3 C j !x infl ^ 1 x3 4x 3 xD2 p2 3 3 p2 3 xD 2 x y yD x/2 .2 x3 .6;2=27/ 2 . 10; 0:144/ p 6C2 3 p 6 2 3 Fig. 4.6-25 x 26. x x yD 2 D , x Cx 2 .2 C x/.x 1/ 2 .x C 2/ 2.x 3 C 6x C 2/ 00 y0 D , y D . .x C 2/2 .x 1/2 .x C 2/3 .x 1/3 From y: Intercepts: .0; 0/. Asymptotes: y D 0 (horizontal), x D 1, x D 2 (vertical). Other points: . 3; 34 /, .2; 12 /. From y 0 : No critical point. y0 Fig. 4.6-24 y & Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) ASY 2 j & ASY 1 j & !x 141 lOMoARcPSD|6566483 www.konkur.in SECTION 4.6 (PAGE 255) ADAMS and ESSEX: CALCULUS 9 y From y 00 : y 00 D 0 if f .x/ D x 3 C 6x C 2 D 0. Since f 0 .x/ D 3x 2 C 6 6, f is increasing and can only have one root. Since f .0/ D 2 and f . 1/ D 5, that root must be between 1 and 0. Let the root be r. y 00 y _ ASY 2 j C ^ r j infl _ ASY 1 j yD . 1;3/ x3 3x 2 C 1 x3 yD1 C ^ !x x .1; 1/ y yD xD 2 x x2 C x Fig. 4.6-27 2 28. .2;1=2/ x r . 3; 3=4/ xD1 y D x C sin x, y 0 D 1 C cos x, y 00 D sin x. From y: Intercept: .0; 0/. Other points: .k; k/, where k is an integer. Symmetry: odd. From y 0 : Critical point: x D .2k C 1/, where k is an integer. f0 C f % Fig. 4.6-26 CP j CP j C % % CP 3 j C % !x From y 00 : y 00 D 0 at x D k, where k is an integer. 27. 3x 2 C 1 1 3 C 3 D1 x3 x x 3 3 3.x 2 1/ y0 D 2 D 4 x x x4 12 2 x2 6 C 5 D6 y 00 D 3 5 x x x From y W Asymptotes: y D 1, x D 0. Symmetry: none. Intercepts: since limx!0C y D 1, and limx!0 y D 1, there are intercepts between 1 and 0, between 0 and 1, and between 2 and 3. 1 Points: . 1; 3/, .1; 1/, .2; 38 /, .3; /. 27 From y 0 : CP: x D ˙1. yD x3 CP ASY y0 C 1 0 j j loc y % max & & p From y 00 : y 00 D 0 at x D ˙ 2. y 00 C y ^ 142 Telegram: @uni_k p 2 j infl y 00 C y ^ 2 C 0 C 2 j j j j j !x infl _ infl ^ infl _ infl ^ infl _ y 2 ASY p 0 C 2 j j !x _ ^ infl _ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) y D x C sin x CP 1 C j !x loc % min Fig. 4.6-28 2 x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 29. y D x C 2 sin x; y D 0 if x D 0 y 0 D 1 C 2 cos x; SECTION 4.6 (PAGE 255) y 00 D 1 From y 00 : y 00 D 0 at x D ˙ p . 2 2 sin x: 1 2 , i.e., x D ˙ ˙ 2n 2 3 y 00 D 0 if x D ˙n From y:Asymptotes: (none). odd. Symmetry: p p 2 8 2 8 Points: ˙ ; ˙ C 3 , ˙ ;˙ C 3 , 3 3 3 3 4 p 4 ˙ ;˙ 3 . 3 3 2 ˙ 2n. From y 0 : CP: x D ˙ 3 y 0 D 0 if x D CP y0 y & CP 8 3 4 3 C j loc % min 00 CP 2 3 C j loc & max C j loc % min CP CP 2 3 4 3 j loc & max j loc % min y ^ 1 p 2 j infl 31. y yDe x CP 1 j abs min 2. & y 00 y 2 !x 2 x y D xe x ; y 0 D e x .1 C x/; y 00 D e x .2 C x/: From y: Asymptotes: y D 0 (at x D 1). Symmetry: .0;0/. none.Intercept 1 2 Points: 1; , 2; , e e2 0 From y : CP: x D 1. From y 00 : y 00 D 0 at x D 4 3 ^ Fig. 4.6-30 y C p1 2 y0 1 3 _ p1 2 C !x 00 yDx 1 p 2 j infl 1 2 C 0 C 2 j j j j j !x infl _ infl ^ infl _ infl ^ infl _ y ^ C y From y : y D 0 at x D ˙n. y 00 C y 00 C % C 2 j infl _ ^ x !x !x y y D x ex y D x C 2 sin x x Fig. 4.6-29 2 2 30. y D e x , y 0 D 2xe x , y 00 D .4x 2 2/e x . From y: Intercept: .0; 1/. Asymptotes: y D 0 (horizontal). Symmetry: even. From y 0 : Critical point: x D 0. y 0 y C % CP 0 j abs max 1; y & !x 1 e y D e x sin x .x 0/, y 0 D e x .cos x sin x/, y 00 D 2e x cos x. From y: Intercept: .k; 0/, where k is an integer. Asymptotes: y D 0 as x ! 1. C k, where k is an From y 0 : Critical points: x D 4 integer. 0 y 0 j C % Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 e2 Fig. 4.6-31 32. 2 2; Downloaded by ted cage (sxnbyln180@questza.com) CP 4 j abs max & CP 5 C 4 j abs min % CP 9 4 j loc max & !x 143 lOMoARcPSD|6566483 www.konkur.in SECTION 4.6 (PAGE 255) ADAMS and ESSEX: CALCULUS 9 y From y 00 : y 00 D 0 at x D .k C 12 /, where k is an integer. . 1;1=e/ y 00 0 j y 2 j infl _ 3 2 j infl C ^ 5 2 j infl _ C ^ a !x p =4 = 2 4 ;e CP 2 j loc max & From y 00 : y 00 D 0 at x D 2˙ y0 C % y 3 2 2 x y 00 C Fig. 4.6-32 y 33. y D x2e x 2 y 00 D e x .2 2x 3 / D 2x.1 6x 2 2x.2x x 2 /e x y % Telegram: @uni_k p 2C 2 j infl _ C ^ !x . 2;4e 2/ 2 x yDx e 2x 3 // CP 1 j abs max & CP 0 C j abs min % 2 a j infl _ p p 2C 2 2 Fig. 4.6-34 35. CP 1 j !x abs & max From y 00 : y 00 D 0 if 2x 4 5x 2 C 1 D 0 p 5 ˙ 25 8 x2 D p4 5 ˙ 17 : D 4 s s p 5 C 17 5 so x D ˙a D ˙ , x D ˙b D ˙ 4 144 ^ % !x 2 y0 C ^ j infl C 2. 2 2 D .2 10x 2 C 4x 4 /e x From y: Asymptotes: y D 0. Intercept: Symmetry: even. .0; 0/. 1 Points ˙1; e From y 0 : CP x D 0, x D ˙1. y p 2 p CP 0 j abs min y 2 C x 2 y 0 D e x .2x y 00 a b b C b j j infl ^ infl _ ln x ; x 1 ln x y0 D x2 1 x2 .1 ln x/2x 2 ln x 3 x y 00 D D x4 x3 From y: Asymptotes: x D 0, y D 0. Symmetry: 0/. Intercept: .1; none. 3 1 , e 3=2 ; 3=2 . Points: e; e 2e From y 0 : CP: x D e. yD y0 p y 17 4 . 2 y D x 2 e x , y 0 D .2x C x 2 /e x D x.2pC x/e x , p 2/.x C 2 C 2/e x . y 00 D .x 2 C 4x C 2/e x D .x C 2 From y: Intercept: .0; 0/. Asymptotes: y D 0 as x ! 1. From y 0 : Critical point: x D 0, x D 2. y D e x sin x 5 4 b y D x2e x Fig. 4.6-33 34. y .1;1=e/ ASY 0 j C % CP e j abs max & !x From y 00 : y 00 D 0 at x D e 3=2 . a C j !x infl ^ y 00 y Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) ASY 0 j _ e 3=2 j infl C ^ !x x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.6 (PAGE 255) y 37. .e;1=e/ e 3=2 1 yD x ln x x ln x .x > 0/, x2 6 ln x 5 1 2 ln x , y 00 D . y0 D 3 x x4 From y: Intercepts: .1; 0/. Asymptotes: y D 0, since ln x ln x D 0, and x D 0, since lim D 1. lim x!1 x 2 x!0C x 2 From y 0 : Critical point: x D e 1=2 . y C % CP p e j abs max & y yD p y y 0 j 1=2 2 x Fig. 4.6-37 5=6 j infl 38. C ^ p . e;.2e/ yD p x x2 C 1 , y 0 D .x 2 C 1/ 3=2 , y 00 D 3x.x 2 C 1/ 5=2 . From y: Intercept: .0; 0/. Asymptotes: y D 1 as x ! 1, and y D 1 as x ! 1. Symmetry: odd. From y 0 : No critical point. y 0 > 0 and y is increasing for all x. From y 00 : y 00 D 0 at x D 0. !x y 1/ e 5=6 1 x2 2 e _ 1 4 !x From y 00 : y 00 D 0 at x D e 5=6 . 00 ASY 2 j y0 36. y D 0 j D .4 x 2 / 1=2 4 x2 x 1 .4 x 2 / 3=2 . 2x/ D y0 D 2 .4 x 2 /3=2 3 .4 x 2 /3=2 x .4 x 2 /1=2 . 2x/ 00 2 y D .4 x 2 /3 4 C 2x 2 D .4 x 2 /5=2 From y: Asymptotes: x D ˙2. Domain 2 < x < 2. Symmetry: even. Intercept: .0; 21 /. From y 0 : CP: x D 0. CP ASY 0 C 2 j j !x abs y & % min 00 00 00 From y : y D 0 nowhere, y > 0 on . 2; 2/. Therefore, y is concave up. Fig. 4.6-35 y0 1 yD p x y 00 C y ^ 0 j infl _ !x y yD1 ln x yD 2 x yD 1 Fig. 4.6-36 x x x2 C 1 Fig. 4.6-38 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k yD p Downloaded by ted cage (sxnbyln180@questza.com) 145 lOMoARcPSD|6566483 www.konkur.in SECTION 4.6 (PAGE 255) 39. ADAMS and ESSEX: CALCULUS 9 y D .x 2 1/1=3 2 y 0 D x.x 2 1/ 2=3 3 2 2 y 00 D Œ.x 2 1/ 2=3 x.x 2 1/ 5=3 2x 3 3 x2 2 2 .x 1/ 5=3 1 C D 3 3 From y: Asymptotes: none. Symmetry: even. Intercepts: .˙1; 0/, .0; 1/. From y 0 : CP: x D 0. SP: x D ˙1. SP 1 j y0 y & & CP 0 j abs min SP 1 j C % C % From f : Intercept: .0; 0/, .˙1; 0/. Asymptotes: none. Symmetry: odd. 1 From f 0 : CP: x D ˙ . SP: x D 0. e f f 0 C % y C 1 j infl _ !x & CP 1 e j loc min & C 0 j infl _ _ % !x ^ !x !x 1 1 ; e e 1/1=3 1 1 e; e 1 C y y y D .x 2 0 j f 00 f 1 j infl ^ SP From f 00 : f 00 is undefined at x D 0. From y 00 : y 00 D 0 nowhere. y 00 CP 1 e j loc max x y D x ln jxj 1 x Fig. 4.6-40 1 sin x . 1 C x2 Curve crosses asymptote at infinitely many points: x D n .n D 0; ˙1; ˙2; : : :/. y sin x yD 1 C x2 41. y D 0 is an asymptote of y D Fig. 4.6-39 40. According to Theorem 5 of Section 4.4, yD 1 1Cx 2 lim x ln x D 0: x!0C Thus, x lim x ln jxj D lim x ln x D 0: x!0 x!0C If f .x/ D x ln jxj for x ¤ 0, we may define f .0/ such that f .0/ D lim x ln jxj D 0. Then f is continuous on x!0 yD the whole real line and f 0 .x/ D ln jxj C 1; 146 Telegram: @uni_k f 00 .x/ D 1 sgn .x/: jxj Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) Fig. 4.6-41 1 1Cx 2 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.8 (PAGE 267) Section 4.7 Graphing with Computers (page 261) 1. > plot([exp(x)*ln(1+1/exp(x)), > 0.71*exp(x-35.7)], x=33..38, > y =0..2, style=point, symbol=[circle, point], > color = [red, black], numpoints = 1500); (b) The problem is with g.x/. When x grows large enough, the argument of the square root is evaluated as 22x , as the computer discards the 1. When this happens, the argument of the logarithm vanishes and the computer could be expected to return 1. However we now arrive at the case of the previous exercise. The computer will return a variety of complex numbers, infinite values, and finite real values, produced by the computer evaluation of the logarithm of the expression in question 2. The computer only plots the finite real values, but all of them are completely spurious. will produce the curve as shown in the given figure (in red), and also the exponential curve (black) conforming to the rightmost stripe as shown in this figure: (c) The argument of the square root is 22x .1 2 2x /. The computer will begin to encounter serious difficulties for values of x beginning where 2 2x , that is, 2x D 52 or x D 26. This is evident in the figure. The longest (rightmost) of the exponential stripes in the given figure seems to begin at about .0:71; 35:7/. Accordingly, the plot command 2 1.8 1.6 1.4 1.2 y 1 0.8 0.6 0.4 0.2 4. 1111111111 D 34 35 x 36 37 2 52 2 1023 D 2 52 1023 D 10 1075 log10 2 10 324 : 38 5. Fig. 4.7-1 (The red curves appear gray here.) Other exponential stripes can be handled similarly As in the previous exercise, there are 10 bits available for the exponent, so the largest possible exponent is 1111111111 in base 2, or 1023 in base 10. The largest possible mantissa is 0:111 111 (52 digits) 1 1 1 D C 2 C C 52 2 2 2 1 D1 1: 252 2. The square of any number will tend to require an increased number of digits to represent it—especially for squares of numbers with large numbers of digits to begin with. However on a computer the number of digits is fixed so the least significant digits are discarded. The resulting number is different from the square of the original number. Consequently the square root will not be quite the same and the expression cannot be expected to vanish exactly. Thus the largest positive floating-point number is approximately 21023 D 101023 log10 2 10308 . Section 4.8 Extreme-Value Problems (page 267) 3. (a) We have g.x/ D ln 2x D ln D p 22x 1 1. 1 p 22x 1 p 2x C 22x 1 p p ln .2x 22x 1/.2x C 22x p ln 2x C 22x 1 D f .x/: 2x 1/ ! Let the numbers be x and 7 x. Then 0 x 7. The product is P .x/ D x.7 x/ D 7x x 2 . P .0/ D P .7/ D 0 and P .x/ > 0 if 0 < x < 7. Thus maximum P occurs at a CP: 0D dP D7 dx 2x ) x D 7 : 2 The maximum product is P .7=2/ D 49=4. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 C 2 C 22 C C 29 ; that is, 1023 in base 10. The smallest positive mantissa is 0:000 001 D 2 52 , so the smallest positive binary floating-point number is 0 33 D Since there are 64 52 2 D 10 bits left to represent the exponent the absolute value of the exponent, the smallest possible exponent is Downloaded by ted cage (sxnbyln180@questza.com) 147 lOMoARcPSD|6566483 www.konkur.in SECTION 4.8 (PAGE 267) ADAMS and ESSEX: CALCULUS 9 8 where x > 0. Their sum is 2. Let the numbers be x and x 8 S D x C . Since S ! 1 as x ! 1 or x ! 0C, the x minimum sum must occur at a critical point: 7. P D P .x/ D 2x C p 8 ) x D 2 2: x2 dS 0D D1 dx 0D 3. Let the numbers be x and 60 x. Then 0 x 60. Let P .x/ D x 2 .60 x/ D 60x 2 x 3 . Clearly, P .0/ D P .60/ D 0 amd P .x/ > 0 if 0 < x < 60. Thus maximum P occurs at a CP: dP D 120x dx 3x 2 D 3x.40 8. x/: 3 D x .16 x/5 5x 3 .16 4 8x/: x/ .48 0D x/4 0 x 10: S.0/ D 100 and S.10/ D 1; 000. For CP: 0 D S 0 .x/ D 3x 2 2.10 x/ D 3x 2 C 2x 20: p The only positive CP is x D . 2 C 4 C 240/=6 2:270. Since S.2:270/ 71:450, the minimum value of S is about 71.45. 6. If the numbers are x and n sum of their squares is x, then 0 x n and the S.x/ D x 2 C .n p A D xh D x y 2 2.n Telegram: @uni_k x2 D x s x/2 : x/ D 2.2x P 2 P 2 x 2 x2: P P2 2P x P x D 0, or x D . Thus y D P =3 and 2 6 the triangle is equilateral since all three sides are P =3. i.e., y y h n/ ) x D n=2: Since S.n=2/ D n2 =2, this is the smallest value of the sum of squares. 148 2x ) x D Evidently, y x so 0 x P =4. If x D 0 or x D P =4, then A D 0. Thus the maximum of A must occur at a CP. For max A: s Px dA P2 D Px r 0D ; dx 4 P2 Px 2 4 Observe that S.0/ D S.n/ D n2 . For critical points: 0 D S 0 .x/ D 2x dA DP dx Let the dimensions of the isosceles triangle be as shown. Then 2x C 2y D P (given constant). The area is x. We want to minimize x/2 ; x 2 D A D xy ) x D y: ) P Hence, the width and the length are and 2 P P / D . Since the width equals the length, it .P 2 2 is a square. 9. S.x/ D x 3 C .10 2A x2 Let the width and the length of a rectangle of given perimeter 2P be x and P x. Then the area of the rectangle is A.x/ D x.P x/ D P x x 2 : 5 The critical points are 0, 6 and 16. Clearly, P .0/ D P .16/ D 0, and P .6/ D 216 105 . Thus, P .x/ is maximum if the numbers are 6 and 10. 5. Let the numbers be x and 10 dP D2 dx Since A.x/ ! 1 as x ! ˙1 the maximum must occur at a critical point: 4. Let the numbers be x and 16 x. Let P .x/ D x .16 x/ . Since P .x/ ! 1 as x ! ˙1, so the maximum must occur at a critical point: 2 .0 < x < 1/: Thus min P occurs for x D y, i.e., for a square. Therefore, x D 0 or 40. Max must correspond to x D 40. The numbers are 40 and 20. 0 D P 0 .x/ D 3x 2 .16 2A ; x Evidently, minimum P occurs at a CP. For CP: p p 8 Thus, the smallest possible sum is 2 2 C p D 4 2. 2 2 0D Let the dimensions of a rectangle be x and y. Then the area is A D xy and the perimeter is P D 2x C 2y. Given A we can express Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x x Fig. 4.8-9 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.8 (PAGE 267) 10. Let the various dimensions be as shown in the figure. Since h D 10 sin and b D 20 cos , the area of the triangle is A. / D 21 bh D 100 sin cos D 50 sin 2 for 0 < < 12. Let x be as shown in the figure. The perimeter of the rectangle is p P .x/ D 4x C 2 R2 : 2 must be at a critial point: 0 D A0 . / D 100 cos 2 ) 2 D .0 x R/: For critical points: dP 2x D4C p dx R2 x 2 p 2R )2 R2 x 2 D x ) x D p : 5 0D , the maximum 2 Since A. / ! 0 as ! 0 and ! x2 ) D : 2 4 Since 2R2 d 2P D <0 2 2 dx .R x 2 /3=2 Hence, the largest possible area is D 50 m2 : A.=4/ D 50 sin 2 4 therefore P .x/ is concave down on Œ0; R, so it must have 2R an absolute maximum value at x D p . The largest 5 perimeter is therefore (Remark: alternatively, we may simply observe that the largest value of sin 2 is 1; therefore the largest possible area is 50.1/ D 50 m2 .) P 2R p 5 2R D4 p 5 C s 10R 4R2 D p units: 5 5 R2 p .x; R2 x 2 / 10 10 h R x b=2 b=2 Fig. 4.8-10 11. Let the corners of the rectangle be as shown. p The area of the rectangle is A D 2xy D 2x R2 x 2 (for 0 x R). If x D 0 or x D R then A D 0; otherwise A > 0. Thus maximum point: pA must occur at a2 critical x dA R2 x 2 p ) R2 2x 2 D 0. 0D D2 dx R2 x 2 R Thus x D p and the maximum area is 2 s R 2p 2 R2 R2 D R2 square units. 2 y .x;y/ R x Fig. 4.8-11 x Fig. 4.8-12 13. Let the upper right r corner be .x; y/ as shown. Then x2 x 0 and y D b 1 , so x a. a2 The area of the rectangle is A.x/ D 4xy D 4bx 1 x2 ; a2 .0 x a/: Clearly, A D 0 if x D 0 or x D a, so maximum A must occur at a critical 0 point: 1 s 2x 2 B C 2 x2 dA C D 4b B ra 1 0D @ A 2 dx a x2 2 1 a2 b a x2 x2 D 0 and x D p . Thus y D p . Thus 1 a2 a2 2 2 a b The largest area is 4 p p D 2ab square units. 2 2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k s Downloaded by ted cage (sxnbyln180@questza.com) 149 lOMoARcPSD|6566483 www.konkur.in SECTION 4.8 (PAGE 267) ADAMS and ESSEX: CALCULUS 9 y x2 y2 C D1 a2 b 2 15. .x;y/ NEED FIGURE If the sides of the triangle are 10 cm, 10 cm, andp2x cm, then the area of the triangle is A.x/ D x 100 x 2 cm2 , where 0 x 10. Evidently A.0/ D A.10/ D 0 and A.x/ > 0 for 0 < x < 10. Thus A will be maximum at a critical point. For a critical point x 0 0 D A .x/ D p 100 100 D p x x2 x2 100 x2 1 p 2 100 x2 . 2x/ : 2 Thus the p critical point is given by 2x D 100, so 50. The maximum area of the triangle is x p D A. 50/ D 50 cm2 . Fig. 4.8-13 14. x2 See the diagrams below. a) The area of the rectangle is A D xy. Since y a x D b b.a x/ )yD : a a 16. Thus, the area is A D A.x/ D bx .a a x/ .0 x a/: A. / D For critical points: 0 D A0 .x/ D b .a a a : 2 2x/ ) x D NEED FIGURE If the equal sides of the isosceles triangle are 10 cm long and the angles opposite these sides are , then the area of the triangle is 2b < 0, A must have a maximum Since A00 .x/ D a a value at x D . Thus, the largest area for the rectan2 gle is ab a b a square units; D a a 2 2 4 that is, half the area of the triangle ABC . A C 1 .10/.10 sin / D 50 sin cm2 ; 2 which is evidently has maximum value 50 cm2 when D =2, that is, when the triangle is right-angled. This solution requires no calculus, and so is easier than the one given for the previous problem. 17. Let the width and the height of the billboard be w and h m respectively. The area of the board is A D wh. The printed area is .w 8/.h 4/ D 100. 100w 100 and A D 4w C ; .w > 8/. Thus h D 4 C w 8 w 8 Clearly, A ! 1 if w ! 1 or w ! 8C. Thus minimum A occurs at a critical point: b x C dA 100 100w D4C dw w 8 .w 8/2 100w D 4.w 2 16w C 64/ C 100w 0D y a x a B A D B w Fig. 4.8.14(a) Fig. 4.8.14(b) (b) This part has the same answer as part (a). To see this, let CD ? AB, and solve separate problems for the largest rectangles in triangles ACD and BCD as shown. By part (a), both maximizing rectangles have the same height, namely half the length of CD. Thus, their union is a rectangle of area half of that of triangle ABC . 150 Telegram: @uni_k 2 wD 16w 16 ˙ 800 136 D 0 p p 800 D 8 ˙ 10 2: 2 p Since w > 0 we must have w D 8 C 10 2. p 100 Thus h D 4 C p D 4 C 5 2. 10 2 p p The billboard should be 8 C 10 2 m wide and 4 C 5 2 m high. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.8 (PAGE 267) For maximum profit: w 2 0D (Since h 4 w 8 20. 2 Fig. 4.8-17 18. Let x be the side of the cut-out squares. Then the volume of the box is V .x/ D x.70 2x/.150 2x/ If the manager charges $.40 C x/ per room, then .80 2x/ rooms will be rented. The total income will be $.80 2x/.40 C x/ and the total cost will be $.80 2x/.10/ C .2x/.2/. Therefore, the profit is P .x/ D .80 0 D V 0 .x/ D 4.2625 220x C 3x 2 / D 4.3x 175/.x 15/ 175 : ) x D 15 or 3 21. T D p 0D 70 dT 1 1 x p D dx 15 122 C x 2 39 p ) 13x D 5 122 C x 2 ) .132 Fig. 4.8-18 D 4.500; 000 C 500x 13 5 T .0/ C D 0:9949 < T .10/: 15 39 (Or note that 2 1 d T D dt 2 15 x, so the total monthly profit D x/ D 4.500 C x/.1000 x 2 /: x/ p 122 C x 2 p x2 122 C x 2 122 C x 2 122 >0 15.122 C x 2 /3=2 so any critical point is a local minimum.) To minimize travel time, head for point 5 km east of A. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 52 /x 2 D 52 122 ) x D 5 T .5/ D 19. Let the rebate be $x. Then number of cars sold per month is x 2000 C 200 D 2000 C 4x: 50 P D .2000 C 4x/.1000 10 x 122 C x 2 C : 15 39 T .0/ D 150 2x The profit per car is 1000 is for x > 0: 10 12 C D 1:0564 hrs 15 39 p 244 D 1:0414 hrs T .10/ D 15 For critical points: 150 70 2x 2x/.10/ C .2x/.2/ Head for point C on road x km east of A. Travel time is We have 30/ D 72; 000 cm3 : x x 2x Œ.80 2 If P 0 .x/ D 16 4x D 0, then x D 4. Since P 00 .x/ D 4 < 0, P must have a maximum value at x D 4. Therefore, the manager should charge $44 per room. The only critical point in Œ0; 35 is x D 15. Thus, the largest possible volume for the box is 30/.150 2x/.40 C x/ D 2400 C 16x .0 x 35/: Since V .0/ D V .35/ D 0, the maximum value will occur at a critical point: V .15/ D 15.70 2x/ ) x D 250: d 2P D 8 < 0 any critical point gives a local dx 2 max.) The manufacturer should offer a rebate of $250 to maximize profit. 4 h 4 dP D 4.500 dx Downloaded by ted cage (sxnbyln180@questza.com) 151 lOMoARcPSD|6566483 www.konkur.in SECTION 4.8 (PAGE 267) A ADAMS and ESSEX: CALCULUS 9 C x (b) Since E t0 D a`w 2 .2w C 3u/=.w C u/2 > 0 for all w > 0, E t is an increasing function of w. Thus its smallest value occurs for the smallest value of w that permits flight, namely w D s. The minimum energy for the tailwind trip is E t D k`s 3 =.s C u/. B 10 x 39 km/h 12 15 km/h p (c) The minimum energy for the tailwind part of the round trip in part (b) is independent of u. However, for the headwind part, the minimum Eh D v D 3u=2 only applies as long as v s. Otherwise the plane cannot fly. If u > 2s=3, then v D 3u=2 and the least total energy for the trip is k`.s 3 =.s C u/ C 27u2 =4/. If u < 2s=3, then the minimum value for Eh at v D 3u=2 implies a speed too slow to stay airborne. As Eh0 > 0 when v > 3u=2 the the least value for Eh that is admissible happens for v D s. Thus the total energy becomes 2k`s 4 =.s 2 u2 /. 122 Cx 2 P Fig. 4.8-21 22. This problem is similar to the previous one except that the 10 in the numerator of the second fraction in the expression for T is replaced with a 4. This has no effect on the critical point of T , namely x D 5, which now lies outside the appropriate interval 0 x 4. Minimum T must occur at an endpoint. Note that 25. Use x m for the circle and 1 of areas is x 2 C 4 2 x/2 A D r 2 C s2 D 2 D 4 12 C D 0:9026 T .0/ D 15 39 1p 2 T .4/ D 12 C 42 D 0:8433: 15 E D 3000k x .1 C 4 42 1 x 4 2 .0 x 1/ 1 1 , A.1/ D > A.0/. For CP: 16 4 1 x 1 x 1 1 dA D )x C : D )xD 0D dx 2 8 2 8 8 4C Now A.0/ D The minimum travel time corresponds to x D 4, that is, to driving in a straight line to B. 23. The time for the trip is 3000=.v needed for the trip is x m for square. The sum Since 100/, so the total energy for A. v3 ; v 100 d 2A 1 1 D C > 0, the CP gives local minimum dx 2 2 8 a) For max total area use none of wire for the square, i.e., x D 1. 4 D m b) For minimum total area use 1 4C 4C for square. where k is a constant. Evidently we must have v > 100 or the trip would be impossible. The minimum value of E will occur at a critical point where 1 metre 3 0D dE 2v D dv .v 2 300v : 100/2 s The minimum thus occurs at v D 150 knots, and the time for the flight at this speed would be 3000=50 D 60 hours, or 2.5 days. This is much slower than commercial airliners can travel; the time for the flight at that speed is not much shorter than a fast ship could cross the Atlantic ocean. r s xDC D2 r 1 xDP D4s Fig. 4.8-25 24. (a) As shown in the previous exercise, the energy for a flight with airspeed v into the headwind is Eh D k`v 3 =.v u/. Similarly, the energy for a flight of the same distance with airspeed w and a tailwind of speed u is E t D k`w 3 =.w C u/. 152 Telegram: @uni_k 1 x x 26. Let the dimensions of the rectangle be as shown in the figure. Clearly, Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x D a sin C b cos ; y D a cos C b sin : lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.8 (PAGE 267) y Therefore, the area is A. / D xy D .a sin C b cos /.a cos C b sin / D ab C .a2 C b 2 / sin cos 1 D ab C .a2 C b 2 / sin 2 2 for 0 Y : 2 . Since 4 A00 . / D 2.a2 C b 2 / sin 2 < 0 when 0 , 2 therefore A. / must have a maximum value at D . 4 Hence, the area of the largest rectangle is 9 If A0 . / D .a2 C b 2 / cos 2 D 0, then D A 4 1 D ab C .a2 C b 2 / sin 2 2 1 1 D ab C .a2 C b 2 / D .a C b/2 2 2 sq. units. b a (Note: x D y D p C p indicates that the rectangle 2 2 containing the given rectangle with sides a and b, has largest area when it is a square.) p .9; 3/ p 3 X x Fig. 4.8-27 28. The longest beam will have length equal to the minimum of L D x C y, where x and y are as shown in the figure below: b a ; yD : xD cos sin Thus, a b L D L. / D C 0< < : cos sin 2 x a x a y y b b Fig. 4.8-28 Fig. 4.8-26 27. Let the line have intercepts x, y as shown. Let be angle shown. The length of line is p 3 9 C LD cos sin .0 < < /: 2 . Clearly, L ! 1 if ! 0C or ! 2 Thus the minimum length occurs at a critical point. For CP: p 3 cos dL 9 sin 1 3 3 p 0D D ) tan D d cos2 sin2 3 ) D 6 Shortest line segment has length LD p 9 3=2 C p 0 If L . / D 0, then a sin b cos D0 cos2 sin2 3 a sin b cos3 , D0 cos2 sin2 , a sin3 b cos3 D 0 b , tan3 D a b 1=3 , tan D 1=3 : a Clearly, L. / ! 1 as ! 0C or ! . Thus, the 2! 1=3 b . Using the minimum must occur at D tan 1 a1=3 b 1=3 , it follows that a1=3 a1=3 b 1=3 cos D p ; sin D p : a2=3 C b 2=3 a2=3 C b 2=3 triangle above for tan D p 3 D 8 3 units: 1=2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 153 lOMoARcPSD|6566483 www.konkur.in SECTION 4.8 (PAGE 267) ADAMS and ESSEX: CALCULUS 9 Hence, the minimum is a L. / D a b !C 1=3 b 1=3 p p a2=3 C b 2=3 a2=3 C b 2=3 3=2 D a2=3 C b 2=3 units. Thus, the maximum height of the fence is ! p p 32=3 1 2 32=3 h. / D 6 1=3 3 ! D 2.32=3 31. 29. If the largest beam that can be carried horizontally around the corner is l m long (by Exercise 26, l D .a2=3 C b 2=3 /2=3 m), then at the point of maximum clearance, one end of the beam will be on the floor at the outer wall of one hall, and the other will be on the ceiling at the outer wall of the second hall. Thus the horizontal projection of the beam will be l. So the beam will have length 32. 2 tan Let .x; y/ be a point on x 2 y 4 D 1. Then x 2 y 4 D 1 and the square of distance from .x; y/ to .0; 0/ is 1 S D x 2 C y 2 D 4 C y 2 , (y ¤ 0) y Clearly, S ! 1 as y ! 0 or y ! ˙1, so minimum S must occur at a critical point. For CP: dS 4 D 5 C 2y ) y 6 D 2 ) y D ˙21=6 dy y 1 ) x D ˙ 1=3 2 Thus the shortest distance from origin to curve is r r 1 3 31=2 1=3 SD C 2 D D units. 22=3 22=3 21=3 30. Let be the angle of inclination of the ladder. The height of the fence is h. / D 6 sin 1/3=2 2:24 m: 0D p l 2 C c 2 D Œ.a2=3 C b 2=3 /3 C c 2 1=2 units. 1 The square of the distance from .8; 1/ to the curve y D 1 C x 3=2 is S D .x 0< < : 2 D .x 3 8/2 C .y 1/2 8/2 C .1 C x 3=2 Dx Cx 2 1/2 16x C 64: Note that y, and therefore also S, is only defined for x 0. If x D 0 then S D 64. Also, S ! 1 if x ! 1. For critical points: dS D 3x 2 C 2x dx ) x D 83 or 2: 0D 6 m 33. Fig. 4.8-30 For critical points: 0 D h0 . / D 6 cos 2 sec2 )3 cos D sec2 ) 3 cos3 D 1 1=3 ) cos D 31 : 4 sec2 tan < 0 for 0 < < , 1=3 2 therefore h. / must be maximum at D cos 1 13 . Then Since h00 . / D sin D 154 Telegram: @uni_k 6 sin p 32=3 1 ; 31=3 2/ Only x D 2 is feasible. At x D 2 we p have SpD 44 < 64. Therefore the minimum distance is 44 D 2 11 units: h 2 m 16 D .3x C 8/.x tan D p 32=3 1: Let the cylinder have radius r and height h. By symmetry, the centre of the cylinder is at the centre of the sphere. Thus h2 D R2 : r2 C 4 The volume of cylinder is h2 ; .0 h 2R/: V D r 2 h D h R2 4 Clearly, V D 0 if h D 0 or h D 2R, so maximum V occurs at a critical point. For CP: dV h2 2h2 D R2 0D dh 4 4 2R 4 )hD p ) h2 D R2 3 3 r 2 R: )r D 3 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.8 (PAGE 267) 2R The largest cylinder has height p units and radius 3 units. r 2 R 3 35. Let the box have base dimensions x m and height y m. Then x 2 y D volume D 4. Most economical box has minimum surface area (bottom and sides). This area is r S D x 2 C 4xy D x 2 C 4x h=2 R D x2 C 4 x2 16 ; x .0 < x < 1/: h Clearly, S ! 1 if x ! 1 or x ! 0C. Thus minimum S occurs at a critical point. For CP: 0D dS D 2x dx 16 ) x 3 D 8 ) x D 2 ) y D 1: x2 Most economical box has base 2 m 2 m and height 1 m. Fig. 4.8-33 34. Let the radius and the height of the circular cylinder be r and h. By similar triangles, h R r D H H.R r/ )hD : R R y Hence, the volume of the circular cylinder is V .r/ D r 2 h D D H r 2 r 2 H.R r/ R r3 for 0 r R: R Since V .0/ D V .R/ D 0, the maximum value2 of V must dV 3r be at a critical point. If D H 2r D 0, then dr R 2R . Therefore the cylinder has maximum volume if r D 3 2R its radius is r D units, and its height is 3 hD H R 2R 3 R D x x Fig. 4.8-35 36. x H units. 3 s x x 2 ft r H h R 2 ft Fig. 4.8-34 Fig. 4.8-36 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 155 lOMoARcPSD|6566483 www.konkur.in SECTION 4.8 (PAGE 267) ADAMS and ESSEX: CALCULUS 9 From the figure, if the side of the square base of the pyramid is 2x, then p the slant height of triangular walls of the pyramid is s D 2 x. The vertical height of the pyramid is q q p p p p 2x: h D s 2 x 2 D 2 2 2x C x 2 x 2 D 2 1 38. V D r 2 h C 43 r 3 : h Thus the volume of the pyramid is p q p 4 2 2 V D 1 2x; x 3 p for 0 x 1= 2. V D 0 at both endpoints, so the maximum will occur at an interior critical point. For CP: p 2 # p " q p 4 2 2x dV 2x D 2x 1 p 0D p dx 3 2 1 2x p 2 p 2x/ D 2x 4x.1 p p 4x D 5 2x 2 ; x D 4=.5 2/: p p p V .4=.5 2// D p 32 2=.75 p 5/.3 The largest volume of such a pyramid is 32 2=.75 5/ ft . r Fig. 4.8-38 If the cylindrical wall costs $k per unit area and the hemispherical wall $2k per unit area, then the total cost of the tank wall is C D 2 rhk C 8 r 2 k 37. Let the dimensions be as shown. The perimeter is x C x C 2y D 10. Therefore, 2 1C x C 2y D 10; or .2 C /x C 4y D 20: 2 The area of the window is 1 x 2 .2 C /x x2 A D xy C Cx 5 D : 2 2 8 4 D 2 rk D 2C x 4 0D 2C 20 x)xD 4 4C 10 : )yD 4C To admit greatest amount of light, let width D and height (of the rectangular part) be 4 3 3 r C 8 r 2 k r2 2V k 16 C r 2k r 3 .0 < r < 1/: dC D dr 2V kr 2 C 32 rk 3 , rD r3 D 20 m 4C 4 1 3 r2h C r3 ) r D h 16 3 4 10 m. 4C ) h D 4r D 4 3V 16 39. Telegram: @uni_k 1=3 Let D 0 be chosen so that mirror AB is the right bisector of DD 0 . Let CD 0 meet AB at X. Therefore, the travel time along CXD is x 156 1=3 : Hence, in order to minimize the cost, the radius and length 3V 1=3 of the cylindrical part of the tank should be 16 3V 1=3 and 4 units respectively. 16 y Fig. 4.8-37 3V 16 Since V D r 2 h C 34 r 3 , x=2 y V Since C ! 1 as r ! 0C or r ! 1, the minimum cost must occur at a critical point. For critical points, To maximize light admitted, maximize the area A. For CP: x dA D C5 0D dx 4 Let h and r be the length and radius of the cylindrical part of the tank. The volume of the tank is TX D CX C XD CX C XD 0 CD 0 D D : speed speed speed Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) : lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.8 (PAGE 267) p The stiffness is S D wh3 D h3 4R2 h2 for .0 h 2R/. We have S D 0 if h D 0 or h D 2R. For maximum stiffness: If Y is any other point on AB, travel time along C YD is TY D C Y C YD C Y C YD 0 CD 0 D > : speed speed speed 0D (The sum of two sides of a triangle is greater than the third side.) Therefore, X minimizes travel time. Clearly, XN bisects †CXD. p dS D 3h2 4R2 dh h2 p h4 4R2 h2 : p Thus 3.4R2 h2 / D h2 so h D 3R, andpw D R. The stiffest beam has width R and depth 3R. D N C w=2 A h=2 B R Y X h D0 w Fig. 4.8-39 Fig. 4.8-41 40. If the path of the light ray is as shown in the figure then the time of travel from A to B is T D T .x/ D p a2 C x 2 C v1 p b 2 C .c v2 x/2 42. : c A a i x c x r 43. b B Fig. 4.8-40 The curve y D 1 C 2x x 3 has slope m D y 0 D 2 3x 2 . Evidently m is greatest for x D 0, in which case y D 1 and m D 2. Thus the tangent line with maximal slope has equation y D 1 C 2x. dQ D kQ3 .L Q/5 .k; L > 0/ dt Q grows at the greatest rate when f .Q/ D Q3 .L maximum, i.e., when 0 D f 0 .Q/ D 3Q2 .L To minimize T , we look for a critical point: 0D D Q2 .L 1 1 x c x dT D p p 2 2 2 dx v1 a C x v2 b C .c x/2 1 1 D sin i sin r: v1 v2 Q/5 5Q3 .L Q/4 Q/4 .3L 8Q/ ) Since f .0/ D f .L/ D 0 and f most rapidly when Q D 3L . 8 3L 8 Q/5 is Q D 0; L; 3L : 8 > 0, Q is growing Thus, v1 sin i D : sin r v2 44. Let h and r be the height and base radius of the cone and R be the radius of the sphere. From similar triangles, 41. Let the width be w, and the depth be h. Therefore 2 w 2 h C D R2 : 2 2 p r ) Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) D R h R 2r 2 R hD 2 r R2 h2 C r 2 .r > R/: 157 lOMoARcPSD|6566483 www.konkur.in SECTION 4.8 (PAGE 267) ADAMS and ESSEX: CALCULUS 9 which simplifies to t 2 C t C 1 D t 2 et : h h R p Both sides of this equation are increasing functions but the left side has smaller slope than the right side for t > 0. Since the left side is 1 while the right side is 0 at t D 0, there will exist a unique solution in t > 0. Using a graphing calculator or computer program we determine that the critical point is approximately t D 1:05032. For this value of t we have x 750:15, so the movement of cars will be optimized by loading 750 cars for each sailing. h2 Cr 2 R R r 46. Fig. 4.8-44 Then the volume of the cone is V D 1 2 2 r4 r h D R 2 3 3 r R2 .R < r < 1/: Clearly V ! 1 if r ! 1 or r ! RC. Therefore to minimize V , we look for a critical point: 2 .r R2 /.4r 3 / r 4 .2r/ dV 2 D R dr 3 .r 2 R2 /2 5 3 2 5 4r 4r R 2r D 0 p r D 2R: 0D , , To maximize (i.e., to get the best view of the mural), we can maximize tan D f .x/. Since f .0/ D 0 and f .x/ ! 0 as x ! 1, we look for a critical point. 2 x C 24 2x 2 0 D f 0 .x/ D 10 ) x 2 D 24 .x 2 C 24/2 p )xD2 6 p Stand back 2 6 ft ( 4:9 ft) to see the mural best. Hence, the smallest possible volume of a right circular cone which can contain sphere of radius R is V D 2 Let distances and angles be as shown. Then tan ˛ D , x 12 tan. C ˛/ D x 2 tan C 12 tan C tan ˛ x D D 2 x 1 tan tan ˛ 1 tan x 12 24 2 tan D tan C x x 2 x 10 10x 24 ; so tan D 2 D f .x/: tan 1 C 2 D x x x C 24 4R4 2 8 R D R3 cubic units. 3 2R2 R2 3 45. If x cars are loaded, the total time for the trip is 10 x T Dt C1C 1;000 1;000 t where x D f .t / D t : e Ct 2 We can minimize the average time per car (or, equivalently, maximize the number of cars per hour). The average time (in hours) per car is AD T e t Ct e t Ct 1 D C C x 1;000 1;000t 1;000 1 1 t D e Ct 1C C1 : 1;000 t Fig. 4.8-46 This expression approaches 1 as t ! 0C or t ! 1. For a minimum we should look for a positive critical point. Thus we want 1 0D 1;000 158 Telegram: @uni_k e t C1 1 1C t e t Ct 1 t2 ˛ x 47. Let r be the radius of the circular arc and be the angle shown in the left diagram below. Thus, ; 2r D 100 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) ) rD 50 : lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.8 (PAGE 267) y r yDtan x The largest cone has volume V yDx wall =2 cu. units. r ! 8 2R3 p D 3 9 3 fence 2 r r R Fig. 4.8.47(a) R Fig. 4.8.47(b) h R The area of the enclosure is 2 r 2 .r cos /.r sin / AD 2 502 502 sin 2 D 2 2 1 sin 2 D 502 2 2 for 0 < . Note that A ! 1 as ! 0C, and for D we are surrounding the entire enclosure with fence (a circle) and not using the wall at all. Evidently this would not produce the greatest enclosure area, so the maximum area must correspond to a critical point of A: dA 2 2 .2 cos 2 / sin 2.4 / 1 0D D 502 d 2 4 4 1 cos 2 sin 2 , C D 2 2 3 2 , 2 cos D 2 sin cos , cos D 0 or tan D : Observe that tan D has no solutions in .0; . (The graphs of y D tan and y D cross at D 0 but nowhere else between 0 and .) Thus, the greatest enclo sure area must correspond to cos D 0, that is, to D . 2 The largest enclosure is thus semicircular, and has area 5000 2 2 .50/2 D m . 48. Let the cone have radius r and height h. Let sector of angle from disk be used. R Then 2 r D R so r D . 2 r p Rp 2 R2 2 Also h D R2 r 2 D R2 D 4 2 2 4 2 The cone has volume R2 2 R p 2 r2h 4 D V D 3 3 4 2 2 3 p R D f . / where f . / D 2 4 2 2 .0 2/ 24 2 V .0/ D V .2/ D 0 so maximum V must occur at a critical point. For CP: p 3 0 D f 0 . / D 2 4 2 2 p 4 2 2 8 2 2 2 ) 2.4 /D ) 2 D 2: 3 R Fig. 4.8-48 49. Let the various distances be as labelled in the diagram. h a x y a y h h Fig. 4.8-49 From the geometry of the various triangles in the diagram we have x 2 D h2 C .a y 2 D a2 C .y x/2 ) h2 D 2ax h/2 ) h2 D 2hy a2 a2 hence hy D ax. Then a2 x 2 h2 2 2 2ax 3 a x D D x2 C 2 2ax a 2ax a2 L2 D x 2 C y 2 D x 2 C a a < x a. Clearly, L ! 1 as x ! C, and 2 p 2 L.a/ D 2a. For critical points of L2 : for 0D d.L2 / .2ax a2 /.6ax 2 / .2ax 3 /.2a/ D dx .2ax a2 /2 2a2 x 2 .4x 3a/ D : .2ax a2 /2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x L Downloaded by ted cage (sxnbyln180@questza.com) 159 lOMoARcPSD|6566483 www.konkur.in SECTION 4.8 (PAGE 267) ADAMS and ESSEX: CALCULUS 9 a 3a ; a is x D . Since The only critical point in 2 4 p 3 3a 3a D < L.a/, therefore the least possible L 4 4 p 3 3a cm. length for the fold is 4 Section 4.9 Linear Approximations (page 274) 1. so that x D 1=100. The edge length must decrease by about 0:01 cm in to decrease the volume by 12 cm3 . 13. The circumference C and radius r of the orbit are linked by C D 2 r. Thus C D 2 r. If r D 10 mi then C 2 r D 20. The circumference of the orbit will decrease by about 20 62:8 mi if the radius decreases by 10 mi. Note that the answer does not depend on the actual radius of the orbit. 14. a D gŒR=.R C h/2 implies that f .x/ D x 2 , f 0 .x/ D 2x, f .3/ D 9, f 0 .3/ D 6. Linearization at x D 3: L.x/ D 9 C 6.x 3/. 2. f .x/ D x 3 , f 0 .x/ D 3x 4 , f .2/ D 1=8, f 0 .2/ D 3=16. 3 .x 2/. Linearization at x D 2: L.x/ D 81 16 p p 3. f .x/ D 4 x, f 0 .x/ D 1=.2 4 x/, f .0/ D 2, f 0 .0/ D 1=4. Linearization at x D 0: L.x/ D 2 41 x. p p 4. f .x/ D 3 C x 2 , f 0 .x/ D x= 3 C x 2 , f .1/ D 2, f 0 .1/ D 1=2. Linearization at x D 1: L.x/ D 2 C 21 .x 1/. a If h D 0 and h D 10 mi, then 15. 1. 0 8. f .x/ D cos.2x/, p f .x/ D 2 sin.2x/, f .=3/ D 1=2, f 0 .=3/ D 3. p 3 x 3 . Linearization at x D =3: L.x/ D 12 2 9. f .x/ D sinp x, f 0 .x/ D 2 sin x cos x, f .=6/ D 1=4, f 0 .=6/ D 3=2. p Linearization at x D =6: L.x/ D 41 C . 3=2/ x 6 . 10. f .x/ D tan x, f 0 .x/ D sec2 x, f .=4/ D 1, f 0 .=4/ D 2. Linearization at x D =4: L.x/ D 1 C 2 x 4 . 11. If A and x are the area and side length of the square, then A D x 2 . If x D 10 cm and x D 0:4 cm, then dA x D 2x x D 20.0:4/ D 8: A dx The area increases by about 8 cm2 . 12. If V and x are the volume and side length of the cube, then V D x 3 . If x D 20 cm and V D 12 cm3 , then 12 D V 160 Telegram: @uni_k 20 32 0:16 ft/s2 : 3960 1 f .x/ D x 1=2 f 0 .x/ D x 1=2 f 00 .x/ D 2 p 50 D f .50/ f .49/ C f 0 .49/.50 49/ 99 1 D 7:071: D7C 14 14 1 3=2 x 4 p 99 f 00 .x/ < 0 on Œ49; 50, so error is negative: 50 < 14 1 1 1 00 jf .x/j < D D 0:00073 D k on 4 73 1372 4 493=2 .49; 50/. 1 k D 0:00036. We have Thus jerrorj .50 49/2 D 2 2744 6. f .x/ D x 1=2 , f 0 .x/ D . 1=2/x 3=2 , f .4/ D 1=2, f 0 .4/ D 1=16. 1 Linearization at x D 4: L.x/ D 21 16 .x 4/. f .x/ D sin x, f 0 .x/ D cos x, f ./ D 0, f 0 ./ D Linearization at x D : L.x/ D .x /. 20g D R a 5. f .x/ D .1 C x/ 2 , f 0 .x/ D 2.1 C x/ 3 , f .2/ D 1=9, f 0 .2/ D 2=27. 2 .x 2/. Linearization at x D 2: L.x/ D 91 27 7. da 2 h: h D gR2 dh .R C h/3 99 14 16. p 99 1 50 ; 2744 14 p i.e., 7:071064 50 7:071429 p Let f .x/ D x, then f 0 .x/ D 21 x 1=2 and 1 00 f .x/ D 4 x 3=2 . Hence, p 47 D f .47/ f .49/ C f 0 .49/.47 49/ 1 48 D7C . 2/ D 6:8571429: 14 7 Clearly, if x 36, then jf 00 .x/j 1 1 D D K: 4 63 864 Since f 00 .x/ < 0, f is concave down. Therefore, the error p 48 < 0 and E D 47 7 dV x D 3x 2 x D 1; 200 x; dx jEj < Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) K .47 2 49/2 D 1 : 432 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL Thus, 48 7 17. SECTION 4.9 (PAGE 274) f 00 .0/ < 0 on Œ45ı ; 46ı so p 1 48 < 47 < 432 p 7 6:8548 < 47 < 6:8572: 1 2 0:0001: jErrorj < p 2 2 180 3 1 x 7=4 f .x/ D x 1=4 ; f 0 .x/ D x 3=4 ; f 00 .x/ D 4 16 p 4 85 D f .85/ f .81/ C f 0 .81/.85 81/ 1 82 4 D3C D 3:037: D 3C 4 27 27 27 p 82 f 00 .x/ < 0 on Œ81; 85 so error is negative: 4 85 < . 27 3 1 jf 00 .x/j < D D k on Œ81; 85. 16 37 11; 664 k Thus jErrorj .85 81/2 D 0:00069. 2 82 27 We have 1 p 1 2 20. 1 2 D 0:0038772: jEj < p 2 2 30 Thus, 0:5906900 21. 5 5 < 0:5906900 < 0:5906900: Let f .x/ D sin x, then f 0 .x/ D cos x and f 00 .x/ D sin x. The linearization at x D gives: sin.3:14/ sin Ccos .3:14 / D 3:14 0:001592654: 0:49925 > 0 Since f 00 .x/ < 0 between 3:14 and , the error E in the above approximation is negative: sin.3:14/ < 0:001592654. For 3:14 t , we have 1 .0:003/2 D 0:000001125: 8 jf 00 .t /j D sin t sin.3:14/ < 0:001592654: Thus the error satisfies 1 < 0:49925 C 0:000001125 2:003 1 0:49925 < < 0:499251125: 2:003 jEj 0:49925 < f .x/ D cos x; f 0 .x/ D sin x; f 00 .x/ D C cos 46ı D cos 4 180 sin cos 4 4 180 1 D p 1 0:694765: 180 2 cos x 0:001592654 .3:14 2 /2 < 0:000000002: Therefore 0:001592652 < sin.3:14/ < 0:001592654. 22. Let f .x/ D sin x, then f 0 .x/ D cos x and f 00 .x/ D sin x. The linearization at x D 30ı D =6 gives sin.33ı / D sin 6 C 60 sin C cos 6p 6 60 3 1 0:545345: D C 2 2 60 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 0:0038772 < sin 0:5868128 < sin Thus, 19. 180 Let f .x/ D sin x, then f 0 .x/ D cos x and f 00 .x/ D sin x. Hence, C Df f Cf0 sin 5 6p 30 6 6 30 3 1 0:5906900: D C 2 2 30 1 2 and f 00 .x/ D 3 . x2 x If x 2, then jf 00 .x/j 82 D 14 . Since f 00 .x/ > 0 for x > 0, f is concave up. Therefore, the error jEj < 1 < cos 46ı < p 1 2 1 , then jf 00 .x/j p . Since f 00 .x/ < 0 on 4 2 0 < x 90ı , f is concave down. Therefore, the error E is negative and 1 D f .2:003/ f .2/ C f 0 .2/.0:003/ 2:003 1 1 .0:003/ D 0:49925: D C 2 4 and If x p or 3:036351 4 85 3:037037 1 ED 2:003 2 2 1802 i.e., 0:694658 cos 46ı < 0:694765. p 1 82 4 < 85 < ; 1458 27 1 18. Let f .x/ D , then f 0 .x/ D x Hence, 180 Downloaded by ted cage (sxnbyln180@questza.com) 161 lOMoARcPSD|6566483 www.konkur.in SECTION 4.9 (PAGE 274) ADAMS and ESSEX: CALCULUS 9 Since f 00 .x/ < 0 between 30ı and 33ı , the error E in the above approximation is negative: sin.33ı / < 0:545345. For 30ı t 33ı , we have for 47 x 49. Thus, on that interval, M f 00 .x/ N , where M D 0:000776 and N D 0:000729. By Corollary C, jf 00 .t /j D sin t sin.33ı / < 0:545345: N M .47 49/2 f .47/ L.47/ C .47 2p 2 6:855591 47 6:855685: L.47/ C Thus the error satisfies jEj 0:545345 2 < 0:000747: 2 60 Using p the midpoint of this interval as a new approximation for 47 ensures that the error is no greater than half the length of the interval: Therefore 0:545345 p ı 0:000747 < sin.33 / < 0:545345 0:544598 < sin.33ı / < 0:545345: 25. 23. From the solution to Exercise 15, the linearization to f .x/ D x 1=2 at x D 49 has value at x D 50 given by L.50/ D f .49/ C f 0 .49/.50 M N .85 81/2 f .85/ L.85/ C .85 2 2 3:036351 851=4 3:036405: L.85/ C 49/2 L.47/ D f .49/ C f 0 .49/.47 49/ 6:8571429: p Also, 6:8548 p 47 6:8572, and, since f 00 .x/ D 1=.4. x/3 /, 1 1 1 p f 00 .x/ 3 3 4.6:8548/3 4.7/ 4. 47/ 162 Telegram: @uni_k 851=4 3:036378; jerrorj 0:000006: From the solution to Exercise 16, the linearization to f .x/ D x 1=2 at x D 49 has value at x D 47 given by 81/2 Using the midpoint of this interval as a new approximation for 851=4 ensures that the error is no greater than half the length of the interval: Using p the midpoint of this interval as a new approximation for 50 ensures that the error is no greater than half the length of the interval: 24. 81/ 3:037037: for 81 x 85. Thus, on that interval, M f 00 .x/ N , where M D 0:000086 and N D 0:000079. By Corollary C, for 49 x 50. Thus, on that interval, M f 00 .x/ N , where M D 0:000729 and N D 0:000707. By Corollary C, 50 7:071070; From the solution to Exercise 17, the linearization to f .x/ D x 1=4 at x D 81 has value at x D 85 given by 3 3 3 f 00 .x/ 16.3/7 16.3:037037/7 16.851=4 /7 1 1 1 f 00 .x/ p 3 3 4.7/3 4.7:071429/ 4. 50/ p jerrorj 0:000047: Also, 3:036351 851=4 3:037037, and, since f 00 .x/ D 3=.16.x 1=4 /7 /, p 50 7:071429, and, since Also, 7:071064 p f 00 .x/ D 1=.4. x/3 /, M N .50 49/2 f .50/ L.50/ C .50 2p 2 7:071064 50 7:071075: 47 6:855638; L.85/ D f .81/ C f 0 .81/.85 49/ 7:071429: L.50/ C 49/2 26. jerrorj 0:000028: From the solution to Exercise 22, the linearization to f .x/ D sin x at x D 30ı D =6 has value at x D 33ı D =6 C =60 given by L.33ı / D f .=6/ C f 0 .=6/.=60/ 0:545345: Also, 0:544597 sin.33ı / 5:545345, and, since f 00 .x/ D sin x, sin.33ı / f 00 .x/ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) sin.30ı / lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 4.10 for 30ı x 33ı . Thus, on that interval, M f 00 .x/ N , where M D 0:545345 and N D By Corollary C, If j j 17ı D 17=180, then 0:5. N M .=60/2 sin.33ı / L.33ı / C .=60/2 L.33ı / C 2 2 0:544597 sin.33ı / 0:544660: jE. /j 1 j j 2 27. f .2/ D 4, f 0 .2/ D 0 1, 31. jerrorj 0:000031: 0 f 00 .x/ 1 if x > 0. x 1. 2/: Thus L.3/ D 3. Also, since 1=.2x/ f 00 .x/ 1=x for x > 0, we have for 2 x 3, .1=6/ f 00 .x/ .1=2/. Thus 3C 1 2 1 .3 6 2/2 f .3/ 3 C 1 2 1 .3 2 2/2 : The best approximation for f .3/ is the midpoint of this interval: f .3/ 3 61 . 3. 29. The linearization of g.x/ at x D 2 is L.x/ D g.2/ C g 0 .2/.x 2/ D 1 C 2.x 2/: Thus L.1:8/ D 0:6. If jg 00 .x/j 1 C .x 2/2 for x > 0, then jg 00 .x/j < 1 C . 0:2/2 D 1:04 for 1:8 x 2. Hence 1 g.1:8/ 0:6 with jerrorj < .1:04/.1:8 2/2 D 0:0208: 2 30. If f . / D sin , then f 0 . / D cos and f 00 . / D sin . Since f .0/ D 0 and f 0 .0/ D 1, the linearization of f at D 0 is L. / D 0 C 1. 0/ D . If 0 t , then f 00 .t / 0, so 0 sin . If 0 t , then f 00 .t / 0, so 0 sin . In either case, j sin t j j sin j j j if t is between 0 and . Thus the error E. / in the approximation sin satisfies j j 2 j j3 jE. / j j D : 2 2 xC x2 2Š x3 x4 C : 3Š 4Š If f .x/ D cos x, then f 0 .x/ D sin x, f 00 .x/ D cos x, and f 000 .x/ D sinpx. In particular, f .=4/ D f 000 .=4/ p D 1= 2 and f 0 .=4/ D f 00 .=4/ D 1= 2. Thus 1 P3 .x/ D p 1 2 x f .x/ D ln x 1 f 0 .x/ D x 1 f 00 .x/ D 2 x 2 f 000 .x/ D 3 x 6 f .4/ .x/ D 4 x Thus f .2/ D ln 2 1 f 0 .2/ D 2 1 00 f .2/ D 4 2 f 000 .2/ D 8 6 .4/ f .2/ D 16 4 1 x 2 2 1 C x 4 6 3 : 4 1 1 1 1 P4 .x/ D ln 2C .x 2/ .x 2/2 C .x 2/3 .x 2/4 : 2 8 24 64 4. f .x/ D sec x f 0 .x/ D sec x tan x f 00 .x/ D 2 sec3 x f 000 .x/ D .6 sec2 x sec x 1/ sec x tan x f .0/ D 1 f 0 .0/ D 0 f 00 .0/ D 1 f 000 .0/ D 0 Thus P3 .x/ D 1 C .x 2 =2/. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 0:044: If f .x/ D e x , then f .k/ .x/ D . 1/k e x , so f .k/ .0/ D . 1/k . Thus P4 .x/ D 1 2. .x 2 V D 34 r 3 ) V 4 r 2 r If r D 20:00 and r D 0:20, then V 4.20:00/2 .0:20/ 1005. The volume has increased by about 1005 cm2 . 2/2 . 28. The linearization of f .x/ at x D 2 is 2/ D 4 17 180 Section 4.10 Taylor Polynomials (page 283) f .3/ f .2/ C f .2/.3 2/ D 4 1 D 3. f 00 .x/ 0 ) error 0 ) f .3/ 3. 1 1 1 jf 00 .x/j if 2 x 3, so jErrorj .3 x 2 4 Thus 3 f .3/ 3 41 L.x/ D f .2/ C f 0 .2/.x Thus the percentage error is less than 5%. Using the midpoint of this interval as a new approximation for sin.33ı / ensures that the error is no greater than half the length of the interval: sin.33ı / 0:544629; (PAGE 283) Downloaded by ted cage (sxnbyln180@questza.com) 163 lOMoARcPSD|6566483 www.konkur.in SECTION 4.10 (PAGE 283) 5. f .x/ D x 1=2 1 f 0 .x/ D x 1=2 2 1 3=2 f 00 .x/ D x 4 3 f 000 .x/ D x 5=2 8 Thus ADAMS and ESSEX: CALCULUS 9 Evidently f .2n/ .=2/ D 0 and f .2n 1/ .=2/ D . 1/n 22n 1 . Thus f .4/ D 2 1 f 0 .4/ D 4 3 25 23 C x x 2 3Š 2 5Š 22n 1 2n 1 C C . 1/n x : .2n 1/Š 2 1 32 3 f 000 .4/ D 256 f 00 .4/ D 1 P3 .x/ D 2 C .x 4 1 .x 64 4/ P2n 1 .x/ D 4/2 C 1 .x 512 f .x/ D .1 x/ 1 f 00 .x/ D 2.1 x/ 3 f .x/ D 3Š.1 :: : x/ f .n/ .x/ D nŠ.1 x/ .nC1/ 0 f .x/ D .1 x/ 000 f .0/ D 1 f 0 .0/ D 1 f 00 .0/ D 2 f 000 .0/ D 3Š :: : 2 4 1 2Cx 1 f 0 .x/ D .2 C x/2 2Š f 00 .x/ D .2 C x/3 3Š f 000 .x/ D .2 C x/4 :: : . 1/n nŠ f .n/ .x/ D .2 C x/nC1 f .n/ .0/ D nŠ f .1/ D 0 < Error 1 3 8. 1 3 1 .x 9 1/ C 2 sin.2x/ 000 3 f .x/ D f .x/ D 1 .x 27 2 2 cos.2x/ f .4/ .x/ D 24 sin.2x/ D 24 f .x/ f .5/ .x/ D 24 f 0 .x/ :: : 164 Telegram: @uni_k f 00 .8/ .x 2 8/2 5 < 0:000241: 81 256 Thus 2:07986 < 91=3 < 2:08010. 1 f .1/ D 9 2Š f 00 .1/ D 27 3Š f 000 .1/ D 4 3 :: : . 1/n nŠ f .n/ .1/ D 3nC1 f .x/ D sin.2x/ f 0 .x/ D 2 cos.2x/ 00 8/ C 0 p 10. Since f .x/ D x, then f 0 .x/ D 12 x 1=2 , f 00 .x/ D 41 x 3=2 and f 000 .x/ D 38 x 5=2 . Hence, p Thus Pn .x/ D f .x/ f .8/ C f 0 .8/.x 1 1 D 2 C .x 8/ .x 8/2 12 9 32 1 1 2:07986 91=2 2 C 12 288 f 000 .c/ 10 1 for some c in Error D .9 8/3 D 3Š 27 6 X 8=3 Œ8; 9. For 8 c 9 we have c 8=3 88=3 D 28 D 256 so Pn .x/ D 1 C x C x 2 C x 3 C C x n : f .x/ D 1 f .x/ D x 1=3 , f 0 .x/ D x 2=3 , 3 10 8=3 2 5=3 x , f 000 .x/ D x . f 00 .x/ D 9 27 aD8W Thus 7. 5 2 4/3 : 9. 6. 2 x 1/2 C . 1/n .x 3nC1 f .=2/ D 0 f 0 .=2/ D 2 f 00 .=2/ D 0 f 000 .=2/ D 23 f .4/ .=2/ D 0 f .5/ .=2/ D :: : 25 1/n : 1 61 f .64/ C f 0 .64/.61 64/ C f 00 .64/.61 64/2 2 1 1 1 D 8 C . 3/ . 3/2 7:8103027: 16 2 2048 f 000 .c/ .61 64/3 for 3Š some c between 61 and 64. Clearly R2 < 0. If t 49, and in particular 61 t 64, then The error is R2 D R2 .f I 64; 61/ D jf 000 .t /j 38 .49/ 5=2 D 0:0000223 D K: Hence, jR2 j K j61 3Š 64j3 D 0:0001004: Since R2 < 0, therefore, 7:8103027 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 0:0001004 < 7:8102023 < p p 61 < 7:8103027 61 < 7:8103027: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 11. SECTION 4.10 1 1 , , f 0 .x/ D x x2 6 2 f 00 .x/ D 3 , f 000 .x/ D 4 . x x 2 a D 1 W f .x/ 1 .x 1/ C .x 1/2 2 1 1 .0:02/ C .0:02/2 D 0:9804. 1:02 f 000 .c/ 1 Error D .0:02/3 where .0:02/3 D 3Š X4 1 c 1:02. 1 0:9804 < 0, Therefore, .0:02/3 1:02 1 i.e., 0:980392 < 0:980400. 1:02 14. f .x/ D Since f .x/ D sin x, then f 0 .x/ D cos x, f 00 .x/ D sin x and f 000 .x/ D cos x. Hence, C sin.47ı / D f 4 90 2 1 f Cf0 C f 00 4 4 90 2 4 90 1 1 2 1 p D p Cp 2 2 90 2 2 90 0:7313587: f 000 .c/ 3 for some c between 45ı 3Š 90 and 47ı . Observe that R2 < 0. If 45ı t 47ı , then The error is R2 D 12. Since f .x/ D tan 1 x, then f 0 .x/ D jf 000 .t /j j 2 2x 2 C 6x 1 ; f 00 .x/ D ; f 000 .x/ D : 1 C x2 .1 C x 2 /2 .1 C x 2 /3 1 cos 45ı j D p D K: 2 Hence, K 3 < 0:0000051: 3Š 90 Since R2 < 0, therefore jR2 j Hence, tan 1 .0:97/ f .1/ C f 0 .1/.0:97 1/ C 21 f 00 .1/.0:97 1 1 D C . 0:03/ C . 0:03/2 4 2 4 D 0:7701731: 1/2 f 000 .c/ . 0:03/3 for some c between The error is R2 D 3Š 0.97 and 1. Note that R2 < 0. If 0:97 t 1, then jf 000 .t /j f 000 .1/ D Hence, jR2 j 15. 2C6 < 0:5232 D K: .1:97/3 K j0:97 3Š 0:7701731 f .x/ D sin x f 0 .x/ D cos x f 00 .x/ D sin x f 000 .x/ D cos x f .4/ .x/ D sin x a D 0I n D 7: x3 x5 C0C 0 3Š 5Š x5 x7 x3 C C R7 .x/ 3Š 5Š 7Š sin x D 0 C x 1j3 < 0:0000024: Dx 0:0000024 < tan 1 .0:97/ < 0:7701731 16. f .k/ .x/ D e x for k D 1; 2; 3 : : : 2 aD0W f .x/ 1 C x C e 0:5 1 0:5 C x 2 jErrorj < and 0:020833 < e 0:5 0:604 < e 0:5 < 0:625. .0:5/3 < 0:020834; 6 0:625 < 0, or x7 C R7 ; 7Š sin c 8 x for some c between 0 and x. 8Š For f .x/ D cos x we have f 0 .x/ D sin x f .4/ .x/ D cos x .0:5/2 D 0:625 2 000 f .c/ ec Error D .0:5/3 D . 0:05/3 for some c between 6 6 0:5 and 0. Thus 0 where R7 .x/ D 0:7701707 < tan 1 .0:97/ < 0:7701731: 13. f .x/ D e x , 0:0000051 < sin.47ı / < 0:7313587 0:7313536 < sin.47ı / < 0:7313587: 0:7313587 Since R2 < 0, f 00 .x/ D f .5/ .x/ D cos x sin x f 000 .x/ D sin x f .6/ .x/ D cos x: The Taylor’s Formula for f with a D 0 and n D 6 is cos x D 1 x4 x2 C 2Š 4Š x6 C R6 .f I 0; x/ 6Š where the Lagrange remainder R6 is given by R6 D R6 .f I 0; x/ D f .7/ .c/ 7 sin c 7 x D x ; 7Š 7Š for some c between 0 and x. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k (PAGE 283) Downloaded by ted cage (sxnbyln180@questza.com) 165 lOMoARcPSD|6566483 www.konkur.in SECTION 4.10 (PAGE 283) ADAMS and ESSEX: CALCULUS 9 f .x/ D sin x aD ; nD4 4 1 1 2 1 1 p x sin x D p C p x 4 4 2 2 2 2Š 1 1 3 1 1 4 p x Cp x C R4 .x/ 4 4 2 3Š 24Š 5 1 where R4 .x/ D .cos c/ x 5Š 4 and x. for some c between 4 1 18. Given that f .x/ D , then 1 x 20. 17. f 0 .x/ D 1 .1 x/2 ; f 00 .x/ D 2 .1 x/3 Given that f .x/ D tan x, then f 0 .x/ D sec2 x f 00 .x/ D 2 sec2 x tan x f .3/ .x/ D 6 sec4 x f .4/ .x/ D 8 tan x.3 sec4 x tan x D f .0/ C f 0 .0/x C DxC The Lagrange remainder is Since a D 0, f .n/ .0/ D nŠ. Hence, for n D 6, the Taylor’s Formula is 1 1 x D f .0/ C 6 X f .n/ .0/ nD1 2 nŠ R3 .f I 0; x/ D n x C R6 .f I 0; x/ D 1 C x C x C x 3 C x 4 C x 5 C x 6 C R6 .f I 0; x/: f .7/ .c/ 7 x7 x D 7Š .1 c/8 22. For e u , P4 .u/ D 1 C u C 2 f .x/ D ln x 1 f 0 .x/ D x 1 x2 2Š f 000 .x/ D 3 x 3Š f .4/ .x/ D 4 x 4Š f .5/ .x/ D 5 x 5Š .6/ f .x/ D 6 x 6Š f .7/ D 7 x a D 1; n D 6 23. 166 Telegram: @uni_k For sin2 x D P4 .x/ D 24. 25. 1 1 2 For uD / D .x 3 / 3Š sin.x / .x /5 5Š D x2 x4 : 3 1 1 at u D 0, P3 .u/ D 1 C u C u2 C u3 . Let u 1 2x 2 . Then for at x D 0, 1 C 2x 2 P6 .x/ D 1 26. x6 x8 C : 3Š 4Š x4 2Š .2x/4 .2x/2 C 2Š 4Š 1 / C .x x2. cos.2x/ at x D 0, we have sin x D sin C .x P5 .x/ D 1/ 1 1 2 u3 u4 u2 C C . Let u D 2Š 3Š 4Š x2 C P8 .x/ D 1 2Š 1 .x 1/2 C .x 1/3 2Š 3Š 3Š 4Š 5Š .x 1/4 C .x 1/5 .x 1/6 C R6 .x/ 4Š 5Š 6Š .x 1/3 .x 1/4 .x 1/2 C D .x 1/ 2 3 4 .x 1/5 .x 1/6 C C R6 .x/ 5 6 1 where R6 .x/ D 7 .x 1/7 for some c between 1 and x. 7c ln x D 0 C 1.x x4 e 3x D e 3.xC1/ e 3 9 9 P3 .x/ D e 3 1 C 3.x C 1/ C .x C 1/2 C .x C 1/3 : 2 2 Then for e x : f 00 .x/ D sec2 C / 21. for some c between 0 and x. 19. tan c.3 sec4 X f .4/ .c/ 4 x D 4Š 3 for some c between 0 and x. The Langrange remainder is R6 .f I 0; x/ D f 00 .0/ 2 f 000 .0/ 3 x C x C R3 .f I 0; x/ 2Š 3Š 2 3 x C R3 .f I 0; x/ 3Š 1 2 D x C x3 C x5: 3 15 : nŠ : x/.nC1/ .1 sec2 x/: Given that a D 0 and n D 3, the Taylor’s Formula is In general, f .n/ .x/ D 4 sec2 x cos.3x / D P8 .x/ D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2x 2 C 4x 4 8x 6 : cos.3x/ 1C 32 x 2 2Š 34 x 4 36 x 6 C 4Š 6Š 38 x 8 : 8Š lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 27. SECTION 4.10 Since x 3 D 0 C 0x C 0x 2 C x 3 C 0x 4 C we have Pn .x/ D 0 if 0 n 2; Pn .x/ D x 3 if n 3 32. 28. 29. P2nC1 .x/ D x For ln.1 x3 3 1 ln.1 C x/ 2 1 ln.1 2 x/, P2nC1 .x/ D For tanh 1 x D x P2nC1 .x/ D x C sin.1/ 1 cos c x 2nC3 .2n C 3/Š 1 1 C 3Š 5Š 1 0:84147 7Š correct to five decimal places. 1/2 , f 0 .x/ D 2.x 1/, 2 2 2x C x D 1 2x C x 2 2 f .x/ D .x f 00 .x/ D 2. f .x/ 1 Error = 0 g.x/ D x 3 C 2x 2 C 3x C 4 Quadratic approx.: g.x/ 4 C 3x C 2x 2 x 2nC1 : 2n C 1 Error D x 3 g 000 .c/ 3 Since g 000 .c/ D 6 D 3Š, error D x 3Š 1 in the error formula for the quadratic so that constant 3Š approximation cannot be improved. x5 x 2nC1 x3 C C C : 3 5 2n C 1 34. 31. f .x/ D e xn e x if n is even f .x/ D e x if n is odd x2 x3 x5 e x D1 xC C C . 1/n C Rn .x/ 2Š 3Š nŠ nC1 X for some X between 0 where Rn .x/ D . 1/nC1 .n C 1/Š and x. For x D 1, we have 1 1 1 1 D1 1C C C . 1/n C Rn .1/ e 2Š 3Š nŠ e X x nC1 where Rn .1/ D . 1/nC1 for some X between .n C 1/Š 1 and 0. 1 . We want Therefore, jRn .1/j < .n C 1/Š jRn .1/j < 0:000005 for 5 decimal places. 1 Choose n so that < 0:000005. n D 8 will do .n C 1/Š since 1=9Š 0:0000027. 1 1 1 1 1 1 1 1 C C C Thus e 2Š 3Š 4Š 5Š 6Š 7Š 8Š 0:36788 (to 5 decimal places). 1 x nC1 D .1 1 .n/ 1 x x/.1 C x C x 2 C x 3 C C x n /. Thus D 1 C x C x2 C x3 C C xn C x nC1 : 1 x If jxj K < 1, then j1 xj 1 K > 0, so ˇ nC1 ˇ ˇ ˇx 1 nC1 ˇ ˇ j D O.x nC1 / ˇ 1 x ˇ 1 K jx as x ! 0. By Theorem 11, the nth-order Maclaurin polynomial for 1=.1 x/ must be Pn .x/ D 1 C x C x 2 C x 3 C C x n . 35. Differentiating 1 1 x D 1 C x C x2 C x3 C C xn C x nC1 1 x with respect to x gives 1 x/2 .1 D 1 C 2x C 3x 2 C C nx n 1 C n C 1 nx n x : .1 x/2 Then replacing n with n C 1 gives 1 .1 x/2 D 1C2xC3x 2 C C.nC1/x n C Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x 2nC1 C R2nC2 ; .2n C 1/Š for some c between 0 and x. In order to use the formula to approximate sin.1/ correctly to 5 decimal places, we need jR2nC2 .f I 0; 1/j < 0:000005. Since j cos cj 1, it is sufficient to have 1=.2n C 3/Š < 0:000005. n D 3 will do since 1=9Š 0:000003. Thus 33. x2 2 C . 1/n R2nC2 .f I 0; x/ D . 1/nC1 x 2nC1 C : 2n C 1 x/ at x D 0 we have x5 x3 C 3Š 5Š where 30. For ln.1 C x/ at x D 0 we have x3 x2 C 2 3 In Taylor’s Formulas for f .x/ D sin x with a D 0, only odd powers of x have nonzero coefficients. Accordingly we can take terms up to order x 2nC1 but use the remainder after the next term 0x 2nC2 . The formula is sin x D x 1 sinh x D .e x e x / 2 x 2nC1 x2 1 C C 1CxC P2nC1 .x/ D 2 2Š .2n C 1/Š 2 1 x 2nC1 x C 1 xC 2 2Š .2n C 1/Š x5 x 2nC1 x3 C C C : Dx C 3Š 5Š .2n C 1/Š (PAGE 283) Downloaded by ted cage (sxnbyln180@questza.com) n C 2 .n C 1/x nC1 x : .1 x/2 167 lOMoARcPSD|6566483 www.konkur.in SECTION 4.10 (PAGE 283) If jxj K < 1, then j1 xj 1 ADAMS and ESSEX: CALCULUS 9 because jxj 2p 1 . Clearly 2 t is the smallest value when added to 1 that will not be discarded thus K > 0, and so ˇ ˇ ˇ n C 2 .n C 1/x nC1 ˇ ˇ ˇ n C 2 jx nC1 j D O.x nC1 / x ˇ ˇ .1 K/2 2 .1 x/ as x ! 0. By Theorem 11XXX, the nth-order Maclaurin polynomial for 1=.1 x/2 must be 2 Pn .x/ DD 1 C 2x C 3x C C .n C 1/x n . jF .x/ 4. a) The result of question 3 suggests that computer produces a number C.1 C ˛/ for one expression and C.1 C ˇ/ for the other. ˛ and ˇ will both be discarded positive numbers that are both less than by assumption. Thus for each expression, individually the computer returns C , however the difference between these two expressions is .˛ ˇ/C , where j.˛ ˇ/j < . Thus the computer returns a value for the difference which is less than C but not necessarily equal to 0. Section 4.11 Roundoff Error, Truncation Error, and Computers (page 287) 1. Since the normal rules of algebra (commutativity, associativity, distributivity, etc.) don’t apply to floating-point calculations, we should not expect the plots of mathematically equivalent expressions to be the same in all cases. b) Yes. In certain cases, internal errors may grow sufficiently that the computer returns a value different from C , meaning that either or both of ˛ and ˇ may not be entirely discarded. The algorithm evaluating such expression may be in need of improvement in such a case. 2. Since f .x/ P4 .x/ D O.jxj5 / as x approaches 0, on this very small interval centred at zero we would expect the graph to be the horizontal line through the origin. Instead, there is a band of points having a peculiar structure. The plot can vary between different implementations of Maple on different operating systems, but some of the four horizontal lines (actually envelop curves) proposed in the exercise seem to provide natural boundaries for most of the points. c) Yes. In certain cases internal errors can be negligible or zero, or they can cancel one another. Consider p for example the expression. 1 - 12 , which will not challenge machine precision at all. 3. As noted in section 4.7, a real number x can be represented in binary form as, x D ˙0:d1 d2 : : : d t d tC1 d tC2 : : : 2p Review Exercises 4 (page 287) 1. Since dr=dt D 2r=100 and V D .4=3/ r 3 , we have 4 2 dr 2 6V dV D 3r D 3V D : dt 3 dt 100 100 where each of the base two digits di is either 0 or 1, but d1 D 1, and p is the appropriate power of 2. Consider the floating point number, y (having the same sign as x) given by Hence The volume is increasing at 6%/min. 2. a) Since F must be continuous at r D R, we have mgR2 D mkR; R2 y D ˙0:d1 d2 : : : d t 10p ; which has only t significant binary digits. The distance between x and y thus satisfies jx It follows that jx 168 Telegram: @uni_k F .x/j 2p t 1 2 t jxj or k D g : R b) The rate of change of F as r decreases from R is yj D 0:d tC1 d tC2 2p t : Since d tC1 may or may not be 1, this distance is less than or equal to 2p t . Thus F .x/, the nearest floating point number to x, can be no further than half that distance away, or 1 jx F .x/j 2p t D 2p t 1 : 2 xj jxj: ˇˇ d ˇ .mkr/ ˇ ˇ dr rDR D mk D mg : R The rate of change of F as r increases from R is d mgR2 dr r 2 ˇˇ ˇ ˇ ˇ rDR D 2mgR2 D R3 2 mg : R Thus F decreases as r increases from R at twice the rate at which it decreases as r decreases from R. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES 4 (PAGE 287) Evidently P 0 if x 75, and P ! 0 as x ! 1. P will therefore have a maximum value at a critical point in .75; 1/. For CP: 3. 1=R D 1=R1 C 1=R2 . If R1 D 250 ohms and R2 D 1; 000 ohms, then 1=R D .1=250/ C .1=1; 000/ D 1=200, so R D 200 ohms. If dR1 =dt D 100 ohms/min, then 1 dR 1 dR1 1 dR2 D R2 dt R12 dt R22 dt dR2 1 1 1 dR D .100/ C : 2002 dt 2502 1; 0002 dt 1; 0002 100 D 2502 x2 dP D 4:5 106 dx from which we obtain x D 150. She should charge $150 per bicycle and order N.150/ D 200 of them from the manufacturer. 1; 600: R h R2 is decreasing at 1,600 ohms/min. R b) If R is increasing at 10 ohms/min, then then dR=dt D 10, and dR2 D 1; 0002 dt 10 2002 100 2502 .x 75/2x ; x4 7. a) If R remains constant, then dR=dt D 0, so dR2 D dt 0D D h R r 1; 350: Fig. R-4-7 R2 is decreasing at 1,350 ohms/min. Let r, h and V denote the radius, height, and volume of the cone respectively. The volume of a cone is one-third the base area times the height, so 4. If pV D 5:0T , then dV dT dp V Cp D 5:0 : dt dt dt V D 1 r 2 h: 3 From the small right-angled triangle in the figure, a) If T D 400 K, d T =dt D 4 K/min, and V D 2:0 m3 , then d V =dt D 0, so dp=dt D 5:0.4/=2:0 D 10. The pressure is increasing at 10 kPa/min. 3 b) If T D 400 K, d T =dt D 0, V D 2 m , and d V =dt D 0:05 m3 /min, then p D 5:0.400/=2 D 1; 000 kPa, and 2 dp=dt C 1; 000.0:05/ D 0, so dp=dt D 25. The pressure is decreasing at 25 kPa/min. 5. If x copies of the book are printed, the cost of printing each book is C D 10; 000 C 8 C 6:25 10 7 x 2 : x .h Thus r 2 D R2 0 D V 0 .h/ D 10; 000 C 12:5 10 7 x; x2 so x 3 D 8 109 and x D 2 103 . 2,000 books should be printed. 6. If she charges $x per bicycle, her total profit is $P , where P D .x 75/N.x/ D 4:5 10 6 x 75 x2 2 h R 3 .4Rh 3 hD0 h3 : 3h2 / D or h D h.4R 3 3h/; 4R : 3 V 0 .h/ > 0 if 0 < h < 4R=3 and V 0 .h/ < 0 if 4R=3 < h < 2R. Hence h D 4R=3 does indeed give the maximum value for V . The volume of the largest cone can be inscribed in a sphere of radius R is ! 4R 3 4R 2 4R D 2R V 3 3 3 3 : Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k R/2 D 2Rh2 3 .h The height of any inscribed cone cannot exceed the diameter of the sphere, so 0 h 2R. Being continuous, V .h/ must have a maximum value on this interval. Since V D 0 when h D 0 or h D 2R, and V > 0 if 0 < h < 2R, the maximum value of V must occur at a critical point. (V has no singular points.) For a critical point, Since C ! 1 as x ! 0C or x ! 1, C will be minimum at a critical point. For CP: dC 0D D dx R/2 and .h V D V .h/ D R/2 C r 2 D R2 : Downloaded by ted cage (sxnbyln180@questza.com) D 32 R3 cubic units: 81 169 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 4 (PAGE 287) ADAMS and ESSEX: CALCULUS 9 Thus x D 10 or x D 100=3. The latter CP is not in the interval Œ0; 25, so the maximum occurs at x D 10. The maximum volume of the box is V .10/ D 9; 000 cm3 . 8. C 10. If x more trees are planted, the yield of apples will be .x; C.x// slope = C.x/ = average cost x Y D .60 C x/.800 10x/ D 10.60 C x/.80 x/ x D 10.4; 800 C 20x Fig. R-4-8 This is a quadratic expression with graph opening downward; its maximum occurs at a CP: a) For minimum C.x/=x, we need 0D x 2 /: xC 0 .x/ C.x/ d C.x/ ; D dx x x2 0D 0 so C .x/ D C.x/=x; the marginal cost equals the average cost. b) The line from .0; 0/ to .x; C.x// has smallest slope at a value of x which makes it tangent to the graph of C.x/. Thus C 0 .x/ D C.x/=x, the slope of the line. c) The line from .0; 0/ to .x; C.x// can be tangent to the graph of C.x/ at more than one point. Not all such points will provide a minimum value for the average cost. (In the figure, one such line will make the average cost maximum.) dY D 10.20 dx 2x/ D 20.10 x/: Thus 10 more trees should be planted to maximize the yield. 11. 9. y side flap 2 km side bottom side top 50 cm Fig. R-4-11 side flap 80 cm Fig. R-4-9 It was shown in the solution to Exercise 41 in Section 3.2 that at time t s after launch, the tracking antenna rotates upward at rate d 800t D D f .t /; say: dt 4002 C t 4 If the edge of the cutout squares is x cm, then the volume of the folded box is V .x/ D x.50 D 2x 3 2x/.40 Observe that f .0/ D 0 and f .t / ! 0 as t ! 1. For critical points, x/ 130x 2 C 2; 000x; and is valid for 0 x 25. Since V .0/ D V .25/ D 0, and V .x/ > 0 if 0 < x < 25, the maximum will occur at a CP: 0 D V 0 .x/ D 6x 2 170 Telegram: @uni_k ) 260x C 2; 000 D 2.3x 2 130x C 1; 000/ D 2.3x 100/.x 10/: .4002 C t 4 / 4t 4 .4002 C t 4 /2 4 3t D 4002 ; or t 15:197: 0 D f 0 .t / D 800 The maximum rate at which the antenna must turn is f .15:197/ 0:057 rad/s. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES 4 (PAGE 287) At t D 0, we have x D 0 and y D 1; 000. Thus the position of the ball at time t is given by 12. The narrowest hallway in which the table can be turned horizontally through 180ı has width equal to twice the greatest distance from the origin (the centre of the table) to the curve x 2 C y 4 D 1=8 (the edge of the table). We maximize the square of this distance, which we express as a function of y: S.y/ D x 2 C y 2 D y 2 C 1 8 y4; 200t xD p ; 2 dS D 2y dy 4y 3 D 2y.1 yD 2y 2 /: 13. Let the ball have radius r cm. Its weight is proportional to the volume of metal it contains, so the condition of the problem states that 16 15. 2x 2 1; 000 C x C 1; 000 D : 2002 1 C .x=500/2 The percentage error in the approximation .g=L/ sin .g:L/ is ˇ ˇ ˇ sin ˇ ˇ D 100 100 ˇˇ sin ˇ sin Graphing the left side of this latter equation with a graphics calculator shows a root between 9 and 10. A “solve routine” or Newton’s Method then refines an initial guess of, say, r D 9:5 to give r D 9:69464420373 cm for the radius of the ball. y 100 trajectory 1;000 1C.x=500/2 1 2:06%: 1 1 2 1 D 1 2 cos.2x/ 24 x 4 26 x 6 22 x 2 8 C C O.x / 1 2Š 4Š 6Š x4 2x 6 D x2 C C O.x 8 / 3 45 3 sin2 x 3x 2 C x 4 lim x!0 x6 2 3x 2 x 4 C x 6 C O.x 8 / 3x 2 x 4 15 D lim x!0 x6 2 2 C O.x 2 / D : D lim x!0 15 15 17. f .x/ D tan 1 x, f 0 .x/ D f 000 .x/ D 6x 2 2 . .1 C x 2 /3 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k =9 sin.=9/ sin2 x D Fig. R-4-14 200 32t C p : 2 16. x If the origin is at sea level under the launch point, and x.t / and y.t / are the horizontal and vertical coordinates of the cannon ball’s position at time t s after it is fired, then d 2y d 2x D 0; D 32: dt 2 dt 2 p At t D 0, we have dx=dt D dy=dt D 200= 2, so 1 : Since lim!0 =.sin / D 1, the percentage error ! 0 as ! 0. Also, = sin grows steadily larger as j j increases from 0 towards =2. Thus the maximum percentage error for j j 20ı D =9 will occur at D =9. This maximum percentage error is 14. dy D dt 2x 2 C x C 1; 000: 2002 Graphing both sides of this equation suggests a solution near x D 1; 900. Newton’s Method or a solve routine then gives x 1; 873. The horizontal range is about 1,873 ft. 1 4 3 4 3 4 r .r 2/3 D r 3 3 2 3 r 3 12r 2 C 24r 16 D 0: dx 200 D p ; dt 2 16 The cannon ball strikes the ground when The CPs are given by y D 0 (already considered), p and y 2 D 1=2, where S.y/ D 3=8. Since 3=8 > 1= 8, this is the maximum p value of S. The hallway must therefore be at least 2 3=8 1:225 m wide. yD 200t 16t 2 C p C 1; 000: 2 We can obtain the Cartesian equation for the path of the cannon ball by solving the first equation for t and substituting into the second equation: .0 y .1=8/1=4 /: p Note that S.0/ D 1=8 and S..1=8/1=4 / D 1= 8 > S.0/. For CP: 0D yD Downloaded by ted cage (sxnbyln180@questza.com) 1 2x , f 00 .x/ D , 1 C x2 .1 C x 2 /2 171 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 4 (PAGE 287) ADAMS and ESSEX: CALCULUS 9 x 1 .x 1/2 C . 4 2 4 1 1 0:832898: On Œ1; 1:1, Thus tan 1 .1:1/ C 4 20 400 we have This line passes through .1; 1/ if 1 D e a .2 a/. A solve routine gives a 1:1461932. The corresponding value of e a is about 0.3178444. The car is at .a; e a /. About x D 1, P2 .x/ D jf 000 .x/j 6.1:1/2 2 D 0:6575: .1 C 1/3 Thus the error does not exceed in absolute value. 0:6575 .1:1 3Š Challenging Problems 4 (page 289) 1/3 :00011 1. 18. The second approximation x1 is the x-intercept of the tangent to y D f .x/ at x D x0 D 2; it is the x-intercept of the line 2y D 10x 19. Thus x1 D 19=10 D 1:9. dV D kx 2 .V0 dt V /. a) If V D x 3 , then 3x 2 k dx D .V0 dt 3 19. y y D cos x kx 2 y D cos x and y D .x 1/2 intersect at x D 0 and at a point x between x D 1 and x D =2 1:57. Starting with an initial guess x0 D 1:3, and iterating the Newton’s Method formula 20. The square of the distance from .2; 0/ to .x; ln x/ is S.x/ D .x 2/2 C .ln x/2 , for x > 0. Since S.x/ ! 1 as x ! 1 or x ! 0C, the minimum value of S.x/ will occur at a critical point. For CP: Telegram: @uni_k 2C ln x x : k .V0 3 If the car is at .a; e a /, then its headlight beam lies along the tangent line to y D e x there, namely a/ D e a .1 C x x3/ D kV0 ; 6 that is, if x D .V0 =2/1=3 . Then V D V0 =2. 2. dy D ky. dt kt Thus, y D C e . Given that at t D 0, y D 4, then 4 D y.0/ D C . Also given that at t D 10, y D 2, thus, Let the speed of the tank be v where v D 2 D y.10/ D 4e 10k ) k D 1 10 ln 2: 1 dy ln 2/t 10 and v D 1 D. ln 2/y: The dt 10 dy 1 slope of the curve xy D 1 is m D . Thus, the D dx x 2 1 equation of the tangent line at the point ; y0 is y0 Hence, y D 4e . We solve this equation using a TI-85 solve routine; x 1:6895797. The minimum distance from the origin to p y D e x is S.x/ 0:6094586. 172 x3/ < 0 c) Initially, x grows at rate kV0 =3. The rate of growth of x will be half of this if .xn 1/2 cos xn ; 2.xn 1/ C sin xn we get x4 D x5 D 1:40556363276. To 10 decimal places the two roots of the equation are x D 0 (exact), and x D 1:4055636328. y D e a C e a .x k2 2 x .V0 3 dx D dt for 0 < x < x0 . Thus the edge length is increasing at a decreasing rate. Fig. R-4-19 21. x 3 /: 1/2 x 1 0 D S 0 .x/ D 2 x x 3 /, so b) The rate of growth of the edge is .k=3/.V0 x 3 /, 1=3 which is positive if 0 x < x0 D V0 . The time derivative of this rate is y D .x xnC1 D xn dV dx D D kx 2 .V0 dt dt y D y0 a/: 1 1 y0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2 x 1 ; y0 i.e., y D 2y0 xy02 : lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 4 (PAGE 289) y 3. yD 1 x .1=y0 ;y0 / y x x b) Starting with x0 D 20, we iterate xnC1 D f .xn /. The first three iterations give Fig. C-4-2 x1 11:03; 2 Hence, the x-intercept is x D and the y-intercept is y0 y D 2y0 . Let be the angle between the gun and the y-axis. We have 2 1 4 x y0 D 2 D 2 tan D D y 2y0 y y0 d 8 dy 2 sec D 3 : dt y dt ) a) If q D 0:99, the number of tests required is T D N ..1=x/ C 1 0:99x /. T is a decreasing function for small values of x because the term 1=x dominates. It is increasing for large x because 0:99x dominates. Thus T will have a minimum value at a critical point, provided N is sufficiently large that the CP is in .0; N /. For CP: 1 dT x DN 0:99 ln.0:99/ 0D dx x2 x .0:99/ x2 D ln.0:99/ .0:99/ x=2 xD p D f .x/; say: ln.0:99/ 4. x3 10:51: This suggests the CP is near 10.5. Since x must be an integer, we test x D 10 and x D 11: T .10/ 0:19562 and T .11/ 0:19557. The minimum cost should arise by using groups of 11 individuals. p P D 2 L=g D 2L1=2 g 1=2 . a) If L remains constant, then dP g D L1=2 g 3=2 g dg P L1=2 g 3=2 1 g g D : 1=2 1=2 P 2 g 2L g P Now sec2 D 1 C tan2 D 1 C x2 10:54; If g increases by 1%, then g=g D 1=100, and P =P D 1=200. Thus P decreases by 0:5%. y 4 C 16 16 D ; 4 y y4 b) If g remains constant, then so d D dt 8y dy y 4 C 16 dt The maximum value of y2 y 4 C 16 D .y 4 C 16/2y y 2 .4y 3 / .y 4 C 16/2 5 2y D 32y; or y D 2. Therefore the maximum rate of rotation of the gun turret must be 8k 22 24 C 16 D kD P occurs at a critical point: 0D , dP L D L 1=2 g 1=2 L dL 1 L P L 1=2 g 1=2 L D : P 2 L 2L1=2 g 1=2 8ky 2 : y 4 C 16 1 ln 2 0:0693 rad/m, 10 and occurs when your tank is 2 km from the origin. If L increases by 2%, then L=L D 2=100, and P =P D 1=100. Thus P increases by 1%. 5. dV D dt p k y; dy dV dy kp p D D k y, so D y. dt dt dt A kt 2 p y0 , then y.0/ D y0 , and b) If y.t / D 2A k dy kt p D2 y0 dt 2A 2A kp D y.t /: A a) A Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k V D Ay. Downloaded by ted cage (sxnbyln180@questza.com) 173 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 4 (PAGE 289) ADAMS and ESSEX: CALCULUS 9 A D 0 if x D 0 or x D P =2 and A > 0 between these values of x. The maximum area will therefore occur at a critical point. Thus the given expression does solve the initial-value problem for y. kT p p c) If y.T / D 0, then D y0 , so k D 2A y0 =T . 2A Thus y.t / D p p 2A y0 t 2 D y0 1 2AT y0 t T 2 P .P x/.P 4x/ x.P dA D dx 4 .P x/2 0 D P 2 5P x C 4x 2 C P x 2x 2 0D : 2x 2 t1 T 2 D y0 : 2 p .1= 2//. Thus t1 D T .1 6. If the depth of liquid in the tank at time t is y.t /, then the surface of the liquid has radius r.t / D Ry.t /=H , and the volume of liquid in the tank at that time is V .t / D 3 Ry.t / H 2 y.t / D 8. 3 R2 y.t / : 2 3H dV R2 2 dy 3y D D 2 3H dt dt The slope of y D x 3 C ax 2 C bx C c is y 0 D 3x 2 C 2ax C b; which ! 1 as x ! ˙1. The quadratic expression y 0 takes each of its values at two different points except its minimum value, which is achieved only at one point given by y 00 D 6x C 2a D 0. Thus the tangent to the cubic at x D a=3 is not parallel to any other tangent. This tangent has equation p k y. Thus By Torricelli’s law, d V =dt D p k y; a3 a3 ab C Cc 27 9 3 2 a 2a2 a C Cb xC 3 3 3 a2 a3 CcC b D x: 27 3 yD k y 3=2 , where k1 D kH 2 =.R2 /. 1 t 2=5 , then y.0/ D y0 , y.T / D 0, and If y.t / D y0 1 T or, dy=dt D 2 dy D y0 1 dt 5 t T 3=5 1 T D k1 y 3=2 ; 9. where k1 D 2y0 =.5T /. Thus this function y.t / satisfies the conditions of the problem. 7. P A .P 2 2 x 2 p x2 C y2 2 y/ D x C y P C x C y 2 C 2xy y.2P 2x/ D P 2 P .P 2x/ : yD 2.P x/ 2 2P x 2P x 174 h C Fig. C-4-9 a) The total resistance of path AP C is 2P y D x 2 C y 2 The area of the triangle is xy P Px AD D 2 4 P B If p the triangle has legs x and y and hypotenuse x 2 C y 2 , then P DxCyC Telegram: @uni_k 4P x C P 2 D 0: This quadratic has two roots, but the only one in Œ0; P =2 is p 4P 16P 2 8P 2 1 xD DP 1 p : 4 2 2 This value of x gives A.x/ D 21 P 2 1 p1 un2 for 2 the maximum area of the triangle. (Note that the maximal triangle is isosceles, as we might have guessed.) d) Half the liquid drains out in time t1 , where y0 1 2x/. 1/ kjP C j kjAP j C 2 r r22 1 h csc L h cot C : Dk r12 r22 RD We have 2 dR csc D kh d r12 2x 2 : x Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) csc cot r22 ; lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 4 (PAGE 289) r2 csc D 12 , that cot r2 is, cos D .r2 =r1 /2 or D cos 1 ..r2 =r1 /2 /. This CP will give the minimum resistance if it is in the interval of possible values of , namely Œtan 1 .h=L/; =2; otherwise the minimum will occur for P D A. Thus, for large L, P should be chosen to make cos D .r2 =r1 /2 . For a given depth y, R will be maximum if h.y is maximum. This occurs at the critical point h D y=2 of the quadratic Q.h/ D h.y h/. so the CP of R is given by h/ c) By the result of part (c) of Problem 3 (with y replaced by y h, the height of the surface of the water above the drain in the current problem), we have t 2 ; for 0 t T . y.t / h D .y0 h/ 1 T b) This is the same problem as that in (a) except that r1 and r2 are replaced with r12 and r22 , respectively. Thus the minimum resistance corresponds to choosing P so that cos D .r2 =r1 /4 . This puts P closer to B than it was in part (a), which is reasonable since the resistance ratio between the thin and thick pipes is greater than for the wires in part (a). As shown above, the range of the spurt at time t is s r 2 R.t / D k h y.t / h : g Since R D R0 when y D y0 , we have R0 : 2p h.y0 h/ g r h y.t / h D R0 1 Therefore R.t / D R0 p h.y0 h/ kD s 10. 11. y t . T x h 25 x 25 2x 25 2x R x 25 cm x y Fig. C-4-10 25 cm a) Let the origin be at the point on the table directly under the hole. If a water particle leaves the tank with horizontal velocity v, then its position .X.t /; Y.t //, t seconds later, is given by d 2X D0 dt 2 dX Dv dt X D vt d 2Y D dt 2 dY D dt g gt 1 2 gt C h: 2 Y D The range R of the particle (i.e., of the spurt)pis the value of X when Y D 0, that is, at time t D 2h=g. p Thus R D v 2h=g. p b) Since v D k y h, the range R is a function of y, the depth of water in the tank. RDk s 2p h.y g h/: Fig. C-4-11 Note that the vertical back wall of the dustpan is perpendicular to the plane of the top of the pan, not the bottom. The volume of the pan is made up of three parts: a triangular prism (the centre part) having height x, width 25 2x, and depth y (all distances 2 2 2 in cm), p where y C xp D .25 x/ , and so y D 625 50x D 5 25 2x, and two triangular pyramids (one on each side) each having height x and a right-triangular top with dimensions x and y. The volume of the pan is, therefore, 1 1 1 V D xy.25 2x/ C 2 xy x 2 3 2 1 2 D xy 25 2x C x 2 3 5 p D x 25 2x.75 4x/ D V .x/: 6 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 175 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 4 (PAGE 289) The appropriate values for x are 0 x 25=2. Note that V .0/ D V .25=2/ D 0 and V .x/ > 0 in .0; 25=2/. The maximum volume will therefore occur at a critical point: 0D 176 Telegram: @uni_k dV D dx 25 4x 2 85x C 375 p 6 25 2x ADAMS and ESSEX: CALCULUS 9 (after simplification). The quadratic in the numerator factors to .x 15/.4x 25/, so the CPs are x D 15 and x D 25=4. Only x D 25=4 is in the required interval. The maximum volume of the dustpan is V .25=4/ 921 cm3 . Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHAPTER 5. SECTION 5.1 (PAGE 295) INTEGRATION 16. mC6 m X kD 5 Section 5.1 Sums and Sigma Notation (page 295) 17. n X iD1 1. 4 X i 3 D 13 C 23 C 33 C 43 100 X j 1 2 3 100 D C C C C j C1 2 3 4 101 n X 3i D 3 C 32 C 33 C C 3n n X1 . 1/i D1 i C1 iD1 2. j D1 3. iD1 4. iD0 5. n X . 2/j D .j 2/2 n X j2 j D1 7. n3 D 1 1 C 2 3 20. 4 2 2 C 2 12 2 2 . 1/ 2 C C 32 .n 2/2 22. 1 4 9 n2 C 3 C 3 C C 3 3 n n n n 9 X 32 C 42 x C x2 1 1 C 4 9 52 C 992 D n 1 . 1/ n2 200 X 100 X j D0 sin j D 100 X sin.i .2i i 2 / D 2nC1 2 ln m D ln 1 C ln 2 C C ln n D ln.nŠ/ n X e i=n D e .nC1/=n 1 e 1=n 1 2 C 2 C C 2 (200 terms) equals 400 ( x nC1 if x ¤ 1 1 x nC1 if x D 1 ( 1 C x 2nC1 if x ¤ x 3 C C x 2n D 1Cx 2n C 1 if x D 2 n 1 x C x2 1 Let f .x/ D 1 C x C x 2 C C x 100 D Then ix i 1 1 1 x 101 1 if x ¤ 1. x 1 f 0 .x/ D 1 C 2x C 3x 2 C C 100x 99 iD1 d x 101 1 100x 101 101x 100 C 1 D : dx x 1 .x 1/2 D iD0 2n X . 1/i x i 27. iD0 22 D n X . 1/i 1 D i2 D iD1 28. 32 C 42 52 C C 982 992 Œ.2k/2 .2k C 1/2 D Œ4k 2 49 X kD1 49 X Œ4k C 1 D 4 kD1 Let s D 49 X 4k 2 4k 1 kD1 49 50 2 49 D 4; 949 2 3 n 1 C C C C n : Then 2 4 8 2 1/ s 1 2 3 n D C C C C nC1 : 2 4 8 16 2 iD1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 6 n.n C 1/.2n C 1/ 2 25. 26. 3n n X iD1 . 1/i i 2 X i 1 2 3 4 n C C C C C n D 2 4 8 16 2 2i 99 X . n 1/ 1 3/ D xi iD1 15. . k 1 C x C x C C x D n 14. 2.1; 000/.1; 001/ C 3; 000 D 1; 004; 000 2 24. iD2 n X x 3 C C x 2n D C 23. iD5 99 X n X iD0 i 1 C x C x2 C x3 C C xn D 13. 1 n X n.n C 1/ n.n C 1/.2n C 7/ n.n C 1/.2n C 1/ C2 D 6 2 6 .2j C 3/ D mD1 n n 10. 1 C 2x C 3x 2 C 4x 3 C C 100x 99 D 12. 1 21. 5 8. 2 C 2 C 2 C C 2 (200 terms) equals 11. 1;000 X iD1 . 1/n 1 n C 3 5C6C7C8C9D 9. 22 .i 2 C2i / D j D1 19. 1 6/2 C 1 iD1 kD1 j D3 6. 18. X 1 D k2 C 1 ..i Downloaded by ted cage (sxnbyln180@questza.com) 177 lOMoARcPSD|6566483 www.konkur.in SECTION 5.1 (PAGE 295) ADAMS and ESSEX: CALCULUS 9 Subtracting these two sums, we get 1 1 1 1 s D C C C C n 2 2 4 8 2 1 1 .1=2n / n D nC1 2 1 .1=2/ 2 nC2 : D1 2nC1 n 2nC1 Thus s D 2 C .n C 2/=2n . 29. n X f .i C 1/ n X f .i / D f .i C 1/ iDm D iDm nC1 X f .j / j DmC1 n X Fig. 5.1-34 f .i / iDm n X f .i / 35. n X iDm D f .n C 1/ f .m/; because each sum has only one term that is not cancelled by a term in the other sum. It is called “telescoping” because the sum “folds up” to a sum involving only part of the first and last terms. 30. 10 X m X iD1 iD n.n C 1/ ; 2 we write n copies of the identity .k C 1/2 k 2 D 2k C 1; one for each k from 1 to n: .n4 nD1 31. To show that .2j j D1 .n 1/4 D 104 2j 1 / D 2m 04 D 10; 000 20 D 2m 22 32 42 1 .n C 1/2 12 D 2.1/ C 1 22 D 2.2/ C 1 32 D 2.3/ C 1 :: : n2 D 2.n/ C 1: Adding the left and right sides of these formulas we get 32. 2m X iDm 33. m X j D1 1 i 1 i C1 D m X 1 D j.j C 1/ j D1 1 m 1 j 1 mC1 D 2m C 1 m.2m C 1/ 1 j C1 D1 n 1 D nC1 nC1 Hence, 36. 34. The number of small shaded squares is 1 C 2 C P C n. Since each has area 1, the total area shaded is niD1 i . But this area consists of a large right-angled triangle of area n2 =2 (below the diagonal), and n small triangles (above the diagonal) each of area 1/2. Equating these areas, we get iD1 178 Telegram: @uni_k iD n2 1 n.n C 1/ Cn D : 2 2 2 n X Pn iD1 i D 1 2 .n C 2n C 1 2 1 iD1 i C n: n/ D n.n C 1/ . 2 P The formula niD1 i D n.n C 1/=2 holds for n D 1, since it says 1 D 1 in this case. Now assume that it holds for P n D some number k 1; that is, kiD1 i D k.k C 1/=2. Then for n D k C 1, we have kC1 X iD1 n X 12 D 2 .n C 1/2 iD k X iD1 i C.kC1/ D .k C 1/.k C 2/ k.k C 1/ C.kC1/ D : 2 2 Thus the formula also holds for n D k C 1. By induction, it holds for all positive integers n. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 5.1 (PAGE 295) P 37. The formula niD1 i 2 D n.n C 1/.2n C 1/=6 holds for n D 1, since it says 1 D 1 in this case. Now assume that it for n D some number k 1; that is, Pholds k 2 i D k.k C 1/.2k C 1/=6. Then for n D k C 1, we iD1 have kC1 X iD1 i2 D k X iD1 i 2 C .k C 1/2 40. k.k C 1/.2k C 1/ D C .k C 1/2 6 kC1 2 D Œ2k C k C 6k C 6 6 kC1 D .k C 2/.2k C 3/ 6 .k C 1/..k C 1/ C 1/.2.k C 1/ C 1/ : D 6 iD1 ri 1 D k X iD1 n X iD1 Thus the formula also holds for n D k C 1. By induction, it holds for all positive integers n. P 38. The formula niD1 r i 1 D .r n 1/=.r 1/ (for r ¤ 1) holds for n D 1, since it says 1 D 1 in this case. Now assume that it holds for n D some number k 1; that is, Pk i 1 r D .r k 1/=.r 1/. Then for n D k C 1, we iD1 have kC1 X that L-shaped region has area i 3 . The sum of the areas of the n L-shaped regions is the area of the large square of side n.n C 1/=2, so ri 1 C rk D rk r 1 r kC1 1 C rk D : 1 r 1 Thus the formula also holds for n D k C 1. By induction, it holds for all positive integers n. i3 D n.n C 1/ 2 2 : To show that n X j D1 j 3 D 13 C 23 C 33 C C n3 D n2 .n C 1/2 ; 4 we write n copies of the identity .k C 1/4 k 4 D 4k 3 C 6k 2 C 4k C 1; one for each k from 1 to n: 24 34 44 .n C 1/4 14 D 4.1/3 C 6.1/2 C 4.1/ C 1 24 D 4.2/3 C 6.2/2 C 4.2/ C 1 34 D 4.3/3 C 6.3/2 C 4.3/ C 1 :: : n4 D 4.n/3 C 6.n/2 C 4.n/ C 1: Adding the left and right sides of these formulas we get .n C 1/4 14 D 4 D4 39. n X j D1 n X j D1 j3 C 6 j3 C n X j D1 j2 C 4 n X j D1 j Cn 4n.n C 1/ 6n.n C 1/.2n C 1/ C C n: 6 2 Hence, n 4 n X j D1 :: : so 3 3 Fig. 5.1-39 1 2 n The L-shaped region with short side i is a square of side i.i C 1/=2 with a square of side .i 1/i=2 cut out. Since i.i C 1/ 2 2 D .i 1/i n X 2 i 4 C 2i 3 C i 2 j3 D 2n.n C 1/ n n2 .n C 1/2 : 4 kC1 X i3 D k X iD1 2 i 3 C .k C 1/3 .k C 1/2 2 k .k C 1/2 C .k C 1/3 D Œk C 4.k C 1/ 4 4 2 .k C 1/ D .k C 2/2 : 4 D .i 4 4 2i 3 C i 2 / D i 3; Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k n.n C 1/.2n C 1/ P 41. The formula niD1 i 3 D n2 .n C 1/2 =4 holds for n D 1, since it says 1 D 1 in this case. Now assume that it holds for Pkn D3 some2 number2k 1; that is, iD1 i D k .k C 1/ =4. Then for n D k C 1, we have iD1 2 1 D n2 .n C 1/2 j D1 2 1 j 3 D .n C 1/4 Downloaded by ted cage (sxnbyln180@questza.com) 179 lOMoARcPSD|6566483 www.konkur.in SECTION 5.1 (PAGE 295) ADAMS and ESSEX: CALCULUS 9 Thus the formula also holds for n D k C 1. By induction, it holds for all positive integers n. P 42. To find jnD1 j 4 D 14 C 24 C 34 C C n4 , we write n copies of the identity .k C 1/5 32 33 3n 1 3 C C C C n!1 n n n n n 3 D lim 2 .1 C 2 C 3 C C n/ n!1 n 3 n.n C 1/ 3 D lim 2 D sq. units: n!1 n 2 2 A D lim k 5 D 5k 4 C 10k 3 C 10k 2 C 5k C 1; one for each k from 1 to n: 25 35 45 y 15 D 5.1/4 C 10.1/3 C 10.1/2 C 5.1/ C 1 25 D 5.2/4 C 10.2/3 C 10.2/2 C 5.2/ C 1 35 D 5.3/4 C 10.3/3 C 10.3/2 C 5.3/ C 1 :: : n5 D 5.n/4 C 10.n/3 C 10.n/2 C 5.n/ C 1: .n C 1/5 .1;3/ yD3x Adding the left and right sides of these formulas we get .n C 1/5 15 D 5 n X j D1 j 4 C 10 n X j D1 j 3 C 10 n X j D1 j2 C 5 n X j D1 1 n j C n: Substituting the known formulas for all the sums except Pn 4 j D1 j , and solving for this quantity, gives n X j D1 j4 D n.n C 1/.2n C 1/.3n2 C 3n 30 1/ : 43. iD1 n X iD1 n X iD1 n X iD1 1 1 5 i D n6 C n5 C n4 6 2 12 1 2 n 12 1 1 1 i D n7 C n6 C n5 7 2 2 1 3 1 n C n 6 42 5 6 i7 D 7 1 8 1 7 n C n C n6 8 2 12 i8 D 1 9 1 8 n C n C 9 2 iD1 i 10 D n 1 n D1 n n 4 n x 2. This is similar to #1; the rectangles now have width 3=n and the i th has height 2.3i=n/C1, the value of 2xC1 at x D 3i=n. The area is n X 3 3i A D lim 2 C1 n!1 n n iD1 n 3 18 X iC n n!1 n2 n D lim 3. 1 11 1 10 n C n C : 11 2 iD1 18 n.n C 1/ D lim 2 C 3 D 9 C 3 D 12sq. units: n!1 n 2 7 4 1 n C n2 24 12 We would guess (correctly) that n X 3 n Fig. 5.2-1 Of course we got Maple to do the donkey work! n X 2 n This is similar to #1; the rectangles have width .3 1/=n D 2=n and the i th has height the value of 2x at x D 1 C .2i=n/. The area is n X 2i 2 2C2 n!1 n n A D lim iD1 1 n Section 5.2 Areas as Limits of Sums (page 301) 1. The area is the limit of the sum of the areas of the rectangles shown in the figure. It is 180 Telegram: @uni_k 8 X 2 iC n n!1 n2 n D lim iD1 8 n.n C 1/ C 2 D 4 C 2 D 6sq. units: D lim 2 n!1 n 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 5.2 (PAGE 301) A D lim n!1 n X 3 iD1 3C3 n 3i C4 n D n 27 X 3 iC n n!1 n2 n D lim a 3 n.n C 1/.2n C 1/ a C .n/ 6 n a3 .n C 1/.2n C 1/ D C a: 6 n2 3 a Area D lim Sn D C asq. units: n!1 3 4. This is similar to #1; the rectangles have width .2 . 1//=n D 3=n and the i th has height the value of 3x C 4 at x D 1 C .3i=n/. The area is n y iD1 yDx 2 C1 27 n.n C 1/ 27 33 D lim 2 C3D C3D sq. units: n!1 n 2 2 2 5. The area is the limit of the sum of the areas of the rectangles shown in the figure. It is 2 A D lim n!1 n " " 2 1C n 2 # 4 2 2n 2 C 1C C C 1 C n n 4 4 8 16 C 2 C1C C 2 n n n n # 4n2 4n C 2 C C 1 C n n 8 n.n C 1/.2n C 1/ 8 n.n C 1/ C 3 D lim 2 C 2 n!1 n 2 n 6 8 26 D2C4C D sq. units: 3 3 D lim 2 n!1 n 1C y yDx 2 x1 x2 a x Fig. 5.2-6 7. The required area is (see the figure) " 3 3 3 2 A D lim C2 1C 1C C3 n!1 n n n 6 2 6 C 1C C2 1C C3 n n # 3n 3n 2 C2 1C C3 C C 1C n n " 3 6 6 32 D lim C 2 2C C3 1 n!1 n n n n 2 6 12 12 C 2 2C C3 C 1 n n n # 6n 9n2 6n 2C C 2 C3 C C 1 n n n 27 n.n C 1/.2n C 1/ D lim 6 C 3 n!1 n 6 D 6 C 9 D 15sq. units: y 1 1C 2 n 2n 1C n D3 x yDx 2 C2xC3 Fig. 5.2-5 a 6. Divide Œ0; a into n equal subintervals of length x D n ia by points xi D , .0 i n/. Then n # " n X ia 2 a C1 Sn D n n iD1 D n a 3 X n iD1 n i2 C aX .1/ n 1 1C iD1 (Use Theorem 1(a) and 1(c).) Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 3 n 1C x 3n D2 n Fig. 5.2-7 181 lOMoARcPSD|6566483 www.konkur.in SECTION 5.2 (PAGE 301) ADAMS and ESSEX: CALCULUS 9 8. y 1 x 1 A 10. y D x2 y 1 Fig. 5.2-8 The region in question lies between x D 1 and x D 1 and is symmetric about the y-axis. We can therefore double the area between x D 0 and x D 1. If we divide this interval into n equal subintervals of width 1=n and use the distance 0 .x 2 1/ D 1 x 2 between y D 0 and y D x 2 1 for the heights of rectangles, we find that the required area is n X 1 i2 A D 2 lim 1 n!1 n n2 iD1 n X i2 1 D 2 lim n!1 n n3 iD1 n n.n C 1/.2n C 1/ D2 D 2 lim n!1 n 6n3 A y D x2 The height of the region at position x is 0 .x 2 2x/ D 2x x 2 . The “base” is an interval of length 2, so we approximate using n rectangles of width 2=n. The shaded area is n X 2 4i 2 2i 2 n!1 n n n2 iD1 n X 8i 2 8i D lim n!1 n2 n3 iD1 8 n.n C 1/ 8 n.n C 1/.2n C 1/ D lim n!1 n2 2 n3 6 4 8 D sq. units: D4 3 3 A D lim 4 4 D sq. units: 6 3 y 2 2x Fig. 5.2-10 9. 4 x A yD1 x 2 x 11. y Fig. 5.2-9 The height of the region at position x is 0 .1 x/ D x 1. The “base” is an interval of length 2, so we approximate using n rectangles of width 2=n. The shaded area is n X 2i 2 1 2C n!1 n n iD1 n X 2 4i D lim C 2 n!1 n n iD1 n.n C 1/ 2n C4 D 2 C 2 D 4 sq. units: D lim n!1 n 2n2 A D lim 182 Telegram: @uni_k Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) y D 4x x2 C 1 A 4 Fig. 5.2-11 x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 5.2 (PAGE 301) The height of the region at position x is 4x x 2 C 1 1 D 4x x 2 . The “base” is an interval of length 4, so we approximate using n rectangles of width 4=n. The shaded area is Now we can use l’H^opital’s rule to evaluate 22=n 1 0 1 n!1 0 n 2 22=n ln 2 n2 D lim 1 n!1 n2 D lim 2.2=n/C1 ln 2 D 2 ln 2: lim n.22=n 1/ D lim n!1 n X 4 16i 2 4i 4 n!1 n n n2 iD1 n X 64i 2 64i D lim n!1 n2 n3 iD1 64 n.n C 1/ 64 n.n C 1/.2n C 1/ D lim n!1 n2 2 n3 6 32 64 D sq. units: D 32 3 3 A D lim 12. Divide Œ0; b into n equal subintervals of length x D by points xi D Sn D ib , .0 i n/. Then n n!1 y b n yD2x n n X b .ib=n/ b X .b=n/ i e e D n n iD1 iD1 i 1 X b D e .b=n/ e .b=n/ n 1 n 1C (Use Thm. 6.1.2(d).) iD1 .b=n/n b e 1 D e .b=n/ .b=n/ n e 1 b .b=n/ e b 1 : D e n e .b=n/ 1 b Let r D : n Area D lim Sn D .e b n!1 D .e b 3 square units. 2 ln 2 Thus the area is 14. r!0C 1 1/.1/ lim r D e b r!0C e r r!0C e r 1 1C 4 6 1C n n 1C x 2n D1 n Fig. 5.2-13 " 3 3 # 2b nb b b 3 C C C Area D lim n!1 n n n n D lim 1/ lim e r lim 2 n b4 n!1 n4 4 .13 C 23 C 33 C C n3 / b4 b n2 .n C 1/2 D sq. units: n!1 n4 4 4 0 0 D lim y 1sq. units: yDx 3 13. The required area is i 2 h 1C.2=n/ 2 C 2 1C.4=n/ C C 2 1C.2n=n/ n!1 n 2 n 1 22=n 2=n 2=n 2=n 1C 2 C 2 C C 2 D lim n!1 n n 1 22=n 22=n D lim n!1 n 22=n 1 1 D lim 22=n 3 2=n n!1 n.2 1/ 1 D 3 lim : n!1 n.22=n 1/ A D lim b n 3b n .n 1/b nb Db n n x Fig. 5.2-14 15. 1=n b Let t D and let a x0 D a; x1 D at; x2 D at 2 ; : : : ; xn D at n D b: Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2b n Downloaded by ted cage (sxnbyln180@questza.com) 183 lOMoARcPSD|6566483 www.konkur.in SECTION 5.2 (PAGE 301) ADAMS and ESSEX: CALCULUS 9 The i th subinterval Œxi 1 ; xi has length xi D 1 at i 1 .t 1/. Since f .xi 1 / D i 1 , we form the sum at n X 1 at i 1 .t 1/ Sn D at i 1 iD1 1=n b D n.t 1/ D n 1 : a 17. y A1 b 1 and c D . The area under the curve is n a cr 1 0 A D lim Sn D lim n!1 r!0C r 0 b c r ln c D ln c D ln square units. D lim r!0C 1 a 1 2 x Fig. 5.2-17 2i represents a sum of areas of n rectn n iD1 angles each of width 2=n and having heights equal to the height to the graph y D 1 x at the points x D 2i=n. Half of these rectangles have negative height, and limn!1 Sn is the difference A1 A2 of the areas of the two triangles in the figure above. It has the value 0 since the two triangles have the same area. y sn D This is not surprising because it follows from the definition of ln. y 1 x x A2 Let r D yD yD1 18. n X 2 1 y D 2 C 3x A a x1 x2 b x 1 Fig. 5.2-18 Fig. 5.2-15 n X 1 3i 2 C represents a sum n2 n n iD1 iD1 of areas of n rectangles each of width 1=n and having heights equal to the height to the graph y D 2 C 3x at the points x D i=n. Thus limn!1 Sn is the area of the trapezoid in the figure above, and has the value 1.2 C 5/=2 D 7=2. 16. y 2 y D 2.1 x x/ sn D n X 2n C 3i Sn D n X 1 D A 1 x Fig. 5.2-16 19. n X i 2 1 represents a sum of areas of n rectsn D n n iD1 angles each of width 1=n and having heights equal to the height to the graph y D 2.1 x/ at the points x D i=n. Thus limn!1 Sn is the area A of the triangle in the figure above, and therefore has the value 1. 184 Telegram: @uni_k j D1 n s 1 2 j n D sum of areas of rectangles in the figure. Thus the limit of Sn is the area of a quarter circle of unit radius: lim Sn D : n!1 4 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 5.3 (PAGE 307) y 5. f .x/ D sin x on Œ0; ; n D 6: 2 5 ; ; P6 D 0; ; ; ; 6 3 2 3 6 # " p p 3 3 1 1 L.f; P6 / D C C C0 0C C 6 2 2 2 2 p D .1 C 3/ 1:43; 6 " # p p 1 1 3 3 C C1C1C C U.f; P6 / D 6 2 2 2 2 p D .3 C 3/ 2:48: 6 6. f .x/ D cos x on Œ0; 2; n D 4: 2 3 L.f; P4 / D cos C cos C cos C cos D : 4 2 2 2 3 U.f; P4 / D C cos 2 D : cos 0 C cos C cos 4 2 2 p yD 1 x 2 1 n 2 n n 1 n D1 n n x Fig. 5.2-19 Section 5.3 The Definite Integral (page 307) 1. y yDcos x f .x/ D x on Œ0; 2; n D 8: 1 1 3 5 3 7 P8 D 0; ; ; ; 1; ; ; ; 1 4 2 4 4 2 4 2 0 1 3 5 3 7 1 7 L.f; P8 / D 0C C C C1C C C D 8 4 2 4 4 2 4 4 1 3 5 3 7 9 2 0 1 C C C1C C C C2 D U.f; P8 / D 8 4 2 4 4 2 4 4 =2 3=2 3. f .x/ D x 2 on Œ0; 4; n D 4: 4 0 L.f; P4 / D Œ0 C .1/2 C .2/2 C .3/2 D 14: 4 4 0 U.f; P4 / D Œ.1/2 C .2/2 C .3/2 C .4/2 D 30: 4 f .x/ D e x on Œ 2; 2; n D 4: e4 1 4:22 e 2 .e 1/ e4 1 11:48: U.f; P4 / D 1.e 1 C e 0 C e 1 C e 2 / D e.e 1/ L.f; P4 / D 1.e 2 C e 1 C e 0 C e 1 / D ˚ f .x/ D x on Œ0; 1. Pn D 0; n1 ; n2 ; : : : ; n n 1 ; nn : We have 2 n 1 1 1 0 C C C C n n n n 1 .n 1/n n 1 D 2 D ; n 2 2n 1 1 2 3 n U.f; Pn / D C C C C n n n n n 1 n.n C 1/n nC1 D 2 D : n 2 2n L.f; Pn / D Thus limn!1 L.f; Pn / D limn!1 U.f; Pn / D 1=2. If P is any partition of Œ0; 1, then L.f; P / U.f; Pn / D 4. f .x/ D ln x on Œ1; 2; n D 5: 7 8 9 6 2 1 ln 1 C ln C ln C ln C ln L.f; P5 / D 5 5 5 5 5 0:3153168: 2 1 6 7 8 9 ln C ln C ln C ln C ln 2 U.f; P5 / D 5 5 5 5 5 0:4539462: nC1 2n for every n, so L.f; P / limn!1 U.f; Pn / D 1=2. Similarly, U.f; P / 1=2. If there exists any number I such that L.f; P / I U.f; P / for all P , then I cannot be less than 1/2 (or there would exist a Pn such that L.f; Pn / > I ), and, similarly, I cannot be greater than 1/2 (or there would exist a Pn such that U.f; Pn / < I ). Thus R1 I D 1=2 and 0 x dx D 1=2. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 x Fig. 5.3-6 7. 2. Downloaded by ted cage (sxnbyln180@questza.com) 185 lOMoARcPSD|6566483 www.konkur.in SECTION 5.3 (PAGE 307) ADAMS and ESSEX: CALCULUS 9 ˚ x on Œ0; 2. Pn D 0; n2 ; n4 ; : : : ; 2nn 2 ; 2n n : We 8. f .x/ D 1 have 2 L.f; Pn / D 1 n D 2 n n 2 n n 4 X n2 C 1 4 n C C 1 2n n By l’H^opital’s Rule, i 4 n.n C 1/ D 2 n 2 2 0 U.f; Pn / D 1 C 1 n n n 1 4 X 2 i D n n n2 Thus 2 ! 0 as n ! 1; n 2 2n 2 C C 1 n n lim L.f; Pn / D lim U.f; Pn / D e 3 n!1 11. Thus Z 2 lim n!1 iD0 4 .n 1/n 2 D ! 0 as n ! 1: n2 2 n D2 12. lim n!1 .1 0 x/ dx D 0. 13. ˚ 9. f .x/ D x 3 on Œ0; 1. Pn D 0; n1 ; n2 ; : : : ; n n 1 ; nn : We have (using the result of Exercise 51 (or 52) of Section 6.1) 1 L.f; Pn / D n 1/ D lim n!1 iD1 D2 e 3=n 1 n!1 1=n 3=n e . 3=n2 / 3e 3=n D lim D lim D 3: 2 n!1 n!1 1=n 1 lim n.e 3=n 3 3 1 n 1 0 C C C n n n ! 3 n!1 14. lim n!1 15. lim n!1 n 1 1 X 3 1 .n 1/2 n2 D 4 i D 4 n n 4 iD0 2 1 1 n 1 ! as n ! 1; D 4 n 4 ! 3 3 n 3 2 1 1 C C C U.f; Pn / D n n n n lim n!1 n X 1 iD1 n n X 1 iD1 n i D n r i n X sin n iD1 Z 1 p D Z 1 p i D n Z sin x dx 1 n iD1 n X 1 iD1 e x dx: 0 x dx 0 n tan 1 x dx 0 0 2i 1 2n D Z 1 tan 1 x dx 0 1 i ; . n n Z 1 n n X X 1 1 dx n D lim D lim 2 n!1 n!1 n2 C i 2 n 1 C .i=n/2 0 1Cx 2i 1 is the midpoint of 2n i iD1 iD1 17. Z 3 Z 2 2i ln 1 C D ln.1 C x/ dx n n 0 n X 2 Note that 16. r 1D b a and xi D a C i x where 1 i n n Since f is continuous and nondecreasing, Let x D n 1 X 3 1 n2 .n C 1/2 i D 4 4 n n 4 iD1 2 1 1 nC1 ! as n ! 1: D 4 n 4 D L.f; Pn / D f .a/x C f .x1 /xC f .x2 /x C C f .xn 1 /x n X1 b a f .a/ C f .xi / ; D n iD1 Z 1 U.f; Pn / D f .x1 /x C f .x2 /x C C f .xn 1 /x C f .b/x n 1 b a X f .xi / C f .b/ : D n 1 Thus x 3 dx D . 4 0 ˚ 10. f .x/ D e x on Œ0; 3. Pn D 0; n3 ; n6 ; : : : ; 3nn 3 ; 3n n : We have (using the result of Exercise 51 (or 52) of Section 6.1) 3 e 0=n C e 3=n C e 6=n C C e 3.n 1/=n n 3.e 3 1/ 3 e 3n=n 1 D ; D 3=n n e 1 n.e 3=n 1/ 3 3=n U.f; Pn / D e C e 6=n C e 9=n C C e 3n=n D e 3=n L.f; Pn /: n L.f; Pn / D 186 Telegram: @uni_k iD1 Thus, U.f; Pn / D D b n .b L.f; Pn / nX1 a f .xi / C f .b/ iD1 a/.f .b/ n Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) f .a// : f .a/ n X1 iD1 f .xi / 1. lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 5.4 (PAGE 312) Since 3. lim ŒU.f; Pn / L.f; P n!1 n/ D 0; R2 2 .x C 2/ dx D 1 .4/.4/ D 8 2 y .2;4/ therefore f must be integrable on Œa; b. 18. P D fx0 < x1 < < xn g, P 0 D fx0 < x1 < < xj 1 < x 0 < xj < < xn g. Let mi and Mi be, respectively, the minimum and maximum values of f .x/ on the interval Œxi 1 ; xi , for 1 i n. Then L.f; P / D U.f; P / D n X iD1 n X mi .xi xi 1 /; Mi .xi xi 1 /: 2 yDxC2 2 2 4. iD1 R2 1 0 .3x C 1/ dx D shaded area D 2 .1 C 7/.2/ D 8 y yD3xC1 If mj0 and Mj0 are the minimum and maximum values of f .x/ on Œxj 1 ; x 0 , and if mj00 and Mj00 are the corresponding values for Œx 0 ; xj , then mj0 mj ; mj00 mj ; Mj0 Mj ; x Fig. 5.4-3 Mj00 Mj : Therefore we have mj .xj Mj .xj xj 1 / mj0 .x 0 xj 1 / Mj0 .x 0 xj 1 / C mj00 .xj xj 1 / C Mj00 .xj If P 00 is any refinement of P we can add the new points in P 00 to those in P one at a time, and thus obtain L.f; P / L.f; P 00 /; Fig. 5.4-4 x 0 /: Hence L.f; P / L.f; P 0 / and U.f; P / U.f; P 0 /. 5. Rb a x dx D a2 2 b2 2 y U.f; P 00 / U.f; P /: yDx Section 5.4 Properties of the Definite Integral (page 312) 1. Z b a f .x/ dx C Z c D 2. Z 2 0 D D Zb c 3f .x/ dx C Z 2 .3 0 Z 3 C .3 0 f .x/ dx a Z 3 Zc c a 3f .x/ dx 1 a f .x/ dx f .x/ dx D 0 Z 3 6. R2 1 .1 2x/ dx D A1 A2 D 0 y 2f .x/ dx 0 A1 1 2 2 1 x A2 2 yD1 3/f .x/ dx 2x 1 Fig. 5.4-6 2/f .x/ dx 2 f .x/ dx 7. R p2 p p 2 2 p t 2 dt D 21 . 2/2 D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x b Fig. 5.4-5 3f .x/ dx 1 Z 2 2/f .x/ dx C .3 C 3 Z 1 Z 3 f .x/ dx C Z a x 2 x 0 /; Downloaded by ted cage (sxnbyln180@questza.com) 187 lOMoARcPSD|6566483 www.konkur.in SECTION 5.4 (PAGE 312) ADAMS and ESSEX: CALCULUS 9 p yD 2 t 2 y 15. Z 1p x 2 dx D area A1 in figure below 4 0 p t p 2 2 1 area of circle area A2 4 (see #14 below) p ! 1 2 3 2 D .2 / 4 3 2 p 3 D C 3 2 D Fig. 5.4-7 8. 9. Z 0 p 2 Z p x 2 dx D quarter disk D 2 1 p 2 . 2/ D 4 2 y sin.x 3 / dx D 0. (The integrand is an odd function and the interval of integration is symmetric about x D 0.) Ra 10. jsj/ ds D shaded area D 2. 21 a2 / D a2 a .a P yD p x2 4 y a yDa jsj A1 A2 a a s 11. R1 5 1 .u 3u3 C / du D O R1 y Z 2p x 2 dx D area A2 in figure above 4 1 D area sector POQ area triangle POR 1 1 p D .22 / .1/ 3 6 p 2 2 3 D 3 2 1 1 u Fig. 5.4-11 17. p Let y D 2x x 2 ) y 2 C .x 1/2 D 1: Z 2p 1 2x x 2 dx D shaded area D .1/2 D : 2 2 0 y Z 2 6x 2 dx D 6 Z 3 .x 2 0 18. Z 2 D x 19. Z 2 .4 2 Z 4 .e x 4 14. Z 3 3 p .2 C t / 9 188 Telegram: @uni_k e x / dx D 0 (odd function, symmetric interval) t 2 dt D 2 Z 3p Z 3 p t 2 dt C t 9 9 3 3 1 2 3 C 0 D 9 D2 2 t 2 dt 20. Z 2 .v 2 0 21. Z 1 0 33 D 16 3 Z 3 Z 2 x 2 dx 0 3 3 3 23 3 .x 2 C p 1 23 3 22 2 D 2 3 Downloaded by ted cage (sxnbyln180@questza.com) 7 3 t 2 / dt 32 23 D 3 3 13 1 C .12 / 3 4 1 D C 3 4 x 2 / dx D Copyright © 2018 Pearson Canada Inc. 4D .4 0 D 2 2.4/ v/ dv D x 2 dx 0 Z 2 t 2 / dt D 2 Fig. 5.4-12 13. x 2 dx D 6 0 4/ dx D 2 p yD 2x x 2 2 x 2 Fig. 5.4-15 16. 12. 1 1 du D 2 1 Q R Fig. 5.4-10 4.3 2/ lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 22. Z 6 6 x 2 .2 C sin x/ dx D Z 6 6 D4 23. Z 2 1 dx D ln 2 x Z 4 Z 4 1 24. 2 25. 26. Z 1 28. 29. 0 y p yD 4 x 2 x 2 sin x dx 6 2 x dx C 0 D 4 3 .6 / D 288 3 1 1=3 t Z 1=3 dt D Z 3 1 1 ds D 1=4 s Z 3 1 Fig. 5.4-31 32. 1 dt D t Z 1=4 1 ds s 1 ln 1 D ln 3 3 1 ds s 33. 1 ln D ln 3 C ln 4 D ln 12 4 2 2 ln 2 D 3 R2 1 sgn x dx D 2 1D1 y Average D 1 2 a Fig. 5.4-33 Z b 34. Let 1 3 0 Average value D 0 1 f .x/ D 1Cx 2 Z 2 3 f .x/ dx D area.1/ C area.2/ y Z 2 x 2 /1=2 dx .4 2 0 0 1 D .shaded area/ 2 1 1 .2/2 D D 2 4 2 1 .2/.2/ D 2 12 : 2 yD2 (1) (2) 3 (3) 1 3 D3 3 3 area.3/ D .2 2/ C 21 .1/.1/ 3 x 2 dx D if x < 0 if x 0. Then Z 1 Average D .1 C sin t / dt . / Z Z 1 1 dt C sin t dt D 2 1 Œ2 C 0 D 1 D 2 Z 3 x 1 .x C 2/ dx a 1 2 1 .b a2 / C 2.b a/ D b a 2 1 4CaCb D .b C a/ C 2 D 2 2 b yDsgn x 1 .x C 2/ dx 4 0 0 1 1 2 D .4 / C 2.4/ D 4 4 2 Average D Z 2 1 ds 1=2 s 1 4 ln D ln 2 2 3 1 1 2 x yDxC1 . 3; 2/ Fig. 5.4-34 35. Z 2 0 g.x/ dx D D Z 1 0 3 x 2 dx C Z 2 x dx 1 22 12 11 1 C D 3 2 6 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 .1=2/ Average value D Z 4 1 x 2 1 1 1 dt D dt dt t t 1 1 t D ln 4 ln 2 D ln.4=2/ D ln 2 30. Average D 31. Z 6 Z 6 Z 2 D ln 3 27. 2x 2 dx C SECTION 5.4 (PAGE 312) Downloaded by ted cage (sxnbyln180@questza.com) 189 lOMoARcPSD|6566483 www.konkur.in SECTION 5.4 (PAGE 312) 36. Z 3 0 j2 xj dx D Z 2 Z 3 x/ dx C .x 2/ dx 2 ˇ ˇ2 ˇ3 ˇ 2 ˇ x 2 ˇˇ x 2x ˇˇ C 2 ˇˇ 2 ˇ .2 0 D 2x D4 37. I D Z 2p ADAMS and ESSEX: CALCULUS 9 9 2 0C 3 6 2C4D .jx C 1j jx 1j C jx C 2j/ dx D area A1 area A2 1 5 5C8 D .5/ C .3/ 2 3 2 2 0 2 Z 4 5 2 1C2 .1/ 2 1C2 .1/ 2 1 1 41 .1/ D 2 3 2 40. 4 x 2 sgn.x y 1/ dx yD 0 D area A1 area A2 : p p Area A1 D 16 22 12 .1/. 3/ D 32 12 3. p Area A2 D 41 22 area A1 D 31 C 21 3. p 3. Therefore I D .=3/ x2 jx x 1j A1 y p yD 4 x 2 p 2 3 A2 3 A1 =3 1 2 x A2 Fig. 5.4-40 Z 3 x2 x dx 1j 0 jx D area A1 area A2 1 7 1C3 .2/ .1/.1/ D D 2 2 2 Fig. 5.4-37 38. R 3:5 0 Œx dx D shaded area D 1 C 2 C 1:5 D 4:5: y Z 1 2 jx C 1jsgn x dx 4 2 Z 2 Z 0 1 .x C 1/ dx jx C 1j dx D 4 2 0 1 1C3 1 D 2 2 11 4 2 2 3 1 D : D1 4 4 yDbxc 41. 1 2 Average D x 3:5 Fig. 5.4-38 y 39. y .2;3/ yDxC1 yD .xC1/ 1 2 A2 A1 1 A3 2 x Z b f dx 1 A1 y D jx C 1j jx . 2; 1/ 1j C jx C 2j x A2 42. Z b f .x/ a Fig. 5.4-41 Z b f dx D f .x/ dx Fig. 5.4-39 190 Telegram: @uni_k Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) a D .b a/f D .b a/f f a Z b dx a .b a/f D 0 x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 43. Z b a D D f .x/ Z b a Z b a D .b k 2 dx 10. 2 f .x/ dx 2k Z b a 2 f .x/ dx Z b a This is minimum if k D f . 1. 2. 2 Z b a/f C k 2 .b 2 f .x/ dx dx a 11. a/ .b a/f 2 12. 14. ˇ1 1 ˇˇ 1 dx D ˇ 2 x ˇ 1=2 x 1=2 D 1 6. 1 7. Z 2 9. x 2 dx D 1 x2 Z 2 ˇ2 4 ˇ x 8 ˇ ˇ D ˇ 15. x 16. Z 1 17. 18. .x 2 C 3/2 dx D 2 19. =4 D 0 ˇ1 2x ˇˇ 2 x 2 dx D ˇ D ln 2 ˇ ln 2 1 1 9=8 20. 1 3 D 2 ln 2 2 ln 2 ˇ1 ˇ dx 1 ˇ D tan x ˇ D 2 ˇ 1 C x 2 1 Z 1 Z 1=2 1 p dx 1 x2 1 D sin ˇ1=2 ˇ ˇ xˇ D ˇ 6 0 ˇ1 Z 1 ˇ dx 1 xˇ D sin p ˇ 2 2ˇ 1 4 x 1 1 sin 1 2 2 D D 6 6 3 ˇ 0 Z 0 ˇ 1 dx 1 1 xˇ D tan tan 1 . 1/ D ˇ D0 2 ˇ 4 C x 2 2 2 8 2 1 1 2 21. ˇ1 R1 4 1 x 5 ˇˇ Area R D 0 x dx D ˇ D sq. units: 5 ˇ 5 0 y .1;1/ yDx 4 R 1 x p 2 1 2 1 Cp D p 2 2 2 2 Fig. 5.5-21 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k e e x / dx D 0 (odd function, symmetric interval) D sin 0 ˇ =6 Z =6 ˇ ˇ cos x dx D sin x ˇ ˇ =4 0 ˇe ae 1 ax ˇˇ a dx D ˇ D ln a ˇ ln a 0 0 1 .x 4 C 6x 2 C 9/ dx ˇˇ2 5 x ˇ C 2x 3 C 9x ˇ D2 ˇ 5 0 32 404 D2 C 16 C 18 D 5 5 ˇ ! 9 Z 9 p ˇˇ p 1 2 x p dx D x 3=2 2 x ˇ ˇ 3 x 4 4 # " # " p p 2 3=2 32 2 3=2 D .9/ 2 9 .4/ 2 4 D 3 3 3 2 8. 2 x3 3 ˇ2 ˇ ˇ cos u/ˇ D 2 ˇ ˇ Z ˇ x xˇ e dx D e ˇ D e ˇ Z e 1 Z 2 =4 .1 C sin u/ du D .u .e x . 2/ D 1 ˇˇ 1 Z 1 1 1 1 1 ˇ dx D C 2 ˇ 4. ˇ x2 x3 x 2x 2 2 1 1 7 1 C D D1C 2 2 8 8 ˇ2 Z 2 ˇ ˇ 2 3 2 5. .3x 4x C 2/ dx D .x 2x C 2x/ˇ D 9 ˇ 1 ˇ=3 p ˇ 2 1 ˇ D cos ˇ ˇ 2 sin d D 2 0 Z 1 0 0 ˇ4 2 3=2 ˇˇ 16 x dx D x ˇ D ˇ 3 3 ˇ=3 ˇ p ˇ sec d D tan ˇ D tan D 3 ˇ 3 2 Z 2 3 p Z 2 0 ˇ2 x 4 ˇˇ 16 0 x dx D D4 ˇ D ˇ 4 4 0 Z 4 Z =3 =4 13. Z 2 Z =3 0 Section 5.5 The Fundamental Theorem of Calculus (page 318) 0 3. f .x/ dx C k 2k.b f /2 C a/.k SECTION 5.5 (PAGE 318) Downloaded by ted cage (sxnbyln180@questza.com) 191 lOMoARcPSD|6566483 www.konkur.in SECTION 5.5 (PAGE 318) 22. Area D Z e2 e D ln e 2 ADAMS and ESSEX: CALCULUS 9 ˇe 2 ˇ 1 ˇ dx D ln x ˇ ˇ x 25. For intersection of y D x 2 x2 e ln e D 2 1 D 1 sq. units. y Thus x D 1 or x D 2. The indicated region has area Area R D 1 e2 x e D1 Fig. 5.5-22 Area R D D D D1 .x 2 0 x3 3 64 3 y 3x C 2 D 0 2/.x 1/ D 0: x .x Area 23. 3x C 3 D 1 2 1 yD x Z 4 3x C 3 and y D 1, we have 4x/ dx ˇˇ4 2 ˇ 2x ˇ ˇ 0 32 32 D sq. units: 3 Z 2 .x 2 3x C 3/ dx ˇˇ2 3x 2 ˇ C 3x ˇ ˇ 2 1 1 3 1 6C6 C3 D sq. units: 3 2 6 1 x3 3 8 3 y yDx 2 3xC3 1 4 x R yDx 2 4x p x x intersect where x D , that Since y D x and y D 2 2 is, at x D 0 and x D 4, thus, 1 2x 3x 2 / dx D 2 Z 1 .5 0 D 2.5x D 2.5 Z 4 Z 4 x x dx dx 0 0 2 ˇ4 ˇ4 ˇ 2 x 2 ˇˇ ˇ D x 3=2 ˇ ˇ ˇ 3 4 ˇ Area D Since y D 5 2x 3x 2 D .5 C 3x/.1 x/, therefore y D 0 at x D 53 and 1, and y > 0 if 53 < x < 1. Thus, the area is .5 p 0 0 16 4 D sq. units. 4 3 16 D 3 3x 2 / dx ˇ1 ˇ ˇ x 3 /ˇ ˇ y p yD x 0 1/ D 8 sq. units. A .4;2/ yDx=2 y x yD5 2x 3x 2 Fig. 5.5-26 27. Area 1 1 Fig. 5.5-24 192 Telegram: @uni_k x p Fig. 5.5-23 Z 1 2 Fig. 5.5-25 26. 24. yD1 R x Area R D 2 shaded area Z 1 1 D2 x 2 dx 2 0 1 1 1 D sq. units: D2 2 3 3 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 5.5 (PAGE 318) y y .1;1/ xDy 2 yDe x A R x a yDx 2 x Fig. 5.5-30 Fig. 5.5-27 31. Area R D 28. The two graphs intersect at .˙3; 3/, thus Z 3 .12 D 2.36 9/ Area D 2 Z 2 0 D .x 0 y Z 3 0 D 2 12x cos x/ dx ˇ2 ˇ sin x/ˇ D 2sq. units: .1 x 2 / dx 2 x dx 0 ˇ 3 ˇ3 1 2 ˇˇ 1 3 ˇˇ x ˇ 2 x ˇ ˇ ˇ 3 2 0 yD1 cos x 0 R 9 D 45 sq. units. 2x y yD12 x 2 Fig. 5.5-31 Area yDjxj .3;3/ 32. Area D x 1=3 1 3 D .27/2=3 2 x Fig. 5.5-28 29. Z 27 1 3 D 12 sq. units. 2 y Z 1 Z 1 x 1=3 dx x 1=2 dx 0 0 ˇ1 ˇ1 2 3=2 ˇˇ 3 2 1 3 4=3 ˇˇ x ˇ D D sq. units: D x ˇ ˇ ˇ 4 3 4 3 12 Area R D ˇ27 3 2=3 ˇˇ dx D x ˇ ˇ 2 0 1=3 yDx Area 0 1 y 27 x Fig. 5.5-32 .1;1/ yDx 1=3 33. R Z 3=2 0 yDx 1=2 j cos xj dx D Z =2 cos x dx 0 ˇ=2 ˇ ˇ D sin x ˇ ˇ 0 x 34. 30. Area D ae x dx D ˇ0 ˇ xˇ e ˇ D ea ˇ Z 3 1 sgn .x 2/ dx D x2 1 sq. units. Z 2 1 ˇ2 1 ˇˇ D ˇ xˇ 1 a Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k cos x dx =2 ˇ3=2 ˇ ˇ sin x ˇ ˇ =2 D1C1C1D3 Fig. 5.5-29 R0 Z 3=2 Downloaded by ted cage (sxnbyln180@questza.com) Z 3 dx dx C 2 2 x 2 x ˇ3 1 ˇˇ 1 ˇ D xˇ 3 2 193 lOMoARcPSD|6566483 www.konkur.in SECTION 5.5 (PAGE 318) 35. 36. ADAMS and ESSEX: CALCULUS 9 Average value Z 1 2 D .1 C x C x 2 C x 3 / dx 2 0 ˇ2 x3 x 4 ˇˇ x2 1 C C xC D ˇ 2 2 3 4 ˇ 0 16 8 1 : 2C2C C4 D D 2 3 3 Z 2 1 Average value D e 3x dx 2 . 2/ 2 ˇ2 1 1 3x ˇˇ e D ˇ ˇ 4 3 44. 1 37. Avg. D 1= ln 2 Z 1= ln 2 0 38. Since g.t / D e 6 45. /: 1 46. Z 3 1 39. 40. 41. Z x d dx Z 0 sin t dt D x2 t D 42. d 2 x dx 194 Telegram: @uni_k Z x2 0 Z x2 sin u du u Z x2 d sin t dt dx 0 t sin x 2 sin x 2 2x 2 D 2 x x Z x2 sin u d sin u du C x 2 du u dx 0 u 0 Z x2 2x sin x 2 sin u du C x 2 D 2x u x2 0 Z x2 sin u D 2x du C 2x sin.x 2 / u 0 Z t d cos y cos t dy D 2 dt 1 C t2 1Cy D 2x 43. 1/ D sin x sin t dt D t x 2 # " Z Z 3 t sin x sin x d d dx D dx D dt t x dt x 3 d dx 1 sin x2 1 dx cos sin2 1 Z sin a 1 1 x2 dx # 1 D csc sec cos Z t F .t / D cos.x 2 / dx Z px cos.u2 / du 0 p d cos x 1 F . x/ D cos x p D p dx 2 x 2 x Z x2 p H.x/ D 3x e t dt Z x2 p t e 4 Z 4 p e 4 t dt C 3x.2xe jxj / dt C 3.2/.4e 2 / 24 e2 D 3.0/ C 24e 2 D 47. 1 1 .3 3 D H 0 .2/ D 3 # D 1 sin cos2 1 H 0 .x/ D 3 if 0 t 1, if 1 < t 3, ˇ3 # " ˇ 1 ˇ 0 C tˇ .0/ dt C 1 dt D ˇ 3 0 1 "Z cos a D dx x2 4 0 the average value of g.t / over [0,3] is 1 3 "Z 1 1 0 ˇ1= ln 2 2x ˇˇ 2 dx D .ln 2/ De ˇ ln 2 ˇ 0; 1; sin p F . x/ D x Z cos d D d 2 1 6 D .e 12 d d f .x/ D C Z x f .t / dt 1 f 0 .x/ D f .x/ ÷ f .x/ D C e x D f .1/ D C e ÷ C D e 2 : 3 48. f .x/ D e .x 1/ : Z x f .x/ D 1 f .t / dt 0 f 0 .x/ D f .x/ 1 D f .0/ D C f .x/ D e x : sin t t 49. 50. 51. ÷ f .x/ D C e x The function 1=x 2 is not defined (and therefore not continuous) at x D 0, so the Fundamental Theorem of Calculus cannot be applied to it on the interval Œ 1; 1. Since Z 1 dx 1=x 2 > 0 wherever it is defined, we would expect 2 1 x to be positive if it exists at all (which it doesn’t). Z x sin x sin t dt , then F 0 .x/ D and If F .x/ D 2 1 C t 1 C x2 17 F .17/ D 0. R 2x x 2 1 F .x/ D 0 cos dt . 1 C t2 Note that 0 < 1 1 for all t , and hence 1 C t2 0 < cos.1/ cos Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 1 C t2 1: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 5.6 (PAGE 326) y The integrand is continuous for all t , so F .x/ is defined and differentiable for all x. Since limx!˙1 .2x x 2 / D 1, therefore limx!˙1 F .x/ D 1. Now 0 F .x/ D .2 1 2x/ cos 1 C .2x x 2 /2 yD D0 only at x D 1. Therefore F must have a maximum value at x D 1, and no minimum value. 52. 1 lim n!1 n " 1 1C n 5 2 C 1C n 5 1 n # n 5 C C 1 C n D area below y D x 5 , above y D 0, between x D 1 and x D 2 ˇ2 Z 2 21 1 1 6 ˇˇ 5 D x dx D x ˇ D .26 1/ D ˇ 6 6 2 1 1. 2 n C C sin sin C sin n!1 n n n n D lim sum of areas of rectangles shown in figure n!1 ˇ Z ˇ ˇ D sin x dx D cos x ˇ D 2 ˇ 0 2. 0 yDsin x 3. :::::: .n 1/ n x 4. Fig. 5.5-53 54. n n n n C C C C n!1 n2 C 1 n2 C 4 n2 C 9 2n2 2 2 2 n n n n2 1 C C C C D lim n!1 n n2 C 1 n2 C 4 n2 C 9 2n2 1 0 lim C 1 1 1 C n 2 C 2 C 2 C C A 1 2 1C 1C 1C n n n 1 D area below y D , above y D 0, 1 C x2 between x D 0 and x D 1 ˇ1 Z 1 ˇ 1 1 ˇ dx D tan x ˇ D D 2 ˇ 4 0 1Cx 1B B B n!1 n @ 5. D lim 0 Z D y 2 3 n n x 6. Z e 5 2x dx Let u D 5 2x du D 2 dx Z 1 u 1 e u du D e CC D 2 2 1 5 2x e C C: 2 cos.ax C b/ dx Let u D ax C b du D a dx Z 1 1 D cos u du D sin u C C a a 1 D sin.ax C b/ C C: a Z p 3x C 4 dx Let u D 3x C 4 du D 3 dx Z 2 1 2 1=2 D u du D u3=2 C C D .3x C 4/3=2 C C: 3 9 9 Z e 2x sin.e 2x / dx Let u D e 2x du D 2e 2x dx Z 1 1 sin u du D cos u C C D 2 2 1 D cos.e 2x / C C: 2 Z x dx Let u D 4x 2 C 1 .4x 2 C 1/5 du D 8x dx Z 1 1 1 5 u du D C C: u 4 CC D D 8 32 32.4x 2 C 1/4 p Z p sin x dx Let u D x p x dx du D p 2 x Z D 2 sin u du D 2 cos u C C p D 2 cos x C C: Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 Fig. 5.5-54 lim n 2 n Section 5.6 The Method of Substitution (page 326) 1 53. 1 1Cx 2 Downloaded by ted cage (sxnbyln180@questza.com) 195 lOMoARcPSD|6566483 www.konkur.in SECTION 5.6 (PAGE 326) 7. 8. Z ADAMS and ESSEX: CALCULUS 9 2 xe x dx Let u D x 2 du D 2x dx Z 1 1 1 2 u D e du D e u C C D e x C C: 2 2 2 Z 3 x 2 2x C1 dx Let u D x 3 C 1 du D 3x 2 dx Z 1 2u 1 2u du D CC D 3 3 ln 2 15. 16. 9. 10. 11. 12. 13. 14. 2x C1 C C: 3 ln 2 Z cos x dx Let u D sin x 4 C sin2 x du D cos x dx Z du D 4 C u2 1 1 1 u 1 1 D tan C C D tan sin x C C: 2 2 2 2 Z sec2 x p dx Let u D tan x 1 tan2 x du D sec2 x dx du D p 1 u2 D sin 1 u C C D sin 1 .tan x/ C C: Z x e C1 dx ex 1 Z x=2 e C e x=2 D dx Let u D e x=2 e x=2 e x=2 e x=2 du D 21 e x=2 C e x=2 dx Z du D 2 ln juj C C D2 ˇ ˇ ˇ ˇ u ˇ ˇ ˇ ˇ D 2 lnˇe x=2 e x=2 ˇ C C D lnˇe x C e x 2ˇ C C: Z 18. 19. 20. ln t dt t Let u D ln t dt t Z 1 2 1 D u du D u C C D .ln t /2 C C: 2 2 Z ds Let u D 4 5s p 4 5s du D 5 ds Z 1 du D p 5 u 2p 2 1=2 u CC D 4 5s C C: D 5 5 Z xC1 dx Let u D x 2 C 2x C 3 p x 2 C 2x C 3 du D 2.x C 1/ dx Z p p 1 1 p du D u C C D x 2 C 2x C 3 C C D 2 u 196 Telegram: @uni_k 17. du D 21. 22. p 4 Z t t4 dt Let u D t 2 du D 2t dt du 1 p 2 4 u2 2 u 1 t 1 D sin 1 C C D sin 1 C C: 2 2 2 2 Z x2 dx Let u D x 3 2 C x6 du D 3x 2 dx Z du 1 u 1 1 D p tan CC p D 3 2 C u2 3 2 2 3 1 x D p tan 1 p C C: 3 2 2 Z Z dx e x dx Let u D 1 C e x D x e C1 1Ce x du D e x dx Z du D ln juj C C D ln.1 C e x / C C: D u Z Z dx e x dx D Let u D e x ex C e x e 2x C 1 du D e x dx Z du D tan 1 u C C D u2 C 1 D tan 1 e x C C: Z tan x ln cos x dx Let u D ln cos x du D tan x dx Z 2 1 1 2 u CC D ln cos x C C: D u du D 2 2 Z xC1 p dx 2 Z Z 1 x dx x dx C Let u D 1 x 2 p p D 1 x2 1 x2 du D 2x dx in the first integral only Z p 1 du p C sin 1 x D u C sin 1 x C C D 2 u p D 1 x 2 C sin 1 x C C: Z Z dx dx D Let u D x C 3 x 2 C 6x C 13 .x C 3/2 C 4 du D dx Z 1 du 1 u D tan CC D u2 C 4 2 2 1 xC3 D tan 1 C C: 2 2 Z dx dx p D p Let u D 1 x 4 C 2x x 2 5 .1 x/2 du D dx Z u du 1 D sin CC D p p 5 5 u2 1 x x 1 D sin 1 p C C D sin 1 p C C: 5 5 D 3 D Z Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 23. Z sin3 x cos5 x dx Z D sin x.cos5 x cos7 x/ dx D D 24. .u7 u8 8 u5 / du u6 cos8 x CC D 6 8 Let u D cos x du D sin x dx cos6 x C C: 6 sin4 t cos5 t dt Z D sin4 t .1 sin2 t /2 cos t dt Z .u4 2u6 C u8 / du D D 1 5 sin t 5 Z sin ax cos2 ax dx D D 26. Z 28. Z D 25. SECTION 5.6 (PAGE 326) u5 5 29. Let u D sin t du D cos t dt 2u7 u9 C CC 7 9 30. 2 7 1 sin t C sin9 t C C: 7 9 Z 1 u2 du a u3 CC D 3a Let u D cos ax du D a sin ax dx 31. 1 cos3 ax C C: 3a Z sin 2x 2 sin2 x cos2 x dx D dx 2 Z x sin 4x 1 cos 4x 1 dx D C C: D 4 2 8 32 Z Z Œ1 C cos.2x/2 cos4 x dx D dx 4 Z 1 D Œ1 C 2 cos.2x/ C cos2 .2x/ dx 4 Z x sin.2x/ 1 D C C 1 C cos.4x/ dx 4 4 8 x sin.2x/ x sin.4x/ D C C C CC 4 4 8 32 sin.2x/ sin.4x/ 3x C C C C: D 8 4 32 Z sec5 x tan x dx Let u D sec x du D sec x tan x dx Z 5 u sec5 x D u4 du D CC D C C: 5 5 Z sec6 x tan2 x dx Z D sec2 x tan2 x.1 C tan2 x/2 dx Let u D tan x du D sec2 x dx Z 1 2 1 D .u2 C 2u4 C u6 / du D u3 C u5 C u7 C C 3 5 7 2 1 1 3 5 7 D tan x C tan x C tan x C C: 3 5 7 Z p tan x sec4 x dx Z p tan x.1 C tan2 x/ sec2 x dx Let u D tan x D du D sec2 x dx Z D u1=2 C u5=2 du 2u7=2 2u3=2 C CC 3 7 2 2 D .tan x/3=2 C .tan x/7=2 C C: 3 7 Z sin 2=3 x cos3 x dx Let u D sin x du D cos x dx Z 3 7=3 1 u2 1=3 du D 3u D u CC 7 u2=3 3 7=3 sin x C C: D 3 sin1=3 x 7 Z cos x sin4 .sin x/ dx Let u D sin x du D cos x dx Z Z 1 cos 2u 2 D sin4 u du D du 2 Z 1 C cos 4u 1 1 2 cos 2u C du D 4 2 3u sin 2u sin 4u D C CC 8 4 32 1 1 3 sin.2 sin x/ C sin.4 sin x/ C C: D sin x 8 4 32 D 32. 27. Z Z 1 cos 2x 3 sin6 x dx D dx 2 Z 1 D .1 3 cos 2x C 3 cos2 2x cos3 2x/ dx 8 Z 3 x 3 sin 2x .1 C cos 4x/ dx C D 8 16 Z 16 1 cos 2x.1 sin2 2x/ dx Let u D sin 2x 8 du D 2 cos 2x dx Z 5x 3 sin 2x 3 sin 4x 1 D C .1 u2 / du 16 16 64 16 3 sin 4x sin 2x sin3 2x 5x 3 sin 2x C C CC D 16 16 64 16 48 3 3 sin 4x sin 2x 5x sin 2x C C C C: D 16 4 64 48 Z 33. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 197 lOMoARcPSD|6566483 www.konkur.in SECTION 5.6 (PAGE 326) 34. 35. 36. Z sin3 .ln x/ cos3 .ln x/ dx x Let u D sin.ln x/ cos.ln x/ dx du D x Z 1 1 6 D u3 .1 u2 / du D u4 u CC 4 6 1 6 1 sin .ln x/ C C: D sin4 .ln x/ 4 6 Z sin2 x dx 4 Zcos x D tan2 x sec2 x dx Let u D tan x du D sec2 x dx Z 3 u 1 D u2 du D C C D tan3 x C C: 3 3 Z Z 3 sin x dx D tan3 x sec x dx 4x cos Z D .sec2 x 1/ sec x tan x dx Let u D sec x du D sec x tan x dx Z D 37. ADAMS and ESSEX: CALCULUS 9 .u2 1/ du D 13 u3 Z .u8 41. 39. Telegram: @uni_k sin. ln x/ dx x D 1 .0 Let u D ln x du D dx x ˇ=2 Z =2 ˇ 1 1 ˇ D sin u du D cos uˇ ˇ 0 Z =2 1 D 4 0 Z =2 cos 2x 2 1 0 2 1 C cos 4x dx 2 ˇ=2 ˇ=2 sin 2x ˇˇ sin 4x ˇˇ 3 : C D ˇ ˇ 4 ˇ 32 ˇ 16 Z =2 1 0 Z 1 1/ D : sin4 x dx D ˇ=2 3x ˇˇ D ˇ 8 ˇ 42. 2 cos 2x C 0 0 sin5 x dx =4 D 1/2 dx Z .1 cos2 x/2 sin x dx =4 Let u D cos x du D sin x dx ˇ1=p2 2 3 1 5 ˇˇ u C u ˇ 2u C u / du D u D p .1 3 5 ˇ 1= 2 1 ! 1 43 1 1 2 1 8 D p D p C : p C p 1C 3 5 15 2 3 2 20 2 60 2 Z Let u D csc x du D csc x cot x dx 2u6 C u4 / du 2u7 u5 u9 C CC 9 7 5 1 2 1 D csc9 x C csc7 x csc5 x C C: 9 7 5 Z Z cos4 x dx D cot4 x csc4 x dx 8 sin x Z D cot4 x.1 C cot2 x/ csc2 x dx Let u D cot x du D csc2 x dx Z 5 7 u u D u4 .1 C u2 / du D CC 5 7 1 1 D cot5 x cot7 x C C: 5 7 Z 4 x 3 .x 2 C 1/ 1=2 dx Let u D x 2 C 1; x 2 D u 1 0 du D 2x dx Z 1 17 1=2 .u 1/u du D 2 1 ˇˇ17 1 2 3=2 1=2 ˇ D u 2u ˇ ˇ 2 3 1 p p p 2 17 17 1 14 17 . 17 1/ D C : D 3 3 3 198 1 0 D 38. Z pe 0 uCC D 31 sec3 x sec x C C: Z csc5 x cot5 x dx Z D csc x cot x csc4 x.csc2 x D 40. 43. Z e2 2 4 dt t ln t Let u D ln t dt du D t Z 2 ˇ2 du ˇ D ln uˇ D ln 2 D 1 u 1 e 44. 1 Z 2 =9 2sin p x p ln 1 D ln 2: p Let u D sin x p cos x du D p dx 2 x ˇp3=2 Z p3=2 2.2u / ˇˇ D 2 p 2u du D ˇ ln 2 ˇ p 1= 2 2 =16 cos p x x dx 1= 2 2 p3=2 D .2 ln 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2 p 1= 2 /: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 45. Z =2 r x 2 cos2 dx 2 0 ˇ=2 p Z =2 p x x ˇˇ D 2 cos dx D 2 2 sin ˇ D 2: 2 2ˇ 0 Z =2 0 p 1 C cos x dx D 1 sin x dx Z =2 q 1 cos 2 D x dx 0 D Z 0 p 47. We start with the addition formulas cos.x C y/ D cos x cos y sin x sin y cos.x y/ D cos x cos y C sin x sin y 1 Let u D 2 du D dx and take half their sum and half their difference to obtain 1 cos.x C y/ C cos.x y/ 2 1 cos.x y/ cos.x C y/ : sin x sin y D 2 x cos x cos y D cos u du =2 ˇ=2 p u ˇˇ du D 2 2 cos D 2 sin ˇ 2 2 ˇ 0 0 p p D 2 C 2 2 D 2. 2 1/: Z 2 x dx Let u D x 2 C 16 Area D 2 C 16 x 0 du D 2x dx ˇ20 Z 20 ˇ du 1 1 ˇ D ln uˇ D ˇ 2 16 u 2 16 5 1 1 sq: units: D .ln 20 ln 16/ D ln 2 2 4 Z 2 x dx Area R D Let u D x 2 4 0 x C 16 du D 2x dx ˇ4 Z 4 ˇ du 1 1 1 uˇ D D tan sq. units: ˇ D 2 0 u2 C 16 8 4ˇ 32 Z =2 r 46. 49. 0 Z =2 p 0 SECTION 5.6 (PAGE 326) Similarly, taking half the sum of the formulas 2 u 0 y yD x x 4 C16 R x Fig. 5.6-47 48. The area bounded by the ellipse .x 2 =a2 / C .y 2 =b 2 / D 1 is s Z a x2 dx Let x D au b 1 4 a2 0 dx D adu Z 1p 2 D 4ab 1 u du: 0 The integral is the area of a quarter circle of radius 1. Hence .1/2 D ab sq: units: Area D 4ab 4 sin.x C y/ D sin x cos y C cos x sin y sin.x y/ D sin x cos y cos x sin y; we obtain sin x cos y D 50. y/ : We have Z cos ax cos bx dx Z 1 D Œcos.ax bx/ C cos.ax C bx/ dx 2Z Z 1 1 D cosŒ.a b/x dx C cosŒ.a C b/x dx 2 2 Let u D .a b/x, du D .a b/ dx in the first integral; let v D .a C b/x, dv D .a C b/ dx in the second integral. Z Z 1 1 cos u du C cos v dv D 2.a b/ 2.a C b/ 1 sinŒ.a b/x sinŒ.a C b/x D C C C: 2 .a b/ .a C b/ Z sin ax sin bx dx Z 1 Œcos.ax bx/ cos.ax C bx/ dx D 2 1 sinŒ.a b/x sinŒ.a C b/x D C C: 2 .a b/ .a C b/ Z sin ax cos bx dx Z 1 D Œsin.ax C bx/ C sin.ax bx/ dx 2 Z Z 1 D Œ sinŒ.a C b/x dx C sinŒ.a b/x dx 2 cosŒ.a b/x 1 cosŒ.a C b/x C C C: D 2 .a C b/ .a b/ Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 sin.x C y/ C sin.x 2 Downloaded by ted cage (sxnbyln180@questza.com) 199 lOMoARcPSD|6566483 www.konkur.in SECTION 5.6 (PAGE 326) 51. ADAMS and ESSEX: CALCULUS 9 For m D 0 we have Z Z f .x/ cos mx dx D f .x/ dx Z a0 dx D 2 k X C .an cos.nx/ C bn sin.nx// dx If m and n are integers, and m ¤ n, then ) Z ( cos mx cos nx dx sin mx sin nx Z 1 D cos.m n/x ˙ cos.m C n/x dx 2 ˇ sin.m C n/x ˇˇ 1 sin.m n/x ˙ D ˇ ˇ 2 m n mCn nD1 a0 D .2/ C 0 C 0 D a0 ; 2 D 0 ˙ 0 D 0: Z sin mx cos nx dx Z 1 sin.m C n/x C sin.m n/x dx D 2 ˇ cos.m n/x ˇˇ 1 cos.m C n/x C D ˇ ˇ 2 mCn m n so the formula for am holds for m D 0 also. Section 5.7 Areas of Plane Regions (page 331) 1. D 0 (by periodicity). Area of R D D If m D n ¤ 0 then Z sin mx cos mx dx Z 1 sin 2mx dx D 2 ˇ ˇ 1 ˇ D cos 2mx ˇ D 0 (by periodicity). ˇ 4m Z 1 .x 0 x2 2 x 2 / dx ˇ1 x 3 ˇˇ 1 ˇ D 3 ˇ 2 1 1 D sq. units. 3 6 0 y .1;1/ yDx R yDx 2 52. If 1 m k, we have Z Z a0 cos mx dx f .x/ cos mx dx D 2 Z k X C an cos nx cos mx dx C nD1 k X Z nD1 bn x Fig. 5.7-1 2. Area of R D D sin nx cos mx dx: 2 3=2 x 3 Z 1 p . x x 2 / dx 0 ˇ1 1 3 ˇˇ 1 2 1 x ˇ D D sq. units. ˇ 3 3 3 3 0 y By the previous exercise, all the integrals on the right side are zero except the one in the first sum having n D m. Thus the whole right side reduces to Z Z 1 C cos.2mx/ dx am cos2 .mx/ dx D am 2 am D .2 C 0/ D am : 2 Thus am D 1 Z Telegram: @uni_k R .1;1/ yDx 2 x Fig. 5.7-2 f .x/ cos mx dx: A similar argument shows that Z 1 f .x/ sin mx dx: bm D 200 p yD x 3. Area of R D 2 Z 2 0 .8 D 16x Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2x 2 / dx ˇ2 4 3 ˇˇ 64 x ˇ D sq. units. ˇ 3 3 0 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL y SECTION 5.7 (PAGE 331) y yD3 x 2 2yD4x x 2 2 2 6 x R 1 x R 2yC3xD6 yDx 2 5 Fig. 5.7-5 Fig. 5.7-3 4. 6. For intersections: 2 2 x 2x D 6x x ) 2x 8x D 0 i:e:; x D 0 or 4: Z 4h i Area of R D 6x x 2 .x 2 2x/ dx 0 Z 4 D .8x 2x 2 / dx 0 ˇ4 2 3 ˇˇ 64 2 x ˇ D sq. units. D 4x ˇ 3 3 1 0 y yD6x For intersections: 7 C y D 2y 2 y C 3 ) 2y 2 2y 4 D 0 2.y 2/.y C 1/ D 0 ) i:e:; y D 1 or 2: Z 2 Area of R D Œ.7 C y/ .2y 2 y C 3/ dy 1 Z 2 D2 .2 C y y 2 / dy 1 ˇ2 1 2 1 3 ˇˇ y ˇ D 9 sq. units. D 2 2y C y ˇ 2 3 2 y x2 .4;8/ .9;2/ xD2y 2 yC3 R x yD7 R yDx 2 2x x .6; 1/ x Fig. 5.7-6 Fig. 5.7-4 5. For intersections: 7. 2y D 4x x 2 2y C 3x D 6 ) Area of R D 2 4x x 2 D 6 3x x 2 7x C 6 D 0 .x 1/.x 6/ D 0 D2 Z 1 Thus intersections of the curves occur at x D 1 and x D 6. We have Area of R D D Z 6 2x 1 7x 2 4 245 D 4 x2 2 x3 6 1 36 C 6 3x 3C 2 ˇˇ6 ˇ 3x ˇ ˇ x2 2 x 3 / dx ˇ1 x 4 ˇˇ 1 ˇ D sq. units. 4 ˇ 2 0 y .1;1/ yDx R dx R yDx 3 x . 1; 1/ 1 125 15 D sq. units. 12 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k .x 0 Downloaded by ted cage (sxnbyln180@questza.com) Fig. 5.7-7 201 lOMoARcPSD|6566483 www.konkur.in SECTION 5.7 (PAGE 331) 8. Z 1 Shaded area D D 1 3 x 3 ADAMS and ESSEX: CALCULUS 9 y .x 2 x 3 / dx 0 ˇ1 1 1 4 ˇˇ x ˇ D sq. units. ˇ 4 12 .4;2/ xD2y 2 y 2 xDy 2 A 0 y x .1; 1/ .1;1/ yDx 2 Fig. 5.7-10 yDx 3 x 11. Fig. 5.7-8 1 5 2x DyD . x 2 2 Thus 2x 5x C 2 D 0, i.e., .2x 1/.x 2/ D 0. The graphs intersect at x D 1=2 and x D 2. Thus For intersections: Area of R D 9. Area of R D D Z 1 D p . x 0 2 3=2 x 3 y x 3 / dx ˇ1 x 4 ˇˇ 5 sq. units. ˇ D 4 ˇ 12 Z 2 0 2x 2 1=2 x2 2 5x 2 15 D 8 y 5 1 dx x ˇ ˇ2 ˇ ln x ˇ ˇ 1=2 2 ln 2 sq. units. 1 ;2 2 2xC2yD5 xDy 2 p yD x R R .1;1/ yD1=x 1 2; 2 yDx 3 x x Fig. 5.7-11 Fig. 5.7-9 12. D2 10. x2/ .x 0 2 1 x / dx D 2 x 3 3 1 5 x 5 2 A 1 D 1 Œ2 C y 9 sq. units. 2 202 1 y dy D 2y C y 2 2 2 .x 2 1 3 y 3 ˇˇ2 ˇ ˇ ˇ 0 A x 1 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) Fig. 5.7-12 1/2 dx ˇˇ1 4 ˇ sq. units. ˇ D ˇ 15 yD1 x 2 y D 2y y 2)y y 2D0 .y 2/.y C 1/ D 0 ) i:e:; y D 1 or 2: Z 2 Œy 2 .2y 2 y 2/ dy Area of R D D y yD.x 2 1/2 2 Z 2 0 4 For intersections: 2 Telegram: @uni_k Z 1 Z 1 Œ.1 Area of shaded region D 2 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 13. The curves y D Thus SECTION 5.7 (PAGE 331) 1 x2 and y D intersect at x D ˙1. 2 1 C x2 Area of R D 2 Z 1 0 y R 2 x2 yD 2 16. Area A D R D 1 1 1 x 2 Fig. 5.7-15 1 yD 2 x C1 1 R 1 sq. units. 3 0 y 4 x2 yD5 x 2 x2 dx 2 ˇ1 x 3 ˇˇ ˇ D 6 ˇ 2 1 1 C x2 D 2 tan 1 x yD Z .y 2 .sin y 2 // dy ˇ 4 3 y 3 ˇˇ sq. units. ˇ D ˇ 3 3 cos y C 2 y y x xDsin y Fig. 5.7-13 A 14. For intersections: 4x D 1 ) x 2 4x C 3 D 0 3 C x2 i:e:; x D 1 or 3: # Z 3" 4x Shaded area D 1 dx 3 C x2 1 ˇ3 ˇ ˇ 2 D Œ2 ln.3 C x / xˇ D 2 ln 3 2 sq. units. ˇ x xDy 2 2 Fig. 5.7-16 17. 1 Area of R D y D 4x yD 3Cx 2 yD1 .1;1/ D .3;1/ Z 5=4 .sin x cos x/ dx =4 ˇ5=4 ˇ .cos x C sin x/ˇˇ p 2C y p =4 p 2 D 2 2 sq. units. yDsin x x 5=4 R x =4 yDcos x Fig. 5.7-14 15. 4 and y D 5 x 2 intersect where x2 4 2 x 5x C 4 D 0, i.e., where .x 2 4/.x 2 1/ D 0. Thus the intersections are at x D ˙1 and x D ˙2. We have Fig. 5.7-17 The curves y D Area of R D 2 Z 2 5 1 D 2 5x 4 dx x2 ˇ 2 4 ˇˇ 4 x3 C ˇ D sq. units. 3 x ˇ 3 x2 1 18. Area D Z =2 .1 1 C cos.2x/ dx 2 0 ˇ =2 sin.2x/ ˇˇ D sq. units. D xC ˇ ˇ 2 2 D2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k sin2 x/ dx =2 Z =2 Downloaded by ted cage (sxnbyln180@questza.com) 0 203 lOMoARcPSD|6566483 www.konkur.in SECTION 5.7 (PAGE 331) ADAMS and ESSEX: CALCULUS 9 y y yDtan x yD yD1 A 4 ;1 4x yDsin2 x x x 2 Fig. 5.7-21 Fig. 5.7-18 19. Area A D D Z =2 22. 2 sin x/ dx .sin x 0 cos x C sin x cos x 2 y ˇ=2 x ˇˇ D1 ˇ ˇ 0 For intersections: x 1=3 D tan.x=4/. Thus x D ˙1. Z 1 Area A D 2 x 1=3 sq. units. 4 0 D2 D 3 4=3 x 4 x dx 4 ˇ1 4 ˇˇ x ˇˇ ˇˇ ln ˇsec ˇ ˇ ˇ 4 tan 0 3 8 p ln 2 D 2 3 2 yDsin x 4 ln 2 sq. units. y yDsin2 x A A x 2 yDtan.x=4/ x 1 A Fig. 5.7-19 yDx 1=3 20. Area A D 2 D2 Z =4 0 Z =4 0 .cos2 x sin2 x/ dx ˇ=4 ˇ ˇ cos.2x/ dx D sin.2x/ˇ D 1 sq. units. ˇ 0 Fig. 5.7-22 23. For intersections: sec x D 2. Thus x D ˙=3. y Area A D 2 Z =3 0 yDcos2 x D .4x yDsin2 x A D 4 3 yD2 Fig. 5.7-20 A 4x D tan x ) x D 0 or : For intersections: 4 Z =4 4x Area D tan x dx 0 ˇˇ=4 1 2 2 ˇ D x ln j sec xj ˇ ln 2 sq. units. D ˇ 8 2 0 204 Telegram: @uni_k ˇ=3 ˇ ˇ 2 ln j sec x C tan xj/ˇ ˇ 0 p 2 ln.2 C 3/ sq. units. y x 4 21. sec x/ dx .2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) yDsec x 3 3 Fig. 5.7-23 x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 24. p For intersections: jxj D Area A D 2 2 cos 0 p 8 2 x sin 4 8 D y 2 cos.x=4/. Thus x D ˙1. Z 1 p D SECTION 5.7 (PAGE 331) x dx !ˇ1 ˇ 2 ˇ x ˇ ˇ x 4 yDxC2 A x1 0 1 sq. units. Z 1p x2 x 4 dx D4 Z 1 p x 1 x 2 dx D2 Z 1 Area of R D 4 p x yD 2 cos 4 A yDjxj 1 x2 x Fig. 5.7-26 27. y yDe x x 0 Let u D 1 x 2 du D 2x dx 0 1=2 u 0 ˇ1 4 4 3=2 ˇˇ du D u ˇ D sq. units. ˇ 3 3 0 y Fig. 5.7-24 y 2 Dx 2 x 4 R R x 25. For intersections: x D sin.x=2/. Thus x D ˙1. Z 1 x sin 2 0 4 x D cos 2 Area A D 2 4 D x dx ˇˇ1 2 ˇ x ˇ ˇ 0 1 sq. units. Fig. 5.7-27 28. Loop area D 2 D2 y Z p2 .u2 0 1 D 4 u7 7 A 1 Z 0 2 p x 2 2 C x dx 2/2 u.2u/ du D 4 4 5 4 3 u C u 5 3 Z Let u2 D 2 C x 2u du D dx p 2 .u6 p p ˇˇ 2 256 2 ˇ sq. units: ˇ D ˇ 105 0 y x x yDsin 2 yDx 2 A x Fig. 5.7-25 y 2 Dx 4 .2Cx/ x 26. For intersections: e D x C2. There are two roots, both of which must be found numerically. We used a TI-85 solve routine to get x1 1:841406 and x2 1:146193. Thus Area A D D Z x2 x1 .x C 2 x2 C 2x 2 e x / dx ˇˇx2 x ˇ e ˇ ˇ Fig. 5.7-28 29. The tangent line to y D e x at x D 1 is y or y D ex. Thus Area of R D x1 1:949091 sq. units. Z 1 0 .e x D ex Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 4u4 C 4u2 / du 0 Downloaded by ted cage (sxnbyln180@questza.com) ex/ dx ˇ1 ex 2 ˇˇ e ˇ D 2 ˇ 2 e D e.x 1/, 1 sq. units. 0 205 lOMoARcPSD|6566483 www.konkur.in SECTION 5.7 (PAGE 331) ADAMS and ESSEX: CALCULUS 9 y yDe x 3. xi D 1 C .2i=n/, .i D 0; 1; 2; : : : ; n/, xi D 2=n. Z 3 .1;e/ 1 R yDex f .x/ dx D lim n!1 30. The tangent line to y D x 3 at .1; 1/ is y 1 D 3.x 1/, or y D 3x 2. The intersections of y D x 3 and this tangent line occur where x 3 3x C 2 D 0. Of course x D 1 is a (double) root of this cubic equation, which therefore factors to .x 1/2 .x C 2/ D 0. The other intersection is at x D 2. Thus D .x 3 2 4 x 4 4. Z 1 p 1 C x dx ˇ1 p ˇ 4 2 2 3=2 ˇ D .1 C x/ ˇ D ˇ 3 3 n!1 2 0 2 : 0 5. .1;1/ x R n p Pn Rn D p iD1 .1=n/ 1 C .i=n/ is a Riemann sum for f .x/ D 1 C x on the interval Œ0; 1. Thus lim Rn D y yDx 3 iD1 2 n 4i 2 4i 4i C 2 C3 1C 2C n n n iD1 n 4 2X 2 C 2 i2 D lim n!1 n n iD1 4 8 n.n C 1/.2n C 1/ D lim nC 3 n!1 n n 6 20 8 D4C D 3 3 3 27 C6C2C4D sq. units. 2 4 15 4 D 3x C 2/ dx ˇˇ1 3x 2 ˇ C 2x ˇ ˇ 2 2xi C 3/ 2X n!1 n Fig. 5.7-29 Z 1 .xi2 D lim x Area of R D n X 6. yD3x 2 7. . 2; 8/ Z Z .2 p 5p 5 0 R3 1 1 Z sin x/ dx D 2.2/ sin x dx D 4 0 D 4 x 2 dx D 1/4 of the area of a circle of radius D 5 1 p 2 . 5/ D 4 4 x dx D area A1 2 area A2 D 0 y yD1 x 2 A1 Fig. 5.7-30 1 3 A2 x Review Exercises 5 (page 332) Fig. R-5-7 1. 1 j2 n X j D1 1 j 2 C 2j C 1 j 2 2j C 1 D D 2 2 .j C 1/ j 2 .j C 1/2 j .j C 1/2 n X 1 1 2j C 1 D j 2 .j C 1/2 j 2 .j C 1/2 j D1 D 1 12 2. The number of balls is D iD1 206 Telegram: @uni_k i.i C 10/ D 0 cos x dx D area A1 y yDcos x A2 x Fig. R-5-8 Z 1 1 .2 sin.x 3 // dx D Œ2.2/ 2 2 Z 1 5 5 1 3 jx 2j dx D D (via #9) 10. h D 3 0 3 2 6 9. .30/.31/.61/ .30/.31/ C 10 D 14;105: 6 2 area A2 D 0 A1 1 n2 C 2n D 2 .n C 1/ .n C 1/2 40 30 C 39 29 C C 12 2 C 11 1 30 X 8. R f D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 0 D 2 p 5 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 11. Z t f .t / D 14. Z sin x p 13 Z 1 4s g. / D Z ecos g 0 .s/ D p 1 C sin2 x.cos x/ ln x dx //e cos . sin / 2f .x/ C 1 D 3 0 Z 1 .ln.e sin //e sin cos f .t / dt x 21. y D sin x and y D cos.2x/ intersect at x D =6, but nowhere else in the interval Œ0; =6. The area between the curves in that interval is ˇ=6 Z =6 ˇˇ 1 .cos.2x/ sin x/ dx D 2 sin.2x/ C cos x ˇ ˇ 0 0 p p p 3 3 3 3 D C 1D 1 sq. units.: 4 2 4 22. y D 5 x 2 and y D 4=x 2 meet where 5 is, where x 4 5x 2 C 4 D 0 3x=2 u/ D sin u) .x 2 0 0 17. 2 Z x2 .2 Cx x 2 / dx D 2x C 2 1 y 0 18. The area bounded by y D .x Z 1 0 .x 2 1/ dx D .x x 3 A A 1 ˇ1 1 1/3 ˇˇ ˇ D sq. units.: ˇ 3 3 2 x Fig. R-5-22 23. Z x 2 cos.2x 3 C 1/ dx Let u D 2x 3 C 1 du D 6x 2 dx Z 1 sin u sin.2x 3 C 1/ D cos u du D CC D CC 6 6 6 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x2 1 1/2 , y D 0, and x D 0 is 0 4 yD 2 x yD5 ˇ2 9 ˇ ˇ D sq. units.: ˇ 2 3 ˇ 4/ D 0: 1 f .sin x/ dx: y D 2 C x x 2 and y D 0 intersect where 2 C x x 2 D 0, that is, where .2 x/.1 C x/ D 0, namely at x D 1 and x D 2. Since 2 C x x 2 0 on Œ 1; 2, the required area is Z 2 1/.x 2 x 2 D 4=x 2 , that There are four intersections: x D ˙1 and x D ˙2. By symmetry (see the figure) the total area bounded by the curves is ˇ2 Z 2 4 ˇˇ 4 4 x3 2 2 C 5 x dx D 2 5x ˇ D sq. units. x2 3 x ˇ 3 1 Now, solving for I , we get xf .sin x/ dx D I D y D 4x x 2 and y D 3 meet where x 2 4x C 3 D 0, that is, at x D 1 and x D 3. Since 4x x 2 3 on Œ1; 3, the required area is ˇˇ3 Z 3 x3 4 ˇ 2 2 .4x x 3/ dx D 2x 3x ˇ D sq. units. ˇ 3 3 1 1 2f .x/ D 3f .x/ ÷ f .x/ D C e 2f .1/ C 1 D 0 1 1 3=2 D f .1/ D C e 3=2 ÷ C D e 2 2 1 .3=2/.1 x/ f .x/ D e : 2 Z I D xf .sin x/ dx Let x D u 0 dx D du Z 0 . u/f .sin. u// du (but sin. D Z Z D f .sin u/ du uf .sin u/ du 0 Z0 D f .sin x/ dx I: Z x D y y 4 and x D 0 intersect where y y 4 D 0, that is, at y D 0 and y D 1. Since y y 4 0 on Œ0; 1, the required area is 2 ˇ1 Z 1 y y 5 ˇˇ 3 4 .y y 0/ dy D sq. units. ˇ D 2 5 ˇ 10 0 0 4e sin.4s/ sin cos .e cos C e sin / D 16. f 0 .x/ D 20. e sin cos g 0 . / D .ln.e 15. 1 C t 2 dt; e sin u du; 19. f 0 .t / D sin.t 2 / 13 12. f .x/ D 13. g.s/ D sin.x 2 / dx; REVIEW EXERCISES 5 (PAGE 332) Downloaded by ted cage (sxnbyln180@questza.com) 207 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 5 (PAGE 332) 24. Z e ADAMS and ESSEX: CALCULUS 9 ln x dx x Since 1=.1 C t 2 / > 0 for all t , F .x/ will be minimum when x 2 2x D .x 1/2 1 Let u D ln x 1 du D dx=x ˇ1 Z 1 u2 ˇˇ 1 D u du D ˇ D ˇ 2 2 0 is minimum, that is, when x D 1. The minimum value is 0 25. Z 4p 9t 2 C t 4 dt 0 Z 4 p t 9 C t 2 dt D F .1/ D 2 Let u D 9 C t du D 2t dt ˇ25 Z 25 p 1 1 3=2 ˇˇ 98 D u du D u ˇ D ˇ 2 9 3 3 9 Z sin3 .x/ dx Z D sin.x/ 1 cos2 .x/ dx Let u D cos.x/ du D sin.x/ dx Z 1 .1 u2 / du D 1 u3 1 1 D u CC D cos3 .x/ cos.x/ C C 3 3 Z ln 2 eu du Let v D e u 4 C e 2u 0 dv D e u du Z 2 dv D 2 1 4Cv ˇ2 ˇ vˇ 1 1 1 tan 1 D tan 1 ˇ D ˇ 2 2 8 2 2 27. 1 0 ˇ 1 ˇ dt 1 ˇ D tan t ˇ D 2 ˇ 1Ct : 4 0 F has no maximum value; F .x/ < =2 for all x, but F .x/ ! =2 if x 2 2x ! 1, which happens as x ! ˙1. 0 26. Z 32. f .x/ D 4x x 2 0 if 0 x 4, and f .x/ < 0 Rb otherwise. If a < b, then a f .x/ dx will be maximum if Œa; b D Œ0; 4; extending the interval to the left of 0 or to the right of 4 will introduce negative contributions to the integral. The maximum value is Z 4 .4x 0 33. x / dx D 2x x3 3 2 ˇˇ4 32 ˇ : ˇ D ˇ 3 0 The average value of v.t / D dx=dt over Œt0 ; t1 is 1 t1 34. 2 t0 Z t1 t0 ˇt1 ˇ x.t1 / 1 dx ˇ dt D x.t /ˇ D ˇ dt t1 t0 t1 x.t0 / D vav : t0 t0 If y.t / is the distance the object falls in t seconds from its release time, then 1 28. Z 4pe y 00 .t / D g; 2 tan . ln x/ dx x Let u D ln x 1 du D .=x/ dx Z Z 1 =4 1 =4 2 tan u du D .sec2 u D 0 0 ˇ=4 ˇ 1 1 1 ˇ D D .tan u u/ˇ ˇ 4 29. 1/ du The average height during the time interval Œ0; T is Z 1 2t 30. cos sin dt D dt sin2 5 5 4 5 Z 1 4t D 1 cos dt 8 5 1 4t 5 D sin t CC 8 4 5 Z x 2 2x 1 dt . 31. F .x/ D 1 C t2 0 208 Telegram: @uni_k 1 T p p sin 2s C 1 p ds Let u D 2s C 1 p 2s C 1 du D ds= 2s C 1 Z p D sin u du D cos u C C D cos 2s C 1 C C Z 2 t 2 t and y 0 .0/ D 0: Antidifferentiating twice and using the initial conditions leads to 1 y.t / D gt 2 : 2 0 Z y.0/ D 0; 35. Z T 0 1 2 g T3 gT 2 gt dt D D Dy 2 2T 3 6 T p : 3 Let f .x/ D ax 3 C bx 2 C cx C d so that Z 1 a b c C C C d: 4 3 2 0 We want this integral to be f .x1 / C f .x2 / =2 for all choices of a, b, c, and d . Thus we require that f .x/ dx D a.x13 C x23 / C b.x12 C x22 / C c.x1 C x 2 / C 2d Z 1 2b a C c C 2d: D2 f .x/ dx D C 2 3 0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 5 (PAGE 332) It follows that x1 and x2 must satisfy 2. 1 x13 C x23 D .3/ n X 2 3 1 6 D sin.jt / j D1 2 sin. 12 t / j D1 1 : 2 2 sin. 12 t / D L.f; Pn / D D 1 .2i=n .i 1/=n 2 iD1 n X .21=n iD1 n X iD1 n X 1 2i=n .1 iD1 .2 i=n 2 lim n.2 n!1 .i 1/=n i cos. 12 t / i i cos .n C 12 /t : n X sin j D1 j 2n 2n 1 cos n!1 2n 2 sin.=.4n// 4n =.4n/ lim cos D lim n!1 sin.=.4n// n!1 4n D 1 cos 0 cos D 1: 2 D lim 3. / 2 1=n / D U.f; Pn / : 21=n 21=x 1 0 1/ D lim x!1 1=x 0 1=x 2 2 ln 2. 1=x / D lim D ln 2: x!1 1=x 2 Thus limn!1 U.f; Pn / D limn!1 L.f; Pn / D ln s. a) sin .j C 21 /t .2n C 1/ 4n .2n C 1/ cos 4n cos sin .j 1 2 /t D sin.jt / cos. 12 t / C cos.jt / sin. 12 t / sin.jt / cos. 12 t / C cos.jt / sin. 12 t / D 2 cos.jt / sin. 12 t /: Therefore, we obtain a telescoping sum: n X cos.jt / j D1 D D 1 n h X sin .j C 21 /t 2 sin. 12 t / j D1 1 2 sin. 12 t / Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 2 /t 1/ Now, by l’H^opital’s rule, 1=n cos .j sin x dx 0 2.i 1/=n / 2 1=n / D n.1 cos .n C 21 /t h 1 cos. 12 t / 2 sin. 12 t / Z =2 (page 332) 1/ D n.21=n cos .j C 12 /t 2 3 b) Let Pn D f0; 2n ; 2n ; 2n ; : : : n 2n g be the partition of Œ0; =2 into n subintervals of equal length x D =2n. Using t D =2n in the formula obtained in part (a), we get xi D 2i=n , 0 i n, f .x/ D 1=x on Œ1; 2. Since f is decreasing, f is largest at the left endpoint and smallest at the right endpoint of any interval Œ2.i 1/=n ; 2i=n of the partition. Thus n X h 1 D D n h X 1 D n!1 U.f; Pn / D sin.jt / sin. 12 t / D Therefore, we obtain a telescoping sum: D lim 1. 2 sin.jt / sin. 12 t /: .2/ x1 x2 C x22 / D 1 Challenging Problems 5 1 2 /t sin.jt / sin. 12 t / cos.jt / cos. 12 t / At first glance this system may seem overdetermined; there are three equations in only two unknowns. However, they do admit a solution as we now show. Squaring equation (3) and subtracting equation (2) we get 2x1 x2 D 1=3. Subtracting this latter equation from equation (2) p then gives .x2 x1 /2 D 1=3, so that x2 x1 D 1= 3 (the positive square root since we want x1 < x2 ). Adding and subtracting this equation and p equation (3) thenpproduces the p p values x2 D . 3 C 1/=.2 3/ and x1 D . 3 1/=.2 3/. These values also satisfy equation (1) since x13 C x23 D .x2 C x2 /.x12 cos .j D cos.jt / cos. 12 t / .1/ 2 2 2 2 x1 C x2 D 3 x2 C x2 D 1: a) cos .j C 12 /t Downloaded by ted cage (sxnbyln180@questza.com) h sin .n C 12 /t sin .j i sin. 12 t / : 1 2 /t i 209 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 5 (PAGE 332) ADAMS and ESSEX: CALCULUS 9 2 3 b) Let Pn D f0; 3n ; 3n ; 3n ; : : : n 3n g be the partition of Œ0; =3 into n subintervals of equal length x D =3n. Using t D =3n in the formula obtained in part (a), we get Z =2 (obtained by the Binomial Theorem) for j D 1; 2; : : : ; n. The left side telescopes, and we get .n C 1/mC2 cos x dx 0 D lim n!1 n X cos j D1 j 3n 3n D D Thus Z 2 1 1 x0 n X 1 .xi 1 xi 1 xi 1 / 1 xi 1 D1 xn j mC1 1 1 D : 2 2 b) Using the technique of Example 2 in Section 6.2 and the result above, Z a 0 j D1 n 1 n!1 nkC1 nk nkC1 C C Pk 1 .n/; kC1 2 j D Da .m C 2/.m C 1/ m j C C1 2 kC1 lim n!1 akC1 : D kC1 1. n1C1 n n.n C 1/ D C C P0 .n/; 2 1C1 2 .j C1/mC2 j mC2 D .mC2/j mC1 C Telegram: @uni_k j D1 D akC1 lim where P0 .n/ D 0 certainly has degree 0. Now assume that the formula above holds for k D 1; 2; 3; : : : ; m. We will show that it also holds for k D m C 1. To this end, sum the the formula 210 a X a j n!1 n n x k dx D lim 5. We want to prove that for each positive integer k, n X m C 2 mC1 n 2 so that the formula is also correct for k D m C 1. Hence it is true for all positive integers k by induction. (telescoping) where Pk 1 is a polynomial of degree at most k First check the case k D 1: j mC1 j D1 1 nmC2 C .m C 2/nmC1 C mC2 nmC1 nmC2 C C D mC2 2 iD1 jk D n X j D1 n j D1 j D1 where the represent terms of degree m or lower in the variable n. Solving for the remaining sum, we get X dx 1 D lim f .ci / xi D . 2 n!1 x 2 n X j D1 D xi 1 xi iD1 n X iD1 n .m C 2/.m C 1/ nmC1 C C ; 2 mC1 so xi 1 < ci < xi . We have iD1 j D1 n nmC2 C .m C 2/nmC1 C D .m C 2/ xi2 1 < xiD1 xi D ci2 < xi2 ; n X j mC1 Expanding the binomial power on the left and using the induction hypothesis on the other terms we get sin 6n sin 6n 4. f .x/ D 1=x 2 , 1 D x0 < x1 < x2 < < xn D 2. If p ci D xi 1 xi , then f .ci / xi D n X X .m C 2/.m C 1/ X m j C C 1: C 2 .2n C 1/ 1 sin D lim n!1 3n 2 sin.=.6n// 6n =.6n/ .2n C 1/ D lim lim sin n!1 sin.=.6n// n!1 6n p 3 sin 0 D : D 1 sin 3 2 n X 1mC2 D .m C 2/ 6. n X jk j D1 1 1 Pk 1 .n/ C C kC1 2n nkC1 Let f .x/ D ax 3 C bx 2 C cx C d . We used Maple to calculate the following: The tangent to y D f .x/ at P D .p; f .p// has equation y D g.x/ D ap 3 Cbp 2 Ccp Cd C.3ap 2 C2bp Cc/.x p/: This line intersects y D f .x/ at x D p (double root) and at x D q, where 2ap C b qD : a Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 5 (PAGE 332) Similarly, the tangent to y D f .x/ at x D q has equation 8. y D h.x/ D aq 3 C bq 2 C cq C d C .3aq 2 C 2bq C c/.x q/; y D g.x/ D ap 4 Cbp 3 Ccp 2 CdpCeC.4ap 3 C3bp 2 C2cpCd /.x p/; and intersects y D f .x/ at x D q (double root) and x D r, where rD and intersects y D f .x/ at x D p (double root) and at the two points 2aq C b 4ap C b D : a a 1 12 D .f .x/ q 4 3 D 81a4 p 4 C 108a3 bp 3 C 54a2 b 2 p 2 C 12ab 3 p C b 4 a3 : 81a4 p 4 C 108a3 bp 3 C 54a2 b 2 p 2 C 12ab 3 p C b 4 a3 bC b qD ; b : 3a uD qD p 3b 2 p4a 3b 2 4a 3b 4 b2 c C 3 256a 16a2 vD b C 3ap and r a 4 81 4 4 3 3 8ac 8ac D b : 2a p 3 b bd CeC 4a 8a2 bc Cd 2a b xC 4a p p 2 3b 2 4a p p b C 2 3b 2 4a 8ac 8ac ; : 2.b C 3ap/ a 2 2 2 3 81a p C 108a bp C 54a b p C 12ab p C b a3 4 V S T B Q ; which is 16=27 of the area between the curve and its tangent at P . This leaves 11=27 of that area to lie between the curve, QR, and the tangent, so QR divides the area between y D f .x/ and its tangent at P in the ratio 16/11. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k b 2 D 0; b have the same sign, R must lie between Q and P on the curve y D f .x/. The line QR has a rather complicated equation y D k.x/, which we won’t reproduce here, but the area between this Rline and the curve y D f .x/ is q the absolute value of r .f .x/ k.x// dx, which Maple evaluates to be : and it intersects y D f .x/ at the points U and V with x-coordinates Since rD 8a2 p 2 4abp (Both roots exist and are distinct provided 3b 2 > 8ac.) The point T corresponds to x D t D .p C q/=2 D b=4a. The tangent to y D f .x/ at x D t has equation y D h.x/ D We continue with the calculations begun in the previous problem. P and Q are as they were in that problem, but R D .r; f .r// is now the inflection point of y D f .x/, given by f 00 .r/ D 0. Maple gives p 4ac 2a which has two solutions, which we take to be p and q: h.x/ dx rD b2 8a2 p 2 C 4abp C 4ac pD which is 16 times the area between y D f .x/ and the tangent at P . 7. p If these latter two points coincide, then the tangent is a “double tangent.” This happens if The area between y D f .x/ and the tangent line at Q D .q; f .q// is the absolute value of Z r b˙ 2ap xD The area between y D f .x/ and the tangent line at P is the absolute value of Z q .f .x/ g.x/ dx p Let f .x/ D ax 4 C bx 3 C cx 2 C dx C e. The tangent to y D f .x/ at P D .p; f .p// has equation Downloaded by ted cage (sxnbyln180@questza.com) A U R P Fig. C-5-8 211 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 5 (PAGE 332) a) The areas between the curve y D f .x/ and the lines PQ and U V are, respectively, the absolute values of Z q Z v A1 D .f .x/ g.x// dx and A2 D .h.x/ f .x// dx: p ADAMS and ESSEX: CALCULUS 9 The region bounded by RS and the curve y D f .x/ is divided into three parts by A and B. The areas of these three regions are the absolute values of u Maple calculates theseptwo integrals and simplifies the ratio A1 =A2 to be 1= 2. A1 D b) The two inflection points A and B of f have xcoordinates shown by Maple to be p 3.3b 2 8ac/ 3b and ˛D p12a 3b C 3.3b 2 8ac/ ˇD : 12a It then determines the four points of intersection of the line y D k.x/ through these inflection points and the curve. The other two points have x-coordinates p 15.3b 2 8ac/ 3b and rD p 12a 2 3b C 15.3b 8ac/ sD : 12a 212 Telegram: @uni_k A2 D A3 D Z ˛ .k.x/ f .x// dx .f .x/ k.x// dx .k.x/ f .x// dx: r Z ˇ Z˛s ˇ The expressions calculated by Maple for k.x/ and for these three areas are very complicated, but Maple simplifies the rations A3 =A1 and A2 =A1 to 1 and 2 respectively, as was to be shown. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.1 (PAGE 339) CHAPTER 6. TECHNIQUES OF INTEGRATION 5. Z x cos x dx U Dx d U D dx Z D x sin x d V D cos x dx V D sin x 6. sin x dx 3. Z .x C 3/e 2x dx U DxC3 d U D dx 1 D .x C 3/e 2x 2 1 D .x C 3/e 2x 2 Z x 2 cos x dx U D x2 d U D 2x dx d V D e 2x dx V D 21 e 2x Z 1 e 2x dx 2 1 2x e C C: 4 d V D cos x dx sin x V D Z 2 x 2 sin x x sin x dx U Dx d V D sin x dx cos x d U D dx V D Z x 2 sin x 2 x cos x 1 D C cos x dx 2 2 1 sin x C C: D x 2 sin x C 2 x cos x 3 Z .x 2 2x/e kx dx D 4. U D x 2 2x d U D .2x 2/ dx d V D e kx 1 V D e kx k Z 7. x.ln x/3 dx D I3 where Z In D x.ln x/n dx Z tan 1 x dx U D tan 1 x dx dU D 1 CZx 2 D x tan 1 x D x tan 1 x 8. 1 1 D .x 2 2x/e kx .2x 2/e kx dx k k U D x 1 d V D e kx dx 1 d U D dx V D e kx k Z 1 2 1 1 D .x 2 2x/e kx .x 1/e kx e kx dx k k k k 2 kx 2 1 2 kx kx .x 1/e C 3 e C C: 2x/e D .x k k2 k Z d V D x dx U D .ln x/n n 1 n 1 dx d U D .ln x/ V D x2 x 2 Z 1 2 n n n 1 D x .ln x/ x.ln x/ dx 2 2 n 1 In 1 D x 2 .ln x/n 2 2 1 3 I3 D x 2 .ln x/3 I2 2 2 1 3 1 2 2 D x 2 .ln x/3 x .ln x/2 I1 2 2 2 2 3 3 1 2 1 1 2 3 2 2 x .ln x/ C x .ln x/ I0 D x .ln x/ 2 4 2 2 2 Z 1 3 2 3 3 D x 2 .ln x/3 x dx x .ln x/2 C x 2 .ln x/ 2 4 4 4 3 3 3 x2 .ln x/2 C .ln x/ .ln x/3 C C: D 2 2 2 4 D x sin x C cos x C C: 2. x 3 ln x dx U D ln x d V D x 3 dx dx x4 dU D V D x Z 4 1 4 1 3 D x ln x x dx 4 4 1 1 4 D x 4 ln x x C C: 4 16 Section 6.1 Integration by Parts (page 339) 1. Z x dx 1 C x2 1 ln.1 C x 2 / C C: 2 Z x 2 tan 1 x dx D x3 tan 1 x 3 U D tan 1 x d V D x 2 dx dx x3 dU D V D 1 C x2 3 Z x3 1 x3 1 dx tan x D 3 3 1 C x2 ! Z x3 1 x D dx tan 1 x x 3 3 1 C x2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k d V D dx V Dx Downloaded by ted cage (sxnbyln180@questza.com) x2 1 C ln.1 C x 2 / C C: 6 6 213 lOMoARcPSD|6566483 www.konkur.in SECTION 6.1 (PAGE 339) 9. Z ADAMS and ESSEX: CALCULUS 9 For n D 5 we have x sin 1 x dx d V D x dx U D sin 1 x dx x2 dU D p V D 2 1 x2 Z 2 x dx 1 1 2 D x sin 1 x Let x D sin p 2 2 1 x2 dx D cos d Z 1 1 2 2 1 sin d D x sin x 2 2 1 1 D x 2 sin 1 x . sin cos / C C 2 4 1 2 1 1 p D x sin 1 x C x 1 x 2 C C: 2 4 4 10. Z Z =4 0 12. D sec x tan x D sec x tan x d V D xe dx U D x 2n 2 d U D 2nx .2n 1/ dx V D 12 e x Z 1 2n x 2 2 D x e C n x .2n 1/ e x dx 2 1 2n x 2 D x e C nIn 1 2 " # Z 1 2 x2 1 4 x2 x2 x e x e C2 C xe dx I2 D 2 2 In D Z .1 C tan2 x/ sec x dx ln j sec x C tan xj I D Z I 1 2 ln j sec x C tan xj C C: e 2x sin 3x dx U D e 2x d V D sin 3x dx d U D 2e 2x dx V D 13 cos 3x Z 1 2x 2 D e 2x cos 3x dx e cos 3x C 3 3 secn x dx Therefore Telegram: @uni_k sec3 x dx d V D cos 3x dx U D e 2x d U D 2e 2x dx V D 13 sin 3x 2 1 2x 2 1 2x e cos 3x C e sin 3x I D 3 3 3 3 1 2x 2 2x 13 I D e cos 3x C e sin 3x C C1 9 3 9 1 2x I D e .2 sin 3x 3 cos 3x/ C C: 13 U D secn 2 x d V D sec2 x dx n 2 d U D .n 2/ sec x tan x dx V D tan x ˇ=4 Z =4 ˇ ˇ D tan x secn 2 x ˇ .n 2/ secn 2 x tan2 x dx ˇ 0 0 p D . 2/n 2 .n 2/.In In 2 /: p .n 1/In D . 2/n 2 C .n 2/In 2 : 214 13. 0 p . 2/n 2 n In D C n 1 n d V D sec x tan x dx V D sec x Thus; I D 21 sec x tan x 1 x2 4 e .x C 2x 2 C 2/ C C: 2 Z =4 tan2 x sec x dx D sec x tan x 2 x2 11. Z U D tan x d U D sec2 xZ dx x 5 e x dx D I2 where Z 2 In D x .2nC1/ e x dx D I D p 2 2 3 sec5 x dx D I5 D C I3 4 4 ! p p 2 2 3 1 C C I1 D 2 4 2 2 p ˇ=4 7 2 3 ˇ D C ln j sec x C tan xjˇ 0 8 8 p p 3 7 2 C ln.1 C 2/: D 8 8 2 In 2 ; 1 .n 2/: 14. I D Z xe p x dx Let x D w 2 dx D 2w dw Z D 2 w 3 e w dw D 2I3 where Z In D w n e w dw U D wn d U D nw n 1 dw D w n e w nIn 1 : d V D e w dw V D ew I D 2I3 D 2w 3 e w 6Œw 2 e w 2.we w I0 / p p p D e x .2x x 6x C 12 x 12/ C C: Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 15. Z 1 sin 1 x dx x2 1=2 D D D Z 1 19. dx dV D 2 U D sin 1 x x dx 1 dU D p V D 1 x2 x ˇ1 Z 1 ˇ dx 1 ˇ C p Let x D sin sin 1 x ˇ ˇ x 1=2 x 1 x2 1=2 dx D cos d Z =2 C C csc d 2 3 =6 ˇ=2 ˇ ln j csc C cot jˇ =6 6 p p ln 1 C ln.2 C 3/ D ln.2 C 3/ : 6 6 D 16. SECTION 6.1 (PAGE 339) p x sin. x/ dx Z 1 I D 20. 2 Let x D w dx D 2w dw w 2 sin.w/ dw d V D sin.w/ dw cos.w/ V D ˇ1 Z ˇ 2 2 4 1 ˇ w cos.w/ˇ C w cos.w/ dw ˇ 0 1 d V D dx U D cos.ln x/ sin.ln x/ V Dx dx dU D x ˇe " # ˇ ˇ D e sin.1/ x cos.ln x/ˇ C I ˇ 0 d V D cos.w/ dw sin.w/ V D #ˇ1 " Z 1 ˇ 4 w 2 4 ˇ sin.w/ ˇ sin.w/ dw D C ˇ 2 0 0 ˇ1 ˇ 2 2 4 4 8 2 ˇ D C 3 cos.w/ˇ D C 3 . 2/ D : ˇ 3 0 Z x sec2 x dx D x tan x 18. 1 Thus; I D 21. d V D sec2 x dx V D tan x x sin 2x 4 1 cos 2x C C: 8 1 Œe sin.1/ 2 Z ln.ln x/ dx x D Z Let u D ln x dx du D x U D ln u du dU D Zu du D u ln u Z 4 p xe p x dx 0 D2 Z 2 0 uCC ln x C C: Let x D w 2 dx D 2w dw w 2 e w dw D 2I2 See solution #16 for the formula R In D w n e w dw D w n e w nIn 1 . ˇ2 ˇ ˇ ˇ2 2 wˇ 2 wˇ I0 2I1 D 8e 4 we ˇ D2 w e ˇ 0 D 8e 2 2 8e C 4 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k d V D du V Du D .ln x/.ln.ln x// 22. e cos.1/ C 1: ln u du D u ln u tan x dx D x tan x ln j sec xj C C: Z Z 1 .x x cos 2x/ dx x sin2 x dx D 2 Z x2 1 D x cos 2x dx 4 2 U Dx d V D cos 2x dx d U D dx V D 21 sin 2x " # Z x2 1 1 1 D x sin 2x sin 2x dx 4 2 2 2 x2 D 4 sin.ln x/ dx U D sin.ln x/ d V D dx cos.ln x/ V Dx dU D dx x D x cos.ln x/ C x sin.ln x/ I 1 I D x cos.ln x/ C x sin.ln x/ C C: 2 Z e I D sin.ln x/ dx 0 U Dx d U D dx Z d V D dx V Dx d V D dx U D sin.ln x/ cos.ln x/ V Dx dx dU D ˇe x Z ˇ e ˇ D x sin.ln x/ˇ cos.ln x/ dx ˇ 1 U Dw d U D dw 17. U D cos.ln x/ sin.ln x/ dU D dx Zx D x cos.ln x/ C U D w2 d U D 2w dw D cos.ln x/ dx 1 p 0 D2 Z Downloaded by ted cage (sxnbyln180@questza.com) Z 2 0 0 w e dw D 4.e 2 1/: 215 lOMoARcPSD|6566483 www.konkur.in SECTION 6.1 (PAGE 339) 23. Z cos 1 x dx Z x sec 1 x dx ADAMS and ESSEX: CALCULUS 9 Z 27. d V D dx U D cos 1 x dx V Dx dU D p 2 1 x Z x dx D x cos 1 x C p 2 p 1 x D x cos 1 x 1 x 2 C C: 24. d V D x dx U D .tan 1 x/2 x2 2 tan 1 x dx V D dU D 2 1 C x2 Z 2 1 x tan x x2 dx Let u D tan 1 x .tan 1 x/2 D 2 1 C x2 dx du D 1 C x2 Z x2 D .tan 1 x/2 u tan2 u du 2 Z x2 .tan 1 x/2 C .u u sec2 u/ du D 2 Z x2 u2 D .tan 1 x/2 C u sec2 u du 2 2 U Du d V D sec2 u du d U D du V D tan u Z 1 2 D .x C 1/.tan 1 x/2 u tan u C tan u du 2 1 D .x 2 C 1/.tan 1 x/2 x tan 1 x C ln j sec uj C C 2 1 1 D .x 2 C 1/.tan 1 x/2 x tan 1 x C ln.1 C x 2 / C C 2 2 U D sec 1 x d V D x dx 1 dx V D x2 dU D p 2 2 jxj x 1 Z 1 2 1 jxj 1 p dx D x sec x 2 2 x2 1 p 1 1 D x 2 sec 1 x sgn .x/ x 2 1 C C: 2 2 25. Z 2 sec 1 x dx Z 2 1 D cos 1 x 1 1 U D cos dU D 26. : 1 1 x 1 r 1 x2 1 x2 d V D dx V Dx dx 1 ˇ2 Z 2 1 ˇˇ dx Let x D sec D x cos 1 ˇ p xˇ x2 1 1 1 dx D sec tan d Z =3 2 sec d 0 D 3 0 ˇ=3 2 ˇ D ln j sec C tan jˇ 0 3 p 2 D ln.2 C 3/: 3 Z .sin 1 x/2 dx Let x D sin dx D cos d Z 2 cos d D U D 2 d U D 2 d Z D 2 sin 2 U D d U D d 2 D sin d V D cos d V D sin sin d d V D sin d V D cos Z 2. cos C Telegram: @uni_k 28. By the procedure used in Example 4 of Section 7.1, Z Z e x cos x dx D 21 e x .sin x C cos x/ C C I e x sin x dx D 21 e x .sin x cos x/ C C: Now Z xe x cos x dx d V D e x cos x dx V D 21 e x .sin x C cos x/ Z D 12 xe x .sin C cos x/ 21 e x .sin x C cos x/ dx U Dx d U D dx D 12 xe x .sin C cos x/ cos d / D 2 sin C 2 cos 2 sin C C p D x.sin 1 x/2 C 2 1 x 2 .sin 1 x/ 216 x.tan 1 x/2 dx D 21 xe 1 x e .sin x 4 x cos x C sin x C cos x/ C C .sin x C cos x/ 2x C C: Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 x 2 e sin x C C: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 29. Z Area D A D SECTION 6.1 (PAGE 339) e x sin x dx 32. 0 d V D sin x dx U De x x d U D eˇ dx V D cos x Z ˇ x ˇ e x cos x dx D e cos x ˇ U De x d U D ex dx 30. The tangent line to y D ln x at x D 1 is y D x 1 .1/.1/ C .1/.e 2 2/ ˇe ˇ ˇ x/ˇ ˇ De 3 2 .x ln x De 3 2 eCeC0 Z e d V D cos x dx U D xn 1 d U D .n 1/x n 2 dx V D sin x ˇ=2 " # Z ˇ =2 ˇ n 1 n 2 Dn x sin x ˇ .n 1/ x sin x dx ˇ 0 0 n 1 n.n 1/In 2 ; .n 2/: Dn 2 ˇ =2 Z =2 ˇ ˇ I0 D sin x dx D cos x ˇ D 1: ˇ 0 0 ( ) 5 3 I6 D 6 6.5/ 4 4.3/ 2 2.1/I0 2 2 2 1, Hence, ln x dx 1 D 3 5 16 33. yD1 1 In D Z sinn x dx x D nIn D Fig. 6.1-30 In D sinn 1 x cos x C .n .ln x/n dx U D .ln x/n In D x.ln x/n I4 D x.ln x/ 4I3 4 x.ln x/3 4 D x.ln x/ D x.ln x/4 3I2 4x.ln x/3 C 12 x.ln x/2 4x.ln x/3 C 12x.ln x/2 24 x ln x x C C D x .ln x/4 4.ln x/3 C 12.ln x/2 In / sin x cos x C .n 1/In 2 1 n 1 n 1 sin x cos x C In 2 : n n cos x C C . Hence 5 1 5 sin x cos x C I4 6 6 5 1 3 3 1 5 sin x cos x C sin x cos x C I2 D 6 6 4 4 5 1 5 3 sin x cos x sin x cos x D 6 24 5 1 1 C sin x cos x C I0 8 2 2 5 5 1 5 sin x cos x sin3 x cos x sin x cos x D 6 24 16 5 C xCC 16 ! sin5 x 5 sin3 x 5 sin x 5x cos x C C C C: D 16 6 24 16 I6 D d V D dx V Dx nIn 1 : 4 D x.ln x/4 dx x 1/.In 2 n 1 Note: I0 D x C C , I1 D d U D n.ln x/n 1 2I1 24 ln x C 24 C C: Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k .n 2/ U D sinn 1 x d V D sin x dx d U D .n 1/ sinn 2 x cos x dx V D cos x Z D sinn 1 x cos x C .n 1/ sinn 2 x cos2 x dx .e;1/ yDln x In D 720: 5 sq. units: 2 1De yDx 1 31. 15 3 C 360 1 y Z x n sin x dx 0 0 d V D cos x dx Vˇ D sinx ˇ x De C1 e sin x ˇˇ C A 0 1Ce Thus Area D A D units2 2 Shaded area D Z =2 d V D sin x dx U D xn V D cos x d U D nx n 1 dx ˇ=2 Z =2 ˇ ˇ D x n cos x ˇ Cn x n 1 cos x dx ˇ 0 0 0 In D Downloaded by ted cage (sxnbyln180@questza.com) 217 lOMoARcPSD|6566483 www.konkur.in SECTION 6.1 (PAGE 339) ADAMS and ESSEX: CALCULUS 9 6 1 6 sin x cos x C I5 7 7 1 6 6 4 1 4 sin x cos x C sin x cos x C I3 7 7 5 5 6 1 6 4 sin x cos x sin x cos x 7 35 1 2 2 24 sin x cos x C I1 C 35 3 3 1 6 6 8 sin x cos x sin4 x cos x sin2 x cos x 7 35 35 16 cos x C C 35 ! sin6 x 6 sin4 x 8 sin2 x 16 cos x C C C C C: 7 35 35 35 I7 D D D D D 35. U Dx d U D dx D 34. We have Z secn x dx .n 3/ U D sec x d V D sec2 x dx d U D .n 2/ secn 2 x tan x dx V D tan x Z D secn 2 x tan x .n 2/ secn 2 x tan2 x dx Z D secn 2 x tan x .n 2/ secn 2 x.sec2 x 1/ dx n 2 D secn 2 x tan x .n 2/In C .n 2/In 2 C C n 2 1 .secn 2 x tan x/ C In 2 C C: In D n 1 n 1 Z I1 D Z sec x dx D ln j sec x C tan xj C C I sec2 x dx D tan x C C: 1 4 1 2 I6 D .sec4 x tan x/ C sec2 x tan x C I2 C C 5 5 3 3 4 8 1 4 2 sec x tan x C tan x C C: D sec x tan x C 5 15 " 15 5 1 1 sec3 x tan xC I7 D .sec5 x tan x/ C 6 6 4 # 3 1 1 CC sec x tan x C I1 4 2 2 I2 D 5 15 1 sec3 x tan x C sec x tan xC D sec5 x tan x C 6 24 48 15 ln j sec x C tan xj C C: 48 1 In 1 a2 Telegram: @uni_k x 2 C a2 x 2 dx .x 2 C a2 /n Z x2 1 dx 2 2 a .x C a2 /n dV D x dx .x 2 C a2 /n V D 2.n 1 1/.x 2 C a2 /n 1 x 2.n 1/.x 2 C a2 /n 1 ! Z 1 dx C : 2.n 1/ .x 2 C a2 /n 1 1 a2 36. 2.n Given that f .a/ D f .b/ D 0. Z b .x a/.b x/f 00 .x/ dx a U D .x a/.b x/ d V D f 00 .x/ dx d U D .b C a 2x/ dx V D f 0 .x/ ˇb Z ˇ b ˇ D .x a/.b x/f 0 .x/ˇ .b C a 2x/f 0 .x/ dx ˇ a a D0 D U DbCa d U D 2dx " .b C a 2 Z b d V D f 0 .x/ dx V D f .x/ ˇb # Z b ˇ ˇ 2x/f .x/ˇ C 2 f .x/ dx ˇ a 2x a f .x/ dx: a 37. Given: f 00 and g 00 are continuous on Œa; b, and f .a/ D g.a/ D f .b/ D g.b/ D 0. We have Z b f .x/g 00 .x/ dx a U D f .x/ d V D g 00 .x/ dx d U D f 0 .x/ dx V D g 0 .x/ ˇb Z ˇ b ˇ D f .x/g 0 .x/ˇ f 0 .x/g 0 .x/ dx: ˇ a a 218 Z x 2n 3 C In 1 : 1/a2 .x 2 C a2 /n 1 2.n 1/a2 x 1 Now I1 D tan 1 , so a a 3 x C 2 I2 I3 D 4a2 .x 2 C a2 /2 4a x 3 1 x D C I C 1 4a2 .x 2 C a2 /2 4a2 2a2 .x 2 C a2 / 2a2 3x x 3 x C 4 2 C 5 tan 1 C C: D 4a2 .x 2 C a2 /2 8a .x C a2 / 8a a In D In D Z dx 1 D 2 2 2 n .x C a / a Z dx 1 D 2 a .x 2 C a2 /n 1 In D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.2 (PAGE 348) Multiplying the expression for I2nC1 by =2 and dividing by the expression for I2n , we obtain, by part (b), Similarly, ˇb ˇ ˇ f .x/g.x/ dx D f .x/g.x/ˇ ˇ a Z b 00 Z b 0 f 0 .x/g 0 .x/ dx: a a 2n 2n 2n C 1 2n lim 1 2n n!1 2n 2n 2n Thus we have Z b Z b f .x/g 00 .x/ dx a f 00 .x/g.x/ dx a D f .x/g 0 .x/ 2 4 2 1 5 3 2 D 1 D ; 3 3 1 2 2 2 4 2 2 or, rearranging the factors on the left, ˇ ˇb ˇ 0 f .x/g.x/ ˇ D 0 ˇ lim a 2 2 4 4 2n 2n D : 3 3 5 2n 1 2n C 1 2 n!1 1 by the assumptions on f and g. Thus Z b a f .x/g 00 .x/ dx D Z b f 00 .x/g.x/ dx: a Section 6.2 Integrals of Rational Functions (page 348) This equation is also valid for any (sufficiently smooth) functions f and g for which f .b/g 0 .b/ f 0 .b/g.b/ D f .a/g 0 .a/ f 0 .a/g.a/: Examples are functions which are periodic with period b a, or if f .a/ D f .b/ D f 0 .a/ D f 0 .b/ D 0, or if instead g satisfies such conditions. Other combinations of conditions on f and g will also do. 38. In D Z =2 1. Z 2. Z 3. Z 4. Z cosn x dx. 0 a) For 0 x =2 we have 0 cos x 1, and so 0 cos2nC2 x cos2nC1 x cos2n x. Therefore 0 I2nC2 I2nC1 I2n . b) Since In D n 1 In 2 , we have I2nC2 D n Combining this with part (a), we get 2n C 1 I2n . 2n C 2 2n C 1 I2nC2 I2nC1 D 1: 2n C 2 I2n I2n The left side approaches 1 as n ! 1, so, by the Squeeze Theorem, lim n!1 I2nC1 D 1: I2n c) By Example 6 we have, since 2n C 1 is odd and 2n is even, 2n 2n 2n C 1 2n 2n 1 2n I2n D 2n 2n I2nC1 D 2 4 2 1 5 3 3 3 1 : 2 4 2 2 2 dx D ln j2x 2x 3 5 1 ln j5 4 dx D 4x x2 x 4 dx D Z xC4C 16 x 4 ! x2 C 4x C 16 ln jx 2 dx 4j C C: A B C x 3 xC3 Ax C 3A C Bx 3B D x2 9 1 ACB =0 )AD ; BD ) 3.A B/ =1 6 Z Z Z dx dx 1 1 dx D x2 9 6 x 3 6 xC3 1 D ln jx 3j ln jx C 3j C C 6 ˇ ˇ 1 ˇ x 3 ˇˇ D ln ˇˇ C C: 6 x C 3ˇ 1 x2 9 D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 4xj C C: Z x dx x C 2 2 1 D dx x C 2 x C 2 x 2 D ln jx C 2j C C: 2 D 5. 3j C C: Downloaded by ted cage (sxnbyln180@questza.com) 1 : 6 219 lOMoARcPSD|6566483 www.konkur.in SECTION 6.2 (PAGE 348) 6. 7. 8. 9. A B Cp 5 x 5Cx p .A C B/ 5 C .A B/x D 5 x2 ( 1 1 ) A C B D p5 ) A D B D p : 2 5 A B ZD 0 Z 1 1 1 1 p dx D p Cp dx 5 x2 2 5 5 x 5Cx p p 1 ln j 5 xj C ln j 5 C xj C C D p 2 5 ˇ ˇ ˇ p5 C x ˇ 1 ˇ ˇ D p ln ˇ p ˇ C C: 2 5 ˇ 5 xˇ 1 5 x2 1 D p A B C a x aCx Aa C Ax C Ba Bx D a2 x 2 n 1 Aa C Ba D 1 ) )ADBD : A BD0 2a Thus Z Z Z dx dx 1 1 dx D C 2 2 a x 2a a x 2a aCx 1 D ln ja xj C ln ja C xj C C 2a ˇ ˇ ˇa C x ˇ 1 ˇ C C: D ln ˇˇ 2a a xˇ a2 x2 10. 11. D A B 1 D C a2 x 2 b ax b C ax .A C B/b C .A B/ax D b 2 a2 x 2 1 )A D B D 2b Z Z 1 1 1 dx D C dx b 2 a2 x 2 2b b ax b C ax 1 ln jb axj ln jb C axj D C CC 2b a a ˇ ˇ ˇ b C ax ˇ 1 ˇ C C: D ln ˇ 2ab ˇ b ax ˇ 12. b2 x 2 dx x2 C x 2 x 2 dx: 2 x Cx 2 A B Ax A C Bx C 2B x 2 D C D , If 2 x Cx 2 xC2 x 1 x2 C x 2 then A C B D 1 and A C 2B D 2, so that A D 4=3 and B D 1=3. Thus Z Z 1 Z Dx x 2 dx Dx x2 C x 2 Dx 220 Z Z dx dx 4 1 C 3 xC2 3 x 1 1 4 ln jx C 2j C ln jx 1j C C: 3 3 x 3x 2 C 8x A B C 3x 1 xC3 .A C 3B/x C .3A B/ D 3x 2 C 8x 3 n 3 1 A C 3B D 1 ; BD : ) )AD 3A B D 0 10 10 Z Z x dx 1 3 1 D C dx 3x 2 C 8x 3 10 3x 1 xC3 3 1 ln j3x 1j C ln jx C 3j C C: D 30 10 D 1 A Bx C C D C 2 x 3 C 9x x x C9 Ax 2 C 9A C Bx 2 C C x D x 3 C 9x ( ACB D0 1 1 )AD ; B D ) C D0 ; C D 0: 9 9 9A D 1 Z Z 1 1 x dx D dx x 3 C 9x 9 x x2 C 9 1 1 D ln jxj ln.x 2 C 9/ C K: 9 18 Z 14. Z 15. 3 x 2 A B Ax C A C Bx D C D x2 C x x xC1 x2 C x n ACB D1 ) ) A D 2; B D 3: AD Z Z 2 Z x 2 dx dx dx D 3 2 x2 C x xC1 x D 3 ln jx C 1j 2 ln jxj C C: 13. x 2 dx D 2 x Cx 2 Z Telegram: @uni_k ADAMS and ESSEX: CALCULUS 9 1 dx D 6x C 9x 2 Z .1 1 dx D C C: 3x/2 3.1 3x/ Z x x dx D dx Let u D 3x C 1 2 2 C 6x C 9x .3x C 1/2 C 1 du D 3 dx Z Z Z u 1 u 1 1 1 1 du D du du 9 u2 C 1 9 u2 C 1 9 u2 C 1 1 1 D ln.u2 C 1/ tan 1 u C C 18 9 1 1 ln.2 C 6x C 9x 2 / tan 1 .3x C 1/ C C: D 18 9 Z 1 x2 C 1 dx D 2 6x 9x 9 D Now Z 9x 2 6x C 6x C 9 dx 2 Z6x 9x x 2x C 3 1 C dx: 9 9 x.2 3x/ A B 2A 3Ax C Bx 2x C 3 D C D x.2 3x/ x 2 3x x.2 3x/ )2A D 3; 3A C B D 2 13 3 BD : )A D ; 2 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.2 (PAGE 348) Z Therefore we have Z x2 C 1 dx 6x 9x 2 Z Z x 1 13 dx dx D C C 9 6 x 18 2 3x x 1 13 D C ln jxj ln j2 3xj C C: 9 6 54 16. First divide to obtain 37x C 85 x3 C 1 Dx 7C 2 x C 7x C 12 .x C 4/.x C 3/ A B 37x C 85 D C .x C 4/.x C 3/ xC4 xC3 .A C B/x C 3A C 4B D x 2 C 7x C 12 n A C B D 37 ) ) A D 63; B D 26: 3A C 4B D 85 Now we have ! Z Z x3 C 1 26 63 dx D x 7C dx 12 C 7x C x 2 xC4 xC3 D 17. x2 2 7x C 63 ln jx C 4j 1 19. a2 / x3 A Bx C C C 2 x a x C ax C a2 Ax 2 C Aax C Aa2 C Bx 2 Bax C C x D x 38 a3 ( < A D a=3 ACB D0 ) Aa Ba C C D 0 ) B D a=3 : C D 2a2 =3: Aa2 C a D a3 a3 D Ca 20. x3 Here the expansion is 1 A Bx C C D C 2 x 3 C 2x 2 C 2x x x C 2x C 2 A.x 2 C 2x C 2/ C Bx 2 C C x D x 3 C 2x 2 C 2 ( ACB D0 1 ) 2A C C D 0 ) A D B D ; C D 2 2A D 1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 2a dx x C a x 2 C a2 x 1 tan 1 C K: 2a3 a Z a3 D 1 C dx x 3 a3 x 3 a3 Z Z a dx x C 2a a DxC 3 x a 3 Z x 2 C ax C a2 2x C a C 3a a a D x C ln jx aj 3 6 x 2 C ax C a2 a a D x C ln jx aj ln.x 2 C ax C a2 / 3 6 Z dx a2 a 2 3 2 2 C a xC 2 4 a a D x C ln jx aj ln.x 2 C ax C a2 / 3 6 a2 2 1 x C .a=2/ p tan p CK 2 3a . 3a/=2 a a ln.x 2 C ax C a2 / D x C ln jx aj 3 6 a 2x C a C K: p tan 1 p 3 3a D a3 / a3 Z 26 ln jx C 3j C C: 18. The partial fraction decomposition is 1 A B Cx C D D C C 2 x 4 a4 x a xCa x C a2 A.x 3 C ax 2 C a2 x C a3 / C B.x 3 ax 2 C a2 x D x 4 a4 C.x 3 a2 x/ C D.x 2 a2 / C x 4 a4 8̂ ACB CC D0 < aA aB C D D 0 ) 2 2 2 :̂ a A C a B a C D 0 3 3 2 a A a B a DD1 1 1 1 ; BD ; C D 0; D D : )A D 4a3 4a3 2a2 x4 Z 1 1 a 4a3 x a ˇ ˇ ˇx aˇ 1 ˇ ˇ D ln 4a3 ˇ x C a ˇ D 4 Therefore we have A B C C C x x a xCa Ax 2 Aa2 C Bx 2 C Bax C C x 2 C ax D x.x 2 a2 / ( ACB CC D0 A D 1=a2 ) B C D0 ) B D C D 1=.2a2 /: Aa2 D 1 Thus we have Z dx x.x 2 a2 / Z Z Z dx dx dx 1 C C 2 D 2a2 x x a xCa 1 D . 2 ln jxj C ln jx aj C ln jx C aj/ C K 2a2 jx 2 a2 j 1 ln C K: D 2 2a x2 x.x 2 dx Downloaded by ted cage (sxnbyln180@questza.com) 1; 221 lOMoARcPSD|6566483 www.konkur.in SECTION 6.2 (PAGE 348) ADAMS and ESSEX: CALCULUS 9 so we have Z 23. Z Z 1 dx dx 1 xC2 D dx x 3 C 2x 2 C 2x 2 x 2 x 2 C 2x C 2 Let u D x C 1 du D dxZ 1 1 uC1 D ln jxj du 2 2 u2 C 1 1 1 1 D ln jxj ln.u2 C 1/ tan 1 u C K 2 4 2 1 1 1 ln.x 2 C 2x C 2/ tan 1 .x C 1/ C K: D ln jxj 2 4 2 21. 1 A C B D C C C x 1 .x 1/2 xC1 .x C 1/2 D 2 A.x 1/.x C 1/2 C B.x C 1/2 2 .x 1/ C C.x C 1/.x 1/2 C D.x 1/2 8̂ 8̂ ACC D0 1 <A D < ACB C CD D0 4 ) ) :̂ A C 2B C 2D D 0 :̂ B D C D D D 1 : ACB CC CD D1 4 .x 2 Z dx .x 2 1/2 Z Z dx dx 1 C D 4 x 1 .x 1/2 ! Z Z dx dx C C xC1 .x C 1/2 1 1 ln jx C 1j ln jx 1j D 4 x 1 ˇ ˇ 1 ˇ x C 1 ˇˇ x D ln ˇˇ C K: 4 x 1 ˇ 2.x 2 1/ 24. 22. Here the expansion is CK A B C D D C C C 4/ x 1 xC1 x 2 xC2 1 1 x2 D D A D lim x!1 .x C 1/.x 2 4/ 2. 3/ 6 1 1 x2 D D B D lim x! 1 .x 1/.x 2 4/ 2. 3/ 6 4 1 x2 D D C D lim 2 x!2 .x 1/.x C 2/ 3.4/ 3 4 1 x2 D D lim D D : x! 2 .x 2 1/.x 2/ 3. 4/ 3 1 ; 3 Z Telegram: @uni_k x2 1/.x 2 .x 2 Z Z x2 C 1 dx 7x 4 5 1 dx D C dx x3 C 8 12 xC2 12 .x 1/2 C 3 Let u D x 1 du D dx Z 5 1 7u C 3 D ln jx C 2j C du 12 12 u2 C 3 7 5 ln jx C 2j C ln.x 2 2x C 4/ D 12 24 1 x 1 C p tan 1 p C K: 4 3 3 1 xC1 The expansion is so we have 222 D Thus 1 A B C D C C x 3 4x 2 C 3x x x 1 x 3 A.x 2 4x C 3/ C B.x 2 3x/ C C.x 2 x/ D x 3 4x 2 C 3x ( ACB CC D0 ) 4A 3B C D 0 3A D 1 1 1 1 )AD ; BD ; C D : 3 2 6 Therefore we have Z dx x 3 4x 2 C 3x Z Z Z dx 1 dx dx 1 1 C D 3 x 2 x 1 6 x 3 1 1 1 D ln jxj ln jx 1j C ln jx 3j C K: 3 2 6 A Bx C C x2 C 1 D C 2 x3 C 8 xC2 x 2x C 4 A.x 2 2x C 4/ C B.x 2 C 2x/ C C.x C 2/ D x3 C 8 ( ACB D1 7 5 BD ; C D ) 2A C 2B C C D 0 ) A D 12 12 4A C 2C D 1 1/2 1 Therefore Z .x 2 x2 1/.x 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 4/ dx D 1 ln jx 6 1 ln jx 3 1j C 2j 1 ln jx C 1jC 6 1 ln jx C 2j C K: 3 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 25. 1 x4 3x 3 Therefore Z 1 x 3 .x 3/ C D B A D C 2 C 3 C x x x x 3 A.x 3 3x 2 / C B.x 2 3x/ C C.x D x 3 .x 3/ 8̂ ACD D0 < 3A C B D 0 ) :̂ 3B C C D 0 3C D 1 8̂ A D 1=27 < B D 1=9 ) :̂ C D 1=3 D D 1=27: SECTION 6.2 (PAGE 348) D 27. 3/ C Dx 3 dx 3x 3 Z Z Z Z 1 1 dx 1 dx 1 dx dx C D 27 ˇ x ˇ 9 x2 3 x3 27 x 3 1 1 1 ˇˇ x 3 ˇˇ ln C 2 C K: C D 27 ˇ x ˇ 9x 6x D .t Z Z .t dt 1/.t 2 1/2 dt 1/3 .t C 1/2 Let u D t du D dt 1 du u3 .u C 2/2 B E 1 A C D C D C 2 C 3 C 3 2 u .u C 2/ u u u uC2 .u C 2/2 4 3 2 3 2 A.u C 4u C 4u / C B.u C 4u C 4u/ D u3 .u C 2/2 2 C.u C 4u C 4/ C D.u4 C 2u3 / C Eu3 u3 .u C 2/2 8̂ ACD D0 ˆ ˆ < 4A C B C 2D C E D 0 ) 4A C 4B C C D 0 ˆ ˆ 4B C 4C D 0 :̂ 4C D 1 3 1 1 3 1 )AD ; BD ; C D ; DD ; ED : 16 4 4 16 8 Z du u3 .u C 2/2 Z Z Z 3 1 du 1 du du D C 16 u 4 u2 4 u3 Z Z 3 1 du du 16 uC2 8 .u C 2/2 1 1 3 ln jt 1j C D 16 4.t 1/ 8.t 1/2 3 1 ln jt C 1j C C K: 16 8.t C 1/ D 28. 29. Z x5 C x3 C 1 1/.x 2 1/.x 3 2/2 Let u D e x du D e x dx 1/ D A x 1 C B .x 1/2 C 3 .x 1/3 C D Ex C F C : x C 1 x2 C x C 1 Since .x 4 16/2 D .x 2/2 .x C 2/2 .x 2 C 4/2 , and the numerator has degree less than the denominator, we have 123 x 7 A B C D D C C C 4 2 2 .x 16/ x 2 .x 2/ xC2 .x C 2/2 Ex C F Gx C H C 2 : C 2 x C4 .x C 4/2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k dx .e x Since .x 1/.x 2 1/.x 3 1/ D .x 1/3 .x C1/.x 2 Cx C1/, and the numerator has degree less than the denominator, we have .x 30. e 2x Z dx D 4e x C 4 du u.u 2/2 A B C 1 D C C 2 u.u 2/ u u 2 .u 2/2 2 2 A.u 4u C 4/ C B.u 2u/ C CC u D u.u 2/2 ( ACB D0 1 1 1 ; C D : ) 4A 2B C C D 0 ) A D ; B D 4 4 2 4A D 1 Z Z Z Z du 1 du du 1 du 1 C D u.u 2/2 4 u 4 u 2 2 .u 2/2 1 1 1 1 D ln juj ln ju 2j CK 4 4 2 .u 2/ 1 x 1 ln je x 2j C K: D 4 4 2.e x 2/ Z d Let u D sin cos .1 C sin / du D Z cos d Z du du D D .1 u2 /.1 C u/ .1 u/.1 C u/2 1 A B C D C C .1 u/.1 C u/2 1 u 1Cu .1 C u/2 A.1 C 2u C u2 / C B.1 u2 / C C.1 u/ D .1 u/.1 C u/2 ( A BD0 1 1 1 ) 2A C D 0 )AD ; B D ; C D : 4 4 2 ACB CC D1 Z du .1 u/.1 C u/2 Z Z Z du du du 1 1 1 C C D 4 1 u 4 1Cu 2 .1 C u/2 ˇ ˇ ˇ ˇ 1 1 C sin ˇ 1 D ln ˇˇ C C: 4 1 sin ˇ 2.1 C sin / D x4 26. We have Z Z Downloaded by ted cage (sxnbyln180@questza.com) 223 lOMoARcPSD|6566483 www.konkur.in SECTION 6.2 (PAGE 348) ADAMS and ESSEX: CALCULUS 9 31. First expand the denominator .x 2 4/.x C 2/2 D x 4 C 4x 3 16x 16. Now divide this into x 5 to obtain the quotient polynomial x 4 and a remainder of degree less than 4. Since the denominator factors to .x C 2/3 .x 2/, the required decomposition is .x 2 Section 6.3 Inverse Substitutions (page 355) 1. A C D B x5 D x 4C C C C : 2 2 3 4/.x C 2/ x C 2 .x C 2/ .x C 2/ x 2 32. We have .x 2 CkxC4/.x 2 kxC4/ D x 4 C.8 k 2 /x 2 C16 D x 4 C4x 2 C16 provided 8 k 2 D 4. Since k > 0, we have k D 2. Accordingly, ln Q.x/ D ln.x a1 /.x a2 / .x a1 / C ln.x an /; Z 4. Z d 1 1 1 Q 0 .x/ D Œln Q.x/ D C C C Q.x/ dx x a1 x a2 x an 1 1 1 1 1 D 0 C C C : Q.x/ Q .x/ x a1 x a2 x an Since P .x/ 1 1 1 C C C Q 0 .x/ x a1 x a2 x an A1 A2 An D C C C : x a1 x a2 x an a1 and get x a1 x a1 P .x/ C C 1 C Q 0 .x/ x a2 x an A2 .x a1 / An .x a1 / D A1 C C C : x a2 x an Now let x D a1 and obtain Similarly, Aj D 224 Telegram: @uni_k P .a1 / D A1 . Q 0 .a1 / P .aj / for 1 j n. Q 0 .aj / dx Let u D 2x du D 2 dx du 1 1 p D sin 1 u C C D sin 1 .2x/ C C: 2 2 2 1 u 4x 2 1 Z x 2 dx p Let 2x D sin u 1 4x 2 2 dx D cos u du Z sin2 u cos u du 1 D 8 Z cos u u sin 2u 1 .1 cos 2u/ du D CC D 16 16 32 1 1 D sin 1 2x sin u cos u C C 16 16 p 1 1 sin 1 2x x 1 4x 2 C C: D 16 8 x 2 dx p Let x D 3 sin 9 x2 dx D 3 cos d Z 9 sin2 3 cos d D 3 cos 9 D . sin cos / C C 2 x 1 p 9 x 9 x 2 C C: D sin 1 2 3 2 p dx x 1 Z 4x 2 Let x D 12 sin dx D 12 cos d Z cos d p D csc d D sin 1 sin2 ˇ ˇ p ˇ 1 1 4x 2 ˇˇ ˇ D ln j csc cot j C C D ln ˇ ˇCC ˇ ˇ 2x 2x ˇ ˇ ˇ 1 p1 4x 2 ˇ ˇ ˇ D ln ˇ ˇ C C1 : ˇ ˇ x we have Multiply both sides by x 1 2 3. and, differentiating both sides, P .x/ A1 A2 An D C C C ; Q.x/ x a1 x a2 x an D Z an /, we have a2 / C C ln.x p 2. x4 x 4 C 4x C 16 4x 16 D x 4 C 4x 2 C 16 x 4 C 4x 2 C 16 4x 2 C 16 D1 .x 2 C 2x C 4/.x 2 2x C 4/ Cx C D Ax C B C 2 : D1C 2 x C 2x C 4 x 2x C 4 33. Since Q.x/ D .x Z 5. Z x2 Z dx p 9 x2 Let x D 3 sin dx D 3 cos d 3 cos d 2 9 sin 3 cos Z 1 2 csc d D 9 p 1 9 x2 1 cot C C D C C: D 9 9 x D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 6. 7. 8. Z p dx Let x D 3 sin x 9 x2 dx D 3 cos Z Z d 3 cos d 1 D D csc d 3 sin 3 cos 3 ˇ 1 1 ˇˇ 3 D ln j csc cot j C C D ln ˇ 3 3 ˇx ˇ ˇ p ˇ ˇ 9 x2 ˇ 1 ˇ3 D ln ˇ ˇ C C: ˇ ˇ 3 x Z Z xC1 x dx dx D C p p 9 x2 9 x2 p x 9 x 2 C sin 1 C C: D 3 Z p SECTION 6.3 (PAGE 355) Z 11. p 9 ˇ x 2 ˇˇ ˇCC ˇ x a x p dx 9 x2 9Cx 2 p 9 C x 2 / C C1 : 12. x 3 Fig. 6.3-8 Z dx Let x D a tan dx D a sec2 d Z Z 2 a sec2 d a sec d D D 2 2 2 3=2 a3 sec3 .a C a tan / Z 1 1 x D 2 C C: cos d D 2 sin C C D p a a a2 a2 C x 2 .a2 C x 2 /3=2 p Z x 3 dx p Let u D 9 C x 2 9 C x2 du D 2x Zdx Z 1 .u 9/ du 1 D .u1=2 9u 1=2 / du D p 2 2 u 1 D u3=2 9u1=2 C C 3 p 1 D .9 C x 2 /3=2 9 9 C x 2 C C: 3 x a Fig. 6.3-12 9 C x2 dx x4 Let x D 3 tan dx D 3 sec2 d Z .3 sec /.3 sec2 / d D 81 tan4 Z Z 1 1 sec3 cos d Let u D sin D d D 4 9 tan 9 sin4 du D cos d Z 1 1 1 du D CC D D CC 9 u4 27u3 27 sin3 .9 C x 2 /3=2 C C: D 27x 3 a2 Cx 2 13. Z x 2 dx Let x D a sin .a2 x 2 /3=2 dx D a cos d Z 2 2 a sin a cos d D a3 cos3Z Z D tan2 d D D tan D p x CC a2 x2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k a2 x 2 dx Let x D 3 tan dx D 3 sec2 d Z Z 3 sec2 d D sec d D 3 sec Z p p 9 C x2 p 10. dx Let x D a sin .a2 x 2 /3=2 dx D Z a cos d Z 1 a cos d sec2 d D 2 D a3 cos3 a x 1 1 C C: D 2 tan C C D 2 p 2 a a a x2 Fig. 6.3-11 D ln j sec C tan j C C D ln.x C 9. Z Downloaded by ted cage (sxnbyln180@questza.com) sin .sec2 1/ d (see Fig. s6-5-17) 1 x a C C: 225 lOMoARcPSD|6566483 www.konkur.in SECTION 6.3 (PAGE 355) 14. Z ADAMS and ESSEX: CALCULUS 9 dx .1 C 2x 2 /5=2 1 Let x D p tan 2 1 dx D p sec2 d 2 Z Z 1 1 sec2 d D p D p cos3 d 2 Z .1 C tan2 /5=2 2 1 .1 sin2 / cos d Let u D sin D p 2 du D cos d Z 1 1 3 1 2 .1 u / du D p u D p u CC 3 2 2 1 1 p sin3 C C D p sin 2 3 2 !3 p p 2x 2x 1 p D p p CC p 2 1 C 2x 2 3 2 1 C 2x 2 D 17. Z dx D x 2 C 2x C 10 18. Z dx D x2 C x C 1 19. Z 4x 3 C 3x C C: 3.1 C 2x 2 /3=2 p 1C2x 2 p 2x 1 Fig. 6.3-14 15. Z p Z 1 xC1 dx D tan 1 C C: .x C 1/2 C 9 3 3 Z dx 1 p 2 Let u D x C 2 1 2 3 du D dx C xC 2 2 Z 2 du 2 1 p u CC D p 2 D p tan 3 3 3 u2 C 2 2 2x C 1 D p tan 1 C C: p 3 3 dx .4x 2 C 4x C 5/2 Z dx D 2 Let 2x C 1 D 2 tan .2x C 1/2 C 4 2 dx D 2 sec2 d Z Z 1 sec2 d cos2 d D D 16 sec4 16 1 D C sin cos 32 2x C 1 1 2x C 1 1 tan 1 C C C: D 32 2 16 4x 2 C 4x C 5 dx p Let x D 2 sec .x > 2/ x x2 4 dx D 2 sec tan d Z 2 sec tan d D 2 sec 2 tan Z 1 x 1 d D C C D sec 1 C C: D 2 2 2 2 4x 2 C4xC5 2xC1 2 Fig. 6.3-19 p x2 4 x 20. Z 21. Z 2 Fig. 6.3-15 16. Z dx p Let x D a sec .a > 0/ x 2 x 2 a2 dx D a sec tan d Z a sec tan d D aZ2 sec2 a tan 1 1 cos d D 2 sin C C D 2 a a p 1 x 2 a2 C C: D 2 a x p x 2 a2 x a Fig. 6.3-16 226 Telegram: @uni_k Z x dx .x 1/ C 1 D dx Let u D x x 2 2x C 3 .x 1/2 C 2 du D dx Z Z du u du C D u2 C 2 u2 C 2 1 1 u D ln.u2 C 2/ C p tan 1 p CC 2 2 2 1 1 1 2 1 x 2x C 3/ C p tan p C C: D ln.x 2 2 2 x dx p 2ax x2 Z x dx Let x a D a sin p D a2 .x a/2 dx D a cos d Z .a C a sin /a cos d D a cos D a. cos / C C x a p D a sin 1 2ax x 2 C C: a Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.3 (PAGE 355) 24. a x a p 2ax x 2 Fig. 6.3-21 22. Z p dx .4x x 2 /3=2 Z dx Let 2 x D 2 sin u D Œ4 .2 x/2 3=2 Z Z dx D 2 cos u du 2 cos u du 1 D sec2 u du D 8 cos3 u 4 1 1 x 2 D tan u C C D p C C: 4 4 4x x 2 2 u Z D .3 Z Z x dx 2x x 2 /3=2 x dx 4 .x C 1/2 1/2 cos d 3 8 cos Z Z 1 1 sec tan d sec2 d D 2 4 1 1 D sec tan C C 2 4 xC1 1 1 CC D p p 2 4 3 2x x 2 3 2x x 3 x 1 D p C C: 4 3 2x x 2 D .2 sin Let x C 1 D 2 sin dx D 2 cos d 3=2 xC1 1 Fig. 6.3-24 25. p 4x x 2 x 2 C2xC2 u 2 x Fig. 6.3-22 23. Z dx dx D Let x C 1 D tan u .x 2 C 2x C 2/2 Œ.x C 1/2 C 12 dx D sec2 u du Z Z 2 sec u du D cos2 u du D sec4 u Z 1 u sin 2u D .1 C cos 2u/ du D C CC 2 2 4 1 1 D tan 1 .x C 1/ C sin u cos u C C 2 2 1 xC1 1 C C: D tan 1 .x C 1/ C 2 2 x 2 C 2x C 2 Z Z dx Let x D tan .1 C x 2 /3 dx D sec2 d Z Z 2 sec D d D cos4 d sec6 Z 1 C cos 2 2 D d 2 Z 1 1 C cos 4 D d 1 C 2 cos 2 C 4 2 3 sin 2 sin 4 D C C CC 8 4 32 sin cos sin 2 cos 2 3 C C CC D 8 2 16 3 sin cos 1 D C C sin cos .2 cos2 1/ C C 8 2 8 3 1 1 x x 2 D tan 1 x C C 1 CC 8 2 1 C x2 8 1 C x2 1 C x2 3 1 3 x x D tan 1 x C C CC 8 8 1 C x2 4 .1 C x 2 /2 3x 3 C 5x 3 C C: D tan 1 x C 8 8.1 C x 2 /2 p 2 1Cx 2 xC1 p x 1 3 2x x 2 Fig. 6.3-23 Fig. 6.3-25 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 227 lOMoARcPSD|6566483 www.konkur.in SECTION 6.3 (PAGE 355) 26. ADAMS and ESSEX: CALCULUS 9 Z x 2 dx Let x D tan u .1 C x 2 /2 dx D sec2 u du Z Z 2 tan u sec2 u du tan2 u du D D 4 sec u Z sec2 u Z 1 .1 cos 2u/ du D sin2 u du D 2 u sin u cos u D CC 2 2 1 1 x D tan 1 x C C: 2 2 1 C x2 p 1Cx 2 28. x 29. 30. Z p 1 x2 dx Let x D sin x3 dx D cos d Z cos2 d D I; where D 3 Z sin I D cot2 csc d D csc cot d V D cot csc d V D csc csc3 d csc cot Z csc d I: Therefore 1 1 I D csc cot C ln j csc C cot j C C 2p 2 ˇ ˇ p 1 ˇˇ 1 1 1 x2 1 x 2 ˇˇ C ln ˇ C D ˇCC ˇ 2 x2 2 ˇx x p p 1 1 x2 1 1 C C: ln jxj D ln.1 C 1 x 2 / 2 2 2 x2 228 Telegram: @uni_k D 9 sec tan 9 D 9 sec tan 9 31. d V D sec2 d V D tan sec tan2 d Z Z D 9 sec tan C 9 Fig. 6.3-26 D Z U D sec d U D sec tanZ d 1 U D cot d U D csc2 Z d Let x D 3 tan dx D 3 sec2 d 3 sec 3 sec2 d Z D 9 sec3 d D u 27. Z p I D 9 C x 2 dx sec .sec2 sec d 1/ d 9 Z sec3 d D 9 sec tan C 9 ln j sec C tan j I ˇp ˇ " p ! # x ˇˇ 9 C x2 x 9 ˇˇ 9 C x 2 9 C ˇCC C ln ˇ I D 2 3 3 2 ˇ 3 3ˇ p p 9 1 D x 9 C x 2 C ln 9 C x 2 C x C C1 : 2 2 9 ln 3) (where C1 D C 2 Z dx p Let x D u2 2C x dx 2u du Z D Z 2 2u du D2 du 1 D 2Cu uC2 p p D 2u 4 ln ju C 2j C C D 2 x 4 ln.2 C x/ C C: Z dx Let x D u3 1 C x 1=3 dx D 3u2 du Z 2 u du D3 Let v D 1 C u 1Cu dv D du Z Z 2 1 v 2v C 1 dv D 3 dv v 2C D3 v v 2 v 2v C ln jvj C C D3 2 3 D .1 C x 1=3 /2 6.1 C x 1=3 / C 3 ln j1 C x 1=3 j C C: 2 Z 1 C x 1=2 . dx Let x D u6 I D 1 C x 1=3 5 dx D 6u du Z 8 Z u C u5 1 C u3 5 6u du D 6 du: D 1 C u2 1 C u2 Division is required to render the last integrand as a polynomial with a remainder fraction of simpler form: observe that u8 D u8 C u6 6 u5 D u5 C u3 3 2 u6 D .u C 1/.u 2 D .u C 1/.u Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) u3 u4 C u4 C u2 4 2 u Cu uCu u/ C u: 1/ C 1 u2 1C1 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.3 (PAGE 355) Thus 35. 8 5 u Cu D u6 u2 C 1 Therefore Z I D6 u6 u7 D6 7 u4 C u3 C u2 u4 C u3 C u2 u u5 u4 u3 C C 5 4 3 u2 2 ! 1C u 1C 1 uC1 : u2 C 1 uC1 u2 C 1 D du 32. 36. 6x 1=6 p x 2 x2 p dx Let u2 D x 2 C 1 x2 C 1 2u du D 2x dx Z p u 3 u2 du D u Z p p D 3 u2 du Let u D 3 sin v p du D 3 cosZv dv Z p p D . 3 cos v/ 3 cos v dv D 3 cos2 v dv Z 3 .v C sin v cos v/ C C 2 p 3 u 3 u2 u 3 C CC D sin 1 p 2 2 3 3 0s 1 x2 C 1 3 1p 2 AC .x C 1/.2 D sin 1 @ 2 3 2 Z 0 e 2x dx dx .x C 1/2 C 1 Z 2 dx p 9 x2 Let x D 3 sin u dx D 3 cos u du Z xD2 Z 3 cos u du 1 xD2 2 D D csc u du 2 9 xD1 xD1 9 sin u.3 cos u/ ˇxD2 ! ˇxD2 p ˇ 1 1 9 x 2 ˇˇ ˇ D D . cot u/ˇ ˇ ˇ ˇ 9 9 x xD1 xD1 p ! p p p 5 8 5 1 2 2 : D D 9 2 1 9 18 x2 3 u p 9 x2 x 2 / C C: A Bt C C Dt C E t D C 2 C 2 2 2 .t C 1/.t C 1/ t C1 t C1 .t C 1/2 1 D A.t 4 C 2t 2 C 1/ C B.t 4 C t 3 C t 2 C t / .t C 1/.t 2 C 1/2 C C.t 3 C t 2 C t C 1/ C D.t 2 C t / C E.t C 1/ 8̂ 8̂ ACB D0 BD A ˆ ˆ ˆ ˆ <B C C D 0 <C D A ) 2A C B C C C D D 0 ) D D 2A ˆ ˆ ˆB C C C D C E D 1 ˆ 2D D 1 :̂ :̂ ACC CE D0 E D 2A: Thus A D 1=4 D C , B D 1=4, D D E D 1=2. We have Z t dt .t C 1/.t 2 C 1/2 Z Z Z 1 1 1 dt .t 1/ dt .t C 1/ dt D C C 4 t C1 4 t2 C 1 2 .t 2 C 1/2 1 1 1 ln jt C 1j C ln.t 2 C 1/ tan 1 t D 4 8 Z 4 dt 1 1 Let t D tan C 2 2 4.t C 1/ 2 .t C 1/2 dt D sec2 d 1 1 1 D ln jt C 1j C ln.t 2 C 1/ tan 1 t 4 8 Z 4 1 1 cos2 d C 4.t 2 C 1/ 2 1 1 1 ln jt C 1j C ln.t 2 C 1/ tan 1 t D 4 8 4 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x Fig. 6.3-36 37. Let e x D sin e x dx D cos d Z =2 ˇ=2 1 ˇ cos2 d D . C sin cos /ˇ D =6 2 =6 p ! p 1 3 3 : D D 2 3 4 6 8 Z =2 cos x p dx Let u D sin x 0 1 C sin2 x du D cos x dx Z 1 du p Let u D tan w D 1 C u2 0 du D sec2 w dw Z =4 Z =4 2 sec w dw D D sec w dw sec w 0 0 ˇ=4 ˇ ˇ D ln j sec w C tan wjˇ ˇ 0 p p D ln j 2 C 1j ln j1 C 0j D ln. 2 C 1/: ln 2 34. p ex 1 1 1 D 33. Z p3 1 0 u 6 7=6 6 5=6 3 2=3 x x C x C 2x 1=2 3x 1=3 7 5 2 C 3 ln.1 C x 1=3 / C 6 tan 1 x 1=6 C C: dx x 2 C 2x C 2 Let u D x C 1 du D dx ˇp3 Z p3 ˇ du ˇ D D tan 1 uˇ D : ˇ u2 C 1 3 0 1 C ln.u2 C 1/ C tan 1 u C C 2 D Z p3 1 Downloaded by ted cage (sxnbyln180@questza.com) 229 lOMoARcPSD|6566483 www.konkur.in SECTION 6.3 (PAGE 355) ADAMS and ESSEX: CALCULUS 9 1 1 C . C sin cos / C K 4.t 2 C 1/ 4 1 1 1 ln jt C 1j C ln.t 2 C 1/ tan 1 t D 4 8 4 1 1 1 t C tan 1 t C CK 4.t 2 C 1/ 4 4 t2 C 1 1 t 1 1 1 D ln jt C 1j C ln.t 2 C 1/ C K: 4 t2 C 1 4 8 1 u2 /.4 u2 / B C D A C C C D 1 u 1 C u ˇ2 u 2Cu ˇ 1 1 ˇ AD D ˇ .1 C u/.4 u2 / ˇ 6 ˇ uD1 ˇ 1 1 ˇ D BD ˇ 2 .1 u/.4 u / ˇ 6 ˇuD 1 ˇ 1 1 ˇ C D D ˇ .1 u2 /.2 C u/ ˇ 12 ˇ uD2 ˇ 1 1 ˇ DD : D ˇ 2 .1 u /.2 u/ ˇ 12 .1 38. We have Z x dx D .x 2 x C 1/2 Z Z Z uD 2 h x dx .x 1 2 / C 34 2 i2 Let u D x du D dx du u du 1 C 2 .u2 C 43 /2 .u2 C 43 /2 p 3 tan v, Let u D 2 p 3 sec2 v dv in the second integral. du D 2 p 3 ! Z sec2 v dv 1 1 1 2 D C 9 2 u2 C 43 2 sec4 v Z 16 1 4 D cos2 v dv C p 2.x 2 x C 1/ 3 3 2 1 C p .v C sin v cos v/ C C D 2.x 2 x C 1/ 3 3 D p v 4u2 C3 p 2u 3 1 2 Z du du 1 C 1 u 6 1Cu ! Z Z du du 1 1 12 2 u 12 2Cu ˇ ˇ ˇ ˇ 1 ˇˇ 1 u ˇˇ 1 ˇˇ 2 C u ˇˇ ln D ln ˇ C CK 6 1 C u ˇ 12 ˇ 2 u ˇ ˇ ˇ ˇ ˇ p p 1 ˇˇ 2 C 1 x 2 ˇˇ 1 x 2 ˇˇ 1 ˇˇ 1 ln ˇ p p D ln ˇ ˇC ˇCK 6 ˇ 1 C 1 x 2 ˇ 12 ˇ 2 1 x2 ˇ p p 1 x 2 /2 1 .1 1 .2 C 1 x 2 /2 D ln C C K: ln 2 6 x 12 3 C x2 I D 40. 2.x 1 2 2x 1 2 C p tan 1 p C p p 2 2.x x C 1/ 3 3 3 3 3 .2 x 2 2x 1 x 2 2 C C C: D p tan 1 p 3.x 2 x C 1/ 3 3 3 D Thus p 1 / 2 3 x C 1/2 CC 1 6 Z Z dx Let x D sec x 2 .x 2 1/3=2 dx D sec tan d Z Z sec tan d cos3 d D D 2 3 sec tan sin2 Z 2 1 sin D cos d Let u D sin sin2 du D cos d Z 1 1 u2 du D uCC D u2 u 1 C sin C C D sin ! p x x2 1 C C: C D p x x2 1 Fig. 6.3-38 p x 39. I Z dx p Let 1 x 2 D u2 x.3 C x 2 / 1 x 2 2x dx D 2u du Z du : D .1 u2 /.4 u2 / 230 Telegram: @uni_k 1 Fig. 6.3-40 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x2 1 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 41. I D Z SECTION 6.3 (PAGE 355) Z Z Z du du 1 du 1 du D .1 u2 /u2 u2 2 1 u 2 1Cu 1 1 1 ln j1 C uj C K D C ln j1 uj u 2 2 ˇ ˇ p ˇ 1 x 2 ˇˇ 1 ˇ1 1 C ln ˇ p D p ˇCK 2 ˇ 1 C 1 x2 ˇ 1 x2 p 1 1 x2 ln jxj C K: D p C ln 1 1 x2 dx x.1 C x 2 /3=2 Let x D tan dx D sec2 d Z Z 2 cos2 d sec d D D 3 tan sec sin Z cos2 sin d D Let u D cos sin2 du D sin d Z Z 2 du u du DuC : D 1 u2 u2 1 We have 1 u2 1 D 1 2 1 u 1 1 : uC1 Thus 43. ˇ ˇ 1 ˇ u 1 ˇˇ I D u C ln ˇˇ CC 2 u C 1ˇ ˇ ˇ 1 ˇ cos 1 ˇˇ D cos C ln ˇˇ CC 2 cos C 1 ˇ ˇ ˇ 1 ˇ ˇ ˇp 1ˇ ˇ 1 1 ˇˇ 1 C x 2 ˇCC D p C ln ˇ ˇ 1 2 ˇ 1 C x2 ˇ p C 1 ˇ ˇ 1 C x2 ! p 1 1 C x2 1 1 C ln p C C: D p 2 1 C x2 1 C x2 C 1 p 44. d 2 C sin Z =2 0 1Cx 2 Let x D tan.=2/; 2 dx 2x ; d D sin D 1 C x2 1 C x2 d 1 C cos C sin Fig. 6.3-41 42. We have dx x 2 /3=2 Let u2 D 1 x 2 2u du Z D 2x dx du u du D D .1 u2 /u3 .1 u2 /u2 C A B D 1 D C 2 C C u2 .1 u2 / u u 1 u 1Cu A.u u3 / C B.1 u2 / C C.u2 C u3 / C D.u2 D u2 .1 u2 / 8̂ ACC D D0 < B CC CD D0 ) :̂ A D 0 BD1 1 1 ) A D 0; B D 1; C D ; D D : 2 2 0 45. u3 / Z d 3 C 2 cos Let x D tan.=2/; 1 x2 2 dx cos D ; d D 1 C x2 1 C x2 2 dx Z 2 dx 1 C x2 D D 2 5 C x2 2 2x 3C 2 1Cx x 2 tan.=2/ 2 p D p tan 1 p C C D p tan 1 C C: 5 5 5 5 Z Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 dx, , d D 2 1 C x2 1 x2 2x cos D , sin D . 2 1 C x 1 C x2 Let x D tan 2 dx 1 C x2 D 1 x2 2x 0 1C C 1 C x2 1 C x2 Z 1 Z 1 dx dx D2 D 2 C 2x 1 Cx 0 0 ˇ1 ˇ ˇ D ln j1 C xjˇ D ln 2: ˇ Z 1 1 x.1 Z Z 2 dx Z dx 1 C x2 D D 2x 1 C x C x2 2C 1 C x2 Z 2x C 1 dx 2 D CC D p tan 1 p 3 3 1 2 3 xC C 2 4 2 2 tan.=2/ C1 p D p tan 1 C C: 3 3 Z Z Z x I D I D Downloaded by ted cage (sxnbyln180@questza.com) 231 lOMoARcPSD|6566483 www.konkur.in SECTION 6.3 (PAGE 355) 46. Area D Z 1 D Z 0 dx D 2x x 2 Let u D x 1 du D dx 1=2 Z 1 1=2 du p 1 dx 1 .x ˇ0 ˇ ˇ uˇ ˇ D sin 1 u2 sq: units: D 6 6 1=2 D0 47. p ADAMS and ESSEX: CALCULUS 9 p For intersection of y D Z 1 4 dx 2 4 0 .x 4x C 8/3=2 Z 4 dx 1 D 4 0 Œ.x 2/2 C 43=2 Let x 2 D 2 tan u dx D 2 sec2 u du Z 1 =4 2 sec2 u du D 4 =4 8 sec3 u ˇ=4 Z =4 ˇ 1 1 ˇ D cos u du D sin uˇ ˇ 16 =4 16 =4 p 2 1 1 1 p Cp D : D 16 16 2 2 48. Average value D 49. Area of R Z pa2 b2 p a2 D2 1/2 1=2 9 and y D 1 we have x 4 C 4x 2 C 4 x 4 C 4x 2 C 4 D 9 x 4 C 4x 2 2 .x C 5/.x 2 5D0 1/ D 0; so the intersections are at x D ˙1. The required area is Z 1 9 dx 1 dx x 4 C 4x 2 C 4 0 Z 1 p dx D 18 2 Let x D 2 tan 2 2 p 0 .x C 2/ dx D 2 sec2 Z xD1 p 2 sec2 d 2 D 18 4 sec4 xD0 Z xD1 9 D p cos2 d 2 2 xD0 ˇxD1 ˇ 9 ˇ 2 D p C sin cos ˇ ˇ 2 2 xD0 !ˇ1 p 9 2x ˇˇ 1 x D p tan p C 2 2 ˇ x C2 ˇ 2 2 2 0 1 1 9 units2 D p tan 1 p 2 2 2 2 AD2 y yD 9 x 4 C4x 2 C4 x2 0 D2 Z xDpa2 b2 b dx a2 cos2 d Let x D a sin dx D a cos d p 2b a2 b2 xD0 ˇxDpa2 b2 ˇ D a . C sin cos /ˇˇ p 2b a2 2 ˇp 2 2 ˇ a b p p x ˇ D a2 sin 1 C x a2 x 2 ˇ 2b a2 b 2 ˇ a 0 s 2 p p b D a2 sin 1 1 C b a2 b 2 2b a2 b 2 2 a p 2 1 b b a2 b 2 units2 D a cos a y p . a2 b 2 ;b/ R R yDb b yD1 a x x 2 Cy 2 Da2 1 1 x Fig. 6.3-49 Fig. 6.3-47 232 Telegram: @uni_k b2 xD0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.3 (PAGE 355) 50. The circles intersect at x D 41 , so the common area is A1 C A2 where A1 D 2 D2 Z 1 p x 2 dx 1 Let x D sin u dx D cos u du 1=4 Z xD1 cos2 u du 1 p 51. xD1=4 ˇxD1 ˇ ˇ D .u C sin u cos u/ˇ ˇ xD1=4 D .sin xCx 1 p ˇxD1 ˇ ˇ x 2 /ˇ ˇ xD1=4 1 15 sin 1 sq. units. D 2 4 16 Z 1=4 p A2 D 2 4 .x 2/2 dx Let x 2 D 2 sin v 0 dx D 2 cos v dv Z xD1=4 2 cos v dv D8 Z 4 p 12 Required area D 25 x 2 dx x 3 Z 4p Z 4 12 D 25 x 2 dx dx 3 3 x Let x D 5 sin u, dx D 5 cos u du in the first integral. ˇ4 Z xD4 ˇ ˇ 2 D 25 cos u du 12 ln x ˇ ˇ xD3 3 ˇxD4 ˇ 25 4 ˇ .u C sin u cos u/ˇ D 12 ln ˇ 2 3 xD3 ˇ4 ˇ 25 x 1 p 4 ˇ D sin 1 C x 25 x 2 ˇ 12 ln ˇ 2 5 2 3 3 4 25 1 4 1 3 D sin sin 12 ln sq. units. 2 5 5 3 y xyD12 .3;4/ .4;3/ xD0 ˇxD1=4 ˇ ˇ D 4.v C sin v cos v/ˇ ˇ xD0 " #ˇxD1=4 p x 2 x 2 4x x 2 ˇˇ 1 D 4 sin C ˇ ˇ 2 2 2 xD0 " # p 7 7 15 1 D 4 sin C 8 64 2 p 7 7 15 D 4 sin 1 C 2 sq. units. 8 16 s 2 Cy 2 D25 x Fig. 6.3-51 52. Shaded area D 2 5 A1 C A2 D 2 p 15 2 1 sin 1 4 4 sin 1 7 sq. units: 8 x 2 a c dx Let x D a sin u dx D a cos u du cos2 u du ˇx a ˇ ˇ D ab.u C sin u cos u/ˇ ˇ xDc ˇˇa p x b ˇ 1 2 2 D ab sin C x a x ˇ ˇ a a c cb p 2 1 c sin a c 2 sq. units. D ab 2 a a D 2ab Hence, the common area is Z xDa Z a r b 1 xDc y xDc .x 2/2 Cy 2 D4 y 1 xD 4 A x 2 Cy 2 D1 x x x2 y2 C D1 a2 b 2 Fig. 6.3-50 Fig. 6.3-52 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 233 lOMoARcPSD|6566483 www.konkur.in SECTION 6.3 (PAGE 355) 53. Area of R p Z ADAMS and ESSEX: CALCULUS 9 Section 6.4 Other Methods for Evaluating Integrals (page 362) Yp 1CY2 x 2 1 dx 2 1 Let x D sec dx D sec tan d Z tan 1 Y Yp D 1CY2 sec tan2 d 2 0 Z tan 1 Y Yp 1CY2 sec3 d D 2 0 Z tan 1 Y C sec d 0 Yp 1 sec tan D 1CY2 C 2 2 D 1CY 2 p 1. We try I D I 0 D 3Ae 3x si n.4x/ C 4Ae 3x cos.4x/ C 3Be 3x cos.4x/ provided 3A 4B D 1 and 4A C 3B D 0. This pair of equations has solution a D 3=25, b D 4=25, so we have Z 2. . p 1CY 2 ;Y / 1 1 D 2 a x C a x2 a2 s x2 a2 p x 2 a2 D C C1 : a2 x 234 Telegram: @uni_k x e x sin x dx D .a0 Ca1 x/e x sin xC.b0 Cb1 x/e x cos x: a1 Z a sinh u du a2 cosh2 u a sinh u 1 1 sech2 u du D 2 tanh u C C D 2 a a x 1 CC D 2 tanh cosh 1 a a s x C a Z provided that x Fig. 6.3-53 D 4 3x e cos.4x/: 25 dI dx D a1 e x sin x .a0 C a1 x/e x sin x C .a0 C a1 x/e x cos x C b1 e x cos x .b0 C b1 x/e x cos x .b0 C b1 x/e x sin x D xe x sin x R a2 3 3x e sin.4x/ 25 Differentiating this formula gives x 2 y 2 D1 dx p 2 Zx e 3x sin.4x/ dx D Here we try I D y x2 4Be 3x si n.4x/ D e 3x sin.4x/ 1 t 1 Area D ln.sinh t C cosh t / D ln e t D units2 2 2 2 54. e 3x sin.4x/ dx D Ae 3x sin.4x/CBe 3x cos.4x/CC for some constants A, B and C . Differentiation gives ˇˇtan 1 Y 1 ˇ ln j sec C tan j C ln j sec C tan j ˇ ˇ 2 0 p p p Y Y 1 D 1CY2 1 C Y 2 C ln.Y C 1 C Y 2 / 2 2 2 p 1 2 2 D ln.Y C 1 C Y / units 2 If Y D sinh t , then we have Z Z x a 1 s x2 a2 x a 1 s x2 a2 1 1C a0 a1 b0 D 0 b1 D 1 a0 C b1 a1 b0 D 0 b1 D 0: This system has solution a0 D 0, a1 D b0 D b1 D Therefore, Z 3. 1 CC x e x sin x dx D 1 xe x sin x 2 1=2. 1 .1Cx/e x cos x CC: 2 It is useful to make the change of variable u D x 2 , du D 2x dx before guessing the form of the integral. Z 2 x 5 e x dx Z 1 D u2 e u du 2 D .a2 u2 C a1 u C a0 /e u C C: I D 1 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.4 (PAGE 362) We will have dI D .2a2 u C a1 C a2 u2 C a1 u C a0 /e u du 1 2 u D u e 2 One suspects it has forgotten to use absolute values in some of the logarithms. 7. I D provided a2 D 1=2, 2a2 C a1 D 0, and a1 C a0 D 0. Thus a1 D 1 and a0 D 1, so 1 2 u C u 1 eu C C I D 2 Z 1 4 2 2 x 5 e x dx D x C x2 C 1 e x C C 2 4. As an alternative to the direct method of Example 3, we begin with the change of variable u D ln x, or, equivalently, x D e u , so that dx D e u du. I D Z x 2 .ln x/4 dx D Z Neither the author’s version of Maple nor his version of Mathematics would do Z p t5 3 2t 4 dt as presented. Both did an integration by parts and left an unevaluated integral. Both managed to evaluate the integral after the substitution u D t 2 was made. (See Exercise 4.) However, Derive had no trouble doing the integral in its original form, giving as the answer p 2 p 6t 3 2 sin 1 16 3 8. p t 2 3 2t 4 : 8 Maple, Mathematica, and Derive readily gave u4 e 3u du Z 1 D a4 u4 C a3 u3 C a2 u2 C a1 u C a0 e 3u C C: 0 3 1 1 dx D C : .x 2 C 1/3 32 4 We will have dI D 4a4 u3 C 3a3 u2 C 2a2 u C a1 e 3u du C 3 a4 u4 C a3 u3 C a2 u2 C a1 u C a0 e 3u 9. Z 4 3u Du e provided 3a4 D 1, 3a3 C 4a4 D 0, 3a2 C 3a3 D 0, 3a1 C 2a2 D 0, and 3a0 C a1 D 0. Thus a4 D 1=3, a3 D 4=9, a2 D 4=9, a1 D 8=27, and a0 D 8=81. We now have 8 8 1 4 4 3 4 2 u u C u uC e 3u C C I D 3 9 9 27 81 Z x 2 .ln x/4 dx 4.ln x/2 8 ln x 8 .ln x/4 4.ln x/3 C C CC D x3 3 9 9 27 81 6. According to Maple Z 1 C x C x2 dx 1/.x 4 16/2 ln.x 1/ ln.x C 1/ 7 D 300 900 15;360.x 2/ 613 1 79 ln.x 2/ C ln.x C 2/ 460;800 5;120.x C 2/ 153;600 47 ln.x 2 C 1/ C ln.x 2 C 4/ 900 115;200 23 6x C 8 tan 1 .x=2/ 25;600 15;360.x 2 C 4/ .x 4 xp 2 x 2 dx D p x 2 2 x 2 p 2 C ln jx C Z p 11. .x 2 C 4/3 dx D x 2 ˙ a2 . x2 p 10. Use the last integral in the list involving 2j C C x 2 ˙ a2 . p p x 2 .x C10/ x 2 C 4C6 ln jxC x 2 C 4jCC 4 Use the 8th integral in the list involving p making the change of variable x D 3t . p x 2 ˙ a2 after Z p dt p Let x D 3t p t 2 3t 2 C 5 dx D 3 dt Z 3 dx D p p 3 x2 x2 C 5 p p p x2 C 5 3t 2 C 5 D 3 CC D CC 5x 5t 12. Use the 8th integral in the miscellaneous algebraic set. Z p dt t 3t Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k p Use the 6th integral in the list involving Downloaded by ted cage (sxnbyln180@questza.com) 2 D p tan 1 5 5 r 3t 5 5 CC 235 lOMoARcPSD|6566483 www.konkur.in SECTION 6.4 (PAGE 362) ADAMS and ESSEX: CALCULUS 9 13. The 5th and 4th integrals in the exponential/logarithmic set give 17. Use the last integral in the miscellaneous algebraic set. Z 1 dx x 2 p p D CC 5 4 2 3 x .ln x/ 4 4 . 4x x / 4x x 2 x 4 .ln x/4 dx D x 4 .ln x/3 dx 5 5 Z x 5 .ln x/4 4 x 5 .ln x/3 3 D x 4 .ln x/2 dx 5 5 5 2 18. Use the last integral in the miscellaneous algebraic set. 3 4 Then complete the square, change variables, and use the 4.ln x/ .ln x/ D x5 second last integral in the elementary list. 5 25 5 Z 2 Z 2 12 x .ln x/ dx x 5 ln x dx C p 25 5 5 . 4x x 2 /4 .ln x/4 4.ln x/3 12.ln x/2 24 ln x 24 Z dx x 2 p 1 D x5 C C C C: . 4x x 2 / 2 C D 5 25 125 625 3;125 8 8 4x x 2 Z dx x 2 1 D Let u D x 2 C 2/ 8.4x x 8 4 .x 2/2 14. We make a change of variable and then use the first two du D dx Z integrals in the exponential/logarithmic set. x 2 1 du D C 8.4x x 2 / 8 4 u2 Z ˇ ˇ 2 7 x 2 ˇu C 2ˇ x 2 1 x e dx Let u D x ˇ ˇCC D C ln 8.4x x 2 / 32 ˇ u 2 ˇ du D 2x dx Z 1 x 2 1 ˇˇ x ˇˇ D u3 e u du C ln ˇ D ˇCC 2 8.4x x 2 / 32 x 4 Z 1 D u3 e u 3 u2 e u du 2 Z Z 1 u3 e u 3 19. Since J .x/ D cos.x sin t / dt we have, differentiat2 u u 0 u e 2 ue du D 0 2 2 ing with respect to x through the integral, 3 u 3u2 C 3.u 1/ e u C C D Z 2 2 1 h 6 x sin2 t cos.x sin t / sin t sin.x sin t / xy 00 C y 0 C xy D 4 x 3x 0 2 x2 i D C 3x 3 e CC 2 2 C x cos.x sin t / dt Z i 1 h D x cos2 t cos.x sin t / sin t sin.x sin t / dt 0 15. Use integrals 14 and 12 in the miscellaneous algebraic set. Z i 1 d h Z p D cos t sin.x sin t / dt 0 dt x 2x x 2 dx iˇˇ 1h Z D cos t sin.x sin t / ˇˇ D 0 3 p .2x x 2 /3=2 0 2 C D 2x x dx 3 3 . x 1p .2x x 2 /3=2 1 C D 2x x 2 C sin 1 .x 1/ C C 20. 3 2 2 d 2 2 (a) Since erf.x/ D p e x by the Fundamental dx 16. Use integrals 17 and 16 in the miscellaneous algebraic set. Theorem of Calculus, we have p p Z Z 2x x 2 x2 dx e dx D erf.x/ C C: x2 2 p Z .2x x 2 /3=2 1 2x x 2 dx D p Z 1 2 x 1 x 2 2 x2 (b) lim erf.x/ D p p e dx D D 1. 2 3=2 x!1 p .2x x / 0 2 1 2 D 2x x sin .x 1/ C C 2 x2 Since e x is an even function, erf.x/ must be an odd function, so limx! 1 erf.x/ D 1 Z 236 Telegram: @uni_k Z Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.5 (PAGE 370) (c) If P .x/ is a polynomial of degree m > 0, then P 0 .x/ has degree m 1 and d 2 P .x/ e x D P 0 .x/ dx 2. 2 2xP .x/ e x ; which cannot equal erf.x/ because P 0 .x/ 2xP .x/ must have degree m C 1 and so cannot equal p 2 .2= /P .x/ which has degree m. Thus P .x/e x cannot be an antiderivative of erf.x/. Z 3. 2 erf.x/ dx D P .x/erf.x/ C Q.x/ e x C C; where P and Q are polynomials to be determined. Then dJ D P 0 .x/ erf.x/ dx 2 C p P .x/ C Q 0 .x/ D erf.x/: Hence we must have P .x/ D 1 and 2 p P .x/ C Q 0 .x/ 2xQ.x/ D 0. The first of these DEs says that P .x/ D x C k; without loss of generality we can take the constant k to be zero. The second DE says that 0 Q .x/ 2xQ.x/ D 4. 2 2xQ.x/ e x 0 Z Section 6.5 Improper Integrals 2x ˇ1 ˇ dx 1=3 ˇ D lim 3.x C 1/ ˇ 2=3 c! 1C 1 .x C 1/ c 1=3 D lim 3 2 .1 C c/1=3 c! 1C p 3 D 3 2: This integral converges. Z 1 6. Z a 0 2x p : 1 2 erf.x/ dx D xerf.x/ C p e x C C: ˇR e 2x ˇˇ e dx D lim ˇ R!1 2 ˇ 0 0 1 1 D lim R!1 2 2e R 1 D : This integral converges. 2 Z 1 Z 1 dx dx D lim 2 R! 1 R x2 C 1 1 x C1 D lim tan 1 . 1/ tan 1 .R/ R! 1 D : D 4 2 4 This integral converges. Z 1 5. dx a2 x D lim 2 C !a Z C 0 ˇ ˇˇC dx a2 x2 ˇ ˇa C x ˇ ˇ 1 ˇˇ D lim ln ˇˇ C !a 2a a x ˇˇ 0 The right side has degree 1 and so must the left side. 0 Thus Q must have p degree zero. Hence Q .x/ D 0 and Q.x/ D 1= . Therefore J D 1 dx Let u D 2x 1 .2x 1/2=3 3 du D 2 dx Z Z R 1 1 du 1 lim u 2=3 du D D 2 5 u2=3 2 R!1 5 ˇR ˇ 1 1=3 ˇ D lim 3u ˇ D 1 (diverges) ˇ 2 R!1 5 (d) Let us try the form J D Z 1 1 aCC D lim ln D 1: C !a 2a a C The integral diverges to infinity. 7. Z 1 (page 370) dx x/1=3 Let u D 1 x du D dx Z 1 Z 1 du du D D lim 1=3 1=3 c!0C u u 0 c ˇ1 ˇ 3 3 ˇ D lim u2=3 ˇ D ˇ c!0C 2 2 0 .1 c 1. Z 1 1 Let u D x 1 du D dx Z R Z 1 du du D lim D 3 3 R!1 u 1 u 1 ˇR 1 ˇˇ 1 1 1 D lim D lim D ˇ R!1 2u2 ˇ R!1 2 2R2 2 2 .x 1/3 dx 1 8. Z 1 dx p x 1 x Let u2 D 1 x 2u du D dx Z 1 Z c 2u du du D D 2 lim c!1 u2 /u u2 0 .1 0 1 ˇ ˇˇˇc 1 ˇˇ u C 1 ˇˇˇ ln ˇ D 2 lim ˇ D 1 (diverges) c!1 2 u 1 ˇˇ 0 0 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 237 lOMoARcPSD|6566483 www.konkur.in SECTION 6.5 (PAGE 370) 9. Z =2 cos x dx .1 sin x/2=3 Let u D 1 sin x 0 du D cos x dx ˇ1 Z 1 ˇ u 2=3 du D lim 3u1=3 ˇˇ D 3: D c!0C 0 10. ADAMS and ESSEX: CALCULUS 9 15. 0 D c The integral converges. Z 1 xe x dx 0 Z R D lim xe x dx 16. R!1 0 d V D e x dx V D e x 1 ˇR Z ˇ R ˇ x x D lim @ xe ˇ C e dx A ˇ R!1 0 0 1 R C 1 D 1: D lim R!1 eR eR The integral converges. Z 1=2 Z 1 dx dx p s D2 x.1 x/ 0 0 1 x 4 Z 1=2 dx s D 2 lim c!0C c 1 2 1 x 4 2 ˇ1=2 ˇ D : D 2 lim sin 1 .2x 1/ˇˇ U Dx dU D 0 dx 11. c!0C 12. 13. 1 2 2 D lim C !.=2/ lim This integral diverges to infinity. 238 Telegram: @uni_k 21. ˇC ˇ ˇ ln j sec x C tan xjˇ ˇ 0 ln j sec C C tan C j D 1: ln 1 D 1: 0 The integral converges. C !.=2/ 0 This integral diverges to infinity. Z 1 dx Let u D ln x x.ln x/ e dx du D x ˇln R Z ln R ˇ du ˇ D lim D lim ln jujˇ ˇ R!1 1 R!1 u R!1 c sec x dx D c!.=2/ c!.=2/ ln sec c D 1: ˇc ˇ ln j sec xjˇˇ This integral diverges to infinity. Z e dx 17. p Let u D ln x 1 x ln x du D dx=x ˇ Z 1 p ˇ1 du p D lim 2 uˇˇ D 2: D c!0C u 0 c This integral converges. Z 1 dx 18. Let u D ln x x.ln x/2 e dx du D x Z ln R du 1 D lim D lim C 1 D 1: R!1 R!1 1 u2 ln R The integral converges. Z 0 Z 1 Z 1 x dx D 19. I D C D I1 C I2 2 1 1 1Cx 0 Z 1 x dx I2 D Let u D 1 C x 2 1 C x2 0 du D 2x dx Z 1 R du D 1 (diverges) D lim R!1 2 1 u Z 0 Z 1 Z 1 x dx 20. I D D C D I1 C I2 4 1 1 1Cx 0 Z 1 x dx Let u D x 2 I2 D 1 C x4 0 du D 2x dx ˇR Z 1 ˇ 1 du 1 ˇ D D lim tan 1 uˇ D 2 ˇ 2 0 1Cu 2 R!1 4 This integral diverges to infinity. Z 1 x dx Let u D 1 C 2x 2 .1 C 2x 2 /3=2 0 du D 4x dx ˇR Z 1 1 du 2 ˇˇ 1 D p ˇ lim D 4 1 u3=2 4 R!1 u ˇ 0 lim lim 1 1 14. tan x dx D D lim ln.ln R/ The integral converges. Z R Z 1 x x dx D lim dx 2 R!1 1 C 2x 1 C 2x 2 0 0 ˇR ˇ 1 ˇ D lim ln.1 C 2x 2 /ˇ ˇ R!1 4 0 # " 1 1 2 ln.1 C 2R / ln 1 D 1: D lim R!1 4 4 1 D : 2 Z =2 Z =2 Similarly, I1 D . Therefore, I D 0. 4 Z 1 Z 0 Z 1 2 I D xe x dx D C D I1 C I2 1 1 0 Z 1 2 xe x dx Let u D x 2 I2 D 0 du D 2x dx ˇR Z 1 ˇ 1 1 1 u uˇ e du D lim e ˇ D D ˇ 2 0 2 R!1 2 0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 1 . Therefore, I D 0. 2 Similarly, I1 D 22. I D I2 D Z 1 1 Z 1 e jxj dx D x e 0 SECTION 6.5 (PAGE 370) Z 0 1 e x dx C 26. Z 1 0 e x dx D I1 C I2 0 dx D 1 Z 1 Area of R D 0 D .0 D1 ln x dx D lim .x ln x c!0C 1/ C lim .c ln c ˇ1 ˇ x/ˇˇ C !0C C C !0C 1=C D lim .e e C !0C Z R Z I2 D lim x 2 e 1=x dx D lim 1=R 1 0 c/ c!0C 2 1 D I1 C I2 : 1 Then let u D and du D x 2 dx in both I1 and I2 : x Z 1 Z 1 I1 D lim x 2 e 1=x dx D lim e u du Similarly, I1 D 1. Therefore, I D 2. 23. The required area is Z 1 Area D x 2 e 1=x dx 0 Z 1 Z 1 D x 2 e 1=x dx C x 2 e 1=x dx 0 D 1 units 1=C 1 /D : e R!1 1 R!1 1 : e Hence, the total area is I1 C I2 D 1 square unit. y D lim .e 1=R R!1 1 e u du 1 x R 27. yDln x e 1/ D 1 First assume that p ¤ 1. Since a > 0 we have ˇR Z 1 x pC1 ˇˇ p x dx D lim ˇ R!1 p C 1 ˇ a a a pC1 1 D C lim R!1 .1 1 p p/Rp 1 ( 1 if p > 1 D .p 1/ap 1 1 if p < 1 ˇa Z a pC1 ˇ x ˇ x p dx D lim ˇ c!0C p C 1 ˇ 0 Fig. 6.5-23 24. Area of shaded region D D lim R!1 D lim R!1 Z 1 1 e x C e 2x 2 e R .e x e 2x / dx 0 c ˇˇR ˇ ˇ ˇ 0 1 C e 2R C 1 2 1 2 y D c1 p a pC1 C lim D c!0C p 1 p 1 8 < a1 p D 1 p if p < 1 : 1 if p > 1. 1 sq. units. 2 1 yDe x 28. yDe 2x x Fig. 6.5-24 25. Z 1 4 2 dx 2x C 1 x C 2 1 ˇˇR D lim 2 ln.2x C 1/ ln.x C 2/ ˇˇ R!1 1 2R C 1 0 D 2 ln 2 sq. units. D lim 2 ln R!1 RC2 Area D 29. If p D 1 both integrals diverge as shown in Examples 2 and 6(a). Z 1 Z 0 Z 1 x sgn x x x dx D dx C dx 1 xC2 1 xC2 0 xC2 ! ! Z 1 Z 0 2 2 dx C dx 1 D 1C x C 2 x C 2 0 1 ˇ0 ˇ1 ˇ ˇ 16 ˇ ˇ D . x C 2 ln jx C 2j/ˇ C .x 2 ln jx C 2j/ˇ D ln : ˇ ˇ 9 1 0 Z 2 x 2 sgn .x 1/ dx 0 Z 2 Z 1 x 2 dx x 2 dx C D 1 0 ˇ1 ˇ2 x 3 ˇˇ x 3 ˇˇ 1 8 1 D C D 2: ˇ C ˇ D 3 ˇ 3 ˇ 3 3 3 0 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 1 239 lOMoARcPSD|6566483 www.konkur.in SECTION 6.5 (PAGE 370) 30. Since x2 x5 C 1 ADAMS and ESSEX: CALCULUS 9 1 for all x 0, therefore x3 35. Z 1 x2 dx x5 C 1 0 Z 1 Z 1 x2 x2 dx C dx D 5 5 x C1 1 0 x C1 Z 1 Z 1 x2 dx dx C 5 x3 0 x C1 1 D I1 C I2 : I D Since I1 is a proper integral (finite) and I2 is a convergent improper integral, (see Theorem 2), therefore I converges. 31. The given integral diverges to infinity. 36. sin x 1. Then x Z Z Z sin x sin x I D dx D lim dx .1/ dx D : !0C x x 0 0 Since sin x x for all x 0, thus Hence, I converges. 2x on Œ0; =2, we have Z 1 Z =2 j sin xj sin x dx dx x2 x2 0 0 Z 2 =2 dx D 1: 0 x 37. Since sin x 1 1 p p on Œ1; 1/. 1 C Zx 2 x Z 1 1 dx dx p . Since p diverges to infinity, so must 1C x 1 1 Z x 1 dx Therefore p also diverges to infinity. 1C x 0 32. Since e 1 ex on Œ 1; 1. Thus xC1 xC1 Z 1 Z ex 1 1 dx dx D 1: e 1 xC1 1 xC1 The given integral diverges to infinity. y p x x 1 p for all x > 1, therefore 2 x 1 x yDsin x I D Z 1 2 p Z 1 dx x x dx p D I1 D 1: 2 x 1 x 2 2x yD Since I1 is a divergent improper integral, I diverges. 33. Z 1 Z 1 x3 Z 1 2 x3 C e dx. e dx D 0 1 Z 1 3 Now e x dx is a proper integral, and is therefore 0 0 finite. Since x 3 x on Œ1; 1/, we have Z 1 1 Thus Z 1 3 e x dx Z 1 1 e x dx D 38. 1 : e 39. 3 e x dx converges. 0 34. On [0,1], p Thus, 1 1 1 1 p . On Œ1; 1/, p 2. x x C x2 x x C x2 Z 1 Z 1 dx dx p 2 x C x x Z0 1 Z 01 dx dx p : x2 x C x2 1 1 Telegram: @uni_k On .0; =2, sin x < x, and so csc x 1=x. Thus Z =2 Z =2 dx csc x dx > D 1: x 0 0 Z =2 Therefore csc x dx must diverge. (It is of the form =2 40. Since bothZof these integrals are convergent, therefore so is 1 dx their sum p . x C x2 0 240 Fig. 6.5-37 p p 2 p x x x 2 D , Since 0 1 cos x D 2 sin2 2 2 2 Z 2 Z 2 dx dx for x 0, therefore , which p 2 x 1 cos x 0 0 diverges to infinity. 1 p x 1.) Since ln x grows more slowly than any positive power of x, therefore we have ln x kx 1=4 for some constant k 1 1 for x 2 and every x 2. Thus, p 3=4 x ln x kx Z 1 dx p diverges to infinity by comparison with and x ln x 2 Z 1 dx 1 . k 2 x 3=4 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 41. Z 1 0 dx D xe x Z 1 0 C Z 1 0 Z 1 1 SECTION 6.5 (PAGE 370) dx . But xe x dx 1 xe x e Z 1 0 44. E./ D dx D 1: x Z 1 0 2 e x dx D 21 p . 0 2 x 2 e x dx D lim Z R R!1 0 2 x 2 e x dx 0 2 x 4 e x dx D lim Z R R!1 0 45. 2 Z b Thus lim c!aC c so E./ D O. pC1 / as ! 0C. f .x/ dx D Z b f .x/ dx. a : b Z c 46. .x/ D Z 1 a f .x/ dx D Z b f .x/ dx. a t x 1 e t dt . 0 a) Since lim t!1 t x 1 e t=2 D 0, there exists T > 0 such that t x 1 e t=2 1 if t T . Thus Z 1 Z 1 0 t x 1 e t dt e t dt D 2e T =2 T and Z 1 T t x 1 e t dt converges by the comparison T theorem. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k dx D I1 ./ C I2 ./ x C x2 Similarly ˇ ˇZ Z b ˇ ˇ c ˇ ˇ f .x/ dx f .x/ dx ˇ ˇ ˇ ˇ a a ˇZ c ˇ ˇ ˇ D ˇˇ f .x/ dx ˇˇ K.b c/ ! 0 as c ! b c!b 0 1= p Since f is continuous on Œa; b, there exists a positive constant K such that jf .x/j K for a x b. If a < c < b, then ˇ ˇZ Z b ˇ ˇ b ˇ ˇ f .x/ dx ˇ f .x/ dx ˇ ˇ ˇ c a ˇZ a ˇ ˇ ˇ D ˇˇ f .x/ dx ˇˇ K.c a/ ! 0 as c ! a C : Thus lim K pC1 ; pC1 Z 1 c 2 d V D xe x dx U D x3 2 2 d U D 3x dx V D 12 e x ˇR " # Z 1 3 x 2 ˇˇ 3 R 2 x2 D lim x e x e dx ˇ C ˇ R!1 2 2 0 0 Z 1 3 1 2 x2 2 D x e lim R3 e R C dx 2 R!1 2 0 3 1p 3p D0C D : 2 4 8 ˇZ ˇ ˇ ˇ ˇ jE./ D ˇ f .x/ dx ˇˇ 0 Z K x p dx D dx x C x2 I1 ./ > .1=2/ k ! 1 as ! 0C, and I1 ./ is not k O. k /. x 4 e x dx 43. Since f .x/ D O.x p / as x ! 0C, we have jf .x/ Kx p for all x in the interval .0; a/ for some constants K and a 1. If 0 < < a it follows that p p Thus b) Similarly, Z 1 Now we must show that no value of k larger than 1/2 will work. If k > 1=2, then Z Z p dx dx > p D : I1 ./ D p x C x2 0 2 x 0 2 U Dx d V D xe x dx 2 d U D dx V D 12 e x ˇR # " Z ˇ 1 R x2 1 x2 ˇ e xe dx D lim ˇ C ˇ R!1 2 2 0 0 Z 1 1 1 x2 2 D lim Re R C e dx 2 R!1 2 0 1p 1p D : D0C 4 4 0 Z Z 1= ! Thus I1 ./ D O. 1=2 / as ! 0C, and I2 ./ D O./ as ! 0C. Since < 1=2 (for 0 < < 1), both I1 ./ and I2 ./ are O. 1=2 / as ! 0C, and so, therefore, is E./. a) First we calculate Z 1 Z 1 dx C x C x2 0 Z p dx jI1 ./j < p D2 x 0 Z 1 dx jI2 ./j < D : 2 1= x D Thus the given integral must diverge to infinity. 42. We are given that We have Downloaded by ted cage (sxnbyln180@questza.com) 241 lOMoARcPSD|6566483 www.konkur.in SECTION 6.5 (PAGE 370) ADAMS and ESSEX: CALCULUS 9 The approximations are If x > 0, then 0 Z T t x 1 e t dt < 0 Z T t x 1 1 1 1 9 5 T4 D C 1C C .1 C 1/ C 1 C C D 4:75 2 2 4 4 2 1 1 9 25 49 M4 D 1C C 1C C 1C C 1C 2 16 16 16 16 D 4:625 1 T8 D .T4 C M4 / D 4:6875 2" 1 9 25 1 M8 D C 1C C 1C 1C 4 64 64 64 49 81 121 C 1C C 1C C 1C 64 64 64 # 169 225 D 4:65625 C 1C C 1C 64 64 dt 0 converges by Theorem 2(b). Thus the integral defining .x/ converges. Z 1 b) .x C 1/ D t x e t dt 0 Z R t x e t dt D lim c!0C R!1 c d V D e t dt U D tx x 1 dU D dx V D e t 0 xt 1 ˇR Z R ˇ ˇ D lim @ t x e t ˇ C x t x 1 e t dt A c!0C ˇ c R!1 c Z 1 D0Cx t x 1 e t dt D x.x/: T16 D 0 c) .1/ D Z 1 0 1 .T8 C M8 / D 4:671875: 2 The exact errors are e t dt D 1 D 0Š. I T4 D I T8 D I T16 D By (b), .2/ D 1.1/ D 1 1 D 1 D 1Š. In general, if .k C 1/ D kŠ for some positive integer k, then .k C 2/ D .k C 1/.k C 1/ D .k C 1/kŠ D .k C 1/Š. Hence .n C 1/ D nŠ for all integers n 0, by induction. Z 1 1 D t 1=2 e t dt Let t D x 2 d) 2 0 dt DZ2x dx Z 1 1 p 1 x2 2 e x dx D e 2x dx D 2 D x 0 0 1 1 1p 3 D D : 2 2 2 2 0:0833333I 0:0208333I 0:0052083: I I M4 D 0:0416667I M8 D 0:0104167I If f .x/ D 1 C x 2 , then f 00 .x/ D 2 D K, and K.2 0/ 1 D . Therefore, the error bounds are 12 3 2 1 0:0833333I 2 2 1 1 0:0208333I T8 j 3 4 1 1 2 0:0052083: T16 j 3 8 2 1 1 M4 j 0:0416667I 6 2 1 1 2 0:0104167: M8 j 6 4 Trapezoid W jI T4 j jI jI Midpoint W jI jI Section 6.6 The Trapezoid and Midpoint Rules (page 377) 1 3 Note that the actual errors are equal to these estimates since f is a quadratic function. 1. The exact value of I is 2. The exact value of I is ˇ2 x 3 ˇˇ I D .1 C x / dx D x C ˇ 3 ˇ 0 Z 2 D2C 242 Telegram: @uni_k 2 0 I D 8 4:6666667: 3 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) Z 1 D1 0 e x dx D e ˇ1 ˇ ˇ ˇ xˇ 1 0:6321206: e 0 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.6 (PAGE 377) The approximations are The approximations are 3 1 C 0 C sin C sin C sin 0:9871158 8 8 4 8 2 3 5 7 C sin C sin C sin sin 1:0064545 M4 D 8 16 16 16 16 1 T8 D .T4 C M4 / 0:9967852 2 3 5 C sin C sin sin M8 D 16 32 32 32 T4 D 41 . 12 e 0 C e 1=4 C e 1=2 C e 3=4 C 21 e 1 / T4 D 0:6354094 M4 D 41 .e 1=8 C e 3=8 C e 5=8 C e 7=8 / 0:6304774 T8 D 12 .T4 C M4 / 0:6329434 M8 D 18 .e 1=16 C e 3=16 C e 5=16 C e 7=16 C e 9=16 C e 11=16 C e 13=16 C e 15=16 / 0:6317092 9 11 7 C sin C sin 32 32 ! 32 13 15 C sin C sin 1:0016082 32 32 C sin T16 D 21 .T8 C M8 / 0:6323263: The exact errors are I T4 D I T8 D I T16 D 0:0032888I 0:0008228I 0:0002057: I I T16 D M4 D 0:0016432I M8 D 0:0004114I The actual errors are I T4 0:0128842I I T8 0:0032148I I T16 0:0008033: If f .x/ D e x , then f .2/ .x/ D e x . On [0,1], jf .2/ .x/j 1. Therefore, the error bounds are: Trapezoid W jI jI jI jI Midpoint W jI jI jI 1 .T8 C M8 / 0:9991967: 2 1 1 2 12 n 1 1 0:0052083I T4 j 12 16 1 1 T8 j 0:001302I 12 64 1 1 0:0003255: T16 j 12 256 1 1 2 Mn j 24 n 1 1 0:0026041I M4 j 24 16 1 1 M8 j 0:000651: 24 64 M4 M8 I I 0:0064545I 0:0016082I If f .x/ D sin x, then f 00 .x/ D sin x, and jf 00 .x/j 1 D K. Therefore, the error bounds are: Tn j Trapezoid W jI jI jI Midpoint W jI jI 2 1 0 0:020186I 12 2 8 2 1 0 0:005047I T8 j 12 2 16 1 2 T16 j 0 0:001262: 12 2 32 2 1 M4 j 0 0:010093I 24 2 8 2 1 0 0:002523: M8 j 24 2 16 T4 j Note that the actual errors satisfy these bounds. Note that the actual errors satisfy these bounds. 4. 3. The exact value of I is I D Z =2 0 sin x dx D 1: The exact value of I is I D Z 1 0 ˇ1 ˇ dx 1 ˇ D tan x ˇ D 0:7853982: 2 ˇ 1Cx 4 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 0 243 lOMoARcPSD|6566483 www.konkur.in SECTION 6.6 (PAGE 377) ADAMS and ESSEX: CALCULUS 9 The approximations are 1 1 16 4 16 1 1 T4 D .1/ C C C C 4 2 17 5 25 2 2 0:7827941 64 64 64 1 64 C C C M4 D 4 65 73 89 113 0:7867001 1 T8 D .T4 C M4 / 0:7847471 2" 1 256 256 256 256 M8 D C C C C 8 257 265 281 305 # 256 256 256 256 C C C 337 377 425 481 0:7857237 1 T16 D .T8 C M8 / 0:7852354: 2 The exact errors are I T4 D 0:0026041I I I T8 D 0:0006511I I I T16 D 0:0001628: M4 D M8 D jI jI jI Midpoint W jI jI jI 4 1 2 12 n 4 1 T4 j 0:0208333I 12 16 1 4 0:0052083I T8 j 12 64 4 1 T16 j 0:001302: 12 256 2 4 1 Mn j 24 n 4 1 M4 j 0:0104167I 24 16 4 1 M8 j 0:0026042: 24 64 Tn j 2 Œ3 C 2.5 C 8 C 7/ C 3 D 46 2 1 T8 D Œ3 C 2.3:8 C 5 C 6:7 C 8 C 8 C 7 C 5:2/ C 3 D 46:7 2 T4 D 244 Telegram: @uni_k 2 T4 D 100 Œ0 C 2.5:5 C 5 C 4:5/ C 0 D 3; 000 km2 2 1 T8 D 100 Œ0 C 2.4 C 5:5 C 5:5 C 5 C 5:5 C 4:5 C 4/ C 0 2 D 3; 400 km2 8. M4 D 100 2.4 C 5:5 C 5:5 C 4/ D 3; 800 km2 9. We have T4 D 0:4 12 .1:4142/ C 1:3860 C 1:3026 C 1:1772 C 21 .0:9853/ 2:02622 M4 D .0:4/.1:4071 C 1:3510 C 1:2411 C 1:0817/ 2:03236 T8 D .T4 C M4 /=2 2:02929 M8 D .0:2/.1:4124 C 1:3983 C 1:3702 C 1:3285 C 1:2734 C 1:2057 C 1:1258 C 1:0348/ 2:02982 T16 D .T8 C M8 /=2 2:029555: 10. The approximations for I D The exact errors are much smaller than these bounds. In part, this is due to very crude estimates made for jf 00 .x/j. 5. 7. M4 D 2.3:8 C 6:7 C 8 C 5:2/ D 47:4 0:0013019I 0:0003255I 2x 1 , then f 0 .x/ D and Since f .x/ D 1 C x2 .1 C x 2 /2 6x 2 2 f 00 .x/ D . On [0,1], jf 00 .x/j 4. Therefore, .1 C x 2 /3 the error bounds are Trapezoid W jI 6. Z 1 2 e x dx are 0 1 e 1=256 C e 9=256 C e 25=256 C e 49=256 C 8 81=256 121=256 169=256 225=256 e Ce Ce Ce M8 D 0:7473 1 1 .1/ C e 1=256 C e 1=64 C e 9=256 C e 1=16 C T16 D 16 2 e 25=256 C e 9=64 C e 49=256 C e 1=4 C e 81=256 C e 25=64 C e 121=256 C e 9=16 C e 169=256 C e 49=64 C 1 1 225=256 e C e 2 0:74658: 2 2 Since f .x/ D e x , we have f 0 .x/ D 2xe x , 2 2 f 00 .x/ D 2.2x 2 1/e x , and f 000 .x/ D 4x.3 2x 2 /e x . Since f 000 .x/ ¤ 0 on (0,1), therefore the maximum value of jf 00 .x/j on Œ0; 1 must occur at an endpoint of that interval. We have f 00 .0/ D 2 and f 00 .1/ D 2=e, so jf 00 .x/j 2 on Œ0; 1. The error bounds are jI Mn j 2 24 2 1 ) jI n jI Tn j 2 12 2 1 ) jI n Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2 1 24 64 0:00130: 1 2 T16 j 12 256 0:000651: M8 j lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 6.6 (PAGE 377) Thus the constant in the error estimate for the Midpoint Rule cannot be improved; no smaller constant will work for f .x/ D x 2 . According to the error bounds, Z 1 0 11. 2 e x dx D 0:747; 14. accurate to two decimal places, with error no greater than 1 in the third decimal place. Z =2 sin x sin x dx. Note that lim D 1. I D x!0 x x 0 " 1 16 8 16 3 4 C sin C sin C sin C sin 16 2 16 8 3 16 4 # 5 8 3 16 7 1 2 16 sin C sin C sin C C 5 16 3 8 7 16 2 T8 D 1:3694596 " 32 3 32 5 32 7 32 sin C sin C sin C sin M8 D 16 32 3 32 5 32 7 32 C 32 9 32 11 32 13 sin C sin C sin 9 32 11# 32 13 32 32 15 C sin 1:3714136 15 32 T16 D .T8 C M8 /=2 1:3704366; I 1:370: 12. The exact value of I is ˇ1 1 x 3 ˇˇ I D x dx D ˇ D : ˇ 3 3 0 Z 1 2 0 The approximation is 1 1 1 T1 D .1/ .0/2 C .1/2 D : 2 2 2 The actual error is I T1 D 16 . However, since f .x/ D x 2 , then f 00 .x/ D 2 on [0,1], so the error estimate here gives 1 2 .1/2 D : jI T1 j 12 6 Since this is the actual size of the error in this case, the constant “12” in the error estimate cannot be improved (i.e., cannot be made larger). 2 Z 1 1 1 1 .1/ D . The actual x 2 dx D . M1 D 13. I D 3 2 4 0 1 1 1 D . error is I M1 D 3 4 12 Since the second derivative of x 2 is 2, the error estimate is 2 1 jI M1 j .1 0/2 .12 / D : 24 12 Let y D f .x/. We are given that m1 is the midpoint of Œx0 ; x1 where x1 x0 D h. By tangent line approximate in the subinterval Œx0 ; x1 , f .x/ f .m1 / C f 0 .m1 /.x The error in this approximation is E.x/ D f .x/ f 0 .m1 /.x f .m1 / m1 /: If f 00 .t / exists for all t in Œx0 ; x1 and jf 00 .t /j K for some constant K, then by Theorem 11 of Section 4.9, K .x 2 m1 /2 : f 0 .m1 /.x m1 /j jE.x/j Hence, jf .x/ f .m1 / K .x 2 m1 /2 : We integrate both sides of this inequlity. Noting that x1 m1 D m1 x0 D 21 h, we obtain for the left side ˇZ Z x1 ˇ x1 ˇ f .x/ dx f .m1 / dx ˇ ˇ x0 x0 ˇ Z x1 ˇ ˇ 0 f .m1 /.x m1 / dx ˇ ˇ x0 ˇZ ˇx1 ˇ ˇ x1 .x m1 /2 ˇˇ ˇˇ ˇ 0 Dˇ f .x/ dx f .m1 /h f .m1 / ˇ ˇ ˇ x0 ˇ ˇ 2 x0 ˇ Z x1 ˇ ˇ ˇ ˇ ˇ Dˇ f .x/ dx f .m1 /hˇ : x0 Integrating the right-hand side, we get ˇx1 K .x m1 /3 ˇˇ K 2 .x m1 / dx D ˇ ˇ 2 3 x0 2 x0 3 K h h3 K 3 D C h : D 6 8 8 24 Z x1 Hence, ˇZ x1 ˇ ˇ f .x/ dx ˇ x0 ˇZ x1 ˇ D ˇˇ Œf .x/ x0 K 3 h : 24 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k m1 /: Downloaded by ted cage (sxnbyln180@questza.com) ˇ ˇ f .m1 /hˇˇ f .m1 / f 0 .m1 /.x ˇ ˇ m1 / dx ˇˇ 245 lOMoARcPSD|6566483 www.konkur.in SECTION 6.6 (PAGE 377) ADAMS and ESSEX: CALCULUS 9 A similar estimate holds on each subinterval Œxj 1 ; xj for 1 j n. Therefore, ˇZ b ˇ ˇ f .x/ dx ˇ a ˇ ˇ n Z x j ˇ ˇX f .x/ dx Mn ˇˇ D ˇˇ because nh D b xj j D1 ˇZ x n X j j D1 n X j D1 ˇ ˇ ˇ xj 1 f .x/ dx 1 I D Z 2 0 ˇ ˇ f .mj /hˇˇ 3. Z =2 0 S8 D a. .1 C x 2 / dx D (page 382) D1 e x dx D e 4. I D 0 0:7853981: ˇ ˇ 0 The actual errors are I The actual errors are Telegram: @uni_k 0 ˇ1 ˇ dx 1 ˇ D tan x ˇ D 0:7853982: 2 ˇ 1Cx 4 0:7853922 " 1 64 16 64 S8 D 1C4 C2 C4 C 24 65 17 73 # 64 16 64 1 4 C4 C2 C4 C 2 5 89 25 113 2 ˇ1 ˇ xˇ 1 0 .e C 4e 1=4 C 2e 1=2 C 4e 3=4 C e 1 / 12 0:6321342 1 0 S8 D .e C 4e 1=8 C 2e 1=4 C 4e 3=8 C 24 2e 1=2 C 4e 5=8 C 2e 3=4 C 4e 7=8 C e 1 / 0:6321214: 246 Z 1 " # 16 4 16 1 1 1C4 C2 C4 C S4 D 12 17 5 25 2 1 0:6321206: e 0:0000136I I 0:0000083. The approximations are 5. 6. S8 D 0:0000008: S4 D 0:0000060I I S8 D 0:0000001; accurate to 7 decimal places. These errors are evidently much smaller than the corresponding errors for the corresponding Trapezoid Rule approximation. S4 D S4 D S8 0:0001346; I The exact value of I is The approximations are I S4 Errors: I 2. The exact value of I is 0 3 0 C 4 sin C 2 sin C 4 sin C 2 sin 48 16 8 16 4 ! 3 7 5 C 2 sin C 4 sin C sin C 4 sin 16 8 16 2 1:0000083: The errors are zero because Simpson approximations are exact for polynomials of degree up to three. Z 1 sin x dx D 1. 3 C sin 0 C 4 sin C 2 sin C 4 sin 24 8 4 8 2 1:0001346 14 4:6666667 3 " 1 9 1 1C4 1C C 2.1 C 1/ C 4 1 C S4 D 6 4 4 # 14 C .1 C 4/ D 3 " 1 1 1 9 S8 D 1C4 1C C2 1C C4 1C 12 16 4 16 25 9 C 2 .1 C 1/ C 4 1 C C2 1C 16 4 # 49 14 C4 1C C .1 C 4/ D 16 3 I D I D S4 D K 3 K 3 K.b a/ 2 h D nh D h 24 24 24 Section 6.7 Simpson’s Rule 1. ˇ ˇ f .mj /h ˇˇ These errors are evidently much smaller than the corresponding errors for the corresponding Trapezoid Rule approximations. 1 Œ3 C 4.3:8 C 6:7 C 8 C 5:2/ C 2.5 C 8 C 7/ C 3 3 46:93 S8 D 1 S8 D 100 Œ0 C 4.4 C 5:5 C 5:5 C 4/ C 2.5:5 C 5 C 4:5/ C 0 3 3; 533 km2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 7. SECTION 6.7 (PAGE 382) If f .x/ D e x , then f .4/ .x/ D e x , and jf .4/ .x/j 1 on Œ0; 1. Thus 1.1 0/ 1 4 0:000022 jI S4 j 180 4 1.1 0/ 1 4 jI S8 j 0:0000014: 180 8 If f .x/ D sin x, then f .4/ .x/ D sin x, and jf .4/ .x/j 1 on Œ0; =2. Thus 1..=2/ 0/ 4 jI S4 j 0:00021 180 8 4 1..=2/ 0/ 0:000013: jI S8 j 180 16 Z b 8. Let I D f .x/ dx, and the interval Œa; b be subdivided a into 2n subintervals of equal length h D .b a/=2n. Let yj D f .xj / and xj D a C j h for 0 j 2n, then 1 b a S2n D y0 C 4y1 C 2y2 C 3 2n C 2y2n 2 C 4y2n 1 C y2n D 1 3 b a 2n and 1 T2n D 2 Tn D 1 2 2n X1 y0 C 4 yj 2 j D1 j D1 b a 2n b y2j C y2n 2n X1 y0 C 2 yj C y2n j D1 Tn , then Tn C 2.2T2n Tn / 4T2n Tn Tn C 2Mn D D 3 3 3 2T2n C 2T2n Tn 4T2n Tn 2T2n C Mn D D : 3 3 3 Hence, Tn C 2Mn 2T2n C Mn 4T2n Tn D D : 3 3 3 Using the formulas of T2n and Tn obtained above, 4T2n Tn 3" 2n X1 1 4 b a y0 C 2 yj C y2n D 3 2 2n j D1 # n X1 1 b a y0 C 2 y2j C y2n 2 n D 1 3 j D1 b D S2n : 2n 2n X1 a y0 C 4 yj j D1 2 n X1 j D1 y2j C y2n 9. Tn C 2Mn 2T2n C Mn 4T2n Tn D D : 3 3 3 We use the results of Exercise 9 of Section 7.6 and Exercise 8 of this section. I D Z 1:6 f .x/ dx 0 0:4 .1:4142 C 4.1:3860/ C 2.1:3026/ C 4.1:1772/ 3 C 0:9853/ 2:0343333 S8 D .T4 C 2M4 /=3 2:0303133 S16 D .T8 C 2M8 /=3 2:0296433: S4 D S8 D n X1 y0 C 2 y2j C y2n : Since T2n D 12 .Tn C Mn / ) Mn D 2T2n S2n D 10. The approximations for I D j D1 a n n X1 Hence, 2 e x dx are 0 1 1 C 4 e 1=64 C e 9=64 C e 25=64 C 8 49=64 e C 2 e 1=16 C e 1=4 C e 9=16 C e 1 1 3 0:7468261 1 1 S16 D 1 C 4 e 1=256 C e 9=256 C e 25=256 C 3 16 e 49=256 C e 81=256 C e 121=256 C e 169=256 C 225=256 e C 2 e 1=64 C e 1=16 C e 9=64 C e 1=4 C 25=64 9=16 49=64 1 e Ce Ce Ce 0:7468243: 2 2 If f .x/ D e x , then f .4/ .x/ D 4e x .4x 4 12x 2 C 3/. On [0,1], jf .4/ .x/j 12, and the error bounds are 12.1/ 1 4 180 n 12 1 4 0:0000163 jI S8 j 180 8 4 1 12 0:0000010: jI S16 j 180 16 jI Sn j Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Z 1 Downloaded by ted cage (sxnbyln180@questza.com) 247 lOMoARcPSD|6566483 www.konkur.in SECTION 6.7 (PAGE 382) ADAMS and ESSEX: CALCULUS 9 Comparing the two approximations, I D Z 1 0 Another possibility: Z 1 x Z 0 e dx e x dx C D I1 C I2 : I D p p 1 x2 1 x2 0 1 2 e x dx D 0:7468; In I1 put 1 C x D u2 ; in I2 put 1 accurate to 4 decimal places. 11. Z 1 x 4 dx 4 5 1 4 1 C 14 D S2 D . 0 C4 6 2 24 D I " 0 If f .x/ D x 4 , then f .4/ .x/ D 24. 1 24.1 0/ 1 4 . D Error estimate: jI S2 j 180 2 120 ˇ ˇ ˇ1 ˇ 5ˇ 1 Actual error: jI S2 j D ˇˇ D . 5 24 ˇ 120 Thus the error estimate cannot be improved. 4. 12. The exact value of I is ˇ1 x 4 ˇˇ 1 I D x dx D ˇ D : 4 ˇ 4 0 Z 1 3 0 The approximation is 1 S2 D 3 # 3 " 1 1 1 3 3 C1 D : 0 C4 2 2 4 5. The actual error is zero. Hence, Simpson’s Rule is exact for the cubic function f .x/ D x 3 . Since it is evidently exact for quadratic functions f .x/ D Bx 2 C C x C D, it must also be exact for arbitrary cubics f .x/ D Ax 3 C Bx 2 C C x C D. Section 6.8 Other Aspects of Approximate Integration (page 388) 1. 2. Z 1 dx Let x D u3 1=3 .1 C x/ 0 x Z 1 Z 1 u du u2 du : D 3 D3 3/ u.1 C u 1 C u3 0 0 Z 1 0 D ex 1 x 248 Telegram: @uni_k e x dx p D 1 x2 1 Z =2 dx D x4 C 1 Z 1 0 dx C x4 C 1 Z 1 1 dx D I1 C I2 : x4 C 1 Hence, Z 1 3. One possibility: let x D sin and get Z 1 Let Z 1 Z 1 1 dt Let x D and dx D in I2 , then t t2 Z 1 Z 0 dt t2 1 dt: D I2 D 4 2 4 t 1 0 1Ct 1 C1 t dx I D 6. 0 Let t 2 D 1 x 2t dt D dx Z 1 Z 0 1 t2 e 2 e 1 t dt: 2t dt D 2 t 0 1 p Z 1 u2 1 2 e du 2e u 1 u du p D2 p u 2 u2 2 u2 0 0 Z 1 1 u2 Z 1 1 u2 e du 2e u du p D2 p I2 D 2 0 u 2 u 2 u2 0 Z 1 u2 1 2 e C e1 u du. p so I D 2 2 u2 0 Z 1 dx 1 p Let x D 2 t x2 C x C 1 1 2 dt dx D t3 Z 0 1 2 dt D 2 r t3 1 1 1 C1 C 2 2 t t Z 1 t dt : D2 4 3 0 t Ct C1 Z =2 dx Let sin x D u2 p p sin x 0 2u du D cos x dx D 1 u4 dx Z 1 u du D2 p u4 0 u 1 Z 1 du Let 1 u D v 2 D2 p 0 .1 u/.1 C u/.1 C y 2 du D 2v dv Z 1 v dv p D4 v 2 /.1 C .1 v 2 /2 / 0 v .1 C 1 Z 1 dv p : D4 .2 v 2 /.2 2v 2 C v 4 / 0 I1 D 1 . D # 5 x D u2 : 0 e sin d: =2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) Z 1 x2 1 C x4 C 1 1 C x4 0 Z 1 2 x C1 dx: D 4 0 x C1 dx D x4 C 1 dx lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 7. I D Z 1 0 p x dx D SECTION 6.8 (PAGE 388) 2 0:666667: 3 Hence, 1 1 0 C 4.4e 4 / C e 1 3 2 0:1101549 " 1 1 0 C 4.16e 16 / C 2.4e 4 / S4 D 3 4 # 16 16=9 1 e Ce C4 9 S2 D ! 1 1 C 0:603553 2 2 r ! r 1 3 1 C 2T2 C 0:643283 T4 D 4 4 4 r r r ! r 1 3 5 7 1 C C C 4T4 C 0:658130 T8 D 8 8 8 8 8 r r r r 1 3 5 7 1 T16 D C C C 8T8 C 16 16 16 16 16 r r r ! r 9 11 13 15 C C C 0:663581: C 16 16 16 16 1 T2 D 0C 2 r 0:1430237 " 64 64 1 1 0 C 4 64e 64 C e 64=9 C e 64=25 C S8 D 3 8 9 25 # 64 64=49 16 16=9 16 4 1 e C 2 16e C 4e C e Ce 49 9 0:1393877: The errors are Hence, I 0:14, accurate to 2 decimal places. These approximations do not converge very quickly, because the 2 fourth derivative of e 1=t has very large values for some values of t near 0. In fact, higher and higher derivatives behave more and more badly near 0, so higher order methods cannot be expected to work well either. I T2 0:0631 I T4 0:0234 I T8 0:0085 I T16 0:0031: Observe that, although these errors are decreasing, they are not decreasing like 1=n2 ; that is, 1 T2n j >> jI 4 jI Tn j: This is because the second derivative of f .x/ D x is f 00 .x/ D 1=.4x 3=2 /, which is not bounded on Œ0; 1. 8. Let Z 1 e x dx D Z 0 2 Let x D 1 e 1 .1=t/2 1 t dt t2 dx D Z 1 1=t 2 1 e dt D dt: 2 t t2 0 Observe that 2 t 2 e 1=t D lim 2 t!0C t!0C t e 1=t 2 lim D lim D lim 1 t!0C e 1=t 2 h1i 1 2t 3 t!0C e 1=t 2 . Referring to Example 5, we have ex D 1 C x C p I D 9. 2t 3 / D 0: where Rn .f I 0; x/ D x. Now e X x nC1 , for some X between 0 and .n C 1/Š jRn .f I 0; x 2 /j x 2nC2 .n C 1/Š if 0 x 1 for any x, since x 2 X 0. Therefore ˇZ 1 ˇ Z 1 ˇ ˇ 1 2 ˇ ˇ x 2nC2 dx R .f I 0; x / dx n ˇ ˇ .n C 1/Š 0 0 1 : D .2n C 3/.n C 1/Š This error will be less than 10 4 if .2n C 3/.n C 1/Š > 10; 000. Since 15 7Š > 10; 000, n D 6 will do. Thus we use seven terms of the series (0 n 6): Z 1 2 e x dx 0 Z 1 x4 x6 x 8 x 10 x 12 1 x2 C C C dx 2Š 3Š 4Š 5Š 6Š 0 1 1 1 1 1 1 C C C D1 3 5 2Š 7 3Š 9 4Š 11 5Š 13 6Š 0:74684 with error less than 10 4 . Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k xn x2 C C C Rn .f I 0; x/; 2Š nŠ Downloaded by ted cage (sxnbyln180@questza.com) 249 lOMoARcPSD|6566483 www.konkur.in SECTION 6.8 (PAGE 388) 10. We are given that previous exercise Z 1 Z 01 0 Z 1 e x2 1 11. ADAMS and ESSEX: CALCULUS 9 2 e x dx D 12 p 2 Dividing p the first two equations gives u D 3=5, so u D 3=5. Then 3A=5 D 1=3, so A D 5=9, and finally, B D 8=9. and from the 2 e x dx D 0:74684. Therefore, Z 1 dx D e x2 dx 0 Z 1 14. e x2 dx Z 1 0 1p D 2 D 0:139 0:74684 1 (to 3 decimal places). 1 Z 1 f .x/ dx D 2 0 .bx 2 C d / dx D 2 Af . u/ C Af .u/ D 2A.bu2 C d /: b Cd 3 2 r ! r !6 3 6 5 3 3 5 C 0 D 0:24000 x 6 dx 4 C 9 5 5 1 Z 1 Error D x 6 dx 0:24000 0:04571 1 " r !# r ! Z 1 3 3 5 8 cos x dx cos C cos C 9 5 5 9 1 12. For any function f we use the approximation 1 p Z 1 f .x/ dx f . 1= 3/ C f .1= 3/: x 4 dx 1 p 3 4 C 15. 1:68300 0:00006 Z 1 d b C Cf .bx C dx C f / dx D 2 F .x/ dx D 2 5 3 0 1 AF . u/ C BF .0/ C AF .u/ D 2A.bu4 C du2 C f / C Bf: 4 2 These two expressions are identical provided Au4 D 1 ; 5 Au2 D 1 ; 3 AC B D 1: 2 Z 1 2 e x dx 0 1 0 1 1 e C e 0:6839397 2 2 1 1 1 0 e C e 1=4 C e 1 0:7313703 T10 D T2 D 2 2 2 1 T20 D T4 D 2T2 C e 1=16 C e 9=16 0:7429841 4 1 0 4T4 C e 1=64 C e 9=64 C e 25=64 C e 49=64 T3 D T8 D 8 0:7458656 T11 D S2 D R1 D 13. If F .x/ D ax 5 C bx 4 C cx 3 C dx 2 C ex C f , then, by symmetry, Z 1 I D T00 D T1 D R0 D .1/ 1 Telegram: @uni_k 1 1 1 4 2 D p 9 3 1 Z 1 2 2 2 D 0:17778 x 4 dx Error D 9 5 9 1 Z 1 1 1 C cos p 1:67582 cos x dx cos p 3 3 1 Z 1 cos x dx 1:67582 0:00712 Error D 1 Z 1 p p e x dx e 1= 3 C e 1= 3 2:34270 1 Z 1 Error D e x dx 2:34270 0:00771: 250 1:68300 Z 1 Error D cos x dx p p 3=5 C e 3=5 2:35034 e x dx e 1 Z 1 Error D e x dx 2:35034 0:00006: p We have Z 1 i 8 p p 5h f. 3=5/ C f . 3=5/ C f .0/: 9 9 Z 1 These two expressionspare identical provided A D 1 and u2 D 1=3, so u D 1= 3. Z 1 f .x/ dx We have If f .x/ D ax 3 C bx 2 C cx C d , then, by symmetry, Z 1 For any function f we use the approximation T21 D S4 D T31 D S8 D 4T20 4T10 3 T10 3 4T30 T20 3 T00 0:7471805 0:7468554 0:7468261 16T21 T11 0:7468337 15 16T31 T21 0:7468242 T32 D 15 2 64T3 T22 T33 D R3 D 0:7468241 63 I 0:746824 to 6 decimal places. T22 D R2 D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 16. From Exercise 9 in Section 7.6, for I D SECTION 6.8 (PAGE 388) Z 1:6 Thetransformation is not suitable because the derivative of 1 1 1 cos sin is , which has very large values at t t2 t some points close to 0. In order to approximate the integral I to an desired degree of accuracy, say with error less than in absolute value, we have to divide the integral into two parts: Z 1 sin x I D dx 1 C x2 Z 1 Z t sin x sin x dx C dx D 2 1 C x2 t 1Cx D I1 C I2 : f .x/ dx, 0 T00 D T1 D 1:9196 T10 D T2 D 2:00188 T20 D T4 D 2:02622 T30 D T8 D 2:02929: Hence, R1 D T11 D T21 D 4T10 T00 3 4T20 T10 3 D 2:0346684 16T21 R2 D T22 D T11 4T30 T20 15 T31 D T32 D 3 16T31 If t tan D 2:0343333 D S4 , then Z 1 Z 1 sin x dx dx < 1 C x2 1 C x2 t t ˇ1 ˇ ˇ 1 tan 1 .t / : D tan .x/ˇ D ˇ 2 2 D 2:0346684 D 2:0303133 D S8 T21 2 t D 2:0300453 Now let ZA be a numerical approximation to the proper t sin x dx, having error less than =2 in absointegral 2 1Cx lute value. Then 15 2 64T T22 3 R3 D T33 D D 2:0299719: 63 jI 17. 2h y0 C 4y2 C y4 3 h 1 T2 D S4 D y0 C 4y1 C 2y2 C 4y3 C y4 3 16T21 T11 R2 D T22 D 15 2h 16h .y C 4y 0 1 C 2y2 C 4y3 C y4 / 3 .y0 C 4y2 C y4 / D 3 15 h D 14y0 C 64y1 C 24y2 C 64y3 C 14y4 45 2h 7y0 C 32y1 C 12y2 C 32y3 C 7y4 D 45 T11 D S2 D 18. Let I D Z 1 Hence, A is an approximation to the integral I with the desired accuracy. 19. sin x ; x 2 x .cos x x cos c sin x ; x2 x sin x cos x/ .x cos x f 00 .x/ D x4 2 x sin x 2x cos x C 2 sin x D : x3 Now use l’H^opital’s Rule to get f .x/ D f 0 .x/ D lim f 00 .x/ 2x sin x x 2 cos x cos x D 3 1 : 3 x!0 sin x dx 1 C x2 Let x D 1 t dt t2 sin x/2x x!0 D lim dx D 1 Z 0 sin 1 t dt D 1 t2 1= 1C 2 t 1 Z 1= sin t dt: D 1 C t2 0 D lim x!0 20. 2 cos x C 2x sin x C 2 cos x 3x 2 If t is time and E is error, then for the trapezoid rule, t 2 E is approximately constant. Since E D 6 10 16 when t D 175:777 seconds, the time we would expect our computer to achieve an error of 10 32 is about s 10 16 175:7772 6 seconds; 10 32 which is about 1,365 years. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Aj D jI1 C I2 Aj jI1 Aj C jI2 j C D : 2 2 Downloaded by ted cage (sxnbyln180@questza.com) 251 lOMoARcPSD|6566483 www.konkur.in SECTION 6.8 (PAGE 388) 21. ADAMS and ESSEX: CALCULUS 9 If t is time and E is error, then t 4 E is approximately constant for the Simpson’s Rule case. Since E D 3:15 10 30 when t D 175:777 seconds, we would expect our computer to achieve quadruple precision in time 1=4 10 30 175:7774 3:15 10 32 5. seconds; 3 A B D C 4x 2 1 2x 1 2x C 1 2Ax C A C 2Bx B D 4x 2 1 n 3 2A C 2B D 0 ) )AD B D A BD3 2 Z 3 dx 3 dx D 2 4x 1 2 2x 1 ˇ ˇ 3 ˇˇ 2x 1 ˇˇ D ln ˇ C C: 4 2x C 1 ˇ or about 12 minutes if it were using Simpson’s Rule. Since our computer did the calculation more than 5,000 times faster than that, it must have been using an even higher-order method. Review Exercises on Techniques of Integration (page 390) 1. 6. A B x D C 2 2x C 5x C 2 2x C 1 xC2 Ax C 2A C 2Bx C B D 2x 2 C 5x C 2 n A C 2B D 1 ) 2A C B D 0 Thus A D D Let u D x 1 du dx Z D uC1 1 1 du D C du u3 u2 u3 1 1 1 1 CC D C C: 2 u 2u x 1 2.x 1/2 D 3. 1/3 sin3 x cos3 x dx Z D sin3 x.1 sin2 x/ cos x dx Z 1 .x 2 C x 3 .u3 d V D sin 3x V D 13 cos 3x Z 1 2/ cos 3x C 3 .2x C 1/ cos 3x dx U D 2x C 1 d V D cos 3x dx d U D 2 dx V D 31 sin 3x 2 1 3 .x C xZ D 2 9 2 1 3 .x C x D u5 / du D u4 4 Let u D sin x du D cos x dx u6 CC 6 7. Z p D Z 2/ cos 3x C 91 .2x C 1/ sin 3x sin 3x dx 2/ cos 3x C 91 .2x C 1/ sin 3x 2 cos 3x C C: 27 1 x2 dx x4 cos2 Let x D sin dx D cos d d 4 Z sin D csc2 cot2 d D D Z Let v D cot dv D csc2 d v3 CC 3 !3 p cot3 1 1 x2 CC D C C: 3 3 x v 2 dv D 1 4 1 6 sin x sin x C C: 4 6 p Z p .1 C x/1=3 dx Let u D 1 C x p x dx du D p 2 x Z D 2 u1=3 du D 2. 43 /u4=3 C C p D 32 .1 C x/4=3 C C: D 4. 252 Telegram: @uni_k 2/ sin 3x dx dx Z D dx 2x C 1 U D x2 C x 2 d U D .2x C 1/ dx C x .x Z .x 2 C x 1=3 and B D 2=3. We have Z Z 1 2 x dx dx dx D C 2x 2 C 5x C 2 3 2x C 1 3 xC2 1 2 ln j2x C 1j C C: D ln jx C 2j 3 6 2. Z D Z Z Z Z 1 x p 1 x2 Fig. RT-7 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 8. Z x 3 cos.x 2 / dx D 21 Z REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390) Let w D x 2 dw D 2x dx 13. 10. 11. p 1C4x 2 x dx Let u D 5x 3 2 .5x 3 2/2=3 du D 15x 2 dx Z 1 1 u 2=3 du D u1=3 C C D 15 5 1 D .5x 3 2/1=3 C C: 5 1 Fig. RT-13 dx Let x D 2 tan .4 C x 2 /2 dx D 2 sec2 d Z Z 1 2 sec2 d D cos2 d D 16 sec4 8 1 D . C sin cos / C C 16 1 x 1 x D tan 1 C C C: 16 2 8 4 C x2 Z 15. Z D D 16. 2 Fig. RT-11 Z Dx .1 C sin 2x/ dx 1 2 cos 2x C C: Z Z .u3 C u5 / du D Let u D tan x du D sec2 x dx u6 u4 C CC 4 6 1 1 tan4 x C tan6 x C C: 4 6 We have q Let x D 35 tan u q dx D 35 sec2 u du q Z . 3 tan2 u/. 3 sec2 u/ du 5 5 x 2 dx .3 C 5x 2 /3=2 .3/3=2 sec3 u Z 1 D p .sec u cos u/ du 5 5 1 D p .ln j sec u C tan uj sin u/ C C 5 5 ˇp ! p p ˇˇ ˇ 5x 2 C 3 5x ˇ 5x 1 ˇ p CC D p ln ˇ C p ˇ p ˇ 5 5 3 3 ˇ 5x 2 C 3 p p 1 x D p ln 5x 2 C 3 C 5x p C C0 ; 5 5 5 5x 2 C 3 p 1 where C0 D C p ln 3: 5 5 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k tan3 x sec4 x dx tan3 x.1 C tan2 x/ sec2 x dx D 4Cx 2 .sin x C cos x/ dx D sin3 x dx D 7 Zcos x Z 2 cos x dx Let u D sin x 1 C sin2 x du D cos x dx Z du D D tan 1 u C C 1 C u2 D tan 1 .sin x/ C C: D x 12. 14. Z p 2x 1 A B .A C B/x C .5A 3B/ D C D x 2 C 2x 15 x 3 xC5 x 2 C 2x 15 n 1 1 ACB D0 ) )AD ; B D : 5A 3B D 1 8 Z Z 8 Z dx dx 1 1 dx D x 2 Cˇ 2x 15 8 x 3 8 x C5 ˇ 1 ˇˇ x 3 ˇˇ D ln ˇ C C: 8 x C 5ˇ Z Let 2x D tan 2x ln 2 dx D sec2 d D D 21 x 2 sin.x 2 / C 12 cos.x 2 / C C: 9. p 2x 1 C 4x dx Z 1 sec3 d ln 2 1 sec tan C ln j sec C tan j C C D 2 ln 2 p 1 xp D 2 1 C 4x C ln.2x C 1 C 4x / C C: 2 ln 2 w cos w dw U Dw d V D cos w dw d U D dw Z V D sin w D 21 w sin w 12 sin w dw Z Z Downloaded by ted cage (sxnbyln180@questza.com) 253 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390) p 20. p 5x 5x 2 C3 u p 3 Fig. RT-16 17. I D Z e x sin 2x dx U De x d U D e x dx 1 x e cos 2x 2 D 1 2 d V D sin 2x dx 1 V D cos 2x 2 Z e x 21. cos 2x dx 1 x e cos 2x 2 I D 18. 19. d V D cos 2x dx 1 V D sin 2x 2 ! 1 1 x 1 e sin 2x C I 2 2 2 Z 2x 2 C 4x 3 dx D x 2 C 5x # " Z 6x C 3 dx D 2 x.x C 5/ I D Z 2x 2 C 10x 6x x 2 C 5x 6x C 3 A B .A C B/x C 5A D C D x.x C 5/ x xC5 x.x C 5/ n 27 3 ACB D6 : ) )AD ; B D 5A D 3 5 Z Z Z 5 3 dx 27 dx I D 2 dx 5 x 5 xC5 3 27 D 2x ln jxj ln jx C 5j C C: 5 5 Z I D cos.3 ln x/ dx U D cos.3 ln x/ 3 sin.3 ln x/ dx dU D Zx D x cos.3 ln x/ C 3 d V D dx V Dx Z x ln.1 C x 2 / dx 1 C x2 3 22. Z 23. Z dx sin2 x cos4 x dx Z 1 D .1 cos 2x/Œ 12 .1 C cos 2x/2 dx 2 Z 1 D .1 C cos 2x cos2 2x cos3 2x/ dx 8 Z 1 1 1 .1 C cos 4x/ dx sin 2x D xC 8 Z 16 16 1 .1 sin2 2x/ cos 2x dx 8 1 x 1 1 x sin 2x sin 4x sin 2x D C 8 16 16 64 16 1 C sin3 2x C C 48 sin 4x sin3 2x x C C C: D 16 64 48 p x 2 dx p Let x D 2 sin p 2 x2 dx D 2 cos d Z D 2 sin2 d D sin cos C C p x 2 x2 x C C: D sin 1 p 2 2 sin.3 ln x/ dx p d V D dx U D sin.3 ln x/ 3 cos.3 ln x/ dx V Dx dU D x D x cos.3 ln x/ C 3 x sin.3 ln x/ 3I I D 254 Telegram: @uni_k Let u D ln.1 C x 2 / 2x dx du D 1 C x2 D 1 x 1 1 x e cos 2x e sin 2x I 2 4 4 2 1 e x cos 2x C sin 2x C C: 5 5 D 1 A Bx C C D C 4x 3 C x x 4x 2 C 1 A.4x 2 C 1/ C Bx 2 C C x D 4x 3 C x 4A C B D 0 ) ) B D 4: C D 0; A D 1 Z Z Z 1 dx x dx dx D 4 4x 3 C x x 4x 2 C 1 1 ln.4x 2 C 1/ C C: D ln jxj 2 Z u2 1 u du D CC 2 4 2 1 ln.1 C x 2 / C C: D 4 U De x d U D e x dx D ADAMS and ESSEX: CALCULUS 9 2 x 1 3 x cos.3 ln x/ C x sin.3 ln x/ C C: 10 10 p 2 x2 Fig. RT-23 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 24. We have Z I D tan4 x sec x dx 27. 3 U D tan x d U D 3 tan2 x Zsec2 x dx D tan3 x sec x 3 D tan3 x sec x 3 3 D tan Z x sec x 3I J D REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390) d V D tan x sec x dx V D sec x tan2 x sec3 x dx Z D tan x sec x D tan x sec x tan2 x.tan2 x C 1/ sec x dx D 3J where D d V D tan x sec x dx V D sec x 28. I D .tan2 x C 1/ sec x dx I D 41 tan3 x sec x 3 tan x sec x 8 3 C 8 ln j sec x C tan xj C C: Z x 2 dx Let u D 4x C 1 .4x C 1/10 du D 4 dx Z u 1 2 1 1 D du 4 4 u10 Z 1 .u 8 2u 9 C u 10 / du D 64 1 1 1 D u 7C u 8 u 9CC 448 256 576 1 1 1 1 D C C C: 64 7.4x C 1/7 4.4x C 1/8 9.4x C 1/9 26. We have Z x dx x sin 1 2 x U D sin 1 d V D x dx 2 x2 dx V D dU D p 2 4 x2 Z x 1 x 2 dx x2 sin 1 Let x D 2 sin u D p 2 2 2 4 x2 dx D 2 cos u du Z x2 1 x 2 sin 2 sin u du D 2 2 Z x x2 D sin 1 .1 cos 2u/ du 2 2 x x2 D sin 1 u C sin u cos u C C 2 2 2 x 1 p x 1 sin 1 C x 4 x 2 C C: D 2 2 4 Z 1 .1 2u2 C u4 / du 4 2 3 1 5 1 u C u CC u 4 3 5 1 1 1 cos 4x C cos3 4x cos5 4x C C: 4 6 20 We have D tan x sec x J ln j sec x C tan xj C C J D 21 tan x sec x 12 ln j sec x C tan xj C C: 25. Z dx D 2x 3 C x 29. Z 30. Let x5 Z Z dx 2 C ex Z e x dx D Let u D 2e x C 1 2e x C 1 du D 2e x dx Z du 1 1 D ln.2e x C 1/ C C: D 2 u 2 In D D I0 D Z x n 3x dx U D xn d U D nx n 1 dx x n 3x Zln 3 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Let u D cos 4x du D 4 sin 4x dx x dx Let u D x 2 x 6 2x 4 C x 2 du D 2x dx Z 1 1 du du D D 2 u3 2u2 C u 2 u.u 1/2 1 A B C D C C u.u 1/2 u u 1 .u 1/2 A.u2 2u C 1/ C B.u2 u/ C C u D u3 2u2 C u ( ACB D0 ) 2A B C C D 0 ) A D 1; B D 1; C D 1: A D1 Z Z Z du du 1 du 1 1 D 2 u3 2u2 C u 2 u 2 u 1 Z du 1 C 2 .u 1/2 1 1 1 1 ln ju 1j CK D ln juj 2 2 2u 1 2 1 x 1 D ln 2 C K: 2 jx 1j 2.x 2 1/ sec3 x dx Z sin5 .4x/ dx Z D .1 cos2 4x/2 sin 4x dx D tan2 x sec x dx U D tan x d U D sec2 xZ dx Z Downloaded by ted cage (sxnbyln180@questza.com) d V D 3x dx 3x V D ln 3 n In 1 : ln 3 3x 3x dx D C C: ln 3 255 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390) Hence, I3 D 34. Z We have Z 3 x x 3 dx !# " 31. Z sin2 x cos x dx Let u D sin x 2 sin x du D cos x dx Z 2 u du D Let 2 u D v 2 u du D dv Z Z 4 4v C v 2 4 D dv D C 4 v dv v v v2 CC D 4 ln jvj C 4v 2 1 D 4 ln j2 uj C 4.2 u/ .2 u/2 C C 2 1 2 sin x C C1 : D 4 ln.2 sin x/ 2 sin x 2 35. 32. We have ! Z x2 C 1 2x C 1 dx D dx 1 x 2 C 2x C 2 x 2 C 2x C 2 Z 2x C 1 dx Let u D x C 1 Dx .x C 1/2 C 1 du D dx Z 2u 1 Dx du u2 C 1 D x ln ju2 C 1j C tan 1 u C C D x ln.x 2 C 2x C 2/ C tan 1 .x C 1/ C C: Z 33. Z dx p Let x D sin x2 1 x2 dx D Z cos d Z cos d D csc2 d D sin2 cos p 1 x2 D cot C C D C C: x p 1 x2 Fig. RT-33 256 Telegram: @uni_k x 3 dx p Let 2x D sin 1 4x 2 2 dx D cos d Z Z 1 sin3 cos d 1 D .1 cos2 / sin d D 16 cos 16 1 1 D cos C cos3 C C 16 3 1p 1 1 4x 2 C C: .1 4x 2 /3=2 D 48 16 2x p 1 4x 2 Fig. RT-35 37. x Z 1 36. 1 x 3 .ln x/2 dx d V D x 3 dx U D .ln x/2 1 2 V D x4 d U D ln x dx 4 x Z 1 4 1 x 3 ln x dx D x .ln x/2 4 2 U D ln x d V D x 3 dx 1 1 d U D dx V D x4 x 4 Z 1 1 4 1 D x 4 .ln x/2 x 3 dx x ln x C 4 8 8 " # 1 1 x4 ln x C .ln x/2 C C: D 4 2 8 3 x 2 3x 2 x3x 1 x 3 3x C C1 I0 ln 3 ln 3 ln 3 ln 3 ln 3 ln 3 " # 3 6x 6 3x 2 x x D3 C C1 : C ln 3 .ln 3/2 .ln 3/3 .ln 3/4 D ADAMS and ESSEX: CALCULUS 9 Z e 1=x dx x2 D Z Z Let u D du D e u du D 1 x 1 dx x2 eu C C D e 1=x C C: xC1 p dx x2 C 1 Z p dx D x2 C 1 C p Let x D tan x2 C 1 dx D sec2 d Z p D x 2 C 1 C sec d p D x 2 C 1 C ln j sec C tan j C C p p D x 2 C 1 C ln.x C x 2 C 1/ C C: Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390) 42. p 1Cx 2 x 1 Fig. RT-37 38. Z e .x D3 1=3 / Z Let x D u3 dx D 3u2 du u2 e u du D 3I2 43. See solution to #16 of Section 6.6 for Z un e u dx D un e u In D D 3Œu2 e u De 39. I D .x 1=3 / Z 2.ue u .3x 2=3 x3 x3 nIn 1 . e u / C C Assume that x 1 and let x D sec u and dx D sec u tan u du. Then Z x 2 dx p x2 1 Z Z sec3 u tan u du D sec3 u du D tan u 1 1 D sec u tan u C ln j sec u C tan uj C C 2 2 p 1 p 1 D x x 2 1 C ln jx C x 2 1j C C: 2 2 Differentiation shows that this solution is valid for x 1 also. Z Z .x C 1 1/ dx x dx D Let u D x C 1 I D x 2 C 2x 1 .x C 1/2 2 du D dx Z Z du u 1 1 2 D du D ln ju 2j : u2 2 2 u2 2 B A p C p D u 2 uC 2 p p Au C 2A C Bu 2B D 2 u 2 ACB D0 p ) 2.A B/ D 1 1 )AD B D p : 2 2 1 u2 6x 1=3 C 6/ C C: Z 9x 3 3 dx D dx: 1C 3 9x x 9x 9x 3 A B C D C C x 3 9x x x 3 xC3 Ax 2 9A C Bx 2 C 3Bx C C x 2 3C x D x 3 9x 8 ( < A D 1=3 ACB CC D0 ) B D 4=3 ) 3B 3C D 9 : C D 5=3: 9A D 3 Thus we have Z Z 4 dx dx 1 C 3 x 3 x 3 1 4 D x C ln jxj C ln jx 3j 3 3 40. 41. p Z 10 xC2 dx p xC2 D .1 Z 5 dx 3 xC3 5 ln jx C 3j C K: 3 p xC2 dx du D p 2 xC2 Z p 2 2 u 10u C C D 10 xC2 C C: D 2 10 du D ln 10 ln 10 Z sin5 x cos9 x dx Z D .1 cos2 x/2 cos9 x sin x dx Let u D cos x du D sin x dx Z Let u D 2u2 C u4 /u9 du u12 u14 u10 C CC 10 6 14 12 10 cos x cos x cos14 x D C C: 6 10 14 D ˇ ˇ ˇ u p2 ˇ 1 ˇ ˇ 2j p ln ˇ p ˇCK 2 2 ˇu C 2ˇ ˇ ˇ ˇ x C 1 p2 ˇ 1 1 ˇ ˇ 2 D ln jx C 2x 1j p ln ˇ p ˇ C K: 2 2 2 ˇx C 1 C 2ˇ 1 I D ln ju2 2 Thus we have I DxC 44. 45. Z 2x 3 Let u D 4 3x C x 2 du D . 3 C 2x/ dx Z p p du D p D 2 u C C D 2 4 3x C x 2 C C: u Z x 2 sin 1 2x dx p 4 3x C x 2 dx U D sin 1 2x d V D x 2 dx 2 dx x3 dU D p V D 3 1 4x 2 Z 3 2 x3 x dx sin 1 2x D Let v D 1 4x 2 p 3 3 1 4x 2 dvD 8x dx Z 1 v 2 1 x3 1 sin 2x dv D 3 3 8 4v 1=2 Z x3 1 D sin 1 2x C v 1=2 v 1=2 dv 3 48 1p x3 1 3=2 sin 1 2x C v CC D v 3 24 72 3 p x 1 1 D sin 1 2x C .1 4x 2 /3=2 C C: 1 4x 2 3 24 72 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 Downloaded by ted cage (sxnbyln180@questza.com) 257 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390) 46. Let p 3x D sec u and Z p 3x 2 x 1 p 3 dx D sec u tan u du. Then 48. dx D 21 cos u du Z cos2 u du D 18 u C 81 sin u cos u C C p D 81 sin 1 .2x 1/ C 14 .2x 1/ x x 2 C C: 1 2 u p x x x2 Fig. RT-48 Z dx p Let x D u2 .4 C x/ x dx DZ 2u du Z 2u du du D D2 .4 C u2 /u 4 C u2 p x u 2 C C: D tan 1 C C D tan 1 2 2 2 258 Telegram: @uni_k x2 2 x2 D 2 1 1 2 D 2 sin u 1 2 52. dx x 3 d V D x dx x2 V D 2 3 dx 9 C x2 x 3 Z x2 x2 dx tan 1 D 2 3 2 9 C x2 ! 3Z x2 9 1 x D dx 1 tan 2 3 2 9 C x2 x 3x x 9 x2 tan 1 C tan 1 C C: D 2 3 2 2 3 Z x4 1 dx I D x 3 C 2x 2 Z 4 3 x C 2x 2x 3 4x 2 C 4x 2 1 D dx x 3 C 2x 2 Z 2 4x 1 D dx: x 2C 3 x C 2x 2 I D Let x 3 U D tan 1 Thus Z p D 41 49. 51. Z 1 sin4 2x dx cos4 x sin4 x dx D 16 Z 1 D .1 cos 4x/2 dx 64 Z 1 1 C cos 8x D 1 2 cos 4x C dx 64 2 sin 8x 1 3x sin 4x C CC D 64 2 2 16 1 sin 8x D 3x sin 4x C C C: 128 8 1 2 2 / dx x dU D Z x x 2 dx Z q 1 D .x 4 x tan 1 dx 1 Z tan u p sec u tan u du 3 D 1 p sec u 3 Z Z 2 D tan u du D .sec2 u 1/ du p p D tan u u C C D 3x 2 1 sec 1 . 3x/ C C p 1 D 3x 2 1 C sin 1 p C C1 : 3x 47. 50. Z ADAMS and ESSEX: CALCULUS 9 A C B 4x 2 1 D C 2 C x 3 C 2x 2 x x xC2 Ax 2 C 2Ax C Bx C 2B C C x 2 D x 3 C 2x82 ( < A D 1=4 ACC D4 ) 2A C B D 0 ) B D 1=2 : C D 15=4: 2B D 1 Z Z Z 15 1 dx 1 dx dx C 4 x 2 x2 4 xC2 1 1 15 2x C ln jxj C C ln jx C 2j C K: 4 2x 4 2x C Let u D x 2 and du D 2x dx; then we have Z Z Z du 1 x dx dx D D : I D 2 2 2 2 2 x.x C 4/ x .x C 4/ 2 u.u C 4/2 Since 1 A B C D C C 2 u.u C 4/ u uC4 .u C 4/2 2 2 A.u C 8u C 16/ C B.u C 4u/ C C u D u.u C 4/2 ( ACB D0 1 1 ; BD ; C D ) 8A C 4B C C D 0 ) A D 16 16 16A D 1 therefore Z Z Z 1 1 1 du du du I D 32 u 32 uC4 8 .u C 4/2 ˇ ˇ 1 ˇˇ u ˇˇ 1 1 D ln C CC 32 ˇ u C 4 ˇ 8 u C 4 ˇ ˇ 1 1 ˇˇ x 2 ˇˇ ln C C: C D 32 ˇ x 2 C 4 ˇ 8.x 2 C 4/ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 ; 4 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 53. Z REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390) sin.2 ln x/ dx x therefore Let u D 2 ln x 2 du D dx x Z 1 1 D cos u C C sin u du D 2 2 1 D cos.2 ln x/ C C: 2 I D Z 8x xC 2 x 1 x2 C 4C p D 2 7 D 54. Since Z sin.ln x/ dx x2 57. dx dV D 2 U D sin.ln x/ x cos.ln x/ 1 dx dU D V D x x Z cos.ln x/ sin.ln x/ dx C D x x2 dx U D cos.ln x/ dV D 2 x sin.ln x/ 1 dU D dx V D x x sin.ln x/ cos.ln x/ D I; x x 58. I D i 1 h sin.ln x/ C cos.ln x/ C C: 2x 55. 1 e 2 tan x dx 1 C x2 Let u D 2 tan 1 x 2 dx du D 1 C x2 Z 1 1 1 1 e u du D e u C C D e 2 tan x C C: D 2 2 2 61. I D Z x3 C x 2 dx D x2 7 D Z Z x3 7x C 8x 2 dx x2 7 ! 8x 2 dx: xC 2 x 7 Since p B .A C B/x C .B A/ 7 2 A p C p D D 7 x2 7 xC 7 x 7 ( ACB D8 1 1 ) B A D p2 ) A D 4 C p ; B D 4 p ; 7 7 7 8x x2 u2 /3 du D .1 ! x 1 p ln jx 7 dx p 7 p 7j C C: .1 3u2 C 3u4 u6 / du u3 C 53 u5 1 7 7u C C 3 3 sin x C 5 sin5 x 71 sin7 x C C: D sin x Z sin 1 .x=2/ dx .4 x 2 /1=2 Let u D sin 1 .x=2/ dx dx du D p D p 2 2 1 .x =4/ 4 x2 Z 2 2 u 1 D u du D CC D sin 1 .x=2/ C C: 2 2 We have Z Z tan4 .x/ dx D tan2 .x/Œsec2 .x/ 1 dx Z Z D tan2 .x/ sec2 .x/ dx Œsec2 .x/ 1 dx 1 tan3 .x/ 3 1 tan.x/ C x C C: Z .x C 1/ dx p 2 Z x C 6x C 10 .x C 3 2/ dx D p Let u D x C 3 .x C 3/2 C 1 du D dx Z .u 2/ du p D u2 C 1 Z p du D u2 C 1 2 p Let u D tan u2 C 1 du D sec2 d Z p D x 2 C 6x C 10 2 sec d p D x 2 C 6x C 10 2 ln j sec C tan j C C p p D x 2 C 6x C 10 2 ln x C 3 C x 2 C 6x C 10 C C: Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k !Z Let u D ln.3 C x 2 / 2x dx du D 3 C x2 Z 2 u2 1 1 u du D CC D ln.3 C x 2 / C C: D 2 4 4 Z Z cos7 x dx D .1 sin2 x/3 cos x dx Let u D sin x du D cos x dx Z Z D 56. We have 1 p 7 ln.3 C x 2 / x dx 3 C x2 Du 60. Z Z D 59. therefore ! dx p C 4 xC 7 p 1 x C 4 C p ln jx C 7j C 4 2 7 2 I D ! 2 dx 7 !Z Downloaded by ted cage (sxnbyln180@questza.com) 259 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390) p ADAMS and ESSEX: CALCULUS 9 p x 2 C6xC10 x 1 63. Z e x .1 Z p x 3 dx Let x D 2 tan p .x 2 C 2/7=2 dx D 2 sec2 d p Z p 3 2 2 tan 2 sec2 d D p 7 Z 8 2 sec 1 sin3 cos2 d D p 2 2Z 1 D p .1 cos2 / cos2 sin d Let u D cos 2 2 sin d 5 du3 D Z 1 u 1 u 4 2 CC D p .u u / du D p 3 2 20 2 2 5 !5 !3 1 p p 1 @1 1 2 2 ACC D p p p 3 2 2 5 2 C x2 2 C x2 2 5.2 C x 2 /5=2 260 Telegram: @uni_k 2 1 C C: 3.2 C x 2 /3=2 ! Z 3 1 1C 2 dx D dx 2x 2 3 2 2x 3 ! p Z 1 x 3 1 p p dx D C p p 2 4 2x 3 2x C 3 ˇ ˇ p ˇ p p x 3 ˇˇ 2x 3ˇ D C p ln ˇ p p ˇ C C: ˇ 2 4 2 2x C 3 ˇ 64. Z 65. Z 66. We have e 2x /5=2 dx Let e x D sin u e x dx D cos u du 3 Z Z 1 D cos6 u du D .1 C cos 2u/3 du 2 Z 1 D .1 C 3 cos 2u C 3 cos2 2u C cos3 2u/ du 8 Z 3 3 u sin 2u C .1 C cos 4u/ duC D C 8 16 16 Z 1 .1 sin2 2u/ cos 2u du 8 5u 3 3 sin 2u D C sin 2u C sin 4u C 16 16 64 16 1 3 sin 2u C C 48 5 1 D sin 1 .e x / C sinŒ2 sin 1 .e x /C 16 4 1 3 1 x sinŒ4 sin .e / sin3 Œ2 sin 1 .e x / C C 64 48 5 1 p D sin 1 .e x / C e x 1 e 2x 16 2 3 xp C e 1 e 2x 1 2e 2x 16 3=2 1 3x e 1 e 2x C C: 6 D p Fig. RT-63 Fig. RT-61 62. 2Cx 2 xC3 x2 x 1=2 dx Let x D u6 1 C x 1=3 dx D 6u5 du Z 8 u D6 du u2 C 1 Z 8 u C u6 u6 u4 C u4 C u2 u2 1 C 1 du D6 u2 C 1 Z 1 D6 du u6 u4 C u2 1 C 2 u C1 7 u5 u3 u C u C tan 1 u C C D6 7 5 3 p 6 6 5=6 D x 7=6 x C 2 x 6x 1=6 C 6 tan 1 x 1=6 C C: 7 5 Z dx x.x 2 C x C 1/1=2 Z dx D 1 2 xŒ.x C 2 / C 43 1=2 p p 1 3 Let x C D tan 2 p2 3 sec2 d dx D 2 3 sec2 d 2 D p p 3 3 1 tan sec 2 2 2 Z Z d 2 sec d D2 p p D 3 tan 1 3 sin cos Z p 3 sin C cos d D2 3 sin2 cos2 Z p Z cos d sin d D2 3 C 2 2 2 2 3 sin cos 3 sin cos2 Z p Z sin d cos d D2 3 C2 3 4 cos2 4 sin2 1 Let u D cos , du D sin d in the first integral; Z Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390) let v D sin , dv D cos d in the second integral. Z p Z du dv C 2 D 2 3 3 4u2 4v 2 1 p Z Z 3 1 du du D 3 1 2 2 2 u v2 4 p ˇ ˇ 4 ˇ ˇ p ˇ cos C 3 ˇ ˇ ˇ 3 1 2 2 p ln ˇˇ p ˇˇ D 2 2 3 3ˇ ˇ ˇ cos ˇ 2 ˇ ˇ 1ˇ ˇ ˇ sin C ˇ 1 1 ˇ 2 ˇˇ C C .2/ ln ˇ 1ˇ ˇ 2 2 ˇ sin ˇ 2 ˇ p ˇ ˇ 1 ˇˇ 3 ˇ cos sin ˇ 1 ˇˇ 2 2 ˇ p D ln ˇ ˇ C C: ˇ 2 ˇ ˇ cos C 3 sin C 1 ˇ ˇ 2 2 ˇ 69. 70. p 3 2x C 1 and cos D p , Since sin D p 2 x2 C x C 1 2 x2 C x C 1 therefore Z 67. 68. ˇ ˇ p 1 ˇˇ .x C 2/ 2 x 2 C x C 1 ˇˇ dx D ln ˇ p ˇCC: 2 ˇ .x C 2/ C 2 x 2 C x C 1 ˇ x.x 2 C x C 1/1=2 71. Z 1Cx p dx Let x D u2 1C x dx D 2u du Z u.1 C u2 / du D2 1Cu Z 3 u C u2 u2 u C 2u C 2 2 D2 du 1Cu Z 2 D2 u2 u C 2 du 1Cu 3 u2 u C 2u 2 ln j1 C uj C C D2 3 2 p p 2 D x 3=2 x C 4 x 4 ln.1 C x/ C C: 3 Z x dx Let u D x 2 4x 4 C 4x 2 C 5 du D 2x dx Z 1 du D 2 4u2 C 4u C 5 Z 1 du Let w D 2u C 1 D 2 .2u C 1/2 C 4 dw D 2du Z w dw 1 1 D tan 1 CC D 2 4 w C4 8 2 1 1 C C: D tan 1 x 2 C 8 2 72. Z x dx Let u D x 2 4 .x 2 4/2 du D 2x dx Z 1 1 du D D CC 2 u2 2u 1 1 D CC D C C: 2.x 2 4/ 2x 2 8 Use the partial fraction decomposition A Bx C C 1 D C 2 x3 C x2 C x x x CxC1 A.x 2 C x C 1/ C Bx 2 C C x D x3 C x2 C x ( ACB D0 ) A C C D 0 ) A D 1; B D 1; C D 1: AD1 Therefore, Z dx x3 C x2 C x Z Z xC1 dx dx Let u D x C 21 D x x2 C x C 1 du D dx Z u C 12 D ln jxj du u2 C 43 2x C 1 1 1 2 C C: ln x C x C 1 p tan 1 p D ln jxj 2 3 3 Z U D tan 1 x d V D x 2 dx dx x3 dU D V D 1 C x2 3 Z x3 1 x 3 dx 1 D tan x 3 3 1 C x2 Z 3 3 x Cx x 1 x tan 1 x dx D 3 3 x2 C 1 x3 1 2 1 D tan 1 x x C ln.1 C x 2 / C C: 3 6 6 Z e x sec.e x / dx Let u D e x du D e x dx Z D 73. x 2 tan 1 x dx sec u du D ln j sec u C tan uj C C D ln j sec.e x / C tan.e x /j C C: Z x 2 dz dx Let z D tan ; dx D I D 4 sin x 3 cos x 2 1 C z2 2z 1 z2 ; sin x D cos x D 1 C z2 1 C z2 2 dz Z 1 C z2 D 8z 3 3z 2 2 1 C z2 Z Z1Cz dz dz D2 D2 : 2 3z C 8z 3 .3z 1/.z C 3/ Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 261 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390) .3z Thus 74. 75. 1 A B D C 1/.z C 3/ 3z 1 zC3 Az C 3A C 3Bz B D .3z 1/.z C 3/ n A D 3=10 A C 3B D 0 ) ) B D 1=10: 3A B D 1 Z Z 3 dz dz 1 I D 5 3z 1 5 zC3 1 1 D ln j3z 1j ln jz C 3j C C 5 ˇ 5 ˇ 1 ˇ 3 tan 1 .x=2/ 1 ˇˇ C C: D ln ˇˇ 5 tan 1 .x=2/ C 3 ˇ Z dx x 1=3 1 D 3 1=3 .x 2 Let x D .u C 1/3 dx D 3.u C 1/2 du ! Z Z 1 .u C 1/2 uC2C D3 du D 3 du u u ! u2 C 2u C ln juj C C D3 2 Z 1/2 C 6.x 1=3 dx tan x C sin x Z cos x dx D sin x.1 C cos x/ 1/ C 3 ln jx 1=3 Let z D tan.x=2/; 1 z2 ; cos x D 1 C z2 x sin2 x 2 D 1 cos x ; tan2 D x 2 1 C cos x 2 cos 2 the answer can also be written as ˇ ˇ 1 ˇˇ 1 cos x ˇˇ 1 1 cos x ln ˇ C C: 4 1 C cos x ˇ 4 1 C cos x Telegram: @uni_k 77. dx D 1 z 2 2 dz 1 C z2 1 C z2 D 1 z2 2z 1 C 1 C z2 1 C z2 Z Z 2 .1 z / dz 1 z2 1 D D dz z.1 C z 2 C 1 z 2 / 2 z 1 z2 D ln jzj CC 2 4 ˇ ˇ xˇ 1 x 2 1 ˇ tan C C: D ln ˇtan ˇ 2 2 4 2 Remark: Since p D 1 4 D 1j C C: 2 dz 1 C z2 2z sin x D 1 C z2 Z D 78. Z 262 76. 79. Z ADAMS and ESSEX: CALCULUS 9 x dx 3 Z 4x 4x 2 D Z p 4 x dx .2x C 1/2 Let u D 2x C 1 du D 2 dx u 1 du p 4 u2 u 1 1p 4 u2 sin 1 CC 4 4 2 ! 1 1p 1 1 2 3 4x 4x xC sin C C: 4 4 2 p x dx Let x D u2 1Cx dx D 2udu Z Z 1 u2 du du D 2 1 D2 1 C u2 1 C u2 p p D 2 u tan 1 u C C D 2 x 2 tan 1 x C C: Z p 1 C e x dx Let u2 D 1 C e x 2u du D e x dx! Z Z 2 2 2u du D du 2C 2 D u2 1 u 1 ! Z 1 1 D 2C du u 1 uC1 ˇ ˇ ˇu 1ˇ ˇCC D 2u C ln ˇˇ u C 1ˇ ˇp ˇ ˇ 1 C ex 1 ˇ p ˇ ˇ x D 2 1 C e C ln ˇ p ˇ C C: ˇ 1 C ex C 1 ˇ I D Z x 4 dx D x3 8 Z xC 8x x3 8 dx: 8x A Bx C C D C 2 x3 8 x 2 x C 2x C 4 Ax 2 C 2Ax C 4A C Bx 2 2Bx C C x D x 3 8( ( ACB D0 BD A ) 2A 2B C C D 8 ) C D 2A 6A D 8 4A 2C D 0 Thus A D 4=3, B D 2C 4=3, C D 8=3. We have Z Z dx x 2 4 4 x2 C dx 2 3 x 2 3 x 2 C 2x C 4 Z x2 4 4 xC1 3 D C ln jx 2j dx 2 3 3 .x C 1/2 C 3 x2 4 2 D C ln jx 2j ln.x 2 C 2x C 4/ 2 3 3 4 xC1 C p tan 1 p C K: 3 3 I D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL OTHER REVIEW EXERCISES 6 (PAGE 391) 80. By the procedure used in Example 4 of Section 7.1, Z e x cos x dx D 21 e x .sin x C cos x/ C C; Z e x sin x dx D 12 e x .sin x cos x/ C C: Now Z 3. Z =2 csc x dx D lim c!0C 0 c!0C xe x cos x dx Z 1 dx D lim R!1 x C x3 Z 1 p 1 D 12 xe x .sin C cos x/ D 21 xe .sin x C cos x/ 1 x e sin x C C: 2 5. i d xh e .ax C b/ cos x C .cx C d / sin x dx h D e x .ax C b/ cos x C .cx C d / sin x C a cos x C c sin x i .ax C b/ sin x C .cx C d / cos x h D e x .a C c/x C b C a C d cos x i C .c a/x C d C c b sin x If a C c D 1, b C a C d D 0, c a D 0, and d C c b D 0, then a D c D d D 1=2 and b D 0. Thus Z i ex h x cos x C .x 1/ sin x C C: I D xe x cos x dx D 2 If a C c D 0, b C a C d D 0, c a D 1, and d C c b D 0, then b D c D a D 1=2 and d D 0. Thus Z i ex h J D xe x sin x dx D x sin x .x 1/ cos x C C: 2 2. c!0C Let x D u2 dx D 2u du x ln x dx Z 1 0 c 6. D 4 lim c 3 ln c 3 c!0C Z 1 p 0 dx x 1 Therefore 7. x r e x dx Z R x r e x dx D lim c 8. x2 Z 1 > 1 x Z 1 0 p 4 9 dx D 1 (diverges) x dx 1 c3/ D x2 diverges: Z 1 R 60 Volume = 0 0 A.x/ dx. The approximation is 10 h 10; 200 C 2.9; 200 C 8; 000 C 7; 100 2 i C 4; 500 C 2; 400/ C 100 364; 000 m3 . Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 4 .1 9 Z 1 Z 1 dx D I1 C I2 C p x D xe 1 0 0 Z 1 Z 1 dx dx I1 D p x < p D2 xe x 0 0 Z 1 Z 1 dx 1 e x dx D I2 D p x < e xe 1 1 Thus I converges, and I < 2 C .1=e/. I D T6 D because limR!1 Rr e R D 0 for any r. In order to ensure that limc!0C c r e c D 0 we must have limc!0C c r D 0, so we need r > 0. dx U D ln u d V D u2 du du u3 dU D V D u 3 0 1 ˇ1 Z 1 ˇ u3 1 ˇ D 4 lim @ ln uˇ u2 duA ˇ c!0C 3 3 c 0 d V D e x dx U D xr r 1 d U D rx dr V D e x ˇR Z 1 ˇ ˇ D lim x r e x ˇ C r x r 1 e x dx c!0C ˇ 0 R!1 c Z 1 r c D lim c e C r x r 1 e x dx 0 Z 1 c!0C R!1 x 1 C x2 u.2 ln u/2u du Z 1 D4 u2 ln u du Other Review Exercises 6 (page 391) 1. 1 1 x ˇˇR 1 ˇ 2 ln.1 C x / ˇ ˇ R!1 2 1 ln 2 R2 1 C ln 2 D ln D lim R!1 2 1 C R2 2 0 D Z R D lim ln jxj d V D e x cos x dx V D 21 e x .sin x C cos x/ Z D 21 xe x .sin C cos x/ 12 e x .sin x C cos x/ dx U Dx d U D dx cos x C sin x C cos x/ C C c D lim ln j csc c C cot cj D 1 (diverges) 4. 1 x 4 e .sin x x ˇ=2 ˇ ˇ ln j csc x C cot xjˇ ˇ Downloaded by ted cage (sxnbyln180@questza.com) 263 lOMoARcPSD|6566483 www.konkur.in OTHER REVIEW EXERCISES 6 (PAGE 391) 9. S6 D ADAMS and ESSEX: CALCULUS 9 10 h 10; 200 C 4.9; 200 C 7; 100 C 2; 400/ 3 i C 2.8; 000 C 4; 500/ C 100 Integrating both sides over Œ0; 1 leads at once to Z 1 3 10. 367; 000 m Z 1p 2 C sin.x/ dx I D 0 h p p 1 p 2 C 2. 2 C sin.=4/ C 2 C sin.=2/ T4 D 8 p p i C 2 C sin.3=4/ C 2 0 13. 0 Z 1 I x 4 .1 x/4 dx > : x2 C 1 2 0 Thus I > .22=7/ 4T8 T4 D 5:504. 3 c) Yes, S4 D S8 suggests that Sn may be independent of n, which is consistent with a polynomial of degree not exceeding 3. > I =2, or 22 7 T8 D b) If T8 D 5:5095, then S8 D : on .0; 1/, we have 1:626765 I 1:6 12. 22 7 x 4 .1 x/4 22 > 0 on .0; 1/, > 0, and so x2 C 1 7 22 . < 7 Z 1 x 4 .1 x/4 dx, then since 1 < x 2 C 1 < 2 b) If I D I > 1 .T4 C M4 / 1:617996 2 1 S8 D .T4 C 2M4 / 1:62092 3 I 1:62 Z 1 x2 dx Let x D 1=t I D 5 3 1=2 x C x C 1 dx D .1=t 2 / dt Z 2 Z 2 4 t dt .1=t / dt D D 5 / C .1=t 3 / C 1 5 C t2 C 1 .1=t t 0 0 T4 0:4444 M4 0:4799 T8 0:4622 M8 0:4708 S8 0:4681 S16 0:4680 I 0:468 to 3 decimal places 0:730 2:198 C 1:001 C 1:332 C 1:729 C a) T4 D 1 2 2 D 5:526 1 S4 D 0:730 C 2:198 C 4.1:001 C 1:729/ C 2.1:332/ 3 D 5:504: 4tan 1 1 D Since 1:609230 p 1 hp M4 D 2 C sin.=8/ C 2 C sin.3=8/ 4 i p p 2 C sin.5=8/ C 2 C sin.7=8/ 11. x 4 .1 x/4 22 dx D x2 C 1 7 c) I D R1 0 .x 8 4x 7 C 6x 6 22 7 2. a) In D Z I << 22 7 I : 2 4x 5 C x 4 / dx D 1 22 < < 630 7 1 . Thus 630 1 : 1260 x 2 /n dx .1 d V D dx U D .1 x 2 /n d U D 2nx.1 x 2 /n 1 dx V Dx Z D x.1 x 2 /n C 2n x 2 .1 x 2 /n 1 dx Z D x.1 x 2 /n 2n .1 x 2 1/.1 x 2 /n 1 dx D x.1 x 2 /n 2nIn C 2nIn 1 ; so 2n 1 x.1 x 2 /n C In 1 : In D 2n C 1 2n C 1 Z 1 b) Let Jn D .1 x 2 /n dx. Observe that J0 D 1. By 0 (a), if n > 0, then we have ˇ1 2n x.1 x 2 /n ˇˇ 2n Jn 1 D Jn 1 : Jn D ˇ C 2n C 1 ˇ 2n C 1 2n C 1 0 Challenging Problems 6 (page 391) Therefore, 1. a) Long division of x 2 C 1 into x 4 .1 x/4 D x 8 4x 7 C 6x 6 4x 5 C x 4 yields x 4 .1 x/4 D x6 x2 C 1 264 Telegram: @uni_k 4x 5 C 5x 4 4x 2 C 4 2n 2n 2 4 2 J0 2n C 1 2n 1 5 3 Œ.2n/.2n 2/ .4/.2/2 22n .nŠ/2 D D : .2n C 1/Š .2n C 1/Š Jn D 4 : x2 C 1 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 6 (PAGE 391) c) From (a): 5. In 1 D 2n C 1 In 2n 1 x.1 2n x 2 /n : Thus Z x 2 / 3=2 dx D I 3=2 .1 2. 1=2/ C 1 I 1=2 1 x : D p 1 x2 D 3. 1 x.1 1 x 2 / 1=2 4. Im;n D n x .ln x/ dx 0 Z 1 0 0 In D nŠ @1 1 n 1X 1 A e jŠ j D0 holds for n D 0 by part (b). Assume that it holds for some integer n D k 0. Then by (b), 0 1 k X 1 1 1 A 1 D .k C 1/kŠ @1 IkC1 D .k C 1/Ik e e jŠ e j D0 0 1 k X 1 1 1 A D .k C 1/Š @1 e j Š e.k C 1/Š j D0 1 0 kC1 X 1 1 A: D .k C 1/Š @1 e jŠ j D0 Thus the formula holds for all n 0, by induction. d) Since limn!1 In D 0, we must have 0 1 n 1X 1 A D 0: lim @1 n!1 e jŠ j D0 Thus e D lim n!1 6. Z 1 . 1/n un e u du .m C 1/n 0 . 1/n .n C 1/ (see #50 in Section 7.5) D .m C 1/n . 1/n nŠ D : .m C 1/n if n 1 c) The formula I D Z 1 e 0 Kx n X 1 j D0 jŠ . ˇ1 1 e Kx ˇˇ 1 dx D ˇ D K ˇ K 0 1 . eK For very large K, the value of I is very small (I < 1=K). However, D 1 1 .1 C / > 100 100 1 1 S100 D .1 C / > 300 300 1 1 .e K=200 C / < : M100 D 100 100 T100 D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 e D nIn 1 t Let u D .m C 1/t du D .m C 1/ dt x n dx D 0 Let x D e dx D e t dt e mt . t /n e t dt 0 Z 1 t n e .mC1/t dt D . 1/n D Z 1 d V D e x dx U D xn n 1 d U D nx dx V D e x ˇ1 Z ˇ 1 ˇ D xn e x ˇ C n x n 1 e x dx ˇ 0 p p 2xC1/.x 2 C 2xC1/. b) x 4 C1 D .x 2 C1/2 2x 2 D .x 2 Thus Z 2 Z x C1 x2 C 1 p p dx D 4 x C1 .x 2 2x C 1/.x 2 C 2x C 1/ Z 1 1 1 p C p dx D 2 2 2 2x C 1 x C 2x C 1 1 0x Z 1 B 1 1 C D C A dx @ 2 2 2 1 1 3 3 p p x xC C4 C4 2 2 p p ! 2 2 1 2x 2x C 1 1 p p D p tan C C: C tan 3 3 3 m x n e x dx < 0 a) x 4 Cx 2 C1 D .x 2 C1/2 x 2 D .x 2 xC1/.x 2 CxC1/. Thus Z Z x2 C 1 x2 C 1 D dx x4 C x2 C 1 .x 2 x C 1/.x 2 C x C 1/ Z 1 1 1 D C 2 dx 2 x2 x C 1 x CxC1 ! Z 1 1 1 C dx D 2 2 2 x 21 C 43 x C 21 C 34 1 2x C 1 2x 1 C tan 1 p D p tan 1 p C C: 3 3 3 Z 1 Z 1 1 , nC1 0 0 x because 0 < e < 1 on .0; 1/. Thus limn!1 In D 0 by the Squeeze Theorem. ˇ1 Z 1 ˇ 1 x xˇ e dx D e ˇ D 1 b) I0 D ˇ e 0 0 Z 1 In D x n e x dx a) 0 < In D Downloaded by ted cage (sxnbyln180@questza.com) 265 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 6 (PAGE 391) In each case the represent terms much less than the first term (shown) in the sum. Evidently M100 is smallest if k is much greater than 100, and is therefore the best approximation. T100 appears to be the worst. 7. ADAMS and ESSEX: CALCULUS 9 8. a) f 0 .x/ < 0 on Œ1; 1/, and limx!1 f .x/ D 0. Therefore Z 1 a) Let f .x/ D Ax 5 C Bx 4 C C x 3 C Dx 2 C Ex C F . Then 5 Z h Bh Dh3 C C Fh : f .x/ dx D 2 5 3 h 1 f 0 .x/ dx Z R lim f 0 .x/ dx 1 D R!1 1 D lim .f .1/ R!1 Also h i 2h af . h/ C bf . h=2/ C cf .0/ C bf .h=2/ C af .h/ D 2 a 2Bh5 C 2Dh3 C 2F 2Bh5 2Dh3 Cb C C 2F C cF h : 16 4 These expressions will be identical if the coefficients of like powers of h on the two sides are identical. Thus Z 1 jf 0 .x/j dx D f .R// D f .1/: Thus ˇZ 1 ˇ Z 1 ˇ ˇ 0 ˇ f .x/ cos x dx ˇˇ jf 0 .x/j dx ! 0 as R ! 1: ˇ R R Z R Thus lim R!1 1 b) Z 1 f 0 .x/ cos x dx exists. f .x/ sin x dx 1 2a C 1 2b D ; 16 5 2a C 2b 1 D ; 4 3 2a C 2b C c D 1: Solving these equations, we get a D 7=90, b D 16=45, and c D 2=15. The approximation for the integral of any function f on Œm h; m C h is " Z mCh 7 16 f .x/ dx 2h f .m h/ C f .m 21 h/ 90 45 m h # 16 7 2 1 C f .m/ C f .m C 2 h/ C f .m C h/ : 15 45 90 b) If m D h D 1=2, we obtain " Z 1 7 0 16 1=4 2 x e dx 1 e C e C e 1=2 90 45 15 0 # 7 1 16 3=4 C e C e 45 90 0:63212087501: 1 the integral converges. c) f .x/ D 1=x satisfies the conditions of part (a), so Z 1 1 16 3=8 7 e C e 1=2 45 45 # 2 16 7 16 C e 5=8 C e 3=4 C e 7=8 C e 1 45 15 45 90 Telegram: @uni_k converges by part (b). Similarly, it can be shown that Z 1 1 Z 1 1 cos.2x/ dx x converges: j sin xj dx x Z 1 1 cos.2x//, we have 1 cos.2x/ : 2x R1 The latter integral diverges R 1 because 1 .1=x/ dx diverges to infinity while 1 .cos.2x//=.2x/ dx converges. Therefore Z 1 1 0:63212055883: 266 sin x dx x But since j sin xj sin2 x D 12 .1 With two intervals having h D 1=4 and m D 1=4 and m D 3=4, we get " Z 1 1 7 0 16 1=8 2 x e dx e C e C e 1=4 2 90 45 15 0 C U D f .x/ d V D sin x dx d U D f 0 .x/ dx V D cos x ˇR Z ˇ 1 ˇ D lim f .x/ cos x ˇ C f 0 .x/ cos x dx ˇ R!1 1 1 Z 1 D f .1/ cos.1/ C f 0 .x/ cos x dxI Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) j sin xj dx x diverges to infinity. lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.1 (PAGE 401) CHAPTER 7. APPLICATIONS OF INTEGRATION 3. By slicing: V D Section 7.1 Volumes by Slicing—Solids of Revolution (page 401) 1. D By slicing: Z 1 D 2 x2 2 x 4 / dx ˇ1 x 5 ˇˇ 3 cu. units. ˇ D 5 ˇ 10 0 4 V D 2 By shells: V D 2 .x 0 By shells: cu. units. x dx D V D 5 0 Z 1 Z 1 y.1 y/ dy !ˇ1 2y 5=2 ˇˇ cu. units. ˇ D ˇ 5 5 y2 2 p y. y 0 2y 5=2 5 D 2 p 0 Z 1 0 y p yD x 0 y y 2 / dy !ˇ1 y 4 ˇˇ 3 cu. units. ˇ D 4 ˇ 10 .1;1/ yDx 2 yDx 2 x x x x Fig. 7.1-3 4. Fig. 7.1-1 Slicing: 2. Slicing: V D Z 1 0 D y Shells: V D .1 y/ dy ˇ1 1 2 ˇˇ y ˇ D cu: units: ˇ 2 2 0 Z 1 x 3 dx 0 4 ˇˇ1 x ˇ cu: units: D 2 ˇ D ˇ 4 2 V D 2 D Z 1 .y 0 1 2 y 2 Shells: V D 2 D 2 0 Z 1 y y 4 / dy ˇ1 1 5 ˇˇ 3 y ˇ D cu: units: ˇ 5 10 0 x.x 1=2 0 2 5=2 x 5 x 2 / dx ˇ1 1 4 ˇˇ 3 x ˇ D cu: units: ˇ 4 10 0 y .1;1/ p yD x yDx 2 yDx 2 1 x Fig. 7.1-4 Fig. 7.1-2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x Downloaded by ted cage (sxnbyln180@questza.com) 267 lOMoARcPSD|6566483 www.konkur.in SECTION 7.1 (PAGE 401) ADAMS and ESSEX: CALCULUS 9 y 5. a) About the x-axis: .1;1/ V D D D Z 2 x 2 .2 x/2 dx yDx 0 Z 2 .4x 2 0 yDx 2 4x 3 C x 4 / dx ˇ2 16 x 5 ˇˇ 4 x C cu. units. ˇ D 5 ˇ 15 4x 3 3 x Fig. 7.1-6 0 b) About the y-axis: 7. V D 2 D 2 Z 2 x 2 .2 0 2x 3 3 x/ dy ˇ2 x 4 ˇˇ 8 cu. units. ˇ D ˇ 4 3 a) About the x-axis: V D 2 y b) About the y-axis: yD2x x 2 y 2 y/ dy ˇ3 27 y 4 ˇˇ cu. units. ˇ D 4 ˇ 2 y.4y 0 D 2 y 3 0 (a) Z 3 0 Z 3h i .4y y 2 /2 y 2 dy 0 Z 3 D .15y 2 8y 3 C y 4 / dy 0 ˇ3 y 5 ˇˇ 108 D 5y 3 2y 4 C cu. units. ˇ D 5 ˇ 5 V D 2 x 0 y y (b) yD2x x 2 .3;3/ xDy 2 x xD4y y 2 Fig. 7.1-5 x 6. Rotate about Fig. 7.1-7 a) the x-axis V D D Z 1 .x 2 0 1 3 x 3 b) the y-axis V D 2 D 2 268 Telegram: @uni_k Z 1 x 4 / dx ˇ1 1 5 ˇˇ 2 x ˇ D cu: units: ˇ 5 15 x.x 0 1 3 x 3 8. Rotate about a) the x-axis 0 x 2 / dx ˇ1 1 4 ˇˇ x ˇ D cu: units: ˇ 4 6 0 V D Z 0 Œ.1 C sin x/2 0 .2 sin x C sin2 x/ dx D Z D 2 cos x C x 2 1 D 4 C 2 cu: units: 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 dx sin 2x 4 ˇˇ ˇ ˇ ˇ 0 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.1 (PAGE 401) y b) the y-axis .1=3;3/ V D 2 Z x sin x dx 3xC3yD10 0 U Dx d V D sin x dx d U D dx ˇ V D cos x ˇ Z ˇ D 2 x cos x ˇ C cos x dx ˇ 0 1 yD x x 0 Fig. 7.1-10 D 2 2 cu: units: 9. a) About the x-axis: Z 1 V D 4 0 D 4 D 4 11. 1 .1 C x 2 /2 Z =4 0 Z =4 Let x D tan dx D sec2 d dx sec2 d sec4 .3;1=3/ Z 1 V D 2 2 .2 x/.1 x/ dx 0 Z 1 D 4 .2 3x C x 2 / dx 0 ˇ1 3x 2 x 3 ˇˇ 10 D 4 2x C cu. units. ˇ D 2 3 ˇ 3 0 cos2 d y 0 ˇ=4 ˇ . C sin cos /ˇˇ D 4 2 0 2 15 2 D 4 D cu. units. 8 4 4 8 b) About the y-axis: Z 1 1 V D 2 dx x 2 1 C x2 0 ˇˇ1 1 ˇ D 2 x 2 ln.1 C x 2 / ˇ ˇ 2 0 1 ln 2 D 2 ln 2 cu. units. D 2 1 2 y xCyD1 xD2 x x Fig. 7.1-11 12. V D Z 1 D x y 1 Œ.1/2 .x 2 /2 dx ˇ1 1 5 ˇˇ x ˇ ˇ 5 1 8 cu: units: D 5 yD2 y yD 1 1Cx 2 yD1 x2 yD1 x 2 x 1 dx x x Fig. 7.1-9 10. By symmetry, rotation about the x-axis gives the same volume as rotation about the y-axis, namely Z 3 1 10 x dx V D 2 x 3 x 1=3 ˇ ˇ3 5 2 1 3 ˇ D 2 x x x ˇ ˇ 3 3 1=3 512 cu: units: D 81 Fig. 7.1-12 13. The volume remaining is Z 2 p x 4 Let u D 4 x 2 du D 2x dx ˇ3 Z 3 ˇ p p 4 3=2 ˇ u du D D 2 u ˇ D 4 3 cu. units. ˇ 3 0 V D 2 2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x Downloaded by ted cage (sxnbyln180@questza.com) x 2 dx 1 0 269 lOMoARcPSD|6566483 www.konkur.in SECTION 7.1 (PAGE 401) ADAMS and ESSEX: CALCULUS 9 32 4 3 2 D cu. units., 3 3 p 32 4 3 cu. units. therefore the volume removed is 3 The percentage removed is Since the volume of the ball is p 4 3 32 3 The volume remaining is Z b V D 2 xh 1 x dx b a ˇ 2 b x x 3 ˇˇ D 2h ˇ 2 3b ˇ a a3 2 2 2 h b 2 a / D h.b 3 b 3 1 2a D h b 2 3a2 C cu. units. 3 b p ! 3 3 35: 8 100 D 100 1 32 3 15. About 35% of the volume is removed. y y p yD 4 x 2 h dx 1 2 x x x y C D1 b h xDa dx b x Fig. 7.1-13 14. The radius of the hole is the remaining volume is V D q 16. Z L=2 " R2 D 2 x2 L=2 L2 x 4 Fig. 7.1-15 1 2 L . 4 R2 1 3 x 3 !ˇL=2 ˇ ˇ ˇ ˇ R2 Thus, by slicing, L2 4 # dx 0 D L3 cu. units (independent of R). 6 y x Let a circular disk with radius a have centre at point .a; 0/. Then the disk is rotated about the y-axis which is one of its tangent lines. The volume is: Z 2a p V D 2 2 x a2 .x a/2 dx Let u D x a 0 du D dx Z a p 2 2 D 4 .u C a/ a u du Z aa p Z ap u a2 u2 du C 4a D 4 a2 u2 du a a 1 D 0 C 4a a2 D 2 2 a3 cu: units: 2 (Note that the first integral is zero because the integrand is odd and the interval is symmetric about zero; the second integral is the area of a semicircle.) p yD R2 x 2 y R q L 2 R2 L2 4 .x a/2 Cy 2 Da2 x 2a a L Fig. 7.1-16 Fig. 7.1-14 270 Telegram: @uni_k Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 17. Volume of the smaller piece: Z a V D .a2 x 2 / dx b ˇa x 3 ˇˇ 2 D a x ˇ 3 ˇ b a3 2 D a .a b/ D .a 3 D .a 3 b/Œ3a 2 SECTION 7.1 (PAGE 401) 19. The volume of the ellipsoid is V D 2 b3 3 2 b 1 2 x 0 D 2b x2 dx a2 ˇa x 3 ˇˇ 4 ˇ D ab 2 cu. units. 3a2 ˇ 3 Z a 2 0 y .a C ab C b 2 / yDb b b/2 .2a C b/ cu. units. q 1 x2 a2 y p yD a2 x 2 dx a x dx b x a x x Fig. 7.1-19 20. Fig. 7.1-17 18. Let the centre of the bowl be at (0, 30). Then the volume of the water in the bowl is Z 20 h i V D 302 .y 30/2 dy 0 Z 20 D 60y y 2 dy 0 ˇ20 1 3 ˇˇ D 30y 2 y ˇ ˇ 3 Z ah p .b C a2 y 2 /2 Z aa p 4b a2 y 2 dy D 2 V D .b p a2 i y 2 /2 dy 0 a2 D 8b D 2 2 a2 b cu. units.: 4 We used the area of a quarter-circle of radius a to evaluate the last integral. 0 29322 cm3 : y The cross-section at height p y is an annulus (ring) having inner radius b a2 y 2 and outer radius p 2 2 bC a y . Thus the volume of the torus is 21. a) Volume of revolution about the x-axis is Z 1 e 2x dx ˇR e 2x ˇˇ cu. units. D lim ˇ D R!1 2 ˇ 2 V D 30 20 0 0 b) Volume of revolution about the y-axis is x 2 C.y 30/2 D302 x V D 2 Fig. 7.1-18 Z 1 D 2 lim . xe x R!1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k xe x dx 0 Downloaded by ted cage (sxnbyln180@questza.com) ˇR ˇ e x /ˇˇ D 2 cu. units. 0 271 lOMoARcPSD|6566483 www.konkur.in SECTION 7.1 (PAGE 401) ADAMS and ESSEX: CALCULUS 9 y 25. 1 x yDe dx x x The volume os S is still the constant cross-sectional area a2 =2 times the height b, that is, V D a2 b=2 cm3 . Fig. 7.1-21 26. 22. The volume is V D Z 1 x 2k 1 ˇR x 1 2k ˇˇ dx D lim ˇ R!1 1 2k ˇ R1 2k C : R!1 1 2k 2k 1 2k < 0; that is; k> 1 : 2 D 3072 D 3072 D 3072 y 27. k x.4 x 2=3 /3=2 dx Z =2 sin5 u cos4 u du 0 D 3072 R1 23. The volume is V D 2 1 x 1 k dx. This improper integral converges if 1 k < 1, i.e., if k > 2. The solid has finite volume only if k > 2. yDx Z 8 D 3072 In order for the solid to have finite volume we need 1 The region is symmetric about x D y so has the same volume of revolution about the two coordinate axes. The volume of revolution about the y-axis is V D 2 1 D lim Since all isosceles right-angled triangles having leg length a cm are congruent, S does satisfy the condition for being a prism given in early editions. It does not satisfy the condition in this edition because one of the line segments joining vertices of the triangular cross-sections, namely the x-axis, is not parallel to the line joining the vertices of the other end of the hypotenuses of the two bases. 0 Z =2 1 Z 1 Telegram: @uni_k v 2 /2 v 4 dv .1 Z 1 Let v D cos u dv D sin u du .v 4 0 1 5 2v 6 C v 8 / dv 2 1 8192 C D cu. units. 7 9 105 4 R3 . Expressing this volume 3 as the “sum” (i.e., integral) of volume elements that are concentric spherical shells having thickness dr and varying radius r, and therefore having surface area kr 2 and volume kr 2 dr, we obtain The volume of the ball is 4 R3 D 3 Z R 0 kr 2 dr D k 3 R : 3 Thus k D 4. A solid consisting of points on parallel line segments between parallel planes will certainly have congruent crosssections in planes parallel to and lying between the two base planes, any solid satisfying the new definition will certainly satisfy the old one. But not vice versa; congruent cross-sections does not imply a family of parallel line segments giving all the points in a solid. For a counterexample, see the next exercise. Thus the earlier, incorrect definition defines a larger class of solids than does the current definition. However, the formula V D Ah for the volume of such a solid is still valid, as all congruent cross-sections still have the same area, A, as the base region. 272 cos2 u/2 cos4 u sin u du 0 x Fig. 7.1-23 24. .1 0 dx x Let x D 8 sin3 u dx D 24 sin2 u cos u du Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) R dr r Fig. 7.1-27 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.1 (PAGE 401) R D sin ˛, so R D .x C h/ sin ˛. xCh Using the result of Exercise #17, the volume of liquid displaced by the ball is 28. Using heights f .x/ estimated from the given graph, we obtain Z 9 2 V D f .x/ dx 1 h 2 3 C 4.3:8/2 C 2.5/2 C 4.6:7/2 C 2.8/2 3 i C 4.8/2 C 2.7/2 C 4.5:2/2 C 32 938 cu. units. Note that V D .R 3 x/2 .2R C x/: We would like to consider V as a function of x for 2R x R since V D 0 at each end of this interval, and V > 0 inside the interval. However, the actual interval of values of x for which the above formulation makes physical sense is smaller: x must satisfy R x h tan2 ˛. (The left inequality signifies nonsubmersion of the ball; the right inequality signifies that the ball is tangent to the glass somewhere below the rim.) We look for a critical point of V , considered as a function of x. (As noted above, R is a function of x.) We have 29. Using heights f .x/ estimated from the given graph, we obtain Z 9 V D 2 xf .x/ dx 1 h 2 1.3/ C 4.2/.3:8/ C 2.3/.5/ C 4.4/.6:7/ C 2.5/.8/ 3 i C 4.6/.8/ C 2.7/.7/ C 4.8/.5:2/ C 9.3/ 1537 cu. units. " dV 0D D 2.R dx 3 30. Using heights f .x/ estimated from the given graph, we obtain Z 9 V D 2 .x C 1/f .x/ dx 1 h 2 2.3/ C 4.3/.3:8/ C 2.4/.5/ C 4.5/.6:7/ C 2.6/.8/ 3 i C 4.7/.8/ C 2.8/.7/ C 4.9/.5:2/ C 10.3/ 1832 cu. units. C .R x/ dR dx 1 .2R C x/ # dR C1 x/ 2 dx 2 dR .4R C 2x C 2R dx Thus 2x/ D 4R C 2x .R x/: R 6R sin ˛ D 3.R C x/ D 3 R C h sin ˛ 2R sin2 ˛ D R sin ˛ C R h sin ˛ h sin ˛ h sin ˛ RD : D cos 2˛ C sin ˛ 1 2 sin2 ˛ C sin ˛ 31. Let the ball have radius R, and suppose its centre is x units above the top of the conical glass, as shown in the figure. (Clearly the ball which maximizes liquid overflow from the glass must be tangent to the cone along some circle below the top of the cone — larger balls will have reduced displacement within the cone. Also, the ball will not be completely submerged.) This value of R yields a positive value of V , and corresponds to x D R.2 sin ˛ 1/. Since sin ˛ sin2 ˛, RxD h sin ˛.2 sin ˛ 1/ h sin2 ˛ D h tan2 ˛: 2 cos2 ˛ 1 C sin ˛ 2 sin ˛ Therefore it gives the maximum volume of liquid displaced. x R 32. h h sec ˛ .hCx/ cos ˛ ˛ Let P be the point .t; 25 t /. The line through P perpendicular to AB has equation y D x C 52 2t , and meets the curve xy D 1 at point Q with x-coordinate s equal to the positive root of s 2 C . 52 2t /s D 1. Thus, " 1 2t sD 2 Fig. 7.1-31 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 5 C 2 s 5 2 2t 2 # C4 : 273 lOMoARcPSD|6566483 www.konkur.in SECTION 7.1 (PAGE 401) ADAMS and ESSEX: CALCULUS 9 y A.1=2;2/ yD Section 7.2 More Volumes by Slicing (page 405) p 2 dt 1 x P xCyD Q 5 2 1. B.2;1=2/ 2. s t x ˇ2 3 2 ˇˇ V D 3x dx D x ˇ D 6 m3 2 ˇ 0 Z 2 A horizontal slice of thickness dz at height a has volume d V D z.h z/ dz. Thus the volume of the solid is V D Fig. 7.1-32 The volume element at P has radius p PQ D 2.t s/ 3 2 s 2 p 5 1 5 2t C 45 D 24 4 2 2 3. 274 Telegram: @uni_k Z h Z 1 p z 1 V D 5. Z 1 0 z3 3 ˇ ˇh 3 ˇ ˇ D h units3 : ˇ 6 ˇ 0 p let u 1 V D Z 6 0 z2 ˇ1 ˇ 2 3=2 ˇˇ u du D u ˇ D units3 : 2 3 3 ˇ 0 2 1 .2 C z/.8 Z 6 z/ dz D .16 C 6z 0 ˇ 6 z 3 ˇˇ ˇ D 132 ft3 3 ˇ z 2 / dz 0 p The area of anequilateral triangle of edge x is p p p p A.x/ D 21 x 23 x D 43 x sq. units. The volume of the solid is V D 7. hz 2 2 ˇ3 x 3 ˇˇ 26 V D x dx D cu. units ˇ D 3 ˇ 3 1 Z 3 D 16z C 3z 2 6. z 2 dz 0 D 2 4. z/ dz D .z.h 0 A horizontal p slice of thickness dz at height a has volume d V D z 1 z 2 dz. Thus the volume of the solid is p and thickness 2 dt . Hence, the volume of the solid is s #2 2 Z 2 "p p 5 1 5 V D 2 2 dt 2t C 4 4 2 2 1=2 1 0s 2 2 p Z 2 25 5 5 @ 4 2t C 4A C D 2 2 16 4 2 1=2 # 2 1 5 2t C 4 dt Let u D 2t 52 4 2 du D 2!dt p Z 3=2 41 5 p u2 du D 2 u2 C 4 C 4 4 3=2 16 !ˇ3=2 ˇ p 41 1 ˇ D 2 u C u3 ˇ ˇ 16 12 3=2 p Z 3=2 p 5 2 u2 C 4 du Let u D 2 tan v 4 3=2 du D 2 sec2 v dv p Z 1 tan .3=4/ p 33 2 5 2 sec3 v dv D 4 tan 1 . 3=4/ p p Z tan 1 .3=4/ 3 33 2 10 2 sec v dv D 4 0 p p 33 2 5 2 sec v tan vC D 4 ˇ 1 ˇtan .3=4/ ˇ ln j sec v C tan vj ˇ ˇ 0 " # p 33 15 D 2 5 C ln 2 0 ln 1 4 16 p 57 5 ln 2 cu: units: D 2 16 0 ˇ4 p p 3 3 2 ˇˇ 15 3 x dx D x ˇ D cu. units. ˇ 4 8 8 Z 4p 1 1 The area of cross-section at height y is A.y/ D 2.1 .y= h// .a2 / D a2 1 2 y sq. units. h The volume of the solid is V D Z h 0 a2 1 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) a2 h y cu. units. dy D h 2 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.2 (PAGE 405) 8. Since V D 4, we have ˇ2 x 4 ˇˇ 4D kx dx D k ˇ D 4k: 4 ˇ 0 Z 2 Thus k D 1. 3 y 0 p 3y x 2y 9. The volume between height 0 and height z is z 3 . Thus z3 D Z z x 2 Cy 2 Dr 2 A.t / dt; 0 Fig. 7.2-12 where A.t / is the cross-sectional area at height t . Differentiating the above equation with respect to z, we get 3z 2 D A.z/. The cross-sectional area at height z is 3z 2 sq. units. 10. This is similar to Exercise 7. We have 4z D Z z 13. The cross-section at distance y from the vertex of the partial cone is a semicircle of radius y=2 cm, and hence area y 2 =8 cm2 . The volume of the solid is V D A.t / dt , 0 Z 12 0 123 2 y dy D D 72 cm3 . 8 24 so A.z/ D 4. Thus the square cross-section at height z has side 2 units. 11. V D2 D8 Z r p 2 r2 Z r 0 0 .r 2 y2 2 r z dy y 2 / dy D 8 r 2 y z p 2 y3 3 y ˇˇr 16r 3 ˇ cu. units. ˇ D ˇ 3 x 0 .12; 12; 0/ Fig. 7.2-13 r 2 y2 14. y p x 12 y xD r2 y2 The volume of a solid of given height h and given crosssectional area A.z/ at height z above the base is given by Z h V D A.z/ dz: 0 If two solids have the same height h and the same area function A.z/, then they must necessarily have the same volume. Fig. 7.2-11 12. The area triangle of base 2y is p of an equilateral p 2 1 3y . Hence, the solid has volume 2 .2y/. 3y/ D V D2 Z rp 0 3.r 2 p D 2 3 r 2x x 2 / dx ˇr 1 3 ˇˇ x ˇ ˇ 3 0 4 D p r 3 cu: units: 3 15. Let the x-axis be along the diameter shown in the figure, with the origin at the centre of the base. The cross-section perpendicular to the x-axis at x is a rectangle having base p aCb a b C x: Thus the 2 r 2 x 2 and height h D 2 2 volume of the truncated cylinder is V D D Z r p .2 r 2 r Z r r p .a C b/ r 2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x2/ Downloaded by ted cage (sxnbyln180@questza.com) aCb a b C x dx 2 2r 2 r .a C b/ cu. units. x 2 dx D 2 275 lOMoARcPSD|6566483 www.konkur.in SECTION 7.2 (PAGE 405) ADAMS and ESSEX: CALCULUS 9 17. Cross-sections of the wedge removed perpendicular to the x-axis are isosceles, right triangles. The volume of the wedge removed from the log is Z 20 1 p . 400 x 2 /2 dx 2 0 ˇ20 16; 000 x 3 ˇˇ cm3 : D 400x ˇ D ˇ 3 3 V D2 h 0 x y yD p r x2 r2 x z Fig. 7.2-15 16. The plane z D k meets the ellipsoid in the ellipse 2 k a b c 2 y2 x " 2 # C " 2 # D 1 k k 2 2 a 1 b 1 c c x 2 that is, C y 2 D1 45ı 20 x x Fig. 7.2-17 which has area " A.k/ D ab 1 2 # k : c The volume of the ellipsoid is found by summing volume elements of thickness d k: 18. The solution is similar to that of Exercise 15 except that the legs of the right-triangular cross-sections are y 10 p p instead of y, and x goes from 10 3 to 10 3 instead of 20 to 20. The volume of the notch is V D2 2 # k dk V D ab 1 c c #ˇc " 1 3 ˇˇ k ˇ D ab k ˇ 3c 2 Z c " D c A.k/ b y p 20 400 0 x 2 dx 4; 000 1; 007 cm3 : 3 One eighth of the region lying inside both cylinders is shown in the figure. If the region is sliced by a horizontal plane at height z, then the intersection is a rectangle with area p p A.z/ D b 2 z 2 a2 z 2 : The volume of the whole region is x V D8 Fig. 7.2-16 Telegram: @uni_k 10/2 dx 20. 2 2 x2 C yb2 C zc 2 D1 a2 k x2 500 x2 The hole has the shape of two copies of the truncated cylinder of Exercise 15, placed base to base, p with a C b D 3 2 in and r D 2 in. Thus the volume of wood p volume of the hole) is p removed (the V D 2.22 /.3 2=2/ D 12 2 in3 . (one-eighth of the solid is shown) a 0 Z 10p3 1 p . 400 2 19. z c Z 10p3 p D 3; 000 3 4 D abc cu: units: 3 276 y p yD 400 x 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) Z bp 0 b2 p z 2 a2 z 2 dz: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.3 (PAGE 412) z 5. b z p b2 z 2 A.z/p a2 z 2 a y 6. x Fig. 7.2-20 21. By the result given in Exercise 18 with a D 4 cm and b D 2 cm, the volume of wood removed is V D8 Z 2p z2 4 p z 2 dz 97:28 cm3 : 16 0 (We used the numerical integration routine in Maple to evaluate the integral.) 7. Section 7.3 Arc Length and Surface Area (page 412) 2 1=3 x ; 3 p 9x 2=3 C 4 4 2=3 dx D ds D 1 C x dx 9 3jxj1=3 p Z 1 9x 2=3 C 4 dx Let u D 9x 2=3 C 4 LD2 3x 1=3 0 du D 6x 1=3 dx Z 13 3=2 p 1 2.13 / 16 units. D u du D 9 4 27 y D x 2=3 ; r y0 D q 2.x C 1/3 D 3.y 1/2 ; y D 1 C 23 .x C 1/3=2 q y 0 D 32 .x C 1/1=2 ; r r 3x C 5 3x C 3 ds D 1 C dx D dx 2 2 ˇ0 p Z 0 ˇ p 2 1 3=2 ˇ 3x C 5 dx D .3x C 5/ ˇ LD p ˇ 9 2 1 1 p 2 3=2 5 23=2 units. D 9 1 x2 1 x3 C ; y0 D 2 12 x 4 x s 2 2 2 x x 1 1 ds D 1 C dx dx D C 4 x2 4 x2 ˇ 3 4 Z 4 2 x 1 1 ˇˇ x C 2 dx D LD ˇ D 6 units. 4 x 12 x ˇ 1 yD 1 1. y D 2x 1; y 0 D 2; ds D Z 3p p LD 5 dx D 2 5 units. p 1 C 22 dx 8. 1 2. y D ax C b, A x B, y 0 D a. The length is LD 3. Z Bp A 1 C a2 dx D p 1 C a2 .B 9. 0 4. 3p y 2 D .x 1/3 ; y D .x 1/3=2 ; y 0 D x 1 2 Z 2p Z 2r 1 9 LD 1 C .x 1/ dx D 9x 5 dx 4 2 1 1 ˇ2 ˇ 133=2 8 1 ˇ .9x 5/3=2 ˇ D units. D ˇ 27 27 1 4x 2 2 1 1 2 dx D x C dx 4x 2 4x 2 ˇ 2 3 Z 2 1 ˇˇ 59 x 1 units. LD x 2 C 2 dx D ˇ D 4x 3 4x ˇ 24 1 x2 1 x ; y0 D 4 2x 2 1 1 x 2 x ds D 1 C dx D C dx 2x 2 2x 2 ˇ e Z e x ln x x 2 ˇˇ 1 C C dx D LD ˇ 2x 2 2 4 ˇ 1 yD ln x s2 1 1 e2 1 e2 C 1 D C D units. 2 4 4 10. If y D x 2 1 ln x then y 0 D 2x 8 1 and 8x 1 2 : 1 C .y 0 /2 D 2x C 8x Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k y 0 D x2 1 A/ units. p p y D 32 x 3=2 ; y 0 D x; ds D 1 C x dx ˇ8 Z 8 ˇ p 52 2 3=2 ˇ units. LD 1 C x dx D .1 C x/ ˇ D ˇ 3 3 0 1 x3 C ; 3 4x s ds D 1 C x 2 yD Downloaded by ted cage (sxnbyln180@questza.com) 277 lOMoARcPSD|6566483 www.konkur.in SECTION 7.3 (PAGE 412) ADAMS and ESSEX: CALCULUS 9 Thus the arc length is given by s Z 2 1 2 1 C 2x sD dx 8x 1 Z 2 1 D 2x C dx 8x 1 ˇ ˇ2 1 1 ˇ D x 2 C ln x ˇ D 3 C ln 2 units: ˇ 8 8 15. 1 11. Z ap Z a 1 C sinh2 x dx D cosh x dx 0 0 ˇa ˇ ea e a ˇ units: D sinh x ˇ D sinh a D ˇ 2 sD ex 1 ; 2x4 ex C 1 e x C 1 .e x C 1/e x .e x y0 D x e 1 .e x C 1/2 2e x D 2x : e 1 The length of the curve is y D ln LD D 13. sD =6 0 Let 2x D tan 2 dx D sec2 d Z 1 xD2 3 sec D 2 xD0 ˇˇxD2 1 D sec tan C ln j sec C tan j ˇˇ 4 xD0 ˇˇ2 p 1 p 2x 1 C 4x 2 C ln.2x C 1 C 4x 2 / ˇˇ D 4 0 p 1 p D 4 17 C ln.4 C 17/ 4 p p 1 D 17 C ln.4 C 17/ units. 4 278 Telegram: @uni_k 16. We have sD ˇ=4 ˇ ˇ D sec x dx D ln j sec x C tan xjˇ ˇ =6 =6 p 2 1 D ln. 2 C 1/ ln p C p 3 3 p 2C1 units: D ln p 3 y D x 2 ; 0 x 2; y 0 D 2x. Z 2p length D 1 C 4x 2 dx 4e 2x dx 1/2 .e 2x Z er 1 1 C 2 dx x Z ep 2 x C1 D dx Let x D tan u, dx D sec2 u du x 1 Z xDe Z xDe du sec3 u du D D 2 tan u xD1 cos u sin u xD1 Z xDe sin u du D v D cos u, dv D sin u du 2 2 xD1 cos u sin u Z xDe dv : D 2 v2/ xD1 v .1 1 C tan2 x dx Z =4 14. 1C 1 units 1=2 Z =4 p 2 Z 4 s e 2x C 1 dx 2x 1 2 e ˇ4 Z 4 x ˇ e Ce x x x ˇ D dx D ln e je j ˇ x e x 2 e 2 1 1 4 2 D ln e ln e e4 e2 8 e4 C 1 e 1 e2 D ln units. D ln 4 4 e e 1 e2 0 2x . 12. y D ln.1 x 2 /, 12 x 21 , y 0 D 1 x2 s Z 1=2 4x 2 length D 1C dx .1 x 2 /2 1=2 Z 1=2 1 C x2 dx D x2 1=2 1 Z 1=2 2 dx D 1C 1 x2 1=2 ˇ1=2 1 C x ˇˇ D x C ln D 2 ln 3 ˇ 1 x ˇ Z 4 1/e x 1 Since 1 C A B D D C 2C C v 2 .1 v 2 / v v 1 v 1Cv A.v v 3 / C B.1 v 2 / C C.v 2 C v 3 / C D.v 2 D v 2 .1 v 2 / 8̂ ACC D D0 < B CC CD D0 ) :̂ A D 0 BD1 1 ) A D 0; B D 1; C D D D ; 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) v3/ lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.3 (PAGE 412) therefore, 19. # 1 1 1 1 C C dv v2 2 1 v 1Cv xD1 ˇ ˇ ˇxDe 1 1 ˇˇ 1 C v ˇˇ ˇˇ 1 ln D / .but v D cos u D p ˇ 2 C1 v 2 ˇ1 v ˇˇ x xD1 #ˇxDe " p 1 .1 C v/2 ˇˇ 2 x C1 ln D ˇ 2 j1 v 2 j ˇ xD1 3 2 20. 2 1 ˇe ˇ 1C p C 2 p 6 ˇ 1 x C17 x2 C 1 2 7ˇ D6 4 x C 1 2 ln 5ˇ 1 ˇ 1 1 x2 C 1 #ˇ " p e p 1 2 C x 2 C 2 1 C x 2 ˇˇ ln x2 C 1 D ˇ ˇ 2 x2 1 p 2 p 2 p p 1 2 C e C 2 1 C e 1 2 C ln.3 C 2 2/ ln D e2 C 1 2 e2 2 p p p 1 .3 C 2 2/e 2 2 D e C1 2 C ln p units: 2 2 C e2 C 2 1 C e2 sD 17. Z xDe " x 2=3 C y 2=3 D x 2=3 . By symmetry, the curve has congruent arcs in the four quadrants. For the first quadrant arc we have 21. 3=2 y D a2=3 x 2=3 1=2 2 3 2=3 x 1=3 : a x 2=3 y0 D 2 3 y D x 1=3 ; 1 x 2; y 0 D Length D have LD4 D 4a 0 1=3 s 1C Z a x ds D 22. 18. The required length is Using a calculator we calculate some Simpson’s Rule approximations as described in Section 7.2: S2 1:59921 S8 1:60025 S4 1:60110 S16 1:60023: s 9x 2 dx D 1C 3 3x 2 4 Z 1 0 3 C 6x 2 dx: 3 3x 2 s 3 C 6x 2 dx 8:73775 units 3 3x 2 (with a little help from Maple’s numerical integration routine.) For the ellipse x 2 C 2y 2 D 2, we have 2x C 4yy 0 D 0, so y 0 D x=.2y/. Thus s x2 dx D 1C 4 2x 2 s 4 x2 dx 4 2x 2 23. s 4 x2 dx 1:05810 units 4 2x 2 (with a little help from Maple’s numerical integration routine). Z 2 p S D 2 jxj 1 C 4x 2 dx Let u D 1 C 4x 2 0 du D 8x dx ˇ17 Z 2 3=2 ˇˇ 17 p u du D D u ˇ ˇ 4 1 4 3 1 p D .17 17 1/ sq: units: 6 p y D x 3 , 0 x 1. ds D 1 C 9x 4 dx. The area of the surface of rotation about the x-axis is S D 2 D To four decimal places the length is 1.6002 units. 18 Z 1 0 p x 3 1 C 9x 4 dx Z 10 1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k s The circumference of the ellipse is 0 0 0 M4 D 1:03363 M8 D 1:03374 M16 D 1:00376: For the ellipse 3x 2 C y 2 D 3, we have 6x C 2yy 0 D 0, so y 0 D 3x=y. Thus Z 1 0 0 1 . We 9x 4=3 p The length of the short arc from .0; 1/ to .1; 1= 2/ is 1=3 Z 1p Z 1p 3 2 LD 1 C .4x / dx D 1 C 16x 6 dx: 1C Thus the length is approximately 1.0338 units. ds D a2=3 x 2=3 dx x 2=3 dx ˇa ˇ 3 D 4a1=3 x 2=3 ˇˇ D 6a units. 2 1 f .x/ dx, where f .x/ D T4 D 1:03406 T8 D 1:03385 T16 D 1:03378 Thus the length of the whole curve is Z a R2 1 2=3 x . 3 r Downloaded by ted cage (sxnbyln180@questza.com) p u du D Let u D 1 C 9x 4 du D 36x 3 dx .103=2 27 1/ sq. units. 279 lOMoARcPSD|6566483 www.konkur.in SECTION 7.3 (PAGE 412) 24. ADAMS and ESSEX: CALCULUS 9 q y D x 3=2 , 0 x 1. ds D 1 C 94 x dx. The area of the surface of rotation about the x-axis is S D 2 D D D Z 1 x 3=2 0 128 243 128 243 128 243 Z 3=2 0 r 9x 1C dx 4 p u4 1 C u2 du Z tan 1 .3=2/ 25. Z 1 r 9x S D 2 dx x 1C 4 0 2 Let 9x D 4u 9 dx D 8u du Let u D tan v du D sec2 v dv D 32 D 81 .sec7 v 64 D 81 0 2 sec5 v C sec3 v/ dv: At this stage it is convenient to use the reduction formula 26. Z secn v dv D 1 n 1 secn 2 v tan v C n n 2 1 Z secn 2 v dv (see Exercise 36 of Section 7.1) to reduce the powers of secant down to 3, and then use Z a 0 sec3 v dv D 1 .sec a tan a C ln j sec a C tan aj: 2 Z a 0 .sec7 v 2 sec5 v C sec3 v/ dv Substituting a D arct an.3=2/ now gives the following value for the surface area: p ! p 8 3 C 13 28 13 C ln SD sq. units. 81 243 2 Telegram: @uni_k 9 dx 4 p 1/ u du .u 2 5=2 u 5 9x 4 2 3=2 u 3 .13=4/5=2 5 1 ˇˇ13=4 ˇ ˇ ˇ 1 .13=4/3=2 3 1 ! sq. units. We have Z 1 ex p Let e x D tan e x dx D sec2 d Z xD1 Z xD1 p D 2 1 C tan2 sec2 d D 2 sec3 d xD0 xD0 ˇˇxD1 ˇ D sec tan C ln j sec C tan j ˇ : ˇ S D 2 0 1 C e 2x dx xD0 Since p 1 C e2 ; p x D 0 ) tan D 1; sec D 2; ˇa Z a Z a 5 sec5 v tan v ˇˇ 2 sec5 v dv C sec3 v dv D ˇ C ˇ 6 6 0 0 0 ˇa # " Z sec5 a tan a 7 sec3 v tan v ˇˇ 3 a 3 D sec v dv C ˇ ˇ 6 6 4 4 0 0 Z a sec3 v dv C 0 27. Z 1 a sec5 a tan a 7 sec3 a tan a C sec3 v dv D 6 24 8 0 sec a tan a C ln j sec a C tan aj sec5 a tan a 7 sec3 a tan a C : D 6 24 16 280 du D 1 Let u D 1 C x D 1 ) tan D e; sec D We have I D Z 13=4 32 81 tan4 v sec3 v dv 0 Z tan 1 .3=2/ If y D x 3=2 , 0 x 1, is rotated about the y-axis, the surface area generated is therefore p p p p 2 ln j 2 C 1j S D e 1 C e 2 C ln j 1 C e 2 C ej " # p p p 1 C e2 C e 2 sq: units: D e 1Ce 2 C ln p 2C1 If y D sin x; 0 x , is rotated about the x-axis, the surface area generated is S D 2 Z 0 p sin x 1 C cos2 dx Let u D cos x du D sin x dx Z 1p 1 C u2 du Let u D tan du D sec2 d Z =4 Z =4 D 2 sec3 d D 4 sec3 d D 2 1 =4 0 ˇˇ=4 D 2 sec tan C ln j sec C tan j ˇˇ 0 p p 2 C ln.1 C 2/ sq. units. D 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 28. SECTION 7.3 (PAGE 412) 2 2 2 2 x x 1 1 D C 1 C .y 0 /2 D 1 C 4 x2 4 x2 2 Z 4 3 1 x 1 x C C 2 dx S D 2 12 x 4 x 1 Z 4 5 x x 1 C C 3 dx D 2 48 3 x 1 ˇˇ4 6 2 x 1 x ˇ C D 2 ˇ 288 6 2x 2 ˇ 32. dy x D p , and dx 2 4 x2 1 D S D 2 2 Z 2p 0 2 D Z 2p 16 3x 2 dx D Z =3 275 sq: units: 8 1 x3 C , 1 x 4, we have 29. For y D 12 x 2 1 x C 2 dx. ds D 4 x The surface generated by rotating the curve about the yaxis has area Z 4 2 x 1 x S D 2 C 2 dx 4 x 1 ˇˇ4 4 x ˇ C ln jxj ˇ D 2 ˇ 16 1 255 C ln 4 sq. units. D 2 16 30. The area of the cone obtained by rotating the line y D .h=r/x, 0 x r, about the y-axis is ˇr p Z r p r 2 C h2 x 2 ˇˇ 2 S D 2 x 1 C .h=r/ dx D 2 ˇ r 2 ˇ 0 0 p D r r 2 C h2 sq. units. b/2 C y 2 D a2 we have 31. For the circle .x dy D0 dx b/ C 2y 2.x Thus s ds D The top half of x 2 C 4y 2 D 4 is y D 1C .x b/2 y2 dx D ) dy D dx a dx D p y a2 x b y a b/2 dx .x (if y > 0). The surface area of the torus obtained by rotating the circle about the line x D 0 is Z bCa a dx Let u D x b S D 2 2 xp 2 a .x b/2 b a du D dx Z a uCb p D 4a du 2 u2 Zaa a du p by symmetry D 8ab a2ˇ u2 0 a u ˇˇ D 8ab sin 1 ˇ D 4 2 ab sq. units. aˇ 0 0 x 2 , so 2 x p dx 2 4 x2 r 16 Let x D sin 3 r 16 dx D cos d 3 1C 4 .4 cos / p cos d 3 Z 16 =3 cos2 d D p 3 0 ˇˇ=3 8 ˇ D p C sin cos ˇ ˇ 3 0 p 2.4 C 3 3/ p sq: units: D 3 3 33. For the ellipse x 2 C 4y 2 D 4 we have 2x dx C 8y D 0 dy ) dx D dy y 4 : x The arc length element on the ellipse is given by ds D s 1C s 1C dx dy 2 dy 16y 2 1p dy D 4 C 12y 2 dy: 2 x x If the ellipse is rotated about the y-axis, the resulting surface has area Z 1 1p S D 2 2 x 4 C 12y 2 dy x 0 Z 1p p 1 C 3y 2 dy Let 3y D tan D 8 p 0 3dy D sec2 d Z =3 8 D p sec3 d 3 0 ˇˇ=3 8 D p sec tan C ln j sec C tan j ˇˇ 2 3 0 p 8 p D p 2 3 C ln.2 C 3/ 2 3 p ! ln.2 C 3/ sq. units. p D 8 1 C 2 3 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k s 0 D : x2 4 1p 4 2 Downloaded by ted cage (sxnbyln180@questza.com) 281 lOMoARcPSD|6566483 www.konkur.in SECTION 7.3 (PAGE 412) ADAMS and ESSEX: CALCULUS 9 Thus, the surface area depends only on the radius R of the sphere, and the distance .b a/ between the parellel planes. 34. As in Example 4, the arc length element for the ellipse is ds D s 1C dy dx 2 y v u u 2 ta dx D a 2 a2 2 b 2 x a2 dx: x2 To get the area of the ellipsoid, we must rotate both the upper and lower semi-ellipses (see the figure for Exercise 20 of Section 8.1): S D 2 2 Z a" c r 1 v u Z au t a2 x 2 b 1 0 cCb D 8c r x 2 a !# a2 a ! C Fig. 7.3-36 k 37. If the curve y D x , 0 < x 1, is rotated about the y-axis, it generates a surface of area x2 p R =2 p a2 b 2 and E."/ D 0 1 a defined in Example 4. where " D Z 1 p x 1 C k 2 x 2.k 1/ dx 0 Z 1p D 2 x 2 C k 2 x 2k dx: S D 2 0 "2 sin t dt as If k Z =2 0 Z =2 s s 1C 2 cos2 t dt 4 1C 0 36. Let the equation of the sphere be x 2 C y 2 D R2 . Then the surface area between planes x D a and x D b . R a < b R/ is S D 2 D 2 D 2R 282 Telegram: @uni_k R2 x2 R2 x2 p a Z bp a Z b a s 1C dy dx R R2 dx D 2R.b x2 x k dx, which is infinite. 0 Z 1 p x 1 C k 2 x 2.k 1/ dx 0 Z 1p D 2 x 2 2k C k 2 x k dx 0 Z 1 p x k dx < 1: < 2 1 C k 2 38. Z bp Z 1 S D 2 2 2 sin2 t dt 4 4 0 s Z =2 2 5p 4 C 2 1 D sin2 t dt 4 C 2 0 5p : D 4 C 2E p 4 C 2 10 D 1, we have S 2k If k 0, the surface area S is finite, since x k is bounded on .0; 1 in that case. Hence we need only consider the case 1 < k < 0. In this case 2 < 2 2k < 4, and 35. From Example 3, the length is 10 sD x x 2 Cy 2 DR2 dx 0 1 D 8c of the circumference of the ellipse 4 D 8caE."/ a2 b ds b2 a2 x2 a 2 Thus the area is finite if and only if k > 1. Z 1 r 1 S D 2 jxj 1 C 2 dx x 0 Z 1p D 2 x 2 C 1 dx Let x D tan 0 dx D sec2 d Z =4 D 2 sec3 d 0 ˇ ˇ=4 ˇ D sec tan C ln j sec C tan j ˇ ˇ 0 p p D Œ 2 C ln. 2 C 1/ sq: units: dx dx a/ sq: units: 39. a) Volume V D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) R 1 dx 1 x2 D cu. units. lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.4 (PAGE 419) b) The surface area is r 1 1 1 C 4 dx x x 1 Z 1 dx D 1: > 2 x 1 S D 2 Z 1 3. c) Covering a surface with paint requires applying a layer of paint of constant thickness to the surface. Far to the right, the horn is thinner than any prescribed constant, so it can contain less paint than would be required to cover its surface. The mass of the plate is m D 0 area D The moment about x D 0 is MxD0 D Z a D 0 2 Z a2 p u du 0 0 The mass of the wire is Z L Z L s ds mD ı.s/ ds D sin L 0 0 ˇL s ˇˇ L 2L cos : D ˇ D Lˇ 0 a3 4 4a MxD0 D D . By symmetry, 2 m 3 0 a 3 y D x. Thus the centre of mass of the plate is 4a 4a ; . 3 3 Thus x D y 0 p yD a2 x 2 Since ı.s/ is symmetric about s D L=2 (that is, ı..L=2/ s/ D ı..L=2/ C s/), the centre of mass is at the midpoint of the wire: s D L=2. 2. A slice of the wire of width dx at x has volume d V D .a C bx/2 dx. Therefore the mass of the whole wire is Z L mD ı0 .a C bx/2 dx 0 Z L D ı0 .a2 C 2abx C b 2 x 2 / dx 0 1 2 3 2 2 D ı0 a L C abL C b L : 3 Its moment about x D 0 is Z L MxD0 D xı0 .a C bx/2 dx 0 Z L D ı0 .a2 x C 2abx 2 C b 2 x 3 / dx 0 1 1 2 2 2 a L C abL3 C b 2 L4 : D ı0 2 3 4 Thus, the centre of mass is 1 1 2 2 2 a L C abL3 C b 2 L4 ı0 2 3 4 xD 1 ı0 a2 L C abL2 C b 2 L3 3 1 2 2 1 2 2 L a C abL C b L 2 3 4 D : 1 a2 C abL C b 2 L2 3 dx x a x Fig. 7.4-3 4. p A vertical strip has area dA D a2 x 2 dx. Therefore, the mass of the quarter-circular plate is Z a p .0 x/ a2 Let u D a2 x 2 du D 2x dx ˇa2 Z a2 p 2 3=2 ˇˇ 1 1 1 u u du D 0 D 0 ˇ D 0 a3 : ˇ 2 2 3 3 0 mD x 2 dx 0 0 The moment about x D 0 is MxD0 D Z a p 0 x 2 a2 0 D 0 a4 Z =2 x 2 dx Let x D a sin dx D a cos d sin2 cos2 d 0 Z 0 a4 =2 2 sin 2 d 4 0 Z 0 a4 0 a4 =2 .1 cos 4 / d D : D 8 16 0 D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Let u D a2 x 2 du D 2x dx x 2 dx 0 ˇa 2 0 2 3=2 ˇˇ 0 a3 D u ˇ D : ˇ 2 3 3 Section 7.4 Mass, Moments, and Centre of Mass (page 419) 1. p x0 a2 0 a2 . 4 Downloaded by ted cage (sxnbyln180@questza.com) 283 lOMoARcPSD|6566483 www.konkur.in SECTION 7.4 (PAGE 419) ADAMS and ESSEX: CALCULUS 9 The moment about y D 0 is Z a 1 MyD0 D 0 2 x.a 2 0 2 2 a x 1 0 2 2 D The moment about x D 0 is Z 3 MxD0 D 10 h2 2 x / dx !ˇa x 4 ˇˇ 1 ˇ D a4 0 : 4 ˇ 8 0 3 h D 10 3 0 mD2 Z 4 0 D 2k D 2k p ky 4 Z 4 0 ˇ4 2 5=2 ˇˇ 256k u : ˇ D ˇ 5 15 8 3=2 u 3 45 3 2 Thus, x D D and y D 15 2 of mass is located at . 32 ; 21 /. u/u1=2 du .4 0 Let u D 4 y du D dy y dy 0 y By symmetry, MxD0 D 0, so x D 0. 2 MyD0 D 2 Z 4 0 D 2k D 2k p ky 2 4 Z 4 1=2 .16u 0 32 3=2 u 3 3=2 8u Cu / du ˇ4 16 5=2 2 7=2 ˇˇ 4096k u C u : ˇ D ˇ 5 7 105 4096k 15 16 D . The centre of mass of the 105 256k 7 plate is .0; 16=7/. y p xD 4 y density ky 2 0 .5h/ 2 2 h D 10 2 284 Telegram: @uni_k x 3 7. The mass of the plate is Z a ka3 : mD kx a dx D 2 0 y a 6. A vertical strip at h has area dA D .2 mass of the plate is 2 3 h/ dh. Z 3 2 h dh D 10 h 3 0 ˇ 3 h3 ˇˇ ˇ D 15 kg: 9 ˇ 1 . The centre 2 Fig. 7.4-6 Fig. 7.4-5 Z 3 D 2a 2 ka4 Thus x D D . The centre of mass of the 3 3 3 ka 2a a plate is ; . 3 2 x mD By symmetry, y D a=2. Z a ka4 : kx 2 a dx D MxD0 D 3 0 4 2 15 2 15 dh h 5=2 0 Thus y D 2 yD2 3 x Let u D 4 y du D dy y dy 0 The moment about y D 0 is Z 3 1 1 2 2 MyD0 D 10 h h h dh 2 3 3 0 2 Z 3 2 2 1 3 h C h dh D 10 h 3 9 0 ˇ3 2 3 4 ˇ h 2h h ˇ 15 D 10 C kg-m: ˇ D 2 9 36 ˇ 2 3 3 a and y D a. Hence, the centre of mass 16 8 3 3 is located at . a; a/. 16 8 5. The mass of the plate is Thus, x D h3 dh 3 ˇ 3 h4 ˇˇ 45 kg-m. ˇ D ˇ 12 2 h2 3 Thus, the density kx dh a Fig. 7.4-7 0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.4 (PAGE 419) a 8. A vertical strip has area dA D 2 p r dr. Thus, the 2 mass is Z a=p2 a r dr kr 2 p mD2 2 0 Z a=p2 a k D 4k p r r 2 dr D p a3 g: 2 3 2 0 50 .8000/k 40 Thus, x D 3 D . Since the density is inde10000k 3 5 pendent of y and z, y D and z D 5. Hence, the centre 2 of mass is located on the 20 cm long central axis of the brick, two-thirds of the way from the least dense 10 5 face to the most dense such face. y 5 Since the mass is symmetric about the y-axis, and the plate is symmetric about both the x- and y-axis, therefore the centre of mass must be located at the centre of the square. y a p 2 dx a x yD p 2 a p dr 2 r x x z 20 10 Fig. 7.4-10 x Fig. 7.4-8 9. mD MxD0 D Z b a Z b a .x/ g.x/ f .x/ dx x.x/ g.x/ f .x/ dx Z b 11. 1 MyD0 D x.x/ .g.x//2 .f .x//2 dx 2 a MxD0 MyD0 ; : Centre of mass: m m y Choose axes through the centre of the ball as shown in the following figure. The mass of the ball is Z R .y C 2R/.R2 y 2 / dy ˇR 8 y 3 ˇˇ 2 D 4R R y ˇ D R4 kg: ˇ 3 3 mD yDg.x/ R 0 density .x/ By symmetry, the centre of mass lies along the y-axis; we need only calculate y. yDf .x/ a b x MyD0 D Fig. 7.4-9 10. The slice of the brick shown in the figure has volume d V D 50 dx. Thus, the mass of the brick is Z 20 ˇ20 ˇ mD kx50 dx D 25kx 2 ˇ D 10000k g: The moment about x D 0, i.e., the yz-plane, is MxD0 D 50k D Z 20 0 50 3 ˇˇ20 kx ˇ x 2 dx D 0 3 50 .8000/k g-cm: 3 y.y C 2R/.R2 Z R y 2 .R2 0 y3 D 2 R2 3 y 2 / dy y 2 / dy ˇR y 5 ˇˇ 4 R5 : ˇ D 5 ˇ 15 0 4R5 R 3 Thus y D D : The centre of mass is on 15 8R4 10 the line through the centre of the ball perpendicular to the plane mentioned in the problem, at a distance R=10 from the centre of the ball on the side opposite to the plane. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k R D 2 0 0 Z R Downloaded by ted cage (sxnbyln180@questza.com) 285 lOMoARcPSD|6566483 www.konkur.in SECTION 7.4 (PAGE 419) ADAMS and ESSEX: CALCULUS 9 y z y a x z yC2R R p a2 z 2 a a y a 2R x Fig. 7.4-13 Fig. 7.4-11 12. A slice at height z has volume d V D y 2 dz and density kz g=cm3 . Thus, the mass of the cone is mD Z b d m D 0 z dz 2 kzy dz Z b D ka2 z 1 0 z b 0 2 z D ka 2 2 dz 1 ka2 b 2 g: 12 ˇˇb ˇ ˇ ˇ MzD0 D ka Z b z 0 2 1 z b 2 dz D 1 ka2 b 3 g-cm: 30 2b . Hence, the centre of mass is on the axis 5 of the cone at height 2b=5 cm above the base. Thus, z D 0 z.a2 2 20 z.a2 D 3 dz z yDa 1 z b y Fig. 7.4-12 13. By symmetry, y D 0. 286 Telegram: @uni_k z2/ p 4 a2 z 2 3 z 2 /3=2 : Thus the mass of the solid is Z 0 a 2 .a z z 3 / dz mD 2 0 ˇa 0 a2 z 2 z 4 ˇˇ 0 a4 D : ˇ D 2 2 4 ˇ 8 0 Also, z b a z 2 /; dMxD0 D d m x D 0 The moment about z D 0 is 2 2 .a 2 and its moment about x D 0 is 2z 3 z4 C 2 3b 4b 2 D A horizontal slice of the solid p at height z with thickness dz is a half-disk of radius a2 z 2 with centre of mass p 4 a2 z 2 at x D , by Exercise 3 above. Its mass is 3 and z D Finally, MzD0 D 0 2 D 0 2 Z a .a2 z 2 0 2 3 a z 3 0 a5 8a 8 D . 15 0 a4 15 z 4 / dz ˇa z 5 ˇˇ 0 a5 ; ˇ D 5 ˇ 15 Z 20 a z.a2 z 2 /3=2 dz 3 0 Z 2 0 a 3=2 u du D 3 0 ˇa2 0 2 5=2 ˇˇ 20 a5 D u ; ˇ D ˇ 3 5 15 MxD0 D 0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 0 Let u D a2 z 2 du D 2z dz lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.4 (PAGE 419) 8 16a 20 a5 D . 4 15 0 a 15 8a 16a ; 0; : The centre of mass is 15 15 3 3a ka4 D . By symmetry, x D 0. 3 2 ka 2 3a Thus, the centre of mass of the plate is 0; . 2 Therefore, y D so x D 14. Assume the cone has its base in the xy-plane and its vertex at height b on the z-axis. By symmetry, the centre of mass lies on the z-axis. A cylindrical shell of thickness dx and radius x about the z-axis has height z D b.1 .x=a//. Since it’s density is constant kx, its mass is x dx: d m D 2bkx 2 1 a Also its centre of mass is at half its height, y shell D b 1 2 ds MzD0 D Z a 0 Z a 2bkx 2 1 bkx 2 0 1 x 2 dx: a x kba3 dx D a 6 x 2 kb 2 a3 dx D a 30 and z D MzD0 =m D b=5. The centre of mass is on the axis of the cone at height b=5 cm above the base. 15. y x 2 Cy 2 Da2 ds d a x Fig. 7.4-16 L The radius of the semicircle is . Let s measure the distance along the wire from the point where it leaves the positive x-axis. Thus, the density at position s is s ı.s/ D sin g=cm. The mass of the wire is L mD s xa Fig. 7.4-15 Consider the area element which is the thin half-ring shown in the figure. We have d m D ks s ds D k s ds: k 3 a . 3 Regard this area element as itself composed of smaller elements at positions given by the angle as shown. Then Z .s sin /s d ks ds dMyD0 D 0 0 s sin ds D L ˇL L s ˇˇ 2L cos g: ˇ D Lˇ 0 Since an arc element ds at position s is at height L s L sin D sin , the moment of the wire about y D L y D 0 is Thus, m D D 2ks 3 ds; Z a ka4 : MyD0 D 2k s 3 ds D 2 0 Z L MyD0 D 2 Z L 0 L 2 s sin ds L Let D s=L d D ds=L Z L 2 2 sin d 0 ˇˇ L2 L2 D sin cos ˇˇ D g-cm: 2 2 2 0 D Since the wire and the density function are both symmetric about the y-axis, we have MxD0 D 0. L . Hence, the centre of mass is located at 0; 4 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k s L dMzD0 D y shell d m D bkx 2 1 mD y x : a Thus its moment about z D 0 is Hence 16. Downloaded by ted cage (sxnbyln180@questza.com) 287 lOMoARcPSD|6566483 www.konkur.in SECTION 7.4 (PAGE 419) 17. Z 1 ADAMS and ESSEX: CALCULUS 9 y 2 C e kr .4 r 2 / dr 0 Z 1 p 2 D 4 C r 2 e kr dr Let u D k r p 0 du D k dr Z 4 C 1 2 u2 u e du D 3=2 k 0 mD dx 2 d V D ue u du 2 V D 12 e u 0 1 ˇR Z R u2 ˇ 4 C ue 1 2 ˇ D 3=2 lim @ e u duA ˇ C ˇ R!1 2 2 0 k 0 Z 1 1 u2 4 C D 3=2 0 C e du 2 0 k p 3=2 5:57C 4 C DC 3=2 : D 3=2 4 k k k U Du d U D du 18. 1 rD m p yD r 2 x 2 r x x r Fig. 7.5-1 2. x D 0. A horizontal strip at y has By symmetry, p p mass d m D 2 9 y dy and moment dMyD0 D 2y 9 y dy about y D 0. Thus, mD2 Z 9p y dy D 9 0 2 .9 2 3 y/ and Z 1 rC e kr 2 2 .4 r / dr Z 1 4 C 2 r 3 e kr dr Let u D kr 2 D C 3=2 k 3=2 0 du D 2kr dr Z 1 4k 3=2 1 u D p ue du 2k 2 0 U Du d V D e u du d U D du 0 V D e u 1 ˇR Z ˇ R 2 ˇ lim @ ue u ˇ C e u duA D p ˇ k R!1 0 0 2 2 0 C lim .e 0 e R D p : D p R!1 k k 0 Section 7.5 Centroids MyD0 D 2 D4 Z 9 p y 9 y dy 0 Z 3 .9u2 Let u2 D 9 y 2u du D dy 0 ˇ3 ˇ 1 5 ˇ u /ˇ 5 u4 / du D 4.3u3 0 ˇ9 ˇ ˇ D 36 ˇ 3=2 ˇ 0 D 648 : 5 648 18 D . Hence, the centroid is at Thus, y D 5 36 5 18 0; . 5 y 9 yD9 x 2 dy y 3 (page 424) 3 x Fig. 7.5-2 3. r2 AD Z4 1. r p x r2 AD Let u D r 2 x 2 0 du D 2x dx ˇr 2 Z r2 3=2 ˇ 1 u ˇ r3 1=2 D u du D ˇ D 2 0 3 ˇ 3 MxD0 D The area and moments of the region are x 2 dx D 0 3 4r 4 r D D y by symmetry: 3 r 2 3 4r 4r ; . The centroid is 3 3 xD 288 Telegram: @uni_k Z 1 0 p Z =4 dx 1 C x2 Let x D tan dx D sec2 d sec d 0 ˇ=4 p ˇ D ln j sec C tan jˇˇ D ln.1 C 2/ MxD0 D Z 1 0 p x dx ˇ1 p p ˇ D 1 C x 2 ˇˇ D 2 1 C x2 0 ˇ1 Z 1 ˇ 1 dx 1 MyD0 D D tan 1 x ˇˇ D : 2 0 1 C x2 2 8 0 0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL Thus x D centroid is p SECTION 7.5 (PAGE 424) 2 1 p , and y D p . The 2/ 8 ln.1 ! C 2/ 1 . p ; p ln.1 C 2/ 8 ln.1 C 2/ 5. ln.1 C p 2 y yD p By symmetry, x D 0. We have Z p3 p 4 AD2 1 0 D2 4 1Cx 2 Z =3 Fig. 7.5-3 0 0 x 2 dx C r3 D p 6 2 Z r 1 2 .r 3 x / p x r= 2 ˇr ˇ 2 3=2 ˇ p r2 p D5 3 p 3 p D4 3 4 r3 8r 8 D p . By symmetry, the p r2 3 2 3 2 p D x. 2 1/. centroid must lie on the line y D x tan 8 p 8r. 2 1/ Thus, y D p . 3 2 x 2 dx p ! p 3 C D3 3 3 4 1 p 4 : 3 p yD 4 x 2 1 p x 3 3 Thus, x D p yD r 2 x 2 4 0 x 2 dx r3 D p : p 3 2 r= 2 y 2 Z p3 p y ˇ ˇ yDx 3 Let x D 2 sin ! dx D 2 cos d p p 9 3 4 3 9 3 4 Thus y D p D p . The 3 4! 3 3 4 3 3 p 9 3 4 centroid is 0; p . 4 3 3 4. The area of the sector is A D 81 r 2 . Its moment about x D 0 is Z r=p2 p cos2 d 0 MxD0 D 1 dx ˇ=3 p ˇ D 4. C sin cos /ˇˇ 2 3 0 p ! p 4 p 3 D4 3 C 2 3D 3 4 3 p Z 3 p 2 1 4 x 2 1 dx MyD0 D 2 2 0 Z p3 p D 5 x 2 2 4 x 2 dx x 1 x2 Fig. 7.5-5 6. By symmetry, x D 0. The area is A D 21 ab. The moment about y D 0 is " 2 # Z a x 2 b 1 dx D b 1 a a 0 ˇ a x 3 ˇˇ 2 D b2 x ˇ D ab 2 : 3a2 ˇ 3 1 MyD0 D 2 Z a 2 x2 dx a2 0 pr Fig. 7.5-4 r 2 x Thus, y D 2 4b 2ab 2 D . 3 ab 3 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 289 lOMoARcPSD|6566483 www.konkur.in SECTION 7.5 (PAGE 424) ADAMS and ESSEX: CALCULUS 9 y 8. yDb q 1 x2 a2 dx a x ax Fig. 7.5-6 7. The quadrilateral consists of two triangles, T1 and T2 , as shown in the figure. The area and centroid of T1 are given by 41 D 2; 2 0C3C4 7 x1 D D ; 3 3 The region is the union of a half-disk and atriangle. The 4 and centroid of the half-disk is known to be at 1; 3 2 2 ; . The area of the semithat of the triangle is at 3 3 and the triangle is 2. Hence, circle is 2 2 3 C 8 MxD0 D I .1/ C .2/ D 2 3 6 4 2 2 MyD0 D C .2/ D : 2 3 3 3 C 2, then Since the area of the whole region is 2 3 C 8 4 xD and y D . 3. C 4/ 3. C 4/ A1 D y 0C1C0 1 y1 D D : 3 3 The area and centroid of T2 are given by 42 D 4; 2 0C2C4 D 2; x2 D 3 y2 D 0 2C0 D 3 2 x Fig. 7.5-8 9. M2;xD0 D 2 4 D 8 2 M2;yD0 D 4D 3 8 : 3 Since areas and moments are additive, we have for the whole quadrilateral A D 2 C 4 D 6; 14 38 MxD0 D C8D ; 3 3 1 2 2 : 3 It follows that 7 14 2D 3 3 2 1 M1;yD0 D 2 D 3 3 1 .x 1/2 yDx 2 A2 D M1;xD0 D p yD MyD0 D 38 19 2 D , and y D D Thus x D 36 9 6 19 1 of the quadrilateral is ; . 9 3 2 3 8 D 3 2: 1 . The centroid 3 A circular strip of the surface between heights y and y C dy has area dS D 2x r dy D 2x dy D 2 r dy: cos x The total surface area is S D 2 r Z r 0 dy D 2 r 2 : The moment about y D 0 is ˇr Z r ˇ 2 ˇ MyD0 D 2 r y dy D r.y /ˇ D r 3 : 0 0 3 r r D . By symmetry, the centroid of the Thus y D 2 r 2 2 hemispherical surface is on the axis of symmetry of the hemisphere. It is halfway between the centre of the base circle and the vertex. y y .3;1/ T1 4 x .x;y/ dS T2 r y x .2; 2/ Fig. 7.5-9 Fig. 7.5-7 290 Telegram: @uni_k Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.5 (PAGE 424) z h 10. By symmetry, x D y D 0. The volume is V D 32 r 3 . A thin slice of the solid at height z will have volume d V D y 2 dz D .r 2 z 2 / dz. Thus, the moment about z D 0 is MzD0 D Z r z.r 2 0 D dz z 2 2 r z 2 z 2 / dz ˇr z 4 ˇˇ r4 : ˇ D 4 ˇ 4 yDr 1 y r4 3r 3 D . Hence, the centroid is on 4 2 r 3 8 the axis of the hemisphere at distance 3r=8 from the base. z p yD r 2 z 2 Fig. 7.5-11 12. A band z with vertical width dz has radius at height z , and has actual (slant) width yDr 1 h s s 2 dy r2 dz D 1 C 2 dz: ds D 1 C dz h Its area is dz y r dA D 2 r 1 x (See the following The cone has volume V D figure.) The disk-shaped slice with vertical width dz has z , and therefore has volume radius y D r 1 h z 2 r2 dz D 2 .h h h D r2 h2 r2 D 2 h Z h z.h z/2 dz .h u/u2 du hu3 3 u4 4 0 Z h 0 Let u D h z du D dz ˇˇh r 2 h2 ˇ : ˇ D ˇ 12 0 r 2 h2 3 h D . The centroid of the Therefore z D 12 r 2h 4 solid cone is on the axis of the cone, at a distance above the base equal to one quarter of the height of the cone. r2 dz: h2 13. p rh r 2 C h2 h 1 p D . By Thus, z D 3 3 r r 2 C h2 symmetry, x D y D 0. Hence, the centroid is on the axis of the conical surface, at distance h=3 from the base. By symmetry, x D . The area and y-moment of the 2 region are given by Z sin x dx D 2 AD 0 Z 1 MyD0 D sin2 x dx 2 0 ˇ ˇ 1 D .x sin x cos x/ˇˇ D : 4 4 0 ; . Thus y D , and the centroid is 8 2 8 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1C 0 z/2 dz: We have r2 h2 s The moment about z D 0 is s Z z r2 h z 1 dz MzD0 D 2 r 1 C 2 h 0 h s ˇh p r 2 z2 z 3 ˇˇ 1 D 2 r 1 C 2 ˇ D rh r 2 C h2 : ˇ h 2 3h 3 1 2 3 r h. MzD0 D z h Thus the area of the conical surface is s Z p r2 h z 1 dz D r r 2 C h2 : A D 2 r 1 C 2 h 0 h Fig. 7.5-10 dV D r2 1 r 0 Thus, z D 11. z h Downloaded by ted cage (sxnbyln180@questza.com) 291 lOMoARcPSD|6566483 www.konkur.in SECTION 7.5 (PAGE 424) ADAMS and ESSEX: CALCULUS 9 y Hence x D r 2 2r 2 D , and the centroid is r y 2r 2r ; . yDsin x x 2 Cy 2 Dr 2 r ds =2 x r Fig. 7.5-13 14. The area of the region is x xCdx ˇ=2 Z =2 ˇ ˇ AD cos x dx D sin x ˇ D 1: ˇ 0 r x Fig. 7.5-15 0 16. The moment about x D 0 is Z =2 MxD0 D x cos x dx 0 U Dx d V D cos x dx d U D dx V D sin x ˇ=2 Z ˇ =2 ˇ sin x dx D D x sin x ˇ ˇ 2 0 By symmetry, the centroid of the solid lies on its vertical axis of symmetry; let us continue to call this the y-axis. We need only determine y S . Since D lies between y D 1 and y D 2, its centroid satisfies yD D 3=2. Also, by Exercise 11, the centroid of the solid cone satisfies y C D 3=4. Thus C and D have moments about y D 0: 1: 0 Thus, x D 2 1. The moment about y D 0 is 1 2 MyD0 D Z =2 cos2 x dx MC;yD0 D 0 ˇˇ=2 1 1 ˇ x C sin 2x ˇ D : D ˇ 4 2 8 Thus, y D y 1 yDcos x . The centroid is 8 2 1; . 8 17. 2 x r . By symmetry, x D y. An 2 element of the arc between x and x C dx has length The arc has length L D r dx r dx dx D D p ds D : sin y r 2 x2 Thus MxD0 D 292 Telegram: @uni_k Z r 0 xr dx p D r 2 x2 p r r2 3 D ; 4 MD;yD0 D .4/ 3 D 6: 2 2 8 3x D 2.1/ C 1 D 3 3 3 2 11 : C1 D 3y D 2 2 3 3 Fig. 7.5-14 15. 4 3 The region in figure (a) is the union of a rectangle of area 2 and centroid .1; 3=2/ and a triangle of area 1 and centroid .2=3; 2=3/. Therefore its area is 3 and its centroid is .x; y/, where dx x Thus MS;yD0 D C 6 D 7, and z S D 7=.16=3/ D 21=16. The centroid of the solid S is on its vertical axis of symmetry at height 21/16 above the vertex of the conical part. 0 The solid S in question consists of a solid cone C with vertex at the origin, height 1, and top a circular disk of radius 2, and a solid cylinder D of radius 2 and height 1 sitting on top of the cone. These solids have volumes VC D 4=3, VD D 4, and VS D VC C VD D 16=3. Therefore, the centroid is .8=9; 11=9/. 18. The p region in figure (b) is the union of a square of area . 2/2 D 2 and centroid .0; 0/ and a triangle of area 1/2 and centroid .2=3; 2=3/. Therefore its area is 5/2 and its centroid is .x; y/, where ˇr ˇ x 2 ˇˇ D r 2 : 0 5 1 x D 2.0/ C 2 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2 1 D : 3 3 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.5 (PAGE 424) Therefore, x D y D 2=15, and the centroid is .2=15; 2=15/. 23. 19. The region in figure (c) is the union of a half-disk of area =2 and centroid .0; 4=.3// (by Example 1) and a triangle of area 1 and centroid .0; 1=3/. Therefore its area is .=2/ C 1 and its centroid is .x; y/, where x D 0 and C2 yD 2 2 4 3 C1 1 3 D y 1 : 3 1 T Therefore, the centroid is .0; 2=Œ3. C 2//. 20. The region in figure (d) is the union of three half-disks, one with area =2 and centroid .0; 4=.3//, and two with areas =8 and centroids . 1=2; 2=.3// and .1=2; 2=.3//. Therefore its area is 3=4 and its centroid is .x; y/, where 1 1 3 .x/ D .0/ C C D0 4 2 8 2 8 2 4 2 2 1 3 .y/ D C C D : 4 2 3 8 3 8 3 2 xD2 24. By symmetry the centroid is .1; 2/. y Fig. 7.5-23 p s 3 . Its centroid is at The altitude h of the triangle is 2 h s height D p above the base side. Thus, by Pappus’s 3 2 3 Theorem, the volume of revolution is p ! s 3s s 3 s V D 2 cu: units: D p 2 2 4 2 3 p h s 3 The centroid of one side is D above the base. 2 4 Thus, the surface area of revolution is p ! p 3s .s/ D s 2 3 sq: units: S D 2 2 4 .1;1/ yD2x x 2 x .1; 2/ x 1 Therefore, the centroid is .0; 2=.3//. 21. The triangle T has centroid 31 ; 13 and area 21 . By Pappus’s Theorem the volume of revolution about x D 2 is 1 5 1 cu. units. D V D 2 2 2 3 3 yD 2 s h s Fig. 7.5-21 22. The line segment from .1; 0/ to .0; 1/ has centroid . 12 ; 12 / p and length 2. By Pappus’s Theorem, the surface area of revolution about x D 2 is A D 2 2 1 p 2 p 2 D 3 2 sq: units: y 1 1 25. For the purpose of evaluating the integrals in this problem and the next, the definite integral routine in the TI-85 calculator p was used. For the region bounded by y D 0 and y D x cos x between x D 0 and x D =2, we have AD r 1 2 Fig. 7.5-24 2 3 x Fig. 7.5-22 Z =2 0 x cos x dx 0:704038 Z 1 =2 3=2 x cos x dx 0:71377 A 0 Z =2 1 x cos2 x dx 0:26053: yD 2A 0 xD Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k p Downloaded by ted cage (sxnbyln180@questza.com) 293 lOMoARcPSD|6566483 www.konkur.in SECTION 7.5 (PAGE 424) ADAMS and ESSEX: CALCULUS 9 29. 26. The region bounded by y D 0 and y D ln.sin x/ between x D 0 and x D =2 lies below the x-axis, so Z =2 AD ln.sin x/ dx 1:088793 0 Z 1 =2 xD x ln.sin x/ dx 0:30239 A 0 Z =2 2 1 yD ln.sin x/ dx 0:93986: 2A 0 27. c Z 2 2 i 1 d h g.y/ f .y/ dy MxD0 D 2 c Z d MyD0 D y g.y/ f .y/ dy: c The area and moments of the region are AD MxD0 D D Z 1 0 Z 1 0 Z 1 0 u 1 D lim R!1 1 MyD0 D 2 The centroid is 1 u3 Z 1 0 30. ˇR ˇ 1 1 dx ˇ D lim ˇ D 3 2 R!1 2.1 C x/ ˇ .1 C x/ 2 x dx .1 C x/3 Let u D x C 1 du D dx du 1 1 C 2 u 2u ˇˇR ˇ ˇ D1 ˇ 1 By analogy with the formulas for the region a x b, f .x/ y g.y/, the region c y d , f .y/ x g.y/ will have centroid .MxD0 =A; MyD0 =A/, where Z d AD g.y/ f .y/ dy Let us take L to be the y-axis and suppose that a plane curve C lies between x D a and x D b where 0 < a < b. Thus, r D x, the x-coordinate of the centroid of C. Let ds denote an arc length element of C at position x. This arc length element generates, on rotation about L, a circular band of surface area dS D 2x ds, so the surface area of the surface of revolution is Z xDb S D 2 x ds D 2MxD0 D 2rs: xDa 1 1 D 2 2 ˇR ˇ dx 1 1 ˇ D lim : ˇ D R!1 10.1 C x/5 ˇ .1 C x/6 10 31. y 0 1; 15 y . 1 yD 1 .x C 1/3 1 t .=4/ L t 1 p 2 x p .=4/ t P x 2 t N Fig. 7.5-27 28. The surface Z 1area ispgiven by 2 2 S D 2 e x 1 C 4x 2 e 2x dx. Since 1 M 2 lim 1 C 4x 2 e 2x D 1, this expression must be bounded Fig. 7.5-31 x!˙1 2 for all x, that is, 1 1 C Z4x 2 e 2x K 2 for some con1 p 2 stant K. Thus, S 2K e x dx D 2K . The 1 integral converges and the surface area is finite. Since the 2 whole curve y D e x lies above the x-axis, its centroid would have to satisfy y > 0. However, Pappus’s Theorem would then imply that the surface of revolution would have infinite area: S D 2y .length of curve/ D 1. The curve cannot, therefore, have any centroid. 294 Telegram: @uni_k We need to find the x-coordinate xLMNP of the centre of buoyancy, that is, of the centroid of quadrilateral LMNP . From various triangles in the figure we can determine the x-coordinates of the four points: xL D sec t; xP D sec t; xM D sec t C .1 C tan t / sin t xN D sec t C .1 tan t / sin t Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.6 (PAGE 431) Triangle LMN has area 1 C tan t , and the x-coordinate of its centroid is xLMN sec t D D 2 sin t 3 sec t C .1 C tan t / sin t C sec t C .1 3 sec t : Triangle LNP has area 1 its centroid is tan t / sin t h dh 6 m tan t , and the x-coordinate of 2 m sec t C sec t C sec t C .1 3 sec t C .1 tan t / sin t D : 3 xLNP D 2 m tan t / sin t Fig. 7.6-1 2. Therefore, 1h .2 sin t sec t /.1 C tan t / 6 i C .sec t C sin t sin t tan t /.1 tan t / i 1h 3 sin t 2 sec t tan t C sin t tan2 t D 6 " # sin2 t 2 sin t C 3 D 6 cos2 t cos2 t i sin t h D 3 cos2 t C sin2 t 2 2 6 cos t i i sin t h sin t h 2 cos2 t 1 D cos.2t / D 2 2 6 cos t 6 cos t xLMNP D A vertical slice of water at position y with thickness dy is in contact withpthe botttom over an area 8 sec dy D 54 101 dy m2 , which is at depth 1 x D 10 y C 1 m. The force exerted on this area is then p 1 y C 1/ 45 101 dy. Hence, the total force dF D g. 10 exerted on the bottom is Z 20 1 y C 1 dy 10 0 ˇˇ20 2 y 4p ˇ Cy ˇ 101 .1000/.9:8/ D ˇ 5 20 4p 101 g F D 5 0 3:1516 106 N: 20 which is positive provided 0 < t < =4. Thus the beam will rotate counterclockwise until an edge is on top. y 1 y dy 3 y xD 10 C1 Section 7.6 Other Physical Applications (page 431) 1. a) The pressure at the bottom is p D 9; 800 6 N/m2 . The force on the bottom is 4 p D 235; 200 N. b) The pressure at depth h metres is 9; 800h N/m2 . The force on a strip between depths h and h C dh on one wall of the tank is dF D 9; 800h 2 dh D 19; 600 h dh N. Thus, the total force on one wall is F D 19; 600 Z 6 0 x Fig. 7.6-2 3. A strip along the slant wall of the dam between depths h and h C dh has area dA D The force on this strip is dF D 9; 800 h dA 2:12 106 h dh N. Thus the total force on the dam is h dh D 19; 600 18 D 352; 800 N. F D 2:12 106 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 26 200 dh D 200 dh: cos 24 Downloaded by ted cage (sxnbyln180@questza.com) Z 24 0 h dh 6:12 108 N. 295 lOMoARcPSD|6566483 www.konkur.in SECTION 7.6 (PAGE 431) ADAMS and ESSEX: CALCULUS 9 5. The unbalanced force is F D 9; 800 5 h dh 2 ˇˇ20 h ˇ D 9; 800 5 ˇ 8:92 106 N. 2 ˇ h hCdh Z 20 6 6 26 24 Fig. 7.6-3 5 m p 4. The height of each triangular p face is 2 3 m and the height of the pyramid is 2 2 m. Let the angle between r 2 the triangular face and the base be , then sin D 3 1 and cos D p . 3 20 m 6 m p 2 2 p 2 3 Fig. 7.6-5 2 4 6. 4 ˇ3 1 2 ˇˇ 9 100 Ncm D W D kx dx D kx ˇ D k: ˇ 2 2 0 Z 3 Fig. 7.6-4 y front view of dy p 10 2 2 p dy sec D 3dy p p xD 2yC10 2 2 x W D ı 60 2 7. Fig. 7.6-4 A vertical slice of water with thickness dy at a distance y from the vertex of the pyramid exerts a force on the shaded strip shown in the front view,pwhich has areap p 2 3y dy m2 and which is at depth 2y C 10 2 2 m. Hence, the force exerted on the triangular face is Z 2 p p p . 2y C 10 2 2/2 3y dy F D g 0 p ˇ2 p 2 ˇˇ p 2 3 y C .5 D 2 3.9800/ 2/y ˇ ˇ 3 0 6:1495 105 N: 296 Telegram: @uni_k Z 4 3 kx dx D 4 side view of one face 0 200 Hence, k D N/cm. The work necessary to compress 9 the spring a further 1 cm is one face 10 The spring force is F .x/ D kx, where x is the amount of compression. The work done to compress the spring 3 cm is 200 9 ˇ4 1 2 ˇˇ 700 x ˇ D Ncm: 2 ˇ 9 3 A layer of water in the tank between depths h and h C dh has weight dF D g d V D 4g dh. The work done to raise the water in this layer to the top of the tank is d W D h dF D 4gh dh. Thus the total work done to pump all the water out over the top of the tank is W D 4g 8. Z 6 0 h dh D 4 9; 800 18 7:056 105 Nm. The horizontal cross-sectional area of the pool at depth h is 160; if 0 h 1; A.h/ D 240 80h; if 1 < h 3. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.6 (PAGE 431) The work done to empty the pool is Z 3 W D g hA.h/ dh 0 Z 1 Z 3 240h 80h2 dh 160h dh C D g 1 0 ˇ1 " ˇ3 # ˇ 80 3 ˇˇ 2ˇ 2 h ˇ D 9800 80h ˇ C 120h ˇ ˇ 3 1 0 10. When the water surface is y m above the centre of the piston face ( R y R), a horizontal strip on the piston face at height z m above thepcentre of the piston face, having height dz has width 2 .R2 z 2 / and so its p area is dA D 2 R2 z 2 dz. The strip is y z m below the surface of the water, and the pressure at that depth is g.y z/ D 9;800.y z/ N/m2 . Thus, the p force of the water on that strip is dF D 19;600.y z/ R2 z 2 dz Newtons. The total force of the water on the piston is D 3:3973 106 Nm: F D R D 19;600 h 1 3 A.h/ 11. Fig. 7.6-8 9. A layer of water between depths y and y C dy has volume d V D .a2 y 2 / dy and weight 2 2 dF D 9; 800.a y / dy N. The work done to raise this water to height h m above the top of the bowl is d W D .h C y/ dF D 9; 800.h C y/.a2 y 2 / dy Nm. Thus the total work done to pump all the water in the bowl to that height is Z a W D 9; 800 .ha2 C a2 y hy 2 y 3 / dy 0 ˇa y 4 ˇˇ a2 y 2 hy 3 2 D 9; 800 ha y C ˇ 2 3 4 ˇ 0 3 2a h a4 D 9; 800 C 3 4 8h 3a C 8h 3 3 D 2450a a C Nm. D 9; 800a 12 3 a Z sin 1 .y=R/ .y z 2 dz (let z D R sin ) R sin / R2 cos2 d =2 Initially the water occupies the bottom half of a cylinder of length X. Half of this water (say, a bottom halfcylinder of length X=2, must be moved to fill the top half of a cylinder of length X=2. By symmetry, we can accomplish this by moving a thin horizontal slice of water of thickness dy at distance y below the central axis of the cylinder to height y above that axis, that is, p X dy of wawe move a volume d V D 2 R2 y 2 2 ter up a distance p 2y. The work required to do this is d W D 2gXy R2 y 2 dy Nm. Thus the work to raise the half-cylinder of water is W D 2gX Z R p y R2 D 9;800X y 2 dy 0 Z R2 p 0 u du D let u D R2 y2 19;600 XR3 Nm: 3 Since no work is lost to friction, the work to push the piston in to distance X=2 is equal to this work needed to raise the water. 12. Let the time required to raise the bucket to height h m be h t minutes. Given that the velocity is 2 m/min, then t D . 2 The weight of the bucket at time t is h 16 kg .1 kg/min/.t min/ D 16 kg. Therefore, 2 the work done required to move the bucket to a height of 10 m is W Dg Z 10 16 0 D 9:8 16h Fig. 7.6-9 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k p z/ R2 ˇsin 1 .y=R/ ˇ R2 y R3 . C sin cos / C cos3 ˇˇ 2 3 =2 2 2 R y yR 1 3=2 y sin 1 C C R2 y 2 D 19;600 N: 2 R 4 3 8 y 19;600.y D 19;600 20 dy Z y Downloaded by ted cage (sxnbyln180@questza.com) h dh 2 ˇ 10 h2 ˇˇ ˇ D 1323 Nm: 4 ˇ 0 297 lOMoARcPSD|6566483 www.konkur.in SECTION 7.6 (PAGE 431) ADAMS and ESSEX: CALCULUS 9 8. Section 7.7 Applications in Business, Finance, and Ecology (page 435) 1. Cost D $4; 000 C D $11; 000: Z 1;000 6 0 2x 6x 2 C 6 3 10 10 V D dx 9. 2. The number of chips sold in the first year was 1; 000 Z 52 0 t e t=10 dt D 100; 000 620; 000e 26=5 that is, about 96,580. 3. The monthly charge is Z x 0 4 p dt 1C t Z px let t D u2 Z px u D8 du D 8 1 1Cu 0 0p p D$8 x ln.1 C x/ : 1 1Cu du 0 11. V D 1;000e 0 ˇ10 1;000 0:02t ˇˇ e dt D ˇ D $9;063:46: ˇ 0:02 V D 7. 0 298 Z 12 2 1;000e ˇ12 1;000 0:08t ˇˇ e dt D ˇ D $5;865:64: ˇ 0:08 2 Z 1 0 1;000e 0:02t dt D 1; 000 D $50; 000: 0:02 After t years, money is flowing at $.1;000 C 100t / per year. The present value of 10 years of payments discounted at 5% is Z 10 0 .10 C t /e 0:05t dt 0 12. After t years, money is flowing at $1;000.1:1/t per year. The present value of 10 years of payments discounted at 5% is Z 10 e t ln.1:1/ e 0:05t dt ˇ10 ˇ 1;000 t.ln.1:1/ 0:05 ˇ D e ˇ D $12; 650:23: ˇ ln.1:1/ 0:05 V D 1;000 0 0:08t The present value of continuous payments of $1,000 per year for all future time at a discount rate of 2% is d V D e 0:05t dt e 0:05t V D 0:05 ˇ10 Z 0:05t ˇ 100 10 0:05t e ˇ e dt C D 100.10 C t / ˇ 0:05 ˇ 0:05 0 0 ˇ10 ˇ 100 0:05t ˇ e D 4261:23 C ˇ D $11; 477:54: 2 ˇ .0:05/ ˇ10 1;000 0:05t ˇˇ 0:05t 1;000e dt D e ˇ D $7;869:39: ˇ 0:05 The present value of continuous payments of $1,000 per year for 10 years beginning 2 years from now at a discount rate of 8% is V D Telegram: @uni_k Z 10 10 U D 10 C t d U D dt 0 6. The present value of continuous payments of $1,000 per year for 10 years at a discount rate of 5% is 1;000e 10 V D 100 400.10 C 5t / dt $4; 750:37: 1 C 0:1t 0:02t ˇ35 1;000 0:05t ˇˇ e dt D ˇ D $8;655:13: ˇ 0:05 0:05t 10. The present value of continuous payments of $1,000 per year beginning 10 years from now and continuing for all future time at a discount rate of 5% is Z 1 1;000 0:5 V D e D $12;130:61: 1;000e 0:05t dt D 0:05 10 5. The present value of continuous payments of $1,000 per year for 10 years at a discount rate of 2% is Z 10 Z 35 V D 4. The price per kg at time t (years) is $10 C 5t . Thus the revenue per year at time t is 400.10C5t /=.1C0:1t / $/year. The total revenue over the year is Z 1 The present value of continuous payments of $1,000 per year for 25 years beginning 10 years from now at a discount rate of 5% is 0 0 13. The amount after 10 years is A D 5; 000 Z 10 e 0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 0:05t ˇ10 5;000 0:05t ˇˇ dt D e ˇ D $64;872:13: ˇ 0:05 0 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 14. SECTION 7.7 (PAGE 435) Let T be the time required for the account balance to reach $1;000;000. The $5; 000.1:1/t dt deposited in the time interval Œt; t C dt grows for T t years, so the balance after T years is For realistic growth functions, the maximum will occur where Q 0 .x/ D 0, that is, where F 0 .x/ D ı. 17. Z T 5; 000.1:1/t .1:06/T t dt D 1; 000; 000 Z T 1; 000; 000 1:1 t dt D D 200 .1:06/T 1:06 5; 000 0 # " 1:1 T .1:06/T 1 D 200 ln.1:1=1:06/ 1:06 We are given L D 80; 000, k D 0:12, and ı D 0:05. According to the analysis in the text, the present value of future harvests will be maximized if the population level is maintained at 0 .1:1/T .1:06/T D 200 ln Let P ./ be the value at time < t that will grow to $P D P .t / at time t . If the discount rate at time is ı./, then d P ./ D ı./P ./; d or, equivalently, dP ./ D ı./ d : P ./ ln P .0/ D Z t 0 dx D 0:02x 1 dt ˇ dx ˇˇ ˇ dt ˇ ı./ d D .t /; 0 .t / D Z t e) The total present value of all future harvesting revenue if the population level is maintained at 75,000 and ı D 0:05 is 0 ıx: ıx: Z 1 0 e 0:05t 7; 500; 000 dt D 7; 500; 000 D $150; 000; 000: 0:05 If we assume that the cost of harvesting 1 unit of population is $C.x/ when the population size is x, then the effective income from 1 unit harvested is $.p C.x//. Using this expression in place of the constant p in the analysis given in the text, we are led to choose x to maximize i h x ıx : Q.x/ D p C.x/ kx 1 L Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k D 0:02.75; 000/.0:5/ D 750 whales: d) At 5%, the interest would be .5=2/.$15; 000/ D $37; 500; 000. ı./ d : If the logistic model dx=dt D kx.1 .x=L// is replaced with a more general growth model dx=dt D F .x/, exactly the same analysis leads us to maximize Q.x/ D F .x/ xDL=2 75; 000.$10; 000/.0:02/ D $15; 000; 000: 19. x L x : 150; 000 c) If the whole population of 75,000 is harvested and the proceeds invested at 2%, the annual interest will be 16. The analysis carried out in the text for the logistic growth model showed that the total present value of future harvests could be maximized by holding the population size x at a value that maximizes the quadratic expression Q.x/ D kx 1 D $11; 900: b) The resulting annual revenue is $750p D $7; 500; 000. The present value of a stream of payments due at a rate P .t / at time t from t D 0 to t D T is where a) The maximum sustainable annual harvest is P .0/ D P .t /e .t/ D P e .t/ : P .t /e .t/ dt; 23; 333:33 80; 000 18. We are given that k D 0:02, L D 150; 000, p D $10; 000. The growth rate at population level x is and, taking exponentials of both sides and solving for P .0/, we get Z T 0:07 L D .80; 000/ D 23; 333:33 2k 0:24 6.0:12/.23; 333:33/ 1 Integrating this from 0 to t , we get ln P .t / ı/ The annual revenue from harvesting to keep the population at this level (given a price of $6 per fish) is 1:1 : 1:06 This equation can be solved by Newton’s method or using a calculator “solve” routine. The solution is T 26:05 years. 15. x D .k Downloaded by ted cage (sxnbyln180@questza.com) 299 lOMoARcPSD|6566483 www.konkur.in SECTION 7.7 (PAGE 435) ADAMS and ESSEX: CALCULUS 9 A reasonable cost function C.x/ will increase as x decreases (the whales are harder to find), and will exceed p if x x0 , for some positive population level x0 . The value of x that maximizes Q.x/ must exceed x0 , so the model no longer predicts extinction, even for large discount rates ı. However, the optimizing population x may be so low that other factors not accounted for in the simple logistic growth model may still bring about extinction whether it is economically indicated or not. Section 7.8 Probability 1. 6. 9 9 0:0225 60 60 9 f .3/ D 2 16 D 0:0500 60 1 9 16 C D 0:0778 f .4/ D 2 60 36 9 2 f .5/ D 2 16 C D 0:1056 60 36 3 9 16 C D 0:1333 f .6/ D 2 60 36 9 4 f .7/ D 2 1160 C D 0:1661 60 36 11 3 f .8/ D 2 16 C D 0:1444 60 36 2 11 16 C D 0:1167 f .9/ D 2 60 36 11 1 f .10/ D 2 16 C D 0:0889 60 36 11 16 D 0:0611 f .11/ D 2 60 11 f .12/ D 1160 D 0:0336: 60 f .2/ D (page 449) The expected winnings on a toss of the coin are $1 0:49 C $2 0:49 C $50 0:02 D $2:47: If you pay this much to play one game, in the long term you can expect to break even. P 2. (a) We need 6nD1 Kn D 1. Thus 21K D 1, and K D 1=21. (b) Pr.X 3/ D .1=21/.1 C 2 C 3/ D 2=7: 3. From the second previous Exercise, the mean winings is D $2:47. Now 2 D 1 0:49 C 4 0:49 C 2;500 0:02 52:45 6:10 D 46:35: 2 The standard deviation is thus $6:81. 4. Since Pr.X D n/ D n=21, we have D 2 D 6 X nD1 6 X nD1 D 21 p D nPr.X D n/ D 1 1 C 2 2 C C 6 6 13 D 4:33 21 3 n2 Pr.X D n/ 2 D 12 C 23 C C 63 21 Similarly, 12 X nD2 n2 f .n/ 57:1783; so the standard deviation of X is 20 1:49: 3 D The expectation of X 2 is 9 1 C.22 C32 C42 C52 / C62 1160 15:7500: 60 6 Hence the standard deviation of X is p 15:75 3:58332 1:7059. 2 29 9 C D 0:4833. Also Pr.X 3/ D 60 6 60 p E.X 2 / 2 2:4124: The mean is somewhat larger than the value (7) obtained for the unweighted dice, because the weighting favours more 6s than 1s showing if the roll is repeated many times. The standard deviation is just a tiny bit smaller than that found for the unweighted dice (2.4152); the distribution of probability is just slightly more concentrated around the mean here. 1 9 C .2 C 3 C 4 C 5/ C 6 1160 3:5833: D1 60 6 Telegram: @uni_k nD2 E.X 2 / D 5. The mean of X is 300 (b) Multiplying each value f .n/ by n and summing, we get 12 X D nf .n/ 7:1665: 2 20 169 D 2:22 9 9 E.X 2 / D 12 (a) Calculating as we did to construct the probability function in Example 2, but using the different values for the probabilities of “1” and “6”, we obtain 7. (a) The sample space consists of the eight triples .H; H; H /, .H; H; T /, .H; T; H /, .T; H; H /, .H; T; T /, .T; H; T /, .T; T; H /, and .T; T; T /. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.8 (PAGE 449) (b) We have 9. We have f .x/ D C x on Œ0; 3. a) C is given by Pr.H; H; H / D .0:55/3 D 0:166375 Pr.H; H; T / D Pr.H; T; H / D Pr.T; H; H / D .0:55/2 .0:45/ D 0:136125 Pr.H; T; T / D Pr.T; H; T / D Pr.T; T; H / D .0:55/.0:45/2 D 0:111375 Pr.T; T; T / D .0:45/3 D 0:091125: (c) The probability function f for X is given by 1D Hence, C D (d) Pr.X 1/ D 1 Pr.X D 0/ D 0:908875. 8. The number of red balls in the sack must be 0:620 D 12. Thus there are 8 blue balls. (a) The probability of pulling out one blue ball is 8=20. If you got a blue ball, then there would be only 7 blue balls left among the 19 balls remaining in the sack, so the probability of pulling out a second blue ball is 7=19. Thus the probability of pulling out two 8 7 14 blue balls is D . 20 19 95 (b) The sample space for the three ball selection consists of all eight triples of the form .x; y; z/, where each of x; y; z is either R(ed) or B(lue). Let X be the number of red balls among the three balls pulled out. Arguing in the same way as in (a), we calculate 8 7 6 14 Pr.X D 0/ D Pr.B; B; B/ D D 20 19 18 285 0:0491 Pr.X D 1/ D Pr.R; B; B/ C Pr.B; R; B/ C Pr.B; B; R/ 12 8 7 28 D3 D 0:2947 20 19 18 95 Pr.X D 2/ D Pr.R; R; B/ C Pr.R; B; R/ C Pr.B; R; R/ 8 44 12 11 D 0:4632 D3 20 19 18 95 12 11 10 11 Pr.X D 3/ D Pr.R; R; R/ D D 20 19 18 57 0:1930 Thus the expected value of X is D 0 2 0 ˇ3 9 2 4 ˇˇ x ˇ D , the x dx D ˇ 36 2 0 3 0 2 D E.X 2 / 2 D 9 2 4D 1 ; 2 p and the standard deviation is D 1= 2. c) We have Pr. D X C / D . C /2 /2 . 9 D 2 9 Z C x dx 4 0:6285: 9 10. We have f .x/ D C x on Œ1; 2. a) To find C , we have ˇ2 3 C 2 ˇˇ 1D C x dx D x ˇ D C: 2 ˇ 2 1 Z 2 Hence, C D 1 2 . 3 b) The mean is 2 D E.X/ D 3 9 D 1:8: 5 ˇ2 2 3 ˇˇ 14 1:556: x dx D x ˇ D ˇ 9 9 1 Z 2 2 Since E.X / D 3 variance is 2 14 28 44 11 C1 C2 C3 285 95 95 57 ˇ3 9 C 2 ˇˇ x ˇ D C: 2 ˇ 2 ˇ3 2 3 ˇˇ x ˇ D 2: x dx D 27 ˇ 0 Z 3 Z 3 2 Since E.X / D 9 variance is Downloaded by ted cage (sxnbyln180@questza.com) 2 1 ˇ2 1 4 ˇˇ 5 x dx D x ˇ D , the 6 ˇ 2 1 Z 2 2 D E.X 2 / Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 . 9 2 (e) E.X/ D 0f .0/C1f .1/C2f .2/C3f .3/ D 1:6500. E.X/ D 0 0 2 D E.X/ D 9 f .1/ D 3 .0:55/.0:45/2 D 0:334125 f .3/ D .0:55/3 D 0:166375: C x dx D b) The mean is f .0/ D .0:45/3 D 0:911125 f .2/ D 3 .0:55/2 .0:45/ D 0:408375 Z 3 3 1 2 D 5 2 196 13 D 81 162 301 lOMoARcPSD|6566483 www.konkur.in SECTION 7.8 (PAGE 449) ADAMS and ESSEX: CALCULUS 9 Since and the standard deviation is r 13 D 0:283: 162 E.X 2 / D D 11. 2 3 X C / D . C /2 . /2 3 D Z C x dx U Dx d V D cos x dx d"U D dx V D ˇ sin Zx # ˇ 1 2 ˇ C 2 x sin x ˇ sin x dx D ˇ 2 0 a) C is given by 0 ˇ1 Z 1 C 3 ˇˇ C 2 1D C x dx D x ˇ D : 3 ˇ 3 0 D D . C / . r !3 3 3 D C 4 80 3 / 3 4 r D 3 80 x 2 dx Hence, C D 0 C sin x dx D 1 . 2 8 0:684: 4 Z 1 2 X C / D ˇ ˇ ˇ C cos x ˇ D 2C: ˇ 0 We have f .x/ D C.x Z C sin x dx x 2 / on Œ0; 1. a) C is given by 1D Z 1 C.x 0 2 x / dx D C Hence, C D 6. 1 D E.X/ D 2 x2 2 x3 3 ˇˇ1 C ˇ ˇ D : ˇ 6 0 b) The mean, variance, and standard deviation are x sin x dx 0 U Dx d V D sin x dx d U D dx ˇ V D cos x ˇ Z 1 ˇ D x cos x ˇ C cos x dx ˇ 2 0 0 D 1:571: D 2 Telegram: @uni_k 2 i 1h cos. C / cos. / 2 D sin sin D sin 0:632: 13. b) The mean is 302 Pr. 0:668: a) To find C , we calculate 1D s D 12. We have f .x/ D C sin x on Œ0; . Z 2 2 8 D 0:467 4 4 4 2 c) Then !3 2 and the standard deviation is c) We have X C / D 3 4/: 2 D E.X 2 / 2 D b) The mean, variance, and standard deviation are Z 1 3 D E.X/ D 3 x 3 dx D 4 0 Z 1 9 3 9 3 2 D E.X 2 / 2 D 3 x 4 dx D D 16 5 16 80 0 p D 3=80: Z C 1 2 . 2 Hence, the variance is 0 Hence, C D 3. 3 x 2 sin x dx 0 0 4 0:5875: 3 We have f .x/ D C x 2 on Œ0; 1. Pr. Z d V D sin x dx U D x2 d U D 2x dx ˇ VZD cos x ˇ 1 ˇ x 2 cos x ˇ C 2 x cos x dx D ˇ 2 0 c) We have Pr. 1 2 1 .x 2 x 3 / dx D 2 0 Z 1 2 D6 .x 3 x 4 / dx D E.X/ D 6 2 D E.X 2 / Z 1 3 1 1 D D 10 4 20 p D 1=20: Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 0 1 4 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.8 (PAGE 449) c) Finally, c) We have Pr. Z .1=2/C .x X C / D 6 .1=2/ " # Z .1=2/C 1 1 2 D6 dx x 4 2 .1=2/ Let u D x 21 du D dx Z 1 u2 du D 12 D 12 4 4 0 1 12 1 0:626: D p 20 4 60 14. X C / Z C D k2 xe kx dx Let u D kx du D k dx Z k.C / D ue u du x 2 / dx Pr. k. / 3 ue D p e .2C 0:738: 2/ C e .2 p 2/ It was shown in Section 6.1 (p. 349) that Z If In D n x e Z 1 x n dx D x e Since I0 D Z 1 0 n R e R 15. x n 1 x e dx: 1DC if n 1: e x dx D 1, therefore In D nŠ for n 1. x n e kx dx D 1 k nC1 Z 1 0 1 nŠ In D nC1 : k nC1 k un e u du D Now let f .x/ D C xe kx on Œ0; 1/. a) To find C , observe that 1DC Z 1 0 xe kx dx D C : k2 E.X 2 / D k 2 Z 1 0 0 2 e x dx D C 2 Z 1 1 2 e x dx D C p 2 : b) The mean, variance, and standard deviation are ˇ Z 1 2 1 e x ˇˇ 2 1 x2 dx D p ˇ D p D p xe 0 ˇ 0 Z 1 1 2 2 2 D x 2 e x dx Cp 0 2 d V D xe x dx 2 V D 12 e x ˇ1 ! Z 1 2 x x 2 ˇˇ 1 1 x2 D Cp e e dx ˇ C ˇ 2 2 0 p0 1 2 1 1 1 D Cp D 0C 2 2 2 r 1 1 D 0:426: 2 U Dx d U D dx Z C 2 2 e x dx X C / D p p Let x D z= 2 p dx D dz= 2 r Z p 2.C / 2 2 D e z =2 dz: p 2. / p p But 2. / 0:195 and 2. C / 1:40. Thus, if Z is a standard normal random variable, we obtain by interpolation in the table on page 386 in the text, Pr. b) The mean is Since Z 1 c) We have Hence, C D k 2 . D E.X/ D k 2 a) We have p Thus C D 2= . C nIn 1 D nIn 1 Let u D kx; then 0 Cn Z 0 R!1 Z 1 x x n e x dx, then In D lim Z 1 0 x 2 e kx dx D k 2 x 3 e kx dx D k 2 then the variance is 2 D E.X 2 / 2 D 6 k2 6 k4 D D 6 ; k2 2 k3 4 2 D 2 k2 k p 2 : k Pr. X C / D 2Pr.0:195 Z 1:400/ 2.0:919 0:577/ 0:68: Copyright © 2018 Pearson Canada Inc. 303 2 and the standard deviation is D . k Telegram: @uni_k Z k.C / C e u du ˇ ˇ k. / k. / p p p p 2/e .2 2/ .2 C 2/e .2C 2/ C .2 D 3 ˇk.C / ˇ uˇ Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in SECTION 7.8 (PAGE 449) 16. No. The identity 1 constant C . 17. Z 1 ADAMS and ESSEX: CALCULUS 9 1 2 2 p e .x / =2 2 Z 1 1 2 2 mean D p xe .x / =2 dx 2 1 f; .x/ D b) For f .x/ D ke kx on Œ0; 1/, we know that 1 (Example 6). Thus 2 < 0 and D D k 3 C 2 D . We have k 3 Pr.jX j 2/ D Pr X k Z 1 kx Dk e dx 3=k ˇ1 ˇ D e 3 0:050: D e kx ˇˇ C dx D 1 is not satisfied for any Let z D dz D x 1 dx Z 1 1 2 D p . C z/e z =2 dz 2 1 Z 1 2 e z =2 dz D D p 2 1 variance D E .x /2 Z 1 1 2 2 D p .x /2 e .x / =2 dx 2 1 Z 1 1 2 2 z 2 e z =2 dz D Var.Z/ D D p 2 1 18. Since f .x/ D 2 Z 1 0 3=k 1 2 2 p e .x / =2 , which has mean 2 and standard deviation , we have c) For f; .x/ D Pr.jX 2 > 0 on Œ0; 1/ and .1 C x 2 / 2 2 dx 1 D lim tan .R/ D D 1; R!1 1 C x2 2 therefore f .x/ is a probability density function on Œ0; 1/. The expectation of X is D E.X/ D 2 Z 1 0 from the table in this section. 20. The density function for T is f .t / D ke k t on Œ0; 1/, 1 1 D (see Example 6). Then where k D 20 Z 12 1 e t=20 dt D 1 e t=20 dt 20 0 12 ˇ12 ˇ t=20 ˇ D1Ce ˇ D e 12=20 0:549: ˇ x dx 1 C x2 Pr.T 12/ D 1 ln.1 C R2 / D 1: R!1 D lim D 304 Telegram: @uni_k D b a p : 2 3 bCa b a C p > b, and similarly, 2 3 2 < a, therefore Pr.jX j 2/ D 0. Since C 2 D bCa ; 2 Z 1 The probability that the system will last at least 12 hours is about 0.549. 21. a) The density function for the uniform distribution on Œa; b is given by f .x/ D 1=.b a/, for a x b. By Example 5, the mean and standard deviation are given by 1 20 0 No matter what the cost per game, you should be willing to play (if you have an adequate bankroll). Your expected winnings per game in the long term is infinite. 19. j 2/ D 2Pr.X 2/ Z 2 1 2 2 D2 p e .x / =2 dx 2 1 x Let z D 1 dz D dx Z 2 2 z2 e dz D p 2 1 D 2Pr.Z 2/ 2 0:023 D 0:046 If X is distributed normally, with mean D 5; 000, and standard deviation D 200, then Pr.X 5500/ Z 1 1 2 2 p e .x 5000/ =.2200 / dx D 200 2 5500 x 5000 Let z D 200 dx dz D 200 Z 1 1 z 2 =2 D p e dz 2 5=2 D Pr.Z 5=2/ D Pr.Z 5=2/ 0:006 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 7.9 (PAGE 458) from the table in this section. 2. 22. If X is the random variable giving the spinner’s value, then Pr.X D 1=4/ D 1=2 and the density function for the other values of X is f .x/ D 1=2. Thus the mean of X is Z 1 1 3 1 1 1 Pr X D x f .x/ dx D C D : C 4 4 8 4 8 0 D E.X/ D Also, Z 1 1 1 1 19 1 Pr X D C C D x 2 f .x/ dx D 16 4 32 6 96 0 9 11 19 D : 2 D E.X 2 / 2 D 96 64 192 E.X 2 / D 3. p 11=192. Thus D 23. 4. (a) The integral exists for p D 0 as a degenerate case. For p > 0, Z 1 x p S˛ .x/dx 1 D 2 lim y!1 D 2 lim y!1 Z x c˛ x .1C˛/ C O x .1C2˛/ dx p c˛ p ˛ x p ˛ CO x p 2˛ jxDy ! 5. This limit will be finite if p < ˛. The case p D ˛ is logarithmically divergent. (b) The mean will not exist for ˛ 1. The variance will not exist for ˛ < 2. 24. 1 D D FS .x/ D Pr.X > x/ D Z 1 6. x2 dy D 2 ) y 2 dy D x 2 dx dx y y3 x3 D C C1 ; or x 3 y 3 D C 3 3 y D 0 is a constant solution. Otherwise, dy D x2y2 dx Z Z dy D x 2 dx y2 1 1 1 D x3 C C y 3 3 3 : ) yD x3 C C Y D 0 is a constant solution. Otherwise dY dY D tY ) D t dt dt Y t2 2 ln Y D C C1 ; or Y D C e t =2 2 dx D e x sin t dt Z Z e x dx D e S˛ ./d x Z 1n o c˛ .1C˛/ C O .1C2˛/ d ) 7. x h c i c˛ ˛ ˛ ˛ C O 2˛ x ; ˛ ˛ jDx where higher order terms were discarded in the last step. Section 7.9 First-Order Differential Equations (page 458) 1. y D 1=3 is a constant solution. Otherwise 3y 1 dy D dx x Z Z dy dx D 3y 1 x 1 1 ln j3y 1j D ln jxj C ln C 3 3 3y 1 DC x3 1 ) y D .1 C C x 3 /: 3 y D 0 is a constant solution. Otherwise y dy D dx 2x dy dx 2 D y x 2 ln y D ln x C C1 ) y2 D C x 8. x D xD sin t dt cos t C ln.cos t C C /: y D 1 and y D 1 are constant solutions, Otherwise dy dy D dx D 1 y2 ) dx 1 y2 1 1 1 C dy D dx 2 1Cy 1 y ˇ ˇ 1 ˇˇ 1 C y ˇˇ ln D x C C1 2 ˇ1 yˇ 1Cy C e 2x 1 D C e 2x or y D 1 y C e 2x C 1 dy D 1 C y2 dx Z Z dy D dx 1 C y2 tan 1 y D x C C ) y D tan.x C C /: Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 305 lOMoARcPSD|6566483 www.konkur.in SECTION 7.9 (PAGE 458) 9. ADAMS and ESSEX: CALCULUS 9 dy dy D dt D 2 C ey ) dt 2 C ey Z Z y e dy D dt 2e y C 1 1 ln.2e y C 1/ D t C C1 2 2e y C 1 D C2 e 2t ; or y D ln C e 2t 13. 1 2 10. y D 0 and y D 1 are constant solutions. For the other solutions we have dy D y 2 .1 y/ dx Z Z dy D dx D x C K: y 2 .1 y/ Expand the left side in partial fractions: C A B 1 D C 2 C y 2 .1 y/ y y 1 y A.y y 2 / C B.1 y/ C Cy 2 D y 2 .1 y/ ( A C C D 0I ) A B D 0I ) A D B D C D 1: B D 1: Hence, Z Z 1 1 1 dy D C C dy y 2 .1 y/ y y2 1 y 1 D ln jyj ln j1 yj: y Therefore, ˇ ˇ ˇ y ˇ ˇ ln ˇˇ 1 yˇ dy dx ) 306 Telegram: @uni_k yD 1 C C 2: x x We have and R dy Cy D e x . Let D dx D x, then e D e x , dx dy dy d x .e y/ D e x C ex y D ex C y D e 2x dx dx dx Z 1 2x x 2x ) e y D e dx D e C C: 2 Hence, y D 15. 1 D x C K: y 2 y D x2 (linear) x Z 2 1 D exp dx D 2 x x 1 dy 2 yD1 x 2 dx x 3 d y D1 dx x 2 y D x C C; so y D x 3 C C x 2 x2 2y 1 dy . Let C D 12. We have 2 dx x x Z 2 D dx D 2 ln x D ln x 2 , then e D x 2 , and x d 2 dy .x y/ D x 2 C 2xy dx dx 2y 1 dy C D x2 D1 D x2 dx x x2 Z ) x2y D dx D x C C 11. 14. Z dy 2 dx D e 2x C 2y D 3 D exp dx d 2x .e y/ D e 2x .y 0 C 2y/ D 3e 2x dx 3 3 e 2x y D e 2x C C ) y D C C e 2x 2 2 16. 1 x e C Ce x. 2 Z dy 1 dx D e x Cy Dx D exp dx d x .e y/ D e x .y 0 C y/ D xe x dx Z ex y D xe x dx D xe x yDx 1 C Ce x ex C C R dy We have C 2e x y D e x . Let D 2e x dx D 2e x , dx then d 2ex x x dy C 2e x e 2e y e y D e 2e dx dx dy x x D e 2e C 2e x y D e 2e e x : dx Therefore, x e 2e y D Hence, y D Z Let u D 2e x du D 2e x dx Z 1 2ex 1 u e du D e C C: D 2 2 1 x C C e 2e . 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x e 2e e x dx lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 17. dy C 10y D 1; dt Z D 18. 1 10 y 10 dt D 10t 2 D 10 22. dy d 10t .e y/ D e 10t C 10e 10t y D e 10t dt dt 1 10t e 10t y.t / D e CC 10 e e 2e 1 2 D CC ) C D y 10 D 10 ) 10 10 10 1 1 yD C e 1 10t : 10 10 dy C 3x 2 y D x 2 ; y.0/ D 1 dx Z D 23. 24. y D 1=.1 tan 1 x/: y.x/ D 1 C Z x 20. sin x y C .cos x/y D 2xe ; Z D cos x dx D sin x dy D e y; dx ey D x C C y./ D 0 ) 0 D C C ) C D 21. 0 sin x y D ln.x C e 3 /: 25. ÷ 2 2 D0 CC ÷ p y D 4 C x2: 26. C D e3 ab e .b a/k t be .b a/k t ab.0 1/ D b: D 0 a t!1 1 a Since b > a > 0 and k > 0, lim x.t / D lim t!1 2 t!1 D lim ab e .b a/k t 1 .b a/k t be a ab 1 e .a b/k t b ae .a b/k t ab.1 0/ D D a: b 0 y.0/ D 2 27. The solution given, namely C D4 xD Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k y.1/ D 1 y D ln.x C C / ÷ t!1 t dt y.t / ÷ Since a > b > 0 and k > 0, : x dy D ; i.e. y dy D x dx dx y y2 D x2 C C 2 ÷ lim x.t / D lim y.0/ D 1 i.e. e y dy D dx y./ D 0 2 2 /e Z x y.x/ D 2 C y.t / dt t .t C 1/ 3 D y.0/ D ln C 2x dx D x 2 C C y D .x 2 1 t!1 d sin x .e y/ D e sin x .y 0 C .cos x/y/ D 2x dx Z e sin x y D ÷ y dy D ; for x > 0 dx x.x C 1/ dx dx dx dy D D y x.x C 1/ x xC1 x ln y D ln C ln C xC1 Cx ; ÷ 1 D C =2 yD xC1 2x yD : xC1 Z x y.x/ D 3 C e y dt ÷ y.0/ D 3 y.1/ D 3e ) 3 D 1 C C ) C D 2 0 0 .y.t //2 dt 1 C t2 0 x 2 y 0 C y D x 2 e 1=x ; y.1/ D 3e 1 1=x 0 y C 2y D e Zx 1 1 D dx D x2 x d 1=x 1 e y D e 1=x y 0 C 2 y D 1 dx x Z e 1=x y D 1 dx D x C C y D .x C 2/e 1=x : Z x y.x/ D 1 C y2 dy D ; i.e. dy=y 2 D dx=.1 C x 2 / dx 1 C x2 1 D tan 1 x C C y 1D0CC ÷ C D 1 3x 2 dx D x 3 d x3 3 3 dy 3 .e y/ D e x C 3x 2 e x y D x 2 e x dx dx Z 1 3 3 3 e x y D x 2 e x dx D e x C C 3 1 2 y.0/ D 1 ) 1 D C C ) C D 3 3 2 1 3 yD C e x : 3 3 19. SECTION 7.9 (PAGE 458) Downloaded by ted cage (sxnbyln180@questza.com) ab e .b a/k t 1 be .b a/k t a ; 307 lOMoARcPSD|6566483 www.konkur.in SECTION 7.9 (PAGE 458) ADAMS and ESSEX: CALCULUS 9 Let a2 D mg=k, where a > 0. Thus, we have is indeterminate (0/0) if a D b. If a D b the original differential equation becomes Z dx D k.a x/2 ; dt which is separable and yields the solution Z Z dx 1 D k dt D k t C C: D a x .a x/2 1 1 1 Since x.0/ D 0, we have C D , so D kt C . a a x a Solving for x, we obtain kt dv D CC a2 ˇ v 2 ˇ m ˇa C vˇ 1 ˇ D kt C C ln ˇ 2a ˇ a v ˇ m r ˇ ˇ ˇa C vˇ ˇ D 2ak t C C1 D 2 kg t C C1 ln ˇˇ a vˇ m m p aCv 2t kg=m D C2 e : a v a2 k t : 1 C ak t This solution also results from evaluating the limit of solution obtained for the case a ¤ b as b approaches a (using l’H^opital’s Rule, say). xD dv D mg kv has constant solution dt v D mg=k. However this solution does not satisfy the initial condition v.0/ D 0. For other solutions we separate variables: Z Z dv D dt k g v ˇm ˇ k ˇˇ m ˇˇ lnˇg v ˇ D t C C: k m m k Since v.0/ D 0, therefore C D ln g. Also, g v k m remains positive for all t > 0, so g m ln Dt k k v g m k g v m D e k t=m g mg ) v D v.t / D 1 e k t=m : k mg , the constant solution of the Note that lim v.t / D t!1 k differential equation noted earlier. This limiting velocity can be obtained directly from the differential equation by dv setting D 0. dt p 29. y D mg=k is a constant solution of the equation. For other solutions we proceed by separation of variables: Assuming v.0/ D 0, we get C2 D 1. Thus p a C v D e 2t kg=m .a v/ p p v 1 C e 2t kg=m D a e 2t kg=m 1 r mg 2t pkg=m e D 1 k p r mg e 2t kg=m 1 p vD k e 2t kg=m C 1 28. The equationt m dv D mg kv 2 dt k 2 dv Dg v dt m dv D dt k 2 g v Z m Z k kt dv D dt D C C: mg 2 m m v k m 308 Telegram: @uni_k Clearly v ! setting 30. r mg as t ! 1. This also follows from k dv D 0 in the given differential equation. dt The balance in the account after t years is y.t / and y.0/ D 1000. The balance must satisfy dy y2 D 0:1y dt 1; 000; 000 dy 105 y y 2 D dt 106 Z Z dy dt D 105 y y 2 106 Z 1 1 t 1 C 5 dy D 6 105 y 10 y 10 t ln jyj ln j105 yj D C 10 105 y D e C .t=10/ y 105 y D C .t=10/ : e C1 C 105 Since y.0/ D 1000, we have 1000 D y.0/ D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 105 eC C 1 ) C D ln 99; lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL and Hence, 105 yD 99e t=10 C 1 The balance after 1 year is 105 yD 99e 1=10 C 1 REVIEW EXERCISES 7 (PAGE 459) : .500 C t /5 x D 0:12 $1; 104:01: 105 t!1 e .4:60 0:1t/ C 1 t!1 D .500 C t /5 dt D 0:02.500 C t /6 C C ) x D 0:02.500 C t / C C.500 C t / 5 : Since x.0/ D 50, we have C D 1:25 1015 and As t ! 1, the balance can grow to lim y.t / D lim Z x D 0:02.500 C t / C .1:25 1015 /.500 C t / 5 : 105 D $100; 000: 0C1 After 40 min, there will be For the account to grow to $50,000, t must satisfy x D 0:02.540/ C .1:25 1015 /.540/ 5 D 38:023 kg 100; 000 50; 000 D y.t / D 99e t=10 C 1 ) 99e t=10 C 1 D 2 ) t D 10 ln 99 46 years: of salt in the tank. Review Exercises 7 (page 459) 31. The hyperbolas xy D C satisfy the differential equation y Cx dy D 0; dx or dy D dx y : x 1. 3 cm Curves that intersect these hyperbolas at right angles must x dy D , or x dx D y dy, a separated therefore satisfy dx y equation with solutions x 2 y 2 D C , which is also a family of rectangular hyperbolas. (Both families are degenerate at the origin for C D 0.) 1 cm 5 cm 32. Let x.t / be the number of kg of salt in the solution in the tank after t minutes. Thus, x.0/ D 50. Salt is coming into the tank at a rate of 10 g=L 12 L=min D 0:12 kg=min. Since the contents flow out at a rate of 10 L=min, the volume of the solution is increasing at 2 L=min and thus, at any time t , the volume of the solution is 1000 C 2t L. x.t / L. Hence, Therefore the concentration of salt is 1000 C 2t salt is being removed at a rate Fig. R-7-1 The volume of thread that can be wound on the left spool is .32 12 /.5/ D 40 cm3 . The height of the winding region of the right spool at distance r from the central axis of the spool is of the form h D A C Br. Since h D 3 if r D 1, and h D 5 if r D 3, we have A D 2 and B D 1, so h D 2 C r. The volume of thread that can be wound on the right spool is Therefore, Z ˇ3 100 r 3 ˇˇ 2 cm3 : 2 r.2 C r/ dr D 2 r C ˇ D ˇ 3 3 1 Z 3 5x dx D 0:12 dt 500 C t dx 5 C x D 0:12: dt 500 C t 5 dt D 5 ln j500 C t j D ln.500 C t /5 for 500 C t t > 0. Then e D .500 C t /5 , and i dx d h C 5.500 C t /4 x .500 C t /5 x D .500 C t /5 dt dy dx 5x D .500 C t /5 C dy 500 C t D 0:12.500 C t /5 : 1 100 .1; 000/ D 833:33 m of The right spool will hold 3 40 thread. 2. Let A.y/ be the cross-sectional area of the bowl at height y above the bottom. When the depth of water in the bowl is Y , then the volume of water in the bowl is Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 cm 5 cm 3 cm 1 cm 5x.t / x.t / kg=L 10 L=min D kg=min: 1000 C 2t 500 C t Let D 3 cm Downloaded by ted cage (sxnbyln180@questza.com) V .Y / D Z Y A.y/ dy: 0 309 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 7 (PAGE 459) ADAMS and ESSEX: CALCULUS 9 The water evaporates at a rate proportional to exposed surface area. Thus 60ı dV D kA.Y / dt dV dY D kA.Y / d Y dt dY A.Y / D kA.Y /: dt 10 cm Hence d Y =dt D k; the depth decreases at a constant rate. 3. The barrel is generated by revolving x D a by 2 , . 2 y 2/, about the y-axis. Since the top and bottom disks have radius 1 ft, we have a 4b D 1. The volume of the barrel is V D2 Z 2 .a 0 D 2 a2 y DD 2 2a2 Fig. R-7-4 5. by 2 /2 dy ˇˇ2 ˇ ˇ ˇ 0 32 16 ab C b 2 : 3 5 2aby 3 b2 y 5 C 3 5 6. 32 16 b.1 C 4b/ C b 2 3 5 60 2 D 0: 128b C 80b C 15 1 cosh.ax/ from x D 0 to x D 1 is a Z 1q Z 1 sD 1 C sinh2 .ax/ dx D cosh.ax/ dx 0 0 ˇ1 ˇ 1 1 ˇ D sinh.ax/ˇ D sinh a: ˇ a a The arc length of y D 0 Since V D 16 and a D 1 C 4b, we have 2 2.1 C 4b/2 x D 16 Solving this quadratic gives two solutions, b 0:0476 and b 0:6426. Since the second of these leads to an unacceptable negative value for a, we must have b 0:0476, and so a D 1 C 4b 1:1904. 1 We want sinh a D 2, that is, sinh a D 2a. Solving this a by Newton’s Method or a calculator solve function, we get a 2:1773. p The area of revolution of y D x, .0 x 6/, about the x-axis is s 2 Z 6 dy S D 2 dx y 1C dx 0 r Z 6 p 1 x 1C dx D 2 4x 0 r Z 6 1 D 2 x C dx 4 0 ˇ6 4 4 125 1 1 3=2 ˇˇ 62 D sq. units. xC D ˇ D ˇ 3 4 3 8 8 3 0 4. A vertical slice parallel to the top ridge of the solid at distance x to the right of the centre p p is a rectangle of base 2 100 x 2 cm�and height 3.10 x/ cm. Thus the solid has volume V D2 Z 10 p 3.10 p x/2 100 0 p Z 10 p D 40 3 100 0 x 2 dx x 2 dx p Z 10 p 4 3 x 100 0 Let u D 100 x 2 du D 2x dx p 100 p Z 100 p u du D 40 3 2 3 4 0 p 4 D 1; 000 3 cm3 : 3 310 Telegram: @uni_k x 2 dx 7. The region is a quarter-elliptic disk with semi-axes a D 2 and b D 1. The area of the region is A D ab=4 D =2. The moments about the coordinate axes are s Z 2 x2 x2 dx Let u D 1 MxD0 D x 1 4 4 0 x du D dx 2 Z 1 p 4 D2 u du D 3 0 Z 1 2 x2 MyD0 D dx 1 2 0 4 ˇ 2 2 x 3 ˇˇ 1 x D ˇ D : 2 12 ˇ 3 0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 7 (PAGE 459) Thus x D MxD0 =A D 8=.3/ and y D MyD0 =A D 4=.3/. The centroid is or, equivalently, f .a/ C g.a/ D 2a. Thus f and g must satisfy 8=.3/; 4=.3/ . f .x/ C g.x/ D 2x for every x > 0. 8. y 11. 1 Z dy 3y dx dy D ) D3 dx x 1 y x 1 ) ln jyj D ln jx 1j3 C ln jC j ) y D C.x 1/3 : Since y D 4 when x D 2, we have 4 D C.2 the equation of the curve is y D 4.x 1/3 . 3 x 1/3 D C , so 12. The ellipses 3x 2 C 4y 2 D C all satisfy the differential equation Fig. R-7-8 6x C 8y Let the disk have centre (and therefore centroid) at .0; 0/. Its area is 9. Let the hole have centre (and therefore centroid) at .1; 0/. Its area is . The remaining part has area 8 and centroid at .x; 0/, where 13. 5 10. We are told that for any a > 0, Z a h Z a h 2 2 i i f .x/ g.x/ dx D 2 x f .x/ g.x/ dx: 0 0 Differentiating both sides of this equation with respect to a, we get 2 f .a/ 2 h g.a/ D 2a f .a/ i g.a/ ; 0 sin.2 t /e 0:04.2 t/ dt $8; 798:85: Challenging Problems 7 (page 459) 1. a) The nth bead extends from x D .n 1/ to x D n, and has volume Z n Vn D e 2kx sin2 x dx .n 1/ Z n D e 2kx .1 cos.2x// dx 2 .n 1/ Let x D u C .n Zdx D du 1/ 2ku 2k.n 1/ h 1 e e 2 0 Z D e 2k.n 1/ e 2ku .1 2 0 D e 2k.n 1/ V1 : D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Z 2 (We omit the details of evaluation of the integral, which is done by the method of Example 4 of Section 7.1.) We are told that F D 1; 000 N when y D 20 cm. Thus k D 20; 000 Ncm. The work done by the piston as it descends to 5 cm is 20; 000 20 dy D 20; 000 ln 27; 726 Ncm: y 5 3x : 4y The original $8,000 grows to $8; 000e 0:08 in two years. Between t and t C dt , an amount $10; 000 sin.2 t / dt comes in, and this grows to $10; 000 sin.2 t /e 0:04.2 t/ dt by the end of two years. Thus the amount in the account after 2 years is 8; 000e 0:08 C10; 000 k kA D : F .y/ D P .y/A D Ay y Z 20 dy D dx The family is given by y 3 D C x 4 . 9. Let the area of cross-section of the cylinder be A. When the piston is y cm above the base, the volume of gas in the cylinder is V D Ay, and its pressure P .y/ satisfies P .y/V D k (constant). The force exerted by the piston is W D or A family of curves that intersect these ellipses at right 4y dy D . angles must therefore have slopes given by dx 3x Thus Z Z dy dx 3 D4 y x 3 ln jyj D 4 ln jxj C ln jC j: .9/.0/ D .8/x C ./.1/: Thus x D 1=8. The centroid of the remaining part is 1=8 ft from the centre of the disk on the side opposite the hole. dy D 0; dx Downloaded by ted cage (sxnbyln180@questza.com) cos.2u C 2.n cos.2u// du i 1// du 311 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 7 (PAGE 459) ADAMS and ESSEX: CALCULUS 9 e 2k n V1 VnC1 D D e 2k ; which deVn e 2k.n 1/ V1 pends on k but not n. or, equivalently, a.100 k 2 /2 D 4. The volume of the pool is Thus b) VnC1 =Vn D 1=2 if if k D .ln 2/=.2/. 2k D ln.1=2/ D ln 2, that is, VP D 2a D 2a c) Using the result of Example 4 in Section 7.1, we calculate the volume of the first bead: Z 2kx V1 D e .1 cos.2x// dx 2 0 ˇ ˇ e 2kx .2 sin.2x/ 2k cos.2x// ˇˇ e 2kx ˇˇ D ˇ ˇ ˇ 4k ˇ 2 4.1 C k 2 / 0 0 2k 2k .1 e / .k ke / D 4k 4.1 C k 2 / D .1 e 2k /: 4k.1 C k 2 / VH D 2a .1 4k.1 C k 2 / D .1 4k.1 C k 2 / e 2k / k 2 / dr 250; 000 3 1 6 k : 12 2; 500k 2 C 25k 4 Z k 0 r.r 2 100/.r 2 k 2 / dr D 2a 25k 4 1 6 k : 12 These two volumes must be equal, so k 2 D 100=3 and k 5:77 m. Thus a D 4=.100 k 2 /2 D 0:0009. The volume of earth to be moved is VH with these values of a and k, namely " 2.0:0009/ 25 .1 e 2k / 4k.1 C k 2 / h 2 n 1 i 1 C e 2k C e 2k C C e 2k D r 2 /.r 2 r.100 k The volume of the hill is By part (a) and Theorem 1(d) of Section 6.1, the sum of the volumes of the first n beads is Sn D Z 10 3. 100 3 2 1 12 100 3 4 # 140 m3 : y D ax C bx 2 C cx 3 .h; r/ y 1 e 2k n 1 e 2k x e 2k n /: Thus the total volume of all the beads is V D lim Sn D n!1 cu. units.: 4k.1 C k 2 / Fig. C-7-3 f .x/ D ax C bx 2 C cx 3 must satisfy f .h/ D r, f 0 .h/ D 0, and f 0 .x/ > 0 for 0 < x < h. The first two conditions require that ah C bh2 C ch3 D r 2. 10 m a C 2bh C 3ch2 D 0; 1m from which we obtain by solving for b and c, bD 3r 2ah ; h2 cD ah 2r h3 : The volume of the nose cone is then Fig. C-7-2 h.r/ D a.r 2 h0 .r/ D 2ar.r 2 100/.r 2 k 2 /; k 2 / C 2ar.r 2 V .a/ D where 0 < k < 10 100/ D 2ar.2r 2 2 2 100 The deepest point occurs where 2r D 100 C k , i.e., r 2 D 50 C .k 2 =2/. Since this depth must be 1 m, we require 2 k2 k 50 50 D 1; a 2 2 312 Telegram: @uni_k k 2 /: Z h 2 h .13ahr C 78r 2 C 2a2 h2 /: f .x/ dx D 210 0 Solving d V =da D 0 gives only one critical point, a D 13r=.4h/. This is unacceptable, because the condition f 0 .x/ > 0 on .0; h/ forces us to require a 0. In fact f 0 .x/ D a C Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2.3r 2ah/ h2 xC 3.ah 2r/ 2 x h3 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 7 (PAGE 459) To minimize this expression for a > 0 we should take k D 0. This gives f .x/ D ax 2 .1 x/. To minimize V .a; k/ for a < 0 we should take k D 1. This gives f .x/ D ax.1 x/2 . Since we want the maximum value of f to be 2 in either case, we calculate the critical points of these two possible functions. For a > 0 the CP is x D 2=3 and f .2=3/ D 2 gives a D 27=2. The volume in this case is V .27=2; 0/ D .27=60/.78 0/. For a < 0 the CP is x D 1=3 and f .1=3/ D 2 gives a D 27=2. The volume in this case is V . 27=2; 1/ D .27=60/.78 155/ D .27=60/.77/. Thus the minimum volume occurs for f .x/ D .27=2/x.1 x/2 , i.e. b D a D 27=2. is clearly positive for small x if a > 0. Its two roots are x1 D h and x2 D h2 a=.3ah 6r/. a must be restricted so that x2 is not in the interval .0; h/. If a < 2r= h, then x2 < 0. If 2r= h < a < 3r= h, then x2 > h. If a > 3r= h, then 0 < x2 < h. Hence the interval of acceptable values of a is 0 a 3r= h. We have V .0/ D 13 r 2 h ; 35 V 3r h D 9 r 2 h : 14 The largest volume corresponds to a D 3r= h, which is the largest allowed value for a and so corresponds to the bluntest possible nose. The corresponding cubic f .x/ is f .x/ D 4. r .3h2 x h3 3hx 2 C x 3 /: a C bx C cx 2 for 0 x 1 a) If f .x/ D , then 2 for 1 x 3 p C qx C rx b C 2cx for 0 < x < 1 . We require that f 0 .x/ D q C 2rx for 1 < x < 3 aD1 aCbCc D2 b C 2c D m p C 3q C 9r D 0 pCqCr D2 q C 2r D m: The solutions of these systems are a D 1, b D 2 m, c D m 1, p D 32 .1 m/, q D 2m C 1, and r D 12 .1 C m/. f .x; m/ is f .x/ with these values of the six constants. b) The length of the spline is L.m/ D Z 3p Z 1p 1 C .b C 2cx/2 dx C 1 C .q C 2rx/2 dx 1 0 with the values of b, c, q, and r determined above. A plot of the graph of L.m/ reveals a minimum value in the neighbourhood of m D 0:3. The derivative of L.m/ is a horrible expression, but Mathematica determined its zero to be about m D 0:281326, and the corresponding minimum value of L is about p 4:41748. The polygonal line ABC has length 3 2 4:24264, which is only slightly shorter. 5. Let b D ka so that the cross-sectional curve is given by y D f .x/ D ax.1 x/.x C k/: The requirement that f .x/ 0 for 0 x 1 is satisfied provided either a > 0 and k 0 or a < 0 and k 1. The volume of the wall is V .a; k/ D Z 1 0 2.15 C x/f .x/ dx D a .78 C 155k/: 30 6. Starting with V1 .r/ D 2r, and using repeatedly the formula Vn .r/ D p Vn 1 . r 2 x 2 / dx; r Maple gave the following results: V1 .r/ D 2r 4 V3 .r/ D r 3 3 8 2 5 r V5 .r/ D 15 16 3 7 V7 .r/ D r 105 32 4 9 r V9 .r/ D 945 V2 .r/ D r 2 1 V4 .r/ D 2 r 4 2 1 V6 .r/ D 3 r 6 6 1 4 8 V8 .r/ D r 24 1 5 10 r V10 .r/ D 120 It appears that 1 n 2n r ; and nŠ 2n V2n 1 .r/ D n 1 r 2n 1 1 3 5 .2n 1/ 22n 1 .n 1/Š n 1 2n 1 r : D .2n 1/Š V2n .r/ D These formulas predict that V11 .r/ D 211 5Š 5 11 r 11Š and V12 .r/ D 1 6 12 r ; 6Š both of which Maple is happy to confirm. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Z r Downloaded by ted cage (sxnbyln180@questza.com) 313 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 7 (PAGE 459) 7. ADAMS and ESSEX: CALCULUS 9 With y and as defined in the statement of the problem, we have 8. y Q y D f .x/ 0 y 10 and L 0 < : P .x; y/ x .L; 0/ The needle crosses a line if y < 5 sin . The probability of this happening is the ratio of the area under the curve to the area of the rectangle in the figure, that is, Fig. C-7-8 If Q D .0; Y /, then the slope of PQ is y x Pr D 1 10 Z 0 5 sin d D Y dy D f 0 .x/ D : 0 dx Since jPQj D L, we have .y Y /2 D L2 px 2 . Since the slope dy=dx is negative at P , dy=dx D L2 x 2 =x. Thus 1 : yD p Z p 2 L C L2 L x2 dx D L ln x x x2 ! p L2 Since y D 0 when x D L, we have C D 0 and the equation of the tractrix is LC y D L ln p L2 x x2 ! p L2 x2 : Note that the first term can be written in an alternate way: y y D L ln y D 10 L p x L2 x2 p L2 x2: y D 5 sin x Fig. C-7-7 314 Telegram: @uni_k 9. a) S.a; a; c/ is the area of the surface obtained by rotating the ellipse .x 2 =a2 / C .y 2 =c 2 / D 1p(where a > c) about the y-axis. Since y 0 D cx=.a a2 x 2 /, we Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x 2 CC: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 7 (PAGE 459) have S.a; a; c/ D 2 2 Z a x 0 s 1C c2x2 a2 .a2 x2 / b a S.a; b; c/ c c aC c c a a b c c, we use dx Z a p 4 a .a2 c 2 /x 2 dx p x a2 x 2 0 Let x D a sin u dx D a cos u du Z q 4 =2 D a sin u a4 .a2 c 2 /a2 sin2 u du a 0 Z =2 p D 4a sin u a2 .a2 c 2 /.1 cos2 u/ du D c) Since b D 4 a b a S.a; a; c/ C a a b c S.a; c; c/: 0 Let v D cos u dv D sin u du Z 1p D 4a c 2 C .a2 c 2 /v 2 dv: 0 d) We cannot evaluate S.3; 2; 1/ even numerically at this stage. The double integral necessary to calculate it is not treated until a later chapter. (The value is approximately 48.882 sq. units.) However, using the formulas obtained above, This integral can now be handled using tables or computer algebra. It evaluates to aC 2ac 2 ln S.a; a; c/ D 2a C p a2 c 2 2 p a2 c c2 ! : b) S.a; c; c/ is the area of the surface obtained by rotating the ellipse p of part (a) about the y-axis. Since y 0 D cx=.a a2 x 2 /, we have S.a; c; c/ D 2 2 Z a 0 y s 1C c2x2 a2 .a2 x2/ S.3; 3; 1/ C S.3; 1; 1/ S.3; 2; 1/ 2 p 1 6 18 D 18 C p ln.3 C 8/ C 2 C p cos 1 .1=3/ 2 8 8 49:595 sq. units. dx p Z 4c a p 2 a4 .a2 c 2 /x 2 2 dx a x p 2 a a2 x 2 Z0 a p 4c D 2 a4 .a2 c 2 /x 2 dx a 0 s Z a a2 c 2 2 D 4c 1 x dx a4 0 c 2a2 c cos 1 : D 2c 2 C p 2 2 a a c D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 315 lOMoARcPSD|6566483 www.konkur.in SECTION 8.1 (PAGE 472) ADAMS and ESSEX: CALCULUS 9 CHAPTER 8. CONICS, PARAMETRIC CURVES, AND POLAR CURVES Section 8.1 Conics 1. 8. If x 2 C 4y 2 x2 C 4 y2 (page 472) 4y D 0, then yC 1 4 D 1; or 1 2 / 2 1 4 D 1: 1 This represents an ellipse with centre at 0; , semi2 p 3 1 1 major axis 1, semi-minor axis , and foci at ˙ ; . 2 2 2 The ellipse with foci .0; ˙2/ has major axis along the yaxis and c D 2. If a D 3, then b 2 D 9 4 D 5. The ellipse has equation y y2 x2 C D 1: 5 9 x 2 C4y 2 4yD0 1 1 2 2. The ellipse with foci .0; 1/ and .4; 1/ has c D 2, centre .2; 1/, and major axis along y D 1. If D 1=2, then a D c= D 4 and b 2 D 16 4 D 12. The ellipse has equation .x 2/2 .y 1/2 C D 1: 16 12 3. A parabola with focus .2; 3/ and vertex .2; 4/ has a D and principal axis x D 2. Its equation is .x 2/2 D 4.y 4/ D 16 4y. 1 x Fig. 8.1-8 1 4. A parabola with focus at .0; 1/ and principal axis along y D 1 will have vertex at a point of the form .v; 1/. Its equation will then be of the form .y C 1/2 D ˙4v.x v/. The origin lies on this curve if 1 D ˙4. v 2 /. Only the sign is possible, and in this case v D ˙1=2. The possible equations for the parabola are .y C 1/2 D 1 ˙ 2x. 9. If 4x 2 C y 2 4y D 0, then 4x 2 C y 2 4y C 4 D 4 2 2/2 D 4 4x C .y x2 C 2/2 .y 4 D1 This is an ellipse with semi-axes 1 and 2, centred at .0; 2/. y 4 5. The hyperbola with semi-transverse axis a D 1 and foci .0; ˙2/ has transverse axis along the y-axis, c D 2, and b 2 D c 2 a2 D 3. The equation is y2 .y x2 C 1 . 1;2/ 4x 2 Cy 2 4yD0 .1;2/ 2 x2 D 1: 3 x 6. The hyperbola with foci at .˙5; 1/ and asymptotes x D ˙.y 1/ is rectangular, has centre at .0; 1/ and has transverse axis along the line y D 1. Since c D 5 and a D b (because the asymptotes are perpendicular to each other) we have a2 D b 2 D 25=2. The equation of the hyperbola is 25 x 2 .y 1/2 D : 2 7. If x 2 C y 2 C 2x D 1, then .x C 1/2 C y 2 D 0. This represents the single point . 1; 0/. 316 Telegram: @uni_k Fig. 8.1-9 10. If 4x 2 4x 2 y2 4y D 0, then .y 2 C 4y C 4/ D 4; or x2 1 .y C 2/2 D 4 1: This represents a hyperbola with centre at .0; 2/, semitransversepaxis 2, semi-conjugate axis 1, and foci at .0; 2 ˙ 5/. The asymptotes are y D ˙2x 2. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 8.1 (PAGE 472) 3 This is a hyperbola with centre ; 1 , and asymptotes p 2 the straight lines 2x C 3 D ˙2 2.y 1/. y y x 2 4x 2 y 2 4yD0 3 ;1/ 2 . 1 . 3;1/ x 2 2y 2 C3xC4yD2 x Fig. 8.1-10 11. If x 2 C 2x y D 3, then .x C 1/2 y D 4. Thus y D .x C 1/2 4. This is a parabola with vertex . 1; 4/, opening upward. y Fig. 8.1-13 14. If 9x 2 C 4y 2 18x C 8y D 9.x 2 x 13, then 2x C 1/ C 4.y 2 C 2y C 1/ D 0 1/2 C 4.y C 1/2 D 0: ,9.x This represents the single point .1; 1/. x 2 C2x yD3 15. . 1; 4/ If 9x 2 C 4y 2 9.x 2 Fig. 8.1-11 12. If x C 2y C 2y D 1, then , 1/2 .x 3 1 D 2 y2 C y C 4 2 3 1 2 xD 2 yC : 2 2 2x C 1/ C 4.y 2 C 2y C 1/ D 23 C 9 C 4 D 36 1/2 C 4.y C 1/2 D 36 9.x 2 4 x .y C 1/2 D 1: 9 This is an ellipse with centre .1; 1/, and semi-axes 2 and 3. .1;2/ 1 2 /, focus at y . 1; 1/ x 2y 2 C 3x C 4y D 2, then 9 3 2 2.y 1/2 D xC 2 4 3 2 2 xC 2 .y 1/ D1 9 9 8 3 1 ; 2 2 x .3; 1/ .1; 4/ Fig. 8.1-15 16. Fig. 8.1-12 The equation .x y/2 .x C y/2 D 1 simplifies to 4xy D 1 and hence represents a rectangular hyperbola with centre at the origin, asymptotes along the coordinate axes, transverse axis alongy D x, conjugate axis along 1 1 y D x, vertices at 12 ; 12 and ; 2 , semi-transverse 2p and semi-conjugate q axes equal to 1= 2, semi-focal sepa- ration equal to 12 C 21 D 1, and hence foci at the points p p1 ; p1 p1 ; p1 . The eccentricity is and 2. 2 2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k .1; 1/ 9x 2 C4y 2 18xC8yD23 xC2yC2y 2 D1 4 C y This represents a parabola with vertex at . 32 ; ; 12 / and directrix x D 13 . . 11 8 8 13. If x 2 18x C 8y D 23, then Downloaded by ted cage (sxnbyln180@questza.com) 2 2 317 lOMoARcPSD|6566483 www.konkur.in SECTION 8.1 (PAGE 472) ADAMS and ESSEX: CALCULUS 9 p This is a rectangular hyperbola with centre .0; 2/, p p The semi-axes a D b D 2, and eccentricity 2. p semi-focal separation is 2; the 2/. The p foci are at .˙2; asymptotes are u D ˙.v C 2/. In terms of thep original coordinates, the centre is .1; 1/, p the foci are .˙ 2 C 1; ˙ 2 1/, and the asymptotes are x D 1 and y D 1. y x 1 1 2;2 y .x y/2 .xCy/2 D1 xyCx yD2 Fig. 8.1-16 .1; 1/ 17. x The parabola has focus at .3; 4/ and principal axis along y D 4. The vertex must be at a point of the form .v; 4/, in which case a D ˙.3 v/ and the equation of the parabola must be of the form .y 4/2 D ˙4.3 v/.x v/: Fig. 8.1-19 This curve passes through the origin if 16 D ˙4.v 2 3v/. We have two possible equations for v: v 2 3v 4 D 0 and v 2 3v C 4 D 0. The first of these has solutions v D 1 or v D 4. The second has no real solutions. The two possible equations for the parabola are 4/2 D 4.4/.x C 1/ .y .y 2 4/ D 4. 1/.x 4/ or y2 or 2 y 8y D 16x 8y D 4x 18. The foci of the ellipse are .0; 0/ and .3; 0/, so the centre is .3=2; 0/ and c D 3=2. The semi-axes a and b must satisfy a2 b 2 D 9=4. Thus the possible equations of the ellipse are y2 .x .3=2//2 C 2 D 1: 2 .9=4/ C b b 19. For xy C x y D 2 we have A D C D 0, B D 1. We therefore rotate the coordinate axes (see text pages 407– 408) through angle D =4. (Thus cot 2 D 0 D .A C /=B.) The transformation is 1 x D p .u 2 v/; 20. We have x 2 C 2xy C y 2 D 4x 4y C 4 and A D 1, B D 2, C D 1, D D 4, E D 4 and F D 4. We rotate the axes through angle satisfying tan 2 D B=.A C / D 1 ) D . Then A0 D 2, 4 p B 0 D 0, C 0 D 0, D 0 D 0, E 0 D 4 2 and the transformed equation is p 2u2 C 4 2v 4D0 ) u2 D p 2 2 v 1 p 2 which represents a parabola with vertex at .u; v/ D 0; p1 and principal axis along u D 0. 2 The distance p a from thepfocus to the vertex is given by 4a D 2 2, so apD 1= 2 and the focus is at .0; 0/. The directrix is v D 2. 1 1 Since x D p .u v/ and y D p .u C v/, the ver2 2 tex of the parabola in terms of xy-coordinates is . 12 ; 12 /, and the focus is .0; 0/. The directrix is x y D 2. The principal axis is y D x. y yD x 1 y D p .u C v/: 2 x 2 C2xyCy 2 D4x 4yC4 . 1=2;1=2/ The given equation becomes x 1 2 .u 2 u2 u2 u2 2 318 Telegram: @uni_k 1 v 2 / C p .u v/ 2 p v 2 2 2v D 4 p 2 vC 2 D2 p .v C 2/2 D 1: 2 1 p .u C v/ D 2 2 Fig. 8.1-20 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 21. SECTION 8.1 (PAGE 472) For 8x 2 C 12xy C 17y 2 D 20, we have A D 8, B D 12, C D 17, F D 20. Rotate the axes through angle where tan 2 D B A C Thus cos 2 D 3=5, sin 2 D 2 cos2 1 D cos 2 D D 12 D 9 Then A0 D 0, B 0 D 0, C 0 D 5, D 0 D transformed equation is 4 : 3 5v 2 C 4=5, and 3 5 cos2 D ) 4 : 5 p 5u D 0 5, E 0 D 0 and the 1 p u 5 v2 D which represents a parabola with vertex at .u; v/ D .0; 0/, 1 1 p ; 0 . The directrix is u D p and the focus at 4 5 4 5 2 1 principal axis is v D 0. Since x D p u p v and 5 5 1 2 y D p u C p v, in terms of the xy-coordinates, the ver5 5 1 1 ; . The directrix is tex is at .0; 0/, the focus at 10 20 2x C y D 14 and the principal axis is 2y x D 0. 2 1 We may therefore take cos D p , and sin D p . 5 5 The transformation is therefore 1 2 1 2 uD p x p y x D p uC p v 5 5 5 5 1 2 1 2 y D p uC p v vD p xCp y 5 5 5 5 The coefficients of the transformed equation are 4 2 1 A0 D 8 C 12 C 17 D5 5 5 5 B0 D 0 4 2 1 0 12 C 17 D 20: C D8 5 5 5 y x x 2 4xyC4y 2 C2xCyD0 The transformed equation is xD2y 2 5u2 C 20v 2 D 20; or u C v 2 D 1: 4 This is an ellipse with centre p .0; 0/, semi-axes a D 2 and b D 1, and foci at u D ˙ 3, v D 0. In terms of the original coordinates, the centre is .0; 0/, p p ! 2 3 3 the foci are ˙ p ; p . 5 5 Fig. 8.1-22 y 23. 8x 2 C12xyC17y 2 D20 p The distance from P to F is x 2 C y 2 . The distance from P to D is x C p. Thus p x2 C y2 D xCp x 2 C y 2 D 2 .x 2 C 2px C p 2 / x .1 2 /x 2 C y 2 Fig. 8.1-21 2p 2 x D 2 p 2 : y 22. We have x 2 4xy C4y 2 C2x Cy D 0 and A D 1, B D 4, C D 4, D D 2, E D 1 and F D 0. We rotate the axes through angle satisfying tan 2 D B=.A C / D 43 . Then p 5 3 ) cos 2 D sec 2 D 1 C tan2 2 D 3 5 r r 8̂ 1 C cos 2 4 2 ˆ D D p I < cos D 2 5 r r 5 ) ˆ 1 cos 2 1 1 :̂ sin D D D p : 2 5 5 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k ) p Downloaded by ted cage (sxnbyln180@questza.com) P D.x;y/ xD p D F x Fig. 8.1-23 319 lOMoARcPSD|6566483 www.konkur.in SECTION 8.1 (PAGE 472) 24. ADAMS and ESSEX: CALCULUS 9 y Let the equation of the parabola be y 2 D 4ax. The focus F is at .a; 0/ and vertex at .0; 0/. Then the distance from the vertex p to the focus is a. At x D a, y D 4a.a/ D ˙2a. Hence, ` D 2a, which is twice the distance from the vertex to the focus. ` x a c x2 y a2 y2 D1 b2 y 2 D4ax ` Fig. 8.1-26 x .a;0/ 27. S2 Fig. 8.1-24 C2 `2 c2 25. We have 2 C 2 D 1. Thus a b 2 2 ` Db 1 D b2 1 B F2 2 c a2 a2 but c 2 D a2 b2 b2 D b2 2 : 2 a a b2 P F1 S1 C1 Therefore ` D b 2 =a. y b A x2 y2 C D1 a2 b 2 ` c a x V Fig. 8.1-27 Let the spheres S1 and S2 intersect the cone in the circles C1 and C2 , and be tangent to the plane of the ellipse at the points F1 and F2 , as shown in the figure. Let P be any point on the ellipse, and let the straight line through P and the vertex of the cone meet C1 and C2 at A and B respectively. Then PF1 D PA, since both segments are tangents to the sphere S1 from P . Similarly, PF2 D PB. Thus PF1 C PF2 D PA C PB D AB D constant (distance from C1 to C2 along all generators of the cone is the same.) Thus F1 and F2 are the foci of the ellipse. Fig. 8.1-25 y2 x2 D 1. The 2 a b2 vertices p are at .˙a; 0/ and p the foci are at .˙c; 0/ where c D a2 C b 2 . At x D a2 C b 2 , 26. Suppose the hyperbola has equation a2 C b 2 y 2 D1 a2 b2 .a2 C b 2 /b 2 a2 y 2 D a2 b 2 yD˙ Hence, ` D 320 Telegram: @uni_k b2 . a b2 : a 28. Let F1 and F2 be the points where the plane is tangent to the spheres. Let P be an arbitrary point P on the hyperbola in which the plane intersects the cone. The spheres are tangent to the cone along two circles as shown in the figure. Let PAVB be a generator of the cone (a straight line lying on the cone) intersecting these two circles at A and B as shown. (V is the vertex of the cone.) We have PF1 D PA because two tangents to a sphere from Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 8.2 (PAGE 478) a point outside the sphere have equal lengths. Similarly, PF2 D PB. Therefore PF2 PF1 D PB PA D AB D constant, F P Y since the distance between the two circles in which the spheres intersect the cone, measured along the generators of the cone, is the same for all generators. Hence, F1 and F2 are the foci of the hyperbola. C V X Q P F1 A A Fig. 8.1-29 V B Section 8.2 Parametric Curves 1. F2 (page 478) If x D t , y D 1 t , .0 t 1/ then x C y D 1. This is a straight line segment. y Fig. 8.1-28 1 xDt yD1 t .0t1/ x 1 Fig. 8.2-1 2. 29. Let the plane in which the sphere is tangent to the cone meet AV at X. Let the plane through F perpendicular to the axis of the cone meet AV at Y . Then VF D V X, and, if C is the centre of the sphere, F C D XC . Therefore V C is perpendicular to the axis of the cone. Hence YF is parallel to V C , and we have Y V D V X D VF . If P is on the parabola, FP ? VF , and the line from P to the vertex A of the cone meets the circle of tangency of the sphere and the cone at Q, then If x D 2 t and y D t C 1 for 0 t < 1, then y D 2 x C 1 D 3 x for 1 < x 2, which is a half line. y xD2 t yDtC1 .2;1/ FP D PQ D YX D 2V X D 2VF: x Fig. 8.2-2 Since FP D 2VF , FP is the semi-latus rectum of the parabola. (See Exercise 18.) Therefore F is the focus of the parabola. 3. 1 If x D 1=t , y D t 1, .0 < t < 4/, then y D x is part of a hyperbola. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 1. This 321 lOMoARcPSD|6566483 www.konkur.in SECTION 8.2 (PAGE 478) ADAMS and ESSEX: CALCULUS 9 y y 1 ;3 4 tD4 bxDay 1 yD x 1 tD0 a x tD1 x yD 1 bxD ay Fig. 8.2-3 4. If x D Fig. 8.2-6 1 t and y D for 1 C t2 1 C t2 1 < t < 1, then 1 1 C t2 D Dx 2 2 .1 C t / 1 C t2 2 1 1 C y2 D : 2 4 7. x2 C y2 D , x If x D 3 sin t , y D 4 cos t , . 1 t 1/, then y2 x2 C D 1. This is an ellipse. 9 16 y tD0 x2 y2 9 C 16 D1 This curve consists of all points of the circle with centre at . 21 ; 0/ and radius 12 except the origin .0; 0/. y x tD1 tD 1 tD1 tD0 tD 1 x Fig. 8.2-7 xD1=.1Ct 2 / yDt=.1Ct 2 / 8. Fig. 8.2-4 5. If x D 3 sin 2t , y D 3 cos 2t , .0 t =3/, then x 2 C y 2 D 9. This is part of a circle. y tD0 If x D cos sin s and y D sin sin s for 1 < s < 1, then x 2 C y 2 D 1. The curve consists of the arc of this circle extending from .a; b/ through .1; 0/ to .a; b/ where a D cos.1/ and b D sin.1/, traversed infinitely often back and forth. y xDcos sin s yDsin sin s x 2 Cy 2 D9 x tD 1 rad 3 x Fig. 8.2-5 6. If x D a sec t and y D b tan t for x2 a2 y2 D sec2 t b2 < t < , then 2 2 Fig. 8.2-8 tan2 t D 1: 9. The curve is one arch of this hyperbola. 322 Telegram: @uni_k If x D cos3 t , y D sin3 t , .0 t 2/, then x 2=3 C y 2=3 D 1. This is an astroid. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL y SECTION 8.2 (PAGE 478) and tD=2 1 D .cos2 t Csin2 t /2 D cos4 t Csin4 t C2 cos2 t sin2 t: x 2=3 Cy 2=3 D1 Hence, tD0 tD tD2 (b) If x D sec4 t and y D tan4 t , then tD3=2 y/2 D .sec4 t .x Fig. 8.2-9 10. If x D 1 p t 2 and y D 2 C t for .x 1/2 D 4 D sec4 t C tan4 t C 2 sec2 t tan2 t 2 t 2 then .y and 2/2 : 1 D .sec2 t tan2 t /2 D sec4 t Ctan4 t 2 sec2 t tan2 t: The parametric curve is the left half of the circle of radius 4 centred at .1; 2/, and is traced in the direction of increasing y. p t2 Hence, y/2 D 2.sec4 t C tan4 t / D 2.x C y/: 1 C .x y 4 yD2Ct 2t2 xD1 tan4 t /2 D .sec2 t C tan2 t /2 4 t2 D 4 y/2 D 2.cos4 t C sin4 t / D 2.x C y/: 1 C .x (c) Similarly, if x D tan4 t and y D sec4 t , then 1 C .x .1;2/ y/2 D 1 C .y x/2 2 tan2 t /2 C .sec4 t D .sec t tan4 t /2 D 2.tan4 t C sec4 t / D 2.x C y/: These three parametric curves above correspond to different parts of the parabola 1C.x y/2 D 2.x Cy/, as shown in the following diagram. x Fig. 8.2-10 11. y x D cosh t , y D sinh t represents the right half (branch) of the rectangular hyperbola x 2 y 2 D 1. xDtan4 t yDsec4 t The parabola 12. x D 2 3 cosh t , y D 1 C 2 sinh t represents the left half (branch) of the hyperbola 2/2 .x 9 2.xCy/D1C.x y/2 1 xDcos4 t yDsin4 t .y C 1/2 D 1: 4 xDsec4 t yDtan4 t 1 x Fig. 8.2-14 13. x D t cos t , y D t sin t , .0 t 4/ represents two revolutions of a spiral curve winding outwards from the origin in a counterclockwise direction. The point on the curve corresponding to parameter value t is t units distant from the origin in a direction making angle t with the positive x-axis. 14. (a) If x D cos4 t and y D sin4 t , then .x y/2 D .cos4 t sin4 t /2 h D .cos2 t C sin2 t /.cos2 t D .cos2 t sin2 t /2 D cos4 t C sin4 t 15. 16. The slope of y D x 2 at x is m D 2x. Hence the parabola can be parametrized x D m=2, y D m2 =4, . 1 < m < 1/. If .x; y/ is any point on the circle x 2 C y 2 D R2 other than .R; 0/, then the line from .x; y/ to .R; 0/ has slope y . Thus y D m.x R/, and mD x R x 2 C m2 .x i2 sin t / 2 2 cos2 t sin2 t .m2 C 1/x 2 h .m2 C 1/x 2 ) xD 2xRm2 C .m2 i .m2 1/R .x 1/R2 D 0 R/ D 0 .m 1/R or x D R: m2 C 1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k R/2 D R2 Downloaded by ted cage (sxnbyln180@questza.com) 323 lOMoARcPSD|6566483 www.konkur.in SECTION 8.2 (PAGE 478) ADAMS and ESSEX: CALCULUS 9 y The parametrization of the circle in terms of m is given by Y .m2 1/R xD m2 C 1 " .m2 1/R yDm m2 C 1 b # 2Rm m2 C 1 R D P D .x; y/ T a t X x where 1 < m < 1. This parametrization gives every point on the circle except .R; 0/. y .x;y/ slope m x .R;0/ Fig. 8.2-18 x 2 Cy 2 DR2 19. Fig. 8.2-16 If x D 3t 3t 2 , y D , .t ¤ 1 C t3 1 C t3 x3 C y3 D 17. y P D .x; y/ t 27t 3 27t 3 .1 C t 3 / D D 3xy: 3 3 .1 C t / .1 C t 3 /2 As t ! 1, we see that jxj ! 1 and jyj ! 1, but T a 1/, then xCy D X x Thus x C y D 3t .1 C t / D 1 C t3 1 3t ! 1: t C t2 1 is an asymptote of the curve. y tD1 Fig. 8.2-17 t!1 Using triangles in the figure, we see that the coordinates of P satisfy x D a sec t; tD0 y D a sin t: x folium of Descartes The Cartesian equation of the curve is t! 1 y2 a2 C D 1: a2 x2 The curve has two branches extending to infinity to the left and right of the circle as shown in the figure. Fig. 8.2-19 20. 18. The coordinates of P satisfy x D a sec t; The Cartesian equation is 324 Telegram: @uni_k y D b sin t: Let C0 and P0 be the original positions of the centre of the wheel and a point at the bottom of the flange whose path is to be traced. The wheel is also shown in a subsequent position in which it makes contact with the rail at R. Since the wheel has been rotated by an angle , a2 y2 C 2 D 1. 2 b x OR D arc SR D a: Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 8.2 (PAGE 478) y Thus, the new position of the centre is C D .a; a/. Let P D .x; y/ be the new position of the point; then Tt x D OR PQ D a b sin. y D RC C CQ D a C b cos. / D a b sin ; / D a b cos : These are the parametric equations of the prolate cycloid. y P Q P t D.x;y/ b t b S C0 t Ct t A a C x a O x R P0 Fig. 8.2-21 Fig. 8.2-20 If a D 2 and b D 1, then x D 2 cos t , y D 0. This is a straight line segment. If a D 4 and b D 1, then y x D 3 cos t C cos 3t D 3 cos t C .cos 2t cos t sin 2t sin t / D 3 cos t C .2 cos2 t 1/ cos t 2 sin2 t cos t xDa b sin yDa b cos x 2a D 2 cos t C 2 cos3 t 2 cos t .1 sin2 t / D 4 cos3 t y D 3 sin t C sin 3t D 3 sin t sin 2t cos t .cos 2t sin t / D 3 sin t 2 sin t cos2 t .1 2 sin2 t / sin t Fig. 8.2-20 D 2 sin t 2 sin t C 2 sin3 t C 2 sin3 t D 4 sin3 t This is an astroid, similar to that of Exercise 11. 22. 21. Let t and t be the angles shown in the figure below. Then arc AT t D arc T t P t , that is, at D b t . The centre C t of the rolling circle is C t D .a b/ cos t; .a b/ sin t . Thus Since t x y .a .a b/ cos t D b cos. t t / b/ sin t D b sin. t t /: tD a t b tD x D .a y D .a a b b a) From triangles in the figure, x D jTXj D jOT j tan t D tan t y D jOY j D si n 2 t D jOY j cos t D jOT j cos t cos t D cos2 t: y yD1 T t , therefore Y 1 2 .a b/t b/ cos t C b cos b .a b/t b/ sin t b sin : b X P D .x; y/ t O x Fig. 8.2-22 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 325 lOMoARcPSD|6566483 www.konkur.in SECTION 8.2 (PAGE 478) b) ADAMS and ESSEX: CALCULUS 9 y 1 1 . D sec2 t D 1Ctan2 t D 1Cx 2 . Thus y D y 1 C x2 23. x D sin t; y D sin.2t / y x x Fig. 8.2-23 28. 24. x D sin t; y D sin.3t / y x tangent to x 2 C y 2 D 1. If n 2 is an integer, the curve closes after one revolution and has n cusps. The figure shows the curve for n D 7. If n is a rational number but not an integer, the curve will wind around the circle more than once before it closes. y Fig. 8.2-24 25. x D sin.2t /; y D sin.3t / Fig. 8.2-27 1 1 cos t C cos..n 1/t / x D 1C n n 1 1 y D 1C sin..n 1/t / sin t n n represents a cycloid-like curve that is wound around the 2 inside circle x 2 C y 2 D 1 C .2=n/ and is externally y x x Fig. 8.2-25 26. x D sin.2t /; y D sin.5t / Fig. 8.2-28 y Section 8.3 Smooth Parametric Curves and Their Slopes (page 483) x 27. Fig. 8.2-26 1 1 x D 1C cos.nt / cos t n n 1 1 y D 1C sin.nt / sin t n n represents a cycloid-like curve that is wound around the circle x 2 C y 2 D 1 instead of extending along the x-axis. If n 2 is an integer, the curve closes after one revolution and has n 1 cusps. The left figure below shows the curve for n D 7. If n is a rational number, the curve will wind around the circle more than once before it closes. 326 Telegram: @uni_k 1. y D 2t 4 x D t2 C 1 dy dx D 2t D2 dt dt No horizontal tangents. Vertical tangent at t D 0, i.e., at .1; 4/. 2. y D t 2 C 2t x D t 2 2t dy dx D 2t 2 D 2t C 2 dt dt Horizontal tangent at t D 1, i.e., at .3; 1/. Vertical tangent at t D 1, i.e., at . 1; 3/. 3. x D t 2 2t y D t 3 12t dx dy D 2.t 1/ D 3.t 2 4/ dt dt Horizontal tangent at t D ˙2, i.e., at .0; 16/ and .8; 16/. Vertical tangent at t D 1, i.e., at . 1; 11/. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 4. 5. 6. 7. SECTION 8.3 (PAGE 483) y D 2t 3 C 3t 2 x D t 3 3t dy dx D 3.t 2 1/ D 6t .t C 1/ dt dt Horizontal tangent at t D 0, i.e., at .0; 0/. Vertical tangent at t D 1, i.e., at . 2; 5/. At t D 1 (i.e., at .2; 1/) both dx=dt and dy=dt change sign, so the curve is not smooth there. (It has a cusp.) 2 12. At t D 13. 2 x D t e t =2 yDe t dx dy 2 2 D .1 t 2 /e t =2 D 2t e t dt dt Horizontal tangent at t D 0, i.e., at .0; 1/. Vertical tangent at t D ˙1, i.e. at .˙e 1=2 ; e 1 /. y D sin t t cos t x D sin t dy dx D cos t D t sin t dt dt Horizontal tangent at t D n, i.e., at .0; . 1/n n/ (for integers n). Vertical tangent at t D .n C 12 /, i.e. at .1; 1/ and . 1; 1/. 14. 15. 2 9. 10. 11. x D t4 t2 y D t 3 C 2t dy dx D 4t 3 2t D 3t 2 C 2 dt dt dy 3. 1/2 C 2 At t D 1; D D dx 4. 1/3 2. 1/ y D sin t dy D cos t dt cos.=6/ D 2 sin.=3/ 1 C .t 1/ D t y D 2 C 4.t 1/ D 4t 2: 4 1 p 2 1 dx D 1 C sin t D 1 C p dt 2 1 at t D y D 1 sin t D 1 p 4 2 1 dy at t D D cos t D p dt 2 4 1 1 p C 1 C p t, Tangent line: x D 4 2 2 1 t yD1 p p . 2 2 xDt x D t3 cos t D t , y D t 2 is at .0; 1/ at t D 1 and t D 1. Since 2t ˙2 dy D 2 D D ˙1; dx 3t 1 2 the tangents at .0; 1/ at t D ˙1 have slopes ˙1. 16. x D sin t , y D sin.2t / is at .0; 0/ at t D 0 and t D . Since dy 2 cos.2t / 2 if t D 0 D D 2 if t D , dx cos t 1 : 2 5 : 2 17. y D t2 x D t3 dy dx D 3t 2 D 2t both vanish at t D 0. dt dt dy 2 dx 3t D has no limit as t ! 0. D ! 0 as t ! 0, dx 3t dy 2 but dy=dt changes sign at t D 0. Thus the curve is not smooth at t D 0. (In this solution, and in the next five, we are using the Remark following Example 2 in the text.) 18. x D .t 1/4 y D .t 1/3 dx dy D 4.t 1/3 D 3.t 1/2 both vanish at t D 1. dt dt dx 4.t 1/ Since D ! 0 as t ! 1, and dy=dt does not dy 3 change sign at t D 1, the curve is smooth at t D 1 and therefore everywhere. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2; the tangents at .0; 0/ at t D 0 and t D have slopes 2 and 2, respectively. y D 1 t3 x D t3 C t dy dx D 3t 2 C 1 D 3t 2 dt dt 3.1/2 3 dy D D : At t D 1; dx 3.1/2 C 1 4 x D cos.2t / dx D 2 sin.2t / dt dy At t D ; D 6 dx 3 : 2 y D t C t 3 D 2 at t D 1 x D t 3 2t D 1 dy dx D 3t 2 2 D 1 D 1 C 3t 2 D 4 at t D 1 dt dt Tangent line: x D 1 C t , y D 2 C 4t . This line is at . 1; 2/ at t D 0. If you want to be at that point at t D 1 instead, use xD y D sin t x D sin.2t / dy dx D 2 cos.2t / D cos t dt dt Horizontal tangent at t D .n C 21 /, i.e., at .0; ˙1/. p Vertical tangent at t D 12 .n C 21 /, i.e., at .˙1; 1= 2/ and p .˙1; 1= 2/. 3t 3t 8. xD yD 1 C t3 1 C t3 dx dy 3.1 2t 3 / 3t .2 t 3 / D D 3 2 dt .1 C t / dt .1 C t 3 /2 Horizontal tangent at t D 0 and t D 21=3 , i.e., at .0; 0/ and .21=3 ; 22=3 /. Vertical tangent at t D 2 1=3 , i.e., at .22=3 ; 21=3 /. The curve also approaches .0; 0/ vertically as t ! ˙1. y D t e 2t dy D e 2t .1 C 2t / dt dy e 4 .1 4/ 2; D D dx 2e 4 x D e 2t dx D 2e 2t dt Downloaded by ted cage (sxnbyln180@questza.com) 327 lOMoARcPSD|6566483 www.konkur.in SECTION 8.3 (PAGE 483) 19. 20. 21. ADAMS and ESSEX: CALCULUS 9 y D t3 x D t sin t dy dx D sin t C t cos t D 3t 2 both vanish at t D 0. dt dt dy 3t 2 6t lim D lim D lim D 0, t!0 dx t!0 sin t C t cos t t!0 2 cos t t sin t but dx=dt changes sign at t D 0. dx=dy has no limit at t D 0. Thus the curve is not smooth at t D 0. 22. 2t and y D t 2 Both f 0 .t / and g 0 .t / vanish at t D 0. Observe that 6t 2 dy D 2 D : dx 3t t Thus, lim dy t!0C dx D 1; ! # & xDt 3 yD3t 2 1 x 1 Fig. 8.3-22 2 j C C ! " % !t The tangent is horizontal at t D 2, (i.e., .0; 4/), and is vertical at t D 1 (i.e., at . 1; 3/. Observe that d 2 y=dx 2 > 0, and the curve is concave up, if t > 1. Similarly, d 2 y=dx 2 < 0 and the curve is concave down if t < 1. y 3 x D t 3t , y D 2=.1 C t 2 /. Observe that y ! 0, x ! ˙1 as t ! ˙1. dx dy 4t D 3.t 2 1/; D dt dt .1 C t 2 /2 dy 4t D dx 3.t 2 1/.1 C t 2 /2 d 2x d 2y 4.3t 2 1/ D 6t; D dt 2 dt 2 .1 C t 2 /3 2 4.3t 1/ 4t .6t / 3.t 2 1/ 2 3 d 2y .1 C t / .1 C t 2 /2 D 2 2 3 dx Œ3.t 1/ 60t 4 C 48t 2 C 12 D 27.t 2 1/3 .1 C t 2 /3 Directional information: xDt 2 2t yDt 2 4t x tD1 tD2 Fig. 8.3-21 328 Telegram: @uni_k 2 <0 3t 4 y 23. # . 1 for all t , the curve is concave down everywhere. Directional information is as follows: C dy D dx .3t 2 /.6/ .6t /.6t / d 2y D D 2 dx .3t 2 /3 dx dy D 2.t 1/; D 2.t 2/ dt dt 2 2 d y d x D D2 dt 2 dt 2 d 2y 1 d dy D dx 2 dx=dt dt dx 1 1 d t 2 D : D 2.t 1/ dt t 1 2.t 1/3 dx=dt dy=dt x y curve lim t!0 and the curve has a cusp at t D 0, i.e., at .0; 1/. Since 4t , then 1 j 1, then f 0 .t / D 3t 2 ; f 00 .t / D 6t I g 0 .t / D 6t; g 00 .t / D 6: x D t3 y D t sin t dx dy D 3t 2 D 1 cos t both vanish at t D 0. dt dt 3t 2 6t dx D lim D lim D 6 and dy=dt lim t!0 1 t!0 sin t t!0 dy cos t does not change sign at t D 0. Thus the curve is smooth at t D 0, and hence everywhere. If x D t 2 If x D f .t / D t 3 and y D g.t / D 3t 2 dx=dt dy=dt x y curve C C ! " % Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 j 0 j 1 j C " - # . C ! # & !t lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 8.4 (PAGE 487) y xDt 3 3t 2 The tangent is horizontal at t D 0, i.e., .0; 2/, and vertical at t D ˙1, i.e., .˙2; 1/. 1 j 2 d y dx 2 curve 1 j yDt 2 2t 2 tD 1; 2 x tD1 !t C C ^ _ Fig. 8.3-24 ^ 25. y 2 yD 1 C t2 tD0 tD 1 Fig. 8.3-23 If x D f .t / D t 3 3t 2 and y D g.t / D t 2 y D sin t t cos t; .t 0/: dy D tan t dt dx d 2 y dy d 2 x d 2y 2 dt 2 D dt dt dt 3 dx 2 dx x 24. x D cos t C t sin t; dy dx D t cos t; D t sin t; dt dt d 2x D cos t t sin t dt 2 d 2y D sin t C t cos t dt 2 xDt 3 3t tD1 1 tD 2 dt t 2, then 1 D t cos3 t f 0 .t / D 3t 2 3; f 00 .t / D 6t I g 0 .t / D 2t 1; g 00 .t / D 2: Tangents are vertical at t D n C 12 , and horizontal at t D n (n D 0; 1; 2; : : :). 1 27 9 , i.e., at ; . 2 8 4 The tangent is vertical at t D ˙1, i.e., . 4; 2/ and .0; 0/. Directional information is as follows: y The tangent is horizontal at t D t f 0 .t / g 0 .t / x y curve 1 2 1 j C 1 j j C ! # & # . " - tD tD3=2 tD0 C C ! " % ! tD2 Fig. 8.3-25 Section 8.4 Arc Lengths and Areas for Parametric Curves (page 487) For concavity, 3.t 2 d 2y D 2 dx 1/.2/ Œ3.t 2 .2t 1/.6t / D 1/3 2.t 2 t C 1/ 9.t 2 1/3 which is undefined at t D ˙1, therefore t d 2y dx 2 curve 1 j 1 j ! C _ ^ _ 1. y D 2t 3 .0 t 1/ dy D 6t 2 dt Z 1p Length D . .6t /2 C .6t 2 /2 dt 0 Z 1 p D6 t 1 C t 2 dt Let u D 1 C t 2 0 du D 2t dt ˇ2 Z 2 ˇ p p 3=2 ˇ u du D 2u ˇ D 4 2 2 units D3 ˇ 1 x D 3t 2 dx D 6t dt Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x Downloaded by ted cage (sxnbyln180@questza.com) 1 329 lOMoARcPSD|6566483 www.konkur.in SECTION 8.4 (PAGE 487) ADAMS and ESSEX: CALCULUS 9 2. If x D 1 C t 3 and y D 1 t 2 for 1 t 2, then the arc length is Z 2p .3t 2 /2 C . 2t /2 dt sD 1 Z 2 p D jt j 9t 2 C 4 dt 1 Z 1 Z 2 p C D t 9t 2 C 4 dt Let u D 9t 2 C 4 0 0 du D 18t dt Z 13 Z 40 p 1 C D u du 18 4 4 p 1 p 13 13 C 40 40 16 units: D 27 The length of the curve is Z 2 p t 4 C t 2 dt Let u D 4 C t 2 du D 2t dt ˇ4C4 2 Z 4C4 2 1 3=2 ˇˇ 1 1=2 u du D u ˇ D ˇ 2 4 3 4 8 2 3=2 D .1 C / 1 units: 3 0 6. 3. x D a cos3 t , y D a sin3 t , .0 t 2/: The length is Z 2 p 9a2 cos4 t sin2 t C 9a2 sin4 t cos2 t dt 0 Z 2 D3a j sin t cos t j dt 0 Z =2 1 sin 2t dt 2 0 ˇ=2 cos 2t ˇˇ D 6a units: D6a ˇ ˇ 2 D12a 0 7. 0 4. If x D ln.1 C t 2 / and y D 2 tan 1 t for 0 t 1, then dx dy 2t 2 I : D D 2 dt 1Ct dt 1 C t2 The arc length is s Z 1 4t 2 C 4 dt sD .1 C t 2 /2 0 Z 1 dt p Let t D tan D2 1 C t2 0 dt D sec2 d Z =4 D2 sec d 0 ˇ=4 ˇ p ˇ D 2 ln j sec C tan jˇ D 2 ln.1 C 2/ units. ˇ 8. D t 2 .4 C t 2 /: 330 Telegram: @uni_k 4t sin t cos t C t 2 sin2 t x D sin2 t dx D 2 sin t cos t dt y D 2 cos t .0 t =2/ dy D 2 sin t dt Length Z =2 p D 4 sin2 t cos2 t C 4 sin2 t dt 0 Z =2 p D2 sin t 1 C cos2 t dt Let cos t D tan u 0 sin t dt D sec2 u du Z =4 D2 sec3 u du 5. x D t 2 sin t , y D t 2 cos t , .0 t 2/. C 4 cos2 t y D cos t .0 t / x D t C sin t dy dx D 1 C cos t D sin t dt dt Z p 1 C 2 cos t C cos2 t C sin2 t dt Length D 0 Z Z p t 4 cos2 .t =2/ dt D 2 D cos dt 2 0 0 ˇ t ˇˇ D 4 sin ˇ D 4 units: 2ˇ 0 0 dx D 2t sin t C t 2 cos t dt dy D 2t cos t t 2 sin t dt 2 ds D t 2 4 sin2 t C 4t sin t cos t C t 2 cos2 t dt y D sin t t cos t .0 t 2/ x D cos t C t sin t dy dx D t cos t D t sin t dt dt Z 2 p Length D t 2 cos2 t C t 2 sin2 t dt 0 ˇ2 Z 2 t 2 ˇˇ D t dt D ˇ D 2 2 units: 2ˇ 0 0 ˇˇ=4 ˇ D sec u tan u C ln.sec u C tan u/ ˇ ˇ 0 p p D 2 C ln.1 C 2/ units: 9. x D a.t dx D a.1 dt sin t / cos t / Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) y D a.1 cos t / .0 t 2/ dy D a sin t dt lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 8.4 (PAGE 487) Z 2 q a2 .1 2 cos t C cos2 t C sin2 t / dt 0 Z 2 p Z 2 r t Da 2 2 cos t dt D a sin2 dt 2 0 0 ˇ Z t ˇˇ t D 2a sin dt D 4a cos ˇ D 4a units: 2 2ˇ 0 the y-axis is Length D Z 2 Sy D 2 jxj ds 0 Z 2 t D 2 dt .at a sin t /2a sin 2 0 " # Z 2 t t t 2 cos sin dt t 2 sin D 4a 2 2 2 0 Z 2 t dt D 4a2 t sin 2 0 Z 2 t t 8a2 sin2 cos dt 2 2 0 " ˇˇ2 # Z 2 t ˇ t dt 0 D 4a2 2t cos cos C 2 ˇ 2 ˇ 2 0 0 10. If x D at then a sin t and y D a a cos t for 0 t 2, dy dx D a a cos t; D a sin t I dt dt p ds D .a a cos t /2 C .a sin t /2 dt s p p p t D a 2 1 cos t dt D a 2 2 sin2 dt 2 t dt: D 2a sin 2 a) The surface area generated by rotating the arch about the x-axis is Sx D 2 Z 2 Z0 0 D 4a2 Œ4 C 0 D 16 2 a2 sq. units. 11. x D e t cos t dx D e t .cos t sin t / dt Arc length element: y D e t sin t .0 t =2/ dy D e t .sin t C cos t / dt p e 2t .cos t p t D 2e dt: sin t /2 C e 2t .sin t C cos t /2 dt ds D The area of revolution about the x-axis is jyj ds t D 4 dt .a a cos t /2a sin 2 0 Z t D 16a2 dt sin3 2 0 " # Z t t 2 2 sin 1 cos D 16a dt 2 2 0 t Let u D cos 2 t 1 sin dt du D 2 2 Z 0 D 32a2 .1 u2 / du 1 " #ˇ1 1 3 ˇˇ 2 D 32a u u ˇ ˇ 3 Z tD=2 tD0 p Z =2 2t 2y ds D 2 2 e sin t dt 0 ˇ=2 ˇ e 2t ˇ .2 sin t cos t /ˇ D 2 2 ˇ 5 0 p 2 2 .2e C 1/ sq. units. D 5 p 12. The area of revolution of the curve in Exercise 11 about the y-axis is Z tD=2 tD0 p Z =2 2t 2x ds D 2 2 e cos t dt 0 D 64 3 a sq. units. 3 0 ˇ=2 ˇ p e 2t ˇ .2 cos t C sin t /ˇ D 2 2 ˇ 5 0 p 2 2 .e 2/ sq. units. D 5 b) The surface area generated by rotating the arch about Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 331 lOMoARcPSD|6566483 www.konkur.in SECTION 8.4 (PAGE 487) 13. ADAMS and ESSEX: CALCULUS 9 y D 2t 3 .0 t 1/ x D 3t 2 dy dx D 6t D 6t 2 dt dt Arc length p element: p ds D 36.t 2 C t 4 / dt D 6t 1 C t 2 dt . The area of revolution about the y-axis is Z tD1 tD0 2x ds D 36 D 18 Z 1 0 Z 2 p t 3 1 C t 2 dt .u p 1/ u du 1 16. Area of R D 4 D 12a 2 D tD0 2y ds D 24 D 24 Z 1 0 p t 4 1 C t 2 dt Z =4 y xDt Let t D tan u dt D sec2 u du tan4 u sec3 u du 17. Fig. 8.4-16 x D sin4 t , y D cos4 t , 0 t . 2 Z =2 .cos4 t /.4 sin3 t cos t / dt Area D Z =4 D4 D4 Z =2 cos5 t .1 D2 D2 .3t 4 0 3t 5 5 Z 1 .u5 u7 / du D 6 0 xDsin4 t 0t=2 A 4t 2 / dt ˇ2 4t 3 ˇˇ 256 sq. units. ˇ D 3 ˇ 15 0 xDt 3 4t x Fig. 8.4-17 18. If x D cos s sin s D 12 sin 2s and y D sin2 s D 21 for 0 s 21 , then x2 C y A x 1 2 2 D 1 2 cos 2s 1 2 1 1 sin 2s C cos2 2s D 4 4 4 which is the right half of the circle with radius 21 and centre at .0; 12 /. Hence, the area of R is " # 1 2 1 D sq. units. 2 2 8 Fig. 8.4-15 Telegram: @uni_k 1 6 yDcos4 t yDt 2 332 Let u D cos t du D sin t dt 1 1 D sq. units. 8 6 y 4/ dt y cos2 t / sin t dt 0 2 Z 2 x 0 0 t 2 .3t 2 0 a 4t , y D t , . 2 t 2/. Area D # ˇ=2 ˇ ˇ ˇ ˇ a a 2 Z 2 sin3 .2t / 48 R We have omitted the details of evaluation of the final integral. See Exercise 24 of Section 8.3 for a similar evaluation. 15. sin.4t / 64 a .sec7 u 2 sec5 u C sec3 u/ du D 24 0 p p 7 2 C 3 ln.1 C 2/ sq. units. D 2 3 t 16 xDa cos3 t yDa sin3 t The area of revolution of the curve of Exercise 13 about the x-axis is Z tD1 sin4 t cos2 t dt =2 3 2 a sq. units. 8 2 5=2 u D 18 5 1 p 72 D .1 C 2/ sq. units. 15 14. " Z 0 (See Exercise 34 of Section 6.4.) ˇ2 2 3=2 ˇˇ u ˇ ˇ 3 .a sin3 t /. 3a sin t cos2 t / dt =2 12a2 D Let u D 1 C t 2 du D 2t dt Z 0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 8.4 (PAGE 487) y upper half of the hyperbola from that of a right triangle: xDcos s sin s yDsin2 s 1 Shaded area D Area 4ABC 1 D sec t0 tan t0 2 1 D sec t0 tan t0 2 1 D sec t0 tan t0 2 1 2 x Area sector ABC Z t0 tan t .sec t tan t / dt 0 Z t0 .sec3 t sec t / dt 0 1 sec t tan t C 2 ˇˇt0 1 ˇ ln j sec t C tan t j ln j sec t C tan t j ˇ ˇ 2 Fig. 8.4-18 0 D 1 ln j sec t0 C tan t0 j sq. units. 2 y 19. x D .2 C sin t / cos t , y D .2 C sin t / sin t , .0 t 2/. This is just the polar curve r D 2 C sin . Area D D Z 2 0 Z 2 0 Z 2 h .2 C sin t / sin t d .2 C sin t / cos t dt dt .2 sin t C sin2 t /.cos2 t 2 sin t R tD0 x xDsec t yDtan t sin2 t / dt 4 sin2 t C 4 sin3 t C sin4 t i 2 sin t cos2 t sin2 t cos2 t dt Z 2 h i 1 cos 2t . cos 2t / dt D 2.1 cos 2t / C 2 0 Z 2 h i 2 C sin t 4 6 cos t dt D tDt0 0 0 D 4 C 9 C0D sq. units. 2 2 Fig. 8.4-20 21. See the figure below. The area is the area of a triangle less the area under the hyperbola: xD.2Csin t/ cos t y Z t0 1 sinh t sinh t dt cosh t0 sinh t0 2 Z t0 0 1 cosh 2t 1 D sinh 2t0 dt 4 2 0 1 1 1 sinh 2t0 C t0 D sinh 2t0 4 4 2 t0 D sq. units. 2 AD yD.2Csin t/ sin t 0t2 A x y Fig. 8.4-19 .cosh t0 ;sinh t0 / A x 20. To find the shaded area we subtract the area under the Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) Fig. 8.4-21 333 lOMoARcPSD|6566483 www.konkur.in SECTION 8.4 (PAGE 487) ADAMS and ESSEX: CALCULUS 9 22. If x D f .t / D at a sin t and y D g.t / D a a cos t , then the volume of the solid obtained by rotating about the x-axis is V D Z tD2 tD0 Z 2 D 0 D a3 D a3 D a 3 Z tD2 y 2 dx D Z 2 r D 5=.3 sin 4 cos / 3r sin 4r cos D 5 3y 4x D 5 straight line. 4. r D sin C cos tD0 a cos t /2 .a .a Œg.t /2 f 0 .t / dt 3. r 2 D r sin C r cos x2 C y2 D y C x 1 2 1 1 2 C y D x 2 2 2 1 1 1 a circle with centre ; and radius p . 2 2 2 a cos t / dt .1 cos t /3 dt .1 3 cos t C 3 cos2 t 0 Z 2 " 0 2 " 0C 3 2 Z 2 0 # cos3 t / dt # .1 C cos 2t / dt 0 5. r 2 sin 2 D 1 2r 2 sin cos D 1 2xy D 1 a rectangular hyperbola. 3 D a3 2 C .2/ D 5 2 a3 cu. units. 2 y 6. dx x tD0 23. Half of the volume corresponds to rotating x D a cos3 t , y D a sin3 t .0 t =2/ about the x-axis. The whole volume is 8. rD p D 2 0 D 6a3 D 6a3 D 6a3 6 x 2 C 4y 2 D 4 2 a sin t .3a cos t sin t / dt Z =2 9. .1 cos2 t /3 cos2 t sin t dt 0 Z 1 0 1 3 .1 Let u D cos t du D sin t dt 3u2 C 3u4 u6 /u2 du 3 3 1 32a3 C cu. units. D 5 7 9 105 rD r 2. r D 2 csc ) r sin D 2 , yD 2 a horizontal line: 334 Telegram: @uni_k 1 1 cos xD1 x 2 C y 2 D 1 C 2x C x 2 y 2 D 1 C 2x 10. rD a parabola. 2 2 cos r cos D 2 4r 2 D .2 C x/2 Section 8.5 Polar Coordinates and Polar Curves (page 493) r D 3 sec r cos D 3 xD3 vertical straight line. an ellipse. r 2 D .1 C x/2 2r 1. 2 cos2 C 4 sin2 2 2 r cos C 4r 2 sin2 D 4 y 2 . dx/ 2 a parabola. r D sec .1 C tan / r cos D 1 C tan y x D1C x x2 x y D 0 a parabola. 0 Z =2 r sin r cos 7. tD2 Fig. 8.4-22 Z =2 r D sec tan ) r cos D x2 D y xDat a sin t yDa a cos t V D2 r 2 D csc 2 4x 2 C 4y 2 D 4 C 4x C x 2 3x 2 C 4y 2 11. rD r 4x D 4 an ellipse. 2 1 2 sin 2y D 2 x 2 C y 2 D r 2 D 4.1 C y/2 D 4 C 8y C 4y 2 x2 3y 2 8y D 4 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) a hyperbola. lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 12. SECTION 8.5 (PAGE 493) y 2 1 C sin r C r sin D 2 5 D 6 rD r 2 D .2 2 x Cy D4 x D4 6 y/2 2 2 D 4y x 4y C y 2 a parabola. 13. r D 1 C sin (cardioid) rD1 2 sin y Fig. 8.5-16 2 17. r D 2 C cos y x 1 3 x Fig. 8.5-13 14. cos C , then r D 0 at D 4 This is a cardioid. If r D 1 7 and . 4 4 Fig. 8.5-17 y 18. If r D 2 sin 2 , then r D 0 at D 0; ˙ and . 2 y p p . 2; 2/ D rD1 cos.C 4 x / 4 x Fig. 8.5-14 rD2 sin 2 15. r D 1 C 2 cos r D 0 if D ˙2=3. Fig. 8.5-18 y 2=3 19. r D cos 3 (three leaf rosette) r D 0 at D ˙=6, ˙=2, ˙5=6. y 1 3 x =6 x 2=3 Fig. 8.5-15 16. If r D 1 2 sin , then r D 0 at D 5 and . 6 6 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) Fig. 8.5-19 335 lOMoARcPSD|6566483 www.konkur.in SECTION 8.5 (PAGE 493) ADAMS and ESSEX: CALCULUS 9 y 3 5 and 20. If r D 2 cos 4 , then r D 0 at D ˙ ; ˙ ; ˙ 8 8 8 7 ˙ . (an eight leaf rosette) 8 =3 y 3 D 8 x D 8 x Fig. 8.5-23 rD2 cos 4 24. sin y D r sin D ln sin D . ln / Fig. 8.5-20 21. If r D ln , then r D 0 at D 1. Note that !0 as ! 0C. Therefore, the (negative) x-axis is an asymptote of the curve. p r 2 D 4 sin 2 . Thus r D ˙2 sin 2 . This is a lemniscate. r D 0 at D 0, D ˙=2, and D . y y x x rDln Fig. 8.5-24 Fig. 8.5-21 22. If r 2 D 4 cos 3 , then r D 0 at D ˙ ; ˙ and 6 2 5 ˙ . This equation defines two functions of r, namely 6 p r D ˙2 cos 3 . Each contributes 3 leaves to the graph. 25. r D 3 cos , and r D sin both pass through the origin, and so intersect there. Also p p sin D 3 cos ) tan D p 3 ) D =3; 4=3. 3=2; =3. Both of these give the same point Œ p Intersections: the origin and Œ 3=2; =3. 26. r 2 D 2 cos.2 /, r D 1. cos.2 / D 1=2 ) D ˙=6 or D ˙5=6. Intersections: Œ1; ˙=6 and Œ1; ˙5=6. 27. r D 1 C cos , r D 3 cos . Both curves pass through the origin, so intersect there. Also 3 cos D 1Ccos ) cos D 1=2 ) D ˙=3. Intersections: the origin and Œ3=2; ˙=3. 28. Let r1 . / D and r2 . / D C . Although the equation r1 . / D r2 . / has no solutions, the curves r D r1 . / and r D r2 . / can still intersect if r1 .1 / D r2 .2 / for two angles 1 and 2 having the opposite directions in the polar plane. Observe that 1 D n and 2 D .n 1/ are two such angles provided n is any integer. Since y D 6 x r 2 D4 cos 3 Fig. 8.5-22 p r1 .1 / D p 23. r 2 D sin 3 . Thus r D ˙ sin 3 . This is a lemniscate. r D 0 at D 0, ˙=3, ˙2=3, . 336 Telegram: @uni_k n D r2 ..n 1//; the curves intersect at any point of the form Œn; 0 or Œn; . Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 8.6 (PAGE 497) 29. If r D 1= for > 0, then lim y D lim !0C !0C 37. r D C C cos sin.3 / For C < 1 there appear to be 6 petals of 3 different sizes. For C 1 there are only 4 of 2 sizes, and these coalesce as C increases. sin D 1: Thus y D 1 is a horizontal asymptote. 38. y y yD1 rD1= r D ln. / x x Fig. 8.5-29 30. The graph of r D cos n has 2n leaves if n is an even integer and n leaves if n is an odd integer. The situation for r 2 D cos n is reversed. The graph has 2n leaves if n is an odd integer (provided negative values of r are allowed), and it has n leaves if n is even. Fig. 8.5-38 We will have Œln 1 ; 1 D Œln 2 ; 2 if 31. If r D f . /, then 2 D 1 C x D r cos D f . / cos y D r sin D f . / sin : 32. r D cos cos.m / For odd m this flower has 2m petals, 2 large ones and 4 each of .m 1/=2 smaller sizes. For even m the flower has m C 1 petals, one large and 2 each of m=2 smaller sizes. and ln 1 D that is, if ln 1 C ln.1 C / D 0. This equation has solution 1 0:29129956. The corresponding intersection point has Cartesian coordinates .ln 1 cos 1 ; ln 1 sin 1 / . 1:181442; 0:354230/. 39. y 33. r D 1 C cos cos.m / These are similar to the ones in Exercise 32, but the curve does not approach the origin except for D in the case of even m. The petals are joined, and less distinct. The smaller ones cannot be distinguished. r D 1= x 34. r D sin.2 / sin.m / For odd m there are m C 1 petals, 2 each of .m C 1/=2 different sizes. For even m there are always 2m petals. They are of n different sizes if m D 4n 2 or m D 4n. 35. r D 1 C sin.2 / sin.m / These are similar to the ones in Exercise 34, but the petals are joined, and less distinct. The smaller ones cannot be distinguished. There appear to be m C 2 petals in both the even and odd cases. 36. r D C C cos cos.2 / The curve always has 3 bulges, one larger than the other two. For C D 0 these are 3 distinct petals. For 0 < C < 1 there is a fourth supplementary petal inside the large one. For C D 1 the curve has a cusp at the origin. For C > 1 the curve does not approach the origin, and the petals become less distinct as C increases. r D ln. / Fig. 8.5-39 The two intersections of r D ln and r D 1= for 0 < 2 correspond to solutions 1 and 2 of ln 1 D 1 ; 1 ln 2 D 1 : 2 C The first equation has solution 1 1:7632228, giving the point . 0:108461; 0:556676/, and the second equation has solution 2 0:7746477, giving the point . 0:182488; 0:178606/. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k ln 2 ; Downloaded by ted cage (sxnbyln180@questza.com) 337 lOMoARcPSD|6566483 www.konkur.in SECTION 8.6 (PAGE 497) ADAMS and ESSEX: CALCULUS 9 y Section 8.6 Slopes, Areas, and Arc Lengths for Polar Curves (page 497) 1. Area D 1 2 Z 2 0 d D D 3 A .2/2 D 2: 4 x y rDsin 3 p rD Fig. 8.6-4 D0 D2x 5. Fig. 8.6-1 Z 1 =8 cos2 4 d Total area D 16 2 0 Z =8 D4 .1 C cos 8 / d 0 ˇ=8 sin 8 ˇˇ sq. units. D4 C D ˇ ˇ 8 2 ˇ2 Z 1 2 2 3 ˇˇ 4 2. Area D d D ˇ D 3 sq. units. 2 0 6 ˇ 3 0 0 y y rDcos 4 =8 x x A rD Fig. 8.6-5 Fig. 8.6-2 3. Area D 4 D 2a 1 2 Z =4 0 6. ˇ=4 ˇ D a2 sq. units. ˇ ˇ 2 sin 2 ˇ 2 a2 cos 2 d The circles r D a and r D 2a cos intersect at D ˙=3. By symmetry, the common area is 4 .area of sector area of right triangle/ (see the figure), i.e., 0 y r 2 Da2 cos 2 4 " 1 2 a 6 1a 22 x p 3a 2 # D 4 p 3 3 2 a sq. units. 6 y rD2a cos rDa Fig. 8.6-3 A 4. 1 Area D 2 Z =3 0 1 D 4 338 Telegram: @uni_k Z =3 1 sin2 3 d D .1 cos 6 / d 4 0 ˇ ˇ=3 1 ˇ D sin 6 ˇ sq. units. ˇ 6 12 0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) Fig. 8.6-6 x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 7. SECTION 8.6 (PAGE 497) Z 1 .1 cos /2 d Area D 2 2 =2 2 Z 1 C cos 2 D d 1 2 cos C 2 =2 ˇ 3 sin 2 ˇˇ D 2 sin ˇ ˇ 2 2 4 =2 D and D ˙=3. The shaded area is given by 2 2 1 2 2 "Z D =3 y 3 D 2 rD1 .1 C cos / d =3 Z C 2 sq. units. 4 rD1 cos 2 x 4 D 9 2 9 2 cos d =3 1 C 2 cos C Z =2 Z =2 1 C cos 2 2 =3 .1 C cos 2 / d # d ˇ sin 2 ˇˇ C 2 sin C ˇ ˇ 4 =3 ˇ ˇ=2 sin 2 ˇ 9 C ˇ ˇ 2 2 =3 p p ! p 3 3 9 3 C sq. units. D 8 4 2 4 2 3 y Fig. 8.6-7 =3 rD3 cos rD1Ccos x 8. Z 1 2 1 =2 2 a C 2 a .1 sin /2 d 2 2 0 Z =2 a2 1 cos 2 D C a2 d 1 2 sin C 2 2 0 ˇˇ=2 1 a2 ˇ 2 3 Ca C 2 cos sin 2 ˇ D ˇ 2 2 4 0 5 2 2 a sq. units. D 4 Area D Fig. 8.6-9 5 and ˙ , 6 6 the area inside the lemniscate and outside the circle is 10. Since r 2 D 2 cos 2 meets r D 1 at D ˙ y 4 rDa x A 1 2 Z =6 h 2 cos 2 0 ˇ=6 ˇ ˇ D 2 sin 2 ˇˇ ˇ i 12 d p D 3 3 sq. units. 3 0 y rDa.1 sin / rD1 Fig. 8.6-8 A A x r 2 D2 cos 2 9. For intersections: 1 C cos D 3 cos . Thus 2 cos D 1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) Fig. 8.6-10 339 lOMoARcPSD|6566483 www.konkur.in SECTION 8.6 (PAGE 497) 11. ADAMS and ESSEX: CALCULUS 9 r D 0 at D ˙2=3. The shaded area is 2 1 2 Z 14. .1 C 2 cos /2 d 2=3 Z 1 C 4 cos C 2.1 C cos 2 / d 2=3 ˇ ˇ ˇ ˇ ˇ ˇ D3 C 4 sin ˇ C sin 2 ˇ ˇ ˇ 3 2=3 2=3 p p p 3 3 3 D sq. units. D 2 3C 2 2 D D0 ˇ ˇD2 ˇ sec u tan u C ln j sec u C tan uj ˇ D ˇ 2 D0 ˇD2 ˇ i h p p a ˇ D 1 C 2 C ln j 1 C 2 C j ˇ ˇ 2 D0 i p ah p 2 D 2 1 C 4 C ln.2 C 1 C 4 2 / units: 2 a y 2=3 rD1C2 cos 15. 1 Z 2 p a2 C a2 2 d 0 Z 2 p Da 1 C 2 d Let D tan u 0 d D sec2 u d Z D2 sec3 u du Da sD 3 x r 2 D cos 2 dr D 2 sin 2 2r d s dr D d ) sin 2 r p sin2 2 d D sec 2 d cos 2 Z =4 p sec 2 d: Length D 4 ds D 2=3 cos 2 C 0 y Fig. 8.6-11 r 2 Dcos 2 x 12. sD Z 0 s dr d 2 C r 2 d D Z p 4 C 2 d Z p 0 4 2 C 4 d Let u D 4 C 2 0 du D 2 d ˇ4C 2 Z 4C 2 p 1 1 3=2 ˇˇ D u du D u ˇ ˇ 2 4 3 4 h i 1 2 3=2 D .4 C / 8 units: 3 D Fig. 8.6-15 16. If r 2 D cos 2 , then 2r dr D d and ds D dr D ae a . 13. r D e , . /. dp p ds D e 2a C a2 e 2a d D 1 C a2 e a d . The length of the curve is a p Z p 1 C a2 a .e 1 C a2 e a d D a 340 Telegram: @uni_k e a 2 sin 2 ) s cos 2 C dr D d sin 2 p cos 2 d sin2 2 : d D p cos 2 cos 2 a) Area of the surface generated by rotation about the xaxis is Sx D 2 / units. D 2 D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) Z =4 r sin ds 0 Z =4 p 0 cos 2 sin p ˇ=4 ˇ ˇ D .2 2 cos ˇ ˇ 0 p d cos 2 2/ sq. units: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 8.6 (PAGE 497) y b) Area of the surface generated by rotation about the yaxis is Z =4 Sy D 2 r cos ds =4 D 4 Z =4 p cos 2 cos p d cos 2 ˇ=4 ˇ p ˇ D 2 2 sq. units. D 4 sin ˇ ˇ 0 x 0 17. For r D 1 C sin , Fig. 8.6-18 19. r 1 C sin D : dr=d cos p If D =4, then tan D 2 Cp1 and D 3=8. 2 and D =8. If D 5=4, then tan D 1 The line y D x meets the cardioid r D 1 C sin at the origin at an angle of 45ı , and also at first and third quadrant points at angles of 67:5ı and 22:5ı as shown in the figure. tan rD2 cos r 2 D2 sin 2 D y The curves r D 1 cos and r D 1 sin intersect on the rays D =4 and D 5=4, as well as at the origin. At the origin their cusps clearly intersect at right angles. For r D 1 cos , tan p1 D .1 cos /= sin . At D =4, tan 1 D 2 p 1, so 1 D =8. At D 5=4, tan 1 D . 2 C 1/, so 1 D 3=8. For r D 1 sin , tan 2 Dp.1 sin /=. cos /. At D =4, tan 2 D 1p 2, so 2 D =8. At D 5=4, tan 2 D 2 C 1, so 2 D 3=8. At =4 the curves intersect at angle =8 . =8/ D =4. At 5=4 the curves intersect at angle 3=8 . 3=8/ D 3=4 (or =4 if you use the supplementary angle). y rD1Csin rD1 cos D =4 rD1 sin x x Fig. 8.6-17 18. The two curves r 2 D 2 sin 2 and r D 2 cos intersect where 2 sin 2 D 4 cos2 4 sin cos D 4 cos2 .sin cos / cos D 0 , sin D cos or cos D 0; p and P2 D .0; 0/. 2; i.e., at P1 D 4 dr For r 2 D 2 sin 2 we have 2r D 4 cos 2 . At P1 we d p have r D 2 and dr=d D 0. Thus the angle between the curve and the radial line D =4 is D =2. For r D 2 cos we have dr=d D 2 sin , so the angle between this curve ˇ and the radial line D =4 satisfies r ˇˇ tan D D 1, and D 3=4. The two dr=d ˇD=4 3 D . curves intersect at P1 at angle 4 2 4 The Figure shows that at the origin, P2 , the circle meets the lemniscate twice, at angles 0 and =2. Fig. 8.6-19 20. We have r D cos C sin . For horizontal tangents: d dy D cos sin C sin2 d d D cos2 sin2 C 2 sin cos cos 2 D sin 2 , tan 2 D 0D , 3 or : The tangents are horizontal at Thus D 8 8 " # cos sin ; and 8 8 8 " # 3 3 3 cos C sin ; . 8 8 8 For vertical tangent: d 2 dx D cos C cos sin d d D 2 cos sin C cos2 sin2 sin 2 D cos 2 , tan 2 D 1: 0D , Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1: Downloaded by ted cage (sxnbyln180@questza.com) 341 lOMoARcPSD|6566483 www.konkur.in SECTION 8.6 (PAGE 497) ADAMS and ESSEX: CALCULUS 9 Thus There are vertical tangents at " D =8 of 5=8. # C sin ; and cos 8 8 8 " # 5 5 5 cos C sin ; . 8 8 8 There are no points on the curve where cos D 0. Therefore, horizontal tangents occur only where 1 tan2 D 1=3: There are horizontal tangents at p ; ˙ 6 2 1 5 and p ; ˙ . 6 2 For vertical tangents: y rDcos Csin , d r cos D r sin C cos d cos 2 sin D sin 2 cos , sin D 0 or 0D , .cos2 x Fig. 8.6-20 sin2 / sin D sin 2 r 2 sin cos2 3 cos2 D sin2 : There are no points on the curve where tan2 D 3, so the only vertical tangents occur where sin D 0, that is, at the points with polar coordinates Œ1; 0 and Œ1; . y 21. r 2 Dcos 2 r D cot . r D 2 cos . tan D dr=d For horizontal tangents we want tan D tan . Thus we want tan D cot , and so p D ˙=4 or ˙3=4. The tangents are horizontal at Œ 2; ˙=4. For vertical tangents we want tan D cot . Thus we want cot D cot , and so D 0, ˙=2, or . There are vertical tangents at the origin and at Œ2; 0. x Fig. 8.6-22 y D=4 23. rD2 cos sin 2 D 21 tan 2 . 2 cos 2 For horizontal tangents: r D sin 2 . tan D 2 x tan 2 D 2 tan 2 tan D 2 tan 2 1 tan tan 1 C .1 tan2 / D 0 D =4 Fig. 8.6-21 tan .2 tan2 / D 0: p p Thus D 0, , ˙ tan 1 2, ˙ tan 1 2. There are horizontal tangents at the origin and the points dr 22. We have r D cos 2 , and 2r D d zontal tangents: 2 , d r sin D r cos C sin d cos 2 cos D sin 2 sin , cos D 0 or 0D , .cos2 342 Telegram: @uni_k 2 sin 2 . For hori- sin 2 r sin2 / cos D 2 sin2 cos cos2 D 3 sin2 : " p # p 2 2 ; ˙ tan 1 2 3 and " p # p 2 2 ; ˙ tan 1 2 : 3 Since the rosette r D sin 2 is symmetric about x D y, there must be vertical tangents at the origin and at the points # " p 2 2 1 1 ; ˙ tan p 3 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) and " p # 2 2 1 1 ; ˙ tan p : 3 2 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES 8 (PAGE 498) y y rD2.1 sin / x x rDsin 2 Fig. 8.6-25 Fig. 8.6-23 24. We have r D e and dr D e . For horizontal tangents: d d r sin D e cos C e sin d C k; tan D 1 , D 4 0D , where k D 0; ˙1; ˙2; : : :. At the points Œe k =4 ; k =4 the tangents are horizontal. For vertical tangents: d r cos D e cos e sin d tan D 1 $ D C k: 4 0D , At the points Œe cal. 25. r D 2.1 kC=4 sin /, tan ; k C =4 the tangents are vertiD For horizontal tangents tan 1 26. dx dy D f 0 . / cos f . / sin ; D f 0 . / sin C f . / cos d r d 2 2 f 0 . / cos f . / sin C f 0 . / sin C f . / cos d ds D 2 2 D f 0 . / cos2 2f 0 . /f . / cos sin C f . / sin2 1=2 2 2 d C f 0 . / sin2 C 2f 0 . /f . / sin cos C f . / cos2 r 2 2 D f 0 . / C f . / d: Review Exercises 8 (page 498) 1. sin . cos D cot , so sin sin D cos cos cos D 0; or 2 sin D 1: 1 The solutions are D ˙=2, ˙=6, and ˙5=6. D =2 corresponds to the origin where the cardioid has a cusp, and therefore no tangent. There are horizontal tangents at Œ4; =2, Œ1; =6, and Œ1; 5=6. For vertical tangents tan D cot , so sin cos D cos sin sin2 sin D cos2 D 1 1 sin2 x D r cos D f . / cos , y D r sin D f . / sin . 2. x2 C y2 D 1 2 p Ellipse, semi-major axis a D 2, along the x-axis. Semiminor axis b D 1. c 2 D a2 b 2 D 1. Foci: .˙1; 0/. x 2 C 2y 2 D 2 , x2 y2 9x 2 4y 2 D 36 , D1 4 9 Hyperbola, transverse axis along the x-axis. Semi-transverse axis a D 2, semi-conjugate axis b D 3. p c 2 D a2 C b 2 D 13. Foci: .˙ 13; 0/. Asymptotes: 3x ˙ 2y D 0. 3. x C y 2 D 2y C 3 , .y 1/2 D 4 x Parabola, vertex .4; 1/, opening to the left, principal axis y D 1. a D 1=4. Focus: .15=4; 1/. 4. 2x 2 C 8y 2 D 4x 48y 2.x 2 2x C 1/ C 8.y 2 C 6y C 9/ D 74 2 2 sin sin 1 D 0 .sin 1/.2 sin C 1/ D 0 .x The solutions here are D =2 (the origin again), D =6 and D 5=6. There are vertical tangents at Œ3; =6 and Œ3; 5=6. Ellipse, p 3/, major axis along y D 3. p centre .1; a D 37, bpD 37=2, c 2 D a2 b 2 D 111=4. Foci: .1 ˙ 111=2; 3/. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k .y C 3/2 1/2 C D 1: 37 37=4 Downloaded by ted cage (sxnbyln180@questza.com) 343 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 8 (PAGE 498) ADAMS and ESSEX: CALCULUS 9 y 5. x D t , y D 2 t , .0 t 2/. Straight line segment from .0; 2/ to .2; 0/. . 2;4/ xDt 3 3t yDt 3 C3t 6. x D 2 sin.3t /, y D 2 cos.3t /, .0 t 2/ Part of a circle of radius 2 centred at the origin from the point .0; 2/ clockwise to .2 sin 6; 2 cos 6/. x .2; 4/ 7. x D cosh t , y D sinh2 t . Parabola x 2 y D 1, or y D x 2 right. 1, traversed left to Fig. R-8-12 13. 8. x D e t , y D e 2t , . 1 t 1/. Part of the curve x 2 y D 1 from .1=e; e 2 / to .e; 1=e 2 /. 9. x D cos.t =2/, y D 4 sin.t =2/, .0 t /. The first quadrant part of the ellipse 16x 2 C y 2 D 16, traversed counterclockwise. y D t3 x D t 3 3t dy dx D 3.t 2 1/ D 3t 2 dt dt Horizontal tangent at t D 0, i.e., at .0; 0/. Vertical tangent at t D ˙1, i.e., at .2; 1/ and . 2; 1/. t2 dy > 0 if jt j > 1 D 2 < 0 if jt j < 1 dx t 1 Slope ! 1 as t ! ˙1. y Slope 10. x D cos t C sin t , y D cos t sin t , .0 t 2/ The circle x 2 C y 2 D 2, traversed clockwise, starting and ending at .1; 1/. tD1 11. 4 xD y D t 3 3t 1 C t2 dy dx 8t D 3.t 2 1/ D dt dt .1 C t 2 /2 Horizontal tangent at t D ˙1, i.e., at .2; ˙2/. Vertical tangent at t D 0, p i.e., at .4; 0/. Self-intersection at t D ˙ 3, i.e., at .1; 0/. y tD 1 p tD˙ 3 tD0 x Fig. R-8-13 14. y D t 3 12t x D t 3 3t dy dx D 3.t 2 1/ D 3.t 2 4/ dt dt Horizontal tangent at t D ˙2, i.e., at .2; 16/ and . 2; 16/. Vertical tangent at t D ˙1, i.e., at .2; 11/ and . 2; 11/. Slope Fig. R-8-11 y D t 3 C 3t x D t 3 3t dy dx D 3.t 2 1/ D 3.t 2 C 1/ dt dt Horizontal tangent: none. Vertical tangent at t D ˙1, i.e., at .2; 4/ and . 2; 4/. 2 dy t C1 > 0 if jt j > 1 D 2 < 0 if jt j < 1 dx t 1 Slope ! 1 as t ! ˙1. Slope 344 Telegram: @uni_k x t2 4 dy > 0 if jt j > 2 or jt j < 1 D 2 < 0 if 1 < jt j < 2 dx t 1 Slope ! 1 as t ! ˙1. y . 2;16/ tD1 12. tD 1 .2;11/ xDt 3 3t yDt 3 12t . 2; 11/ .2; 16/ Fig. R-8-14 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 15. REVIEW EXERCISES 8 (PAGE 498) The curve x D t 3 t , y D jt 3 j is symmetric about x D 0 since x is an odd function and y is an even function. Its self-intersection occurs at a nonzero value of t that makes x D 0, namely, t D ˙1. The area of the loop is 19. r D ; 3 3 2 2 y rD Z 1 .t 3 t /3t 2 dt . x/ dy D 2 0 tD0 ˇ1 3 4 ˇˇ 1 6 D t C t ˇ D sq. units. ˇ 2 2 AD2 Z tD1 x 0 y Fig. R-8-19 x D t3 t y D jt 3 j 20. r D j j; . 2 2/ y rDjj tD˙1 x tD0 x Fig. R-8-15 16. The volume of revolution about the y-axis is V D Z tD1 D .t 6 D 3 D 3 17. x D et 0 Z 1 .t 8 0 Fig. R-8-20 x 2 dy tD0 Z 1 1 9 21. 2t 4 C t 2 /3t 2 dt r D 1 C cos.2 / y 2t 6 C t 4 / dt 2 1 8 C cu. units. D 7 5 105 rD1Ccos 2 x t , y D 4e t=2 , .0 t 2/. Length is Z 2p .e t 1/2 C 4e t dt 0 Z 2p Z 2 D .e t C 1/2 dt D .e t C 1/ dt 0 0 ˇ2 ˇ t D .e C t /ˇˇ D e 2 C 1 units: LD Fig. R-8-21 22. r D 2 C cos.2 / y rD2Ccos.2/ 0 18. Area of revolution about the x-axis is Z S D 2 4e t=2 .e t C 1/ dt ˇˇ2 2 3t=2 t=2 ˇ e C 2e D 8 ˇ ˇ 3 x Fig. R-8-22 0 16 3 D .e C 3e 3 4/ sq. units. 23. r D 1 C 2 cos.2 / Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 345 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 8 (PAGE 498) ADAMS and ESSEX: CALCULUS 9 y 27. rD1C2 cos 2 x Z =4 p 1 AD2 .1 C 2 2 sin C 2 sin2 / d 2 =2 Z =4 p D .2 C 2 2 sin cos.2 // d =2 Fig. R-8-23 24. r D1 D 2 sin.3 / y p the origin in the directions for r D 1C 2 sin approaches p which sin D 1= 2, that is, D 3=4 and D =4. The smaller loop corresponds to values of between these two values. By symmetry, the area of the loop is D 2 rD1 sin.3/ p 2 2 cos ˇˇ =4 1 ˇ sin.2 / ˇ ˇ 2 =2 1 3 2C D sq. units. 2 2 p rD1C 2 sin y =6 x 3=4 Fig. R-8-27 Fig. R-8-24 28. 25. Area of a large loop: AD2 D 1 2 Z =3 Z =3 0 D 1 2 Z =2 =3 =3 346 Telegram: @uni_k p Only . 2 1/=2 is between value of cos . Let 0 D cos v u u sin 0 D t1 p 1 1 : 1pand 1, so is a possible 2 1 . Then 2 2 1 2 !2 D p p 1C2 2 : 2 By symmetry, the area inside r D 1 C cos to the left of the line x D 1=4 is .1 C 2 cos.2 //2 d Œ1 C 4 cos.2 / C 2.1 C cos.4 // d ˇˇ=2 1 ˇ D 3 C 2 sin.2 / C sin.4 / ˇ ˇ 2 =3 p 3 3 D sq. units. 2 4 1 4 cos 4 cos2 C 4 cos 1 D 0 p p 4 ˙ 16 C 16 ˙ 2 D cos D 8 2 .1 C 2 cos.2 //2 d 26. Area of a small loop: AD2 r cos D x D 1=4 and r D 1 C cos intersect where 1 C cos D Œ1 C 4 cos.2 / C 2.1 C cos.4 // d 0 ˇˇ=3 1 ˇ D 3 C 2 sin.2 / C sin.4 / ˇ ˇ 2 0 p 3 3 sq. units. DC 4 Z =2 x =4 Z 1 C cos.2 / 1 C 2 cos C d C cos 0 sin 0 2 0 ˇ ˇ 3 1 ˇ D . 0 / C 2 sin C sin.2 / ˇ ˇ 2 4 0 p p p . 2 1/ 1 C 2 2 C 4 ! q ! p p p 3 2 1 2 9 1 cos C 1C2 2 sq. units. D 2 2 8 AD2 1 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL y xD1=4 CHALLENGING PROBLEMS 8 (PAGE 498) rD1Ccos 0 C1 x A1 S1 Fig. R-8-28 C Challenging Problems 8 (page 498) P F1 F2 C2 A2 1. The surface of the water is elliptical (see Problem 2 below) whose semi-minor axis is 4 cm, the radius of the cylinder, and whose semi-major axis is 4 sec cm because of the tilt of the glass. The surface area is that of the ellipse x D 4 sec cos t; y D 4 sin t; S2 Fig. C-8-2 Let P be any point on C . Let A1 A2 be the line through P that lies on the cylinder, with A1 on C1 and A2 on C2 . Then PF1 D PA1 because both lengths are of tangents drawn to the sphere S1 from the same exterior point P . Similarly, PF2 D PA2 . Hence .0 t 2/: This area is AD4 D4 Z tD=2 PF1 C PF2 D PA1 C PA2 D A1 A2 ; x dy tD0 which is constant, the distance between the centres of the two spheres. Thus C must be an ellipse, with foci at F1 and F2 . Z =2 .4 sec cos t /.4 cos t / dt Z =2 D 32 sec .1 C cos.2t // dt D 16 sec cm2 : 0 3. 0 4 cm Given the foci F1 and F2 , and the point P on the ellipse, construct N1 PN2 , the bisector of the angle F1 PF2 . Then construct T1 P T2 perpendicular to N1 N2 at P . By the reflection property of the ellipse, N1 N2 is normal to the ellipse at P . Therefore T1 T2 is tangent there. N2 T1 P 4 sec cm T2 N1 F1 F2 Fig. C-8-1 2. Let S1 and S2 be two spheres inscribed in the cylinder, one on each side of the plane that intersects the cylinder in the curve C that we are trying to show is an ellipse. Let the spheres be tangent to the cylinder around the circles C1 and C2 , and suppose they are also tangent to the plane at the points F1 and F2 , respectively, as shown in the figure. Fig. C-8-3 4. Without loss of generality, choose the axes and axis scales so that the parabola has equation y D x 2 . If P is the point .x0 ; x02 / on it, then the tangent to the parabola at P has equation y D x02 C 2x0 .x x0 /; Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 347 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 8 (PAGE 498) ADAMS and ESSEX: CALCULUS 9 which intersects the principal axis x D 0 at .0; x02 /. Thus R D .0; x02 / and Q D .0; x02 /. Evidently the vertex V D .0; 0/ bisects RQ. y y 6. P D Œr; Œa; 0 r Q P D .x0 ; x02 / a L 0 x x V Fig. C-8-6 R a) Let L be a line not passing through the origin, and let Œa; 0 be the polar coordinates of the point on L that is closest to the origin. If P D Œr; is any point on the line, then, from the triangle in the figure, Fig. C-8-4 To construct the tangent at a given point P on a parabola with given vertex V and principal axis L, drop a perpendicular from P to L, meeting L at Q. Then find R on L on the side of V opposite Q and such that QV D VR. Then PR is the desired tangent. a D cos. r rD a : cos. 0 / b) As shown in part (a), any line not passing through the origin has equation of the form r D g. / D 5. y b or 0 /; a D a sec. cos. 0 / 0 /; for some constants a and 0 . We have c g 0 . / D a sec. 0 / tan. g 00 . / D a sec. 2 ft 2 ft C a sec3 . a x D a2 sec2 . 2 2 0 / 0 / tan2 . 0 / 2 2 0 / g. / C 2 g 0 . / 0 / C 2a2 sec2 . 2 g. /g 00 . / 0 / tan2 . 2 0 / 4 a sec . 0 / tan . 0 / a sec . 0 / h i D a sec2 . 0 / 1 C tan2 . 0 / sec4 . 0 / 2 D 0: Fig. C-8-5 x2 y2 C 2 D 1, with a D 2 and foci at 2 a b .0; ˙2/ so that c D 2 and b 2 D a2 C c 2 D 8. The volume of the barrel is c) If r D g. / is the polar equation of the tangent to r D f . / at D ˛, then g.˛/ D f .˛/ and g 0 .˛/ D f 0 .˛/. Suppose that Let the ellipse be Z 2 V D2 x 2 dy D 2 4 1 0 0 ˇ 2 40 3 y 3 ˇˇ ft : D 8 y ˇ D 24 ˇ 3 Z 2 0 348 Telegram: @uni_k 2 y 8 2 2 f .˛/ C 2 f 0 .˛/ f .˛/f 00 .˛/ > 0: By part (b) we have dy 2 2 g.˛/ C 2 g 0 .˛/ g.˛/g 00 .˛/ D 0: Subtracting, and using g.˛/ D f .˛/ and g 0 .˛/ D f 0 .˛/, we get f 00 .˛/ < g 00 .˛/. It follows Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 8 (PAGE 498) that f . / < g. / for values of near ˛; that is, the graph of r D f . / is curving to the origin side of its tangent at ˛. Similarly, if 2 2 f .˛/ C 2 f 0 .˛/ f .˛/f 00 .˛/ < 0; 8. Take the origin at station O as shown in the figure. Both of the lines L1 and L2 pass at distance 100 cos from the origin. Therefore, by Problem 6(a), their equations are 100 cos 100 cos D sin. C / cos 2 100 cos 100 cos D : rD sin. / cos C 2 L1 W then the graph is curving to the opposite side of the tangent, away from the origin. rD L2 W The search area A./ is, therefore, 7. x.t / 0 B A x Z C 1002 cos2 1002 cos2 d sin2 . / sin2 . C / 4 Z C 4 csc2 . / csc2 . C / d D 5; 000 cos2 1 A./ D 2 4 4 D 5; 000 cos cot 4 C 2 2 cot 4 C cot 4 2 # " cos 4 C 2 sin 4 C 2 2 C 2 D 5; 000 cos sin 4 C 2 cos 4 C 2 D 10; 000 cos2 csc 2 C 4 1 2 r R D 10; 000 cos2 .sec.4/ 1/ mi2 : For D 3ı D =60, we have A./ 222:8 square miles. Also A0 ./ D Fig. C-8-7 When the vehicle is at position x, as shown in the figure, the component of the gravitational force on it in the direction of the tunnel is ma.r/ cos D 20; 000 cos sin .sec.4/ When D 3ı , the search area increases at about 8645.=180/ 151 square miles per degree increase in . y mg x: R mgr cos D R L2 L1 By Newton’s Law of Motion, this force produces an acceleration d 2 x=dt 2 along the tunnel given by d 2x m 2 D dt 1/ C 40; 000 cos2 sec.4/ tan.4/ A0 .=60/ 8645: Area A./ 100 mi mg x; R =4 that is d 2x C ! 2 x D 0; dt 2 x O where g ! D : R 2 Fig. C-8-8 This is the equation ofpsimple harmonic motion, with period T D 2=! D 2 R=g. For R 3960 mi 2:09 107 ft, and g 32 ft/s2 , we have T 5079 s 84:6 minutes. This is a rather short time for a round trip between Atlanta and Baghdad, or any other two points on the surface of the earth. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 349 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 8 (PAGE 498) 9. The easiest way to determine which curve is which is to calculate both their areas; the outer curve bounds the larger area. The curve C1 with parametric equations x D sin t; 1 y D sin.2t /; 2 .0 t 2/ ADAMS and ESSEX: CALCULUS 9 The curve C2 with polar equation r 2 D cos.2 / has area 4 A2 D 2 Z =4 0 ˇ=4 ˇ ˇ cos.2 / d D sin.2 /ˇ D 1 sq. units. ˇ 0 C1 is the outer curve, and the area between the curves is 1/3 sq. units. y has area x A1 D 4 D4 Telegram: @uni_k Fig. C-8-9 y dx tD0 Z =2 0 D4 Z =2 D4 Z 1 Let u D cos t du D sin t dt 350 Z tD=2 1 sin.2t / cos t dt 2 sin t cos2 t dt 0 0 u2 du D 4 sq. units. 3 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 9.1 (PAGE 507) CHAPTER 9. SEQUENCES, SERIES, AND POWER SERIES Section 9.1 Sequences and Convergence (page 507) 1. 2. 3. 4. 5. 6. 7. 8. 2 8 9 2n2 D 2 D 1; ; ; : : : is bounded, n2 C 1 n2 C 1 5 5 positive, increasing, and converges to 2. ) ( 2n 4 3 8 D 1; ; ; ; : : : is bounded, positive, den2 C 1 5 5 17 creasing, and converges to 0. . 1/n 7 13 4 D 5; ; ; : : : is bounded, positive, and n 2 3 converges to 4. ( ) 1 1 1 D sin 1; sin ; sin ; : : : is bounded, sin n 2 3 positive, decreasing, and converges to 0. 2 n 1 1 3 8 15 D n D 0; ; ; ; : : : is bounded n n 2 3 4 below, positive, increasing, and diverges to infinity. ) ( n e e e 2 e 3 ; ; ; : : : is bounded, positive, D n decreasing, and converges to 0, since e < . n p e n e p D . Since e= > 1, the sequence n=2 is bounded below, positive, increasing, and diverges to infinity. ) ( 3 . 1/n n 1 2 ; ; : : : is bounded, alternating, ; D en e e2 e3 and converges to 0. 14. 5 5 2n n D lim lim 3n 7 3 15. 4 n n2 4 n D 1: lim D lim 5 nC5 1C n 16. n2 lim 3 D lim n C1 n n 9. f2 =n g is bounded, positive, decreasing, and converges to 0. n 1 2 3 n 1 .nŠ/2 . D 10. .2n/Š nC1 nC2 nC3 2n 2 2 anC1 .n C 1/ 1 Also, D < . Thus the sequence a .2n C 2/.2n C 1/ 2 n .nŠ/2 is positive, decreasing, bounded, and convergent .2n/Š to 0. fn cos.n=2/g D f0; 2; 0; 4; 0; 6; : : :g is divergent. ) ( sin 2 sin 3 sin n ; ; : : : is bounded and conD sin 1; 12. n 2 3 verges to 0. 11. 13. f1; 1; 2; 3; 3; 4; 5; 5; 6; : : :g is divergent. 17. lim. 1/n n2 18. lim 1 19. 20. 21. 22. lim 1 n 1C 1 n3 2 : 3 D 0: 2 1 p 1 p C 2 2 nC1 n n n D lim D 1 1 n 3n2 3 2 n n 1 : 3 en e n 1 e 2n D lim D 1: en C e n 1 C e 2n sin x cos x 1 D lim D lim D 1: x!0C x!0C n x 1 3 n n 3 n D lim 1 C D e 3 by l’H^opital’s lim n n Rule. x n D lim lim x!1 ln.x C 1/ ln.n C 1/ 1 D lim x C 1 D 1: D lim x!1 x!1 1 xC1 lim n sin p lim. n C 1 24. lim n p n/ D lim p nC1 n p D 0: nC1C n n2 .n2 4n/ 4n D lim p n C n2 4n 4 4n r D 2: p D lim D lim 2 4 nC n 4n 1C 1 n p p lim. n2 C n n2 1/ p n2 n2 C n .n2 1/ D lim p p n2 C n C n2 1 nC1 ! D lim r r 1 1 n 1C C 1 n n2 D lim r 1C 1C 1 C n Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k D n D 0: n3 C 1 23. 25. 2 7 n Downloaded by ted cage (sxnbyln180@questza.com) 1 n r 1 1 n2 D 1 : 2 351 lOMoARcPSD|6566483 www.konkur.in SECTION 9.1 (PAGE 507) 26. If an D n 1 nC1 n ADAMS and ESSEX: CALCULUS 9 Thus, fan g is increasing by induction. Observe that a1 < 5 and a2 < 5. If ak < 5 then , then n n lim an D lim n nC1 n , 1 1 n D lim 1 lim 1 C n n D 27. n 1 n e 1 De 2 e akC1 D (by Theorem 6 of Section 3.4). .1 2 3 n/.1 2 3 n/ .nŠ/2 D .2n/Š 1 2 3 n .n C 1/ .n C 2/ 2n n 1 2 3 n 1 D : nC1 nC2 nC3 nCn 2 Thus lim an D 0. aD 32. n2 D 0 since 2n grows much faster than n2 2n p 15 C 2a ) a2 ! 4n lim nŠ ! D 0: n ) 0 < an < .=4/n . Since =4 < 1, 1 C 22n therefore .=4/n ! 0 as n ! 1. Thus lim an D 0. p 30. Let a1 D 1 and anC1p D 1 C 2an for n D 1; 2; 3; : : :. Then we have a2 D 3 > a1 . If akC1 > ak for some k, then p 1 C 2akC1 > p b) Since ln x x p p p p p 1 C 2ak < 1 C 2.3/ D 7 < 9 D 3: 1 C 2a ) a2 2a 1D0)a D1˙ p 352 Telegram: @uni_k 25 D 5: p p 15 C 2akC1 > 15 C 2ak D akC1 : 15 D 0 ) a D 3; or a D 5: 1 1, 1 D1 which implies that ak e for all k. Since fan g is increasing, e is an upper bound for fan g. 33. 2: p 2 < 0, p it is not appropriate. Hence, we Since a D 1 must have lim an D 1 C 2. p 31. Let a1 D 3 and anC1pD 15 C 2an for n D 1; 2; 3; : : :. Then we have a2 D 21 > 3 D a1 . If akC1 > ak for some k, then akC2 D 2a 1 1 ln ak D k ln 1 C k 1C k k 1 C 2ak D akC1 : Therefore, an < 3 for all n, by induction. Since fan g is increasing and bounded above, it converges. Let lim an D a. Then aD p Since f 0 .x/ > 0, f .x/ must be an increasing function. Thus, fan g D fe f .xn / g is increasing. Thus, fan g is increasing by induction. Observe that a1 < 3 and a2 < 3. If ak < 3 then akC1 D 15 C 2.5/ D x ln x f 0 .x/ D ln.x C 1/ C x C 1 xC1 1 D ln x xC1 Z xC1 dt 1 D t xC1 x Z xC1 1 1 dt > xC1 x xC1 1 1 D 0: D xC1 xC1 29. an D akC2 D p Since a > a1 , we must have lim an D 5. 1 1 n so ln an D n ln 1 C . Let an D 1 C n n 1 a) If f .x/ D x ln 1 C D x ln.x C 1/ x ln x, then x 4n and lim D 0 by Theorem 3(b). Hence, nŠ n2 22n n2 n2 2n D lim n D lim n lim nŠ 2 nŠ 2 15 C 2ak < Therefore, an < 5 for all n, by induction. Since fan g is increasing and bounded above, it converges. Let lim an D a. Then an D 28. We have lim p Suppose fan g is ultimately increasing, say anC1 an if n N. Case I. If there exists a real number K such that an K for all n, then lim an D a exists by completeness. Case II. Otherwise, for every integer K, there exists n N such that an > K, and hence aj > K for all j n. Thus lim an D 1. If fan g is ultimately decreasing, then either it is bounded below, and therefore converges, or else it is unbounded below, and therefore diverges to negative infinity. 34. If fjan jg is bounded then it is bounded above, and there exists a constant K such that jan j K for all n. Therefore, K an K for all n, and so fan g is bounded above and below, and is therefore bounded. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 9.2 (PAGE 514) 35. Suppose limn!1 jan j D 0. Given any > 0, there exists an integer N D N./ such that if n > N , then jjan j 0j < . In this case jan 0j D jan j D jjan j 0j < , so limn!1 an D 0. 3. nD5 b) “If lim an D 1 and lim bn D 1, then lim.an C bn / D 0” is FALSE. Let an D 1 C n and bn D n; then lim an D 1 and lim bn D 1 but lim.an C bn / D 1. 1 1 1 C C C .2 C /10 .2 C /12 .2 C /14 1 1 1 D C C 1 C .2 C /10 .2 C /2 .2 C /4 1 1 1 h D D 1 .2 C /10 8 .2 C / .2 C /2 1 .2 C /2 4. D 5. d) “If neither fan g nor fbn g converges, then fan bn g does not converge” is FALSE. Let an D bn D . 1/n ; then lim an and lim bn both diverge. But an bn D . 1/2n D 1 and fan bn g does converge (to 1). e) “If fjan jg converges, then fan g converges” is FALSE. Let an D . 1/n . Then limn!1 jan j D limn!1 1 D 1, but limn!1 an does not exist. 6. (page 514) 9. 1. 1 1 1 1 1 C C C D 1C C 3 9 27 3 3 D 1 3 1 1 1 3 D 2. 3 3 3 C 4 16 3 C D 64 nD1 3 e D1C n e 1 C e D 8e 3 k 3 P1 5000 : 999 2 1 C D e 1 k X 2 e kD0 D 1 1 1 e D e e 1 : 8e 3 8e 4 : D 2 e 2 1 e P1 nD1 j=2 3 C 2n diverges to 1 because 2nC2 3 C1 n 3 C 2n 1 2 lim D lim D > 0: nC2 n!1 2 n!1 4 4 1 : 2 1 X 3 C 2n 1 4 n 1 D 3 1 C 41 D 3nC2 12 : 5 D 1 1X 3 nD0 1 D 3 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k D P cos.j/ D j1D2 . 1/j j=2 diverges because limj !1 . 1/j j=2 does not exist. j D1 nD0 1 1000 1 1 X 2kC3 ! 2 1 C 3 10. 1 X 5 1 X 1 kD0 8. i: 1 X . 5/2 . 5/3 . 5/4 . 5/n D C C C 2n 4 6 8 8 8 88 nD2 52 25 5 C 2 D 4 1 8 64 64 25 1 25 25 D 4 D : D 5 8 64 69 4416 1C 64 nD0 7. 1 # " 2 1 1 5 C D5 1C C 103n 1000 1000 1 X nD0 c) “If lim an D 1 and lim bn D 1, then lim an bn D 1” is TRUE. Let R be an arbitrary, large positive number. Since lim an D p1 R and lim bn D 1, we must have a n p R, for all sufficiently large n. Thus and bn an bn R, and lim an bn D 1. Section 9.2 Infinite Series 1 .2 C /2n D a) “If lim an D 1 and lim bn D L > 0, then lim an bn D 1” is TRUE. Let R be an arbitrary, large positive number. Since lim an D 1, and L > 0, 2R it must be true that an for n sufficiently large. L L for n Since lim bn D L, it must also be that bn 2 2R L sufficiently large. Therefore an bn D R for n L 2 sufficiently large. Since R is arbitrary, lim an bn D 1. 36. 1 X Downloaded by ted cage (sxnbyln180@questza.com) 1 n 1 1 1X 2 n C 3 9 3 nD0 1 1 1 5 1 1 C D C D : 1 2 9 2 3 6 1 3 3 353 lOMoARcPSD|6566483 www.konkur.in SECTION 9.2 (PAGE 514) 11. 1 1 D n.n C 2/ 2 Since 1 n ADAMS and ESSEX: CALCULUS 9 1 , therefore nC2 14. 1 1 1 1 C C C C 13 24 35 n.n C 2/ 1 1 1 1 1 1 1 1 1 D C C C C 2 1 3 2 4 3 5 4 6 1 1 1 1 1 1 C C C n 2 n n 1 nC1 n nC2 1 1 1 1 D 1C : 2 2 nC1 nC2 sn D Thus lim sn D Hence, 1 X 1 3 3 , and D . 4 n.n C 2/ 4 nD1 nD1 1 1 1 1 D C C C : 1/.2n C 1/ 13 35 57 1 1 D .2n 1/.2n C 1/ 2 partial sum is Since 1 sn D 1 2 1 2n 1 1 , the 2n C 1 1 1 1 C C 2 3 5 1 1 1 1 1 C C 2 2n 3 2n 1 2 2n 1 1 1 D 1 : 2 2n C 1 1 3 16. 17. 1 2n C 1 nD1 13. Since .3n fore .2n 1 1 D lim sn D : 1/.2n C 1/ 2 1 1 D 2/.3n C 1/ 3 1 3n 2 1 , there3n C 1 1 1 1 1 C C C C 14 47 7 10 .3n 2/.3n C 1/ 1 1 1 1 1 1 1 C C C D 3 1 4 4 7 7 10 1 1 1 1 C C 3n 5 3n 2 3n 2 3n C 1 1 1 1 : D 1 ! 3 3n C 1 3 Thus 354 Telegram: @uni_k nD1 .3n 1 1 D . 2/.3n C 1/ 3 1 1 D lim sn D : n.n C 1/.n C 2/ 4 1 1 1 1 > D , therefore the partial sums 2n 1 2n 2 n of the given series P exceed half those of the divergent harmonic series .1=2n/. Hence the given series diverges to infinity. 1 X n n diverges to infinity since lim D 1 > 0. nC2 nC2 1 1 Since n 1=2 D p for n 1, we have n n n X kD1 k 1=2 n X 1 kD1 k ! 1; P 1=2 as n ! 1 (harmonic series). Thus n diverges to infinity. 1 X 1 1 1 2 D 2 C C C diverges to infinity 18. nC1 2 3 4 nD1 since it is just twice the harmonic series with the first term omitted. n 1 if n is odd 19. sn D 1 C 1 1 C C . 1/n D . P 0 n if n is even Thus lim sn does not exist, and . 1/ diverges. 20. sn D P1 2 1 C ; nC1 nC2 Since nD1 Hence, 1 X 1 X nD1 15. .2n 1 1 1 D n.n C 1/.n C 2/ 2 n the partial sum is 1 1 1 2 1 1 2 sn D C C 1 C C 2 2 3 2 2 3 4 1 2 1 2 1 1 1 1 C C C C 2 n 1 n nC1 2 n nC1 nC2 1 1 1 1 D C : 2 2 nC1 nC2 12. Let 1 X Since 21. n.n C 1/ Since 1 C 2 C 3 C C n D , the given series 2 P1 2 is nD1 which converges to 2 by the result of n.n C 1/ Example 3 of this section. The total distance is " # 2 3 3 C C2 4 4 # " 2 3 3 3 C 1C C D2C2 2 4 4 2C2 2 D2 C 3 1 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 3 4 D 14 metres. lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 9.3 (PAGE 524) 29. “If an c > 0 for all n, then TRUE. We have P an diverges to infinity” is sn D a1 C a2 C a3 C C an c C c C c C C c D nc; and nc ! 1 as n ! 1. 2 m 30. Fig. 9.2-21 22. The balance at the end of 8 years is 31. h i sn D 1000 .1:1/8 C .1:1/7 C C .1:1/2 C .1:1/ .1:1/8 1 D 1000.1:1/ $12; 579:48: 1:1 1 23. For n > N let sn D n X j D1 aj , and Sn D n X j DN P Then sn D Sn C C , where C D jND11 aj . P an diverges and fbn g is bounded, then an bn 1 1 and bn D . diverges” is FALSE. Let an D n n C 1 P Then an D 1 and 0 bn 1=2. But P P 1 which converges by Example an bn D n.n C 1/ 3. “If P P P 2 “If an > 0 and an converges, then an converges” is TRUE.P Since an converges, therefore lim an D 0. Thus there exists N such that 0 < an 1 for n N . Thus 0 < an2 an for n N . n n X X If Sn D ak2 and sn D ak , then fSn g is increasing kDN aj . Sn sn We have lim sn D lim Sn C C W n!1 n!1 eitherPboth sides exist or neither does. Hence and 1 nDN both converge or neither does. Thus P1 nD1 an If fan g is ultimately positive, then the sequence fsn g of partial sums of the series must be ultimately increasing. By Theorem 2, if fsn g is ultimately increasing, then either it is bounded above, and therefore convergent, or else it Pis not boundedPabove and diverges to infinity. Since an D lim sn , an must either converge when fsn g converges and lim sn D s exists, or diverge to infinity when fsn g diverges to infinity. P 25. If fan g is ultimately negative, then the series an must either converge (if its partial sums are bounded below), or diverge to 1 (if its partial sums are not bounded below). P 26. “If an D 0 forPevery n, then an converge” is TRUE n 0 D 0, for every n, and so because s D n kD0 P an D lim sn D 0. P P 27. “If an converges, then 1=a Pn diverges to infinity” is FALSE. A counterexample is . 1/n =2n . P P 28. “If an and bn both diverge, then so does P 1 .an C bn /” is FALSE. Let an D and n P P 1 bn D , then an D 1 and bn D 1 but n P P .an C bn / D .0/ D 0. 24. kDN and bounded above: 1 X kD1 ak2 converges, and so kDN ak < 1: 1 X ak2 converges. kD1 Section 9.3 Convergence Tests for Positive Series (page 524) 1. 2. X 1 1 since converges by comparison with n2 C 1 n2 1 1 0< 2 < 2. n C1 n X 1 X nD1 n n4 2 converges by comparison with n 4 lim n 3. 1 X 1 nD1 X n2 C 1 1 n3 2 D 1; n3 since and 0 < 1 < 1: diverges to infinity by comparison with n3 C 1 2 1 n C1 > . since 3 n C1 n Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 X Downloaded by ted cage (sxnbyln180@questza.com) X1 n , 355 lOMoARcPSD|6566483 www.konkur.in SECTION 9.3 (PAGE 524) 4. 1 X nD1 p n n2 C n C 1 ADAMS and ESSEX: CALCULUS 9 converges by comparison with nD1 since lim 1 X p n n2 C n C 1 1 n3=2 D 1; and 1 n3=2 0 < 1 < 1: 14. 1 X 1 p diverges to infinity by the integral test, n ln n ln ln n nD3 since Z 1 Z 1 dt du p D p D 1: u t ln t ln ln t 3 ln ln 3 1 X nD2 1 converges by the integral test: n ln n.ln ln n/2 Z 1 5. Since sin x x for x 0, we have ˇ ˇ ˇ ˇ ˇsin 1 ˇ D sin 1 1 ; ˇ n2 ˇ n2 n2 6. 13. a ˇ X ˇˇ X 1 ˇ ˇsin 1 ˇ converges by comparison with so . ˇ n2 ˇ n2 1 X 1 converges by comparison with the geometric n C 5 nD8 1 n X 1 1 1 series since 0 < n < n. C5 15. 16. n D lim 1 D 1, the series n n X 1 converges by comparison with the geometric n n X 1 . series n 1 X 1Cn 1Cn 10. diverges to infinity since lim D 1 > 0. 2Cn 2Cn 9. Since limn!1 n n 1 nD0 11. Since nD1 .n C 1/4 nC1 4 1 .n C 1/Š D 0: D lim lim n nC1 n4 nŠ 19. X nŠ diverges to infinity by the ratio test, since n2 e n D lim 20. 1 X .2n/Š6n nD1 .3n/Š Telegram: @uni_k .n C 1/Š n2 e n 1 n2 D lim D 1: .n C 1/2 e nC1 nŠ e nC1 converges by the ratio test since .2n C 2/Š6nC1 lim .3n C 3/Š n2 p D 1. 1Cn n 356 1 p X 1 1 p D 2 p 2k k kD1 kD1 X 1 1 1 , the series < con2n .n C 1/ 2n 2n .n C 1/ X 1 . verges by comparison with the geometric series 2n 1 X n4 converges by the ratio test since 18. nŠ 17. n2 12. p diverges to infinity since 1Cn n nD1 lim 1 X diverges to infinity. 2 C n5=3 1 X < 1 if ln ln a > 0: X 1 . 1/n converges by comparison with , 4 n n4 n 2 1 . 1/ since 0 4. n4 n The series D2 X 1 C n4=3 diverges to infinity by comparison with the X 1 divergent p-series , since n1=3 , n1=3 C n5=3 1 1 C n4=3 D lim D 1: lim 5=3 1=3 n!1 2 C n n 2 C n5=3 du 2 ln ln a u X1 7. nD1 Z 1 1 X 2 2 2 1 C . 1/n p D 0 C p C 0 C p C 0 C p C n 2 4 6 nD1 nD8 X 1 Since .ln n/3 < n for large n, diverges to .ln n/3 X1 . infinity by comparison with n 1 X 1 8. diverges to infinity by comparison with the ln.3n/ nD1 1 X 1 1 1 since > for n 1. harmonic series 3n ln.3n/ 3n dt D t ln t .ln ln t /2 D lim Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) , .2n/Š6n .3n/Š .2n C 2/.2n C 1/6 D 0: .3n C 3/.3n C 2/.3n C 1/ lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 21. SECTION 9.3 (PAGE 524) p 1 X n converges by the ratio test, since 3n ln n We use the approximation nD2 s sn D sn C p 3n ln n nC1 p D lim nC1 3 ln.n C 1/ n r 1 ln n 1 nC1 D lim lim D < 1: 3 n ln.n C 1/ 3 22. D lim 24. .2n C 2/Š .nŠ/3 .2n C 2/.2n C 1/ D lim D 0 < 1: ..n C 1/Š/3 .2n/Š .n C 1/3 1 X 1 C nŠ diverges by comparison with the harmonic .1 C n/Š nD1 1 X 1 C nŠ nŠ 1 1 since > D . series nC1 .1 C n/Š .1 C n/Š nC1 25. 2n 1 X 1 nD1 n3 D lim 3nC1 2 D lim 3 26. 28. converges by the ratio test since 3n 3n n3 2nC1 3 .n C 1/ 2n 3 3n n 3 n 2 D lim 3 .n C 1/3 1 3 1 1 1 2 3n3 3.n C 1/3 1 .n C 1/3 n3 D 6 n3 .n C 1/3 1 3n2 C 3n C 1 7 D < : 6 n3 .n C 1/3 6n4 n s6 D 1 C 4 1:082 27. 1 n3 2 3n D < 1: 3 .n C 1/3 3nC1 1 1 C 3 73 6 1 is positive, continuous and decreasing x3 on Œ1; 1/, for any n D 1; 2; 3; : : :, we have Since f .x/ D where sn D n X 1 kD1 Z 1 n 1 dx D . If x3 2n2 2 nn e 1 1 n D D lim 1 C < 1: n nŠ n f .x/ D 1=x 4 is positive, continuous, and decreasing on Œ1; 1/. Let k and An D 3 1 sn D sn C .AnC1 C An /, then jsn ˇR Z 1 1 ˇˇ dx 1 D lim An D ˇ D 3: R!1 x4 3x 3 ˇ 3n n AnC1 1 1 D 2 4 n2 1 2n C 1 < 0:001 D 4 n2 .n C 1/2 sn j An 1 .n C 1/2 if n D 8. Thus, the error in the approximation s s8 is less than 0.001. 29. Since f .x/ D 1 x 3=2 is positive, continuous and decreasing on Œ1; 1/, for any n D 1; 2; 3; : : :, we have n sn C AnC1 s sn C An Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k with error less than 0.001 in absolute value: nD1 , 1 1 1 1 1 1 C 4 C 4 C 4 C 4 C 24 3 4 5 6 6 sn C AnC1 s sn C An 1 X nn converges by the ratio test since n nŠ .n C 1/nC1 lim .nC1/ .n C 1/Š : that is, if n4 > 7000=6. Since 64 D 1296 > 7000=6, n D 6 will do. Thus nD1 X We have used 3n2 C 3n C 1 7n2 and n3 .n C 1/3 > n6 to obtain the last inequality. We will have js sn j < 0:001 provided 7 < 0:001; 6n4 converges by the ratio test, since .nŠ/3 1 1 C 3 3.n C 1/3 3n sn j js 1 X n100 2n converges by the ratio test since p nŠ nD0 X .2n/Š The error satisfies , .n C 1/100 2nC1 n100 2n lim p p nŠ .n C 1/Š 100 nC1 1 p D 0: D lim 2 n nC1 23. 1 2 Downloaded by ted cage (sxnbyln180@questza.com) 357 lOMoARcPSD|6566483 www.konkur.in SECTION 9.3 (PAGE 524) n X ADAMS and ESSEX: CALCULUS 9 Z 1 2 dx D p . If n k x 3=2 n kD1 1 1 1 sn D sn C .AnC1 C An / D sn C p C p , then 2 n nC1 where sn D 1 and An D 3=2 31. kD1 An AnC1 sn j 2 2 1 2 p p D 2 n nC1 p p nC1 n 1 D p p D p p p p n nC1 n n C 1. n C n C 1/ 1 < 0:001 < 2n3=2 jsn sn D 0<s 1 If sn D sn C .AnC1 C An /, then 2 jsn where a D tan 1 nC1 2 and b D tan 1 n 2 . Now tan a tan b tan.a b/ D 1C tan atan b n nC1 2 2 D n C 1 n 1C 2 2 2 D 2 n C nC 4 2 , a b D tan 1 : n2 C n C 4 kD1 sn 1 1 1 C nC2 C nC3 C 2nC1 .n C 1/Š 2 .n C 2/Š 2 .n C 3/Š 1 1 1 D nC1 C 2 C 1C 2 .n C 1/Š 2.n C 2/ 2 .n C 2/.n C 3/ # " 2 1 1 1 C 1C C < nC1 2 .n C 1/Š 2.n C 2/ 2.n C 2/ n 32. b/ D 1 tan 1 4 2 n2 C n C 4 2 < tan 0:004 n2 C n C 4 2 , n C n > 2 cot.0:004/ 4 496: 1 2nC1 .n C 1/Š D nC2 < 0:001 2n .n C 1/Š.2n C 3/ Telegram: @uni_k 1 1 2.n C 2/ 1 1 1 1 if n D 4. Thus, s s4 D C 2 C 3 C 4 with 2 2 2Š 2 3Š 2 4Š error less than 0.001. 1 X 1 We have s D and .2k 1/Š kD1 sn D n X kD1 1 .2k 1/Š D 1 1 1 1 C C C C : 1Š 3Š 5Š .2n 1/Š Then 0<s 1 1 1 C C C .2n C 1/Š .2n C 3/Š .2n C 5/Š " 1 1 D C 1C .2n C 1/Š .2n C 2/.2n C 3/ sn D < 0:001 , 358 1 D We want error less than 0.001: 1 .a 4 1 1 1 1 1 D C 2 C 3 C C n : 2 2 2Š 2 3Š 2 nŠ 2k kŠ D 30. Again, we have sn C AnC1 s sn C An where P 1 sn D nkD1 2 and k C4 ˇ Z 1 ˇ1 1 dx 1 1 x ˇ 1 n D tan tan : An D D ˇ x2 C 4 2 2 ˇ 4 2 2 n n X Then if n 63. Thus, the error in the approximation s s63 is less than 0.001. An AnC1 sn j 2 " # 1 1 1 1 n 1 nC1 D tan C tan 2 4 2 2 4 2 2 " # n 1 nC1 1 tan 1 tan 1 D .a b/; D 4 2 2 4 n D 22 will do. The approximation s s22 has error less than 0.001. 1 X 1 and We have s D 2k kŠ 1 C .2n C 2/.2n C 3/.2n C 4/.2n C 5/ " 1 1 C 1C < .2n C 1/Š .2n C 2/.2n C 3/ # 1 C Œ.2n C 2/.2n C 3/2 2 3 6 1 6 .2n C 1/Š 4 7 7 5 1 1 .2n C 2/.2n C 3/ 4n2 C 10n C 6 1 < 0:001 D .2n C 1/Š 4n2 C 10n C 5 D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 # lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 9.3 (PAGE 524) 33. We have s D 0<s n 1 X1 2k X 2k and sn D . Thus .2k/Š .2k/Š kD0 kD0 X 1 D f .n/ converges by the integral test, and 2 1Cn nD1 nD1 its sum is less than =2. 36. Let u D ln ln t , du D Z 1 sn n nC1 nC2 2 2 2 D C C C .2n/Š .2n C 2/Š .2n C 4/Š 2 2n 1C D .2n/Š .2n C 1/.2n C 2/ 22 C C .2n C 1/.2n C 2/.2n C 3/.2n C 4/ # " 2 n 2 2 2 C < C 1C .2n/Š .2n C 1/.2n C 2/ .2n C 1/.2n C 2/ D 1 1 X 1 1 C D 1:175 with error if n D 3. Thus, s s3 D 1 C 3Š 5Š less than 0.001. 2n .2n/Š sn D kD1 kk 1 X nDN CASE I. Suppose < 1. Pick such that < < 1. Then there exists N such that .an /1=n for all n N . Therefore and aN N ; 1 1 1 1 D C 2 C 3 C C n: 1 2 3 n kk Thus Then 0<s 1 1 1 sn D C C C .n C 1/nC1 .n C 2/nC2 .n C 3/nC3 " # 1 1 1 1C < C C nC1 .n C 1/ nC1 .n C 1/2 2 3 D D 1 6 4 .n C 1/nC1 series nDN 1 X error less than 0.001. 1 . Then f is decreasing on Œ1; 1/. 35. Let f .x/ D 1 C x2 1 X Since f .n/ is a right Riemann sum for 38. 0 R!1 ˇR ˇ ˇ xˇ D ; ˇ 2 0 an also converges. nD1 lim n!1 p n an D lim n!1 Since this limit is less than 1, root test. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 X Let an D 2nC1 =nn . Then nD1 f .x/ dx D lim tan n , and 1 1 and bn D 2 . n n ln n Since lim n1=n D 1 (because lim D 0), we have n 1=n 1=n lim.an / D 1 and D 1. That is, D 1 for X lim.bn / X both series. But an diverges to infinity, while bn converges. Thus the case D 1 provides no information on the convergence or divergence of a series. 1 1 1 C 3 C 4 D 1:291 with 22 3 4 1 an converges by comparison with the geometric CASE III. Let an D 1 < 0:001 n.n C 1/n Z 1 aN C2 N C2 ; : : : : CASE II. Suppose > 1. Then .an /1=n 1, and an 1, for all Xsufficiently large values of n. Therefore lim an ¤ 0 and an must diverge. Since an > 0 it diverges to infinity. 7 1 5 nC1 1 1 X aN C1 N C1 ; nDN 1 if n D 4. Thus, s s4 D 1 C 1 n.ln n/.ln ln n/ .lnj n/.lnj C1 n/p 37. Let an > 0 for all n. (Let’s forget the “ultimately” part.) Let D lim.an /1=n . n X 1 kD1 du p ln ln a u converges if and only if p > 1, where N is large enough that lnj N > 1. if n D 4. Thus, s s4 with error less than 0.001. 34. We have s D Z 1 will converge if and only if p > 1. Thus, 1 X 1 will converge if and only if p > 1. n ln n.ln ln n/p nD3 Similarly, 1 1 X 1 dt D t ln t .ln ln t /p a 1 2 .2n C 1/.2n C 2/ 4n2 C 6n C 2 2n D < 0:001 .2n/Š 4n2 C 6n dt and ln ln a > 0; then t ln t Downloaded by ted cage (sxnbyln180@questza.com) 2 21=n D 0: n P1 nD1 an converges by the 359 lOMoARcPSD|6566483 www.konkur.in SECTION 9.3 (PAGE 524) 1 X 39. nD1 since n nC1 n2 " D lim n!1 40. Let an D ADAMS and ESSEX: CALCULUS 9 converges by the root test of Exercise 31 n nC1 n2 #1=n 43. b) Let sN D 1 1 D lim D < 1: n!1 e 1 n 1C n 2nC1 . Then nn X 22n .nŠ/2 .2n/Š , we obtain k.1 .k Since the terms exceed 1, the series diverges to infinity. 1 2 3 4 2n .2n/Š D 22n .nŠ/2 .2 4 6 8 2n/2 1 3 5 .2n 1/ D 2 4 6 .2n 2/ 2n 3 5 7 2n 1 1 1 D 1 > : 2 4 6 2n 2 2n 2n 360 Telegram: @uni_k 1/k 2 k 2k 2 D 0 1/ For given n, the p upper bound is minimal if n2 C 8 nC2 (for n 2). kD 2.n 1/ diverges to infinity by comparison 44. nD1 k.n C 1/ C 1 .n C 2/ ˙ an D 1 . 2n 1 X n .n C 2/k C 1 D 0 p .n C 2/2 4.n kD 2.n 1/ p n C 2 ˙ n2 C 8 D : 2.n 1/ .n with the harmonic series n k/.n C 1/.1 C k/n .1 C k/nC1 .1 k 2 .1 k/2 2 k /.n C 1/ .1 C k/.1 2k/ D 0 k 2 .n C 1/ 42. We have 1 X 1Ck 2 2 . 2n k.1 k/ nD0 Since the maximum value of k.1 k/ is 1=4 (at k D 1=2), the best upper bound we get for s by this method is s 8. 1 1 X 1Ck j 1 X j < c) s sn D 2j k 2 j DnC1 j DnC1 nC1 1 1Ck 1 D 1 Ck k 2 1 2 .1 C k/nC1 G.k/ D D n ; k.1 k/2n 2 .1 C k/nC1 where G.k/ D . For minimum G.k/, look k.1 k/ for a critical point: 22n .nŠ/2 Œ2n.2n 2/ 6 4 22 D .2n/Š 2n.2n 1/.2n 2/ 3 2 1 2n 2n 2 4 2 D > 1: 2n 1 2n 3 3 1 nD1 nD0 Therefore, s D Thus the ratio test provides no information. However, .2n/Š N 1 X k 1 C k N C1 1 2 sn < 1Ck k 1 2 ! 2 2 1 C k N C1 D : 1 k.1 k/ 2 k.1 k/ 22nC2 ..n C 1/Š/2 .2n/Š 4.n C 1/2 2n D lim D 1: 2 .2n C 2/Š 2 .nŠ/ .2n C 2/.2n C 1/ 22n .nŠ/2 2 N X 1 41. Trying to apply the ratio test to 1 X nD0 < n 1 1 r N C1 1 ; rn D k k 1 r nD0 where r D .1 C k/=n. Thus P Thus 1 nD1 an converges by the ratio test. (Remark: the question contained a typo. It was intended to ask that #33 be repeated, using the ratio test. That is a little harder.) Therefore N X n D anC1 nn 2nC2 nC1 D nC1 an .n C 1/ 2 1 2 2 D n D n C 1 n 1 n .n C 1/ 1C nC1 n 1 ! 0 D 0 as n ! 1: e D lim a) If n is a positive integer and k > 0, then 1 .1 C k/n 1 C nk > nk, so n < .1 C k/n . k If s D 1 X kD1 ck D 1 X kD1 1 , then we have k 2 .k C 1/ sn C AnC1 s sn C An Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2k/ D0 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL where sn D An D Z 1 n n X kD1 1 k 2 .k C 1/ dx D x 2 .x C 1/ SECTION 9.4 (PAGE 531) (b) Let Sn D and Z 1 n ˇ1 ˇ 1 ˇ C ln.x C 1/ˇ D ln x ˇ x n ˇ1 ˇ 1 1ˇ D ln 1 C ˇ x xˇ n 1 1 D ln 1 C : n n 1 1 1 C 2 C x x xC1 dx 0< 1 X bi iD1 nD1 1 X nD1 1 X 1 1 D 2n C 1 2n nD1 1 1. (a) We have 2. 1 X Since 210 D 1;024, s10 will approximate s to within 0:001. 5 X 1 X bn nD1 bn nD1 X . 1/n p converges by the alternating series test (since n the terms alternate in sign, decrease in size, and approach 0). However, the convergence is only conditional, since X 1 p diverges to infinity. n 1 X nD1 . 1/n n2 C ln n converges absolutely since ˇ ˇ 1 X ˇ . 1/n ˇ 1 ˇ 1 and ˇ converges. ˇ n2 C ln n ˇ n2 n2 nD1 3. X . 1/n cos.n/ D converges .n C 1/ ln.n C 1/ .n C 1/ ln.n C 1/ by the alternating series test, but only conditionally since X 1 diverges to infinity (by the integral .n C 1/ ln.n C 1/ test). X Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1C Section 9.4 Absolute and Conditional Convergence (page 531) with error less than 0.001. P n 45. s D 1 nD1 1=.2 C 1/. 1 1 1 1 sn D D nC1 C nC2 C nC3 C 2i C 1 2 2 2 iD1 1 1 1 D nC1 1 C C 2 C 2 2 2 1 1 D n < if 2n > 1;000: 2 1;000 1 1 1 1 1 1 D1 2 3 4 5 8 9 1 1 1 1 16 17 32 33 0:765 with error less than 0.001: D 1:6450 0<s 1 provided 4n > 1;000=3. Thus n D 5 will P do (but n D 4 is insufficient). S5 approximates 1 nD1 bn to within 0:001. P n (c) Since 1 nD1 1=2 D 1, we have An 1 D 1 C s8 D 1 C s8 C .A9 C A8 / 2 1 1 1 1 C 2 C 2 C C 2 C D1C 2 2 .3/ 3 .4/ 8 .9/ " # 1 1 10 1 9 ln C ln 2 9 9 8 8 2n C 1 2n 1 < n; 2n .2n C 1/ 4 1 1 C C 4nC1 4 42 1 1 4 < D f rac14nC1 D 3 3 4n 1;000 if n D 8. Thus, n2 1 . Since 2n C 1 Sn D bnC1 C bnC2 C bnC3 C < 1 D < 0:001 2n.n C 1/2 1 X 1 1 2n we have AnC1 2 " # 1 1 1 1 1 D ln 1 C C ln 1 C 2 n n nC1 nC1 # " n2 C 2n 1 1 C ln D 2 2 n.n C 1/ n C 2n C 1 " # n2 C 2n 1 1 C 1 2 n.n C 1/ n2 C 2n C 1 sn j iD1 bi , where bn D 0 < bn D 1 If sn D sn C .AnC1 C An /, then 2 jsn Pn Downloaded by ted cage (sxnbyln180@questza.com) 361 lOMoARcPSD|6566483 www.konkur.in SECTION 9.4 (PAGE 531) 4. 5. 6. ADAMS and ESSEX: CALCULUS 9 1 1 X X 1 . 1/2n D is a positive, convergent geometric n 2 2n nD1 nD1 series so must converge absolutely. 12. X . 1/n .n2 1/ diverges since its terms do not apn2 C 1 proach zero. 1 X . 2/n converges absolutely by the ratio test since nŠ nD1 13. ˇ ˇ ˇ . 2/nC1 nŠ ˇˇ 1 lim ˇˇ D 0: D 2 lim .n C 1/Š . 2/n ˇ nC1 7. X . 1/n converges absolutely, since, for n 1, n n and 8. 1 X X 1 n 10. X 20n2 n 1 converges by the alternating series n3 C n2 C 33 test (the terms are ultimately decreasing in size, and approach zero), but the convergence is only conditional since X 20n2 n 1 diverges to infinity by comparison with n3 C n2 C 33 X1 . n 1 X 100. 1/n converges by the alter2n C 3 2n C 3 nD1 nD1 nating series test but only conditionally since 1 X 100 cos.n/ 1 X and nD1 11. X ˇ ˇ ˇ 100. 1/n ˇ 100 ˇ ˇ ˇ 2n C 3 ˇ D 2n C 3 100 diverges to infinity. 2n C 3 n . 1/k 1 kD1 X k k , and sn D , . 1/k 1 2 2 k C1 k C1 kD1 14. 15. sn j < . 1/n are alter.2n/Š nating in sign and decreasing in size, the size of the error in the approximation s sn does not exceed that of the first omitted term: 1 js sn j < 0:001 .2n C 2/Š 1 1 1 if n D 3. Hence s 1 C ; four terms will 2Š 4Š 6Š approximate s with error less than 0.001 in absolute value. P1 Since the terms of the series s D If s D 1 X nD0 n X k k . 1/k 1 k , then . 1/k 1 k , and sn D 2 2 kD1 kD1 nC1 js sn j < nC1 < 0:001 2 if n D 13, because the series satisfies the conditions of the alternating series test from the second term on. 16. 17. 3n are nŠ alternating in sign and ultimately decreasing in size (they decrease after the third term), the size of the error in the approximation s sn does not exceed that of the first 3nC1 omitted term (provided n 3): js sn j < 0:001 .n C 1/Š if n D 12. Thus twelve terms will suffice to approximate s with error less than 0.001 in absolute value. X xn Applying the ratio test to p , we obtain nC1 ˇ ˇ r p ˇ x nC1 n C 1 ˇˇ nC1 ˇ D lim ˇ p D jxj: ˇ D jxj lim ˇ nC2 xn ˇ nC2 Since the terms of the series s D P1 nD0 . 1/n Hence the series converges absolutely if jxj < 1, that is, if 1 < x < 1. The series converges conditionally for x D 1, but diverges for all other values of x. .x 2/n . Apply the ratio test n2 22n ˇ ˇ ˇ .x 2/nC1 jx 2j n2 22n ˇˇ D lim ˇˇ D <1 .n C 1/2 22nC2 .x 2/n ˇ 4 18. Let an D nŠ nŠ diverges since lim D 1. n . 100/ 100n 362 Telegram: @uni_k D 1 X D nC1 < 0:001 .n C 1/2 C 1 if n D 999, because the series satisfies the conditions of the alternating series test. is a convergent geometric series. . 1/n If s D js nD0 9. 2 then ˇ ˇ ˇ . 1/n ˇ 1 ˇ ˇ ˇ n n ˇ n ; n diverges to 1 since all terms are negative n2 C 1 nD0 1 X n diverges to infinity by comparison with and n2 C 1 nD0 1 X 1 . n 1 X . 1/n converges by the alterln ln n ln ln n nD10 nD10 1 X 1 nating series test but only conditionally since ln ln n nD10 1 X 1 . diverges to infinity by comparison with n nD10 (ln ln n < n for n 10.) 1 X sin.n C 1 / Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 9.4 (PAGE 531) if and only if jx 2j < 4, that is 2 < x < 6. If x D 2, 1 1 X X . 1/n an D , which converges absolutely. If then n2 nD1 nD1 1 1 X X 1 x D 6, then an D , which also converges abson2 nD1 nD1 lutely. Thus, the series converges absolutely if 2 x 6 and diverges elsewhere. 19. Apply the ratio test to X . 1/n .x 1/n : 2n C 3 23. ˇ ˇ ˇ .x 1/nC1 2n C 3 ˇ ˇ D jx ˇ D lim ˇ 2n C 5 .x 1/n ˇ 1j: The series converges absolutely if jx 1j < 1, that is, if 0 < x < 2, and converges conditionally if x D 2. It diverges for all other values of x. 1 3x C 2 n 20. Let an D . Apply the ratio test 2n 1 5 ˇ ˇ ˇ 1 3x C 2 nC1 2n 1 3x C 2 n ˇ ˇ ˇ D lim ˇ ˇ ˇ ˇ 2n C 1 5 1 5 ˇ ˇ ˇ 3x C 2 ˇ ˇ<1 D ˇˇ 5 ˇ 24. Apply the ratio test to X xn : 2n ln n 25. ˇ ˇ ˇ x nC1 2n ln n ˇˇ ln n jxj jxj D lim ˇˇ nC1 lim D : D ˇ n 2 ln.n C 1/ x 2 ln.n C 1/ 2 (The last limit can be evaluated by l’H^opital’s Rule.) The given series converges absolutely if jxj < 2, that is, if 2 < x < 2. By the alternating series test, it converges conditionally if x D 2. It diverges for all other values of x. 22. Let an D .4x C 1/n . Apply the ratio test n3 ˇ ˇ ˇ 3 ˇˇ ˇ The series converges absolutely if ˇx C ˇ < 2, that is, if 2 1 7 < x < . By the alternating series test it converges 2 2 7 conditionally at x D . It diverges elsewhere. 2 n 1 1 . Apply the ratio test 1C Let an D n x ˇ ˇ ˇ ˇ ˇ 1 1 nC1 n 1 n ˇˇ ˇˇ 1 ˇˇ ˇ D lim ˇ 1C 1C ˇ D ˇ1C ˇ < 1 ˇ ˇn C 1 x 1 x x ˇ ˇ ˇ 2ˇ 5 7 if and only if ˇˇx C ˇˇ < , that is < x < 1. If 3 3 3 1 1 X X 7 1 , then , which diverges. an D x D 3 2n 1 nD1 nD1 1 1 X X . 1/n , which converges If x D 1, then an D 2n 1 nD1 nD1 conditionally. Thus, the series converges absolutely if 7 < x < 1, converges conditionally if x D 1 and 3 diverges elsewhere. 21. 1 1 < x < 0. If x D , then 2 2 1 1 n X X . 1/ an D , which converges absolutely. n3 nD1 nD1 1 1 X X 1 , which also conIf x D 0, then an D n3 nD1 nD1 verges absolutely. Thus, the series converges absolutely 1 if x 0 and diverges elsewhere. 2 X .2x C 3/n Apply the ratio test to : n1=3 4n ˇ ˇ ˇ ˇ ˇ .2x C 3/nC1 ˇx C 3 ˇ n1=3 4n ˇˇ j2x C 3j ˇ 2 D lim ˇ D : ˇD ˇ .n C 1/1=3 4nC1 .2x C 3/n ˇ 4 2 if and only if 26. if and only if jx C 1j < jxj, that is, 1 1 1 2 < < 0 ) x < . If x D , then x 2 2 1 1 n X X . 1/ an D , which converges conditionally. n nD1 nD1 1 Thus, the series converges absolutely if x < , con2 1 and diverges elsewhere. It verges conditionally if x D 2 is undefined at x D 0. 1 X 1 1 1 sin.n=2/ D1C0 C0C C0 C 0 C n 3 5 7 nD1 The alternating series test does not apply directly, but does apply to the modified series with the zero terms deleted. Since this latter series converges conditionally, the given series also converges conditionally. If Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 X 10 for every n 1. Hence, an converges 2 n nD1 1 X 10 absolutely by comparison with . n2 then jan j ˇ ˇ ˇ .4x C 1/nC1 ˇ n3 ˇ D j4x C 1j < 1 D lim ˇˇ .n C 1/3 .4x C 1/n ˇ 8̂ 10 < ; if n is even; 2 n an D 1 :̂ ; if n is odd; 10n3 Downloaded by ted cage (sxnbyln180@questza.com) nD1 363 lOMoARcPSD|6566483 www.konkur.in SECTION 9.4 (PAGE 531) 27. ADAMS and ESSEX: CALCULUS 9 P P a) “ an converges implies . 1/n an converges” is . 1/n is a counterexample. FALSE. an D n P P b) “P an converges and . 1/n an converges implies an converges absolutely” is FALSE. The series of Exercise 25 is a counterexample. P P c) “ an converges absolutely implies . 1/n an converges absolutely” is TRUE, because j. 1/n an j D jan j. 29. obtain D lim jxj an D 1 n C 1: y .2n/Š 22n .nŠ/2 ln an D ln 1 3 4 n 1 n x D Fig. 9.4-28 b) Let an D nŠx n . Apply the ratio test nn ˇ ˇ ˇ .n C 1/Šx nC1 nn ˇˇ D lim ˇˇ .n C 1/nC1 nŠx n ˇ jxj jxj D lim <1 D e 1 n 1C n if and only if 1 X an converges absolutely if nD1 and diverges elsewhere. 364 Telegram: @uni_k X an x n , we .2n C 2/.2n C 1/ D jxj: 4.n C 1/2 D 1 : 2n e < x < e 1 2 C ln 1 1 4 C C ln 1 1 1 1 24 2n 1 1 1 1C C C ! 2 2 n 1 2n 1 as n ! 1: Thus lim an D 0, and the given series converges conditionally at x D 1 by the alternating series test. 30. P 1 1 and qn D . Then pn diverges 2n 1 2n P to 1 and qn diverges to 1. Also, the alternating harmonic series is the sum of all the pn s and qn s in a specific order: Let pn D 1 1 X X . 1/n 1 D .pn C qn /: n e < x < e. If x D ˙e, then, by (a), ˇ nˇ ˇ nŠe ˇ ln ˇˇ n ˇˇ D ln.nŠ/ C ln e n ln nn n > .n ln n n C 1/ C n n ln n D 1: ˇ nˇ ˇ nŠe ˇ ) ˇˇ n ˇˇ > e: n Hence, D It is evident that an decreases as n increases. To see whether lim an D 0, take logarithms and use the inequality ln.1 C x/ x: yDln x 2 22n .nŠ/2 1 2 3 4 2n .2 4 6 8 2n/2 1 3 5 .2n 1/ D 2 4 6 .2n 2/ 2n 2n 3 2n 1 1 3 D 2 4 2n 2 2n 1 1 1 1 1 1 D 1 2 4 2n 2 ln.nŠ/ D ln 1 C ln 2 C ln 3 C C ln n D sum of area of the shaded rectangles ˇn Z n ˇ ˇ ln t dt D .t ln t t /ˇ > ˇ 1 1 X .2n/Šx n P Thus an x n converges absolutely if 1 < x < 1, and diverges if x > 1 or x < 1. In Exercise 36 of Section 9.3 1 , so the given series definitely it was shown that an 2n diverges at x D 1 and may at most converge conditionally at x D 1. To see whether it does converge at 1, we write, as in Exercise 36 of Section 9.3, 28. a) We have D n ln n Applying the ratio test to nD1 nD1 a) Rearrange the terms as follows: first add terms of P pn until the sumP exceeds 2. Then add q1 . Then add more terms of pn until the sum exceeds 3. Then add q2 . Continue P in this way; at the nth stage, add new terms from pn until the sum exceeds n C 1, and then add qn . All partial sums after the nth stage exceed n, so the rearranged series diverges to infinity. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 9.5 (PAGE 541) b) Rearrange the terms of the original alternating P harmonic series as follows: first add terms of qn until the sum is less than 2. Then add p1 . The sum will now be greater than P 2. (Why?) Then resume adding new terms from qn until the sum is less than 2 again, and add p2 , which will raise the sum above 2 again. Continue in this way. After the nth stage, all succeeding partial sums will differ from 2 by less than 1=n, so the rearranged series will converge to 2. 1 X . 1/n , which converges. n4 nD1 1 X 1 At x D 4, the series is , which also converges. n4 nD1 Hence, the interval of convergence is Œ 4; 4. At x D 4, the series is 5. 1 X n3 .2x 3/n D nD0 Section 9.5 Power Series 1. 1 X 2. We have 1 X nD0 xD 6. 2nC1 .n C 1/3 Here D 1 X en .4 x/n . The centre of convergence is n3 x D 4. The radius of convergence is We have R D lim 1. The radius of convergence is 3n D 1: 3.n C 1/ 1 e n .n C 1/3 D : n3 e nC1 e 1 X . 1/n 1 , which converges. , the series is e n3 nD1 1 X 1 1 , which also converges. , the series is At x D 4 e n3 nD1 # " 1 1 ;4 C . Hence, the interval of convergence is 4 e e At x D 4 C The series converges absolutely on . 2; 0/ and diverges on . 1; 2/ and .0; 1/. At x D 2, the series is 1 X 3n. 1/n , which diverges. At x D 0, the series is nD0 1 X nD0 nD1 3n.x C 1/n . The centre of convergence is R D lim 3 n 2 . 2n n3 x 1 . The radius of convergence 2 is 1=2; the centre of convergence is 3=2; the interval of convergence is .1; 2/. R D lim (page 541) ˇp ˇ ˇ n C 2ˇ x 2n ˇ ˇ For p we have R D lim ˇ p ˇ D 1. The ˇ n C 1ˇ n C 1 nD0 radius of convergence is 1; the centre of convergence is 0; the interval of convergence is . 1; 1/. (The series does not converge at x D 1 or x D 1.) 2n n3 1 X 3n, which diverges to infinity. Hence, the interval of nD0 convergence is . 2; 0/. 1 X n For 8. We have 2nC1 .n C 1/ 1 xC2 we have R D lim D 2. n 2 2n n nD1 The radius of convergence is 2; the centre of convergence is 2. For x D 4 the series is an alternating harmonic series, so converges. For x D 0, the series is a divergent harmonic series. Therefore the interval of convergence is Œ 4; 0/. 3. For 1 X . 1/n n x . The centre of convergence is n4 22n nD1 x D 0. The radius of convergence is 4. We have ˇ ˇ ˇ . 1/n .n C 1/4 22nC2 ˇ ˇ R D lim ˇˇ 4 2n ˇ n 2 . 1/nC1 ˇ ˇ ˇ nC1 4 ˇ ˇ ˇ 4ˇ D 4: D limˇ ˇ ˇ n 1 C 5n n x we nŠ 1 C 5n .n C 1/Š have R D lim D 1. The radius of nŠ 1 C 5nC1 convergence is infinite; the centre of convergence is 0; the interval of convergence is the whole real line . 1; 1/. P1 7. nD0 1 X .4x nD1 nn D 1 n X 4 n x 1 4 n . The cen- nD1 tre of convergence is x D 41 . The radius of convergence is 4n .n C 1/nC1 nn 4nC1 1 nC1 n .n C 1/ D 1: D lim 4 n R D lim Hence, the interval of convergence is . 1; 1/. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1/n Downloaded by ted cage (sxnbyln180@questza.com) 365 lOMoARcPSD|6566483 www.konkur.in SECTION 9.5 (PAGE 541) ADAMS and ESSEX: CALCULUS 9 9. By Example 5(a), C 3x 2 C 4x 3 x2 C C 3x 2 C 2x 2 x2 C 2x 1 C x C 1 C 2x x 1 1 C 3x Thus C 6x 2 Thus 1 .1 for D x/3 1 C D .1 x3 C D 1 C C C 4x 3 3x 3 2x 3 x3 C C C C C 10x 3 C D 1 X .n C 1/.n C 2/ 2 nD0 x/2 1 12. x 13. 1 .1 x/3 1 x/2 .1 1 xn; .2 1 1 x D 1 X xn 14. 1 X 1 D . 2x/n 1 C 2x nD0 nD0 xCx 2 1 X 1 D . 1/n x n x C D 1Cx 3 15. nD0 0 holds for 1 < x < 1. Since an D 1 and bn D . 1/n for n D 0; 1; 2; : : :, we have Cn D n X n j . 1/ j D0 4 1 C x C x C D 0; if n is odd; D 1; if n is even. 16. nD0 x 2n D 1 1 366 2x C x Let y D x nD0 . 1 < x < 12 /: 2 0 x nC1 2nC1 .n C 1/ 1 X xn x/ D ln 2 nD1 2n n . 2 x < 2/: : 1. Then x D 1 C y and 1 . 1 < y < 1/ nD0 D 1 h X 17. .x nD0 D1 1 C 2x C 3x 2 C 4x 3 C 1 1 2x C x 2 2x x2 2x 4x 2 C 2x 3 C 3x 2 2x 3 C 3x 2 6x 3 C 4x 3 C nD0 1 X X 1 1 D D . y/n x 1Cy 1 1 D x 1Cx 1 x2 1 < x < 1. 2 x/ C ln 2 D ln.2 By long division: 1 Telegram: @uni_k 1 X ln.2 Then the Cauchy product is 2x C 22 x 2 23 x 3 C Z xX 1 dt tn D dt nC1 t 0 2 0 nD0 2 ˇx ˇx 1 ˇ ˇ X t nC1 ˇ ˇ ln.2 t /ˇ D ˇ nC1 ˇ 2 .n C 1/ ˇ D1 Z x and 11. 1 2x 3x 2 D C C C x/2 22 23 24 1 X .n C 1/x n ; . 2 < x < 2/: D 2nC2 nD0 1 C x C x2 C x3 C D for 1 < x < 1. nD0 D 10. We have 2 .n C 1/x n , for 1 1 X x n 1 1 D x 2 x 2 2 2 nD0 1 2 x x2 x3 1 . 2 < x < 2/: D C 2 C 3 C 4 C 2 2 2 2 1 1 1 D 2 x 2 1 x 2 1 x x2 x3 D C 2 C 3 C 4 2 2 2 2 for 2 < x < 2. Now differentiate to get 1 1 < x < 1. 1 1 X D in 1/ .x 1/ C .x 1/2 .for 0 < x < 2/: Let x C 2 D t , so x D t 1/3 C .x .x 1/4 2. Then 1 X .n C 1/t n 1 1 D D 2 2 x .2 t / 2nC2 nD0 D 1 X .n C 1/.x C 2/n nD0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2nC2 ; . 4 < x < 0/: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 18. SECTION 9.5 (PAGE 541) 2 1 x D 1 1Cx 1Cx D 2.1 x C x 2 x 3 C / 1 1 X D1C2 . x/n . 1 < x < 1/: for 1 < x < 1. Therefore, 1 X .n C 3/x n D nD0 .1 3 D .1 19. We have 1 X x3 3 .2x 2 /n D x 1 2x 2 nD0 D 20. Let y D x 1 X nD0 23. 1 1 p <x< p : 2 2 2n x 2nC3 ; 4. Then x D 4 C y and 1 1 1 D D x 4Cy 4 1X D 4 D 42 1 nD0 4/ 4 4/ y 4 n 1C .x .x 1 4 1 C D 1 X y n 4 4 1 nD0 24. 4/3 .x 43 Z x dt D t 1 Z x" 1 D ln 4 C 4 4 Z 4 1 dt C t .t 4/ 42 Z x dt t C 43 4 .t 4/2 x 4/3 .t 44 4 x/2 2x x/2 C xn nD0 3 1 x . 1 < x < 1/: x3 C x4 1 1Cx D # x3 C dt D 1 : .1 C x/2 4x 6 C D x3 : .1 C x/2 3x 3 C 4x 3 Now we multiply by .x 4/3 .x 4/2 C 2 4 24 3 43 .for 0 < x 8/: D ln 4 C x C x2 1 C 2x for 0 < x < 8. Therefore, ln x D x 1 X and differentiate to get C 44 nx n C 3 1 x x2 x3 C C C C 3 4 5 6 1 x3 x4 x5 D 3 C C C x 3 4 5 x2 x3 x4 x2 1 C C C C x D 3 xC x 2 3 4 2 1 x2 D 3 ln.1 x/ x x 2 1 1 1 D ln.1 x/ : . 1 x < 1; x ¤ 0/: x3 x2 2x We start with 1 4/2 .x nD0 D nD1 ! 1 X x3: 2x 4 C 3x 5 Differentiating again we get .x 4/4 C 4 44 3x 2 2 4x 3 C 3 5x 4 4 6x 5 C D x 3 C 3x 2 : .1 C x/3 Finally, we remove the factor x 2 : 21. 1 D 4x C 16x 2 64x 3 C 3 D1 C . 4x/ C . 4x/2 . 4x/3 C 1 1 1 D < x < 14 : D ; 4 1 . 4x/ 1 C 4x All steps are valid for 25. 22. We differentiate the series 1 X nD0 xn D 1 C x C x2 C x3 C D nD0 xC3 : .1 C x/3 1 < x < 1. 1 x2 2x C 4x 3 C 6x 5 C 8x 7 C D x , for 1 < x < 1, .1 2x ; x 2 /2 .1 2 ; x 2 /2 or, on division by x, nx n D x C 2x 2 C 3x 3 C D 2 C 4x 2 C 6x 4 C 8x 6 C D x .1 x/2 for 1 < x < 1. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 4 6x 3 C D Since 1 C x 2 C x 4 C x 6 C D 1 we obtain by differentiation 1 1 and multiply by x to get 1 X 2 4x C 3 5x 2 Downloaded by ted cage (sxnbyln180@questza.com) 367 lOMoARcPSD|6566483 www.konkur.in SECTION 9.5 (PAGE 541) x2 x3 C 2 3 26. Since x therefore for 6 Now let x D 1 < x 1, 1 X x C D ln.1 C x 2 / 4 ( ln.1 C x 2 / x2 1 Finally, multiply by nD1 1 X if 1 x 1, x ¤ 0 if x D 0. From Example 5(a), 1 X Since 1 X . 1/n 1 nD1 x/2 ; . 1 < x < 1/: n 1 9 D D : 1 2 3n 1 4 1 3 nD1 nD1 32. 1 9 3 D . 3 4 4 D 3n nD0 2n 29. From Example 7, nD1 n2 x n 1 D 1 X . 1/k 1 . 1/ n D n2 kC1 nD1 1Cx for .1 x/3 kD0 n nD0 D nD1 1 X 1 < x < 1. nD3 1 X 1 C 1 2 . C 1/ k2 D D : 1 3 k 1 . 1/3 .1 / 1 X nx nD1 n 1 1. D 1 .1 x/2 ; nD2 1 X .n C 1/nx n 1 D nD1 368 Telegram: @uni_k 1/x n 2 D 2 x/3 .1 2 .1 x/3 e 3xC1 Dee 3x ; . 1 < x < 1/ ; . 1 < x < 1/: De D 2. n.n 1 D ln 2 n2n . 1 < x < 1/: Differentiate with respect to x and then replace n by nC1: 1 X 1 2 kC1 D ln 1 1 2 1 2 1 D ln 2 8 5 : 8 Section 9.6 Taylor and Maclaurin Series (page 550) kD1 30. From Example 5(a), 1 X 1 D ln 2 n2n Putting x D 1=, we get 1 X .n C 1/2 1 < x 1, Therefore k 1 1 X 1 1 D D k 2 D 4: 2 1 12 kD1 1 X xn D ln.1 C x/ for n In the series for ln.1 C x/ in Example 5(c), put x D to get 1 X 28. From Example 5(a) with x D 1=2, 1 X nC1 8 : 27 nD1 1 X 1 X n n.n C 1/ D 2n 1 X 1 3 . 1/n 1 D ln 1 C D ln : n2n 2 2 Putting x D 1=3, we get Thus . 1/n therefore 1 .1 n.n C 1/ 16 D : 2n 1 27 1=2: nD1 31. nx n 1 D . 1/n 1 nD1 1 x 1, and, dividing by x 2 , x6 C D 4 1=2: 8 x x C 2 3 x2 x4 1 C 2 3 27. x4 C D ln.1 C x/ for 4 4 x2 ADAMS and ESSEX: CALCULUS 9 1 X .3x/n nŠ nD0 1 n n X nD0 e3 x nŠ ! .for all x/: .2x 3 /4 .2x 3 /6 .2x 3 /2 C C 2Š 4Š 6Š 2 6 4 12 6 18 2 x 2 x 2 x D1 C C 2Š 4Š 6Š 1 X . 1/n 4n D x 6n .for all x/: .2n/Š cos.2x 3 / D 1 nD0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1=2 D ln 2: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 3. 4. SECTION 9.6 (PAGE 550) sin x cos x sin D sin x cos 4 4 4 1 1 2nC1 1 X 1 X x 2n n x p D p . 1/ . 1/n .2n C 1/Š .2n/Š 2 nD0 2 nD0 1 2nC1 2n x x 1 X C .for all x/: D p . 1/n .2n/Š .2n C 1/Š 2 nD0 / D cos.2x 10. 22 x 2 24 x 4 26 x 6 C 2Š 4Š 6Š 1 X . 1/n .2x/2n .2n/Š 1C D 11. .for all x/: 6. cos2 x 2 . 1/n x 2nC3 nD0 32nC1 .2n C 1/Š D1C 7. 1Cx D ln.1 C x/ ln.1 x/ 1 x 1 1 X xn xn X . 1/n 1 D n n D2 .for all x/: 1 D .1 C cos x/ 2 x4 x2 1 C 1C1 D 2 2Š 4Š ln nD1 1 X nD0 x D 3 x6 C 6Š 1 1 X . 1/n 2n x 2 .2n/Š ! 22 x 2 23 x 4 24 x 6 C C C 2Š 3Š 4Š 1 X 2nC1 2n D x .for all x 6D 0/: .n C 1/Š nD0 13. cosh x cos x D D 9. nD0 1 X nD0 14. sinh x nD0 i x 2n 2 6 .2n/Š i x 2nC1 .2n C 1/Š nD0 2 6 x x x 10 C C C D2 2Š 6Š 10Š 1 X x 4nC3 D2 .for all x/: .4n C 3/Š sin x D 1 h X 1 . 1/n nD0 15. 1 C x3 D .1 C x 3 / 1 x 2 C x 4 x 6 C 2 1Cx D 1 x2 C x3 C x4 x5 x6 C x7 C x8 1 X D 1 x2 C . 1/n x 2n 1 C x 2n .jxj < 1/: Let t D x C 1, so x D t 1. We have f .x/ D e 2x D e 2.t 1/ 1 X . 2/n t n D e2 nŠ nD2 D e2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k . 1/n nD0 .5x 2 /7 C 7 1 1 p x p : 5 5 1 x x x 10 C C C 2Š 6Š 10Š 1 4nC2 X x D2 .for all x/: .4n C 2/Š . 1/n 52nC1 4nC2 x .2n C 1/ for 1 h X D2 .for all x/: .5x 2 /3 .5x 2 /5 C 3 5 1 X . 1/n 2 2nC1 D .5x / .2n C 1/ 1 ! D2C .for all x/: tan 1 .5x 2 / D .5x 2 / . 1 < x < 1/: e 2x 1 1 2x 2 e 1 D x2 x2 .2x 2 /2 .2x 2 /3 1 C C D 2 1 C 2x 2 C x 2Š 3Š nD1 1 sin x cos x D sin.2x/ 2 1 X 22n x 2nC1 D . 1/n .2n C 1/Š nD1 nD1 x 2n 1 2n 1 2 12. nD0 8. 1 X . 1/n 1 x 2n n n 2 nD1 p p 2 x 2/: .for nD0 1 X . 1/nC1 n 2n 4 .x/ D .2n/Š 5. x 2 sin # D ln 2 C cos.2x/ D 1 X x2 ln.2 C x 2 / D ln 2 1 C 2 x2 D ln 2 C ln 1 C 2 " 2 3 2 1 x2 1 x2 x C D ln 2 C 2 2 2 3 2 Downloaded by ted cage (sxnbyln180@questza.com) nD0 1 X nD0 . 1/n 2n .x C 1/n nŠ .for all x/: 369 lOMoARcPSD|6566483 www.konkur.in SECTION 9.6 (PAGE 550) 16. Let y D x ADAMS and ESSEX: CALCULUS 9 ; then x D y C . Hence, 2 2 21. D cos y sin x D sin y C 2 y4 y2 C .for all y/ D1 2Š 4Š 2 1 1 4 D1 C x x 2Š 2 4Š 2 1 X . 1/n 2n D .for all x/: x .2n/Š 2 Let t D x .=4/, so x D t C .=4/. Then f .x/ D sin x cos x D sin t C cos t C 4 4 i 1 h D p .sin t C cos t / .cos t sin t / 2 1 p p X t 2nC1 D 2 sin t D 2 . 1/n .2n C 1/Š nD0 1 p X D 2 nD0 nD0 17. Let t D x , so x D t C . Then f .x/ D cos x D cos.t C / D cos t D 1 X . 1/nC1 .x D .2n/Š 1 X . 1/n nD0 /2n t 2n .2n/Š 22. 3; then x D y C 3. Hence, nD1 x 2 ln.2 C x/ D lnŒ4 C .x 2/ D ln 4 1 C 4 x 2 D ln 4 C ln 1 C 4 1 X .x 2/n D ln 4 C . 1/n 1 . 2 < x 6/: n4n nD1 20. Let t D x C 1. Then x D t 1, and e 2xC3 D e 2tC1 D e e 2t 1 X 2n t n (for all t ) De nŠ 2 nD0 D 1 X e2n .x C 1/n nD0 nŠ (for all x). 23. 370 Telegram: @uni_k .for all x/: ; then x D y C . Thus, 8 8 Let y D x 2 y ln x D ln.y C 3/ D ln 3 C ln 1 C 3 1 y 2 1 y 3 1 y 4 y C C D ln 3 C 3 2 3 3 3 4 3 2 3 .x 3/ .x 3/ .x 3/ .x 3/4 D ln 3 C C C 2 3 3 23 33 4 34 1 X . 1/n 1 .x 3/n .0 < x 6/: D ln 3 C n 3n 19. 2nC1 4 cos x D cos y C 8 " # 1 1 C cos 2y C D 2 4 " # 1 1 1 1 C p cos.2y/ p sin.2y/ D 2 2 2 # " 2 1 .2y/4 .2y/ 1 C D C p 1 2 2Š 4Š 2 2 # " .2y/5 .2y/3 1 C p 2y 3Š 5Š 2 2 " 1 .2y/3 .2y/2 1 C D C p 1 2y 2 2Š 3Š 2 2 # 5 4 .2y/ .2y/ C 4Š 5Š " 1 1 22 2 D C p 1 2 x x 2 8 2Š 8 2 2 # 3 4 4 3 2 25 5 2 C x x x C 3Š 8 4Š 8 5Š 8 " 1 1 X 22n 1 1 2n 1 1 . 1/n x D C p C p 2 .2n 1/Š 8 2 2 2 2 nD1 2n # 2n 2 C .for all x/: x .2n/Š 8 .for all x/: nD0 18. Let y D x . 1/n x .2n C 1/Š Let t D x C 2, so x D t Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 2. We have lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL f .x/ D 1 1 D D x2 .t 2/2 D 1 4 1 X 4 1 1 t 2 n 1 nD1 1 X t n n 1 2 SECTION 9.6 (PAGE 550) In the first sum replace n by n 2 xe x D . 2 t < 2/ n.x C 2/ 2n 1 D 1 X .n C 1/.x C 2/n D 4 2n D D 1 4 nD1 1 n 1 . 4 < x < 0/: 1 X e 2 un .n 1/Š 2 C e2 1 X nD1 2 C e2 nD0 24. Let y D x nD1 1 X nD1 1 e2 1 e2 1. 1 X 2e 2 un nŠ nD0 1 .n 1/Š 1 .n 1/Š 27. x4 x2 C 2 24 cos x D 1 . 1 C 1 x2 x4 C 2 24 25. Let u D x 1 2 x ln x D .1 C u/ ln.1 C u/ 1 X un D .1 C u/ . 1/n 1 n . 1 < u 1/ nD1 1 X . 1/n 1 nD1 Replace n by n 1 X 1 X un unC1 C : . 1/n 1 n n C 24 x4 C 4 4 5x 24 x2 5x 4 C C . 2 24 1 in the last sum. 28. 1 . 1/n 1 1 X . 1/n 1/ C .x n.n 1/ n 26. Let u D x C 2. Then x D u x xe D .u D .u 2/e 2, and sec x D 1 C 2/e 2 1 X un nD0 nŠ nŠ Now we can differentiate and obtain sec x tan x D x C (for all u) 1 X 2e 2 un nD0 x6 C 720 x2 5x 4 61x 6 C C C : 2 24 720 u 2 1 X e 2 unC1 nD0 x4 x2 C 2 24 into 1 we obtain .0 x 2/: 1/ nD2 If we divide the first four terms of the series cos x D 1 nD2 D Thus sec x D 1 C nD1 X un un C . 1/n 2 n n 1 nD1 nD2 1 X 1 1 un DuC . 1/n 1 n n 1 D .x x4 x2 C 2 24 x4 x2 2 x2 1. Then x D 1 C u, and D 5x 4 x2 C C 2 24 1 nD1 nŠ : 61x 5 5x 3 C C : 6 120 (Note: the same result can be obtained by multiplying the first three nonzero terms of the series for sec x (from Exercise 25) and tan x (from Example 6(b)).) Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k (for all x). 1; then x D y C 1. Thus, x 1Cy 1 D D1 y 1Cx 2Cy 2 1C 2 1 y y 2 y 3 D1 C 1 C 2 2 2 2 1 y3 y4 y y2 D C C . 1 < y < 1/ 1C 2 2 22 23 24 1 1 1 1 .x 1/2 C 4 .x 1/3 D C 2 .x 1/ 2 2 23 2 1 1 X . 1/n 1 D C .x 1/n .for 0 < x < 2/: 2 2nC1 x ln x D 2 un nŠ 2 .x C 2/n nŠ Downloaded by ted cage (sxnbyln180@questza.com) 371 lOMoARcPSD|6566483 www.konkur.in SECTION 9.6 (PAGE 550) ex 29. tan 1 .e x ADAMS and ESSEX: CALCULUS 9 x3 x2 C C 2 6 x .e 1/3 1/ D .e x 1/ 3 .e x 1/5 C 5 2 3 x x DxC C C 2 6 3 x3 x2 1 C C xC 3 2 6 5 2 x x3 1 xC C C C C 5 2 6 3 2 x 1 3 x C x C C DxC 2 6 3 x2 x3 DxC C 2 6 1DxC 35. 36. x2 x4 x6 C C C 3Š 5Š 7Š 1 ex e x D sinh x D x 2x if x ¤ 0. The sum is 1 if x D 0. 1C 1 1 1 C C C 2" 2Š 4 3Š 8 4Š # 1 1 3 1 1 2 1 C C C D2 2 2Š 2 3Š 2 D 2 e 1=2 1 : 1C 37. P .x/ D 1 C x C x 2 . a) The Maclaurin series for P .x/ is 1 C x C x 2 (for all x). b) Let t D x 30. We have e tan 1 x " 3 5 1 D exp x x x C 3 5 D1C x x5 x3 C 3 5 7 # x C 1 7 1 C x 2Š P .x/ D P .t C 1/ D 1 C t C 1 C .t C 1/2 D 3 C 3t C t 2 : 2 x3 C 3 The Taylor series for P .x/ about 1 is 3 C 3.x 1/ C .x 1/2 . 38. If a ¤ 0 and jx 1 .x /3 C 1 3Š x2 x3 x3 C C C higher degree terms Dx 3 2 6 2 3 x x DxC C : 2 6 p 31. Let 1 C x D 1 C ax C bx 2 C . Then 1 C x D 1 C 2ax C .a2 C 2b/x 2 C , so 2a D 1, and a2 C 2b p D 0. Thus a D 1=2 and b D 1=8. Therefore 1 C x D 1 C .x=2/ .x 2 =8/ C . D 39. 9 34. 21 x a a If a > 0 and t D x D ln a C 1 1 a 1C x C a a a/2 .x a2 .x a/3 a3 C : a, then x D t C a and nD1 1 X . 1/n 1 nD1 .x a/n an .0 < x < 2a/: Since the series converges to ln x on an interval of positive radius .a/, centred at a, ln is analytic at a. 40. If 27 x x x x C C x3 3Š 4 5Š 16 7Š 64 9Š 256 # " 3 5 1 x3 1 x3 x3 C D2 2 3Š 2 5Š 2 3 x .for all x/: D 2 sin 2 372 Telegram: @uni_k 15 1 1 a D t ln x D ln.a C t / D ln a C ln 1 C a 1 n X t . a < t a/ . 1/n 1 n D ln a C a 5y 4 y2 C C csc x D sec y D 1 C 2 24 2 1 5 4 D1C x C x C : 2 2 24 2 .for all x/: a/ The radius of convergence of this series is jaj, and the series converges to 1=x throughout its interval of convergence. Hence, 1=x is analytic at a. 32. csc x does not have a Maclaurin series because limx!0 csc x does not exist. Let y D x . Then x D y C and sin x D cos y. 2 2 Therefore, using the result of Exercise 25, x4 x6 2 C C D ex 2Š 3Š aj < jaj, then 1 1 D x a C .x C 33. 1 C x 2 C 1, so x D t C 1. Then 1=x 2 ; if x 6D 0; f .x/ D e 0; if x D 0; then the Maclaurin series for f .x/ is the identically zero series 0 C 0x C 0x 2 C since f .k/ .0/ D 0 for every k. The series converges for every x, but converges to f .x/ only at x D 0, since f .x/ 6D 0 if x 6D 0. Hence, f cannot be analytic at 0. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 41. ex ey D e xCy D D D 1 X xn nD0 nŠ ! 1 X ym mD0 1 X .x C y/k kD0 1 X j D0 1 X j D0 kŠ D mŠ ! kDj 1 j X x jŠ mD0 We have 1 k X 1 X kD0 kŠ 1 x X yk j jŠ .k j /Š j SECTION 9.6 (PAGE 550) 0 ˇtDx 1 f .kC1/ .t /.x t /kC1 ˇˇ f .x/ D Pk .x/ C @ ˇ ˇ kŠ kC1 tDc 1 Z x .x t /kC1 f .kC2/ .t / A dt C kC1 c kŠ xj yk j j Š.k j /Š j D0 .let k j D m/ f .kC1/ .c/ .x c/kC1 .k C 1/Š Z x 1 .x t /kC1 f .kC2/ .t / dt C .k C 1/Š c D PkC1 .x/ C EkC1 .x/: ym D ex ey : mŠ D Pk .x/ C 42. We want to prove that f .x/ D Pn .x/ C En .x/, where Pn is the nth-order Taylor polynomial for f about c and Z 1 x En .x/ D .x t /n f .nC1/ .t / dt: nŠ c (a) The Fundamental Theorem of Calculus written in the form Z x f .x/ D f .c/ C f 0 .t / dt D P0 .x/ C E0 .x/ c is the case n D 0 of the above formula. We now apply integration by parts to the integral, setting U D f 0 .t /; d U D f 00 .t / dt; d V D dt; V D .x t /: (We have broken our usual rule about not including a constant of integration with V . In this case we have included the constant x in V in order to have V vanish when t D x.) We have ˇtDx Z ˇ x ˇ 0 f .x/ D f .c/ f .t /.x t /ˇ .x t /f 00 .t / dt C ˇ c tDc Z x 0 D f .c/ C f .c/.x c/ C .x t /f 00 .t / dt Thus the formula is valid for n D k C 1 if it is valid for n D k. Having been shown to be valid for n D 0 (and n D 1), it must therefore be valid for every positive integer n for which En .x/ exists. 43. If f .x/ D ln.1 C x/, then 2 1 1 ; f 000 .x/ D ; ; f 00 .x/ D 1Cx .1 C x/2 .1 C x/3 3Š . 1/n 1 .n 1/Š ; : : : ; f .n/ D f .4/ .x/ D 4 .1 C x/ .1 C x/n f 0 .x/ D and f .0/ D 0; f 0 .0/ D 1; f 00 .0/ D f .4/ .0/ D 2 3Š 4 1 2 x C x3 C x C C 2Š 3Š 4Š . 1/n 1 .n 1/Š n x C En .x/ nŠ where (b) We complete the proof for general n by mathematical induction. Suppose the formula holds for some n D k: d U D f .kC2/ .t / dt; Z 1 x .x t /n f .nC1/ .t / dt nŠ 0 Z . 1/n nŠ 1 x dt .x t /n D nŠ 0 .1 C t /nC1 Z x .x t /n D . 1/n dt: nC1 0 .1 C t / En .x/ D t /k f .kC1/ .t / dt: Again we integrate by parts. Let U D f .kC1/ .t /; d V D .x V D t /k dt; 1 .x kC1 t /kC1 : If 0 t x 1, then 1 C t 1 and jEn .x/j Z x 0 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1/Š: f .x/ D x C D P1 .x/ C E1 .x/: f .x/ D Pk .x/ C Ek .x/ Z 1 x .x D Pk .x/ C kŠ c 3Š; : : : ; f .n/ .0/ D . 1/n 1 .n Therefore, the Taylor Formula is c We have now proved the case n D 1 of the formula. 1; f 000 .0/ D 2; Downloaded by ted cage (sxnbyln180@questza.com) .x t /n dt D 1 x nC1 !0 nC1 nC1 373 lOMoARcPSD|6566483 www.konkur.in SECTION 9.6 (PAGE 550) ADAMS and ESSEX: CALCULUS 9 1Ct t3 t5 D 2 tC C C for 1 t 3 5 1 < t < 1. Thus 0 1 1 C 0 < cn cnC1 D .2n C 1/@ 2n C 1 3.2n C 1/3 1 1 C C A 1 5.2n C 1/5 1 1 1 C C < 3 .2n C 1/2 .2n C 1/4 (geometric) 1 1 D 1 3.2n C 1/2 1 .2n C 1/2 1 D 12.n2 C n/ 1 1 1 D : 12 n n C 1 as n ! 1. If 1 < x t 0, then (c) ln ˇ ˇ ˇx t ˇ t x ˇ ˇ ˇ 1 C t ˇ D 1 C t jxj; t x increases from 0 to 1Ct from x to 0. Thus, because jEn .x/j < 1 1Cx Z jxj 0 x D jxj as t increases jxjn dt D jxjnC1 !0 1Cx as n ! 1 since jxj < 1. Therefore, x3 x2 C 2 3 f .x/ D x for 1 X x4 xn C D . 1/n 1 ; 4 n nD1 These inequalities imply that fcn g is de˚ 1 is increasing. Thus creasing and cn 12n 11 1 fcn g is bounded below by c1 12 D 12 and so limn!1 cn D c exists. Since e cn D nŠn .nC1=2/ e n , we have nŠ lim D lim e cn D e c n!1 nnC1=2 e n n!1 p exists. It remains to show that e c D 2. 1 < x 1. 44. We follow the steps outlined in the problem: (d) The Wallis Product, 2n 2n 22446 D lim n!1 1 3 3 5 5 2n 1 2n C 1 2 can be rewritten in the form r 2n nŠ ; lim p D n!1 1 3 5 .2n 2 1/ 2n C 1 Rj (a) Note that ln.j 1/ < j 1 ln x dx < ln j , j D 1; 2; : : :. For j D 0 the integral is improper but convergent. We have n ln n nD Z n ln x dx < ln.nŠ/ < 0 cn 374 Telegram: @uni_k ln x dx or, equivalently, 1 D .n C 1/ ln.n C 1/ (b) If cn D ln.nŠ/ Z nC1 n 1 < .n C 1/ ln.n C 1/ n: 22n .nŠ/2 lim p D n!1 .2n/Š 2n C 1 : 2 Substituting nŠ D nnC1=2 e ne cn and a similar expression for .2n/Š, we obtain n C 12 ln n C n, then nŠ n C 12 ln n .n C 1/Š C n C 23 ln.n C 1/ 1 1 D ln n C 12 ln n nC1 C n C 21 ln.n C 1/ C ln.n C 1/ nC1 D n C 12 ln 1 n 1 1 C 2nC1 1: D n C 21 ln 1 1 2nC1 r 22n n2nC1 e 2n e 2cn lim n!1 22nC1=2 n2nC1=2 e 2n e c2n p D ec e 2c D : c 2e 2 2n p Thus e =2 D 2, and e D 2, which completes the proof of Stirling’s Formula. cnC1 D ln p c 45. 1 c We have: (a) By part (a) of Exercise 44, n < ln.nŠ/ < .n C 1/ ln.n C 1/ n ln n n: Therefore, 1< Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) .n C 1/ ln.n C 1/ ln.nŠ/ < n ln n n n ln n n n : lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 9.7 (PAGE 554) METHOD II. (using Taylor’s Theorem) Since P5 .x/ D P6 .x/ (Maclaurin polynomials for sin have only odd degree terms) we are better off using the remainder E6 . We divide the numerator and denominator of the fraction on the right by n and take the limit as n ! 1 by l’H^opital’s rule: 1 ln.n C 1/ 1 1C n lim n!1 ln n 1 1 .n C 1/ 1 ln.n C 1/ C n2 n nC1 D lim 1 n!1 n ln.n C 1/ D lim C 1 D 1: n!1 n Hence lim ln.nŠ/ n!1 n ln n D 1, and n ˇ ˇ ln.nŠ/ lim ˇˇ n!1 jf .0:2/ jf .0:2/ 2. so that the relative error in the approximation ln.nŠ/ n ln n approaches 0as n ! 1. ln.20Š/ p ln 20 C 10 ln.10/ ln.10Š/ p ln 40 C 20 ln.20/ 10 20 ln.20Š/ 0:000552 3. ln.20Š/ 10 ln.10/ ln.10Š/ 20 ln.20/ ln.20Š/ 10 20 0:057185 4. 1. x5 x3 C . 6 120 METHOD I. (using an alternating series bound) If f .x/ D sin x, then P5 .x/ D x If f .x/ D ln x, then f 0 .x/ D 1=x, f 00 .x/ D 1=x 2 , f 000 .x/ D 2=x 3 , f .4/ .x/ D 6=x 4 , and f .5/ .x/ D 24=x 5 . If P4 .x/ is the Taylor polynomial for f about x D 2, then for some s between 1.95 and 2 we have (using Taylor’s Theorem) e 0:2 1 C 0:2 C P5 .0:2/j .0:2/7 < 2:6 10 9 : 7Š 1 De 1D1 e 1 1 C 1Š 2Š 1 1 C 3Š 4Š which satisfies the conditions for the alternating series test, and the error incurred in using a partial sum to approximate e 1 is less than the first omitted term in absolute 1 < 5 10 5 if n D 7, so value. Now .n C 1/Š 1 1 C 6 24 1 1 C 120 720 1 0:36786 5040 with error less than 5 10 5 in absolute value. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k .0:2/n .0:2/2 C C D sn 2Š nŠ We have 1 1 e 2 jf .0:2/ 24 .0:05/5 s5 5Š 24.0:05/5 < 2:22 10 9 : .1:95/5 120 P4 .1:95/j D .0:2/nC2 .0:2/nC1 sn D C C .n C 1/Š .n C 2/Š .0:2/nC1 .0:2/2 0:2 C C 1C .n C 1/Š nC2 .n C 2/2 nC1 .0:2/ 10n C 20 D < 5 10 5 if n D 4: .n C 1/Š 10n C 18 .0:2/3 .0:2/4 .0:2/2 C C e 0:2 1 C 0:2 C 2Š 3Š 4Š 1:221400 0:137613 Section 9.7 Applications of Taylor and Maclaurin Series (page 554) 1 .0:2/7 < 2:6 10 9 : 7Š 0 < e 0:2 0:000098 Evidently, Stirling’s formula gives a significantly better approximation than the modified version. cos x, so Error estimate: For the modified Stirling formula, ln.nŠ/ n ln n n, these errors are ln.10Š/ P5 .0:2/j < jf .1:95/ (b) From Stirling’s p Formula, ln.nŠ/ ln 2 n C n ln n n. For this approximation we have the following relative errors (using a calculator) ln.10Š/ jf .7/ .s/j .0:2/7 ; 7Š for some s between 0 and 0:2. Now f .7/ .x/ D ˇ n/ ˇˇ ˇD0 .n ln n ln.nŠ/ P5 .0:2/j D jE6 .0:2/j D Downloaded by ted cage (sxnbyln180@questza.com) 375 lOMoARcPSD|6566483 www.konkur.in SECTION 9.7 (PAGE 554) ADAMS and ESSEX: CALCULUS 9 5. e 1:2 D ee 0:2 . From Exercise 1: e 0:2 1:221400, .0:2/5 60 0:000003. Since with error less than 5Š 58 e D 2:718281828 , it follows that e 1:2 3:3201094 , 1 with error less than 3 0:000003 D 0:000009 < . 20; 000 Thus e 1:2 3:32011 with error less than 1/20,000. 10. We have sin 80ı D cos 10ı D cos 18 1 4 1 2 C D1 2Š 18 4Š 18 6. We have Since sin.0:1/ D 0:1 Since .0:1/3 .0:1/5 C 3Š 5Š 8. We have 1 6 D ln 1 C ln 5 5 2 1 1 3 1 1 4 1 1 1 C C : D 5 2 5 3 5 4 5 1 1 n Since < 5 10 5 if n D 6, therefore n 5 1 1 1 2 1 1 3 1 1 4 1 1 5 6 C C ln 5 5 2 5 3 5 4 5 5 5 0:18233 11. ln.0:9/ D ln.1 376 Telegram: @uni_k 5 5 D 180 180 with error less than 1 3Š 5 180 3 0:0871557 55 5 < 0:00000005. Thus 5Š1805 0:996192 cos 65 2 ı p 3.0:0871557/ 0:42262 2 with error less than 0.00005. 12. We have tan 1 .0:2/ D 0:2 Since 0:1 .0:2/5 .0:2/3 C 3 5 .0:2/7 C : 7 .0:2/7 < 5 10 5 , therefore 7 tan 1 .0:2/ 0:2 0:1/ .0:1/2 .0:1/3 .0:1/n 2 3 n .0:1/nC1 .0:1/nC2 jErrorj < C C nC1 nC2 .0:1/nC1 < 1 C 0:1 C .0:1/2 C nC1 .0:1/nC1 10 < 0:00005 if n D 3: D nC1 9 .0:1/2 .0:1/3 ln.0:9/ 0:1 0:10533 2 3 with error less than 0:00005. 5 C 3 180 p 5 3 5 1 sin D cos 2 180 2 180 From Exercise 5, cos.5=180/ 0:996192 with error less than 0:000003. Also cos 65ı D cos sin with error less than 5 10 5 in absolute value. 9. 1 2 0:98477 2Š 18 with error less than 5 10 5 in absolute value. .0:1/3 0:09983 3Š with error less than 5 10 5 in absolute value. 5 cos 5ı D cos D cos 180 36 1 2 1 4 . 1/n 2n 1 C C 2Š 36 4Š 36 .2n/Š 36 2nC2 1 jErrorj < .2n C 2/Š 36 1 < 0:00005 if n D 1: < .2n C 2/Š92nC2 1 2 cos 5ı 1 0:996192 2Š 36 with error less than 0:00005. 1 4 < 5 10 5 , therefore 4Š 18 sin 80ı 1 .0:1/5 D 8:33 10 8 < 5 10 5 , therefore 5Š sin.0:1/ D 0:1 7. .0:1/7 C : 7Š : .0:2/5 .0:2/3 C 0:19740 3 5 with error less than 5 10 5 in absolute value. 13. cosh 1 1 C 1 1 1 C C C with error less than 2Š 4Š .2n/Š 1 1 1 C C 1C .2n C 2/Š .2n C 3/2 .2n C 3/4 1 1 D < 0:00005 if n D 3: 1 .2n C 2/Š 1 .2n C 3/2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 9.7 (PAGE 554) 1 1 1 C 1:54306 with error Thus cosh 1 1 C C 2 24 720 less than 0:00005. 18. L.x/ D D 14. We have 1 4 1 10 0 Z x 20. We have L.0:5/ D 0:5 0 Since J.x/ D D 0 Z x 0 L.0:5/ 0:5 1 t ! t2 t3 t C C C dt 1C 2Š 3Š 4Š 21. x5 x3 C 3Š3 5Š5 1 1 1 C C . 1/n I.1/ 1 3Š3 5Š5 .2n C 1/Š.2n C 1/ 1 jErrorj < 0:0005 if n D 2: .2n C 3/Š.2n C 3/ Z 1Cx ln t dt let u D t 1 t 1 1 Z x ln.1 C u/ D du u 0 Z x u u2 u3 D 1 C C du 2 3 4 0 2 3 4 x x x Dx C 2 C 22 3 42 1 X x nC1 . 1/n D . 1 x 1/ .n C 1/2 From Exercise 15: I.x/ D x nD1 nD0 Thus I.1/ 1 decimal places. 22. 1 1 C 0:946 correct to three 3Š3 5Š5 x 10 x6 C x2 sin.x 2 / 3Š 5Š lim D lim 3 5 x!0 sinh x x!0 x x xC C C 3Š 5Š 9 5 x x C x 3Š 5Š D 0: D lim 4 2 x!0 x x C C 1C 3Š 5Š Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k .0:5/5 0:497 2Š 5 dt x2 x3 x4 C C C 2Š 2 3Š 3 4Š 4 1 X xn D : nŠ n K.x/ D .0:5/13 C : 6Š 13 rounded to 3 decimal places. et DxC 17. .0:5/9 .0:5/5 C 2Š 5 4Š 9 .0:5/4nC1 < 5 10 4 if n D 2, therefore .2n/Š .4n C 1/ nD0 16. ! t 12 C dt 6Š nD0 Z x Z x t8 t4 C 2Š 4Š tan 1 .t 2 / dt t2 0 Z x 4 t t 8 t 12 1 D C C dt 3 5 7 0 9 5 x x 13 x C C Dx 35 5 9 7 13 1 X x 4nC1 D . 1 x 1/ . 1/n .2n C 1/.4n C 1/ M.x/ D 10 1 2 sin t dt t Z x t4 t6 t2 C C dt D 1 3Š 5Š 7Š 0 3 5 x x Dx C 3 3Š 5 5Š 1 X x 2nC1 D for all x: . 1/n .2n C 1/.2n C 1/Š I.x/ D 1 nD0 19. with error less than 5 10 5 in absolute value. 15. Z x x9 x 13 x5 C C 2Š 5 4Š 9 6Š 13 1 X x 4nC1 : D . 1/n .2n/Š .4n C 1/ 4 1 C : 2 1 1 1 n < if n D 11, therefore Since n 2 20000 3 1 1 1 2 1 1 3 C 2 2 2 2 3 2 0:40543 cos.t 2 / dt 0 Dx 3 1 ln D ln 1 C 2 2 1 1 1 2 1 1 3 D C 2 2 2 3 2 ln Z x Downloaded by ted cage (sxnbyln180@questza.com) 377 lOMoARcPSD|6566483 www.konkur.in SECTION 9.7 (PAGE 554) ADAMS and ESSEX: CALCULUS 9 x4 x8 1 1C C 1 cos.x 2 / 2Š 4Š D lim 23. lim 2 2 x!0 x!0 .1 cos x/ x2 x4 1 1C C 2Š 4Š 1 2 C O.x / D 2: D lim 2Š x!0 1 C O.x 2 / 4 24. We have x2 2 x3 x4 C C C x 2 .e 1 x/ 4Š lim D lim 2Š4 3Š6 x!0 x 2 x!0 ln.1 C x 2 / x x x8 C 2 3 4 2 4 2 x x 1 x C 1C C 1 4 3 12 4 D D : D lim 6 8 4 1 x!0 2 x x x C 2 2 3 4 Section 9.8 The Binomial Theorem and Binomial Series (page 559) 1. 1 C x D .1 C x/1=2 2 3 1 1 1 x 1 3 x x C C D1C C 2 2 2 2Š 2 2 2 3Š 1 X x 1 3 5 .2n 3/ n D1C C x . 1/n 1 2 2n nŠ D1C 2 sin 3x 3 sin 2x 5x tan 1 5x 23 x 3 33 x 3 C 3 2x C 2 3x 3Š 3Š D lim x!0 53 x 3 5x 5x C 3 3 9 C 4 C O.x 2 / 53 D : D lim D 125 x!0 125 25 C O.x 2 / 3 26. We have 25. p x C 2 nD2 1 X .2n 2/Š . 1/n 1 2n 1 xn 2 .n 1/ŠnŠ . 1 < x < 1/: nD2 lim x!0 sin.sin x/ x lim x!0 xŒcos.sin x/ 1 1 1 sin3 x C sin5 x x sin x 3Š 5Š D lim i h 1 1 x!0 sin2 x C sin4 x 1 x 1 2Š 4Š 1 3 5 x3 1 x3 x C C C x x 3Š 3Š 3Š 5Š D lim h i 3 2 4 x!0 1 1 x C C x x x 2Š 3Š 4Š 2 3 2 x C higher degree terms 2 3Š 3Š D lim D : D 1 3 1 x!0 3 x C higher degree terms 2Š 2Š 27. lim p x 1 x/1=2 1 1 . 1/2 x 3 x C Dx 2 2 2 2Š 1 1 3 . 1/3 x 4 C C 2 2 2 3Š 1 x 2 X 1 3 5 .2n 3/ nC1 Dx x 2 2n nŠ x D x.1 2 2 Dx x 3. p x 2 x3 C 3Š x2 C 2Š nD2 1 X .2n 2/Š x nC1 . 1/n 1 2n 1 2 .n 1/ŠnŠ . 1 < x < 1/: nD2 r x 4Cx D2 1C 4 2 6 1 x D 26 41 C 2 4 C sinh x x!0 cosh x 378 Telegram: @uni_k sin x cos x x3 C x xC 3Š D lim x!0 x2 1C C 1 2Š x3 C O.x 5 / 3 D lim 2 D 0: x!0 x C O.x 4 / 2. C D2C 1 2 1 2 3Š 1 2 1 2 2Š 3 2 1 x 2 x 3 4 4 3 7 C 7 5 X x 1 3 5 .2n C2 . 1/n 1 4 23n nŠ nD2 1 3/ X x .2n 1/Š D2C C2 xn . 1/n 1 4n 1 4 2 nŠ.n 1/Š nD2 . 4 < x < 4/: Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) xn lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 4. p 1 4 C x2 D r 1 SECTION 9.8 (PAGE 559) " # x 2 1=2 1 x 2 D 2 1 C 2 9. 2 1C 2 1 x 2 1 3 x 4 1 1 1C D C C 2 2 2 2Š 2 2 2 # 1 3 5 x 6 1 C 3Š 2 2 2 2 " i) n 0 n nŠ D 1; 0ŠnŠ D n ii) If 0 k n, then nŠ D 1. nŠ0Š D n n C D k 1 k .k nŠ nŠ C 1/Š.n k C 1/Š kŠ.n k/Š nŠ k C .n k C 1/ D kŠ.n k C 1/Š .n C 1/Š nC1 D : D k kŠ.n C 1 k/Š 3 1 2 35 6 1 x C 7 x4 x C 4 2 2 2 2Š 210 3Š 1 1 X 1 2 3 .2n 1/ 2n D C . 1/n x 2 23nC1 nŠ D nD1 . 2 x 2/: 5. .1 x/ 2 . 2/. 3/ . 2/. 3/. 4/ D 1 2. x/ C . x/2 C . x/3 C 2Š 3Š 1 X D 1 C 2x C 3x 2 C 4x 3 C D nx n 1 . 1 < x < 1/: 10. The formula .a C b/n D . 3/. 4/ 2 . 3/. 4/. 5/ 3 x C x C .1 C x/ 3 D 1 3x C 2Š 3Š .3/.4/ 2 .4/.5/ 3 D 1 3x C x x C 2 2 1 X .n C 2/.n C 1/ n D x . 1 < x < 1/: . 1/n 2 .a C b/mC1 D .a C b/ nD0 7. D Using the Maclaurin series for sin 1 x obtained in this section, we have 2 D 2 cos 1 x D D 2 1 X 1 3 5 .2n nD1 3 x x 6 2n nŠ.2n C 1/ 3 5 x 40 1/ dt p 1 C t2 0 Z x" 1 X 1 3 5 .2n D 1C . 1/n 2n nŠ 0 1/ 2n t nD1 Dx for nD1 3 . 1/n 1 3 5 .2n 2n nŠ.2n C 1/ x 3 5 C x 6 40 1/ k kD0 amC1 k b k C D amC1 C D # mC1 X kD0 m h X m kD1 an k b k am k b k m X m kD0 k am k b kC1 k m i C amC1 k b k C b mC1 k 1 (by #9i)) m X mC1 kD1 k amC1 k b k C b mC1 mC1 amC1 k b k k (by #9(ii)) (by #9(i) again). Thus the formula holds for n D m C 1. By induction it holds for all positive integers n. dt x 2nC1 11. Consider the Leibniz Rule: .fg/.n/ D 1<x<1 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k k kD1 Z x 1 X kD0 k D amC1 C 1<x<1 1 d , we have, using the series sinh 1 x D p 8. Since dx 1 C x2 p for 1 C x 2 , DxC m X m m X m kD0 x 2nC1 for sinh 1 x D n (replace k by k 1 in the latter sum) m mC1 X X m m mC1 k k D a b C amC1 k b k k k 1 sin 1 x x kD0 holds for n D 1; it says a C b D a C b in this case. Suppose the formula holds for n D m, where m is some positive integer. Then nD1 6. Pn Downloaded by ted cage (sxnbyln180@questza.com) n X n kD0 k f .n k/ g .k/ : 379 lOMoARcPSD|6566483 www.konkur.in SECTION 9.8 (PAGE 559) ADAMS and ESSEX: CALCULUS 9 This holds for n D 1; it says .fg/0 D f 0 g C fg 0 in this case. Suppose the formula holds for n D m, where m is some positive integer. Then d .fg/ D .fg/.m/ dx m d X m .m k/ .k/ D f g k dx completing the induction and thus the proof. 13. We want to prove that .mC1/ .f1 f2 fn /.k/ D kD0 D m X m kD0 k f .mC1 k/ g .k/ C m X m kD0 k f .m k/ g .kC1/ D f .mC1/ g .0/ C f mC1 X kD0 k kD1 .mC1 k/ .k/ C m i k 1 g C f .0/ g .mC1/ (by Exercise 9(i)) D f .mC1/ g .0/ C D Observe that the case n D 2 has been proved in Exercise 11. To complete the induction on n, we assume that the formula above holds for some n and all k. If u D f1 f2 fn and v D f1 f2 fn fnC1 D ufnC1 then the Product rule and the induction hypothesis show that kD1 m h X m m X mC1 k kD1 v .k/ D .ufnC1 /.k/ D D f .mC1 k/ g .k/ C f .0/ g .mC1/ (by Exercise 9(ii)) mC1 f .mC1 k/ g .k/ k (by 9(i) again). .x1 Cx2 C Cxn / D jmjDk D kŠ x m1 x m2 xnmn ; m1 Š m2 Š mn Š 1 2 holds for some n 2 and all k, and we apply the Binomial Theorem to k .x1 C C xm C xmC1 /k D .x1 C C xn / C xnC1 D D k X j D0 k X j D0 kŠ k j .x1 C jxn /j xnC1 j Š .k j /Š jm jDk 380 Telegram: @uni_k kŠ mnC1 ; x m1 xnmn xnC1 m1 Š mn Š mnC1 Š 1 X kŠ jŠ .k j / .m / .m / fnC1 f 1 f2 2 fn.mn / : j Š.k j /Š m1 Š mn Š 1 jmjDj k X j D0 X jŠ kŠ .m / .m / f 1 fn.mn / fnC1nC1 j ŠmnC1 Š m1 Š mn Š 1 X jmjDj jm jDk kŠ .m / .m / f 1 fn.mn / fnC1nC1 ; m1 Š mn ŠmnC1 Š 1 Section 9.9 Fourier Series 1. (page 565) f .t / D sin.3t / has fundamental period 2=3 since sin t has fundamental period 2: f t C 2 D sin 3 t C 2 D sin.3t C 2/ 3 3 D sin.3t / D f .t /: jmjDj Now let k j D mnC1 and let m D .m1 ; mn ; mnC1 /, so that jm j D j C .k j / D k. The result above then simplifies to j D0 kŠ .k j / u.j / fnC1 j Š.k j /Š which completes the induction and the proof. X kŠ jŠ k j xnC1 x m1 xnmn : j Š .k j /Š m1 Š mn Š 1 .x1 C C xn C xnC1 /k X D j D0 v .k/ D 12. As suggested in the hint, we assume that X k X k X Now let m D .m1 ; m2 ; : : : ; mn ; mnC1 /, where mnC1 D k j to ensure that jm j D k. Then we have Thus the Rule holds for n D m C 1. By induction, it holds for all positive integers n. k jmjDk kŠ m1 Šm2 Š mn Š holds for all n 2 and all positive integers k. The sum is taken over all multiindices m or order n satisfying jmj D k. i (replace k by k 1 in the latter sum) m mC1 X X m m D f .mC1 k/ g .k/ C f .mC1 k/ g .k/ k k 1 kD0 X 2. g.t / D cos.3 C t / has fundamental period 2 since cos t has fundamental period 2: g.t C 2/ D cos 3 C .t C 2/ D cos.3 C t C 2/ D cos.3 C t / D g.t /: Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 9.9 (PAGE 565) 3. h.t / D cos2 t D 21 .1 C cos 2t / has fundamental period : h.t C / D 7. 1 C cos 2t / 1 C cos.2t C 2/ D D h.t /: 2 2 4. Since sin 2t has periods , 2, 3, : : : , and cos 3t 4 6 has periods 2 D 2, 8 3 , 3 , 3 3 , : : : , the sum k.t / D sin.2t / C cos.3t / has periods 2, 4, : : : . Its fundamental period is 2. 0 . 1/n D : n The Fourier series of f is 1 1 1 X . 1/n 1 1 2 X cos .2n 1/ t / sin.n t /: 4 2 .2n 1/2 n 5. Since f .t / D t is odd on . ; / and has period 2, its cosine coefficients are 0 and its sine coefficients are given by 2 bn D 2 Z 2 t sin.nt / dt D Z t sin.nt / dt: 0 8. 2 2 cos.n/ D . 1/nC1 : n n The Fourier series of f is 1 X . 1/nC1 nD1 2 sin.nt /. n 0 if 0 t < 1 , f has period 2. 1 if 1 t < 2 The Fourier coefficients of f are as follows: 6. f .t / D Z Z a0 1 2 1 2 1 D f .t / dt D dt D 2 2 0 2 1 2 Z 2 Z 2 an D f .t / cos.n t / dt D cos.n t / dt 0 1 ˇ2 ˇ 1 ˇ sin.n t /ˇ D 0; .n 1/ D ˇ n 1 ˇ2 Z 2 ˇ 1 ˇ cos.n t /ˇ bn D sin.n t / dt D ˇ n 1 1 ( n 2 1 . 1/ if n is odd D D n n 0 if n is even The Fourier series of f is 1 2 1 X nD1 2 .2n 1/ sin .2n nD1 nD1 This integral can be evaluated by a single integration by parts. Instead we used Maple to do the integral: bn D 0 if 1 t < 0 , f has period 2. t if 0 t < 1 The Fourier coefficients of f are as follows: Z Z 1 1 1 1 1 a0 D f .t / dt D t dt D 2 1 2 4 1 0 Z 1 Z 1 an D f .t / cos.n t / dt D t cos.n t / dt 1 0 . 1/n 1 2=.n/2 if n is odd D D 2 2 n 0 if n is even Z 1 t sin.n t / dt bn D f .t / D 1/ t : ( t if 0 t < 1 1 if 1 t < 2 , f has period 3. 3 t if 2 t < 3 f is even, so its Fourier sine coefficients are all zero. Its cosine coefficients are Z 2 1 2 3 2 a0 f .t / dt D .2/ D D 2 2 3 0 3 3 Z 2n t 2 3 f .t / cos dt an D 3 0 3 "Z Z 2 1 2n t 2n t 2 dt C dt t cos cos D 3 0 3 3 1 # Z 3 2n t C .3 t / cos dt 3 2 4n 2n 3 1 cos.2n/ C cos : cos D 2n2 2 3 3 The latter expression was obtained using Maple to evaluate the integrals. If n D 3k, where k is an integer, then an D 0. For other integers n we have an D 9=.2 2 n2 /. Thus the Fourier series of f is 1 1 2 9 X 1 2n t 1 X 1 cos C cos.2n t /: 3 2 2 n2 3 2 2 n2 f .t / D nD1 9. The even extension of h.t / D 1 on Œ0; 1 to Œ 1; 1 has the value 1 everywhere. Therefore all the coefficients an and bn are zero except a0 , which is 2. The Fourier series is a0 =2 D 1. 10. The Fourier sine series of g.t / D t on Œ0; has coefficients Z 2 2 . t / sin nt dt D : bn D 0 n The required Fourier sine series is 1 X 2 sin nt: n Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k nD1 Downloaded by ted cage (sxnbyln180@questza.com) nD1 381 lOMoARcPSD|6566483 www.konkur.in SECTION 9.9 (PAGE 565) 11. ADAMS and ESSEX: CALCULUS 9 The Fourier sine series of f .t / D t on Œ0; 1 has coefficients Z 1 . 1/n : bn D 2 t sin.n t / dt D 2 n 0 The required Fourier sine series is 1 X Similarly, 2 an D T 2 D T n 2. 1/ sin.n t /: n nD1 4 D T 12. The Fourier cosine series of f .t / D t on Œ0; 1 has coefficients Z 1 a0 1 D t dt D 2 2 0 Z 1 an D 2 t cos.n t / dt 0 ( 0 if n is even 2. 1/n 2 4 D D 2 2 if n is odd. n n2 2 "Z "Z 0 2n t dt C f .t / cos T T =2 0 Z T =2 0 2n t . dt / C f .t / cos T T =2 Z T =2 0 2n t f .t / cos dt T Z T =2 0 2n t f .t / cos dt T 2n t f .t / cos dt: T The corresponding result for an odd function f states that an D 0 and bn D 4 T Z T =2 f .t / sin 0 2n t dt; T and is proved similarly. The required Fourier cosine series is 1 cos .2n 1/ t X 1 4 : 2 2 .2n 1/2 Review Exercises 9 (page 566) nD1 13. From Example 3, C 2 1 X nD1 4 cos .2n 2 .2n 1/ 1/ t D . 1/n e n D 0. The sequence converges. n!1 nŠ 2. n100 C 2n n100 C D . D lim n!1 n!1 2n 2n The sequence converges. jt j t . Putting t D , we obtain for 1. 1 X 4 . 1/ D 0: C 2 .2n 1/2 lim lim nD1 Thus 1 X nD1 14. 1 .2n 1/2 D D . 2 4 8 3. In the first integral in the line above replace t with t . Since f . t / D f .t / and sine is odd, we get "Z 0 2n t 2 . dt / f .t / sin bn D T T =2 T # Z T =2 2n t C f .t / sin dt T 0 # " Z Z T =2 T =2 2n t 2n t 2 dt C dt f .t / sin f .t / sin D T T T 0 0 382 Telegram: @uni_k ln n lim ln n n!1 =2 D 1: The sequence diverges to infinity. If f is even and has period T , then Z 2n t 2 T =2 f .t / sin dt bn D T T T =2 "Z # Z T =2 0 2 2n t 2n t D dt C dt : f .t / sin f .t / sin T T T T =2 0 D 0: lim n!1 tan 1 n 4. . 1/n n2 . 1/n D lim does not exist. n!1 1 n!1 n.n / .=n/ The sequence diverges (oscillates). 5. Let a1 > lim p 2 and anC1 D an 1 C . 2 an p 1 1 1 x > 0 if x > 2. If f .x/ D C , then f 0 .x/ D x 2 p x2 p2 p p Since f . 2/ D 2, > 2. p we have f .x/ > 2 if x p Therefore, ifpan > 2, then anC1 D f .an / > 2. Thus an > 2 for all n 1, by induction. p an > 2 ) 2 < an2 ) an2 C 2 < 2an2 an2 C 2 < an ) anC1 < an : 2an p Thus fan g is decreasing and an > 2 for all n. ) Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) # # lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES 9 (PAGE 566) p Being decreasing and bounded below by 2, fan g must converge by p the completeness axiom. Let limn!1 an D a. Then a 2, and lim anC1 D lim n!1 n!1 1 a aD C : 2 a an 1 C 2 an 12. nD1 p 13. Thus ln.n C 1/ lim ln ln.nC1/ ln ln n D lim ln D ln 1 D 0: n!1 n!1 ln n 1 1 2 .n 5/=2 D 22 1 C p C C 2 2 nD1 p 4 4 2 D p D p : 1 .1= 2/ 2 1 n 1 1 X 4 1X 4n 1 D . 1/2n 4 . 1/2 1 4 . 1/2 1/2 > 4. 1 since . 9. 1 X nD1 1 n2 D 1 4 1 X 1 X 1 9 4 n2 nD1 D D 1 lim n C 21 N !1 N C 1 2 1 X 1 nD1 1 3 1 2 n " 3 1 n 1 1=2 D 1 1 nD1 D2 10. 14. nD0 1 D 4 3 2 . 1/2 ; 4. 1/2 16 Since 0 1 X nD1 n 1 n3 n 1 n3 n2 p converges by comparison with the .1 C 2n /.1 C n n/ nD1 1 p X n (which converges by the ratio convergent series n 2 nD1 test) because n2 .1 C 2n /.1 C n p 1 n C 32 ! n 2n 15. 1 1 C 5=2 1=2 1 X 32nC1 nŠ nD1 (telescoping) D lim n!1 1 C1 2n 1 1 n3=2 C 1 D 1: converges by the ratio test, because 32.nC1/C1 9 nŠ 2nC1 D lim D 0 < 1: n!1 .n C 1/Š n!1 n C 1 3 1 7=2 1 nD1 16. 1 X nŠ converges by comparison with the con.n C 2/Š C 1 nD1 1 X 1 , because vergent p-series n2 nD1 0 nŠ 1 1 nŠ < D < 2: .n C 2/Š C 1 .n C 2/Š .n C 2/.n C 1/ n Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k p n/ lim X 1 1 for n 1 and converges, 2 n n2 must also converge. 1 X (telescoping) # 1 1 1 1 C C C 3=2 9=2 5=2 11=2 1 2 2 D 2C2C D : 3 3 9 11. n p converges by comparison with the .1 C n/.1 C n n/ nD1 1 X 1 because convergent p-series 3=2 n nD1 n!1 D 2: because 1 X lim ! 3 n p 1 .1 C n/.1 C n n/ D 1: lim D lim 1 n!1 n!1 1 1 C1 n3=2 n n3=2 C 1 1 X nD0 1 n X 2 n C 2n .n=2n / C 1 1 C 3n D lim D 1: lim n!1 .2=3/n n!1 .1=3n / C 1 2. ln.x C 1/ 1=.x C 1/ x lim D lim D lim D 1: x!1 x!1 x!1 ln x 1=x xC1 8. converges by comparison with the convergent nD1 6. By l’H^opital’s Rule, 7. 1 C 3n geometric series Thus a=2 D 1=a, so a2 D 2, and limn!1 an D a D 1 X n C 2n Downloaded by ted cage (sxnbyln180@questza.com) 383 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 9 (PAGE 566) 17. ADAMS and ESSEX: CALCULUS 9 1 X . 1/n 1 converges absolutely by comparison with the 1 C n3 nD1 1 X 1 convergent p-series , because n3 22. nD1 23. 1 X 1 Let s D kD1 ˇ ˇ ˇ . 1/n ˇ ˇ ˇ ˇ 2n n ˇ 1 lim D lim n D 1: 1 n!1 n!1 1 2n 2n 19. kD1 k3 Z 1 dt dt < s s < n 3 t3 nC1 t n 1 1 sn C < s < sn C 2 : 2 2.n C 1/ 2n converges by the alternating series test 1 C n3 nD1 (note that cos.n/ D . 1/n ), but the convergence is only conditional because This error is less than 0.001 if n 8. Hence 1 sn D 2 1 is a divergent harmonic series. 2n nD1 ˇ ˇ ˇ .x 2/nC1 ˇ ˇ ˇ r p ˇ ˇ nC1 n jx 2j jx 2j nC1ˇ ˇ3 D lim D : lim ˇ ˇ n!1 ˇ .x 2/n ˇ n!1 3 nC1 3 ˇ ˇ p ˇ ˇ 3n n 1 X .x jx 2j 2/n < 1, that is, p converges absolutely if 3 n nD1 if 1 < x < 5, and diverges if x < 1 or x > 5. X . 1/n If x D 1 the series is p , which converges condin tionally. X 1 If x D 5 the series is p , which diverges (to 1). n sn C 1 1 C sn C 2 2.n C 1/2 2n Then s sn with error satisfying sn j < js 1 X n2 cos.n/ 1 X s 24. 1 1 2 2n2 1 2.n C 1/2 D sn C D n2 C .n C 1/2 : 4n2 .n C 1/2 2n C 1 : 4n2 .n C 1/2 1 1 1 1 1 1 1 1 C 3 C 3 C 3 C 3 C 3 C 3 C 3 13 2 3 4 5 6 7 8 64 C 81 1:202 C 4.64/.81/ with error less than 0:001. 1 n X X 1 1 Let s D and s D . Then n 4 C k2 4 C k2 kD1 kD1 Z 1 Z 1 dt dt < s sn < 2 4 C t 4 C t2 n nC1 1 n C 1 sn C tan 1 < s < sn C 4 2 2 4 1 n tan 1 : 2 2 Let 3n 384 2xj: . Then Let for n 1, and Telegram: @uni_k n X 1 1 X . 1/n 1 converges by the alternating series test, but ln ln n nD1 1 X 1 dithe convergence is only conditional since ln ln n nD1 verges to infinity by comparison with the divergent har1 X 1 . (Note that ln ln n < n for all n 1.) monic series n ˇ 2 ˇ ˇ n cos.n/ ˇ 1 n2 ˇ ˇ ˇ 1 C n3 ˇ D 1 C n3 2n 21. k and sn D 3 Z 1 nD1 20. n D j5 nC1 2x/n converges absolutely if j5 2xj < 1, that is, n nD1 if 2 < x < 3, and diverges if x < 2 or x > 3. X1 If x D 2 the series is , which diverges. n X . 1/n If x D 3 the series is , which converges condin tionally. ˇ ˇ ˇ . 1/n 1 ˇ ˇ 1 : 0 ˇˇ 3 1 C n ˇ n3 1 X . 1/n converges absolutely by comparison with the 2n n nD1 1 X 1 , because convergent geometric series 2n 2xj 1 X .5 nD1 18. ˇ ˇ ˇ .5 2x/nC1 ˇ ˇ ˇ ˇ ˇ nC1 ˇ ˇ D lim j5 lim n!1 ˇˇ .5 2x/n ˇˇ n!1 ˇ ˇ n sn D sn C 4 1 n nC1 C tan 1 tan 1 : 4 2 2 Then s sn with error satisfying js sn j < Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) nC1 1 tan 1 4 2 tan 1 n : 2 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES 9 (PAGE 566) This error is less than 0.001 if n 22. Hence s 22 X kD1 (Remark: Examining the ln of the absolute value of the nth term at x D 8 shows that this term ! 0 as n ! 1. Therefore the series also converges at x D 8.) 1 23 tan 1 C tan 1 .11/ 0:6605 4 2 1 C 4 C k2 4 with error less than 0:001. 1 1 25. D x 3 x 3 1 3 1 1 X xn 1 X x n D D 3 3 3nC1 nD0 32. . 3 < x < 3/: nD0 26. Replace x with x 2 in Exercise 25 and multiply by x to get x 3 27. x D 2 1 X x 2nC1 nD0 3nC1 . p x2 ln.e C x / D ln e C ln 1 C e 1 X x 2n D ln e C . 1/n 1 n ne 1 1 e 2x D x x D 29. p (Remark: the series also converges at x D 1.) 3/: 33. 1 1 1 X . 2x/n nŠ nD1 1 X . 1/n 1 nD1 2n x n 1 nŠ . p e<x p 1 1 D x C .x D e/: ! D (for all x ¤ 0). 34. nD0 30. . 1/n nD1 22n 1 x 2nC1 .2n/Š (for all x). sin x C D sin x cos C cos x sin 3 3 3 p 1 1 2nC1 X 3 x x 2n 1X . 1/n . 1/n C D 2 .2n C 1/Š 2 .2n/Š nD0 nD0 ! p 2n 1 X 3x x 2nC1 . 1/n C D (for all x). 2 .2n/Š .2n C 1/Š x 1=3 1 1C 2 8 1 4 " 1 1 x 3 3 x 2 D 1 C 2 3 8 2Š 8 4 7 1 # 3 3 3 x 3 C C 3Š 8 1 1 X 1 4 7 .3n 2/ n . 1/n C x 2 2 3n 8n nŠ nD1 nD0 1 X . 1/n nD0 .x /n nC1 35. 2 2n 4 (for all x). 2 e x C2x D e x e 2x 8x 3 4x 2 C C D .1 C x 2 C / 1 C 2x C 2Š 3Š 4 3 2 2 3 D 1 C 2x C 2x C x C x C 2x C 3 10 2 P3 .x/ D 1 C 2x C 3x C x 3 : 3 sin.1 C x/ D sin.1/ cos x C cos.1/ sin x x2 x3 D sin.1/ 1 C C cos.1/ x C 2Š 3Š sin.1/ 2 cos.1/ 3 P3 .x/ D sin.1/ C cos.1/x x x : 2 6 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k .0 < x < 2/: .=4/, so x D u C .=4/. Then nD0 36. . 8 < x < 8/: 1 nD0 .8 C x/ 1=3 D D 1 1 1C x 1 p X . 1/n x D 2 .2n/Š nD0 31. D sin x C cos x D sin u C C cos u C 4 4 1 D p .sin u C cos u/ C .cos u sin u/ 2 1 p p X u2n D 2 cos u D 2 . 1/n .2n/Š x .1 C cos.2x// 2 ! 1 2n X x n .2x/ D 1C . 1/ 2 .2n/Š DxC / x n 1 X . 1/n Let u D x x cos2 x D 1 X . 1 < x < 1/: nD2 2 nD1 28. 3<x< 2 1 1 3 3 1=3 x2 .1 C x/ D1C xC 3 2Š 1 2 5 3 3 3 x3 C C 3Š 1 X x 2 5 8 .3n 4/ n D1C C x . 1/n 1 3 3n nŠ Downloaded by ted cage (sxnbyln180@questza.com) 385 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 9 (PAGE 566) 37. cos.sin x/ D 1 D1 P4 .x/ D 1 38. ADAMS and ESSEX: CALCULUS 9 x 2 x3 C .x /4 3Š C 2Š 4Š x4 1 x4 2 x C C C 2 3 24 1 2 5 x C x4: 2 24 1 1 p 1 2 2 1 C sin x D 1 C sin x C .sin x/2 2Š 2 1 1 3 2 2 2 C .sin x/3 3Š 1 1 3 5 2 2 2 2 C .sin x/4 C 4Š 2 1 x3 x3 1 C C x x D1C 2 6 8 6 1 5 3 4 C .x / .x / C 16 128 x2 x4 x 3 5x 4 x x3 C C C D1C 2 12 8 24 16 128 x x2 x3 x4 P4 .x/ D 1 C C : 2 8 48 384 39. The series 1 X . 1/n x n is the Maclaurin series for cos x .2n/Š nD0 1 X with x 2 replaced by x. For x > 0 the series therefore 1 X p jxjn represents cos x. For x < 0, the series is , .2n/Š nD0 p which is the Maclaurin series for cosh jxj. Thus the given series is the Maclaurin series for p cos p x cosh jxj nx n D nD0 1 X nC1 n nD0 42. 1 X nD0 1 X nD1 n2 D kD0 kŠ 43. .2n/Š ; n2 1 1 1 2 C 1/2 . C x .1 1 1 1 1 D 2 : 1 as in Exercise 23 x/2 x 1Cx d D 2 dx .1 x/ .1 x/3 ln.1 x/ 1 D ne n ln 1 1 e nD1 44. nD1 1 X 45. D1 ln.e 1/: 1 X . 1/n 1 x 2n 1 D sin x .2n 1/Š . 1/n 2n 1 D .2n 1/Š sin D 0 . 1/n 2n 4 1 D 3 .2n 1/Š 0 . 1/ 1Š f .2n 1/ .0/ D 0: x 3 D 1 : 2 Z x sin.t 2 / dt Z x t6 2 C dt D t 3Š 0 x3 x7 D C 3 7 3Š S.x/ D x!0 Telegram: @uni_k x/2 D n nD1 1 X lim 386 x/2 .1 x .1 nx n D 1 X xn for x near 0, we have, for n D 1, 2, 3, : : : f .2n/ .0/ D nD1 1 x.1 C x/ .1 x/3 nD0 1 1 1 C 1 X . C 1/ n2 D : 3 D n . 1/3 1 nD0 1 nD1 1 X xk nx n 1 D n2 x n D 40. Since 1C x 1 X n2 x n 1 D nD0 1 X if x 0 if x < 0. 1 X f .k/ .0/ 1 1 D D nD2 1 X x 2n xn D nx n 1 D nD0 1 X nD0 f .x/ D 1 X 41. 0 3S.x/ D lim x!0 x7 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) x3 x3 C x7 x7 14 D 1 : 14 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 46. CHALLENGING PROBLEMS 9 (PAGE 567) .x tan 1 x/.e 2x 1/ 2x 2 1 C cos.2x/ x5 4x 2 x3 C 2x C C x xC 3 5 2Š D lim 2 4 x!0 4x 16x 2x 2 1 C 1 C 2Š 4Š 2 x4 C 3 D 1: D lim 2 x!0 x4 C 3 Z 1=2 Z 1=2 X 1 . x 4 /n 4 dx e x dx D nŠ 0 0 nD0 ˇ1=2 1 X . 1/n x 4nC1 ˇˇ D ˇ .4n C 1/nŠ ˇ lim 49. x!0 47. D nD0 1 X nD0 bn D The series is 50. 0 1 : 24kC1 .4k C 1/kŠ This is less than 0:000005 if 24kC1 .4k C 1/kŠ > 200; 000, which happens if k 3. Thus, rounded to five decimal places, 0 4 e x dx 1 211 1 1 C 0:49386: 32 5 1 512 9 2 48. If f .x/ D ln.sin x/, then calculation of successive derivatives leads to f .5/ .x/ D 24 csc4 x cot x 8 csc2 cot x: for 1:5 x =2. Therefore, the error in the approximation ln.sin 1:5/ P4 .x/; 2 ˇˇ ˇ1:5 5Š ˇˇ5 ˇ 3 10 8 : 2 2C 4 " 1 2 cos .2n X nD1 .2n 1/t 1/2 sin.nt /. / sin .2n .2 C .2n 1/t 1/ C # sin.2nt / : 2n Challenging Problems 9 (page 567) 1. If an > 0 and anC1 n > for all n, then an nC1 1 a2 > a1 2 a3 2 > a2 3 ) ) a1 2 2a2 a1 a3 > > 3 3 a2 > :: : n 1 an > an 1 n Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 : n 1 if < t 0 has period 2. Its Fourier t if 0 < t coefficients are f .t / D where P4 is the 4th degree Taylor polynomial for f .x/ about x D =2, satisfies jerrorj t / sin.nt / dt D . 0 The required Fourier series is, therefore, Observe that 1:5 < =2 1:5708, that csc x 1 and cot x 0, and that both functions are decreasing on that interval. Thus jf .5/ .x/j 24 csc4 .1:5/ cot.1:5/ 2 n Z t on Œ0; has Z a0 1 D f .t / dt 2 2 Z 0 Z 1 1 dt C t dt D C D 2 2 4 0 Z 0 Z 1 t cos.nt / dt cos.nt / dt C an D 0 Z 1 D .1 C t / cos.nt / dt 0 n . 1/ 1 2=. n2 / if n is odd D D n2 0 if n is even Z 0 Z 1 bn D sin.nt / dt C t sin.nt / dt 0 Z 1 .t 1/ sin.nt / dt D 0 1 C . 1/n . 1/ . 2=. n/ if n is odd D D .1=n/ if n is even. n The series satisfies the conditions of the alternating series test, so if we truncate after the term for n D k 1, then the error will satisfy Z 1=2 2 1 X 2 nD1 . 1/n : 4nC1 2 .4n C 1/nŠ jerrorj The Fourier sine series of f .t / D coefficients Downloaded by ted cage (sxnbyln180@questza.com) ) an > a1 : n 387 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 9 (PAGE 567) ADAMS and ESSEX: CALCULUS 9 If x ¤ m for any integer m, then sin.x=2/ ¤ 0. Using the addition formulas we obtain i 1h sin.nx/ sin.x=2/ D cos .n C 21 /x : cos .n 12 /x 2 Therefore, using the telescoping property of these terms, h i N N cos .n 12 /x cos .n C 21 /x X X sin.nx/ D 2 sin.x=2/ nD1 nD1 cos.x=2/ cos .N C 21 /x D : 2 sin.x=2/ P Therefore, the partial sums of 1 nD1 sin.nx/ are bounded. Since the sequence f1=ng is positive, decreasing, and has P sin.nx/=n limit 0, part (b) of Problem 2 shows that 1 nD1 converges in this case too. Therefore the series converges for all x. (This can be P verified by induction.) Therefore 1 nD1 an diverges by comparison with the harP 1 monic series 1 nD1 . n P 2. a) If sn D nkD1 vk for n 1, and s0 D 0, then vk D sk sk 1 for k 1, and n X kD1 uk vk D n X n X uk sk uk sk 1 : kD1 kD1 In the second sum on the right replace k with k C 1: n X kD1 uk vk D D n X uk sk kD1 n X n X1 ukC1 sk kD0 .uk ukC1 /sk u1 s0 C unC1 sn n X ukC1 /sk : kD1 D unC1 sn C .uk 4. kD1 b) If fun g is positive and decreasing, and limn!1 un D 0, then n X ukC1 / D u1 .uk kD1 Thus D u1 n X u2 C u2 u3 C C un unC1 unC1 ! u1 as n ! 1. ukC1 / is a convergent, positive, tele- .uk scoping series. If the partial sums sn of fvn g are bounded, say jsn j K for all n, then j.un unC1 /sn j K.un unC1 /; P unC1 /sn is absolutely convergent (and so 1 nD1 .un therefore convergent) by the comparison test. Therefore, by part (a), kD1 Z kC1=2 nD1 mD1 f .x/ dx f .k/ D k 1=2 interval Œk kD1 1 X 5. Let an be the nth integer that has no zeros in its decimal representation. The number of such integers that have m digits is 9m . (There are nine possible choices for each of the m digits.) Also, each such m-digit number is greater than 10m 1 (the smallest m-digit number). Therefore the sum of all the terms 1=an for which an has m digits is less than 9m =.10m 1 /. Therefore, 1 1 X X 1 9 m 1 D 90: <9 an 10 uk vk D lim n!1 D 1 X .uk unC1 sn C n X .uk kD1 ukC1 /sk ! ukC1 /sk a) By the Mean-Value Theorem, f 0 k C 23 f 0 k C 21 D 23 f 00 .u/ D f 00 .u/ for some v in Œk 32 ; k 12 . Since f 00 is decreasing and v c u, we have f 00 .u/ f 00 .c/ f 00 .v/, and so 0 f k C 23 f 0 k C 12 f 00 .c/ f 0 k 12 f 0 k 32 : b) If f 00 is decreasing, 0 R1 1 f .x/ dx converges, and NC2 f .x/ ! 0 as x ! 1, then Z 1 1 X f .x/ dx f .n/ 1 D 3. If x DPm for some integer m, then all the terms of the series 1 nD1 .1=n/ sin.nx/ are 0, so the series converges to 0. Telegram: @uni_k 1 2 for some u in Œk C 12 ; k C 23 . Similarly, 1 C 23 f 00 .v/ D f 00 .v/ f 0 k 32 D f 0 k 12 2 converges. 388 1 1 2 ; k C 2 . nDN C1 kD1 f 00 .c/ , for some c in the 24 1 X nDN C1 D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 1 24 0 NC2 @f .n/ 1 X nDN C1 Z nC 1 2 1 n 2 f 00 .cn /; 1 f .x/ dx A lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 9 (PAGE 567) for some numbers cn in Œn 12 ; n C 21 . Using the result of part (a), we see that 1 X 0 f .n C 23 / nDN C1 1 X nDN C1 f 0 .n f 0 .N C 32 / f 0 .N 24 1 / 2 f 0 .n C 12 / 1 X nDN C1 1 X f 0 .n 1 2/ f 00 .cn / nDN C1 3 2/ f 00 .cn / f .n/ 1 X Z 1 j D0 is a difference of two integers and so is an integer. f 0 .N 1 2/ f .x/ dx 1 NC2 nDN C1 b) Suppose e is rational, say e D M=N where M and N are P positive integers. Then N Še is an integer and N Š jND0 .1=j Š/ is an integer (since each j Š is a factor of N Š). Therefore the number 1 0 N X 1 A Q D N Š @e jŠ f 0 .N C 32 / : 24 7. 1 1. By part (b), Q is c) By part (a), 0 < Q < N an integer. This is not possible; there are no integers between 0 and 1. Therefore e cannot be rational. 1 X 22k kŠ Let f .x/ D ak x 2kC1 , where ak D . .2k C 1/Š kD0 c) Let f .x/ D 1=x 2 . Then f 0 .x/ D 2=x 3 ! 0 as x ! 1, f 00 .x/ D 6=x 4 is decreasing, and Z 1 f .x/ dx D 1 NC2 Z 1 1 NC2 a) Since ˇ ˇ ˇa 2kC3 ˇ ˇ ˇ kC1 x lim ˇ ˇ k!1 ˇ ak x 2kC1 ˇ 1 dx D x2 N C 12 22kC2 .k C 1/Š .2k C 1/Š kŠ .2k C 3/Š k!1 22k 4k C 4 2 D jxj lim D0 k!1 .2k C 3/.2k C 2/ D jxj2 lim converges. From part (b) we obtain ˇ ˇ 1 ˇ X 1 ˇ ˇ 2 ˇnDN C1 n ˇ ˇ 1 ˇˇ 1 1ˇ N C 2 ˇ 12 N 1 3 2 for all x, the series for f .x/ converges for all x. Its radius of convergence is infinite. : b) f 0 .x/ D The right side is less than 0:001 if N D 5. Therefore 1 X 1 nD1 n2 D kD0 1 C 1:6454 n2 5:5 6. a) Since e D j D0 n X 1 1 , we have jŠ D1C D The last inequality follows from is, n2 C 2n < n2 C 2n C 1. 1 nC2 < , that .n C 1/2 n x 2kC2 kD1 1 X kD1 22k kŠ 2k x D f 0 .x/: .2k/Š 0 8. Let f be a polynomial and let g.x/ D 1 X . 1/j f .2j / .x/: j D0 This “series” is really just a polynomial since sufficiently high derivatives of f are all identically zero. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k kD0 .2k C 1/Š d x2 2 2 c) e f .x/ D e x f 0 .x/ 2xf .x/ D e x . dx d) Since f .0/ D 0, we have Z x Z x d t2 2 2 e x f .x/ f .0/ D e f .t / dt D e t dt dt 0 Z x 0 x2 t2 f .x/ D e e dt: 1 X 1 jŠ jŠ j DnC1 j D0 1 1 1 D 1C C C .n C 1/Š nC2 .n C 2/.n C 3/ 1 1 1 C C 1C .n C 1/Š nC2 .n C 2/2 1 1 1 nC2 D < : D 1 .n C 1/Š .n C 1/Š.n C 1/ nŠn 1 nC2 0<e 1 X 22kC1 kŠ (replace k with k 1) 1 X 22k 1 .k 1/Š 2k D1C x .2k 1/Š correct to within 0.001. 1 X kD1 1 C 2xf .x/ D 1 C 5 X 1 nD1 1 1 X X 22k kŠ 22k kŠ 2k .2k C 1/x 2k D 1 C x .2k C 1/Š .2k/Š Downloaded by ted cage (sxnbyln180@questza.com) 389 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 9 (PAGE 567) a) By replacing j with j g 00 .x/ D D 1 X j D0 1 X j D1 ADAMS and ESSEX: CALCULUS 9 This constant is an integer because the binomial coef k is an integer and i Š=kŠ is an integer. ficient i k (The other factors are also integers.) Hence f .i/ .0/ is an integer, and so g.0/ is an integer. 1, observe that . 1/j f .2j C2/ .x/ . 1/j 1 f .2j / .x/ D g.x/ f .x/ : e) Observe that f . x/ D f ..m=n/ x/ D f .x/ for all x. Therefore f .i/ ./ is an integer (for each i ), and so g./ is an integer. Thus g./ C g.0/ is an integer, which contradicts the conclusion of part (b). (There is no integer between 0 and 1.) Therefore, cannot be rational. Also d 0 g .x/ sin x g.x/ cos x dx D g 00 .x/ sin x C g 0 .x/ cos x g 0 .x/ cos x C g.x/ sin x D g 00 .x/ C g.x/ sin x D f .x/ sin x: 9. Let x > 0, and let Thus Z 0 ˇ ˇ ˇ g.x/ cos x ˇ D g./ C g.0/: ˇ f .x/ sin x dx D g 0 .x/ sin x f .x/ D x k .m nx/k kŠ k D 1 X kŠ j D0 k mk j . n/j x j Ck : j The sum is just the binomial expansion. For 0 < x < D m=n we have 0 < f .x/ < Thus 0 < R 0 f .x/ sin x dx < so 0 < g./ C g.0/ < 1. c) f .i/ .x/ D 1 k mk < : kŠ 2 k 1 X k kŠ j j D0 1 2 Z 0 sin x dx D 1, and d) Evidently f .i/ .0/ D 0 if i < k or if i > 2k. If k i 2k, the only term in the sum for f .i/ .0/ that is not zero is the term for which j D i k. This term is the constant 390 Telegram: @uni_k k i k iŠ mk j . n/j : 0Š .k C 2/IkC1 : Therefore, Z x e 1=t dt D I0 D x 2 e 1=x 2I1 D x 2 e 1=x 2 x 3 e 1=x 3I2 D e 1=x Œx 2 2Šx 3 C 3Š x 4 e 1=x 4I3 D e 1=x Œx 2 2Šx 3 C 3Šx 4 4Š x 5 e 1=x 0 5I4 :: : mk j . n/j j D0 1 kŠ 0 Ik D x kC2 e 1=x D e 1=x .j C k/.j C k 1/ .j C k i C 1/x j Ck i k .j C k/Š 1 X k mk j . n/j x j Ck i : D j kŠ .j C k i /Š t k e 1=t dt 0 1 d V D 2 e 1=t dt U D t kC2 t d U D .k C 2/t kC1 dt V D e 1=t ˇx Z x ˇ ˇ D t kC2 e 1=t ˇ t kC1 e 1=t dt .k C 2/ ˇ 0 0 b) Suppose that D m=n, where m and n are positive integers. Since limk!1 x k =kŠ D 0 for any x, there exists an integer k such that . m/k =kŠ < 1=2. Let Z x Ik D N X . 1/n .n nD2 C . 1/N C1 N Š 1/Šx n Z x t N 1 e 1=t dt: 0 The Maclaurin series for e 1=t does not exist. The function is not defined at t D 0. For x D 0:1 and N D 5, the approximation I D Z 0:1 0 e 1=t dt e 10 D e 10 .0:1/2 0:00836e 10 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 5 X . 1/n .n 1/Š.0:1/n nD2 2.0:1/3 C 6.0:1/4 24.0:1/5 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 9 (PAGE 567) has error E given by E D . 1/6 5Š Z 0:1 t 4 e 1=t dt: which is about 0:4% of the size of I . For N D 20, the error estimate is 0 Since e 1=t e 10 for 0 t 0:1, we have Z 0:1 t 4 dt 2:4 10 4 e 10 ; jEj 120e 10 0 jEj 20Še 10 Z 0:1 0 t 19 dt 1:2 10 3 e 10 ; which is about 3% of the size of I . For N D 10, the error estimate is Z 0:1 jEj 10Še 10 t 9 dt 3:6 10 5 e 10 ; 0 which is about 15% of the size of I . Observe, therefore, that the sum for N D 10 does a better job of approximating I than those for N D 5 or N D 20. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 391 lOMoARcPSD|6566483 www.konkur.in SECTION 10.1 (PAGE 574) ADAMS and ESSEX: CALCULUS 9 CHAPTER 10. VECTORS AND COORDINATE GEOMETRY IN 3-SPACE 8. If A D .1; 2; 3/, B D .1; 3; 4/, and C D .0; 3; 3/, then p .1 p jAC j D .0 p jBC j D .0 jABj D Section 10.1 Analytic Geometry in Three Dimensions (page 574) 1. 9. 22 C . 1/2 C . 2/2 D 3 units. 2. The distance between . 1; 1; 1/ and .1; 1; 1/ is p p .1 C 1/2 C .1 C 1/2 C .1 C 1/2 D 2 3 units: 3. The distance between .1; 1; 0/ and .0; 2; 2/ is p .0 1/2 C .2 1/2 C . 2 0/2 D p 5. 3/2 C .3 8/2 C . 6 C 1/2 D 5 1/2 C .3 1/2 C .3 p 3 units. a) The shortest distance from .x; y; z/ to the xy-plane is jzj units. 11. Since jABj2 C jAC j2 D 17 C 21 D 38 D jBC j2 , the triangle ABC has a right angle at A. p p 02 C 12 C 12 C C 12 D n 2: 1 units. z 2 zD2 y x Fig. 10.1-12 13. y 1 is the half-space consisting of all points on the plane y D 1 (which is perpendicular to the y-axis at .0; 1; 0/) and all points on the same side of that plane as the origin. z 2/2 C .1 C 1/2 C . 2 C 1/2 D 3 2/2 C . 3 C 1/2 C .1 C 1/2 D 3 p 0/2 C . 3 1/2 C .1 C 2/2 D 26: yD 1 1 By the Cosine Law, x a2 D b 2 C c 2 2bc cos †A 26 D 9 C 9 18 cos †A 26 18 116:4ı : †A D cos 1 18 Telegram: @uni_k 2 12. z D 2 is a plane, perpendicular to the z-axis at .0; 0; 2/. If A D .2; 1; 1/, B D .0; 1; 2/, and C D .1; 3; 1/, then 392 p 2 The point on the x1 -axis closest to .1; 1; 1; : : : ; 1/ is .1; 0; 0; : : : ; 0/. The distance between these points is p p jABj D 32 C . 2/2 C 22 D 17 p p jAC j D 22 C 42 C 12 D 21 p p jBC j D . 1/2 C 62 C . 1/2 D 38: p .0 p b D jAC j D .1 p a D jBC j D .1 4/2 D p p p 12 C 12 C 12 C C 1 D n units. 6. If A D .1; 2; 3/, B D .4; 0; 5/, and C D .3; 6; 4/, then c D jABj D 3/2 C .3 3/2 D p IfA D .1; 1; 0/, B D .1; 0; 1/, and C D .0; 1; 1/, then p jABj D jAC j D jBC j D 2: p Thus the triangle ABC is equilateral with sides 2. Its area is, therefore, r p 1 p 3 1 2 2 D sq. units. 2 2 2 6 units: b) The p shortest distance from .x; y; z/ to the x-axis is y 2 C z 2 units. 7. 2/2 C .3 3/2 D 10. The distance from the origin to .1; 1; 1; : : : ; 1/ in Rn is 4. The distance between .3; 8; 1/ and . 2; 3; 6/ is p . 2 2/2 C .4 All three sides being equal, the triangle is equilateral. The distance between .0; 0; 0/ and .2; 1; 2/ is p 1/2 C .3 y Fig. 10.1-13 14. z D x is a plane containing the y-axis and making 45ı angles with the positive directions of the x- and z-axes. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.1 z (PAGE 574) z .1;0;1/ x 2 Cz 2 D4 2 zDx y x y x Fig. 10.1-14 Fig. 10.1-20 x C y D 1 is a vertical plane (parallel to the z-axis) passing through the points .1; 0; 0/ and .0; 1; 0/. z D y 2 is a “parabolic cylinder” — a surface all of whose cross-sections in planes perpendicular to the x-axis are parabolas. z z 21. 15. xCyD1 zDy 2 1 y x 1 y x Fig. 10.1-15 22. 16. x 2 C y 2 C z 2 D 4 is a sphere centred at the origin and having radius 2 (i.e., all points at distance 2 from the origin). 17. .x 1/2 C .y C 2/2 C .z 3/2 D 4 is a sphere of radius 2 with centre at the point .1; 2; 3/. Fig. 10.1-21 p z x 2 C y 2 represents every point whose distance above the xy-plane is not less than its horizontal distance from the z-axis. It therefore consists of all points inside and on a circular cone with axis along the positive z-axis, vertex at the origin, and semi-vertical angle 45ı . z 18. x 2 C y 2 C z 2 D 2z can be rewritten x 2 C y 2 C .z 1/2 D 1; p 45ı zD and so it represents a sphere with radius 1 and centre at .0; 0; 1/. It is tangent to the xy-plane at the origin. x 2 Cy 2 y x Fig. 10.1-22 z x 2 Cy 2 Cz 2 D2z 23. .0;0;1/ x C 2y C 3z D 6 represents the plane that intersects the coordinate axes at the three points .6; 0; 0/, .0; 3; 0/, and .0; 0; 2/. Only the part of the plane in the first octant is shown in the figure. z .0;0;2/ y x Fig. 10.1-18 .0;3;0/ y 19. y 2 C z 2 4 represents all points inside and on the circular cylinder of radius 2 with central axis along the x-axis (a solid cylinder). 20. x 2 C z 2 D 4 is a circular cylindrical surface of radius 2 with axis along the y-axis. x Fig. 10.1-23 24. xD1 represents the vertical straight line in which the yD2 plane x D 1 intersects the plane y D 2. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k .6;0;0/ Downloaded by ted cage (sxnbyln180@questza.com) 393 lOMoARcPSD|6566483 www.konkur.in SECTION 10.1 (PAGE 574) ADAMS and ESSEX: CALCULUS 9 z 27. yD2 xD1 y x .1;2;0/ Fig. 10.1-24 28. 25. xD1 is the straight line in which the plane z D 1 yDz intersects the plane y D z. It passes through the points .1; 0; 0/ and .1; 1; 1/. x2 C y2 C z2 D 4 is the circle in which the sphere x 2 C y 2 C z 2 D 4z of radius 2 centred at the origin intersects the sphere of radius 2 centred at .0; 0; 2/. (The second equation can be rewritten x 2 C y 2 C .z 2/2 D 4 for easier recognition.) Subtracting the equations of the two spheres we get z D 1, so the circle must lie in the plane z D 1 as well. Thus it is the same circle as in the previous exercise. x 2 C y 2 C z 2 D 4 represents the two circles in x2 C z2 D 1 which the cylinder x 2 C z 2 1 intersects the sphere x 2 C y 2 C z 2 D 4. Subtracting the two equations, we p get y 2 D 3. Thus,pone circle lies in the plane y D 3 and haspcentre .0; 3; 0/ and the p other lies in the plane 3 and has centre .0; 3; 0/. Both circles have y D radius 1. z z xD1 .1;1;1/ 2 1 zDy p y 3 x .1;0;0/ x y Fig. 10.1-25 26. Fig. 10.1-28 29. x 2 C y 2 C z 2 D 4 is the circle in which the horizontal zD1 plane z D 1 intersects the sphere of radius 2 centred at the origin. Thepcircle has centre .0; 0; 1/ and radius p 4 1 D 3. x 2 C y 2 D 1 is the ellipse in which the slanted plane zDx z D x intersects the vertical cylinder x 2 C y 2 D 1. z x 2 Cy 2 D1 z zDx p 3 1 y zD1 .0;0;1/ x 2 y Fig. 10.1-29 x x 2 Cy 2 Cz 2 D4 Fig. 10.1-26 394 Telegram: @uni_k 30. yx is the quarter-space consisting of all points lying zy on or on the same side of the planes y D x and z D y as does the point .0; 1; 0/. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 31. SECTION 10.2 x2 C y2 1 represents all points which are inside or on zy the vertical cylinder x 2 C y 2 D 1, and are also above or on the plane z D y. z 35. S D f.x; y/ W x C y D 1g The boundary of S is S. The interior of S is the empty set. S is closed, but not bounded. There are points on the line x C y D 1 arbitrarily far away from the origin. 36. S D f.x; y/ W jxj C jyj 1g The boundary of S consists of all points on the edges of the square with vertices .˙1; 0/ and .0; ˙1/. The interior of S consists of all points inside that square. S is closed since it contains all its boundary points. It is bounded since all points in it are at distance not greater than 1 from the origin. zDy x y x 2 Cy 2 D1 Fig. 10.1-31 32. 37. S D f.x; y; z/ W 1 x 2 C y 2 C z 2 4g Boundary: the spheres of radii 1 and 2 centred at the origin. Interior: the region between these spheres. S is closed. 38. S D f.x; y; z/ W x 0; y > 1; z < 2g Boundary: the quarter planes x D 0; .y 1; z 2/, y D 1; .x 0; z 2/, and z D 2; .x 0; y 1/. Interior: the set of points .x; y; z/ such that x > 0, y > 1, z < 2. S is neither open nor closed. 39. S D f.x; y; z/ W .x z/2 C .y z/2 D 0g The boundary of S is S, that is, the line x D y D z. The interior of S is empty. S is closed. 40. S D f.x; y; z/ W x 2 C y 2 < 1; y C z > 2g Boundary: the part of the cylinder x 2 C y 2 D 1 that lies on or above the plane y C z D 2 together with the part of that plane that lies inside the cylinder. Interior: all points that are inside the cylinder x 2 C y 2 D 1 and above the plane y C z D 2. S is open. x2 C y2 C z2 1 p represents all points which are inx2 C y2 z side or on the sphere of radius 1 centred at the origin and which are also inside or on the upper half of the circular cone with axis along the z-axis, vertex at the origin, and semi-vertical angle 45ı . z p zD (PAGE 583) x 2 Cy 2 y Section 10.2 Vectors x x 2 Cy 2 Cz 2 D1 1. Fig. 10.1-32 34. S D f.x; y/ W x 0; y < 0g The boundary of S consists of points .x; 0/ where x 0, and points .0; y/ where y 0. The interior of S consists of all points of S that are not on the y-axis, that is, all points .x; y/ satisfying x > 0 and y < 0. S is neither open nor closed; it contains some, but not all, of its boundary points. S is not bounded; .x; 1/ belongs to S for 0 < x < 1. A D . 1; 2/; B D .2; 0/; C D .1; 3/; D D .0; 4/. ! ! (a) AB D 3i 2j (b) BA D 3i C 2j ! (c) AC D 2i 33. S D f.x; y/ W 0 < x 2 C y 2 < 1g The boundary of S consists of the origin and all points on the circle x 2 C y 2 D 1. The interior of S is S, which is therefore open. S is bounded; all points in it are at distance less than 1 from the origin. 5j ! (d) BD D 2i C 4j ! ! ! (e) DA D i 2j (f) AB BC D 4i C j ! ! ! (g) AC 2AB C 3CD D 7i C 20j 5 1 ! ! ! j AB C AC C AD D 2i (h) 3 3 2. uDi j v D j C 2k a) u C v D i C 2k u v D i 2j 2k 2u 3v D 2i 5j 6k Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k (page 583) Downloaded by ted cage (sxnbyln180@questza.com) 395 lOMoARcPSD|6566483 www.konkur.in SECTION 10.2 (PAGE 583) b) juj D jvj D p p 1C1D 1C4D p p ADAMS and ESSEX: CALCULUS 9 Thus MN is parallel to and half as long as BC . 2 C 5 1 c) uO D p .i j/ 2 1 vO D p .j C 2k/ 5 d) u v D 0 1 C 0 D N 1 A e) The angle between u and v is 1 cos 1 p 108:4ı . 10 f) The scalar projection of u in the direction of v is 1 uv D p . jvj 5 g) The vector projection of v along u is .v u/u 1 D .i j/. juj2 2 3. u D 3i C 4j v D 3i 4j 6. We have ! ! ! ! ! ! PQ D PB C BQ D 21 AB C 12 BC D 21 AC I ! ! ! ! ! ! SR D SD C DR D 12 AD C 12 DC D 21 AC : ! ! Therefore, PQ D SR. Similarly, u C v D 6i 10k u v D 8j 2u 3v D 3i C 20j C 5k p p b) juj D 9 C 16 C 25 D 5 2 p p jvj D 9 C 16 C 25 D 5 2 a) ! ! ! ! QR D QC C CR D 12 BDI ! ! ! ! PS D PA C AS D 21 BD: ! ! Therefore, QR D PS . Hence, PQRS is a parallelogram. 1 c) uO D p .3i C 4j 5k/ 5 2 1 vO D p .3i 4j 5k/ 5 2 d) u v D 9 16 C 25 D 18 B g) The vector projection of v along u is 9 .v u/u D .3i C 4j 5k/. juj2 25 4. If a D . 1; 1/, B D .2; 5/ and C D .10; 1/, then ! ! ! ! AB D 3i C 4j and BC D 8i 6j. Since AB BC D 0, ! ! therefore, AB ? BC . Hence, 4ABC has a right angle at B. A Telegram: @uni_k ! AB 2 D ! BC : 2 C R S D Fig. 10.2-6 7. Let the parallelogram be ABCO. Take the origin at O. The position vector of the midpoint of OB is ! ! ! ! ! OB C CB OC C OA OB D D : 2 2 2 The position vector of the midpoint of CA is ! ! ! ! CA ! OA OC D OC C OC C 2 2 ! ! OC C OA : D 2 5. Let the triangle be ABC . If M and N are the midpoints ! ! of AB and AC respectively, then AM D 12 AB, and ! ! AN D 12 AC . Thus 396 Q P e) The angle between u and v is 18 68:9ı . cos 1 50 f) The scalar projection of u in the direction of v is uv 18 D p . jvj 5 2 ! AC ! AM D B Fig. 10.2-5 5k 5k ! ! MN D AN M Thus the midpoints of the two diagonals coincide, and the diagonals bisect each other. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.2 9. B C (PAGE 583) Let i point east and j point north. Let the wind velocity be vwind D ai C bj: Now vwind D vwind rel car C vcar . When vcar D 50j, the wind appears to come from the west, so vwind rel car D i. Thus ai C bj D i C 50j; A O Fig. 10.2-7 8. Let X be the point of intersection of the medians AQ and BP as shown. We must show that CX meets AB in the ! ! ! ! midpoint of AB. Note that PX D ˛ PB and QX D ˇ QA for certain real numbers ˛ and ˇ. Then 1 ! 1 ! 1 ! ! ! ! CX D CB C ˇ QA D CB C ˇ CB C BA 2 2 2 1Cˇ ! ! D CB C ˇ BAI 2 1 ! 1 ! 1 ! ! ! ! CX D CA C ˛ PB D CA C ˛ CA C AB 2 2 2 1C˛ ! ! CA C ˛ AB: D 2 so a D and b D 50. When vcar D 100j, the wind appears to come from the northwest, so vwind rel car D .i-j). Thus ai C bj D .i so a D and b D 100 . Hence 50 D 100 , so D 50.pThus a D b D 50. The wind is from the southwest at 50 2 km/h. 10. Let the x-axis point east and the y-axis north. The velocity of the water is vwater D 3i: If you row through the water with speed 5 in the direction making angle west of north, then your velocity relative to the water will be vboat rel water D Thus, 1Cˇ ! 1C˛ ! ! ! CB C ˇ BA D CA C ˛ AB 2 2 1C˛ ! 1Cˇ ! ! .ˇ C ˛/BA D CA CB 2 2 1C˛ ! 1Cˇ ! ! ! .ˇ C ˛/.CA CB/ D CA CB 2 2 1Cˇ ! 1C˛ ! CA D ˇ C ˛ CB: ˇC˛ 2 2 ! ! Since CA is not parallel to CB, 1C˛ DˇC˛ 2 1 )˛DˇD : 3 ˇC˛ 1Cˇ D0 2 j/ C 100j; 5 sin i C 5 cos j: Therefore, your velocity relative to the land will be vboat rel land D vboat rel water C vwater D .3 5 sin /i C 5 cos j: To make progress in the direction j, choose so that 3 D 5 sin . Thus D sin 1 .3=5/ 36:87ı . In this case, your actual speed relative to the land will be 4 5 cos D 5 D 4 km/h. 5 To row from A to B, head in the direction 36:87ı west of north. The 1=2 km crossing will take .1=2/=4 D 1=8 of an hour, or about 7 12 minutes. B Since ˛ D ˇ, x divides AQ and BP in the same ratio. By symmetry, the third median CM must also divide the other two in this ratio, and so must pass through X and MX D 31 M C . vboat rel water A M vwater P X B i A Q C Fig. 10.2-10 Fig. 10.2-8 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k j Downloaded by ted cage (sxnbyln180@questza.com) 397 lOMoARcPSD|6566483 www.konkur.in SECTION 10.2 (PAGE 583) 11. ADAMS and ESSEX: CALCULUS 9 If this velocity is true easterly, then We use the notations of the solution to Exercise 4. You 1 now want to make progress in the direction ki C j, that 2 is, in the direction making angle D tan 1 100 750 sin D p ; 2 1 2k so 5:41ı . The speed relative to the ground is with vector i. Head at angle upstream of this direction. Since your rowing speed is 2, the triangle with angles and has sides 2 and 3 as shown in the figure. By the 3 2 Sine Law, D , so sin sin 100 p 675:9 km/h: 2 750 cos The time for the 1500 km trip is 1500 2:22 hours. 675:9 y 3 3 1 3 : D p sin D sin D q 2 2 2 k2 C 1 2 4k 2 C 1 4 3 This is only possible if p 1, that is, if 2 4k 2 C 1 p 5 . k 4 3 Head in the direction D sin 1 p upstream of 2 4k 2 C 1 the direction of AC p , as shown in the figure. The trip is not possible if k < 5=4. B 750 x 100 C k Fig. 10.2-12 13. 3 2 1 2 The two vectors are perpendicular if their dot product is zero: .2t i C 4j .10 C t /k/ .i C t j C k/ D 0 2t C 4t 10 t D 0 ) t D 2: 3i The vectors are perpendicular if t D 2. 14. A Fig. 10.2-11 The cube with edges i, j, and k has diagonal iCjCk. The angle between i and the diagonal is cos 1 12. Let i point east and j point north. If the aircraft heads in a direction north of east, then its velocity relative to the air is 750 cos i C 750 sin j: i .i C j C k/ 1 p D cos 1 p 54:7ı : 3 3 z The velocity of the air relative to the ground is 100 p iC 2 100 p j: 2 Thus the velocity of the aircraft relative to the ground is 750 cos 398 Telegram: @uni_k 100 p i C 750 sin 2 100 p j: 2 x Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) iCjCk y i Fig. 10.2-14 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 15. SECTION 10.2 The cube of Exercise 10 has six faces, each with 2 diagonals. The angle between i C j C k and the face diagonal i C j is cos 1 2 .i C j/ .i C j C k/ D cos 1 p 35:26ı : p p 2 3 6 20. If a ¤ 0, then a r D 0 implies that the position vector r is perpenducular to a. Thus the equation is satisfied by all points on the plane through the origin that is normal (perpendicular) to a. 21. If r a D b, then the vector projection of r along a is the constant vector ra a b D 2 a D r0 ; say. jaj jaj jaj Six of the face diagonals make this angle with i C j C k. The face diagonal i j (and five others) make angle cos Thus r a D b is satisfied by all points on the plane through r0 that is normal to a. j/ .i C j C k/ D cos 1 0 D 90ı p p 2 3 1 .i with the cube diagonal i C j C k. In Exercises 22–24, u D 2i C j w D 2i 2j C k. 22. ui u1 16. If u D u1 i C u2 j C u3 k, then cos ˛ D . juj juj u3 u2 and cos D . Similarly, cos ˇ D juj juj Thus, the unit vector in the direction of u is uO D 17. O 2 D 1. D juj If uO makes equal angles ˛ D ˇ D with p the coordinate axes, then 3 cos2 ˛ D 1, and cos ˛ D 1= 3. Thus uO D 23. Let x D xi C yj C zk. Then xuD9 xvD4 xw D 6 24. †ABC D cos 1 2x C y 2z D 9 x C 2y 2z D 4 2x 2y C z D 6: Since u, v, and w all have the same length (3), a vector x D xi C yj C zk will make equal angles with all three if it has equal dot products with all three, that is, if 2x C y 2x C y 2z D x C 2y 2z 2z D 2x 2y C z , , xDyD0 3y 3z D 0: Thus x D y D z. Two unit vectors satisfying this condition are 1 1 1 xD˙ p iC p jC p k : 3 3 3 25. Let uO D u=juj and vO D v=jvj. Then uO C vO bisects the angle between u and v. A unit vector which bisects this angle is u v C uO C vO juj jvj ˇ D ˇ ˇ u v ˇˇ juO C vO j ˇ C ˇ juj jvj ˇ jvju C jujv ˇ: D ˇ ˇ ˇ ˇjvju C jujvˇ Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k , , , This system of linear equations has solution x D 2, y D 3, z D 4. Thus x D 2i 3j 4k. 18. If A D .1; 0; 0/, B D .0; 2; 0/, and C D .0; 0; 3/, then 19. Since r r1 D r1 C .1 /r2 r1 D .1 /.r1 r2 /, therefore r r1 is parallel to r1 r2 , that is, parallel to the line P1 P2 . Since P1 is on that line, so must P be on it. 1 1 If D , then r D .r1 C r2 /, so P is midway between 2 2 P1 and P2 . 2 1 2 If D , then r D r1 C r2 , so P is two-thirds of the 3 3 3 way from P2 towards P1 along the line. If D 1, the r D r1 C 2r2 D r2 C .r2 r1 /, so P is such that P2 bisects the segment P1 P . If D 2, then r D 2r1 r2 D r1 C .r1 r2 /, so P is such that P1 bisects the segment P2 P . 2k, and Vector x D xi C yj C zk is perpendicular to both u and v if u x D 0 , 2x C y 2z D 0 v x D 0 , x C 2y 2z D 0: iCjCk : p 3 ! ! BA BC 4 D cos 1 p p 60:26ı jBAjjBC j 5 13 ! ! CB CA 9 †BCA D cos 1 D cos 1 p p 37:87ı jCBjjCAj 10 13 ! ! 1 AC AB D cos 1 p p 81:87ı : †CAB D cos 1 jAC jjABj 10 5 2k, v D i C 2j Subtracting these equations, we get x y D 0, so x D y. The first equation now gives 3x D 2z. Now x is a unit vector if x 2 C y 2 C z 2 D 1, that is, if x 2 C x 2 C 49 x 2 D 1, p or x D ˙2= 17. The two unit vectors are 2 3 2 xD˙ p iC p jC p k : 17 17 17 u D cos ˛i C cos ˇj C cos k; juj and so cos2 ˛ C cos2 ˇ C cos2 (PAGE 583) Downloaded by ted cage (sxnbyln180@questza.com) 399 lOMoARcPSD|6566483 www.konkur.in SECTION 10.2 (PAGE 583) ADAMS and ESSEX: CALCULUS 9 Similarly, r v D b and r w D c. u uO 31. uO C vO vO Let u D vDw and v wa a, (the vector projection of w along a). Let jaj2 u. Then w D u C v. Clearly u is parallel to a, wa aaDwa jaj2 vaDwa Fig. 10.2-25 w a D 0; so v is perpendicular to a. 26. If u and v are not parallel, then neither is the zero vector, and the origin and the two points with position vectors u and v lie on a unique plane. The equation r D u C v (; real) gives the position vector of an arbitrary point on that plane. 27. w v a 2 a) ju C vj D .u C v/ .u C v/ DuuCuvCvuCvv u D juj2 C 2u v C jvj2 : Fig. 10.2-31 b) If is the angle between u and v, then cos 1, so u v D jujjvj cos jujjvj: 32. c) ju C vj2 D juj2 C 2u v C jvj2 juj2 C 2jujjvj C jvj2 Let nO be a unit vector that is perpendicular to u and lies in the plane containing the origin and the points U , V , and P . Then uO D u=juj and nO constitute a standard basis in that plane, so each of the vectors v and r can be expressed in terms of them: D .juj C jvj/2 : Thus ju C vj juj C jvj. 28. v D s uO C t nO O r D x uO C y n: a) u, v, and u C v are the sides of a triangle. The triangle inequality says that the length of one side cannot exceed the sum of the lengths of the other two sides. Since v is not parallel to u, we have t ¤ 0. Thus O and nO D .1=t /.v s u/ b) If u and v are parallel and point in the same direction, (or if at least one of them is the zero vector), then ju C vj D juj C jvj. 29. u D 53 i C 45 j, v D 54 i 35 j, w D k. q q 9 16 9 a) juj D 25 C 16 D 1, jvj D 25 25 C 25 D 1, jwj D 1, u v D 12 25 12 25 D 0, u w D 0, v w D 0. r D x uO C where D .tx 33. Let jaj2 30. Suppose juj D jvj D jwj D 1, and uv D uw D vw D 0, and let r D au C bv C ww. Then Telegram: @uni_k ys/=.t juj/ and D y=t . 4rst D K 2 , where K > 0. Now D r 2 jxj2 C s 2 jyj2 C 2rsx y K 2 D jaj2 D jrx 4rsx y syj2 (since x y D t ). O for some unit vector u. O Therefore rx sy D K u, Since rx C sy D a, we have 2rx D a C K uO O 2sy D a K u: Thus r u D au u C bv u C cw u D ajuj2 C 0 C 0 D a: 400 O D u C v; s u/ jaj2 D a a D .rx C sy/ .rx C sy/ b) If r D xi C yj C zk, then .r u/u C .r v/v C .r w/w 4 3 4 3 xC y iC j D 5 5 5 5 4 4 3 3 C x y i j C zk 5 5 5 5 16y C 9y 9x C 16x iC j C zk D 25 25 D xi C yj C zk D r: y .v t xD Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) a C K uO ; 2r yD a K uO ; 2s lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.3 p where K D jaj2 4rst, and uO is any unit vector. (The solution is not unique.) (PAGE 592) The length of the cable between x D 0 and x D 10 m is Z 10 q LD 1 C sinh2 .ax/ dx 0 ˇ10 Z 10 ˇ 1 ˇ D cosh.ax/ dx D sinh.ax/ˇ ˇ a 0 34. The derivation of the equation of the hanging cable given in the text needs to be modified by replacing W D ıgsj with W D ıgxj. Thus Tv D ıgx, and the slope of the cable satisfies dy ıgx D D ax dx H 0 1 D sinh.10a/ 12:371 m: a where a D ıg=H . Thus yD 1 2 ax C C I 2 Section 10.3 The Cross Product in 3-Space (page 592) the cable hangs in a parabola. 1 35. If y D cosh.ax/, then y 0 D sinh.ax/, so a Z xq Z x sD 1 C sinh2 .au/ du D cosh.au/ du 0 0 ˇx sinh.au/ ˇˇ 1 D ˇ D sinh.ax/: ˇ a a jTj D q 2. .j C 2k/ . i 3. If A D .1; 2; 0/, B D .1; 0; 2/, and C D .0; 3; 1/, then ! ! AB D 2j C 2k, AC D i C j C k, and the area of triangle ABC is ! ! j jAB AC j D 2 4. The equation sinh.45a/ D 50a has approximate solution a 0:0178541. The vertical distance between the lowest point on the cable and the support point is 1 cosh.45a/ a 1 19:07 m: 1 cosh.ax/. a At the point P where x D 10 m, the slope of the cable is sinh.10a/ D tan.55ı /. Thus 37. The equation of the cable is of the form y D 2j 2 2kj D p 6 sq. units. A vector perpendicular to the plane containing the three given points is A unit vector in this direction is p bci C acj C abk b 2 c 2 C a2 c 2 C a2 b 2 : 1p 2 2 b c C a2 c 2 C a2 b 2 . 2 A vector perpendicular to i C j and j C 2k is The triangle has area 5. ˙.i C j/ .j C 2k/ D ˙.2i 2j C k/; which has length 3. A unit vector in that direction is 2 2 1 ˙ i jC k : 3 3 3 6. A vector perpendicular to u D 2i j 2k and to v D 2i 3j C k is the cross product ˇ ˇ ˇi j k ˇˇ ˇ 1 2 ˇˇ D 7i 6j 4k; u v D ˇˇ 2 ˇ2 3 1 ˇ p which has length 101. A unit vector with positive k component that is perpenducular to u and v is 1 sinh 1 .tan.55ı / 0:115423: 10 p 1 101 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 4i 2j C k . ai C bj/ . ai C ck/ D bci C acj C abk: Z 45 q 1 50 D 1 C sinh2 .ax/ dx D sinh.45a/: a 0 aD j C k/ D 3i ıg cosh.ax/ D ıgy: Th2 C Tv2 D a 1 36. The cable hangs along the curve y D cosh.ax/, and its a length from the lowest point at x D 0 to the support tower at x D 45 m is 50 m. Thus 4k/ D 5i C 13j C 7k .i 0 As shown in the text, the tension T at P has horizontal ıg and vertical components that satisfy Th D H D and a ıg Tv D ıgs D sinh.ax/. Hence a 2j C 3k/ .3i C j 1. Downloaded by ted cage (sxnbyln180@questza.com) uv D p 1 101 .7i C 6j C 4k/: 401 lOMoARcPSD|6566483 www.konkur.in SECTION 10.3 (PAGE 592) ADAMS and ESSEX: CALCULUS 9 Since u makes zero angle with itself, ju uj D 0 and u u D 0. ˇ ˇ ˇ i j k ˇˇ ˇ 8. u v D ˇˇ u1 u2 u3 ˇˇ ˇ v1 v2 v3 ˇ ˇ ˇ ˇ i j k ˇˇ ˇ D ˇˇ v1 v2 v3 ˇˇ D v u: ˇ u1 u2 u3 ˇ 7. 9. 10. Thus u v C w v D 0. Thus u v D w v D v w. By symmetry, we also have v w D w u. 14. u .u v/ ˇ ˇ ˇ u2 u3 ˇ ˇ D u1 ˇˇ v2 v3 ˇ ˇ ˇ u1 u2 ˇˇ v1 ˇ ˇ ˇ u1 u3 ˇˇ ˇ C u 3 ˇ v1 v3 ˇ hD V D ˇ k ˇˇ 0 ˇˇ D .sin ˛ cos ˇ 0ˇ u h h ˇ/ D sin ˛ cos ˇ cos ˛ sin ˇ: w v Fig. 10.3-14 15. cos ˛ sin ˇ/k: Because v is displaced counterclockwise from u, the cross product above must be in the positive k direction. Therefore its length is the k component. Therefore sin.˛ 1 1 Ah D ju .v w/j 3 6 ˇ ˇ ˇu u2 u3 ˇˇ 1 ˇˇ 1 D j ˇ v1 v2 v3 ˇˇ j: 6 ˇw w w ˇ 1 2 3 uv ˇ u2 ˇˇ v2 ˇ But j sin ˇ sin ˛ ju .v w/j : jv wj The volume of the tetrahedron is 12. Both u D cos ˇ i C sin ˇ j and v D cos ˛ i C sin ˛ j are unit vectors. They make angles ˇ and ˛, respectively, with the positive x-axis, so the angle between them is j˛ ˇj D ˛ ˇ, since we are told that 0 ˛ ˇ . They span a parallelogram (actually a rhombus) having area ju vj D jujjvj sin.˛ ˇ/ D sin.˛ ˇ/: ˇ ˇ i ˇ u v D ˇˇ cos ˇ ˇ cos ˛ 1 jv wj: 2 The altitude h of the tetrahedron (measured perpendicular to the plane of the base) is equal to the length of the projection of u onto the vector v w (which is perpendicular to the base). Thus .t v/ u t .v u/ D t .u v/: D u1 u2 v3 u1 v2 u3 u2 u1 v3 C u2 v1 u3 C u3 u1 v2 u3 v1 u2 D 0; v .u v/ D v .v u/ D 0: Telegram: @uni_k The base of the tetrahedron is a triangle spanned by v and w, which has area AD D u w C v w: ˇ ˇ ˇ i j k ˇˇ ˇ .t u/ v D ˇˇ t u1 t u2 t u3 ˇˇ ˇ v1 v2 v3 ˇ ˇ ˇ ˇ i j k ˇˇ ˇ D t ˇˇ u1 u2 u3 ˇˇ D t .u v/; ˇ v1 v2 v3 ˇ 402 Suppose that u C v C w D 0. Then u v C v v C w v D 0 v D 0: ˇ ˇ ˇ ˇ i j k ˇ ˇ .u C v/ w D ˇˇ u1 C v1 u2 C v2 u3 C v3 ˇˇ ˇ w1 w2 w3 ˇ ˇ ˇ ˇ ˇ ˇ i j k ˇˇ ˇˇ i j k ˇˇ ˇ D ˇˇ u1 u2 u3 ˇˇ C ˇˇ v1 v2 v3 ˇˇ ˇ w1 w2 w3 ˇ ˇ w1 w2 w3 ˇ u .t v/ D D 11. 13. 16. The tetrahedron with vertices .1; 0; 0/, .1; 2; 0/, .2; 2; 2/, and .0; 3; 2/ is spanned by u D 2j, v D i C 2j C 2k, and w D i C 3j C 2k. By Exercise 14, its volume is ˇ ˇ ˇ 0 2 0ˇ ˇ 4 1 ˇˇ V D j ˇ 1 2 2 ˇˇ j D cu. units. 6 ˇ 1 3 2ˇ 3 Let the cube be as shown in the figure. The required parallelepiped is spanned by ai C aj, aj C ak, and ai C ak. Its volume is ˇ ˇ ˇa a 0ˇ ˇ ˇ V D j ˇˇ 0 a a ˇˇ j D 2a3 cu. units. ˇa 0 aˇ Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.3 (PAGE 592) z 20. .0;a;a/ If v w ¤ 0, then .v w/ .v w/ ¤ 0. By the previous exercise, there exist constants , and such that .a;0;a/ u D v C w C .v w/: But v w is perpendicular to both v and w, so u .v w/ D 0 C 0 C .v w/ .v w/: If u .v w/ D 0, then D 0, and y x u D v C w: .a;a;0/ Fig. 10.3-16 17. The points A D .1; 1; 1/, B D .0; 3; 2/, C D . 2; 1; 0/, ! ! ! and D D .k; 0; 2/ are coplanar if .AB AC / AD D 0. Now ˇ ˇ i j ! ! ˇˇ 1 2 AB AC D ˇ ˇ 3 0 ˇ k ˇˇ 1 ˇˇ D 2i C 4j C 6k: 1 ˇ Thus the four points are coplanar if 2.k that is, if k D 18. 1/ C 4.0 21. u D i C 2j C 3k v D 2i 3j wDj k u .v w/ D u .3i C 2j C 2k/ D 2i C 7j 4k .u v/ w D .9i C 6j 7k/ w D i C 9j C 9k: u .v w/ lies in the plane of v and w; .u v/ w lies in the plane of u and v. 22. u v w makes sense in that it must mean u .v w/. (.u v/ w makes no sense since it is the cross product of a scalar and a vector.) 1/ C 6.2 C 1/ D 0; uvw makes no sense. It is ambiguous, since .uv/w and u .v w/ are not in general equal. 6. 23. ˇ ˇ ˇ u1 u2 u3 ˇ ˇ ˇ u .v w/ D ˇˇ v1 v2 v3 ˇˇ ˇ w1 w2 w3 ˇ ˇ ˇ ˇ v1 v2 v3 ˇ ˇ ˇ D ˇˇ u1 u2 u3 ˇˇ ˇ w1 w2 w3 ˇ ˇ ˇ ˇ v1 v2 v3 ˇ ˇ ˇ D ˇˇ w1 w2 w3 ˇˇ ˇ u1 u2 u3 ˇ D v .w u/ D w .u v/ v D v1 i; u .v w/ D .u1 i C u2 j C u3 k/ .v1 w2 k/ D u1 v1 w2 i k C u2 v1 w2 j k D u1 v1 w2 i u1 v1 w2 j: But .by symmetry/: x .v w/ D u .v w/ C v .v w/ C w .v w/ D u .v w/: D x .v w/ : u .v w/ Since u .v w/ D v .w u/ D w .u v/, we have, by symmetry, D x .w u/ ; u .v w/ D .u w/v .u v/w D .u1 w1 C u2 w2 /v1 i u1 v1 .w1 i C w2 j/ D u2 v1 w2 i u1 v1 w2 j: x .u v/ : u .v w/ Thus u .v w/ D .u w/v .u v/w. 24. If u, v, and w are mutually perpendicular, then v w is parallel to u, so u .v w/ D 0. In this case, u .v w/ D ˙jujjvjjwj; the sign depends on whether u and v w are in the same or opposite directions. 25. Applying the result of Exercise 23 three times, we obtain u .v w/ C v .w u/ C w .u v/ D .u w/v .u v/w C .v u/w .v w/u C .w v/u .w u/v D 0: Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k w D w1 i C w2 j: Thus v w D v1 w2 i j D v1 w2 k, and 19. If u .v w/ ¤ 0, and x D u C v C w, then Thus As suggested in the hint, let the x-axis lie in the direction of v, and let the y-axis be such that w lies in the xyplane. Thus Downloaded by ted cage (sxnbyln180@questza.com) 403 lOMoARcPSD|6566483 www.konkur.in SECTION 10.3 (PAGE 592) 26. If a D ADAMS and ESSEX: CALCULUS 9 i C 2j C 3k and x D xi C yj C zk, then ˇ ˇ i ˇ a x D ˇˇ 1 ˇ x ˇ k ˇˇ 3 ˇˇ zˇ j 2 y D .2z 3y/i C .3x C z/y D i C 5j 3k; .y C 2x/k provided 2z 3y D 1, 3x C z D 5, and This system is satisfied by x D t , y D 3 for any real number t . Thus x D t i C .3 2t /j C .5 y 2x D 2t , z D 5 The plane through the origin having normal i equation x y C 2z D 0. 4. The plane passing through .1; 2; 3/, parallel to the plane 3x C y 2z D 15, has equation 3z C y 2z D 3 C 2 6, or 3x C y 2z D 1. 5. The plane through .1; 1; 0/, .2; 0; 2/, and .0; 3; 3/ has normal .i j C 2k/ .i 2j 3k/ D 7i C 5j k: It therefore has equation 3. 3t , or 7x C 5y 3t /k Let a D i C 2j C 3k and b D i C 5j. If x is a solution of a x D b, then 0/ D 0; .z z D 12. The plane passing through . 2; 0; 0/, .0; 3; 0/, and .0; 0; 4/ has equation y z x C C D 1; 2 3 4 or 6x 4y 3z D 12. 7. The normal n to a plane through .1; 1; 1/ and .2; 0; 3/ must be perpendicular to the vector i j C 2k joining these points. If the plane is perpendicular to the plane x C 2y 3z D 0, then n must also be perpendicular to i C 2j 3k, the normal to this latter plane. Hence we can use However, a b ¤ 0, so there can be no such solution x. 28. The equation a x D b can be solved for x if and only if a b D 0. The “only if” part is demonstrated in the previous solution. For the “if” part, observe that if a b D 0 and x0 D .b a/=jaj2 , then by Exercise 23, n D .i j C 2k/ .i C 2j 3k/ D i C 5j C 3k: The plane has equation .a a/b .a b/a 1 a .b a/ D D b: 2 jaj jaj2 The solution x0 is not unique; as suggested by the example in Exercise 26, any multiple of a can be added to it and the result will still be a solution. If x D x0 C t a, then 1/ 6. a b D a .a x/ D 0: a x0 D 1/ C 5.y 7.x gives a solution of a x D i C 5j 3k for any t . These solutions constitute a line parallel to a. 27. j C 2k has 3. or x 8. a x D a x0 C t a a D b C 0 D b: 5y .x 1/ C 5.y 3z D 7. 1/ C 3.z 1/ D 0; Since . 2; 0; 1/ does not lie on x 4y C 2z D required plane will have an equation of the form 2x C 3y z C .x 5, the 4y C 2z C 5/ D 0 for some . Thus Section 10.4 Planes and Lines 1. (page 599) a) x 2 Cy 2 Cz 2 D z 2 represents a line in 3-space, namely the z-axis. b) x C y C z D x C y C z is satisfied by every point in 3-space. c) x 2 C y 2 C z 2 D 3-space. 4.x or 4x 404 y 0/ .y 2/ j 2k has 2 C 5/ D 0; so D 3. The required plane is 5x 9. xCy 2 C .y 15. z 3/ D 0: This plane will be perpendicular to 2x C 3y C 4z D 5 if .2/.1/ C .1 C /.3/ 2.z C 3/ D 0; 9y C 5z D A plane through the line x Cy D 2, y z D 3 has equation of the form 1 is satisfied by no points in (real) 2. The plane through .0; 2; 3/ normal to 4i equation Telegram: @uni_k 4 C 1 C . 2 ./.4/ D 0; that is, if D 5. The equation of the required plane is x C 6y 2z D 4. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 5z D 17: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.4 10. Three distinct points will not determine a unique plane through them if they all lie on a straight line. If the points have position vectors r1 , r2 , and r3 , then they will all lie on a straight line if r1 / .r3 .r2 11. r1 / D 0: or in standard form y z 1 xC1 D D : 2 1 7 17. h r1 / .r3 .i C 2j (or if they satisfy any similar such condition that asserts that the tetrahedron whose vertices they are has zero volume). 14. The distance from the planes x C p p r D .1 C 2t /i C .2 3t /j C .3 6j 5k: (vector parametric) (scalar parametric) 5t (standard form). j C 2k/ D 2.i j k/: Since the line passes through .2; 1; 1/, its equations are, in vector parametric form r D .2 C t /i 4k has .1 C t /j .1 C t /k; or in scalar parametric form 4t /k; x D 2 C t; or in scalar parametric form by x D 1 C 2t; zD .i C j/ .i The line through .1; 2; 3/ parallel to 2i 3j equations given in vector parametric form by j C 4k/ D 7i 18. A line parallel to x C y D 0 and to x y C 2z D 0 is parallel to the cross product of the normal vectors to these two planes, that is, to the vector to the origin is 1= 2 C 1 2 D 1. Hence the equation represents the family of all vertical planes at distance 1 from the origin. All such planes are tangent to the cylinder x 2 C y 2 D 1. 15. k/ .2i r D 7t i 6t j 5t k x D 7t; y D 6t; y z x D D 7 6 5 2 y D 1 1 y C 4z D 5 2x Since the line passes through the origin, it has equations 12. x C y C z D is the family of all (parallel) planes normal to the vector i C j C k. 13. x C y C z D is the family of all planes containing the line of intersection of the planes x D 0 and y C z D 1, except the plane y C z D 1 itself. All these planes pass through the points .0; 1; 0/ and .0; 0; 1/. z D 2; is parallel to the vector i r1 / D 0 r1 / .r4 A line parallel to the line with equations x C 2y If the four points have position vectors ri , .1 i 4/, then they are coplanar if, for example, .r2 (PAGE 599) yD .1 C t /; zD .1 C t /; .y C 1/ D .z C 1/: or in standard form yD2 zD3 3t; 4t; x 2D or in standard form by x 1 2 D y 2 z 3 19. 3 : 4 16. The line through . 1; 0; 1/ perpendicular to the plane 2x y C 7z D 12 is parallel to the normal vector 2i j C 7k to that plane. The equations of the line are, in vector parametric form, r D . 1 C 2t /i xD 1 C 2t; yD r D .1 C t /i C .2 C t /j C . 1 C t /k (vector parametric) x D 1 C t; y D 2 C t; z D 1 C t (scalar parametric) x 1Dy 2DzC1 (standard form). t j C .1 C 7t /k; 20. or in scalar parametric form, t; A line making equal angles with the positive directions of the coordinate axes is parallel to the vector i C j C k. If the line passes through the point .1; 2; 1/, then it has equations The line r D .1 form z D 1 C 7t; Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 2t /i C .4 C 3t /j C .9 x 1 2 D y 4 3 D z 4t /k has standard 9 : 4 405 lOMoARcPSD|6566483 www.konkur.in SECTION 10.4 (PAGE 599) 21. The line ( x D 4 5t has standard form y D 3t zD7 x 4 5 22. The line .i ADAMS and ESSEX: CALCULUS 9 D y ; 3 26. p z D 7: 27. x 2y C 3z D 0 is parallel to the vector 2x C 3y 4z D 4 2j C 3k/ .2i C 3j 4k/ D 2y D 0; The distance from .1; 2; 0/ to 3x 28. i C 10j C 7k: 1 8 y 7 D a D .i C j C k/ .2i 29. x1 / y1 / z1 / .r2 .r3 r1 / .r4 r3 / ¤ 0; and h i r1 / .r2 r1 / .r4 r3 / D 0: Telegram: @uni_k 1 j: 3 a1 D 4i a2 D 2i 2j C k 3j C k; we calculate the distance between the two lines by the formula in Section 10.4 as sD D j.r1 j.2i r2 / .a1 a2 /j ja1 a2 j 4j k/ .i 2j ji 2j 8kj 8k/j 18 D p units. 69 30. y C3 z 1 D passes through the point 2 4 .2; 3; 1/, and is parallel to a D i C 2j C 4k. The plane 2y z D 1 has normal n D 2j k. Since a n D 0, the line is parallel to the plane. The distance from the line to the plane is equal to the distance from .2; 3; 1/ to the plane 2y z D 1, so is The line x (Other similar answers are possible.) 406 3k: x C 2y D 3 contains the points .1; 1; 1/ and y C 2z D 3 1 .3; 0; 3=2/, so is parallel to the vector 2i j C k, or to 2 4i 2j Cnk. xCy Cz D6 The line contains the points . 5; 11; 0/ x 2z D 5 and . 1; 5; 2/, and so is parallel to the vector 4i 6j C 2k, or to 2i 3j C k. Using the values The line The point on the line corresponding to t D 1 is the point P3 such that P1 is midway between P3 and P2 . The point on the line corresponding to t D 1=2 is the midpoint between P1 and P2 . The point on the line corresponding to t D 2 is the point P4 such that P2 is the midpoint between P1 and P4 . 25. Let ri be the position vector of Pi (1 i 4). The line P1 P2 intersects the line P3 P4 in a unique point if the four points are coplanar, and P1 P2 is not parallel to P3 P4 . It is therefore sufficient that 1 i 3 r1 D i C j C k r2 D i C 5j C 2k certainly represent a straight line. Since .x; y; z/ D .x1 ; y1 ; z1 / if t D 0, and .x; y; z/ D .x2 ; y2 ; z2 / if t D 1, the line must pass through P1 and P2 . 24. 4i C 7j The distance from the origin to the line is r jr1 aj ji C j C kj 3 D sD D units. p jaj 74 74 23. The equations x D x1 C t .x2 ˆ ˆ ˆ ˆ y D y1 C t .y2 :̂ z D z1 C t .z2 5k/ D j r1 D 4 7 D z; 10 7 8̂ ˆ ˆ ˆ ˆ < 5z D 2 is We need a point on this line: if z D 0 then x C y D 0 and 2x y D 1, so x D 1=3 and y D 1=3. The position vector of this point is 2x C 3y D 4: though, of course, this answer is not unique as the coordinates of any point on the line could have been used. 4y A vector parallel to the line x Cy Cz D 0, 2x y 5z D 1 is The solution of this system is y D 4=7, x D 8=7. A possible standard form for the given line is x 4 units: D p 14 4 12 C 22 C 32 j3 8 0 2j 7 p D p units: 5 2 32 C 42 C 52 We need a point on this line. Putting z D 0, we get x The distance from .0; 0; 0/ to x C 2y C 3z D 4 is 2D DD Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) j 6 p 1 4C1 1j 8 D p units. 5 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.5 31. .1 /.x x0 / D .y y0 / represents any line in the xy-plane passing through .x0 ; y0 /. Therefore, in 3-space the pair of equations .1 /.x x0 / D .y y0 /; 4. 2 aD p 1 .y 1/2 z2 .x C 2/ C C D1 2 2 4 2 .4=3/2 This is an ellipsoid with centre . 2; 1; 0/ and semi-axes 4, 2, and 4/3. 5. 2 i C j C k: All such lines are generators of the circular cone .z z0 /2 D .x x0 /2 C .y 8y D 8 1/ C 9z 2 D 8 C 8 D 16 2 z D z0 x x0 y y0 p D z z0 represents all lines through D 1 2 .x0 ; y0 ; z0 / parallel to the vectors 2 .x C 2/ C 4.y represents all straight lines in the plane z D z0 which pass through the point .x0 ; y0 ; z0 /. 32. x 2 C 4y 2 C 9z 2 C 4x (PAGE 603) z D x 2 C 2y 2 represents an elliptic paraboloid with vertex at the origin and axis along the positive z-axis. Crosssections in p p planes z D k > 0 are ellipses with semi-axes k and k=2. z zDx 2 C2y 2 y0 /2 ; so the given equations specify all straight lines lying on that cone. 33. The equation .A1 x C B1 y C C1 z C D1 /.A2 x C B2 y C C2 z C D2 / D 0 is satisfied if either A1 x C B1 y C C1 z C D1 D 0 or A2 x C B2 y C C2 z C D2 D 0, that is, if .a; y; z/ lies on either of these planes. It is not necessary that the point lie on both planes, so the given equation represents all the points on each of the planes, not just those on the line of intersection of the planes. Fig. 10.5-5 6. Section 10.5 Quadric Surfaces 1. y x (page 603) z D x2 2y 2 represents a hyperbolic paraboloid. z zDx 2 2y 2 x 2 C 4y 2 C 9z 2 D 36 x2 y2 z2 C 2 C 2 D1 2 6 3 2 This is an ellipsoid with centre at the origin and semi-axes 6, 3, and 2. 2. x 2 Cy 2 C4z 2 D 4 represents an oblate spheroid, that is, an ellipsoid with its two longer semi-axes equal. In this case the longer semi-axes have length 2, and the shorter one (in the z direction) has length 1. Cross-sections in planes perpendicular to the z-axis between z D 1 and z D 1 are circles. 3. 2x 2 C 2y 2 C 2z 2 4x C 8y y x Fig. 10.5-6 12z C 27 D 0 2.x 2 2x C 1/ C 2.y 2 C 4y C 4/ C 2.z 2 6z C 9/ D 27 C 2 C 8 C 18 1 .x 1/2 C .y C 2/2 C .z 3/2 D 2 p This is a sphere with radius 1= 2 and centre .1; 2; 3/. 7. x 2 y 2 z 2 D 4 represents a hyperboloid of two sheets with vertices at .˙2; 0; 0/ and circular cross-sections in planes x D k, where jkj > 2. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 407 lOMoARcPSD|6566483 www.konkur.in SECTION 10.5 (PAGE 603) ADAMS and ESSEX: CALCULUS 9 z z x2 y2 z 2 D4 x 2 C4z 2 D4 x y y x Fig. 10.5-10 Fig. 10.5-7 8. x 2 C y 2 C z 2 D 4 represents a hyperboloid of one sheet, with circular cross-sections in all planes perpendicular to the x-axis. 11. x 2 4z 2 D 4 represents a hyperbolic cylinder with axis along the y-axis. z z x 2 Cy 2 Cz 2 D4 y y x x x y x 2 4z 2 D4 Fig. 10.5-11 Fig. 10.5-8 9. z D xy represents a hyperbolic paraboloid containing the x- and y-axes. 12. y D z 2 represents a parabolic cylinder with vertex line along the x-axis. z z yDz 2 y y x zDxy x Fig. 10.5-12 Fig. 10.5-9 10. x 2 C4z 2 D 4 represents an elliptic cylinder with axis along the y-axis. 408 Telegram: @uni_k 13. 1 2 1 represents a parabolic cylinder x D z Cz D z C 2 4 with vertex line along the line z D 1=2, x D 1=4. 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.5 (PAGE 603) z z .z 1/2 D.x 2/2 C.y 3/2 C4 xDz 2 Cz .2;3;1/ y y x x Fig. 10.5-16 Fig. 10.5-13 14. x 2 D y 2 C 2z 2 represents an elliptic cone with vertex at the origin and axis along the x-axis. 17. z x 2 Dy 2 C2z 2 x2 C y2 C z2 D 4 represents the circle of intersection of xCy Cz D1 a sphere and a plane. The circle lies in the plane x pC y C z D 1,pand has centre .1=3; 1=3; 1=3/ and radius 4 .3=9/ D 11=3. z xCyCzD1 x 1 1 1 3;3;3 y x 2 Cy 2 Cz 2 D4 x y Fig. 10.5-14 15. .z 1/2 D .x 2/2 C .y 3/2 represents a circular cone with axis along the line x D 2, y D 3, and vertex at .2; 3; 1/ Fig. 10.5-17 z .z 1/2 D.x 2/2 C.y 3/2 18. x2 C y2 D 1 is the ellipse of intersection of the plane z DxCy z D x C y and the circular cylinder x 2 C y 2 D 1. The centre of the ellipse p is at the p origin, p and the ends of the major axis are ˙.1= 2; 1= 2; 2/. z .2;3;1/ y zDxCy x y x 2 Cy 2 D1 Fig. 10.5-15 x 16. .z 1/2 D .x 2/2 C .y 3/2 C 4 represents a hyperboloid of two sheets with centre at .2; 3; 1/, axis along the line x D 2, y D 3, and vertices at .2; 3; 1/ and .2; 3; 3/. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) Fig. 10.5-18 409 lOMoARcPSD|6566483 www.konkur.in SECTION 10.5 (PAGE 603) 19. ADAMS and ESSEX: CALCULUS 9 z 2 D x 2 C y 2 is the parabola in which the plane z D1Cx z D 1 C x intersects the circular cone z 2 D x 2 C y 2 . (It is a parabola because the plane is parallel to a generator of the cone, namely the line z D x, y D 0.) The vertex of the parabola is . 1=2; 0; 1=2/, and its axis is along the line y D 0, z D 1 C x. z Family 2: The cylinder 2x 2 C y 2 D 1 intersects horizontal planes p in ellipses with semi-axes 1 in the y direction and 1= 2 in the x direction. Tilting the plane in the x direction will cause the shorter semi-axis to increase in length. The plane z D cx intersects the cylinder in an ellipse with principal axes p p through the points .0; ˙1; 0/ and .˙1= 2; 0; ˙c= 2/. The semi-axes will be equal (and the ellipse will be a circle) if .1=2/ C .c 2 =2/ D 1, that is, if c D ˙1. Thus cross-sections of the cylinder perpendicular to the vectors a D i ˙ k are circular. 24. The plane z D cx C k intersects the elliptic cone z 2 D 2x 2 C y 2 on the cylinder zD1Cx c 2 x 2 C 2ckx C k 2 D 2x 2 C y 2 c 2 /x 2 2ckx C y 2 D k 2 2 ck 2k 2 c2k2 2 2 D .2 c 2 / x C y D k C 2 c2 2 c2 2 c2 2 2 y .x x0 / C 2 D 1; a2 b y .2 Fig. 10.5-19 20. x 2 C 2y 2 C 3z 2 D 6 is an ellipse in the plane yD1 y D 1. Its projection onto the xz-plane is the ellipse x 2 C 3z 2 D 4. One quarter of the ellipse is shown in the figure. where x0 D z p 2 yD1 p 3 a; 0; c.x0 a/ C k/ and .x0 ; b; cx0 C k/ x 2 C2y 2 C3z 2 D6 a2 C c 2 a2 D b 2 ; Fig. 10.5-20 21. x2 y2 z2 C 2 D1 2 a b c2 2 2 z y2 x D 1 2 2 2 a x cz x bz y y C D 1C 1 a c a c b b 8 x z y < C D 1C a c b Family 1: : x z D 1 y. a c b 8 x z y < C D 1 a c b : Family 2: : x z D 1C y. a c b 22. z D xy Family 1: 410 Telegram: @uni_k z D x D y: to .x0 C a; 0; c.x0 C a/ C k/; to .x0 ; b; cx0 C k/: The centre of this ellipse is .x0 ; 0; cx0 C k/. The ellipse is a circle if its two semi-axes have equal lengths, that is, if y p 6 2k 2 2k 2 ck , a2 D , and b 2 D . 2 2 2 2 c .2 c / 2 c2 As in the previous exercise, z D cx C k intersects the cylinder (and hence the cone) in an ellipse with principal axes joining the points .x0 x z D y D x: 23. z 2 Dx 2 Cy 2 x n that is, 2k 2 2k 2 D ; 2 2 .2 c / 2 c2 p or 1 C c 2 D 2 c 2 . Thus p c D ˙1= 2. Apvector normal to the plane z D ˙.x= 2/ C k is a D i ˙ 2k. .1 C c 2 / Section 10.6 Cylindrical and Spherical Coordinates (page 607) 1. Cartesian: .2; p2; 1/; Cylindrical: Œ2 2; =4; 1; Spherical: Œ3; cos 1 .1=3/; =4. 2. Cylindrical: p Œ2; =6; 2; p Cartesian: . 3; 1; 2; Spherical: Œ2 2; 3=4; =6. 3. Spherical: Œ4;p =3; 2=3; p 3; 3; 2/; Cylindrical: Œ2 3; 2=3; 2. Cartesian: . Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.7 4. Spherical: Œ1; ; ; Cylindrical: Œr; =4; r. 4. p x D sin cos D r cos =4 D r= 2 p y D sin sin D r sin =4 D r= 2 z D cos D r: Thus x D y, D =4,pand r D sin D cos . Hence D =4, p so r D 1= 2. Finally: x D y D 1=2, z D 1= 2. p Cartesian: .1=2; 1=2; 1= 2/. 5. 5. D =2 represents the half-plane x D 0, y > 0. 6. D 2=3 represents the lower half of the right-circular cone with vertex at the origin, axis along the z-axis, and semi-vertical angle =3. Its Cartesian equation is p zD .x 2 C y 2 /=3. 7. D =2 represents the xy-plane. 8. R D 4 represents the sphere of radius 4 centred at the origin. 9. r D 4 represents the circular cylinder of radius 4 with axis along the z-axis. 10. R D z represents the positive half of the z-axis. 11. a c b d D 1 B 0 AAT D B @0 0 0 4 B3 DB @2 1 0 1 B 0 A2 D B @0 0 0 1 B0 DB @0 0 1 1 0 0 3 3 2 1 1 1 0 0 2 1 0 0 1 1 1 0 2 2 2 1 1 1 1 0 3 2 1 0 10 1 1 B1 1C CB 1A@1 1 1 1 1 1C C 1A 1 10 1 1 B0 1C CB 1A@0 0 1 1 4 3C C 2A 1 w y x z 0 6. 13. If R D 2 cos , then R2 D 2R cos , so x 2 C y 2 C z 2 D 2z x2 C y2 C z2 2z C 1 D 1 x 2 C y 2 C .z 1/2 D 1: Thus R D 2 cos represents the sphere of radius 1 centred at .0; 0; 1/. r D 2 cos ) x 2 C y 2 D r 2 D 2r cos D 2x, or .x 1/2 C y 2 D 1. Thus the given equation represents the circular cylinder of radius 1 with axis along the vertical line x D 1, y D 0. bw C dx by C dz 0 1 1 1 0 0 1 1 1 0 0C C 0A 1 1 1 0 0 1 1 1 0 1 1 1C C 1A 1 0 10 1 0 1 3 0 2 2 1 6 7 3A 1. @ 1 1 2 A @ 3 0 A D @ 5 1 1 1 0 2 1 1 0 10 1 0 1 1 1 1 1 1 1 1 2 3 2. @ 0 1 1 A @ 0 1 1 A D @ 0 1 2 A 0 0 1 0 0 1 0 0 1 a b w x aw C by ax C bz 3. D c d y z cw C dy cx C dz 0 1 0 1 x a p q x D @y A; A D @p b r A z q r c 0 1 0 2 1 x x xy xz T 2 yz A xx D @ y A .x; y; z/ D @ xy y z xz yz z 2 0 1 x T x x D .x; y; z/ @ y A D .x 2 C y 2 C z 2 / z 0 10 1 a p q x T x Ax D .x; y; z/ @ p b r A @ y A q r c z 0 1 ax C py C qz D .x; y; z/ @ px C by C rz A qx C ry C cz D ax 2 C by 2 C cz 2 C 2pxy C 2qxz C 2ryz Section 10.7 A Little Linear Algebra (page 617) 7. ˇ ˇ 2 ˇ ˇ 4 ˇ ˇ 1 ˇ ˇ 2 D 3 3 0 0 0 ˇ 1 0 ˇˇ 2 1 ˇˇ D 1 1 ˇˇ 0 1ˇ ˇ ˇ ˇ 1 1ˇ ˇ 2 ˇˇ 2 1ˇ D 6.3/ C 3.6/ D 36 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k aw C cx ay C cz R D r represents the xy-plane. 12. R D 2x represents the half-cone with vertex at the origin, axis along the positive x-axis, and p semi-vertical angle =3. Its Cartesian equation is x D .y 2 C z 2 /=3. 14. (PAGE 617) Downloaded by ted cage (sxnbyln180@questza.com) ˇ ˇ 4 ˇ 3 ˇˇ 1 ˇ 2 ˇ ˇ 4 1 ˇˇ 2 ˇ 2 1 ˇˇ 1 1 ˇˇ 0 1ˇ ˇ 1 ˇˇ 1ˇ 411 lOMoARcPSD|6566483 www.konkur.in SECTION 10.7 (PAGE 617) 8. ˇ ˇ 1 ˇ ˇ 1 ˇ ˇ 2 ˇ ˇ 3 D D D 9. ADAMS and ESSEX: CALCULUS 9 ˇ 1 1 1 ˇˇ 2 3 4 ˇˇ 0 2 4 ˇˇ 3 2 2ˇ ˇ ˇ ˇ ˇ ˇ1 1 ˇ 1 1 1 ˇ 1 ˇˇ ˇ ˇ ˇ 4 ˇˇ 2 ˇˇ 2 3 4 ˇˇ C 2 ˇˇ 1 2 ˇ 3 2 ˇ3 2ˇ 3 2ˇ ˇ ˇ ˇ1 1 1ˇ ˇ ˇ 4 ˇˇ 1 2 3 ˇˇ ˇ3 3 2ˇ ˇ ˇ ˇ ˇ ˇ1 1 1ˇ ˇ1 1 1 ˇˇ ˇ ˇ ˇ 3 ˇˇ 2 ˇˇ 0 1 2 ˇˇ C 2 ˇˇ 0 1 ˇ0 ˇ0 5 1ˇ 6 5ˇ ˇ ˇ ˇ1 1 1 ˇˇ ˇ 2 ˇˇ 4 ˇˇ 0 1 ˇ0 6 1ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ1 2ˇ ˇ 1 ˇ 1 3 ˇˇ 2 ˇˇ ˇ ˇ ˇ ˇ 2ˇ C 2ˇ 4ˇ 5 1ˇ 6 5ˇ 6 1ˇ D 2. 9/ C 2.13/ 4.11/ D 0 ˇ ˇ ˇ a11 a12 a13 a1n ˇ ˇ ˇ ˇ 0 a22 a23 a2n ˇ ˇ ˇ ˇ 0 0 a33 a3n ˇˇ ˇ :: :: :: ˇ ˇ :: :: ˇ : : : : : ˇˇ ˇ ˇ 0 0 0 ann ˇ ˇ ˇ ˇ a22 a23 a2n ˇ ˇ ˇ ˇ 0 a33 a3n ˇ ˇ ˇ :: :: ˇ D a11 ˇ :: :: ˇ : ˇ : : : ˇ ˇ ˇ 0 0 ann ˇ ˇ ˇ ˇ a33 a3n ˇ ˇ ˇ :: ˇ ˇ : :: D a11 a22 ˇ :: ˇ : : ˇ ˇ ˇ 0 ann ˇ Generalization: ˇ ˇ 1 ˇ ˇ x1 ˇ 2 ˇ x1 ˇ ˇ :: ˇ : ˇ ˇ xn 1 1 11. ˇ ˇ 1 ˇ f .x; y; z/ D ˇˇ x ˇ x2 Telegram: @uni_k a c b ` m w , CD , BD d n p y 12. If A D b a , then AT D d b a c x/.z xi /: x . Then z c , and d bc D det.AT /: det.A/ D ad We generalize this by induction. Suppose det(BT )=det(B) for any .n where n 3. Let 0 1 y y2 a11 B a21 B A D B :: @ : an1 ˇ 1 ˇˇ z ˇˇ ; z2 ˇ x/.z a12 a22 :: : :: : an2 1/ .n 1/ matrix, 1 a1n a2n C C :: C : A ann be an n n matrix. If det(A) is expanded in minors about the first row, and det(AT ) is expanded in minors about the first column, the corresponding terms in these expansions are equal by the induction hypothesis. (The .n 1/.n 1/ matrices whose determinants appear in one expansion are the transposes of those in the other expansion.) Therefore det(AT )=det(A) for any square matrix A. 13. Let A D a c b d and B D x and f .x; y; z/ D .y 412 ˇ 1 ˇ ˇ xn ˇ Y 2 ˇˇ xn .xj ˇD :: ˇ : ˇˇ 1i<j n xnn 1 ˇ :: : Thus .AB/C D A.BC/: then f is a polynomial of degree 2 in z. Since f .x; y; x/ D 0 and f .x; y; y/ D 0, we must have f .x; y; z/ D A.z x/.z y/ for some A independent of z. But ˇ ˇ ˇ 1 1 1 ˇˇ ˇ y 0 ˇˇ D xy.y x/; Axy D f .x; y; 0/ D ˇˇ x ˇ x2 y2 0 ˇ so A D y Let A D 1 x3 x32 :: : x3n 1 w x a` C bn am C bp .AB/C D y z c` C d n cm C dp a`w C bnw C amy C bpy a`x C bnx C amz C bpz D c`w C d nw C cmy C dpy c`x C d nx C cmz C dpz a b `w C my `x C mz A.BC/ D c d nw C py nx C pz a`w C amy C bnw C bpy a`x C amz C bnx C bpz D c`w C cmy C d nw C dpy c`x C cmz C d nx C dpz D a11 a22 a33 ann (or use induction on n) ˇ ˇ ˇ1 1ˇ ˇ D y x. If 10. ˇˇ x yˇ 1 x2 x22 :: : x2n 1 y/: AB D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) w y aw C by cw C dy x . Then z ax C bz cx C dz : lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.7 Therefore, det.A/det.B/ D .ad bc/.wz xy/ D adwz adxy bcwz C bcxy det.AB/ D .aw C by/.cx C dz/ .ax C bz/.cw C dy/ D awcx C awdz C bycx C bydz axwc axdy bzcw bzdy D adwz adxy bcwz C bcxy D det.A/det.B/: 14. If A D A D sin , then cos cos sin cos. / sin. / D 1 0 0 1 D I: sin. / cos. / and A A D cos sin sin cos 0 1 a 1 0 1 1 18. Let A D @ 1 1 0 A, A D @ d g 2 1 3 AA 1 D I we must have 0 a gD1 aCd D0 2a C d C 3g D 0 0 1 D D ; with a similar calculation for the product with a reversed order of factors. 20. 16. Since D D detB D xy 2 x 2 y D xy.y x/, B is nonsingular, and therefore invertible, provided x ¤ 0, y ¤ 0, and x ¤ y. Using the formula for the inverse of a general 2 2 matrix, we have 1 0 2 1 0 y 1 y y B x.y x/ x.y x/ C B D C C B 1 D @ D2 ADB @ A x 1 x x y.y x/ y.y x/ D D 17. 0 1 0 1 1 1 a Let A D @ 0 1 1 A, A 1 D @ d 0 0 1 g AA 1 D I we must have aCd Cg D1 d Cg D0 gD0 bCeChD0 eChD1 hD0 b e h 1 c f A. Since i cCf Ci D0 f Ci D0 i D 1: Thus a D 1, d D g D 0, h D 0, e D 1, b D f D 1, c D 0, and so 0 1 1 1 0 1A: A 1 D @0 1 0 0 1 1 6 5 6 1 6 1 1 6 1 A : 6 1 6 The given system of equations is 1, i D 1, 21. Thus 0 1 1 0 1 0 1 x 2 @y A D A 1 @ 1 A D @2A; 3 z 13 so x D 1, y D 2, and z D 3. If A is the matrix of Exercises 16 and 17 then det.A/ D 6. By Cramer’s Rule, ˇ ˇ 2 1 ˇˇ xD ˇ 1 6 ˇ 13 ˇ ˇ 1 1 ˇˇ yD ˇ 1 6ˇ 2 ˇ ˇ 1 1 ˇˇ zD ˇ 1 6ˇ 2 ˇ 1 ˇˇ 6 0 ˇˇ D D 1 6 3 ˇ ˇ 2 1 ˇˇ 12 1 0 ˇˇ D D2 6 ˇ 13 3 ˇ 0 2 ˇˇ 18 1 1 ˇˇ D D 3: 6 ˇ 1 13 0 1 1 1 1 1 1 1 B1 1 1 1C C ADB @1 1 1 1A 1 1 1 1 ˇ ˇ ˇ0 0 ˇ 0 2 ˇ ˇ ˇ0 0 ˇ 2 0 ˇ det.A/ D ˇˇ 0 0 ˇˇ ˇ0 2 ˇ1 1 1 1ˇ ˇ ˇ ˇ ˇ ˇ0 0 2 ˇ ˇ0 ˇ ˇ ˇ ˇ 0 2 0 D 2ˇ ˇ D 4ˇ1 ˇ1 1 1ˇ 0 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k c i D0 c Cf D0 2c C f C 3i D 1: 0 1 0 1 x 2 A@y A D @ 1 A: z 13 1 ab C ab 1 0 C D D cb C ad A 0 1 D D 2 1 2 1 2 A 1D@ 19. If D D ad bc, we have 1 0 0 ad bc b d a b B C B D D D a A D @ cd cd c d @ c 1 c f A. Since i Solving these three systems of equations, we get Thus A D .A / 1 : 15. b hD0 bCe D1 2b C e C 3h D 0 b e h (PAGE 617) Downloaded by ted cage (sxnbyln180@questza.com) ˇ 2 ˇˇ D8 1ˇ 413 lOMoARcPSD|6566483 www.konkur.in SECTION 10.7 (PAGE 617) ADAMS and ESSEX: CALCULUS 9 ˇ ˇ ˇ0 1 1 1 ˇˇ ˇ 1 ˇ4 1 1 1 ˇˇ x1 D ˇˇ ˇ 6 1 1 1 8ˇ ˇ ˇ2 1 1 1ˇ ˇ ˇ ˇ0 1 1 1 ˇ ˇ ˇ 1 ˇˇ 4 0 0 2 ˇˇ D ˇ 8 ˇ 6 2 0 0 ˇˇ ˇ2 0 0 0 ˇ ˇ ˇ ˇ1 1 1 ˇ ˇ ˇ ˇ 2 ˇˇ ˇ D 4 ˇ1 0 0 2 D 8 ˇˇ 2 0 0 ˇˇ 8 ˇ0 ˇ ˇ1 ˇ 1 ˇˇ 1 x2 D ˇ 8 ˇ1 ˇ1 ˇ ˇ2 ˇ 1 ˇˇ 0 D ˇ 8 ˇ0 ˇ0 ˇ ˇ4 2 ˇˇ D ˇ6 8 ˇ2 0 4 6 2 0 4 6 2 2 0 0 ˇ 1 1 ˇˇ 1 1 ˇˇ 1 1 ˇˇ 1 1ˇ ˇ 0 1 ˇˇ 2 1 ˇˇ 0 1 ˇˇ 0 1ˇ ˇ ˇ 1 ˇˇ 4 ˇˇ 6 1 ˇˇ D 8 ˇ2 1ˇ ˇ ˇ ˇ1 1 0 1 ˇ ˇ ˇ 1 ˇ1 1 4 1 ˇˇ x3 D ˇˇ 1 ˇˇ 8 ˇ1 1 6 ˇ1 1 2 1ˇ ˇ ˇ ˇ0 2 0 1 ˇ ˇ ˇ 1 ˇˇ 1 ˇˇ 0 0 4 D ˇ 1 ˇˇ 8 ˇ0 0 6 ˇ2 2 2 1ˇ ˇ ˇ ˇ2 0 1 ˇ ˇ ˇ 2 ˇˇ 4 ˇˇ 4 ˇ 0 4 1 D ˇ D 8 ˇ6 8 ˇˇ 0 6 1ˇ x4 D .x1 C x2 C x3 / D 2: pa C qc pb C qd x1 x2 ra C sc rb C sd x1 a b p q D x2 c d r s x1 : D GF x2 D 23. ˇ 1 ˇˇ D1 2ˇ Telegram: @uni_k ˇ ˇ1 ˇ D3 D ˇˇ 2 ˇ0 ˇ 1 ˇˇ D2 1ˇ 25. ˇ ˇ1 D2 D ˇˇ 2 D1 D 2 > 0; 26. ˇ 1 ˇˇ D 1ˇ 1 ˇ 2 0 ˇˇ 1 0 ˇˇ D 0 1ˇ ˇ 2 ˇˇ D 1ˇ Thus A is indefinite. 0 1 2 1 1 A D @ 1 2 1 A. Use Theorem 8. 1 1 2 ˇ ˇ2 ˇ D3 D ˇˇ 1 ˇ1 G ı F .x1 ; x2 / D G.y1 ; y2 / ax1 C bx2 p q D cx1 C dx2 r s pax1 C pbx2 C qcx1 C qdx2 D rax1 C rbx2 C scx1 C sdx2 414 Thus, G ı F is represented by the matrix GF. 1 1 A D . Use Theorem 8. D1 D 1 < 0, 1 2 ˇ ˇ ˇ 1 1 ˇ ˇ D 1 > 0. Thus A is negative definite. D2 D ˇˇ 1 2ˇ 0 1 1 2 0 A D @ 2 1 0 A. Use Theorem 8. 0 0 1 D1 D 1 > 0; a b x1 . , where F D c d x 2 p q y1 , where G D . Let G.y1 ; y2 / D G y2 r s If y1 D ax1 C bx2 and y2 D cx1 C dx2 , then 22. Let F .x1 ; x2 / D F 24. ˇ ˇ2 D2 D ˇˇ 1 3 < 0; 3 < 0: ˇ 1 ˇˇ D 3 > 0; 2ˇ ˇ 1 1 ˇˇ 2 1 ˇˇ D 4 > 0: 1 2ˇ Thus A is positive definite. 1 0 ˇ ˇ 1 1 0 ˇ1 1ˇ ˇ D 0, we cannot A D @ 1 1 0 A. Since D2 D ˇˇ 1 1ˇ 0 0 1 use Theorem 8. The corresponding quadratic form is Q.x; y; z/ D x 2 C y 2 C 2xy C z 2 D .x C y/2 C z 2 ; 27. which is positive semidefinite. (Q.1; 1; 0/ D 0.). Thus A is positive semidefinite. 0 1 1 0 1 1 A. Use Theorem 8. A D @0 1 1 1 1 D1 D 1 > 0; ˇ ˇ1 ˇ D3 D ˇˇ 0 ˇ1 Thus A is indefinite. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) ˇ ˇ1 D2 D ˇˇ 0 0 1 1 ˇ 1 ˇˇ 1 ˇˇ D 1 ˇ ˇ 0 ˇˇ D 1 > 0; 1ˇ 1 < 0: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 0 2 28. A D @ 0 1 0 4 1 SECTION 10.8 1 1 11 A. Use Theorem 8. 1 D1 D 2 > 0; ˇ ˇ2 ˇ D3 D ˇˇ 0 ˇ1 ˇ ˇ ˇ2 0ˇ ˇ D 8 > 0; D2 D ˇˇ 0 4ˇ 0 4 1 Thus A is positive definite. ˇ 1 ˇˇ 11 ˇˇ D 2 > 0: 1 ˇ Section 10.8 Using Maple for Vector and Matrix Calculations (page 626) > a := DotProduct(U,(V W),conjugate=false): &x > b := DotProduct(V,(W U),conjugate=false): &x > c := DotProduct(W,(U V),conjugate=false): &x (PAGE 626) > simplify(a-b); simplify(a-c); 0 0 4. These calculations verify the identity: > U := Vector[row](3,symbol=u): V := Vector[row](3,symbol=v): It is assumed that the Maple package LinearAlgebra has been loaded for all the calculations in this section. > W := Vector[row](3,symbol=w): 1. We use the result of Example 9 of Section 10.4. > LHS := (U &x V) &x (U &x W): > RHS := (DotProduct(U,(V &x W),conjugate=false))*U: > r1 := <3|0|2>: v1 := <2,1,-2>: > r2 := <1|2|4>: v2 := <1,3,4>: > v1xv2 := v1 &x v2: > dist := abs((r2-r1).v1xv2)/Norm(v1xv2,2); d i st WD 2 > simplify(LHS-RHS); Œ0; 0; 0 The distance between the two lines is 2 units. 2. The plane P through the origin containing the vectors v1 D i 2j 3k and v2 D 2i C 3j C 4k has normal n D v1 v2 . > n := <1|-2|-3> &x <2|3|4>; n WD Œ1; 10; 7 The angle between v D i sp := (U,V) -> DotProduct(U,Normalize(V,2),conjugate=false) 6. vp := (U,V) -> DotProduct(U,Normalize(V,2), conjugate=false)*Normalize(V,2) 7. ang := (u,v) -> evalf((180/Pi)*VectorAngle(U,V)) 8. unitn := (U,V)->Normalize((U &x V),2) 9. VolT := (U,V,W)->(1/6)*abs(DotProduct(U,(V &x W), conjugate=false)) j C 2k and n (in degrees) is > angle := evalf((180/Pi)*VectorAngle(n,<1,-1,2>)); angvn WD 33:55730975 Since this angle is acute, the angle between v and the plane P is its complement. > 5. angle := 90 - angvn; angle WD 56:44269025 10. dist:=(A,B)->Norm(A-B,2) 3. These calculations verify the identity: > dist(<1,1,1,1>,<3,-1,2,5>); 5 > U := Vector[row](3,symbol=u): V := Vector[row](3,symbol=v): > W := Vector[row](3,symbol=w): 11. We use LinearSolve. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 415 lOMoARcPSD|6566483 www.konkur.in SECTION 10.8 (PAGE 626) ADAMS and ESSEX: CALCULUS 9 > A := Matrix([[1,2,3,4,5], > [6,-1,6,2,-3],[2,8,-8,-2,1], > [1,1,1,1,1],[10,-3,3,-2,2]]): > X := LinearSolve(A,<20,0,6,5,5>,free=t); 2 3 1 6 0 7 6 7 7 X WD 6 6 17 4 3 5 2 The solution is u D 1, v D 0, x D 16. > A := Matrix([[1,1/2,1/3], > [1/2,1/3,1/4],[1/3,1/4,1/5]]): > Ainv := MatrixInverse(A): > Digits := 10: evalf(Eigenvalues(A)); 2 3 1:408318927 4 10 11 I 4 0:00268734034 5:673502694 10 10 I 5 0:1223270659 C 5:873502694 10 10 I > evalf(Eigenvalues(Ainv)); 3 2 372:1151279 2 10 9 I 4 0:710066409 5:096152424 10 8 I 5 8:174805711 C 5:296152424 10 8 I 1, y D 3, z D 2. 12. We use LinearSolve. > B := Matrix([[1,1,1,1,0], > [1,0,0,1,1],[1,0,1,1,0], > [1,1,1,0,1],[0,1,0,1,-1]]): > X := LinearSolve(B,<10,10,8,11,1>,free=t); 2 3 11 2t5 6 7 2 6 7 7 2 C t X WD 6 5 6 7 4 1 C t5 5 t5 The small imaginary parts are due to round-off errors in the solution process. The eigenvalues are real since the matrix and its inverse are real and symmetric. Although they appear in different orders, each eigenvalue of A 1 is the reciprocal of an eigenvalue of A. This is to be expected since A 1 x D x .1=/x D Ax: There is a one-parameter family of solutions: u D 11 2t , v D 2, x D 2 C t , y D 1 C t , z D t , for arbitrary t . 13. 14. 15. > A := Matrix([[1,2,3,4,5], > [6,-1,6,2,-3],[2,8,-8,-2,1], > [1,1,1,1,1],[10,-3,3,-2,2]]): > Determinant(A); 935 > B := Matrix([[1,1,1,1,0], > [1,0,0,1,1],[1,0,1,1,0], > [1,1,1,0,1],[0,1,0,1,-1]]): > Digits := 5: evalf(Eigenvalues(B)); 2 3 0 6 3:3133 0:0000053418I 7 6 7 6 0:8693 C 0:0000073520I 7 6 7 4 1:2728 0:0000025143I 5 7 1:9098 C 5:041 10 I 1. x C 3z D 3 represents a plane parallel to the y-axis and passing through the points .3; 0; 0/ and .0; 0; 1/. 2. y z 1 represents all points on or below the plane parallel to the x-axis that passes through the points .0; 1; 0/ and .0; 0; 1/. 3. x C y C z 0 represents all points on or above the plane through the origin having normal vector i C j C k. 4. x 2y 4z D 8 represents all points on the plane passing through the three points .8; 0; 0/, .0; 4; 0/, and .0; 0; 2/. 5. y D 1 C x 2 C z 2 represents the circular paraboloid obtained by rotating about the y-axis the parabola in the xy-plane having equation y D 1 C x 2 . y D z 2 represents the parabolic cylinder parallel to the x-axis containing the curve y D z 2 in the yz-plane. The tiny imaginary parts are due to roundoff error in the calculations. They should all be 0. Since B is a real, symmetric matrix, its eigenvalues are all real. The eigenvalues, rounded to 5 decimal places are 0, 3.3133, 0.8693, 1:2728, and 1:9098. 6. 7. x D y 2 z 2 represents the hyperbolic paraboloid whose intersections with the xy- and xz-planes are the parabolas x D y 2 and x D z 2 , respectively. > A := Matrix([[1,1/2,1/3], > [1/2,1/3,1/4],[1/3,1/4,1/5]]): > Ainv := MatrixInverse(A); 2 3 9 36 30 180 5 Ai nv WD 4 36 192 30 180 180 8. z D xy is the hyperbolic paraboloid containing the x- and y-axes that results from rotating the hyperbolic paraboloid z D .x 2 y 2 /=2 through 45ı about the z-axis. 416 Telegram: @uni_k Review Exercises 10 (page 627) 9. x 2 C y 2 C 4z 2 < 4 represents the interior of the circular ellipsoid (oblate spheroid) centred at the origin with semiaxes 2, 2, and 1 in the x, y, and z directions, respectively. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES 10 10. x 2 C y 2 4z 2 D 4 represents a hyperboloid of one sheet with circular cross-sections in planes perpendicular to the z-axis, and asymptotic to the cone obtained by rotating the line x D 2z about the z-axis. 11. 21. x 2 y 2 4z 2 D 0 represents an elliptic cone with axis along the x-axis whose cross-sections in planes x D k are ellipses with semi-axes jkj and jkj=2 in the y and z directions, respectively. 14. 15. .x z/2 C y 2 D z 2 represents an elliptic cone with oblique axis along the line z D x in the xz-plane, having circular cross-sections of radius jkj in horizontal planes z D k. The z-axis lies on the cone. 17. or 2x C 5y C 3z D 2. 22. 23. 19. The given line is parallel to the vector a D 2i jC3k. The plane through the origin perpendicular to a has equation 2x y C 3z D 0. 20. A plane through .2; 1; 1/ and .1; 0; 1/ is parallel to b D .2 1/i C . 1 0/j C .1 . 1//k D i j C 2k. If it is also parallel to the vector a in the previous solution, then it is normal to ˇ ˇi ˇ a b D ˇˇ 2 ˇ1 The plane has equation .x x y z D 2. j 1 1 1/ ˇ k ˇˇ 3 ˇˇ D i 2ˇ .y j 0/ 24. 2 C .x C y C z 3z This plane is perpendicular to x 2y normals are perpendicular, that is, if 1.2 C / or 9x C 7y 25. 2.1 C / 5z D 17 if their 5. 3 C / D 0; The line through .2; 1; 1/ and . 1; 0; 1/ is parallel to the vector 3i C j 2k, and has vector parametric equation r D .2 C 3t /i C .1 C t /j 26. 0/ D 0: z D 4. .1 C 2t /k: A vector parallel to the planes x y D 3 and x C 2y C z D 1 is .i j/ .i C 2j C k/ D i j C 3k. A line through .1; 0; 1/ parallel to this vector is x 1 1 D y zC1 D : 1 3 27. The line through the origin perpendicular to the plane 3x 2y C 4z D 5 has equations x D 3t , y D 2t , z D 4t . 28. The vector a D .1 C t /i tj .2 C 2t /k D .1 C t 2s/i .t C s Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 0/ D 0: A plane containing the line of intersection of the planes x C y C z D 0 and 2x C y 3z D 2 has equation 2x C y k: .z C 1/ D 0, or 2 C .x C y C z 3z This plane passes through .2; 0; 1/ if 1 C 3 D 0. In this case, the equation is 7x C 4y 8z D 6. 2 18. x C z 1, x y 0 together represent all points that lie inside or on the circular cylinder of radius 1 and axis along the y-axis and also either on the vertical plane x y D 0 or on the side of that plane containing the positive x-axis. A plane containing the line of intersection of the planes x C y C z D 0 and 2x C y 3z D 2 has equation 2x C y x 2 C y 2 C z 2 D 4, x C y C z D 3 together represent the circle in which the sphere of radius 2 centred at the origin intersects the plane through .1; 1;p1/ with normal i C j C k. Since this plane lies p at distance 3 from the origin, the circle has radius 4 3 D 1. 2 The plane through A D . 1; 1; 0/, B D .0; 4; 1/ and C D .2; 0; 0/ has normal ˇ ˇ ˇi j k ˇˇ ! ! ˇˇ 3 1 0 ˇˇ D i C 3j C 10k: AC AB D ˇ ˇ1 3 1ˇ Its equation is .x 2/C3y C10z D 0, or x C3y C10z D 2. x C2y D 0, z D 3 together represent the horizontal straight line through the point .0; 0; 3/ parallel to the vector 2i j. 16. x C y C 2z D 1, x C y C z D 0 together represent the straight line through the points . 1; 0; 1/ and .0; 1; 1/. A plane perpendicular to x y Cz D 0 and 2x Cy 3z D 2 has normal given by the cross product of the normals of these two planes, that is, by ˇ ˇ ˇi j k ˇˇ ˇ ˇ1 1 1 ˇˇ D 2i C 5j C 3k: ˇ ˇ2 1 3ˇ If the plane also passes through .2; 1; 1/, then its equation is 2.x 2/ C 5.y C 1/ C 3.z 1/ D 0; 12. x 2 y 2 4z 2 D 4 represents a hyperboloid of two sheets asymptotic to the cone of the previous exercise. 13. .x z/2 C y 2 D 1 represents an elliptic cylinder with oblique axis along the line z D x in the xz-plane, having circular cross-sections of radius 1 in horizontal planes z D k. (PAGE 627) Downloaded by ted cage (sxnbyln180@questza.com) 2/j 2si C .s .1 C 2t 2/j 3s/k .1 C 3s/k 417 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 10 (PAGE 627) ADAMS and ESSEX: CALCULUS 9 joins points on the two lines and is perpendicular to both lines if a .i j 2k/ D 0 and a .2i C j 3k/ D 0, that is, if 1Ct 2 C 2t 2s C t C s 2 C 2 C 4t 4s t s C 2 C 3 C 6t 32. The tetrahedron with vertices A D .1; 2; 1/, B D .4; 1; 1/, C D .3; 4; 2/, and D D .2; 2; 2/ has volume 1 1 ! ! ! j.AB AC / ADj D j.9i C 9j C 12k/ .i C k/j 6 6 7 9 C 12 D cu. units: D 6 2 6s D 0 9s D 0; or, on simplification, 33. 6t 7s D 7t 14s D 1 7: This system has solution t D 1, s D 1. We would expect to use a as a vector perpendicular to both lines, but, as it happens, a D 0 if t D s D 1, because the two given lines intersect at .2; 1; 4/. A nonzero vector perpendicular to both lines is ˇ ˇ ˇi j k ˇˇ ˇ ˇ1 1 2 ˇˇ D 5i j C 3k: ˇ ˇ2 1 3ˇ Thus the required line is parallel to this vector and passes through .2; 1; 4/, so its equation is r D .2 C 5t /i a D 1; b D 0; c D 0; d D 0; 30. The points with position vectors r1 , r2 , r3 , and r4 are coplanar if the tetrahedron having these points as vertices has zero volume, that is, if .r2 r1 / .r3 i r1 / .r4 r1 / D 0: (Any permutation of the subscripts 1, 2, 3, and 4 in the above equation will do as well.) 31. The triangle with vertices A D .1; 2; 1/, B D .4; 1; 1/, and C D .3; 4; 2/ has area ˇ ˇi ˇ 1 1 ! ! jAB AC j D j ˇˇ 3 2 2 ˇ2 j 3 2 ˇ k ˇˇ 0 ˇˇ j 3ˇ p 1 3 34 D j9i C 9j C 12kj D sq. units: 2 2 a D 1; e D 2; b D 0; f D 1; c D 0; g D 0; d D 0; h D 0; Thus 0 34. 0 1 B 2 A 1DB @ 1 0 Telegram: @uni_k 0 0 1 0 1 0 0C C: 0A 1 4a C 3e C 2i C m D 0: 4b C 3f C 2j C n D 0: 4c C 3g C 2k C o D 0: 4d C 3h C 2l C p D 1: i D 1; j D 2; k D 1; l D 0; m D 0; n D 1; o D 2; p D 1: 0 1 2 1 1 0 0C C: 0A 1 0 0 1 2 1 0 1 0 1 1 1 1 x1 b1 Let A D @ 2 1 0 A, x D @ x2 A, and b D @ b2 A. 1 0 1 x3 b3 Then Ax D b , x1 C x2 C x3 D b1 2x1 C x2 D b2 x1 x3 D b3 : The sum of the first and third equations is 2x1 C x2 D b1 C b3 , which is incompatible with the second equation unless b2 D b1 C b3 , that is, unless b .i j C k/ D 0: If b satisfies this condition then there will be a line of solutions; if x1 D t , then x2 D b2 2t , and x3 D t b3 , so 0 1 t x D @ b2 2t A t b3 is a solution for any t . 418 0 1 0 0 These systems have solutions r1 / D 0: (Any permutation of the subscripts 1, 2, and 3 in the above equation will do as well.) h 2a C e D 0; 3a C 2e C i D 0; 2b C f D 1; 3b C 2f C j D 0; 2c C g D 0; 3c C 2g C k D 1; 2d C h D 0; 3d C 2h C l D 0; .1 C t /j C . 4 C 3t /k: r1 / .r3 1 0 d 1 B0 hC CDB l A @0 p 0 c g k o Expanding the product on the left we get four systems of equations: 29. The points with position vectors r1 , r2 , and r3 are collinear if the triangle having these points as vertices has zero area, that is, if .r2 The inverse of A satisfies 10 0 1 0 0 0 a b B2 1 0 0CB e f B CB @3 2 1 0A@ i j 4 3 2 1 m n Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 0 3 35. A D @ 1 1 CHALLENGING PROBLEMS 10 1 1 1 A. We use Theorem 8. 2 1 1 1 D1 D 3 > 0; ˇ ˇ 3 ˇ D3 D ˇˇ 1 ˇ 1 1 1 1 Thus A is positive definite. 1 jx2 y3 x2 y1 x1 y3 2 ˇ ˇ ˇ x y1 1 ˇ ˇ 1 ˇˇ 1 D j ˇ x2 y2 1 ˇˇ j: 2 ˇx y 1ˇ 3 3 D ˇ 1 ˇˇ D 2 > 0; 3 ˇ ˇ ˇ 3 D2 D ˇˇ 1 ˇ 1 ˇˇ 1 ˇˇ D 2 > 0: 2 ˇ 4. Challenging Problems 10 (page 628) 1. If d is the distance from P to the line AB, then d is the altitude of the triangle APB measured perpendicular to the base AB. Thus the area of the triangle is ! .1=2/d jBAj D .1=2/d jrA rB j: On the other hand, the area is also given by ! ! .1=2/jPA PBj D .1=2/j.rA rP / .rB j.rA rP / .rB jrA rB j rP /j a) Let Q1 and Q2 be the points on lines L1 and L2 , respectively, that are closest together. As observed in ! Example 9 of Section 1.4, Q1 Q2 is perpendicular to both lines. Therefore, the plane P1 through Q1 having normal ! Q1 Q2 contains the line L1 . Similarly, the plane P2 ! through Q2 having normal Q1 Q2 contains the line L2 . These planes are parallel since they have the same normal. They are different planes because Q1 ¤ Q2 (because the lines are skew). r1 D .1 C t /i C .1 r2 D si C .1 C s/j C .1 C s/k: Now r2 .u v/ .w x/ D Œ.u v/ xw Œ.u v/ wx .u v/ .w x/ D .w x/ .u v/ D Œ.w x/ vu C Œ.w x/ uv: .s .s .u v/ .u x/ D Œ.u v/ xu; .u v/ .u w/ D Œ.u v/ wu: x1 /.y3 y1 / .x3 x1 /.y2 t t 5. 1/i C .s C t /j C .1 C s t /k. 1/ .s C t / C .1 C s 1/ C .s C t / C .1 C s t/ D 0 t / D 0: This problem is similar to Exercise 28 of Section 1.3. The equation a x D b has no solution x unless a b D 0. If this condition is satisfied, then x D x0 C t a is a solution for any scalar t , where x0 D .b a/=jaj2 . y1 /kj Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k t Subtracting these equations gives s C t D 0, so t D s. Then substituting into either equation gives 2s 1 C 1 C 2s D 0, so s D t D 0. Thus Q1 D .1; 1; 0/ and Q2 D .0; 1; 1/, and ! Q1 Q2 D i C k. The required planes are x z D 1 (containing L1 ) and x z D 1 (containing L2 ). or, replacing x with w, 3. The triangle with vertices .x1 ; y1 ; 0/, .x2 ; y2 ; 0/, and .x3 ; y3 ; 0/, has two sides corresponding to the vectors .x2 x1 /i C .y2 y1 /j and .x3 x1 /i C .y3 y1 /j. Thus the triangle has area given by ˇ ˇ ˇ i j k ˇˇ 1 ˇˇ A D j ˇ x2 x1 y2 y1 0 ˇˇ j 2 ˇx x1 y3 y1 0 ˇ 3 r1 D .s To find the points Q1 on L1 and Q2 on L2 for ! which Q1 Q2 is perpendicular to both lines, we solve In particular, if w D u, then, since .u v/ u D 0, we have 1 jŒ.x2 2 t /j C t k: Line L2 through .0; 1; 1/ and .1; 2; 2/ is parallel to i C j C k, and has parametric equation : 2. By the formula for the vector triple product given in Exercise 23 of Section 1.3, D x3 y2 C x3 y1 C x1 y2 j b) Line L1 through .1; 1; 0/ and .2; 0; 1/ is parallel to i j C k, and has parametric equation rP /j: Equating these two expressions for the area of the triangle and solving for d we get dD (PAGE 628) Downloaded by ted cage (sxnbyln180@questza.com) 419 lOMoARcPSD|6566483 www.konkur.in SECTION 11.1 (PAGE 635) ADAMS and ESSEX: CALCULUS 9 CHAPTER 11. VECTOR FUNCTIONS AND CURVES 9. Section 11.1 Vector Functions of One Variable (page 635) 1. 10. Position: r D 3 cos t i C 4 sin t j C t k Velocity: v D p 3 sin t i C 4 cos t j C k p Speed: v D 9 sin2 t C 16 cos2 t C 1 D 10 C 7 cos2 t Acceleration : a D 3 cos t i 4 sin t j D t k r Path: a helix (spiral) wound around the elliptic cylinder .x 2 =9/ C .y 2 =16/ D 1. Position: r D i C t j Velocity: v D j Speed: v D 1 Acceleration : a D 0 Path: the line x D 1 in the xy-plane. Position: r D ae t i C be t j C ce t k Velocity and acceleration: vDaDr p Speed: v D e t a2 C b 2 C c 2 x y z Path: the half-line D D > 0. a b c 12. Position: r D at cos !t i C at sin !t j C b ln t k Velocity: v D a.cos !t !t sin !t /i C a.sin !t C !t cos !t /j C .b=t /k p Speed: v D a2 .1 C ! 2 t 2 / C .b 2 =t 2 / Acceleration: a D a!.2 sin !t C ! cos !t /i 11. 2. Position: r D t 2 i C k Velocity: v D 2t i Speed: v D 2jt j Acceleration : a D 2i Path: the line z D 1, y D 0. 3. Position: r D t 2 j C t k Velocity: v D p2t j C k Speed: v D 4t 2 C 1 Acceleration : a D 2j Path: the parabola y D z 2 in the plane x D 0. C a!.2 cos !t ! sin !t /j .b=t 2 /k Path: a spiral on the surface x 2 C y 2 D a2 e z=b . 13. 4. Position: r D i C t j C t k Velocity: v D pj C k Speed: v D 2 Acceleration : a D 0 Path: the straight line x D 1, y D z. 5. Position: r D t 2 i t 2 j C k Velocity: v Dp 2t i 2t j Speed: v D 2 2t Acceleration: a D 2i 2j Path: the half-line x D y 0, z D 1. Position: r D a cos t i C a sin t j C ct k Velocity: v D p a sin t i C a cos t j C ck Speed: v D a2 C c 2 Acceleration: a D a cos t i a sin t j Path: a circular helix. Telegram: @uni_k et k Position: r D a cos t sin t i C a sin2 t j C a cos t k a a 1 cos 2t j C a cos t k D sin 2t i C 2 2 Velocity: v Dp a cos 2t i C a sin 2t j a sin t k Speed: v D a 1 C sin2 t Acceleration: a D 2a sin 2t i C 2a cos 2t j a cos t k Path: the path lies on the sphere x 2 C y 2 C z 2 D a2 , on the surface defined in terms of spherical polar coordinates by D , on the circular cylinder x 2 C y 2 D ay, and on the parabolic cylinder ay C z 2 D a2 . Any two of these surfaces serve to pin down the shape of the path. 15. The position of the particle is given by r D 5 cos.!t /i C 5 sin.!t /j; 8. Position: r D a cos !t i C bj C a sin !t k Velocity: v D a! sin !t i C a! cos !t k Speed: v D ja!j Acceleration: a D a! 2 cos !t i a! 2 sin !t k Path: the circle x 2 C z 2 D a2 , y D b. 420 t t Position: r D e cos.e t /i C e t sin.e t /j e k Velocity: v D e t cos.e t / C sin.e t / i e t sin.e t / cos.e t / j e t k p Speed: v D 1 C e 2t C e 2t Acceleration: a D .e t e t / cos.e t / C sin.e t / i C .e t e t / sin.e t / cos.e t / j p Path: a spiral on the surface z x 2 C y 2 D 1. 14. 6. Position: r D t i C t 2 j C t 2 k Velocity: v D pi C 2t j C 2t k Speed: v D 1 C 8t 2 Acceleration: a D 2j C 2k Path: the parabola y D z D x 2 . 7. Position: r D 3 cos t i C 4 cos t j C 5 sin t k Velocity: v D p 3 sin t i 4 sin t j C 5 cos t k Speed: v D 9 sin2 t C 16 sin2 t C 25 cos2 t D 5 Acceleration : a D 3 cos t i 4 cos t j 5 sin t k D r Path: the circle of intersection of the sphere x 2 C y 2 C z 2 D 25 and the plane 4x D 3y. where ! D to ensure that r has period 2=! D 2 s. Thus d 2r a D 2 D ! 2 r D 2 r: dt At .3; 4/, the acceleration is 3 2 i 4 2 j. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 11.1 16. When its x-coordinate is x, the particle is at position r D xi C .3=x/j, and its velocity and speed are Thus The particle is at .3; 3; 2/ when u D 1. At this point du=dt D 2=3 and d 2 u=dt 2 D 16=27, and so We know that dx=dt > 0 since the particle is moving to the right. When x D p2, we have 10 D v D .dx=dt / 1 C .9=16/ D .5=4/.dx=dt /. Thus dx=dt D 8. The velocity at that time is v D 8i 6j. 2 .3i C 6uj C 6u2 k/ D 2i C 4j C 4k 3 2 2 16 .6j C 12k/ .3i C 6j C 6k/ C aD 27 3 8 D . 2i j C 2k/: 9 vD The particle moves along the curve z D x 2 , x C y D 2, in the direction of increasing y. Thus its position at time t is r D .2 y/2 k; y/i C yj C .2 where y is an increasing function of time t . Thus 20. i dy h i C j 2.2 y/k dt dy p vD 1 C 1 C 4.2 y/2 D 3 dt vD since the speed p When y D 1, we have p is 3. dy=dt D 3= 6 D 3=2. Thus vD r 3 . iCj 2 2k/: i dx h i C 2xj C 3x 2 k . Since dt dz=dt D 3x 2 dx=dt D 3, when x D 2 we have 12 dx=dt D 3, so dx=dt D 1=4. Thus so its velocity is v D Since u is increasing and the speed of the particle is 6, 6 D jvj D 3 aD 21. r D 3ui C 3u2 j C 2u3 k du vD .3i C 6uj C 6u2 k/ dt 2 d 2u du 2 aD .6j C 12uk/: .3i C 6uj C 6u k/ C dt 2 dt du p du 1 C 4u2 C 4u4 D 3.1 C 2u2 / : dt dt 22. 15.i 2j C 2k/ C 9. 2j C 2k/ D dx dt 2 : 15i C 12j 12k: d 2 d jvj D v v D 2v a. dt dt If v a > 0 then the speed v D jvj is increasing. If v a < 0 then the speed is decreasing. If u.t / D u1 .t /i C u2 .t /j C u3 .t /k v.t / D v1 .t /i C v2 .t /j C v3 .t /k then u v D u1 v2 C u2 v2 C u3 v3 , so du1 dv1 du2 dv2 d uvD v1 C u1 C v2 C u2 dt dt dt dt dt dv3 du3 v3 C u3 C dt dt dv du vCu : D dt dt Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k The left side is 3 when x D 1, so 3.d 2 x=dt 2 / C 48 D 3, and d 2 x=dt 2 D 15 at that point, and the acceleration there is r D xi C x 2 j C x 3 k; 1 v D i C j C 3k: 4 r D xi x 2 j C Cx 2 k dx vD .i 2xj C 2xk/ dt 2 d 2x dx aD . 2j C 2k/: .i 2xj C 2xk/ C 2 dt dt ˇ ˇ p ˇ dx ˇ p dx Thus jvj D ˇˇ ˇˇ 1 C 4x 4 C 4x 4 D 1 C 8x 4 , dt dt since x is increasing. At .1; 1; 1/, x D 1 and jvj D 9, so dx=dt D 3, and the velocity at that point is v D 3i 6j C 6k. Now p 16x 3 d 2x d jvj D 1 C 8x 4 2 C p dt dt 1 C 8x 4 18. The position of the object when its x-coordinate is x is 19. 2 du , and D dt 1 C 2u2 2 du 16u d 2u D 4u : D dt 2 .1 C 2u2 /2 dt .1 C 2u2 /3 dr dx 3 dx D i j dt rdt x 2 dt ˇ ˇ ˇ dx ˇ 9 v D ˇˇ ˇˇ 1 C 4 : dt x vD 17. (PAGE 635) Downloaded by ted cage (sxnbyln180@questza.com) 421 lOMoARcPSD|6566483 www.konkur.in SECTION 11.1 (PAGE 635) 23. 24. 25. ADAMS and ESSEX: CALCULUS 9 ˇ ˇ ˇa a12 a13 ˇˇ d ˇˇ 11 a21 a22 a23 ˇˇ dt ˇˇ a 31 a32 a33 ˇ d h D a11 a22 a33 C a12 a23 a31 C a13 a21 a32 dt i a11 a23 a32 a12 a21 a33 a13 a22 a31 0 0 0 D a11 a22 a33 C a11 a22 a33 C a11 a22 a33 0 0 0 C a12 a23 a31 C a12 a23 a31 C a12 a23 a31 0 0 0 C a13 a21 a32 C a13 a21 a32 C a13 a21 a32 0 0 0 a11 a23 a32 a11 a23 a32 a11 a23 a32 0 0 0 a12 a21 a33 a12 a21 a33 a12 a21 a33 0 0 0 a13 a22 a31 a13 a22 a31 a13 a22 a31 ˇ 0 ˇ ˇ ˇ 0 0 ˇ ˇ a11 a12 ˇ a13 a12 a13 ˇˇ ˇ ˇ ˇ a11 0 0 0 ˇ a22 a23 D ˇˇ a21 a22 a23 ˇˇ C ˇˇ a21 ˇ ˇ a31 a32 a33 ˇ ˇ a31 a32 a33 ˇ ˇ ˇ ˇ a11 a12 a13 ˇ ˇ ˇ C ˇˇ a21 a22 a23 ˇˇ 0 0 0 ˇ a31 a32 a33 ˇ r0 j2 D d .r dt r0 / .r 32. 33. 34. r0 / dr D0 dt implies that jr r0 j is constant. Thus r.t / lies on a sphere centred at the point P0 with position vector r0 . D 2.r r0 / 30. d u .v w/ dt D u0 .v w/ C u .v0 w/ C u .v w0 /: d d u d 2u 2 u dt dt dt 2 du d u d 2u d u d 2u D 2 Cu dt dt dt dt 2 dt 2 3 du d u 3 Cu dt dt d u d 2u d u d 3u du 2 Cu 3 : D dt dt dt dt dt 422 Telegram: @uni_k Since dr D v.t / D 2r.t / and r.0/ D r0 , we have dt The path is the half-line from the origin in the direction of r0 . v 0 r D r0 cos !t C sin !t ! dr D !r0 sin !t C v0 cos !t dt d 2r D ! 2 r0 cos !t !v0 sin !t D ! 2 r dt 2 ˇ d r ˇˇ D v0 : r.0/ D r0 ; ˇ dt ˇ tD0 Observe that r .r0 v0 / D 0 for all t . Therefore the path lies in a plane through the origin having normal N D r0 v0 . Let us choose our coordinate system so that r0 D ai (a > 0) and v0 D !bi C !cj (c > 0). Therefore, N is in the direction of k. The path has parametric equations 26. If r v > 0 then jrj is increasing. (See Exercise 16 above.) Thus r is moving farther away from the origin. If r v < 0 then r is moving closer to the origin. d 2u d 2u d u d 3u d d u d 2u 27. 2 D 2 C 3 2 dt dt dt dt dt dt dt d u d 3u 3: D dt dt d 28. u .v w/ dt D u0 .v w/ C u .v0 w/ C u .v w0 /: 29. i d h .u C u00 / .u u0 / dt D .u0 C u000 / .u u0 / C .u C u00 / .u0 u0 / C .u C u00 / .u u00 / D u000 .u u0 /: i d h .u u0 / .u0 u00 / dt D .u0 u0 / .u0 u00 / C .u u00 / .u0 u00 / C .u u0 / .u00 u00 / C .u u0 / .u0 u000 / D .u u00 / .u0 u00 / C .u u0 / .u0 u000 /: r.t / D r.0/e 2t D r0 e 2t ; dv dr a.t / D D2 D 4r0 e 2t : dt dt d d 2 jrj D r r D 2r v D 0 implies that jrj is constant. dt dt Thus r.t / lies on a sphere centred at the origin. d jr dt 31. x D a cos !t C b sin !t y D c sin !t: The curve is a conic section since it has a quadratic equation: by 2 y 2 1 C 2 D 1: x a2 c c Since the path is bounded (jr.t /j jr0 j C .jv0 j=!/), it must be an ellipse. If r0 is perpendicular to v0 , then b D 0 and the path is the ellipse .x=a/2 C .y=c/2 D 1 having semi-axes a D jr0 j and c D jv0 j=!. 35. d 2r D dt 2 gk r.0/ D r0 ; dr dtˇ d r ˇˇ ˇ dt ˇ c Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) tD0 D v0 : lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL Let w D e ct SECTION 11.2 dr . Then dt 2. dr d 2r dw D ce ct C e ct 2 dt dt dt dr ct ct d r e gk ce ct D ce dt dt D e ct gk Z e ct w.t / D e ct gk dt D gk C C: c e ct Let v.t / be the speed of the tank car at time t seconds. The mass of the car at time t is m.t / D M k t kg. At full power, the force applied to the car is F D Ma (since the motor can accelerate the full car at a m/s2 ). By Newton’s Law, this force is the rate of change of the momentum of the car. Thus d h .M dt .M g k C C, so c Put t D 0 and get v0 D 1 e ct v0 c kv D c!0 D r0 C v0 t D r0 C v0 t gk lim t e ct 2c t c!0 t 2 e ct gk lim c!0 2 1 2 gt k; 2 which is the solution obtained in Example 4. v.0/ D ln 3. Mak t : M kt Mat m/s: M kt dr D k r, r.0/ D i C k. dt Let r.t / D x.t /i C y.t /j C z.t /k. Then x.0/ D z.0/ D 1 and y.0/ D 0. Since k .d r=dt / D k .k r/ D 0, the velocity is always perpendicular to k, so z.t / is constant: z.t / D z.0/ D 1 for all t . Thus dx dy dr iC jD D k r D xj dt dt dt dx D dt 2 2ve , then m.T / D .1=e /m.0/, so e2 1 the rocket must burn fraction of its initial mass to e2 accelerate to twice the speed of its exhaust gases. yi: y; dy D x: dt Therefore, d 2x D dt 2 dy D x; dt and x D A cos t C B sin t . Since x.0/ D 1 and y.0/ D 0, we have A D 1 and B D 0. Thus x.t / D cos t and y.t / D sin t . The path has equation Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Ma D Given: m.0/ ve : m.T / If v.0/ D 0 and v.T / D ve then ln.m.0/=m.T // D 1 and m.T / D .1=e/m.0/. The rocket must therefore burn e 1 fraction of its initial mass to accelerate to the speed e of its exhaust gases. Similarly, if v.T / D 1 ln C k Separating this equation into components, It was shown in the text that v.T / M 2a M kt v.t / D Section 11.2 Some Applications of Vector Differentiation (page 642) 1. kt/ C 1/k: The limit of this solution, as c ! 0, is calculated via l’H^opital’s Rule: t e ct c!0 1 kv D Ma The speed of the tank car at time t (before it is empty) is g .ct C e ct c2 lim r.t / D r0 C v0 lim dv dt At t D 0 we have v D 0, so Ma D C =M . Thus C D M 2 a and Thus we have r D r0 C kt/ i k t /v D Ma dv dt D Ma C kv M kt 1 1 ln.Ma C kv/ D ln.M k k C Ma C kv D : M kt g dr D w D v0 C .1 e ct /k dt c dr g ct D e v0 .1 e ct /k dt c e ct g e ct rD v0 tC kCD c c c g 1 k C D: v0 r0 D r.0/ D c c2 (PAGE 642) Downloaded by ted cage (sxnbyln180@questza.com) r D cos t i C sin t j C k: 423 lOMoARcPSD|6566483 www.konkur.in SECTION 11.2 (PAGE 642) ADAMS and ESSEX: CALCULUS 9 Remark: This result also follows from comparing the given differential equation with that obtained for circular motion in the text. This shows that the motion is a rotation with angular velocity k, that is, rotation about the z-axis with angular speed 1. The initial value given for r then forces r D cos t i C sin t j C k: 6. 4. First observe that d jr dt bj2 D 2.r b/ dr D 2.r dt b/ a .r b/ D 0; I K V D R p : 2 The angular velocity of the earth is D .=12/K. so jr bj is constant; for all t the object lies on the sphere centred at the point with position vector b having radius r0 b. Next, observe that d .r dt r0 / a D a .r We use the fixed and rotating frames as described in the text. Assume the satellite is in an orbit in the plane spanned by the fixed basis vectors I and K. When the satellite passes overhead an observer at latitude 45ı , its position is ICK RDR p ; 2 where R is the radius of the earth, and since it circles the earth in 2 hours, its velocity at that point is The rotating frame with origin at the observer’s position has, at the instant in question, its basis vectors satisfying ID b/ a D 0; JDi 1 1 K D p j C p k: 2 2 so r r0 ? a; for all t the object lies on the plane through r0 having normal a. Hence the path of the object lies on the circle in which this plane intersects the sphere described above. The angle between r b and a must therefore also be constant, and so the object’s speed jd r=dt j is constant. Hence the path must be the whole circle. As shown in the text, the velocity v of the satellite as it appears to the observer is given by V D v C R. Thus vD V R R pi R K p .I C K/ D p .I K/ 12 2 2 R R p J D p .I K/ 2 12 2 R p i: D Rj 12 2 5. Use a coordinate system with origin at the observer, i pointing east, and j pointing north. The angular velocity of the earth is 2=24 radians per hour northward: D j: 12 Because the earth is rotating west to east, the true north to south velocity of the satellite will appear to the observer to be shifted to the west by R=12 km/h, where R is the radius of the earth in kilometres. Since the satellite circles the earth at a rate of radians/h, its velocity, as observed at the moving origin, is vR D Rj v makes angle p ! p 1 R=12 2 tan D tan 1 .1=.12 2/ 3:37ı with R the southward direction. Thus the satellite appears to the observer to be moving in a direction 3:37ı west of south. The apparent Coriolis force is R R p J 2 v D 2 K p .I K 12 2 12 2 2R 1 D p JC I 12 6 2 1 2R p i C p . j C k/ : D 6 2 12 2 R i: 12 R=12 D tan 1 .1=12/ 4:76ı R with the southward direction. Thus the satellite appears to the observer to be moving in a direction 4:76ı west of south. vR makes angle tan 1 The apparent Coriolis force is 2 vR D 2 j 12 7. Rj R i D 12 which is pointing towards the ground. 424 Telegram: @uni_k 2R k; 72 1 1 p jC p k 2 2 The angular velocity of the earth is , pointing due north. For a particle moving with horizontal velocity v, the tangential and normal components of the Coriolis force C, and of , are related by CT D 2N v; Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) CN D 2T v: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 11.3 At the north or south pole, T D 0 and N D . Thus CN D 0 and CT D 2 v. The Coriolis force is horizontal. It is 90ı east of v at the north pole and 90ı west of v at the south pole. Section 11.3 Curves and Parametrizations (page 649) 1. At the equator, N D 0 and T D . Thus CT D 0 and CN D 2 v. The Coriolis force is vertical. 8. We continue with the same notation as in Example 4. Since j points northward at the observer’s position, the angle between the direction vector of the sun, S D cos I C sin J and north satisfies cos D S j D On the first quadrant part of the circle x 2 C y 2 D a2 p we have x D a2 y 2 , 0 y a. The required parametrization is p r D r.y/ D a2 y 2 i C yj; .0 y a/: 2. On the first quadrant part of the circle x 2 C y 2 D a2 p we have y D a2 x 2 , 0 x a. The required parametrization is p r D r.x/ D xi C a2 x 2 j; .0 x a/: 3. From the figure we see that cos cos cos C sin sin : For the sun, D 0 and at sunrise and sunset we have, by Example 4, cos D tan = tan , so that ; 0 2 2 D a sin x D a cos D a cos 2 y D a sin D a sin D a cos : 2 DC tan C sin sin tan 2 cos C sin sin D sin sin sin : D sin cos D cos cos The required parametrization is r D a sin i tan 23:3ı tan 40:8ı cos 1 ı sin 23:3 sin 40:8ı .x;y/ 16 a 4. tan 23:3ı tan 26:5ı s s s x D a sin ; y D a cos ; 0 a a a 2 s a s 0s r D a sin i C a cos j; : a a 2 y sin 23:3ı sin 26:5ı .x;y/ s a 27:6ı a x Fig. 11.3-4 to the east and west of north. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k s a 20 hours between sunrise and sunset. By Exercise 8, the sun will rise and set at an angle cos 1 x Fig. 11.3-3 10. At Umea, D 90ı 63:5ı D 26:5ı . On June 21st, D 23:3ı . By Example 4 there will be 52:7ı to the east and west of north. 24 cos 1 2 : a hours between sunrise and sunset. By Exercise 8, the sun will rise and set at an angle a cos j; y 9. At Vancouver, D 90ı 49:2ı D 40:8ı . On June 21st, D 23:3ı . Ignoring the mountains and the rain, by Example 4 there will be 24 cos 1 (PAGE 649) Downloaded by ted cage (sxnbyln180@questza.com) 425 lOMoARcPSD|6566483 www.konkur.in SECTION 11.3 (PAGE 649) ADAMS and ESSEX: CALCULUS 9 5. z D x 2 , z D 4y 2 . If t D y, then z D 4t 2 , so x D ˙2t . The curve passes through .2; 1; 4/ when t D 1, so x D 2t . The parametrization is r D 2t i C t j C 4t 2 k. 6. z D x 2 , x C y C z D 1. If t D x, then z D t 2 and y D 1 t t 2 . The parametrization is r D t i C .1 t t 2 /j C t 2 k. 12. By symmetry, the centre of the circle C of intersection of the plane x C y C z D 1 and the sphere x 2 C y 2 C z 2 D 1 must lie on the plane and must have its three coordinates equal. Thus the centre has position vector r0 D z D x C y, x 2 C y 2 D 9. One possible parametrization is r D 3 cos t i C 3 sin t j C 3.cos t C sin t /k. p 8. x C y D 1, z D 1 x 2 y 2 . If x D t , then y D 1 t and p p z D 1 t 2 .1 t /2 D 2.t t 2 /. One possible parametrization is 7. p t /j C 2.t r D t i C .1 9. z D x 2 C y 2 , 2x 4y z on the vertical cylinder .x Since C passes through the point .0; 0; 1/, its radius is s 4y 2 0 2 C 0 1 3 2 C 1 v2 D .i C j C k/ .i 1 3 2 D r 2 : 3 j/ D i C j 2k: Two perpendicular unit vectors that are parallel to the plane of C are that is 1; 1 3 Any vector v that satisfies v .i C j C k/ D 0 is parallel to the plane x C y C z D 1 containing C. One such vector is v1 D i j. A second one, perpendicular to v1 , is t 2 /k: 1 D 0. These surfaces intersect x 2 C y 2 D 2x 1 .i C j C k/: 3 2 i j vO 1 D p ; 2 1/ C .y C 2/ D 4: vO 2 D i C j 2k : p 6 One possible parametrization is Thus one possible parametrization of C is x D 1 C 2 cos t y D 2 C 2 sin t z D 1 C 2.1 C 2 cos t / 4. 2 C 2 sin t / D 9 C 4 cos t 8 sin t r D .1 C 2 cos t /i 2.1 sin t /j C .9 C 4 cos t 8 sin t /k: 10. yz C x D 1, xz x D 1. One possible parametrization is x D t , z D .1 C t /=t , and y D .1 t /=z D .1 t /t =.1 C t /, that is, 1Ct t t2 jC k: r D ti C 1Ct t 11. 2 .cos t vO 1 C sin t vO 2 / 3 cos t iCjCk sin t C p .i j/ C .i C j D 3 3 3 r D r0 C 13. z 2 D x 2 C y 2 , z D 1 C x. 2 2 2 a) If t D x, then p z D 1 C t , so 1 C 2t C t D t C y , and y D ˙ 1 C 2t . Two parametrizations are needed to get the whole parabola, one for y 0 and one for y 0. t2 1 2 i C tj C 14. 15. 426 Telegram: @uni_k 2 3 r D t i C t j C t k; .0 t T / p p v D 1 C .2t /2 C 9t 4 D .1 C 3t 2 /2 p if 42 D 6, that is, if D ˙ 3=2. In this case, the length of the curve is s.T / D t2 C 1 k: 2 2 2 c) If t D z, then 2t C 1 C y 2 , p x D t 1 and t D t so y D ˙ 2t 1. Again two parametrizations are needed to get the whole parabola. r D t 2 i C t 2 j C t 3 k; .0 t 1/ p p v D .2t /2 C .2t /2 C .3t 2 /2 D t 8 C 9t 2 Z 1 p t 8 C 9t 2 dt Let u D 8 C 9t 2 Length D 0 du D 18t dt ˇ17 p p ˇ 17 17 16 2 1 2 3=2 ˇ u ˇ D units. D ˇ 18 3 27 2k/: 8 b) If t D y, then x 2 C t 2 D z 2 D 1 C 2x C x 2 , so 2x C 1 D t 2 , and x D .t 2 1/=2. Thus z D 1 C x D .t 2 C 1/=2. The whole parabola is parametrized by rD r Z T 0 .1 C 3t 2 / dt D T C T 3 : Z Tˇ ˇ ˇdrˇ ˇ ˇ dt ˇ dt ˇ 1 s Z T c2 D 4a2 t 2 C b 2 C 2 dt units. t 1 Length D Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 11.3 If b 2 D rthen Z T4ac c 2 Length D dt 2at C t 1 Z T c D dt 2at C t 1 D a.T 2 1/ C c ln T units. Since the first octant part of C lies in the plane y D z, it must be a quarter of a circle of radius 1. Thus the length of all of C is 8 .=2/ D 4 units. If you wish to use an integral, the length is Z =2 r 1 1 sin2 t C cos2 t C cos2 t dt 8 2 2 0 Z =2 dt D 4 units. D8 a 16. x D a cos t sin t D sin 2t , 2 a y D a sin2 t D .1 cos 2t /, 2 z D bt . The curve is a circular helix lying on the cylinder 2 x C y (PAGE 649) 0 z x2 C y2 C z2 D 1 x 2 C 2z 2 D 1 a 2 a2 : D 2 4 C Its length, from t D 0 to t D T , is Z Tp a2 cos2 2t C a2 sin2 2t C b 2 dt 0 p D T a2 C b 2 units. y LD x Fig. 11.3-18 19. 17. r D t cos t i C t sin t j C t k; 0 t 2 v D .cos t t sin t /i C .sin t C t cos t /j C k p p v D jvj D .1 C t 2 / C 1 D 2 C t 2 : The length of the curve is LD Z 2 p 2 C t 2 dt D2 0 Z tD2 x D e t cos t; p Let t D 2 tan p dt D 2 sec2 d sec3 d tD0 ˇˇtD2 D sec tan C ln j sec C tan j ˇˇ tD0 !ˇ2 p p 2 C t2 t 2 C t2 t ˇˇ p D C ln Cp ˇ 2 2 2 ˇ0 p p p D 2 C 4 2 C ln 1 C 2 2 C 2 units. The curve is called a conical helix because it is a spiral lying on the cone x 2 C y 2 D z 2 . 18. One-eighth of the curve C lies in the first octant. That part can be parametrized x D cos t; r yD 1 If C is the curve 1 z D p sin t; .0 t =2/ 2 1 2 1 cos2 t sin t D p sin t: 2 2 z D t; .0 t 2/; then the length of C is s Z 2 2 2 2 dy dz dx C C dt LD dt dt dt 0 Z 2 p e 2t .cos t sin t /2 C e 2t .sin t C cos t /2 C 1 dt D 0 Z 2 p D 2e 2t C 1 dt Let 2e 2t C 1 D v 2 0 2e 2t dt D v dv Z tD2 2 Z tD2 v dv 1 D D dv 1 C v2 1 v2 1 tD0 tD0 ˇ ˇˇtD2 1 ˇ v 1 ˇˇ ˇˇ D v C ln ˇˇ ˇ 2 v C 1ˇ ˇ tD0 ˇ2 p p p 1 2e 2t C 1 1 ˇˇ 3 C ln p D 2e 4 C 1 ˇ 2 2e 2t C 1 C 1 ˇ0 ˇ2 p p p 2e 2t C 1 1 ˇˇ 3 C ln p D 2e 4 C 1 ˇ ˇ 2e t 0 p p p 4 4 D 2e C 1 3 C ln 2e C 1 1 p 2 ln. 3 1/ units. Remark: This answer appears somewhat different from that given in the answers section of the text. The two are, however, equal. Somewhat different simplifications were used in the two. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k y D e t sin t; Downloaded by ted cage (sxnbyln180@questza.com) 427 lOMoARcPSD|6566483 www.konkur.in SECTION 11.3 (PAGE 649) 20. ADAMS and ESSEX: CALCULUS 9 r D t 3i C t 2j v D 3t 2 i C 2t j p p v D jvj D 9t 4 C 4t 2 D jt j 9t 2 C 4 The length L between t D LD Z 0 1 2a cos L 1 and t D 2 is Z 2 p p 2 t 9t 2 C 4 dt: . t / 9t C 4 dt C a a 0 2.a C b/ one revolution Making the substitution u D 9t 2 C 4 in each integral, we obtain Z 13 Z 40 1 u1=2 du C 18 4 4 1 3=2 3=2 D 13 C 40 16 27 LD 21. Fig. 11.3-22 Observe that tan D u1=2 du sin D units. cos D 2a 1 . Therefore cos 2.a C b/ a .a C b/ s a2 D 2 .a C b/2 1 p 2 .a C b/2 .a C b/ a2 : The total length of spool required is r1 D t i C t j, .0 t 1/ represents the straight line segment from the origin to .1; 1/ in the xy-plane. H D L sin C 2a cos p a L C 2 2 .a C b/2 D .a C b/ r2 D .1 t /i C .1 C t /j, .0 t 1/ represents the straight line segment from .1; 1/ to .0; 2/. Thus C D C1 C C2 is the 2-segment polygonal line from the origin to .1; 1/ and then to .0; 2/. L sin 2a cos 23. 22. (Solution due to Roland Urbanek, a student at Okanagan College.) Suppose the spool is vertical and the cable windings make angle with the horizontal at each point. a2 units. r D At i C Bt j C C t k. The arc length from the point where t D 0 to the point corresponding to arbitrary t is s D s.t / D Z tp p A2 C B 2 C C 2 du D A2 C B 2 C C 2 t: 0 p Thus t D s= A2 C B 2 C C 2 : The required parametrization is Asi C Bsj C C sk : rD p A2 C B 2 C C 2 24. H 2a b p r D e t i C p2t j e t k t v D e i C p2j C e t k v D jvj D e 2t C 2 C e 2t D e t C e t : The arc length from the point where t D 0 to the point corresponding to arbitrary t is s D s.t / D a Z t 0 .e u C e u / du D e t Fig. 11.3-22 The centreline of the cable is wound around a cylinder of 2a in radius a C b and must rise a vertical distance cos one revolution. The figure below shows the cable unwound from the spool and inclined at angle . The total length of spool required is the total height H of the cable as shown in that figure. 428 Telegram: @uni_k Thus t D sinh 1 p .s=2/ D ln sC e t D 2 sinh t: ! s2 C 4 , 2 p s2 C 4 . The required parametrization is 2 ! p p s C s2 C 4 p 2k s C s2 C 4 iC 2 ln rD j p : 2 2 s C s2 C 4 and e t D sC Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 25. r D a cos3 t i C a sin3 t j C b cos 2t k; 2 4b sin t cos t k ds D g 0 .t / D jv.t /j > 0 dt on Œa; b, by the Fundamental Theorem of Calculus. Hence g is invertible, and defines t as a function of arc length s: t D g 1 .s/ , s D g.t /: 0 1p 2 9a C 16b 2 sin2 t D K sin2 t D 2 1p 2 where K D 9a C 16b 2 2 r r s s , cos t D 1 , Therefore sin t D K K 2s . cos 2t D 1 2 sin2 t D 1 K The required parametrization is rDa 1 Then 27. s 3=2 s 3=2 iCa Cb 1 K K 2s K Section 11.4 Curvature, Torsion, and the Frenet Frame (page 658) k p p t sin t /i C 3.sin t C t cos t /j C 3 2 t k p v D jvj D 3 1 C t 2 C 2t D 3.1 C t / Z t t2 3.1 C u/ du D 3 t C sD 2 0 r 2s 2s 2 , so t D 1 C 1 C since t 0. Thus t C 2t D 3 3 The required parametrization is the given one with t rep placed by 1 C 1 C .2s/=3. As claimed in the statement of the problem, r1 .t / D r2 u.t / , where u is a function from Œa; b to Œc; d , having u.a/ D c and u.b/ D d . We assume u is differentiable. Since u is one-to-one and orientationpreserving, du=dt 0 on Œa; b. By the Chain Rule: 1. 2. 3. 4. ˇ ˇ Z bˇ Z bˇ Z dˇ ˇˇ ˇ ˇ ˇd ˇd ˇ d ˇ r2 u.t / ˇ du dt D ˇ r1 .t /ˇ dt D ˇ r2 .u/ˇ du: ˇ ˇ ˇ dt ˇ du ˇ dt ˇ du a c a 5. 28. If r D r.t / has nonvanishing velocity v D d r=dt on Œa; b, then for any t0 in Œa; b, the function 6. t0 jv.u/j du; 2t 2 j C 3t 3 k i 4t j C 9t 2 k v TO D D p : v 1 C 16t 2 C 81t 4 and so Z t r D ti v D i 4t j C 9t 2 k p v D 1 C 16t 2 C 81t 4 d du d r1 .t / D r2 .u/ ; dt du dt s D g.t / D r D r2 .s/ D r g 1 .s/ is a parametrization of the curve r D r.t / in terms of arc length. 1p 2 9a C 16b 2 . for 0 s K, where K D 2 p 26. r D 3t cos t i C 3t sin t j C 2 2t 3=2 k; .t 0/ v D 3.cos t r D a sin !t i C a cos !t k v D a! cos !t i a! sin !t k; v D ja!j h i TO D sgn.a!/ cos !t i sin !t k : r D cos t sin t i C sin2 t C cos t k 1 1 D sin 2t i C .1 cos 2t /j C cos t k 2 2 v D cos 2t i C sin 2t j sin t k p v D jvj D 1 C sin2 t 1 TO D p cos 2t i C sin 2t j sin t k : 1 C sin2 t r D a cos t i C b sin t j C t k v D a sin t i C b cos t j C k p v D a2 sin2 t C b 2 cos2 t C 1 a sin t i C b cos t j C k v : TO D D p v a2 sin2 t C b 2 cos2 t C 1 d TO O D 0, so D N ds dr O O O T.s/ D T.0/ is constant. This says that D T.0/, so ds O r D T.0/s C r.0/, which is the vector parametric equation of a straight line. If .s/ D 0 for all s, then If .s/ D 0 for all s, then O dB O D 0, so B.s/ O O D N D B.0/ is constant. Therefore, ds dr O d O O O B.s/ D T.s/ B.s/ D 0: r.s/ r.0/ B.s/ D ds ds Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k (PAGE 658) which gives the (signed) arc length s measured from r.t0 / along the curve, is an increasing function: 0t 3a cos2 t sin t i C 3a sin2 t cos t j p v D 9a2 C 16b 2 sin t cos t Z tp sD 9a2 C 16b 2 sin u cos u du vD SECTION 11.4 Downloaded by ted cage (sxnbyln180@questza.com) 429 lOMoARcPSD|6566483 www.konkur.in SECTION 11.4 (PAGE 658) ADAMS and ESSEX: CALCULUS 9 It follows that O r.s/ r.0/ B.0/ D r.s/ for all s. This says that r.s/ lies in the plane through r.0/ O having normal B.0/. 7. Hence .0/ D 1 and .=2/ D 0. The radius of curvature at x D 0 is 1. The radius of curvature at x D =2 is infinite. O r.0/ B.s/ D0 3. The circle C1 given by r D 2t i C .1=t /j v D 2i .1=t 2 /j 2t k 2k a D .2=t 3 /j 1 1 rD cos C si C sin C sj C C v a D .4=t 3 /i C .4=t 3 /k At .2; 1; 2/, that is, at t D 1, we have is parametrized in terms of arc length, and has curvature C and torsion 0. (See Examples 2 and 3.) If curve C has constant curvature .s/ D C and constant torsion .s/ D 0, then C is congruent to C1 by Theorem 3. Thus C must itself be a circle (with radius 1=C ). D .1/ D p jv aj 4 2 D : v3 27 p Thus the radius of curvature is 27=.4 2/. 8. The circular helix C1 given by r D a cos t i C a sin t j C bt k 4. has curvature and torsion given by .s/ D a ; a2 C b 2 .s/ D v D 3t 2 i C 2t j C k a D 6t i C 2j v.1/ D 3i C 2j C k; a.1/ D 6i C 2j v.1/ a.1/ D 2i C 6j 6k p p 2 19 4 C 36 C 36 D 3=2 .1/ D .9 C 4 C 1/3=2 14 p At t D 1 the radius of curvature is 143=2 =.2 19/. b ; a2 C b 2 by Example 3. if a curve C has constant curvature .s/ D C > 0, and constant torsion .s/ D T ¤ 0, then we can choose a and b so that a D C; a2 C b 2 b D T: a2 C b 2 5. r D t i C t 2 j C 2k v D i C 2t j a D 2j v a D 2k At .1; 1; 2/, where t D 1, we have p TO D v=jvj D .i C 2j/= 5 BO D .v a/=jv aj D k p O D BO TO D . 2i C j/= 5: N 6. r D t i C t 2j C t k v D i C 2t j C k a D 2j v a D 2i C 2k At .1; 1; 1/, where t D 1, we have p TO D v=jvj D .i C 2j C k/= 6 p BO D .v a/=jv aj D .i k/= 2 p O D BO TO D .i j C k/= 3: N T C , and b D 2 .) By C2 C T 2 C CT2 Theorem 3, C is itself a circular helix, congruent to C1 . (Specifically, a D Section 11.5 Curvature and Torsion for General Parametrizations (page 664) 1. For y D x 2 we have .x/ D jd 2 y=dx 2 j 2 D : .1 C .dy=dx/2 /3=2 .1 C 4x 2 /3=2 p Hence .0/ D 2 and . 2/ p D 2=27. The radii of curvature at x D 0 and x D 2 are 1=2 and 27=2, respectively. 2. For y D cos we have .x/ D 430 Telegram: @uni_k jd 2 y=dx 2 j j cos xj : D .1 C .dy=dx/2 /3=2 .1 C sin2 x/3=2 r D t 3i C t 2j C t k Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 7. t3 t2 jC k 2 3 v D i C t j C t 2k da D 2k a D j C 2t k; dt 2 v a D t i 2t j C k p v D jvj D 1 C t 2 C t 4 ; da .v a/ D2 dt SECTION 11.5 9. r D ti C jv aj D p 1 C 4t 2 C t 4 i C t j C t 2k O D v D p T v 1 C t2 C t4 v a t 2 i 2t j C k O D B D p jv aj 1 C 4t 2 C t 4 3 C t /i C .1 t 4 /j C .t 3 C 2t /k O DB O TO D .2t N p .1 C t 2 C t 4 /.1 C 4t 2 C t 4 / p 1 C 4t 2 C t 4 jv aj D D 3 v .1 C t 2 C t 4 /3=2 da .v a/ 2 dt D D : jv aj2 1 C 4t 2 C t 4 8. r D e t cos t i C e t sin t j C e t k v D e t .cos t sin t /i C e t .sin t C cos t /j C e t k a D 2e t sin t i C 2e t cos t j C e t k da D 2e t .cos t C sin t /i C 2e t .cos t sin t /j C e t k dt v a D e 2t .sin t cos t /i e 2t .cos t C sin t /j C 2e 2t k p p jv aj D 6e 2t v D jvj D 3e t ; da .v a/ D 2e 3t dt v .cos t sin t /i C .cos t C sin t /j C k p TO D D v 3 .sin t cos t /i .cos t C sin t /j C 2k v a p D BO D jv aj 6 .cos t C sin t /i .cos t sin t /j O D BO T O D p N 2 p 2 jv aj D t D v3 3e da .v a/ dt D 1 : D 2 jv aj 3e t 10. p r D .2 C 2 cos t /i C .1 sin t /j C .3 C sin t /k p 2 sin t i cos t j C cos t k vD p p v D 2 sin2 t C cos2 t C cos2 t D 2 p aD 2 cos t i C sin t j sin t k p da D 2 sin t i C cos t j cos t k dt p p 2j 2k va D 1 jv aj 2 D D p D p v3 2 2 2 p p da .v a/ 2 cos t C 2 cos t D 0 D dt D 0: p Since D 1= 2 is constant, and D 0, thepcurve is a circle. Its centre is .2; 1; 3/ andpits radius is 2. It lies in O 2B/. a plane with normal j C k.D r D xi C sin xj dx dx vD i C cos x j D k.i C cos xj/ dt dt p v D k 1 C cos2 x dx a D k sin x j D k 2 sin xj dt v a D k 3 sin xk j sin xj jv aj D : D v3 .1 C cos2 x/3=2 The tangential and normal components of acceleration are k dx dv D p 2 cos x/. sin x/ D 2 dt dt 2 1 C cos x k 2 j sin xj : v2 D p 1 C cos2 x 11. k 2 cos x sin x p 1 C cos2 x r D sin t cos t i C sin2 t j C cos t k v D cos 2t i C sin 2t j sin t k a D 2 sin 2t i C 2 cos 2t j cos t k da D 4 cos 2t i 4 sin 2t j C sin t k: dt At t D 0 we have v D i, a D 2j k, da D dt da v a D j C 2k, .v a/ D 0. dtp O D .2j O O ThuspT D i, B D .j C 2k/= 5, N D 5, and D 0. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k (PAGE 664) Downloaded by ted cage (sxnbyln180@questza.com) 4i, p k/= 5, 431 lOMoARcPSD|6566483 www.konkur.in SECTION 11.5 (PAGE 664) ADAMS and ESSEX: CALCULUS 9 1 1 p k, a D 2i p k, 2 2 p 1 1 da D 4j C p k, v a D p i C 2j C 2k, dt 2 2 p da .v a/ D 3 2. dt Thus p O D p1 . 2j k/ T 3 p 1 O D p . i C 2j C 2 2k/ B 13 p O D p1 .6i C j C 2k/ N 39 p p 2 39 6 2 D ; D : 9 13 The rate of change of the speed, dv=dt , is the tangential component of the acceleration, and is due entirely to the tangential component of the gravitational force since there is no friction: At t D =4 we have v D j 12. dv O D g cos D g. j/ T; dt O and j. (See the figwhere is the angle between T ure.) Since the p slope of y D x 2 at .1; 1/ is 2, p we have O D .i C 2j/= 5, and therefore dv=dt D 2g= 5. T y r D a cos t i C b sin t j v D a sin t i C b cos t j a D a cos t i b sin t j v a D abk p v D a2 sin2 t C b 2 cos2 t : .1; 1/ gj dv O T dt The tangential component of acceleration is .a2 b 2 / sin t cos t dv D p ; dt a2 sin2 t C b 2 cos2 t 15. jv aj ab D p : 2 v3 2 a sin t C b 2 cos2 t ex . Therefore, the radius .1 C e 2x /3=2 2x 3=2 .1 C e / . of curvature is D ex The unit normal is ab .a2 sin2 t C b 2 cos2 t /3=2 ab D 3=2 : .a2 b 2 / sin2 t C b 2 x O DB O TO D .v a/ v D p e i C j : N j.v a/ vj 1 C e 2x The centre of curvature is If a > b > 0, then the maximum curvature occurs when sin t D 0, and is a=b 2 . The minimum curvature occurs when sin t D ˙1, and is b=a2 . O rc D r C N D .x 2 D p : 5 5 xD1 432 iC This is the equation of the evolute. 16. Thus the magnitude of the normal p acceleration of the bead at that point is v 2 D 2v 2 =.5 5/. 1 1 j ex e 2x /i C .2e x C e x /j: D xi C e x j C .1 C e 2x / By Example 2, the curvature of y D x 2 at .1; 1/ is ˇ ˇ 2 ˇ D ˇ 2 3=2 .1 C 4x / ˇ jv aj D e x : The curvature is D D Telegram: @uni_k Curve: r D xi C e x j. p Velocity: v D i C e x j. Speed: v D 1 C e 2x . Acceleration: a D e x j. We have v a D e x k; 13. The ellipse is the same one considered in Exercise 16, so its curvature is 14. x Fig. 11.5-14 which is zero if t is an integer multiple of =2, that is, at the ends of the major and minor axes of the ellipse. The normal component of acceleration is v2 D v2 y D x2 O v2 N The curve with polar equation r D f . / is given parametrically by r D f . / cos i C f . / sin j: Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 11.5 Thus we have v D f 0 . / cos f . / sin i C f 0 . / sin C f . / cos j a D f 00 . / cos 2f 0 . / sin f . / cos i C f 00 . / sin C 2f 0 . / cos f . / sin j r 2 2 f 0 . / C f . / v D jvj D h 2 2 i v a D 2 f 0 . / C f . / f . /f 00 . / k: The curvature is, therefore, 2 2 j2 f 0 . / C f . / f . /f 00 . /j : h 2 2 i3=2 f 0 . / C f . / 17. Therefore r.t / D 3a2 .1 2a2 .1 cos / 3 : 3=2 D p 2 2ar cos / r.t / D 19. Given that 20. 1 t C sin t p C1 iC 2 2 For r D a cos t i C a sin t j C bt k, we have, by Example 3 of Section 2.4, cos t i sin t j; D a : a2 C b 2 The centre of curvature rc is given by O O D r C 1 N: rc D r C N Thus the evolute has equation 1 1 O i D T.0/ D p j1 C p k1 2 2 O j D N.0/ D i1 1 1 O k D B.0/ D p j1 C p k1 : 2 2 r D a cos t i C a sin t j C bt k D Solving these equations for i1 , j1 , and k1 in terms of the given basis vectors, we obtain a2 C b 2 .cos t i C sin t j/ a 2 b b2 cos t i sin t j C bt k: a a The evolute is also a circular helix. 21. The parabola y D x 2 has curvature D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k t sin t cos t p k: jC 2 2 2 dr D c r.t /, we have dt O D N for some right-handed basis fi1 ; j1 ; k1 g, and some constant vector r0 . Example 3 of Section 2.4 provides values for O O O T.0/, N.0/, and B.0/, which we can equate to the given values of these vectors: j 1 1 j1 D p i p k 2 2 1 1 k1 D p i C p k: 2 2 Thus jr.t /j D jr.0/j is constant, and r.t / r.0/ c D 0 is constant. Thus r.t / lies on the sphere centred at the origin with radius jr.0/j, and also on the plane through r.0/ with normal c. The curve is the circle of intersection of this sphere and this plane. 1 1 1 cos t i1 C sin t j1 C t k1 C r0 2 2 2 i1 D cos t t sin t p k C r0 : jC 2 2 2 d 2 d jrj D r r D 2r .c r/ D 0 dt dt d dr c D .c r/ c D 0: r.t / r.0/ c D dt dt 18. By Exercise 8 of Section 2.4, the required curve must be a circular helix with parameters a D 1=2 (radius), and b D 1=2. Its equation will be rD t C sin t p i 2 2 1 We also require that r.0/ D i, so r0 D iC j. The required 2 equation is, therefore, If r D a.1 cos /, then r 0 D a sin , and r 00 D a cos . By the result of Exercise 20, the curvature of this cardioid is ˇ 1 ˇ 2 2 D 3=2 ˇ2a sin 2 a2 sin C a2 .1 cos /2 ˇ ˇ C a2 .1 cos /2 a2 .cos cos2 /ˇ D (PAGE 664) Downloaded by ted cage (sxnbyln180@questza.com) 2 ; .1 C 4x 2 /3=2 433 lOMoARcPSD|6566483 www.konkur.in SECTION 11.5 (PAGE 664) ADAMS and ESSEX: CALCULUS 9 by Exercise 18. The normal at .x; x 2 / is perpendicular to the tangent, so has slope 1=.2x/. Since the unit normal points upward (the concave side of the parabola), we have The conditions at x D 1 become A C A C O D p 2xi C j : N 1 C 4x 2 .1 C 4x 2 /3=2 2 p 2xi C j 1 C 4x 2 1 C 4x 2 D xi C x 2 j .1 C 4x 2 /xi C j 2 1 D 4x 3 i C 3x 2 C j: 2 C C C 15 x 8 f .x/ D y .1;1/ 2 , so the radius of .3 sin2 t C 1/3=2 2 3=2 .3 sin t C 1/ curvature is D . We have 2 x yD 1 . 1; 1/ O Dk B Fig. 11.5-23 24. We require f .0/ D 1; f . 1/ D 1; d2 p 1 dx 2 2 D 3 cos3 i 2 3 sin t C 1 .cos t i C 2 sin t j/ 2 As in Example 5, we try 1 follows from the fact that ˇ ˇ ˇ x2 ˇ ˇ xD0 D 1: f .x/ D A C Bx C C x 2 C Dx 3 C Ex 4 C F x 5 f 0 .x/ D B C 2C x C 3Dx 2 C 4Ex 3 C 5F x 4 f .1/ D 1; f . 1/ D 1; 0 f .1/ D 0; f 0 . 1/ D 0; 00 f .1/ D 0; f 00 . 1/ D 0: f 00 D 2C C 6Dx C 12Ex 2 C 20F x 3 : The required conditions force the coefficients to satisfy the system of equations As in Example 5, we try a polynomial of degree 5. However, here it is clear that an odd function will do, and we need only impose the conditions at x D 1. Thus we try f .x/ D Ax C Bx 3 C C x 5 f 0 .x/ D A C 3Bx 2 C 5C x 4 f 00 .x/ D 6Bx C 20C x 3 : Telegram: @uni_k f 00 .0/ D 1; f 00 . 1/ D 0: 3 sin3 t j: 23. We require that 434 f 0 .0/ D 0; f 0 . 1/ D 0; The condition f 00 .0/ D Therefore the evolute has equation r D 2 cos t i C sin t j yD1 yDf .x/ The curvature is D O D N 5=4, and is one possible solution. v D 2 sin t i C cos t j a D 2 cos t i sin t j v a D 2k p p v D 4 sin2 t C cos2 t D 3 sin2 t C 1: 2 sin t i C cos t j p ; 3 sin2 t C 1 cos t i C 2 sin t j p : 3 sin2 t C 1 1 0 0: 5 3 3 5 x C x 4 8 22. For the ellipse r D 2 cos t i C sin t j, we have TO D D D D C 5C 20C This system has solution A D 15=8, B D C D 3=8. Thus Thus the evolute of the parabola has equation r D xi C x 2 j C B 3B 6B A B CC DCE F D1 B 2C C 3D 4E C 5F D 0 2C 6D C 12E 20F D 0 AD1 BD0 2C D 1 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 11.5 which has solution A D 1, B D 0, C D 1=2, D D 3=2, E D 3=2, F D 1=2. Thus we can use a track section in the shape of the graph of f .x/ D 1 1 2 x 2 3 3 x 2 3 4 x 2 1 5 x D1 2 This leads to the values p 2 p 2 cos.t /2 C 2 .cos.t /2 C 1/ 1 2 x .1 C x/3 : 2 and 0 for the curvature and torsion, respectively. Maple doesn’t seem to recognize that the curvature simplifies to 1=.cos2 t C 1/3=2 . The torsion is zero because the curve is lies in the plane z D x. It is the ellipse in which this plane intersects the ellipsoid 2x 2 C y 2 C 2z 2 D 4. The maximum and minimum values of the curvature are 1 and 1=23=2 , respectively, at the ends of the major and minor axes of the ellipse. y . 1;1/ yD1 (PAGE 664) yDf .x/ x 27. > R := t -> <t-sin(t), 1-cos(t), t>; > assume(t::real): > interface(showassumed=0): > V := t -> diff(R(t),t): > A := t -> diff(V(t),t): > v := t -> Norm(V(t),2): > VxA := t -> V(t) &x A(t): > vxa := t -> Norm(VxA(t),2): > Ap := t -> diff(A(t),t): > Curv := t -> > simplify(vxa(t)/(v(t))^3): > Tors := t -> simplify( > (VxA(t).Ap(t))/(vxa(t))^2): > Curv(t); Tors(t); x 2 Cy 2 D1 Fig. 11.5-24 25. Given: a.t / D .t /r.t / C .t /v.t /, v a ¤ 0. We have v a D v r C v v D v r da D 0 r C v C 0 v C a dt D 0 r C . C 0 /v C .r C v/ D .0 C /r C . C 0 C 2 /v: Since v r is perpendicular to both v and r, we have .v a/ After loading the LinearAlgebra and VectorCalculus packages, issue the following commands: da D 0: dt This leads to the values p cos.t /2 C 2 2 cos.t / .3 2 cos.t //3=2 Thus the torsion .t / of the curve is identically zero. It remains zero when expressed in terms of arc length: .s/ D 0. By Exercise 6 of Section 2.4, r.t / must be a plane curve. and 26. After loading the LinearAlgebra and VectorCalculus packages, issue the following commands: > R := t -> <cos(t), 2*sin(t), cos(t)>; > assume(t::real): > interface(showassumed=0): > V := t -> diff(R(t),t): > A := t -> diff(V(t),t): > v := t -> Norm(V(t),2): > VxA := t -> V(t) &x A(t): > vxa := t -> Norm(VxA(t),2): > Ap := t -> diff(A(t),t): > Curv := t -> > simplify(vxa(t)/(v(t))^3): > Tors := t -> simplify( > (VxA(t).Ap(t))/(vxa(t))^2): > Curv(t); Tors(t); 1 2 cos.t /2 C sin.t /2 for the curvature and torsion, respectively. Each of these formulas can be simplified somewhat: p 2 2 cos t C cos2 t .3 2 cos t /3=2 1 Tors.t / D : 2 2 cos t C cos2 t Curv.t / D Since 3 2 cos t > 0 and 2 2 cos t C cos2 t D 1 C .1 cos t /2 > 0 for all t , the curvature and torsion are both continuous for all t . The curve appears to be some sort of helix (but not a circular one) with central axis along the line x D z, y D 1. 28. After loading the LinearAlgebra and VectorCalculus packages, issue the following commands: Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 cos.t / C 1 Downloaded by ted cage (sxnbyln180@questza.com) 435 lOMoARcPSD|6566483 www.konkur.in SECTION 11.5 (PAGE 664) ADAMS and ESSEX: CALCULUS 9 > R := t -> <cos(t)*cos(2*t), cos(t)*sin(2*t), sin(t)>; > assume(t::real): > interface(showassumed=0): > V := t -> diff(R(t),t): > A := t -> diff(V(t),t): > v := t -> Norm(V(t),2): > VxA := t -> V(t) &x A(t): > vxa := t -> Norm(VxA(t),2): > Ap := t -> diff(A(t),t): > Curv := t -> > simplify(vxa(t)/(v(t))^3): > Tors := t -> simplify( > (VxA(t).Ap(t))/(vxa(t))^2): > Curv(t); Tors(t); > simplify(%,trig); This leads to the values p 2 cos.t /2 C cos t C 1 3=2 5 cos.t /2 C 2 cos t 1 Tors.t / D 2.cos.t /2 / C cos t C 1 Curv.t / D This appears to be an elliptical helix with central axis along the line x D y D z 1. 30. evolute := R -> (t -> R(t)+TNBFrame(R)[2](t) *(1/Curvature(R)(t))); 31. tanline := R -> ((t,u) -> R(t)+TNBFrame(R)[1](t)*u); The last line simplifies the rather complicated expression that Tors(t) returns by applying some trigonometric identities. The values for the curvature and torsion are p 17 C 60 cos.t /2 C 48 cos.t /4 Curv.t / D 3=2 4 cos.t /2 C 1 12 cos t .2 cos.t /2 C 3/ : Tors.t / D 17 C 60 cos.t /2 C 48 cos.t /4 Section 11.6 Kepler’s Laws of Planetary Motion (page 673) 1. 436 Telegram: @uni_k r C x D ` 2`x C 2 x 2 2 /x 2 C 2`x C y 2 D `2 2 `2 `2 2 ` 2 2 D C y D ` C .1 2 / x C 1 2 1 2 1 2 2 ` xC y2 1 2 2 D 1: 2 C ` ` p 1 2 1 2 .1 The command simplify(Norm(R(t),2)); gives output 1, indicating that the curve lies on the sphere x 2 C y 2 C z 2 D 1. > R := t -> <t+cos(t), t+sin(t), 1+tcos(t)>; > assume(t::real): > interface(showassumed=0): > V := t -> diff(R(t),t): > A := t -> diff(V(t),t): > v := t -> Norm(V(t),2): > VxA := t -> V(t) &x A(t): > vxa := t -> Norm(VxA(t),2): > Ap := t -> diff(A(t),t): > Curv := t -> > simplify(vxa(t)/(v(t))^3): > Tors := t -> simplify( > (VxA(t).Ap(t))/(vxa(t))^2): > Curv(t); Tors(t); ÷ x 2 C y 2 D r 2 D `2 Plotting the curvature as a function of t , (plot(Curv(t),t=-2*Pi..2*Pi)), shows that the minimum curvature occurs at t D 0 (and p any integer multiple of ). The minimum curvature is 125=53=2 D 1. 29. After loading the LinearAlgebra and VectorCalculus packages, issue the following commands: ` 1 C cos r D ` x rD 2. Position: r D r rO D k rO . Velocity: v D k rPO D k P O ; speed: v D k P . P Acceleration: k R O C k P O D k P 2 rO C k R O . Radial component of acceleration: k P 2 . Transverse component of acceleration: k R D vP (the rate of change of the speed). 3. Position: on the curve r D e . Radial velocity: rP D e P . Transverse velocity: r P D e P . p p P Speed v D 2ep D 1 ÷ P D .1= 2/e . Thus R D .1= 2/e P D e 2 =2. p Radial velocity = transverse velocity D 1= 2. Radial acceleration: rR r P 2 D e P 2 C e R e P 2 D e R D e =2. Transverse acceleration: r R C 2rP P D .e /=2 C e D e =2. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 11.6 4. Path: r D .qThus rP D P , rR D R . p Speed: v D .Pr/2 C .r P /2 D P 1 C r 2 . Transverse acceleration = 0 (central force). Thus r R C 2rP P D 0, or R D 2P 2 =r. Radial acceleration: rR r P 2 D R D The satellite has a circular orbit of radius aS and period TS D 1 day. (If the orbit is in the plane of the equator, the satellite will remain above the same point on the earth.) By Kepler’s third law, TS2 aS3 r P 2 2 C r P 2 D r 8. For r D 2 2 3 P D 2hP D 2 h 2 3 2 D r 2h: : The period T (in years) and radius R (in km) of the asteroid’s orbit satisfies If R is the radius and T is the period of the asteroid’s circular orbit, then almost stopping the asteroid causes it to drop into a very eccentric elliptical orbit with major axis approximately R. (Thus, a D R=2.) The period Te of the new elliptical orbit satisfies Te2 .R=2/3 1 D D : T2 R3 8 2 2h =r . The speed v is given by v 2 D rP 2 C r 2 P 2 D 4h2 2 C .h2 =r 2 /: 3 aM Thus the radius of the asteroid’s orbit is R 150 106 T 2=3 km. 9. , we have rP D Thus rR D r P 2 j: 2 TM T2 T2 12 D D earth : 3 3 R .150 106 /3 Rearth 5. For a central force, r 2 P D h (constant), and the acceleration is wholly radial, so jaj D jRr D Thus aS D 385; 000 .1=27/2=3 42; 788. The satellite’s orbit should have radius about 42,788 km, and should lie in the equatorial plane. .2 C r 2 /v 2 : r.1 C r 2 / The magnitude of the acceleration is, therefore, .2 C r 2 /v 2 . r.1 C r 2 / (PAGE 673) p Thus Te D T =.2 2/. The time the asteroid will p take to fall into the sun is half of Te . Thus it is T =.4 2/. Since the speed is v0 when pD 1 (and so r D 1), we have v02 D 5h2 , and h D v0 = 5. Hence the magnitude of the acceleration at any point on the path is ˇ ˇ h2 jaj D ˇˇ 2 2 r r 6. Let the period and the semi-major axis of the orbit of Halley’s comet be TH D 76 years and aH km respectively. Similar parameters for the earth’s orbit are TE D 1 year and aE D 150 106 km. By Kepler’s third law TH2 3 aH T2 D 3E : aE Thus aH D 150 106 762=3 2:69 109 : The major axis of Halley’s comet’s orbit is 2aH 5:38 109 km. 7. R ˇ v02 2 h2 ˇˇ 1 D C : r4 ˇ 5 r2 r3 Fig. 11.6-9 10. At perihelion, r D a c D .1 /a. At aphelion r D a C c D .1 C /a. Since rP D 0 at perihelion and aphelion, the speed is v D r P at each point. Since r 2 P D h is constant over the orbit, v D h=r. Therefore vperihelion D vaphelion D ; h aM 385; 000 km. a.1 Hence 1 C D 2.1 the orbit is 1/3. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k / h : a.1 C / If vperihelion D 2vaphelion then The period and semi-major axis of the moon’s orbit around the earth are TM 27 days, h a.1 Downloaded by ted cage (sxnbyln180@questza.com) / D 2h : a.1 C / /, and D 1=3. The eccentricity of 437 lOMoARcPSD|6566483 www.konkur.in SECTION 11.6 (PAGE 673) 11. ADAMS and ESSEX: CALCULUS 9 The orbital speed v of a planet satisfies (by conservation of energy) v2 2 k DK r 14. (total energy). If v is constant so must be r, and the orbit will therefore be circular. As in Exercise 12, rP vP D rA vA , where rA D `=.1 / and rP D `=.1 C /, being the eccentricity of the orbit. Thus rA 1C vP D D : vA rP 1 Solving this equation for in terms of vP and vA , we get 2P 12. Since r D h D constant for the planet’s orbit, and since the speed is v D r P at perihelion and at aphelion (the radial velocity is zero at these points), we have rp vp D ra va ; By conservation of energy the speed v at the ends of the minor axis of the orbit (where r D a) satisfies where the subscripts p and a refer to perihelion and aphelion, respectively. Since rp =ra D 8=10, we must have vp =va D 10=8 D 1:25. Also, rp D ` ` D ; 1 C cos 0 1C ra D ` ` D : 1 C cos 1 aD k.1 vP2 D vP2 C D vP2 D vP2 .1 C /R R D : 4.1 2 / 4.1 / D vP2 It follows that 1 D 4 4, so D 3=4. The new elliptical orbit has eccentricity D 3=4. Thus v D 15. c S a vA2 D 2k v 2 D vP2 C 2k H2 R D D : 2/ 4k.1 2 / 4.1 2 / R p Telegram: @uni_k 4k : ` 1 rP 1 rA D 1 a 1 rP 2k 1 2 .1 C / ` 2k .1 C / ` 2 vP vA2 vP vA 1C 2 vP C vA vP vA .2vP / D vP vA : 2 Since the radial line from the sun to the planet sweeps out equal areas in equal times, the fraction of the planet’s period spend on the same side of the minor axis as the sun is equal to the shaded area in the figure to the total area of the ellipse, that is, 1 2 .2bc/ ab 438 vP vA . 1 2 ab Fig. 11.6-13 k : rA Using this result and the parameters of the orbit given in the text, we obtain R , so Similarly, c D a D 4.1 2 / R DcCa D v2 k D A rP 2 The latter equality shows that 13. Let the radius of the circular orbit be R, and let the parameters of the new elliptical orbit be a and c, as shown in the figure. Then R D a C c. At the moment of the collision, r does not change (r D R), but the speed r P is cut in half. Therefore P is cut in half, and so h D r 2 P is cut in half. Let H be the value of r 2 P for the circular orbit, and let h be the value for the new elliptical orbit. Thus h D H=2. We have h2 v2 k D P a 2 v2 2 Thus `=.1C/ D .8=10/`=.1 /, and so 10 10 D 8C8. Hence 2 D 18. The eccentricity of the orbit is D 1=9. H2 RD ; k vP vA : vP C vA D 1 D 2 ab ab 1 D ab 2 where D c=a is the eccentricity of the orbit. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) ; lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 11.6 2 Thus T D p k v02 k 2 r0 3=2 : y a b (PAGE 673) A c a b a c S P x Fig. 11.6-15 Fig. 11.6-16 17. 16. By conservation of energy, we have 1 2 h2 rP C 2 D 2 r k r K where K is a constant for the orbit (the total energy). The term in the parentheses is v 2 , the square of the speed. Thus k 1 2 k 1 2 v D KD v ; r 2 r0 2 0 where r0 and v0 are the given distance and speed. We evaluate K at perihelion. The parameters of the orbit are `D h2 ; k aD h2 k.1 2/ ; bD p h2 k 1 2 ; Let r1 .s/ and r2 .s/ be the distances from the point P D r.s/ on the ellipse E to the two foci. (Here s denotes arc length on E, measured from any convenient point.) By symmetry Z Z E r1 .s/ ds D E r2 .s/ ds: But r1 .s/ C r2 .s/ D 2a for any s. Therefore, Z E r1 .s/ ds C Z E r2 .s/ ds D Z E 2a ds D 2ac.E/: R Hence E r1 .s/ ds D ac.E/, and Z 1 r1 .s/ ds D a: c.E/ E c D a: y P At perihelion P we have r Da c D .1 /a D Since rP D 0 at perihelion, the speed there is v D r P . By Kepler’s second law, r 2 P D h, so v D h=r D k.1 C /= h. Thus k v2 KD r 2 k2 1 k2 D 2 .1 C / .1 C /2 h 2 h2 h i k2 D 2 .1 C / 2 .1 C / 2h k2 k D 2 .1 2 / D : 2h 2a Thus a D k . By Kepler’s third law, 2K T2 D 4 2 4 2 3 a D k k r2 h2 : k.1 C / k 2K 3 F2 F1 Fig. 11.6-17 18. Start with rR h2 D r3 k : r2 1 , where D .t /. Since r 2 P D h u. / (constant), we have Let r.t / D rR D du h du 1 du P D r2 D h u2 d d r 2 d 2 d 2u P h2 d 2 u 2 2d u h 2 D D h u : 2 2 d r d d 2 Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k x E rP D : r1 Downloaded by ted cage (sxnbyln180@questza.com) 439 lOMoARcPSD|6566483 www.konkur.in SECTION 11.6 (PAGE 673) Thus h2 u2 d 2u d 2 ADAMS and ESSEX: CALCULUS 9 ku2 , or Such orbits are bounded away from zero and infinity only if A D 0, in which case they are circular. k d 2u Cu D 2: d 2 h Thus, the only possible orbits which are bounded away from zero and infinity (i.e., which do not escape to infinity or plunge into the sun) in a universe with an inverse cube gravitational attraction are some circular orbits for which h2 D k. Such orbits cannot be considered “stable” since even slight loss of energy would result in decreased h and the condition h2 D k would no longer be satisfied. Now aren’t you glad you live in an inverse square universe? h2 u3 D This is the DE for simple harmonic motion with a constant forcing term (nonhomogeneous term) on the righthand side. It is easily verified that uD k 1 C cos. h2 0 / is a solution for any choice of the constants and 0 . Expressing the solution in terms of r, we have h2 =k 1 C cos. rD 0 / 20. h D r3 k ; r3 where r 2 P D h is constant, since the force is central. Making the same change of variables used in Exercise 18, we obtain d 2u h2 u2 2 h2 u3 D ku3 ; d or d 2 u k h2 u D 0: d 2 h2 There are three cases to consider. d 2u C ! 2 u D 0, where d 2 k/= h2 . This has solution u D A cos !. 0 /. CASE I. If k < h2 the DE is ! 2 D .h2 Thus k K > 0; r k . The orbit is, therefore, bounded. K ` rD , . > 1/. 1 C cos See the following figure. Vertices: At V1 , D 0 and r D `=.1 C /. At V2 , D and r D `=.1 / D `=. 1/. Semi-focal separation: ` ` ` 1 C D 2 : cD 2 1C 1 1 The centre is .c; 0/. Semi-transverse axis: ` ` ` D 2 : aD 2 1 C1 1 Semi-conjugate axis: p ` : b D c 2 a2 D p 2 1 Direction of asymptotes (see figure): b a 1 D tan 1 D cos 1 D cos 1 : a c so r 19. For inverse cube attraction, the equation of motion is rR K by conservation of energy, if K < 0, ; which is an ellipse if jj < 1. 2 1 k D v2 Since r 2 then 1 : A cos !. 0 / . There are no bounded Note that r ! 1 as ! 0 C 2! orbits in this case. rD 21. d 2u ! 2 u D 0, where d 2 ! 2 D .k h2 /= h2 . This has solution u D Ae ! C Be ! . Since u ! 0 or 1 as ! 1, the corresponding solution r D 1=u cannot be both bounded and bounded away from zero. (Note that P D h=r 2 K > 0 for any orbit which is bounded away from zero, so we can be sure ! 1 on such an orbit.) y CASE II. If k > h2 the DE is c F1 V1 b F2 C a d 2u D 0, which has d 2 solutions u D A C B, corresponding to CASE III. If k D h2 the DE is rD 440 Telegram: @uni_k 1 : A C B Fig. 11.6-21 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) V2 x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES 11 22. By Exercise 17, the asymptotes make angle D cos 1 .1=/ with the transverse axis, as shown in the figure. The angle of deviation ı satisfies 2 C ı D , so ı D , and 2 2 ı cos D sin ; 2 (PAGE 675) Therefore 2 v2 a ı ı Dv1 D 1 cot D cot : k k 2 2 ı sin D cos : 2 Review Exercises 11 (page 675) y 1. Given that a r D 0 and a v D 0, we have d jr.t / t v.t /j2 dt D 2 r.t / t v.t / v.t / v.t / t a.t / D 2 r.t / t v.t / a.t / D 0 0 D 0: .c;0/ D 2 rp S a x ı 2. Fig. 11.6-22 r D t cos t i C t sin t j C .2 t /k, .0 t p 2/ is a conical helix wound around the cone z D 2 x 2 C y 2 starting at the vertex .0; 0; 2/, and completing one revolution to end up at .2; 0; 0/. Since By conservation of energy, v D .cos t k v2 D constant D 1 r 2 v2 2 ! p Z 2 p p 2 C 2 C 4 2 2 2 LD 2 C t dt D 2 C 4 Cln p 2 0 ` ; C1 h. C 1/ h D : v D vp D rp P D rp ` a D . 1/a D Since h2 D k`, we have 2 v1 D vp2 units. 3. The position of the particle at time t is 2k rp r D xi C x 2 j C 32 x 3 k; 2 h 2k . C 1/2 . C 1/ `2 ` i kh . C 1/2 2. C 1/ D ` k k 2 1/ D : D . ` a D where x is an increasing function of t . Thie velocity is vD dx i C 2xj C 2x 2 k : dt Since the speed is 6, we have 2 Thus av1 D k. If D is the perpendicular distance from the sun S to an asymptote of the orbit (see the figure) then D D c sin D a sin D a Da cos.ı=2/ ı D a cot : sin.ı=2/ 2 sin cos 6D dx dx p 1 C 4x 2 C 4x 4 D .2x 2 C 1/ ; dt dt so that dx=dt D 6=.2x 2 C 1/. The particle is at .1; 1; 32 / when x D 1. At this time its velocity is v.1/ D 2.i C 2j C 2k/: Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k k; the length of the curve is for all points on the orbit. At perihelion, r D rp D c t sin t /i C .sin t C t cos t /j Downloaded by ted cage (sxnbyln180@questza.com) 441 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 11 (PAGE 675) ADAMS and ESSEX: CALCULUS 9 Also 5. 6 dx 144x .4x/ D .2x 2 C 1/2 dt .2x 2 C 1/3 d 2x .i C 2xj C 2x 2 k/ aD dt 2 dx dx dx C k : 2 j C 4x dt dt dt At x D 1, we have D 16 .i C 2j C 2k/ C 2.4j C 8k/ 3 8 . 2i 3 j C 2k/: 6. 4. The position, velocity, speed, and acceleration of the particle are given by 7. ˇ ˇp ˇ dx ˇ v D ˇˇ ˇˇ 1 C 4x 2 dt 2 2 d x dx aD j: .i C 2xj/ C 2 dt 2 dt ks 2 ks 2 i C sin j 2 2 ks 2 ks 2 a D ks sin i C ks cos j 2 2 v a D ksk: v D cos dx t D p dt 1 C 4x 2 p 4tx dx 1 C 4x 2 p 2 d 2x 1 C 4x dt D : 2 dt 1 C 4x 2 Therefore the curvature at position s is D jv aj=v 3 D ks. 8. p p 4 2 .i C 2 2j/ C 2j: 9 If the particle is moving to the left, so that dx=dt < 0, a similar calculation shows that at t D 3 its acceleration is aD 442 Telegram: @uni_k p p 3C4 2 .i C 2 2j/ C 2j: 9 If r D e , and P D k, then rP D e P D rR D k 2 r. Since r D r rO , we have v D rP rO C r P O D kr rO C kr O a D .Rr Hence the acceleration is 3 t Tangential acceleration: dv=dt e t. pD e 2 Normal acceleration: v D 2. Since v D 2 cosh t , the minimum speed is 2 at time t D 0. Z s Z s kt2 kt2 For x.s/ D dt , y.s/ D dt , we have cos sin 2 2 0 0 Since x.0/ D y.0/ D 0, the arc length along the curve, measured from the origin, is s. Also, Let us assume that the particle is moving to the right, so that dx=dt > 0. Since the speed is t , we have aD 2t j C e t k dx ks 2 dy ks 2 D cos ; D sin ; ds 2 ds 2 so that the speed is unity: s 2 2 dy dx C D 1: vD ds ds r D xi C x 2 j dx vD .i C 2xj/; dt p If the particle is at . 2; 2/ at t D 3, then dx=dt D 1 at that time, and p d 2x 3 4 2 D : dt 2 9 p v D e t i C 2j e t k a D et i C e t k da D et i e t k dt p p t v a D 2e t i 2j 2e k p 2t 2t D et C e t v D e C2Ce p t jv aj D 2.e C e t / p jv aj 2 D D 3 t v .e C e t /2 da p .v a/ 2 dt D D t D : jv aj2 .e C e t /2 d 2x D dt 2 a.1/ D p r D et i C D .k 2 r 9. r P 2 /Or C .r R C 2rP P /O k 2 r/Or C .0 2k 2 r/O D kr, and 2k 2 r O : r D a.t sin t /i C a.1 cos t /j v D a.1 cos t /i C a sin t j p v D a 1 2 cos t C cos2 t C sin2 t p p t D a 2 1 cos t D 2a sin if 0 t 2. 2 The length of the cycloid from t D 0 to t D T 2 is Z T T t units. s.T / D 2a sin dt D 4a 1 cos 2 2 0 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 10. s D 4a 1 cos t 2 REVIEW EXERCISES 11 ) t D 2cos 1 1 s D t .s/. 4a sin t .s/ i C a 1 cos t .s/ j: Thus rQ .t / represents the same cycloid as r.t /, but translated a units to the left and 2a units upward. From Exercise 11, the given cycloid is the evolute of its involute. y The required arc length parametrization of the cycloid is r D a t .s/ 11. Q A From Exercise 9 we have P O / D v D .1 cos t /i C sin t j T.t v 2 sin.t =2/ t t D sin i C cos j 2 2 1 t 1 t cos i sin j O O dT 1 dT 2 2 2 D D 2 t ds v dt 2a sin 2 t 1 cot i j D 4a 2 ˇ ˇ ˇ d TO ˇ 1 ˇ ˇ .t / D ˇ ˇD ˇ ds ˇ 4a sin.t =2/ O x Fig. R-11-12 13. The position vector of P is given by r D R sin cos i C R sin sin j C R cos k: Mutually perpendicular unit vectors in the directions of increasing R, and can be found by differentiating r with respect to each of these coordinates and dividing the resulting vectors by their lengths. They are O 1 dT O / D r.t / C rC .t / D r.t / C .t /N.t ..t //2 ds t 16a2 sin2 .t =2/ cot i j D r.t / C 4a 2 t t t D r.t / C 4a cos sin i 4a sin2 j 2 2 2 D a.t sin t /i C a.1 cos t /j C 2a sin t i 2a.1 cos t /j D a.t C sin t /i a.1 cos t /j (let t D u D a.u sin u /i C a.1 cos u 2/j: O D d r D sin cos i C sin sin j C cos k R dR 1 dr O D D cos cos i C cos sin j sin k R d dr 1 D sin i C cos j: O D R sin d The triad fOr; O ; O g is right-handed. This is the reason for ordering the spherical polar coordinates .R; ; / rather than .R; ; /. ) 14. This is the same cycloid as given by r.t / but translated a units to the right and 2a units downward. 12. Let P be the point with position vector r.t / on the cycloid. By Exercise 9, the arc OP has length 4a 4a cos.t =2/, and so PQ has length 4a - arc OP D 4a cos.t =2/ units. Thus t O ! PQ D 4a cos T.t / 2 t t t sin i C cos j D 4a cos 2 2 2 D 2a sin t i C 2a.1 C cos t /j: By Kepler’s Second Law the position vector r from the origin (the sun) to the planet sweeps out area at a constant rate, say h=2: dA h D : dt 2 As observed in the text, dA=dt D r 2 P =2, so r 2 P D h, and r v D .r rO / .Pr rO C r P O / D r 2 P rO O D hk D h is a constant vector. 15. By Exercise 14, r rP D r v D h is constant, so, by Newton’s second law of motion, r F.r/ D mr rR D m It follows that Q has position vector ! rQ D r C PQ D a.t sin t /i C a.1 cos t /j C 2a sin t i C 2a.1 C cos t /j D a.t C sin t /i C a.1 C cos t C 2/j (let t D u C ) D a.u sin u C /i C a.1 cos u C 2/j: d .r rP / D 0: dt Thus F.r/ is parallel to r, and therefore has zero transverse component: F.r/ D f .r/Or for some scalar function f .r/. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k (PAGE 675) Downloaded by ted cage (sxnbyln180@questza.com) 443 lOMoARcPSD|6566483 www.konkur.in REVIEW EXERCISES 11 (PAGE 675) 16. By Exercise 15, F.r/ D m.Rr r P 2 /Or D given that r D `=.1 C cos /. Thus ADAMS and ESSEX: CALCULUS 9 which is appropriate since is much smaller than g, then p dz dr 2 2 i: dt dt Breaking the DE into its components, we get f .r/Or. We are ` . sin /P .1 C cos /2 ` sin P D .1 C cos /2 h sin 2 P r D sin D ` ` h h2 cos rR D : .cos /P D ` `r 2 rP D p dz d 2x D 2 ; dt 2 dt z.t / D 100 h2 h2 cos 2 `r r3 h2 ` D 2 cos D `r r h2 ; `r 2 mh : `r 2 x This says that the magnitude of the force on the planet is inversely proportional to the square of its distance from the sun. Thus Newton’s law of gravitation follows from Kepler’s laws and the second law of motion. Challenging Problems 11 (page 676) a) The angular velocity of the earth points northward in the direction of the earth’s axis; in terms of the basis vectors defined at a point P at 45ı north latitude, it points in the direction of j C k: jCk 2 b) If v D D 2 rad/s: 24 3;600 vk, then 2v p .j C k/ k D 2 aC D 2 v D p gk C 2 dr dt and the initial conditions r.0/ D 100k, r0 .0/ D 0. If we use the approximation dz dr k; dt dt 444 Telegram: @uni_k x.t / D gt 3 p : 3 2 s 200 4:52; g 2. 2 9:8 p .4:52/3 24 3;600 3 2 0:0155 m: The object strikes the ground about 15.5 cm west of P. 8 < d v D k v 32k dt : v.0/ D 70i a) If v D v1 i C v2 j C v3 k, then k v D v1 j v2 i. Thus the initial-value problem breaks down into component equations as 8 8 8 < dv1 D v < dv2 D v < dv3 D 32 2 1 dt dt dt : : : v1 .0/ D 70 v2 .0/ D 0 v3 .0/ D 0: b) If r D xiCyjCzk denotes the position of the baseball t s after it is thrown, then x.0/ D y.0/ D z.0/ D 0 and we have dz D v3 D dt 2vi: c) If r.t / D x.t /i C y.t /j C z.t /k is the position of the falling object at time t , then r.t / satisfies the DE d 2r D dt 2 y.t / D 0; at which time we have 2 D p ; gt 2 ; 2 tD (because .`=r/ D 1 C cos ). Hence 1. g: Since g 9:8 m/s2 , the time of fall is r P 2 D f .r/ D d 2z D dt 2 Solving these equations (beginning with the last one), using the initial conditions, we get It follows that rR d 2y D 0; dt 2 32t ) z D dv2 d 2 v1 D Also, D dt 2 dt harmonic motion), so v1 .t / D A cos t C B sin t; 16t 2 : v1 (the equation of simple v2 .t / D A sin t B cos t: Since v1 .0/ D 70, v2 .0/ D 0, x.0/ D 0, and y.0/ D 0, we have dx D v1 D 70 cos t dt x.t / D 70 sin t Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) dy D v2 D 70 sin t dt y.t / D 70.1 cos t /: lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 11 d) If d r=dt D v and r.0/ D 0, then v0 .v0 k/k sin.!t / r.t / D ! v0 k C 1 cos.!t / C .v0 k/t k: ! Since the three constant vectors v0 k v0 .v0 k/k ; ; and .v0 k/k ! ! are mutually perpendicular, and the first two have the same length because At time t seconds after it is thrown, the ball is at position r D 70 sin t i C 70.1 cos t /j 16t 2 k: c) At t D 1=5 s, the ball is at about .13:9; 1:40; 0:64/. If it had been thrown without the vertical spin, its position at time t would have been r D 70t i 16t 2 k; jv0 a) !D qB m 4. d dv .v k/ D k D !.v k/ k D 0: dt dt Thus v k D constant D v0 k. dv d jvj2 D 2 v D 2!.v k/ v D 0, Also, dt dt so jvj D constant D jv0 j for all t . b) If w.t / D v.t / .v0 k/k, then w k D 0 by part (a). Also, using the result of Exercise 23 of Section 1.3, we have d 2v dv d 2w D D! k D ! 2 .v k/ k dt 2 dt 2 h dt i D ! 2 .k k/v .k v/k h i D ! 2 v .v0 k/k D ! 2 w; the equation of simple harmonic motion. Also, w.0/ D v0 .v0 k/k w0 .0/ D !v0 k: c) Solving the above initial-value problem for w, we get w D A cos.!t / C B sin.!t /; A D w.0/ D v0 .v0 k/k; !B D w0 .0/ D ! k: where and Therefore, v.t / D w.t / C .v0 k/k h i D v0 .v0 k/k cos.!t / C .v0 k/ sin.!t / C .v0 k/k: The arc length element on x D a. sin /, y D a.cos 1/ is (for ) q ds D a .1 cos /2 C sin2 d p D a 2.1 cos / d D 2a sin.=2/ d: If the bead slides downward from rest at height y.0 / to height y. /, its gravitational potential energy has decreased by h i mg y.0 / y. / D mga.cos 0 cos /: Since there is no friction, all this potential energy is converted to kinetic energy, so its speed v at height y. / is given by 1 2 mv D mga.cos 0 cos /; 2 p and so v D 2ga.cos 0 cos /. The time required for the bead to travel distance ds at speed v is dt D ds=v, so the time T required for the bead to slide from its starting position at D 0 to the lowest point on the wire, D , is Z Z D ds 1 ds T D D d 0 v d D0 v s Z sin.=2/ 2a d D p g 0 cos 0 cos s Z 2a sin.=2/ D p d 2 g 0 2 cos .0 =2/ 2 cos2 .=2/ Let u D cos.=2/ du D 12 sin.=2/ d r Z cos.0 =2/ a du p D2 2 g 0 cos .0 =2/ u2 ˇˇcos.0 =2/ r a u ˇ 1 D2 sin ˇ g cos.0 =2/ ˇ 0 p D ag Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k .v0 k/kj D jv0 j sin D jv0 kj; where is the angle between v0 and k, the curve r.t / is generally a circular helix with axis in the z direction. However, it will be a circle if v0 k D 0, that is, if v0 is horizontal, and it will be a straight line if v0 k D 0, that is, if v0 is vertical. so its position at t D 1=5 s would have been .14; 0; 0:64/. Thus the spin has deflected the ball approximately 1.4 ft to the left (as seen from above) of what would have been its parabolic path had it not been given the spin. 8 < d v D !v k; 3. dt : v.0/ D v0 (PAGE 676) Downloaded by ted cage (sxnbyln180@questza.com) 445 lOMoARcPSD|6566483 www.konkur.in CHALLENGING PROBLEMS 11 (PAGE 676) ADAMS and ESSEX: CALCULUS 9 A which is independent of 0 . vertical section y x D 0 starting point E y horizontal section p g D .g= 2/.i j/ B D D . 2; 2/ .2; 2/ C Fig. C-11-4 5. x a) The curve BCD is the graph of an even function; a fourth degree polynomial with terms of even degree only will enable us to match the height, slope, and curvature at D, and therefore also at C . We have f .x/ D ax 4 C bx 2 C c f 0 .x/ D 4ax 3 C 2bx f 00 .x/ D 12ax 2 C 2b: At D we have x D 2, so we need Fig. C-11-5 6. a) At time t, the hare is at P D .0; vt / and the fox is at Q D x.t /; y.t / , where x and y are such that the slope dy=dx of the fox’s path is the slope of tline y vt dy D . PQ: dx x b) Since 2 D f .2/ D 16a C 4b C c 1 D f 0 .2/ D 32a C 4b 0 D f 00 .2/ D 48a C 2b: These equations yield a D 1=64, b D 3=8, c D 3=4, so the curved track BCD is the graph of 1 y D f .x/ D . x 4 C 24x 2 C 48/. 64 b) Since we are ignoring friction, the p speed v of the car during its drop is given by v D 2gs, where s is the vertical distance dropped. (See the previous solution.) At B the car p has dropped about 7.2 m, so its speed there is v 2.9:8/.7:2/ 11:9 m/s. At C the car p has dropped 10 .c= 2/ 9:47 m, so its speed there is v D 13:6 m/s. At D the car has dropped 10 m, so its speed is v D 14:0 m/s. c) At C we have x D 0, f 0 .0/ D 0, and f 00 .0/ D 2b D 3=4. Thus the curvature of the track at C is jf 00 .0/j 3 D D : 4 .1 C .f 0 .0//2 /3=2 The normal accelerationpis v 2 138:7 m/s2 (or about 14g). Since v D 2gs, we have p p p 2g ds 2g 19:6 dv .13:6/ 9:78 m/s2 ; D p D p v p dt 2 s dt 2 s 2 9:47 so the total acceleration has magnitude approximately p .138:7/2 C .9:78/2 139 m/s2 ; dx d 2 y d y vt D dt dx 2 dt x dx dy v .y vt / x dt dt D 2 x 1 dx 1 dy dx .y vt / v D x dx dt x2 dt 1 dx v 1 dx D 2 .y vt / .y vt / D x dt x x2 dt v d 2y D . dx 2 dx=dt Since the fox’s speed is also v, we have Thus x Telegram: @uni_k dx dt 2 C dy dt 2 D v2: Also, the fox is always running to the left (towards the y-axis from points where x > 0), so dx=dt < 0. Hence v D dx dt s .dy=dt /2 1C D .dx=dt /2 s 1C dy dx and so the fox’s path y D y.x/ satisfies the DE which is again about 14g. 446 d 2 y dx d dy D , we have dt dx dx 2 dt d 2y D x dx 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) s 1C dy dx 2 : 2 ; v : x lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 11 c) If u D dy=dx, then u D 0 and y D 0 when x D a, and p du x D 1 C u2 Z dx Z du dx p D Let u D tan 2 x 1Cu du D sec2 d Z sec d D ln x C ln C ln.tan C sec / D ln.C x/ p u C 1 C u2 D C x: b) Since yourpvelocity at any point has a northward component v= 2, and progress northward is measured along a circle of radius a (a meridian), your colatitude .t / satisfies a p x u 1 C u2 D a x 2 2xu 1 C u2 D 2 C u2 a a x2 2xu D 2 1 a a dy x a DuD dx 2a 2x a x2 ln x C C1 : yD 4a 2 Since y D 0 when x D a, we have C1 D so x 2 a2 a x ln yD 4 2 a T D a a=2 p D p : v= 2 2v vt p : a 2 d v .a sin / D p dt 2 d v vt D p cos p a 2 dt a 2 Z v vt D p sec p dt a 2 a 2 vt vt C C: D ln sec p C tan p a 2 a 2 As D 0 at t D 0, we have C D 0, and so vt vt .t / D ln sec p C tan p : a 2 a 2 p c) As t ! T D a=. 2v/, the expression for .t / ! 1, so your path spirals around the north pole, crossing any meridian infinitely often. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 2 Since p your velocity also has an eastward component v= 2 measured along a parallel of latitude that is a circle of radius a sin , your longitude coordinate satisfies a a C ln a, 4 2 a) Since you are always travelling northeastpat speed v, you are always moving north at rate v= 2. Therefore you will reach the north pole in finite time v p : 2 Since .0/ D =2, it follows that is the path of the fox. 7. d D dt .t / D Since u D 0 when x D a, we have C D 1=a. (PAGE 676) Downloaded by ted cage (sxnbyln180@questza.com) 447 lOMoARcPSD|6566483 www.konkur.in SECTION 12.1 (PAGE 684) ADAMS and ESSEX: CALCULUS 9 z CHAPTER 12. PARTIAL DIFFERENTIATION .2;0;2/ .2;3;2/ Section 12.1 Functions of Several Variables (page 684) 1. zDx xCy . f .x; y/ D x y The domain consists of all points in the xy-plane not on the line x D y. p 2. f .x; y/ D xy. Domain is the set of points .x; y/ for which xy 0, that is, points on the coordinate axes and in the first and third quadrants. x . x2 C y2 The domain is the set of all points in the xy-plane except the origin. 3. f .x; y/ D 4. f .x; y/ D Fig. 12.1-11 12. f .x; y/ D sin x; 0 x 2; 0 y 1 z z D sin x xy . x2 y2 The domain consists of all points not on the lines x D ˙y. p 5. f .x; y/ D 4x 2 C 9y 2 36. The domain consists of all points .x; y/ lying on or outside the ellipse 4x 2 C 9y 2 D 36. p 6. f .x; y/ D 1= x 2 y 2 . The domain consists of all points in the part of the plane where jxj > jyj. 7. f .x; y/ D ln.1 C xy/. The domain consists of all points satisfying xy > 1, that is, points lying between the two branches of the hyperbola xy D 1. y 3 x 2 x 1 y Fig. 12.1-12 13. z D f .x; y/ D y 2 z 1 8. f .x; y/ D sin .x C y/. The domain consists of all points in the strip 1 x C y 1. z D y2 xyz . x2 C y2 C z2 The domain consists of all points in 3-dimensional space except the origin. 9. f .x; y; z/ D y e xyz 10. f .x; y; z/ D p . xyz The domain consists of all points .x; y; z/ where xyz > 0, that is, all points in the four octants x > 0, y > 0, z > 0; x > 0, y < 0, z < 0; x < 0, y > 0, z < 0; and x < 0, y < 0, z > 0. 11. z D f .x; y/ D x 448 Telegram: @uni_k x Fig. 12.1-13 14. f .x; y/ D 4 x2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) y 2 ; .x 2 C y 2 4; x 0; y 0/ lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 12.1 (PAGE 684) z z 4 zD4 x 2 y 2 x 2 y y 2 x Fig. 12.1-14 15. z D f .x; y/ D p Fig. 12.1-17 18. f .x; y/ D 6 x2 C y2 z zD p x2 C y2 x 2y z 6 zD6 x 2y y 3 y 6 x x Fig. 12.1-18 Fig. 12.1-15 19. 16. f .x; y/ D 4 f .x; y/ D x y D C , a family of straight lines of slope 1. x2 y cD 3 cD 2 z zD4 x2 cD 1 cD0 x cD1 cD2 y cD3 x x Fig. 12.1-19 Fig. 12.1-16 20. 17. z D f .x; y/ D jxj C jyj f .x; y/ D x 2 C 2y 2 D C , a family of similar ellipses centred at the origin. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k yDc Downloaded by ted cage (sxnbyln180@questza.com) 449 lOMoARcPSD|6566483 www.konkur.in SECTION 12.1 (PAGE 684) ADAMS and ESSEX: CALCULUS 9 y 23. x 2 C 2y 2 D c x y D C , a family of straight lines through xCy the origin, but not including the origin. f .x; y/ D y cD16 cD9 cD 1 cD4 cD :5 cD1 cD0 x cD:5 cD1 x cD2 cD 2 Fig. 12.1-20 21. x y xCy D c f .x; y/ D xy D C , a family of rectangular hyperbolas with the coordinate axes as asymptotes. Fig. 12.1-23 y cD9 cD4 cD1 24. cD0 f .x; y/ D y x2 C y2 D C. 1 2 D 4C1 2 of circles This is the family x 2 C y 2C passing through the origin and having centres on the yaxis. The origin itself is, however, not on any of the level curves. x cD 1 cD 4 y cD 9 cD1 xy D c cD2 Fig. 12.1-21 cD3 x2 D C , a family of parabolas, y D x 2 =C , 22. f .x; y/ D y with vertices at the origin and vertical axes. x y cD1 cD 3 cD0:5 cD 2 cD2 cD 1 Fig. 12.1-24 x x2 y Dc cD 2 cD 0:5 Fig. 12.1-22 450 Telegram: @uni_k cD 1 25. f .x; y/ D xe y D C . x This is the family of curves y D ln . C Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) y Dc x 2 Cy 2 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 12.1 y 29. cD 1 (PAGE 684) The graph of the function whose level curves are as shown in part (a) of Figure 12.1.29 is a plane containing the yaxis and sloping uphill to the right. It is consistent with, say, a function of the form f .x; y/ D y. (a) (b) y y cD1 C D10 cD2 cD 2 x x cD4 cD 4 xe y D c CD 5 (c) Fig. 12.1-25 26. f .x; y/ D s 1 y x2 D C ) y D y f .x; y/ D s 1 y x2 D C y C D5 C D5 1 . x2 C C 2 x C D0 (d) y C D3 x C D 0:8 C D0 x CD 5 Fig. 12.1-29 C D1 30. The graph of the function whose level curves are as shown in part (b) of Figure 12.1.29 is a cylinder parallel to the x-axis, rising from height zero first steeply and then more and more slowly as y increases. Itpis consistent with, say, a function of the form f .x; y/ D y C 5. 31. The graph of the function whose level curves are as shown in part (c) of Figure 12.1.29 is an upside down circular cone with vertex at height 5 on the z-axis and base circle in the xy-plane. It is consistent with, say, a function of p the form f .x; y/ D 5 x2 C y2. C D2 x Fig. 12.1-26 27. The landscape is steepest at B where the level curves are closest together. 500 A 500 600 C 32. The graph of the function whose level curves are as shown in part (d) of Figure 12.1.29 is a cylinder (possibly parabolic) with axis in the yz-plane, sloping upwards in the direction of increasing y. It is consistent with, say, a function of the form f .x; y/ D y x 2 . 33. The curves y D .x C /2 are all horizontally shifted versions of the parabola y D x 2 , and they all lie in the half-plane y 0. Since each of these curves intersects all of the others, they cannot be level curves of a function f .x; y/ defined in y 0. To be a family of level curves of a function f .x; y/ in a region, the various curves in the family cannot intersect one another in that region. 34. 4z 2 D .x z/2 C .y z/2 . If z D c > 0, we have .x c/2 C .y c/2 D 4c 2 , which is a circle in the plane z D c, with centre .c; c; c/ and radius 2c. N B W E S 400 300 200 100 Fig. 12.1-27 28. C is a “pass” between two peaks to the east and west. The land is level at C and rises as you move to the east or west, but falls as you move to the north or south. Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k Downloaded by ted cage (sxnbyln180@questza.com) 451 lOMoARcPSD|6566483 www.konkur.in SECTION 12.1 (PAGE 684) ADAMS and ESSEX: CALCULUS 9 y 40. cD3 x2 C y2 . z2 The equation f .x; y; z/ D c can be rewritten x 2 C y 2 D C 2 z 2 . The level surfaces are circular cones with vertices at the origin and axes along the z-axis. f .x; y; z/ D cD2 41. f .x; y; z/ D jxj C jyj C jzj. The level surface f .x; y; z/ D c > 0 is the surface of the octahedron with vertices .˙c; 0; 0/, .0; ˙c; 0/, and .0; 0; ˙c/. (An octahedron is a solid with eight planar faces.) cD1 x 42. .x c/2 C .y Fig. 12.1-34 c/2 D 4c 2 The graph of the function z D z.x; y/ 0 defined by the given equation is (the upper half of) an elliptic cone with axis along the line x D y D z, and circular crosssections in horizontal planes. 35. f .x; y; z; t / D x 2 C y 2 C z 2 C t 2 . The “level hypersurface” p f .x; y; z; t / D c > 0 is the “4-sphere” of radius c centred at the originpin R4 . That is, it consists of all points in R4 at distance c from the origin. 43. z 2 2 2 a) f .x; y/ D C p is x C y D C implies that f .x; y/ D x 2 C y 2 . zD b) f .x; y/ D C is x 2 C y 2 D C 4 implies that f .x; y/ D .x 2 C y 2 /1=4 . 1 1 C x2 C y2 c) f .x; y/ D C is x 2 C y 2 D C implies that f .x; y/ D x 2 C y 2 . d) f .x; y/ D Cpis x 2 C y 2 D .ln C /2 implies that f .x; y/ D e x 2 Cy 2 y . 36. If the level surface f .x; y; z/ D C is the plane x Fig. 12.1-43 x y z C C D 1; C3 2C 3 3C 3 that is, x C z y C D C 3 , then 2 3 44. z zD z 1=3 y f .x; y; z/ D x C C : 2 3 cos x 1 C y2 37. f .x; y; z/ D x 2 C y 2 C z 2 . The level surface f .x; y; z/ D c > 0 is a sphere of radius p c centred at the origin. 38. f .x; y; z/ D x C 2y C 3z. The level surfaces are parallel planes having common normal vector i C 2j C 3k. y x 39. f .x; y; z/ D x 2 C y 2 . The level p surface f .x; y; z/ D c > 0 is a circular cylinder of radius c with axis along the z-axis. 452 Telegram: @uni_k Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) 5 x 5; 5y5 Fig. 12.1-44 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 12.2 (PAGE 689) z 45. z zD zD y 1 C x2 C y2 1 xy y y 4x4 4y4 x Fig. 12.1-48 x Fig. 12.1-45 Section 12.2 Limits and Continuity (page 689) 46. z zD .x 2 1. x 1/2 C y 2 2. 3. y xy C x 2 D 2. 1/ C 22 D 2 lim p .x;y/!.0;0/ x2 C y2 D 0 x2 C y2 does not exist. y .x;y/!.0;0/ lim If .x; y/ ! .0; 0/ along x D 0, then If .x; y/ ! .0; 0/ along y x2 C y2 D 1 C x 2 ! 1. y x Fig. 12.1-46 47. lim .x;y/!.2; 1/ 4. x . x2 C y2 Then jf .x; 0/j D j1=xj ! 1 as x ! 0. But jf .0; y/j D 0 ! 0 as y ! 0. Thus lim.x;y/!.0;0/ f .x; y/ does not exist. Let f .x; y/ D z 5. z D xy 6. y lim .x;y/!.1;/ 1 cos.xy/ D x cos y 1 Fig. 12.1-47 x 2 .y 1/2 D 0, because 1/2 .x;y/!.0;1/ x 2 C .y and x 2 ! 0 as .x; y/ ! .0; 1/. ˇ ˇ ˇ y3 ˇ y2 ˇ ˇ ˇ x 2 C y 2 ˇ x 2 C y 2 jyj jyj ! 0 8. lim sin.x y/ .x;y/!.0;0/ cos.x C y/ Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k 1 lim as .x; y/ ! .0; 0/. Thus 48. The graph is asymptotic to the coordinate planes. cos D 1 cos ˇ 2 ˇ ˇ x .y 1/2 ˇ ˇ x2 0 ˇˇ 2 x C .y 1/2 ˇ 7. x D x2 C y2 D y ! 0. y 2 x , then Downloaded by ted cage (sxnbyln180@questza.com) D lim y3 .x;y/!.0;0/ x 2 C y 2 D 0. sin 0 D 0: cos 0 453 lOMoARcPSD|6566483 www.konkur.in SECTION 12.2 (PAGE 689) ADAMS and ESSEX: CALCULUS 9 sin.xy/ . x2 C y2 2 Now f .0; y/ D 0=x D 0 ! 0 as x ! 0. sin x 2 1 However, f .x; x/ D ! as x ! 0. 2x 2 2 Therefore lim f .x; y/ does not exist. However there is no way to define f .x; x/ so that f becomes continuous on y D x, since jf .x; y/j D 1=jx C yj ! 1 as y ! x. 9. Let f .x; y/ D 16. Let f be the function of Example 3 of Section 3.2: .x;y/!.0;0/ f .x; y/ D 10. The fraction is not defined at points of the line y D 2x and so cannot have a limit at .1; 2/ by Definition 4. However, if we use the extended Definition 6, then, cancelling the common factor 2x y, we get 2x 2 .x;y/!.1;2/ 4x 2 lim x 2 x 2 C y 4 . Thus xy 1 x D lim D : y2 4 .x;y/!.1;2/ 2x C y x y y 2 ! 0 as y ! 0. Thus x2 C y4 x!a x3y3 17. lim .x;y/!.0;0/ f .x; y/ D 1 0 D 1. Define f .0; 0/ D 1. 14. For x ¤ y, we have f .x; y/ D 15. y3 D x 2 C xy C y 2 : y The latter expression has the value 3x 2 at points of the line x D y. Therefore, if we extend the definition of f .x; y/ so that f .x; x/ D 3x 2 , then the resulting function will be equal to x 2 C xy C y 2 everywhere, and so continuous everywhere. x y x y f .x; y/ D 2 D . x y2 .x y/.x C y/ Since f .x; y/ D 1=.x C y/ at all points off the line x D y and so is defined at some points in any neighbourhood of .1; 1/, it approaches 1=.1 C 1/ D 1=2 as .x; y/ ! .1; 1/; If we define f .1; 1/ D 1=2, then f becomes continuous at .1; 1/. Similarly, f .x; y/ can be defined to be 1=.2x/ at any point on the line x D y except the origin, and becomes continuous at such points. 454 Telegram: @uni_k x3 x fu .t / D f .a C t u; b C t v/, where u D ui C vj is a unit vector. f .x; y/ may not be continuous at .a; b/ even if fu .t / is continuous at t D 0 for every unit vector u. A counterexample is the function f of Example 4 in this section. Here a D b D 0. The condition that each fu should be continuous is the condition that f should be continuous on each straight line through .0; 0/, which it is if we extend the domain of f to include .0; 0/ by defining f .0; 0/ D 0. (We showed that f .x; y/ ! 0 as .x; y/ ! .0; 0/ along every straight line.) However, we also showed that lim.x;y/!.0;0/ f .x; y/ did not exist. . But ˇ ˇ 3 3 ˇ ˇ ˇ x y ˇ ˇ x2 ˇ 3 3 ˇDˇ ˇ ˇ ˇ x 2 C y 2 ˇ ˇ x 2 C y 2 ˇ jxy j jxy j ! 0 as .x; y/ ! .0; 0/. Thus yDb Similarly, h.y/ D f .a; y/ is continuous at y D b. x2y2 D 0: 2x 4 C y 4 2 2 x y x4 1 If x D y ¤ 0, then D D . 2x 4 C y 4 2x 4 C x 4 3 x2y2 Therefore lim does not exist. .x;y/!.0;0/ 2x 4 C y 4 12. If x D 0 and y ¤ 0, then x2 C y2 if .x; y/ D .0; 0/. lim g.x/ D x!a lim f .x; y/ D f .a; b/: x2y2 D 0. .x;y/!.0;0/ x 2 C y 4 x2 C y2 x3 y3 D1 x2 C y2 if .x; y/ ¤ .0; 0/ If f .x; y/ is continuous at .a; b/, then g.x/ D f .x; b/ is continuous at x D a because lim 13. f .x; y/ D 2xy 2 C y2 x : 0 Let a D b D 0. If g.x/ D f .x; 0/ and h.y/ D f .0; y/, then g.x/ D 0 for all x, and h.y/ D 0 for all y, so g and h are continuous at 0. But, as shown in Example 3 of Section 3.2, f is not continuous at .0; 0/. 2 2 11. 8 < On the other hand, if f .x; y/ is continuous at .a; b/, then f .x; y/ ! f .a; b/ if .x; y/ approaches .a; b/ in any way, in particular, along the line through .a; b/ parallel to u. Thus all such functions fu .t / must be continuous at t D 0. p p 18. Since jxj x 2 C y 2 and jyj x 2 C y 2 , we have ˇ ˇ ˇ x m y n ˇ .x 2 C y 2 /.mCn/=2 ˇ ˇ D .x 2 Cy 2 / pC.mCn/=2 : ˇ .x 2 C y 2 /p ˇ .x 2 C y 2 /p The expression on the right ! 0 as .x; y/ ! .0; 0/, provided m C n > 2p. In this case xm yn lim .x;y/!.0;0/ .x 2 C y 2 /p 19. D 0: Suppose .x; y/ ! .0; 0/ along the ray y D kx. Then f .x; y/ D xy ax 2 C bxy C cy 2 Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) D k : a C bk C ck 2 lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL SECTION 12.3 (PAGE 696) z Thus f .x; y/ has different constant values along different rays from the origin unless a D c D 0 and b ¤ 0. If this condition is not satisfied, lim.x;y/!.0;0/ f .x; y/ does not exist. If the condition is satisfied, then lim.x;y/!.0;0/ f .x; y/ D 1=b does exist. sin x sin3 y cannot be defined at .0; 0/ 1 cos.x 2 C y 2 / so as to become continuous there, because f .x; y/ has no limit as .x; y/ ! .0; 0/. To see this, observe that f .x; 0/ D 0, so the limit must be 0 if it exists at all. However, 20. f .x; y/ D y 2x 2 y zD 4 x C y2 x 4 4 f .x; x/ D 1 sin x sin x D cos.2x 2 / 2 sin2 .x 2 / Fig. 12.2-22 which approaches 1/2 as x ! 0 by l’H^opital’s Rule or by using Maclaurin series. 21. z zD 2xy x2 C y2 23. The graph of a function f .x; y/ that is continuous on region R in the xy-plane is a surface with no breaks or tears in it and that intersects each line parallel to the zaxis through a point .x; y/ of R at exactly one point. 24. (a) We say that limx!a f .x/ D L provided that (i) every open interval containing a contains at least one point of the domain of f different from a, and (ii) if for every > 0 there exists ı > 0 depending on such that if x is in the domain of f and satisfies 0 < jx aj < ı, then jf .x/ Lj < . (b) There are no points in the domain of f to the right of 1 or between 1=2 and 1 so condition (i) of rhe definition is not satisfied and limx!1 f .x/ does not exist. If .a; b/ is any open interval containing 0, then b > 0. If integer n > 1=b, then 1=n < b and so .a; b/ contains a point of the domain of f . If > 0, let ı D . If 1=n < ı, then y x ˇ ˇ ˇf 1 ˇ n Fig. 12.2-21 The graphing software is unable to deal effectively with the discontinuity at .x; y/ D .0; 0/ so it leaves some gaps and rough edges near the z-axis. The surface lies between a ridge of height 1 along y D x and a ridge of height 1 along y D x. It appears to be creased along the z-axis. The level curves are straight lines through the origin. 22. The graphing software is unable to deal effectively with the discontinuity at .x; y/ D .0; 0/ so it leaves some gaps and rough edges near the z-axis. The surface lies between a ridge along y D x 2 , z D 1, and a ridge along y D x 2 , z D 1. It appears to be creased along the z-axis. The level curves are parabolas y D kx 2 through the origin. One of the families of rulings on the surface is the family of contours corresponding to level curves. 1 n ˇ ˇ 1 1ˇˇ D < : n Thus limx!0 f .x/ exists and equals 1. (c) Since every open interval on the real line contains irrational numbers that are not in the domain of f , the conditions of Definition 8 in Section 1.5 are not met, so neither of the limits above can exist under that definition. Section 12.3 Partial Derivatives (page 696) 1. f .x; y/ D x y C 2, f1 .x; y/ D 1 D f1 .3; 2/, f2 .x; y/ D Copyright © 2018 Pearson Canada Inc. Telegram: @uni_k ˇ ˇ ˇ ˇn 1ˇˇ D ˇˇ Downloaded by ted cage (sxnbyln180@questza.com) 1 D f2 .3; 2/. 455 lOMoARcPSD|6566483 www.konkur.in SECTION 12.3 (PAGE 696) ADAMS and ESSEX: CALCULUS 9 2. f .x; y/ D xy C x 2 , f1 .x; y/ D y C 2x, f2 .x; y/ D x, f1 .2; 0/ D 4; f2 .2; 0/ D 2: 3. f .x; y; z/ D x 3 y 4 z 5 , f1 .x; y; z/ D 3x 2 y 4 z 5 ; f2 .x; y; z/ D 4x 3 y 3 z 5 ; f3 .x; y; z/ D 5x 3 y 4 z 4 ; 9. f1 .0; 1; 1/ D 0, f2 .0; 1; 1/ D 0, f3 .0; 1; 1/ D 0. 10. If g.x1 ; x2 ; x3 ; x4 / D xz , yCz 1 z ; g1 .1; 1; 1/ D , g1 .x; y; z/ D yCz 2 1 xz ; g2 .1; 1; 1/ D , g2 .x; y; z/ D .y C z/2 4 1 xy ; g3 .1; 1; 1/ D : g3 .x; y; z/ D .y C z/2 4 z D tan 1 @z D @x 1 y x y D x2 . 1;1/ . 1;1/ 1 x3 C x42 2x2 g2 .x1 ; x2 ; x3 ; x4 / D x3 C x42 x22 x1 g3 .x1 ; x2 ; x3 ; x4 / D .x3 C x42 /2 .x22 x1 /2x4 g4 .x1 ; x2 ; x3 ; x4 / D .x3 C x42 /2 11. 1 : 2 12. @w yze xyz , D @x 1 C e xyz xyz xyz @w xze xye @w ; , D D @y 1 C e xyz @z 1 C e xyz @w @w @w At .2; 0; 1/: D 0; D 1; D 0. @x @y @z 6. w D ln.1 C e xyz /; 7. p f .x; y/ D sin.x y/, p p ; 4 D 1; f1 .x; y/ D y cos.x y/; f1 3 x p ;4 D : f2 .x; y/ D p cos.x y/; f2 2 y 3 24 1 , 8. f .x; y/ D p 2 x C y2 1 2 x , f1 .x; y/ D .x C y 2 / 3=2 .2x/ D 2 2 .x C y 2 /3=2 y By symmetry, f2 .x; y/ D , .x 2 C y 2 /3=2 4 3 , f2 . 3; 4/ D . f1 . 3; 4/ D 125 125 456 Telegram: @uni_k x1 x22 , then x3 C x42 g1 .x1 ; x2 ; x3 ; x4 / D y x2 C y2 y2 x2 1 1 x @z D D 2 2 @y x x C y2 y 1C 2 x ˇ ˇ @z ˇˇ 1 @z ˇˇ ; D D ˇ ˇ @x ˇ 2 @y ˇ 1C ˇ @w ˇˇ D 2e; ˇ @x ˇ .e;2;e/ ˇ @w ˇˇ @w y ln z D e2; D ln x ln z x ; ˇ @y @y ˇ ˇ .e;2;e/ y @w ˇˇ @w y ln z D ln x x ; D 2e: ˇ @z z @z ˇ .e;2;e/ 4. g.x; y; z/ D 5. w D x y ln z , @w D y ln z x y ln z 1 ; @x 13. 14. 1 3 2 g2 .3; 1; 1; 2/ D 3 2 g3 .3; 1; 1; 2/ D 9 8 g4 .3; 1; 1; 2/ D : 9 g1 .3; 1; 1; 2/ D 8 < 2x 3 y 3 f .x; y/ D x 2 C 3y 2 if .x; y/ ¤ .0; 0/ : 0 if .x; y/ D .0; 0/ 2h3 0 f1 .0; 0/ D lim D2 h!0 h.h2 C 0/ 1 k3 0 f2 .0; 0/ D lim D : 3 k!0 k.0 C 3k 2 / 8 < x 2 2y 2 if x ¤ y f .x; y/ D : x y 0 if x D y h 0 f .h; 0/ f .0; 0/ D lim D 1; f1 .0; 0/ D lim h h h!0 h!0 f .0; k/ f .0; 0/ 2k f2 .0; 0/ D lim D lim D 2: k k!0 k!0 k f .x; y/ D x 2 y 2 f . 2; 1/ D 3 f1 .x; y/ D 2x f1 . 2; 1/ D 4 f2 .x; y/ D 2y f2 . 2; 1/ D 2 Tangent plane: z D 3 4.x C 2/ 2.y 1/, or 4x C 2y C z D 3. y 1 z 3 xC2 D D . Normal line: 4 2 1 x y ; f .1; 1/ D 0; f .x; y/ D xCy 1 .x C y/ .x y/ ; f1 .1; 1/ D f1 .x; y/ D .x C y/2 2 .x C y/. 1/ .x y/ 1 f2 .x; y/ D ; f2 .1; 1/ D : .x C y/2 2 Tangent plane to z D f .x; y/ at (1,1) has equation x 1 y 1 , or 2z D x y. zD 2 2 Normal line: 2.x 1/ D 2.y 1/ D z. Copyright © 2018 Pearson Canada Inc. Downloaded by ted cage (sxnbyln180@questza.com) lOMoARcPSD|6566483 www.konkur.in INSTRUCTOR’S SOLUTIONS MANUAL 15. f .x; y/ D cos x y SECTION 12.3 1 f .; 4/ D p 2 20. x 1 1 sin p f1 .; 4/ D y y 4 2 x x f2 .x; y/ D 2 sin f2 .; 4/ D p y y 16 2 The tangent plane at x D , y D 4 is f1 .x; y/ D 1 zD p 1 2 1 .x 4 / C p or 4x y C 16 2z D 16. Normal line: p p 16 2 4 2.x / D .y 4/ D .y 16 z 4/ ; 21. p .1= 2/ . 16. f .x; y/ D e xy ; f1 .x; y/ D ye xy ; f2 .x; y/ D xe xy , f .2; 0/ D 1; f1 .2; 0/ D 0; f2 .2; 0/ D 2. Tangent plane to z D e xy at (2,0) has equation z D 1C2y. Normal line: x D 2, y D 2 2z. 17. x x2 C y2 .x 2 C y 2 /.1/ x.2x/ y2 x2 f1 .x; y/ D D .x 2 C y 2 /2 .x 2 C y 2 /2 2xy f2 .x; y/ D .x 2 C y 2 /2 1 3 4 f .1; 2/ D ; f1 .1; 2/ D ; f2 .1; 2/ D : 5 25 25 The tangent plane at x D 1, y D 2 is f .x; y/ D zD 1 3 C .x 5 25 4 .y 25 1/ 2/; 25z D 10. x 1 y 2 5z 1 Normal line: D D . 3 4 125 or 3x 4y 2 22. 2 z D x 4 4xy 3 C 6y 2 2 @z D 4x 3 4y 3 D 4.x y/.x 2 C xy C y 2 / @x @z D 12xy 2 C 12y D 12y.1 xy/: @y The tangent plane wi