INSTRUCTOR'S SOLUTIONS MANUAL for 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 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 Chris Essex essex@uwo.ca CONTENTS Solutions for Chapter P Solutions for Chapter 23 Solutions for Chapter 39 Solutions for Chapter 81 Solutions for Chapter 108 Solutions for Chapter 177 Solutions for Chapter 213 Solutions for Chapter 267 Solutions for Chapter 316 Solutions for Chapter 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 INSTRUCTOR’S SOLUTIONS MANUAL CHAPTER P SECTION P.1 (PAGE 10) PRELIMINARIES 19 Given: 1=.2 x/ < CASE I If x < 2, then < 3.2 x/ D 3x, so 3x < and x < 5=3 This case has solutions x < 5=3 CASE II If x > 2, then > 3.2 x/ D 3x, so 3x > and x > 5=3 This case has solutions x > Solution: 1; 5=3/ [ 2; 1/ 20 Given: x C 1/=x CASE I If x > 0, then x C 2x, so x CASE II If x < 0, then x C 2x, so x (not possible) Solution: 0; 1 21 Given: x 2x Then x.x 2/ This is only possible if x and x Solution: Œ0; 2 Section P.1 Real Numbers and the Real Line (page 10) D 0:22222222 D 0:2 D 0:09090909 D 0:09 11 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 If x D 3:277777 , then 10x 32 D 0:77777 and 100x 320 D C 10x 32/, or 90x D 295 Thus x D 295=90 D 59=18 1=7 D 0:142857142857 D 0:142857 22 Given 6x 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 > 4x, we have x.x 4/ > This is possible if x < and x < 4, or if x > and x > The possibilities are, therefore, < x < or < x < 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 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 25 x and x define the interval Œ0; 5 x < and x define the interval Œ 3; 2/ x > or x < 10 x 11 defines the union 1; 6/ [ 5; 1/ defines the interval 1; 1 x> 27 If jxj D then x D ˙3 28 If jx 29 If j2t C 5j D 4, then 2t C D ˙4, so t D t D 1=2 30 Ifj1 31 If j8 3sj D 9, then 3s D ˙9, so 3s D s D 1=3 or s D 17=3 Solution: 1; 2/ 14 If 3x C 8, then 3x 1; 1 15 If 5x 1; 5=4 x 3x and x Solution: 3x, then 8x 10 and x 5=4 Solution: , then x 6x and x Solution: 1; 2 16 If 17 Thus 14 7x If 3.2 x/ < 2.3 C x/, then < 5x and x > Solution: 0; 1/ 18 If x < 9, then jxj < and 3; 3/ < x < Solution: < x xC1 CASE I If x > then x 1/.x C 1/ > 0, so that 3.x C 1/ < 2.x 1/ Thus x < There are no solutions in this case CASE II If < x < 1, then x 1/.x C 1/ < 0, so 3.x C 1/ > 2.x 1/ Thus x > 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 < All numbers x < are solutions Solutions: 1; 5/ [ 1; 1/ Given: defines the interval 2; 1/ 2x > 4, then x < x 1C x CASE I If x > 0, then x 2x C 8, so that x 2x 0, or x 4/.x C 2/ This is possible for x > only if x CASE II If x < 0, then we must have x 4/.x C 2/ 0, which is possible for x < only if x Solution: Œ 2; 0/ [ Œ4; 1/ Given: 26 12 x < or x defines the interval 1; 1/, that is, the whole real line 13 If Given x x 2, then x x so x 2/.x C1/ This is possible if x and x or if x and x The latter situation is not possible The solution set is Œ 1; 2 3j D 7, then x t j D 1, then Copyright © 2018 Pearson Canada Inc D ˙7, so x D or x D 10 9=2 or t D ˙1, so t D or t D or 17, and SECTION P.1 (PAGE 10) ADAMS and ESSEX: CALCULUS ˇs ˇ s ˇ ˇ 32 If ˇ 1ˇ D 1, then D ˙1, so s D or s D 2 33 If jxj < 2, then x is in 2; 2/ From A.0:5; 3/ to B.2; 3/, x D y D 3 D jABj D 1:5 Starting point: 2; 3/ Increments x D 4, y D New position is C 4; C 7//, that is, 2; 4/ Arrival point: 2; 2/ Increments x D 5, y D Starting point was 5/; 1/, that is, 3; 3/ x C y D represents a circle of radius centred at the origin p x C y D represents a circle of radius centred at the origin 34 If jxj 2, then x is in Œ 2; 2 35 If js 1j 2, then 36 If jt C 2j < 1, then 3; 1/ s C 2, so s is in Œ 1; 3 < t < C 1, so t is in 37 If j3x 7j < 2, then < 3x < 7C2, so x is in 5=3; 3/ 38 If j2x C 5j < 1, then < 2x < C 1, so x is in 3; 2/ ˇ ˇx x ˇ ˇ 1ˇ 1, then 1 C 1, so x is in Œ0; 4 39 If ˇ 2 ˇ ˇ x ˇ ˇ 40 If ˇ2 ˇ < , then x=2 lies between 1=2/ and 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 is greater than the distance from x to 3, so x must be to the right of the point half-way between and Thus x > 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) From A.0; 3/ to B.4; 0/, xp D D and y D D jABj D 42 C 3/2 D From A 1; 2/ to B.4; 10/, xpD 1/ D and y D 10 D 12 jABj D 52 C 12/2 D 13 From A.3; 2/ to B 1; 2/, x and pD D 4p y D 2 D jABj D 4/2 C 4/2 D 2 11 x C y represents points inside and on the circle of radius centred at the origin y x represents all points lying on or above the parabola y D x 12 y < x 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 is y D 1C1.xC1/, or y D x C 16 Line through 2; 2/ with slope m D 1=2 is y D C 1=2/.x C 2/, or x 2y D 17 Line through 0; b/ with slope m D is y D b C 2x 1/, 45 The triangle inequality jx C yj jxj C jyj implies that 10 x C y D represents the origin 14 43 jaj D a if and only if a It is false if a < .x 13 42 jx 3j < 2jxj , x 6x C D x 3/2 < 4x , 3x C 6x > , 3.x C 3/.x 1/ > This inequality holds if x < or x > 44 The equation jx 1j D x holds if jx 1j D that is, if x 0, or, equivalently, if x 0:5 D 1:5 and 18 Line through a; 0/ with slope m D or y D 2a 2x is y D 2.x a/, 19 At x D 2, the height of the line 2x C 3y D is y D 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 is y D 7/=4 D Thus 3; 1/ lies on the line 21 The line through 0; 0/ and 2; 3/ has slope m D 0/=.2 0/ D 3=2 and equation y D 3=2/x or 3x 2y D 22 The line through 2; 1/ and 2; 2/ has slope m D 1/=.2 C 2/ D 3=4 and equation y D 3=4/.x C 2/ or 3x C 4y D 23 The line through 4; 1/ and 2; 3/ has slope m D 1/= 4/ D or x C 3y D 24 25 1=3 and equation y D The line through 2; 0/ and 0; 2/ has slope m D 0/=.0 C 2/ D and equation y D C x p If m D and p b D 2, then the line has equation y D 2x C Copyright © 2018 Pearson Canada Inc x 4/ 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 1:5x 27 3x C 4y D 12 has x-intercept a D 12=3 D and yintercept b D 12=4 D Its slope is b=a D 3=4 y 2y D x Fig P.2-30 3x C 4y D 12 31 line through 2; 1/ parallel to y D x C is y D x perpendicular to y D x C is y D x C 32 line through 2; 2/ parallel to 2xCy D is 2xCy D line perpendicular to 2x C y D is x 2y D 33 We have x Fig P.2-27 3x C 4y D 2x 3y D 13 28 x C 2y D has x-intercept a D and y-intercept b D 4=2 D 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 and x D 13 9/=2 D The intersection point is 2; 3/ 34 x x C 2y D ÷ 1; line We have 2x C y D 5x 7y D ÷ 14x C 7y D 56 5x 7y D 1: Adding these equations gives 19x D 57, so x D and y D 2x D The intersection point is 3; 2/ 35 If a ¤ and b ¤ 0, then x=a/ C y=b/ D represents a straight line that is neither horizontal nor vertical, and does not pass through the origin Putting y D we get x=a D 1, so the x-intercept of this line is x D a; putting x D gives y=b D 1, so the y-intercept is y D b 36 The line x=2/ y=3/ D has x-intercept a D 2, and y-intercept b D y Fig P.2-28 29 p p p p 2x 3y D has x-intercept a D 2= D p and y-intercept 2= Its slope is p b pD b=a D 2= D 2=3 y p 2x p x x 3y D x y D1 3 Fig P.2-29 Fig P.2-36 30 1:5x 2y D has x-intercept a D 3=1:5 D 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/=.3 2/ D and equation y D 2.x 2/ D 2x Its y-intercept is Copyright © 2018 Pearson Canada Inc SECTION P.2 (PAGE 16) ADAMS and ESSEX: CALCULUS 38 The line through 2; 5/ and k; 1/ has x-intercept 3, so also passes through 3; 0/ Its slope m satisfies k Thus k 3D 0 DmD D 3C2 43 B D 1; 3/; C D 3; 2/ p p jABj D 2/2 C C 1/2 D 17 p p p p jAC j D 2/2 C C 1/2 D 34 D 17 p p jBC j D 1/2 C 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 D and y D D Thus D D C 1; 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 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 F 47 A D 2; 1/; B D 6; 4/; C D 5; 3/ p p jABj D 2/2 C 1/2 D 25 D p p jAC j D 2/2 C 1/2 D 25 D p p p jBC j D 5/2 C C 3/2 D 50 D 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 0/2 C 0/2 D D p p jAC j D 0/2 C 0/2 D D q p p jBC j D 1/2 C 3/2 D D 2: Since jABj D jAC j D jBC j, triangle ABC is equilateral 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 Thus P is 5; 0/ p x 2/2 C y D says that the distance of x; y/ from 2; 0/ is 4, so the equation represents a circle of radius centred at 2; 0/ p p x 2/2 C y D x 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 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 The line is parallel to 4x C y D 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 and 2x 3y D will also satisfy the equation Copyright © 2018 Pearson Canada Inc .x C 2y 3/ C k.2x 3y C 1/ D INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.3 (PAGE 22) because they cause both expressions in parentheses to be 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 C C k.2 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 C 2x 3y C 1/ D 0; 13 Together, x C y > and x C y < represent annulus (washer-shaped region) consisting of all points that are outside the circle of radius centred at the origin and inside the circle of radius centred at the origin 14 Together, x C y and x C 2/2 C y represent the region consisting of all points that are inside or on both the circle of radius centred at the origin and the circle of radius centred at 2; 0/ 15 Together, x Cy < 2x and x Cy < 2y (or, equivalently, x 1/2 C y < and x C y 1/2 < 1) represent the region consisting of all points that are inside both the circle of radius centred at 1; 0/ and the circle of radius centred at 0; 1/ 16 x2 C y2 4x C 2y > can be rewritten x 2/2 C y C 1/2 > This equation, taken together with x C y > 1, represents all points that lie both outside the circle of radius centred at 2; 1/ and above the line x C y D 17 The interior of the circle with centre 1; 2/ and radius p is given by x C 1/2 C y 2/2 < 6, or x C y C 2x 4y < or, on simplification, x D Section P.3 Graphs of Quadratic Equations (page 22) x C y D 16 x C y 2/2 D 4, or x C y 4y D .x C 2/2 C y D 9, or x C y C 4y D .x x2 C y2 x 3/2 C y C 4/2 D 25, or x C y 2 6x C 8y D 2x D 2x C C y D x 1/2 C y D centre: 1; 0/; radius 18 The exterior of the circle with centre 2; 3/ and radius is given by x 2/2 C y C 3/2 > 16, or x C y 4x C 6y > x C y C 4y D 19 2 x C y C 4y C D x2 C y2 x 2x C 4y D 2x C C y C 4y C D x 1/2 C y C 2/2 D centre: 1; 2/; radius x2 C y2 x2 2x yC1D0 2x C C y yC D 2 x 1/2 C y 21 D 14 centre: 1; 1=2/; radius 1=2 2 x1 x 1/2 C y 3/2 < 10 20 x C y > 4; 21 The parabola with focus 0; 4/ and directrix y D equation x D 16y 22 The parabola with focus 0; 1=2/ and directrix y D 1=2 has equation x D 2y 23 The parabola with focus 2; 0/ and directrix x D equation y D 8x 24 The parabola with focus 1; 0/ and directrix x D has equation y D 4x 25 y D x =2 has focus 0; 1=2/ and directrix y D x C y C 2/ D centre: 0; 2/; radius x C y < 2; has 1=2 y x C y > represents all points lying outside the circle of radius centred at the origin yDx =2 0;1=2/ 10 x C y < represents the open disk consisting of all points lying inside the circle of radius centred at the origin 11 has x yD 1=2 x C 1/2 C y represents the closed disk consisting of all points lying inside or on the circle of radius centred at the point 1; 0/ 12 x C y 2/2 represents the closed disk consisting of all points lying inside or on the circle of radius centred at the point 0; 2/ Fig P.3-25 26 yD x has focus 0; 1=4/ and directrix y D 1=4 Copyright © 2018 Pearson Canada Inc SECTION P.3 (PAGE 22) ADAMS and ESSEX: CALCULUS y y Version (c) yD1=4 y D x2 x 3; 3/ Version (b) 0; 1=4/ yD x Fig P.3-26 x Version (d) 4; 2/ 27 xD Version (a) y =4 has focus 1; 0/ and directrix x D y Fig P.3-29 a) has equation y D x xD1 1;0/ b) has equation y D x x c) has equation y D x d) has equation y D x xD y =4 30 Fig P.3-27 28 x D y =16 has focus 4; 0/ and directrix x D y 31 .4;0/ 32 x 33 xD xDy =16 34 Fig P.3-28 8x C 16 3/2 C or y D x 4/2 2, or y D x 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 p 4y D x C p y D 3x=2/ C p y=2/ D 4x C x shifted down 1, left gives y D 35 yD1 36 x2 C y2 D x C 4/2 C y 38 4/2 or y D x 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 .x C 1/2 shifted up 2, left gives 2/2 D 1/2 2/2 shifted down 1, right gives p p y D x shifted down 2, left gives y D x C Copyright © 2018 Pearson Canada Inc SECTION 18.5 (PAGE 1038) ADAMS and ESSEX: CALCULUS y 00 C 2y C 5y D 16 auxiliary eqn r C 2r C D ) r D ˙ 2i y D Ae t cos 2t C Be t sin 2t 10 For y 00 4y C 5y D the auxiliary equation is r 4r C D 0, which has roots r D ˙ 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 C 3y D the auxiliary equation isp r C 2r C D 0, which has solutions r D ˙ 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 C y D 0, hence r C r C D Since a D 1, b D and c D 1, the discriminant is D D b 4ac D < and b=2a/ D 21 and p ! D 3=2 Thus, the general solution is p p 3 1=2/t t C Be 1=2/t sin t y D Ae cos 2 00 < 2y C 5y 3y D 13 y.0/ D : y 0/ D The DE has auxiliary equation 2r C 5y D 0, with roots r D 21 and r D Thus y D Ae t=2 C Be 3t A Now D y.0/ D A C B, and D y 0/ D 3B Thus B D 1=7 and A D 6=7 The solution is y D e t=2 C e 3t 7 14 00 Given that y C 10y C 25y D 0, hence r C 10r C 25 D ) r C 5/2 D ) r D y0 D 5t 5e C Bt e 5t D y.1/ D Ae D y 1/ D 5t : 2t cos t C Be y D 2Ae 5e C Be Given that a > 0, b > and c > 0: Case 1: If D D b 4ac > then the two roots are p b ˙ b 4ac r1;2 D : 2a Since b2 p ˙ b2 p b ˙ b2 A C B/ C Be 2t C Be 2t 2t Ae 2t 4ac < b 4ac < therefore r1 and r2 are negative The general solution is y.t / D Ae r1 t C Be r2 t : y.t / D Ae r1 t C Bt e r2 t : ; sin t / cos t 4ac < b t!1 Now D y.0/ D A ) A D 2, and D y 0/ D 2A C B ) B D Therefore y D e 2t cos t C sin t / 678 17 Case 2: If D D b 4ac D then the two equal roots r1 D r2 D b=.2a/ are negative The general solution is we have A D 2e and B D 2e Thus, y D 2e e 5t C 2t e e 5t D 2.t 1/e 5.t 1/ 00 < y C 4y C 5y D y.0/ D : y 0/ D The auxiliary equation for the DE is r C 4r C D 0, which has roots r D ˙ i Thus which is, along with e t , a solution of the CASE II DE y 00 2y C y D If t ! 1, then e r1 t ! and e r2 t ! Thus, lim y.t / D Since y D Ae !0 Thus, 5t A C Bt / C Be e 1C/t e t !0 e tCh e t D t lim h h!0 d t e D t et Dt dt lim y t / D lim y D Ae 15 The auxiliary equation r 2 C /r C C / factors to r /.r 1/ D and so has roots r D C and r D Thus the DE y 00 C /y C C /y D 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 , C 2Be 2t / sin t: If t ! 1, then e r1 t ! and e r2 t ! at a faster rate than Bt ! Thus, lim y.t / D t!1 Case 3: If D D b y D Ae 4ac < then the general solution is b=2a/t cos.!t / C Be b=2a/t sin.!t / p 4ac b If t ! 1, then the amplitude of 2a both terms Ae b=2a/t ! and Be b=2a/t ! Thus, lim y.t / D where ! D t!1 18 The auxiliary equation ar C br C c D has roots r1 D Copyright © 2018 Pearson Canada Inc b p 2a D ; r2 D p bC D ; 2a INSTRUCTOR’S SOLUTIONS MANUAL SECTION 18.5 (PAGE 1038) where D D p b 4ac Note that a.r2 r1 / D D D 2ar1 C b/ If y D e r1 t u, then y 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 C cy D 0, and simplifying, we obtain e r1 t au00 C 2ar1 u0 C bu0 / D 0; 25 r2 v D r2 26 uD Z C e r2 r1 /t xy 27 r1 /t dt D Be r2 : Hence r1 /t C A; 19 y 000 4y 00 C 3y D Auxiliary: r 4r C 3r D r.r 1/.r 3/ D ) r D 0; 1; General solution: y D C1 C C2 e t C C3 e 3t 28 29 20 y 4/ 2y 00 C y D Auxiliary: r 2r C D 30 y 4/ C 2y 00 C y D Auxiliary: r C 2r C D 31 .r 1/2 D ) r D 1; 1; 1; General solution: y D C1 e t C C2 t e t C C3 e t C C4 t e t 22 y 4/ C 4y 3/ C 6y 00 C 4y C y D Auxiliary: r C 4r C 6r C 4r C D r C 1/ D ) r D 1; 1; 1; General solution: y D e t C1 C C2 t C C3 t C C4 t / 23 If y D e 2t , then y 000 2y 4y D e 2t 4/ D The auxiliary equation for the DE is r 2r D 0, for which we already know that r D is a root Dividing the left side by r 2, we obtain the quotient r C 2r C Hence the other two auxiliary roots are ˙ i General solution: y D C1 e 2t C C2 e t cos t C C3 e t sin t Aux eqn: r r 2/2 r 4/2 D Ce 2t C Bx : ) r D ˙1 Consider x y 00 xy C 5y D Since a D 1, b D 1, and c D 5, therefore b a/2 < 4ac Then k D a b/=2a D and ! D Thus, the general solution is y D Ax cos.2 ln x/ C Bx sin.2 ln x/ x y 00 C xy D aux: r.r 1/ C r D Thus y D A C B ln x: ) r D 0; 0: Given that x y 00 C xy C y D Since a D 1, b D 1, c D therefore b a/2 < 4ac Then k D a b/=2a D and ! D Thus, the general solution is y D A cos.ln x/ C B sin.ln x/ x y 000 C xy y D Trying y D x r leads to the auxiliary equation r.r 1/.r r r 2/ C r 3r C 3r 1D0 1D0 1/3 D ) r D 1; 1; 1: Thus y D x is a solution To find the general solution, try y D xv.x/ Then y D xv C v; Now x y 000 C xy y 00 D xv 00 C 2v ; y 000 D xv 000 C 3v 00 : y D x v 000 C 3x v 00 C x v C xv 2 000 00 xv D x x v C 3xv C v /; and y is a solution of the given equation if v D w is a solution of x w 00 C 3xw C w D This equation has auxiliary equation r.r 1/ C 3r C D 0, that is r C 1/2 D 0, so its solutions are r C 1/2 r 2/2 r 2/2 r C 2/2 D r D 2; 2; 2; 2; 1; 1; 2; 2: The general solution is y D e 2t C1 C C2 t C C3 t C C4 t / C e t C5 C C6 t / x y 00 C xy y D aux: r.r 1/ C r D B y D Ax C : x r C 1/2 D ) r D i; i; i; i General solution: y D C1 cos t C C2 sin t C C3 t cos t C C4 t sin t 24 3y D r.r 1/ r D ) r 2r D ).r 3/.r C 1/ D ) r1 D and r2 D and y D e r1 t u D Ae r1 t C Be r2 t 21 x y 00 Thus; y D Ax r1 /v; which has general solution v D C e r2 2r C D r 1/2 D 0; r D 1; 1: Thus y D Ax C Bx ln x: or, more simply, u00 r2 r1 /u0 D Putting v D u0 reduces this equation to first order: x y 00 xy C y D aux: r.r 1/ r C D 2C3 ln x C2 C x x v D C1 C C2 ln x C C3 ln x/2 : v0 D w D The general solution of the given equation is, therefore, C7 C C8 t /: Copyright © 2018 Pearson Canada Inc y D C1 x C C2 x ln x C C3 x.ln x/2 : 679 SECTION 18.5 (PAGE 1038) 32 Since ADAMS and ESSEX: CALCULUS dx D e t D x, we have dt Since a sum of squares cannot vanish unless each term vanishes, g.t / D for all t dy dx dy dz D Dx ; dt dx dt dx d 2z dx dy d y dx D Cx 2 dt dt dx dx dt dy d y Dx Cx : dx dx 37 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/ Then g is also solution of †/ Also g.0/ D f 0/ A D and g 0/ D f 0/ B! D Thus, g.t / D 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 Accordingly, z D z.t / satisfies d 2z dz C b a/ C cz dt dt d y dy dy Dax C ax C b a/x C cy D 0: dx dx dx a 38 4ac b b and ! D which is 2a 4a2 kt positive for Case III If y D e u, then We are given that k D y D e k t u0 C ku y 00 D e k t u00 C 2ku0 C k u : t 33 By the previous exercise, z.t / D y.e / D y.x/ must satisfy the constant coefficient equation d 2z dt 2 dz C 2z D 0: dt Substituting into ay 00 C by C cy D leads to The auxiliary equation for this equation is r which has roots r D ˙ i Thus D e k t au00 C 2ka C b/u0 C ak C bk C c/u D e k t au00 C C b =.4a/ b =.2a/ C c/u D a e k t u00 C ! u : 2r C D 0, z D C1 e t cos t C C2 e t sin t: Since t D ln x, the given Euler equation has solution Thus u satisfies u00 C ! u D 0, which has general solution y D C1 x cos.ln x/ C C2 x sin.ln x/: u D A cos.!t / C B sin.!t / 34 If y D A cos !t C B sin !t then y 00 C ! y D A! cos !t by the previous problem Therefore ay 00 C by C cy D has general solution B! sin !t C ! A cos !t C B sin !t / D y D Ae k t cos.!t / C Be k t sin.!t /: for all t So y is a solution of (†) 35 If f t / is any solution of †/ then f 00 t / D all t Thus, ! f t / for 2 2 i d h 2 ! f t / C f t / dt D 2! f t /f t / C 2f t /f 00 t / D 2! f t /f t / 39 2! f t /f t / D 2 2 for all t Thus, ! f t / C f t / is constant (This can be interpreted as a conservation of energy statement.) 36 If g.t / satisfies †/ and also g.0/ D g 0/ D 0, then by Exercise 20, 40 2 2 ! g.t / C g t / 2 2 D ! g.0/ C g 0/ D 0: 680 00 < y C 100y D y.0/ D : y 0/ D y D A cos.10t / C B sin.10t / A D y.0/ D 0; 10B D y 0/ D 3 yD sin.10t / 10 Circular frequency = 10, freq = 5=, period = =5, amplitude = 3=10 Because y 00 C 4y D 0, therefore y D A cos 2t C B sin 2t Now y.0/ D ) A D 2; Copyright © 2018 Pearson Canada Inc y 0/ D 5)B D 2: INSTRUCTOR’S SOLUTIONS MANUAL SECTION 18.5 (PAGE 1038) Thus, y D cos 2t 52 sin 2t circular frequency = ! D 2, frequency = ! D 0:318 2 2 D 3:14 period = !q amplitude = 2/2 C 52 /2 ' 3:20 41 y D A cos !.t c/ C B sin !.t therefore, y D A cos 20 t C B sin 20 t and y.0/ D y 0/ D ) B D D : 20 10 c/ (easy tocalculate y 00 C ! 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 45 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/ 42 For y 00 C y D 0, we have y D A sin t C B cos t Since, y.2/ D D A sin C B cos y 2/ D D A cos B sin 2; 46 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 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 cos 2/ sin t C sin C cos 2/ cos t 2/ sin.t 2/: 47 a/ 44 From Example 9, the spring constant is k D 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, k 104 mD D D 22:8 gm: 400 400 < y 00 C 400 y D y.0/ D : y 0/ D t0 / D e k t ŒA1 cos !t C B1 sin !t ; Thus, The motion is determined by t0 /B sin !.t 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 A D sin cos B D sin C cos 2: < y 00 C ! y D 43 y.a/ D A : y a/ D B B y D A cos !.t a/ C sin !.t ! sin 20 t , with y in cm and t Thus, y D cos 20 t C 10 in r second, gives the displacement at time t The amplitude / 1:0005 cm is 1/2 C 10 ! k Frequency D , !2 D (k = spring const, m = mass) 2 m Since the spring does not change, ! m D k (constant) For m D 400 gm, ! D 2.24/ (frequency = 24 Hz) 4 24/2 400/ If m D 900 gm, then ! D 900 2 24 D 32 so ! D 32 Thus frequency = D 16 Hz 2 4 24/2 400 For m D 100 gm, ! D 100 ! so ! D 96 and frequency = D 48 Hz 2 Using the addition identities for cosine and sine, y D e k t ŒA cos !.t therefore y D sin D cos.t 1)AD Expanding the hyperbolic functions in terms of exponentials, y D e k t ŒA cosh !.t t0 /B sinh !.t A !.t t0 / A !.t t0 / D ek t e C e 2 B !.t t0 / B !.t t0 / C e e 2 D A1 e kC!/t C B1 e k t0 / !/t 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 C br C c D 0, so e k˙!/t are independent solutions of ay 00 C by C cy D 0, and our function y must also be a solution Since it involves two arbitrary constants, it is a general solution Copyright © 2018 Pearson Canada Inc 681 SECTION 18.5 (PAGE 1038) ADAMS and ESSEX: CALCULUS 00 < y C 2y C 5y D 48 y.3/ D : y 3/ D This IVP for the equation of simple harmonic motion has solution The DE has auxiliary equation r C 2r C D with roots r D ˙ 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 y0 D c u.x/ k c D c c k a/ cos.kx/ C kb sin.kx/ k c b D cos.kx/ C a cos.kx/ C sin.kx/: k k y.x/ D The initial conditions give 0 D y 3/ D Thus A D 2e and B D solution A e A C 2B/ A=2 D y D e t Œ2 cos 2.t 3/ e The IVP has sin 2.t 3/: 00 < y C 4y C 3y D 49 y.3/ D : y 3/ D The DE has auxiliary equation r C 4r C D with roots r D C D and r D D (i.e k ˙ !, where k D 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 Ce ŒA cosh.t 2t 3/ C B sinh.t ŒA sinh.t 3/ 3/ C B cosh.t 3/: The initial conditions give D y.3/ D e 0 D y 3/ D e A 2A C B/ Thus A D e and B D 2A D 2e The IVP has solution y D e6 50 Let u.x/ D c Also u0 x/ D u00 x/ D 682 2t Œcosh.t kb sin.kx/ so that 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/: D y.3/ D e k a/ cos.kx/ u.x/ D c 3/ C sinh.t 3/: 51 Since x 0/ D and x.0/ D > 1=5, the motion will be governed by x 00 D x C 1=5/ until such time t > when x t / D 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/ D 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 t / D u0 t / D 4=5/ sin t These formulas remain valid until t D when x t / becomes again Note that x./ D 4=5/ C 1=5/ D 3=5/ Since x./ < 1=5/, the motion for t > will be governed by x 00 D x 1=5/ until such time t > when x t / D 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 / D x / D 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 t / D 2=5/ sin t These formulas remain valid for t until t D 2 when x becomes again We have x.2/ D 2=5/ 1=5/ D 1=5 and x 2/ D The conditions for stopping the motion are met at t D 2; the mass remains at rest thereafter Thus k y.x/ Then u.0/ D c k a k y x/, so u0 0/ D k b We have k y 00 x/ D k c k y.x/ D k u.x/ Copyright © 2018 Pearson Canada Inc 84 < cos t C x.t / D cos t :1 if t if < t 2 if t > 2 INSTRUCTOR’S SOLUTIONS MANUAL APPENDIX I (PAGE A-10) APPENDICES Appendix I Complex Numbers (page A-10) zD Re.z/ D C 2i; zD zD 3 and Arg w/ D , then 3 5 arg zw/ D C D , so 4 5 3 Arg zw/ D 2 D 4 5 and Arg w/ D , then 17 If Arg z/ D 5 13 arg z=w/ D D , so 12 13 11 C 2 D Arg z=w/ D 12 12 18 jzj D 2; arg z/ D ) z D 2.cos C i sin / D 16 5; C 2i Im.z/ D y z-plane x zD4 i 19 zD i 20 Fig APX I-1 z D i; Re.z/ D 4; Im.z/ D z D i; Re.z/ D 0; Im.z/ D z D 6; 6; Im.z/ D p jzj D 2; Arg z/ D 3=4 zD C i; p z D cos.3=4/ C i sin.3=4// z D 2; jzj D 2; Arg z/ D z D 2.cos C i sin / z D 3i; jzj D 3; Arg z/ D =2 z D 3.cos.=2/ C i sin.=2// 10 21 Re.z/ D 22 23 24 z D 5i; jzj D 5; Arg z/ D =2 z D 5.cos =2/ C i sin =2// p z D C 2i; jzj D 5; D Arg z/ D tan p z D 5.cos C i sin / p z D C i; jzj D 5; D Arg z/ D tan p z D 5.cos C i sin / z D 4i; jzj D 5; z D 5.cos C i sin / 12 z D 4i; jzj D 5; D Arg z/ D tan 4=3/ z D 5.cos C i sin / p z D i; jzj D 2; Arg z/ D =6 z D 2.cos =6/ C i sin =6// p p zD 3i; jzj D 3; Arg z/ D 2=3 p z D 3.cos 2=3/ C i sin 2=3// 14 15 4 4 C 3i sin 5 4 Arg z/ D C tan 4=3/ jzj D 5; D arg z/ D ) sin D 3=5; cos D 4=5 z D C 3i 3 3 3 jzj D 1; arg z/ D ) z D cos C i sin 4 1 ) zD p Cp i 2 ) z D cos C i sin jzj D ; arg z/ D 6 p ) zD C i 2 jzj D ) z D for any value of arg z/ 1 cos ) zD i sin jzj D ; arg z/ D 3 p ) zD i 4 C 3i D 3i 5i D 26 4i D 4i 27 28 jzj D represents all points on the circle of radius centred at the origin 29 jzj represents all points in the closed disk of radius centred at the origin 30 jz 2i j represents all points in the closed disk of radius centred at the point 2i 31 jz C 4i j represents all points in the closed disk of radius centred at the point 4i 32 arg z/ D =3 represents all points on the ray from the origin in the first quadrant, making angle 60ı with the positive direction of the real axis 33 arg z/ 7=4 represents the closed wedge-shaped region in the third and fourth quadrants bounded by the ray from the origin to on the real axis and the ray from the origin making angle 45ı with the positive direction of the real axis 34 .2 C 5i / C z D cos jzj D 3; 25 .1=2/ 11 13 D Arg z/ D If Arg z/ D C 5i i D2Ci Copyright © 2018 Pearson Canada Inc i / D C 4i 683 APPENDIX I (PAGE A-10) 35 i 2i / C 3i / D i / D 16 37 .1 C i /.2 3i / D C 2i 38 .a C bi /.2a C C i C 2i i D 17 36 .4 C i /.4 ADAMS and ESSEX: CALCULUS 3i 3i D 3i D i bi / D a C bi /.2a C bi / D 2a 49 b) z D 2=z can be rewritten jzj2 D zz D 2, which has no solutions since the square of jzj is nonnegative for all complex z b C 3abi 39 .2 C i / D C 12i C 6i C i D C 11i 40 41 42 43 i D 2Ci i /2 4i D i2 50 .1 C 3i /.2 C i / C 7i C 3i D D i i2 1Ci 1Ci C i / 2i / D D D i.2 C 3i / C 2i 9C4 a) z D 2=z can be rewritten jzj2 D zz D 2, so isp satisfied by all numbers z on the circle of radius centred at the origin 51 5i 13 p If z p D w Dp 1, then zw D 1, so zw D 1.pBut if we use p z D D ip and the same value for w, then p z w D i D Ô zw The three cube roots of D cos C i sin are of the form cos Ci sin where D =3, D , and D 5=3 Thus they are C 2i /.2 3i / 8Ci D D1 i /.3 C 2i / 8Ci p Ci ; 2 44 If z D x C yi and w D u C vi , where x, y, u, and v are real, then z C w D x C u C y C v/i D x C u y C v/i D x 52 yi C u vi D z C w: 45 Using the fact that jzwj D jzjjwj, we have z w 46 47 48 D zw jwj2 D cos C 3i cos sin Thus cos.3 / D cos3 cos sin sin3 D sin sin3 : p i; i: p 3 3 The three cube roots of 1Ci D cos C i sin 4 1=6 are of the form cos C i sin / where D =4, D 11=12, and D 19=12 55 The equation z C p i D hassolutions that are the p 2 2 four fourth roots of C i D cos C i sin 3 1=4 Thus they are of the form cos C i sin /, where D =6, 2=3, 7=6, and 5=3 They are the complex numbers ˙2 i sin cos p The four p fourth roots of D 4.cos C i sin 0/ are of the form 2.cos C i sin / where D 0,p D =2, , p p and p 2, and i D 3=2 Thus they are 2, i 2, 3 cos sin2 D cos3 sin.3 / D cos2 sin 684 : 54 cos.3 / C i sin.3 / D cos C i sin / p 3 3 C i sin are 8i D cos 2 of the form 2.cos C i sin / where D =2, D 7=6, and D 11=6 Thus they are 2i; 53 i The three cube roots of zw zw z D : D D jwj2 ww w p p z D C i D cos C i sin 6 p 2 2 w D C i D cos C i sin 3 p 5 5 zw D cos C i sin 6 p p z D cos C i sin D i w 2 p 3 3 z D C i D cos C i sin 4 w D 3i D cos C i sin 2 p 5 5 zw D cos C i sin D 3i 4 p z cos C i sin D C i D w 4 3 1; 56 1=4 p ! C i ; 2 ˙2 1=4 p ! i : The equation z C a5 D (a > 0) has solutions that are the five fifth roots of a5 D a cos C i sin /; they are of the form a.cos C i sin /, where D =5, 3=5, , 7=5, and 9=5 Copyright © 2018 Pearson Canada Inc INSTRUCTOR’S SOLUTIONS MANUAL APPENDIX II (PAGE A-19) 57 The n nth roots of unity are !1 D 2 2 C i sin n n 4 4 !3 D cos C i sin D !22 n n 6 6 C i sin D !23 !4 D cos n n :: : 2.n 1/ 2.n 1/ C i sin D !2n !n D cos n n The function w D z D x y C 2xyi transforms the line y D to u D x 1, v D 2x, which is the parabola v D 4u C with vertex at w D and opening to the right !2 D cos 10 The function w D 1=z D x yi /=.x C y / transforms the line x D to the curve given parametrically by uD : u2 C v D !1 C !2 C !3 C C !n D C !2 C !22 C C !2n !2n D D 0: D !2 !2 vD y : C y2 This curve is, in fact, a circle, Hence, Appendix II Complex Functions (page A-19) In Solutions 1–12, z D x C yi and w D u C vi , where x, y, u, and v are real The function w D z transforms the closed rectangle x 1, y to the closed rectangle u 1, v The function w D z transforms the line x C y D to the line u v D 13 The function w D z D x y C 2xyi transforms the line x D to u D y , v D 2y, which is the parabola v D 4u with vertex at w D 1, opening to the left f z/ D z D x C yi /2 D x y C 2xyi f z/ D z D x C yi /3 D x 3xy C 3x y u D x2 y2; v D 2xy @u @v @u @v D 2x D ; D 2y D @x @y @y @x @u @v f z/ D Ci D 2x C 2yi D 2z: @x @x 14 y /i u D x 3xy ; v D 3x y y @u @v @u @v D 3.x y / D ; D 6xy D @x @y @y @x @u @v f z/ D Ci D 3.x y C 2xyi / D 3z : @x @x The function w D 1=z D z=jzj2 transforms the closed quarter-circular disk jzj 2, arg z/ =2 to the closed region lying on or outside the circle jwj D 1=2 and in the fourth quadrant, that is, having =2 arg w/ The function w D iz rotates the z-plane 90ı , so transforms the wedge =4 arg z/ =3 to the wedge =4 arg z/ =6 p The function w D z transforms the ray arg z/ D =3 (that is, Arg z/ D 5=3) to the ray arg w/ D 5=6 The function w D e z D e x cos y C i e x sin y transforms the horizontal strip < x < 1, =4 y =2 to the wedge =4 arg w/ =2, or, equivalently, u 0, v u 12 The function w D e iz D e y cos x C i sin x/ transforms the vertical half-strip < x < =2, < y < to the first-quadrant part of the unit open disk jwj D e y < 1, < arg w/ D x < =2, that is u > 0, v > 0, u2 C v < The function w D z transforms the closed annular sector jzj 2, =2 arg z/ 3=4 to the closed annular sector jwj 4, arg w/ 3=2 The function w D z transforms the closed quarter-circular disk jzj 2, arg z/ =2 to the closed threequarter disk jwj 8, arg w/ 3=2 C y2 D u; C y /2 with centre w D 1=2 and radius 1/2 11 1 ; C y2 15 x yi D z x C y2 x y uD ; vD x C y2 x C y2 @u y2 x2 @v @u 2xy @v D D ; D D @x x C y /2 @y @y x C y /2 @x @u @v x y / C 2xyi z/2 f z/ D Ci D D D 2 @x @x x C y / zz/2 f z/ D Copyright © 2018 Pearson Canada Inc : z2 685 APPENDIX II (PAGE A-19) 16 f z/ D e z D e x x2 y2 y2 ADAMS and ESSEX: CALCULUS cos.2xy/ C i sin.2xy// x2 21 y2 uDe cos.2xy/; vDe sin.2xy/ @v @u 2 D e x y 2x cos.2xy/ 2y sin.2xy// D @x @y @v @u 2 D e x y 2y cos.2xy/ C 2x sin.2xy// D @y @x @u @v f z/ D Ci @x @x 2 D e x y Œ2x cos.2xy/ 2y sin.2xy/ C i.2y cos.2xy/ C 2x sin.2xy// D 2x C 2yi /e x 17 y2 yi e Thus cos y D e yi yi Ce e yi and sin y D e e zi C e zi and sin z D D 2i sin y: yi yi e zi cosh z D ez C e z and sinh z D e 20 d cos z dz d sin z dz d cosh z dz d sinh z dz D D D D d dz d dz d dz d dz 686 zi , e 2zi D Œcos.2x/ C i sin.2x/ D By Exercises 20 and 21, cosh z D if and only if cos.iz/ D 0, that is, if and only if z D 2n C 1/ i=2 for integer n Similarly, sinh z D if and only if sin.iz/ D 0, that is, if and only if z D n i for integer n 24 e z D e xCyi D e x cos y C i e x sin y D e x yi D e x cos y e x sin y ex C e x ex e x ez C e z D cos y C i sin y cosh z D 2 D cosh x cos y C i sinh x sin y Re.cosh z/ D cosh x cos y; Im.cosh z/ D sinh x sin y: e ez e z e e iz D e iz De yCxi y xi D ex e x De y y cos y C i ex C e x sin y 2 D sinh x cos y C i cosh x sin y Re.sinh z/ D sinh x cos y; Im.cosh z/ D cosh x sin y: 26 e zi C e zi i e zi e zi D D sin z 2 zi zi zi zi e e ie C e D D cos z 2i 2i z z z z e Ce e e D D sinh z 2 z z z z e e e Ce D D cosh z 2 z sinh z D e iz C e iz D cosh z iz iz e e i sinh.iz/ D D sin z i e z C ez cos.iz/ D D cosh z z z e e e z C ez sin.iz/ D Di D i sinh z 2i cosh.iz/ D 23 z are periodic with period 2 i Thus the only complex zeros of sin z are its real zeros at z D n for integers n 25 ez , sin.2x/ D 0; e 2y cos.2x/ D , y D 0; cos.2x/ D D, y D 0; x D 0; ˙; ˙2; : : : zi are periodic with period 2, and 2y ,e e 2i , e 2zi D Œcos.2x/ C i sin.2x/ D sin z D , e zi D e e zC2 i D e x cos.y C 2/ C i sin.y C 2// D e x cos y C i sin y/ D e z : z Thus e is periodic with period 2 i So is e z D 1=e z Since e i.zC2/ D e ziC2 i D e zi , therefore e zi and also e zi are periodic with period 2 Hence cos z D 19 22 yi e 2i zi e , sin.2x/ D 0; e 2y cos.2x/ D , y D 0; cos.2x/ D 3 D, y D 0; x D ˙ ; ˙ ; : : : 2 Thus the only complex zeros of cos z are its real zeros at z D 2n C 1/=2 for integers n .cos.2xy/ C i sin.2xy// D 2ze z : D cos y; 2y ,e e yi D cos y C i sin y (for real y) Replacing y by y, we get e yi D cos y i sin y (since cos is even and sin is odd) Adding and subtracting these two formulas gives e yi C e 18 cos z D , e zi D cos x C i e D e cos x y sin x y i e sin x e iz C e iz e y C ey e y ey cos z D D cos x C i sin x 2 D cos x cosh y i sin x sinh y Re.cos z/ D cos x cosh y; Im.cos z/ D sin x sinh y e iz e y ey e y C ey D cos x C i sin x 2i 2i 2i D sin x cosh y C i cos x sinh y Re.sin z/ D sin x cosh y; Im.sin z/ D cos x sinh y: sin z D 27 28 e iz z C 2iz D ) z D or z D z 2z C i D ) z Copyright © 2018 Pearson Canada Inc 2i 1/ D i p 7 7 D cos C i sin 4 7 7 ) z D ˙ 21=4 cos C i sin 8 INSTRUCTOR’S SOLUTIONS MANUAL 29 z C 2z C D ) z C 1/2 D ) z D ˙ 2i 30 z2 2iz 31 z3 3iz 33 By long division (details omitted) we discover that z C 3z C 4z C 4z C 3z C D z C 3z C 3z C z2 C D z C 1/3 : D ) z i /2 D ) z D i (double root) 2z D z.z 3iz ) z D or z 3iz D ) z D or z i D ) z D or z D ˙ i 2 ) z D or z D i or z D 2i 32 APPENDIX III (PAGE A-26) 2/ D Thus P z/ has the five zeros: i , 36 i, 1, 1, and Since P z/ D z 2z 8z C 8z C 31z 30 has real coefficients, if z1 D C i is a zero of P z/, then so is z2 D i Now z z2 / D z C 4z C 5: z1 /.z 2z C D ) z 1/2 D p p z2 D i or z2 D C i 5 5 z D cos C i sin ; z D cos C i sin 3 3 p 5 5 z D ˙ cos C i sin ; or 6 p z D ˙ cos C i sin ! ! r r i i zD˙ Cp p ; zD˙ 2 2 By long division (details omitted) we discover that z C D ) z D i or z D i i 1Ci ) zD˙ p ; zD˙ p 1Ci i z C1D z z p p 1Ci i zC p zC p 2 ! ! 2 1 C C D z p zC p 2 2 p p 2 D z 2z C 1/.z C 2z C 1/ Hence P z/ has the five zeros z4 34 Since P z/ D z 4z 3p C 12z 16z C 16 has real coefficients, if z1 D 3i is a zero of P z/, then so is z1 Now z z1 /.z z1 / D z 1/2 C D z 2z C 4: By long division (details omitted) we discover that z4 4z C 12z 16z C 16 D z2 z 2z C 35 Since P z/ D z C 3z C 4z C 4z C 3z C has real coefficients, if z1 D i is a zero of P z/, then so is z2 D i Now z z1 /.z z2 / D z z5 i /.z C i / D z C 1: 2z 8z C 8z C 31z z C 4z C 6z C 11z 6: D z3 Observe that z3 D is a zero of z long division again: z3 6z C 11z z 37 If w D z C z jw 2iz z j D jz D z2 30 6z C 11z 5z C D z C i, By 2/.z 3/: i , 1, 2, and 3 and jzj D 2, then jz j D 16 and 2iz 3j C C D 15 < 16: By the mapping principle described in the proof of Theorem 2, the image in the w-plane of the circle jzj D is a closed curve that winds around the origin the same number of times that the image of z does, namely times Appendix III Continuous Functions (page A-26) To be proved: If a < b < c, f x/ g.x/ for a x c, limx!b f x/ D L, and limx!b g.x/ D M , then L M Proof: Suppose, to the contrary, that L > M Let D L M /=3, so > There exist numbers ı1 > and ı2 > such that if a x b, then jx jx 2z C 4: Thus z1 and z1 are both double zeros of P z/ These are the only zeros Thus if jx bj < ı1 ) jf x/ bj < ı2 ) jg.x/ bj < ı D minfı1 ; ı2 ; b f x/ g.x/ > L M Lj < M j < : a; c DL M bg, then 2 D L M > 0: This contradicts the fact that f x/ g.x/ on Œa; b Therefore L M Copyright © 2018 Pearson Canada Inc 687 APPENDIX III (PAGE A-26) ADAMS and ESSEX: CALCULUS To be proved: If f x/ K on Œa; b/ and b; c, and if limx!b f x/ D L, then L K Proof: If L > K, then let D L K/=2; thus > There exists ı > such that ı < b a and ı < c b, and such that if < jx bj < ı, then jf x/ Lj < In this case L K f x/ > L D L > K; which contradicts the fact that f x/ K on Œa; b/ and b; c Therefore L K x!a a) Let f x/ D C , g.x/ D x Let > be given and let ı D For any real number x, if jx aj < ı, then jf x/ jg.x/ f a/j D jC C j D < ; g.a/j D jx aj < ı D : Thus limx!a f x/ D f a/ and limx!a g.x/ D g.a/, and f and g are both continuous at every real number a A polynomial is constructed by adding and multiplying finite numbers of functions of the type of f and g in Exercise By Theorem 1(a), such sums and products are continuous everywhere, since their components have been shown to be continuous everywhere If P and Q are polynomials, they are continuous everywhere by Exercise If Q.a/ Ô 0, then P a/ P x/ D by Theorem 1(a) Hence P =Q is limx!a Q.x/ Q.a/ continuous everywhere except at the zeros of Q If m and n are integers and n is odd, then x/m=n D cx m=n , where c D 1/m=n is either or depending on the parity of m Since x m=n is continuous at each positive number a, so is cx m=n Thus x/m=n is continuous at each positive number, and x m=n is continuous at each negative number If r D m=n > 0, then limx!0C x r D by Exercise Hence limx!0 x r D 1/r limx!0C x r D 0, also Therefore limx!0 x r D 0, and x r is continuous at x D Thus limx!0C x r D x!a and x m=n is continuous at each positive number < x r < ı r D : ) x!a D a1=n /m D am=n ; Let > be given Let ı D 1=r , (r > 0) Then 0 we have m m lim x m=n D lim x 1=n D lim x 1=n 10 Let > be given Let ı D If a is any real number then ˇ ˇ ˇ ˇ ˇjxj jajˇ jx aj < if jx aj < ı: Thus limx!a jxj D jaj, and the absolute value function is continuous at every real number 11 By the definition of sin, P t D cos t; sin t /, and Pa D cos a; sin a/ are two points on the unit circle x C y D Therefore jt aj D length of the arc from P t to Pa > length of the chord from P t to Pa p D cos t cos a/2 C sin t sin a/2 : If > is given, and jt inequality implies that j cos t j sin t aj < ı D , then the above cos aj jt sin aj jt aj < ; aj < : Thus sin is continuous everywhere Suppose n is a positive integer and a > Let > be given Let b D a1=n , and let ı D minfa.1 n /; b n g If jx aj < ı, then x > a=2n , and if y D x 1=n , then y > b=2 Thus ˇ ˇ 1=n ˇx ˇ ˇ a1=n ˇ D jy bj jy n b n j C y n b C C bn bn aj < D : bn D yn < jx bn Thus limx!a x 1=n D a1=n , and x 1=n is continuous at x D a 688 12 The proof that cos is continuous everywhere is almost identical to that for sin in Exercise 11 n a a o 13 Let a > and > Let ı D ; 2 a If jx aj < ı, then x > , so < whenever t is t a between a and x Thus j ln x ln aj between t D a and t D x t a aj < D : a D area under y D < jx a Thus limx!a ln x D ln a, and ln is continuous at each point a in its domain 0; 1/ Copyright © 2018 Pearson Canada Inc INSTRUCTOR’S SOLUTIONS MANUAL 14 APPENDIX IV (PAGE A-31) Let a be any real number, and let > be given Assume (making smaller if necessary) that < e a Since ln C ln C D ln ea ea 2 e 2a a f c/ f x/ f d / < 0; we have ln C a < ln e ea Let ı D ln C a If jx aj < ı, then e ln < x a < ln C a a e e x a < e < C ea ea x a 1j < a je e je x e a j D e a je x a 1j < : x for all x in Œa; b Since g is increasing, so is its inverse g Therefore for all x in Œa; b, and f is bounded on that interval Appendix IV The Riemann Integral (page A-31) x Thus limx!a e D e and e is continuous at every point a in its domain L.f; P / D 1 U.f; P / D 1 Suppose a xn b for each n, and lim xn D L Then a L b by Theorem Let > be given Since f is continuous on Œa; b, there exists ı > such that if a x b and jx Lj < ı then jf x/ f L/j < Since lim xn D L, there exists an integer N such that if n N then jxn Lj < ı Hence jf xn / f L/j < for such n Therefore lim.f xn / D f L/ t 16 Let g.t / D For t Ô we have C jt j 15 g t / D C jt j t sgn t C jt j jt j D D > 0: C jt j/2 C jt j/2 C jt j/2 h!0 g.h/ g.0/ h D lim h!0 D 1: C jhj f x/ C jf x/j is also continuous there, being the composition of continuous functions Also, h.x/ is bounded on Œa; b, since ˇ ˇ ˇ ˇ ˇg f x/ ˇ jf x/j 1: C jf x/j By assumption in this problem, h.x/ must assume maximum and minimum values; there exist c and d in Œa; b such that g f c/ g f x/ g f d / Then C0C0D1 3 2 C1 C0D1C : 3 Thus g is continuous and increasing on R If f is continuous on Œa; b, then h.x/ D g f x/ D ; C 3 ; 2g Since U.f; P / L.f; P / < , f is integrable on Œ0; 2 Since L.f; P / < < U.f; P / for every , therefore Z f x/ dx D If t D 0, g is also differentiable, and has derivative 1: g 0/ D lim if x if < x Let < < Let P D f0; f x/ D if x D 1=n n D 1; 2; 3; : : :/ otherwise If P is any partition of Œ0; 1 then L.f; P / D Let < Let N be an integer such that N C1 > N A partition P of Œ0; 1 can be constructed so that the first two points of P are and , and such that each of the n D 1; 2; 3; : : : ; n/ lies in a subinterval N points n of P having length at most Since every number 2N n h i with n a positive integer lies either in 0; or one of these other N subintervals of P , and since max f x/ D for these subintervals and max f x/ D for all other subintervals of P , therefore U.f; P / C N D 2N By Theorem 3, f is integrable on Œ0; 1 Evidently f x/ D Z n f x/ D n f x/ dx D least upper bound L.f; P / D 0: 1=n if x D m=n in lowest terms otherwise Copyright © 2018 Pearson Canada Inc 689 APPENDIX IV (PAGE A-31) ADAMS and ESSEX: CALCULUS Clearly L.f; P / D for every partition P of Œ0; 1 Let > be given To show that f is integrable we must exhibit a partition P for which U.f; P / < We can assume < Choose a positive integer N such that 2=N < There are only finitely many integers n such that n N For each such n, there are only finitely many integers m such that m=n Therefore there are only finitely many points x in Œ0; 1 where f x/ > =2 Let P be a partition of Œ0; 1 such that all these points are contained in subintervals of the partition having total length less than =2 Since f x/ on these subintervals, and f x/ < =2 on all other subintervals P , therefore U.f; P / =2/ C =2/ D ; and f is R1 integrable on Œ0; 1 Evidently f x/ dx D 0, since all lower sums are I I Suppose, to the contrary, that I > I Let D , so > By the definition of I and I , there exist partitions P1 and P2 of Œa; b, such that L.f; P1 / I and U.f; P2 / I C By Theorem 2, L.f; P1 / U.f; P2 /, so 3 D I I L.f; P1 / C Therefore U.f C g; P / Hence a b Af x/ dx D A Z b f x/ dx: a It therefore remains to be proved only that the integral of a sum of functions is the sum of the integrals Suppose that Z a b f x/ dx D I; and Z a U.g; P2 / g.x/ dx D J: I < L.f; P1 / C 2 J < L.g; P2 / C : 2 Let P be the common refinement of P1 and P2 Then the above inequalities hold with P replacing P1 and P2 If m1 f x/ M1 and m2 g.x/ M2 on any interval, then m1 C m2 f x/ C g.x/ M1 C M2 there It follows that U.f C g; P / U.f; P / C U.g; P /; L.f; P / C L.g; P / L.f C g; P /: 690 b then there exist partitions P1 of Œa; b, and P2 of Œb; c such that L.f; P1 / I < L.f; P1 / C L.f; P2 / J < L.f; P2 / C (with similar inequalities for upper sums) Let P be the partition of Œa; c formed by combining all the subdivision points of P1 and P2 Then L.f; P / D L.f; P1 / C L.f; P2 / I C J < L.f; P / C : Similarly, U.f; P / < I C J U.f; P / Therefore Z c a f x/ dx D I C J: Let Z b a f x/ dx D I; and Z a b g.x/ dx D J; where f x/ g.x/ on Œa; b We want to show that I J Suppose, to the contrary, that I > J Then there would exist a partition P of Œa; b for which b If > 0, then there exist partitions P1 and P2 of Œa; b such that U.f; P1 / Assume a < b < c; the other cases are similar Let > If Z b Z c f x/ dx D I; and f x/ dx D J; a Since > 0, it follows that This contradiction shows that we must have I I Z Z b f x/ C g.x/ dx D I C J a U.f; P2 / C 2: Multiplying a function by a constant multiplies all its Riemann sums by the same constant If the constant is positive, upper and lower sums remain upper and lower; if the constant is negative upper sums become lower and vice versa Therefore I C J L.f C g; P / C : I < L.f; P / C I J ; and U.g; P / I J < J: I CJ > U.g; P / L.g; P / However, f x/ g.x/ on Œa; b implies that L.f; P / L.g; P / for any partition Thus we have a contradiction, and so I J Thus L.f; P / > Since jf x/j f x/ jf x/j for any x, we can apply the above result to obtain Z b a jf x/j dx Z b a f x/ dx Z a ˇZ ˇ Z ˇ b ˇ b ˇ ˇ Therefore ˇ f x/ dx ˇ jf x/j dx ˇ a ˇ a Copyright © 2018 Pearson Canada Inc b jf x/j dx: INSTRUCTOR’S SOLUTIONS MANUAL We have Z APPENDIX IV (PAGE A-31) By repeated applications of the triangle inequality, a a f x/ dx D D D Z Z a a Z0 a f x/ dx C Z f x/ dx C a f x/ dx Z jf xk a 1/ f a/j D jf xk 1/ f x0 /j < k 1: f x/ dx If x is any point in Œa; b, then x belongs to one of the intervals Œxk ; xk , so, by the triangle inequality again, Œf x/ C f x/ dx: If f is odd, last integral is If f is even, the last Z the a integral is 2f x/ dx jf x/ f a/j jf x/ f xk /jCjf xk / f a/j < k N: Let > be given Let ı D =2 Let x and 0p y If p x < =4 and y < =4 then p p j x yj x C y < If jx yj < ı and either x =4 or y =4 then p jx yj 2 p yj D p D : j x p < xC y p Thus f x/ D x is uniformly continuous on Œ0; 1 10 Suppose f is uniformly continuous on Œa; b Taking D in the definition of uniform continuity, we can find a positive number ı such that jf x/ f y/j < whenever x and y are in Œa; b and jx yj < ı Let N be a positive integer such that h D b a/=N satisfies h < ı If xk D a Ckh, (0 k N ), then each of the subintervals of the partition P D fx0 ; x1 ; : : : ; xN g has length less than ı Thus jf xk / f xk /j 0), and that f is integrable on Œa; b Let > be given, and let ı D =K If x and y belong to Œa; b and jx yj < ı, then jF x/ ˇZ x ˇ Z y ˇ ˇ F y/j D ˇˇ f t / dt f t / dt ˇˇ ˇZax ˇ a ˇ ˇ ˇ Dˇ f t / dt ˇˇ Kjx yj < K D : K y (See Theorem 3(f) of Section 6.4.) Thus F is uniformly continuous on Œa; b Copyright © 2018 Pearson Canada Inc 691 ... 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... Normal has slope a , and equation y or y D a2 x a3 C a 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 D lim h!0 Œ .a C h/ C a C h .a2 ... average rate of change of x over ? ?a; b is b a3 D b C ab C a2 : b a The instantaneous rate of change of x at x D c is x x!0 (page 94) For x near we have jx Thus p 1Ca p 1Ca , r a/ D a a a) lima!0