INSTRUCTOR SOLUTIONS MANUAL - mttk.no

1 INSTRUCTOR S SOLUTIONS MANUAL to accompany ADAMS / ESSEX CALCULUS: A COMPLETE COURSE; CALCULUS: SINGLE VARIABLE; and CALCULUS: SEVERAL VARIABLES Eighth Edition Prepared by Robert A. Adams University of British Columbia Christopher Essex University of Windsor Ontario Toronto Copyright 2014 Pearson Canada Inc., Toronto, Ontario. Pearson Canada. 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, Eighth Edition, by Adams/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.

2 FOREWORDT hese SOLUTIONS are provided for the benefit of instructors using the textbooks:Calculus: A Complete Course (8th Edition),Single-Variable Calculus (8th Edition),andCalculus of Several Variables (8th Edition)by R. A. Adams and Chris Essex, published by Pearson Education Canada. For the most part,the SOLUTIONS are detailed, especially in exercises on core material and techniques. Occasion-ally some details are omitted for example, in exercises on applications of integration, theevaluation of the integrals encountered is not always given with the same degree of detail asthe evaluation of integrals found in those exercises dealing specifically with techniques of may wish to make these SOLUTIONS available to their students. However, studentsshould use such SOLUTIONS with caution. It is always more beneficial for them to attempt ex-ercises and problems on their own, before they look at SOLUTIONS done by others.

3 If they ex-amine SOLUTIONS as study material prior to attempting the exercises, they can lose much ofthe benefit that follows from diligent attempts to develop their own analytical powers. Whenthey have tried unsuccessfully to solve a problem, then looking at a solution can give them a hint for a second attempt. SeparateStudent SOLUTIONS Manualsfor the books are availablefor students. They contain the SOLUTIONS to the even-numbered exercises , A. 2014 Pearson Canada for Chapter P1 SOLUTIONS for Chapter 123 SOLUTIONS for Chapter 240 SOLUTIONS for Chapter 382 SOLUTIONS for Chapter 4109 SOLUTIONS for Chapter 5179 SOLUTIONS for Chapter 6215 SOLUTIONS for Chapter 7270 SOLUTIONS for Chapter 8318 SOLUTIONS for Chapter 9353 SOLUTIONS for Chapter 10393 SOLUTIONS for Chapter 11421 SOLUTIONS for Chapter 12450 SOLUTIONS for Chapter 13494 SOLUTIONS for Chapter 14541 SOLUTIONS for Chapter 15583 SOLUTIONS for Chapter 16614 SOLUTIONS for Chapter 17641 SOLUTIONS for Chapter 18648 SOLUTIONS for Chapter 18 extended669 SOLUTIONS for Appendices699 NOTE: Chapter 18 extended is only needed by users ofCalculus of Several Variables (Eighth Edition)Copyright 2014 Pearson Canada S SOLUTIONS MANUALSECTION (PAGE 10)CHAPTER P.

4 PRELIMINARIESS ection Real Numbers and the Real Line(page 10) , then 99xD12 andxD12/99D4 , then 10x and100x 32/, or 90xD295. ThusxD295/90D59 the same cyclic order of the repeating digits5 different decimal expansions can represent the samenumber. For instance, both represent the number 0 andx 5 define the interval [0;5]. <2 andx 3 define the interval [ 3;2/. > 5 orx< 6 defines the union. ; 6/ . 5; /. 1 defines the interval. ; 1]. > 2 defines the interval. 2; /. <4 orx 2 defines the interval. ; /, that is, thewhole real 2x>4, thenx< 2. Solution:. ; 2 3xC5 8, then 3x 8 5 3 andx 1. Solution:. ;1] 5x 3 7 3x, then 8x 10 andx 5/4. Solution:. ;5/4] x4 3x 42, then 6 x 6x 8. Thus 14 7xandx 2. Solution:. ;2] x/ < , then 0<5xandx>0. Solution.

5 0; <9, then|x|<3 and 3<x<3. Solution:. 3;3 : 1/.2 x/ < I. Ifx<2, then 1< x/D6 3x, so 3x<5andx<5/3. This case has solutionsx<5 II. Ifx>2, then 1> x/D6 3x, so 3x>5andx>5/3. This case has solutionsx> :. ;5/3/ .2; /. :.xC1//x I. Ifx>0, thenxC1 2x, sox II. Ifx<0, thenxC1 2x, sox 1. (notpossible)Solution:.0;1]. :x2 2x 0. 2/ 0. This is onlypossible ifx 0 andx 2. Solution: [0;2]. 6x2 5x 1, 1/.3x 1/ 0, soeitherx 1/2 andx 1/3, orx 1/3 andx 1 latter combination is not possible. The solution set is[1/3;1/2]. >4x, we 4/ >0. This is possibleifx<0 andx2<4, or ifx>0 andx2>4. Thepossibilities are, therefore, 2<x<0 or 2<x< .Solution:. 2;0/ .2; /. x 2, thenx2 x 2 0 2/.xC1/ is possible ifx 2 andx 1 or ifx 2 andx 1. The latter situation is not possible. The solutionset is [ 1;2]. :x2 I. Ifx>0, thenx2 2xC8, so thatx2 2x 8 0, 4/.

6 XC2/ 0. This ispossible forx>0 only ifx II. Ifx<0, then we must 4/.xC2/ 0,which is possible forx<0 only ifx : [ 2;0/ [4; /. :3x 1< I. Ifx>1 1/.xC1/ >0, so < 1/. Thusx< 5. There are no solutionsin this II. If 1<x<1, 1/.xC1/ <0, > 1/. Thusx> 5. In this case allnumbers in. 1;1/are III. Ifx< 1, 1/.xC1/ >0, so < 1/. Thusx< 5. All numbersx< 5are :. ; 5/ . 1;1/. |x|D3 thenxD |x 3|D7, thenx 3D 7, soxD 4 |2tC5| D4, then 2tC5D 4, sotD 9/2 ortD 1 |1 t|D1, then 1 tD 1, sotD0 2014 Pearson Canada (PAGE 10)ADAMS and ESSEX: CALCULUS |8 3s|D9, then 8 3sD 9, so 3sD 1 or 17, andsD 1/3 orsD17 s2 1 D1, thens2 1D 1, sosD0 |x|<2, thenxis in. 2;2/. |x| 2, thenxis in [ 2;2]. |s 1| 2, then 1 2 s 1C2, sosis in [ 1;3]. |tC2|<1, then 2 1<t< 2C1, sotis in. 3; 1/. |3x 7|<2, then 7 2<3x<7C2, soxis ;3/. |2xC5|<1, then 5 1<2x< 5C1, soxis in.]]

7 3; 2/. x2 1 1, then 1 1 x2 1C1, soxis in [0;4]. 2 x2 <12, thenx/2 lies between 2 .1/2 Thusxis ;5/. inequality|xC1|>|x 3|says that the distancefromxto 1 is greater than the distance fromxto 3, soxmust be to the right of the point half-way between 1and 3. Thusx> |x 3|<2|x| x2 3/2<4x2 3x2C6x 9>0 1/ >0. Thisinequality holds ifx< 3 orx> |a|Daif and only ifa 0. It is false ifa< equation|x 1|D1 xholds if|x 1|D .x 1/,that is, ifx 1<0, or, equivalently, ifx< triangle inequality|xCy| |x|C|y|implies that|x| |xCy| |y|.Apply this inequality withxDa bandyDbto get|a b| |a| |b|.Similarly,|a b|D |b a| |b| |a|. Since |a| |b| is equal to either|a| |b|or|b| |a|, depending on thesizes ofaandb, we have|a b| |a| |b| .Section Cartesian Coordinates in thePlane(page 16) ;3 ;0/,1xD4 0D4 and1yD0 3D 3.|A B|D 42C.

8 3 1;2 ; 10/,1xD4 . 1/D5 and1yD 10 2D 12.|A B|D 52C. 12 ;2/toB. 1; 2/,1xD 1 3D 4 and1yD 2 2D 4.|A B|D . 4/2C. 4/2D4 ;3 ;3/,1xD2 and1yD3 3D0.|A B| point:. 2;3/. Increments1xD4,1yD position is. 2C4;3C. 7//, that is,.2; 4/. point:. 2; 2/. Increments1xD 5, point was. 2 . 5/; 2 1/, that is,.3; 3/. represents a circle of radius 1 centred at represents a circle of radius 2 centred atthe 1 represents points inside and on the circle ofradius 1 centred at the represents the x2represents all points lying on or above <x2represents all points lying below the vertical line through. 2;5/3/isxD 2; the hori-zontal line through that point isyD5 vertical line through. 2; 2; thehorizontal line through that point isyD through. 1;1/with slopemD1 , through. 2;2/with slopemD1/2 , orx 2yD ;b/with slopemD2 ;0/with slopemD 2 isyD0 a/, oryD2a , the height of the line 2xC3yD6 4//3D2/3.

9 ;1/lies above the , the height of the linex 4yD7 7//4D 1. ; 1/lies on the line ;0 ;3/has 0//.2 0/D3/2 and line through. 2;1 ; 2/has slopemD. 2 1//.2C2/D 3/4 and equationyD1 .3/4/.xC2/or 3xC4yD line ;1/and. 2;3/has 1//. 2 4/D 1/3 and equationyD1 4 2014 Pearson Canada S SOLUTIONS MANUALSECTION (PAGE 16) line through. 2;0 ;2/has 0//.0C2/D1 and 2 andbD 2, then the line has equationyD 2xC 1/2 andbD 3, then the line has equationyD .1/2/x 3, orxC2yD hasx-interceptaD12/3D4 andy-interceptbD12/4D3. Its slope is b/aD 3 4 hasx-interceptaD 4 andy-interceptbD 4/2D 2. Its slope is b/aD2/. 4/D 1 4 Fig. 2x 3yD2 hasx-interceptaD2/ 2D 2andy-interceptbD 2/ 3. Its slope is b/aD2/ 6D 2 2x 3yD2 Fig. 2yD 3 hasx-interceptaD 3 2 andy-interceptbD 3/. 2/D3/2. Its slope is b/aD3 2yD 3 Fig. ;1/parallel toyDxC2 isyDx 1; lineperpendicular toyDxC2 isyD through.

10 2;2/parallel to 2xCyD4 is2xCyD 2; line perpendicular to 2xCyD4 isx 2yD have3xC4yD 62x 3yD13H 6xC8yD 126x these equations gives 17yD 51, soyD 9//2D2. The intersection point ; 3/. have2xCyD85x 7yD1H 14xC7yD565x these equations gives 19xD57, soxD3 andyD8 2xD2. The intersection point ;2/. andb6D0, representsa straight line that is neither horizontal nor vertical, anddoes not pass through the origin. PuttingyD0 we getx/aD1, so thex-intercept of this line isxDa; puttingxD0 givesy/bD1, so they-intercept .y/3/D1 hasx-interceptaD2, andy-interceptbD 3x2 y3D12 Fig. line ;1 ; 1/has slopemD. 1 1//.3 2/D 2 and equationyD1 2/D5 2x. Itsy-intercept is 2014 Pearson Canada (PAGE 16)ADAMS and ESSEX: CALCULUS line through. 2;5 ;1/hasx-intercept 3, soalso passes ;0/. Its slopemsatisfies1 0k 3 DmD0 53C2D 3D 1, and IfCD5;000 whenxD10;000 andCD6;000 whenxD15;000, then10;000 ACBD5;00015;000 ACBD6;000 Subtracting these equations gives 5;000AD1;000, soAD1/5.