7. INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.1 (PAGE 100)SECTION 2.1 (PAGE 100) ADAMS and ESSEX: CALCULUS 9 42 42 42 42 -20 -30 Fig. 2.1-26 .2; 19/ -3 Fig. 2.1-29 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/.
15. 50 50 50 50 INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.2 (PAGE 107)SECTION 2.2 (PAGE 107) ADAMS and ESSEX: CALCULUS 9 C C 3 D ˇ ˇ D lim p p p p f .x/ f .2/ f .x/ f .2/ h!0 h 1 C x C h 1 C x 1 C x C 1 C x C h x x 1 x 2 x 2 D lim p p p p 1:9 0:26316 2:1 0:23810 h!0 1 C x C h 1 1 C x 1 C x C 1 C x C h 1:99 0:25126 1:999 0:25013 2:01 0:24876 2:001 0:24988 D 2.1 C x/3=2 1 1:9999 0:25001 2:0001 0:24999 dy D 2.1 dx C x/3=2 f 0 .2/ D lim h!0 1 2 h 2 h D lim h!0 2 .2 C h/ h.2 C h/2 24. t2 3 f.t/ D t2 1 .t C h/2 3 t2 3 D lim h!0 1 .2 C h/2 1 D 4 f 0 .t/ lim h!0 h .t C h/2 C 3 t2 C 3 30. The slope of y D 5 C 4x x2 at x D 2 is D lim h!0 Œ.t C h/2 3.t2 C 3/ .t2 3/Œ.t C h/2 C 3 h.t2 C 3/Œ.t C h/2 C 3 dy ˇ dx ˇ D lim h!0 5 C 4.2 C h/ .2 C h/2 9 h D lim 12th C 6h2 D 12t ˇ xD2 h2 h!0 h.t2 C 3/Œ.t C h/2 C 3 .t2 C 3/2 D lim 0: df .t/ D .t2 12t 2 d t h!0 h D C 3/ Thus, the tangent line at x D 2 has the equation y D 9.
16. 51 51 51 51 INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.2 (PAGE 107)SECTION 2.2 (PAGE 107) ADAMS and ESSEX: CALCULUS 9 ˇ p ˇ ˇ 4 4 D : 2 31. y D p x C 6. Slope at .3; 3/ is p 9 C h 3 9 C h 9 1 44. The slope of y D p x at x D x0 is ˇ m D lim h!0 lim h h!0 h 1 p 9 C h C 3 D 6 : dy ˇ dx ˇ ˇ xDx0 1 D 2 p x0 Tangent line is y 3 D 6 .x 3/, or x 6y D 15. Thus, the equation of the tangent line is 32. The slope of y t D t2 2 at t D 2 and y D 1 is y D x0 C 1 2 p x0 .x x0 /, that is, y D x C x0 . 2 p x0 dy ˇ ˇ D lim 1 2 C h . 1/ dt ˇ ˇ tD 2 h!0 h . 2 C h/2 2 ˇ 2 C h C Œ. 2 C h/2 2 3 45. Slope of y 1 1 ˇ 1 D lim h!0 hŒ. 2 C h/2 2 D 2 : D x at x D a is x2 ˇ ˇ xDa 2 D a2 . 1 2 Thus, the tangent line has the equation Normal has slope a , and equation y a D a .x a/, y D 1 3 .t C 2/, that is, y D 3 t 4. 2 3 1 2 33. y D t2 C t 2 Slope at t D a is 2 or y D a x a C a 2 m D lim h!0 D lim .a C h/2 C .a C h/ a2 C a h 2.a2 C a a2 2ah h2 a h/ 46. The intersection points of y D x2 and x C 4y D 18 satisfy 4x2 C x 18 D 0 h!0 D lim hŒ.a C h/2 C a C h.a2 C a/ 4a 2h 2 .4x C 9/.x 2/ D 0: h!0 Œ.a C h/2 C a C h.a2 C a/ 4a C 2 D .a2 C a/2 Therefore x D 9 or x D 2. dy The slope of y D x2 is m1 D D 2x. 2 Tangent line is y D a2 2.2a C 1/ dx .t a/ 9 9 C a 34. f 0.x/ D 17x 18 for x ¤ 0 .a2 C a/2 At x D 4 , m1 D 2 . At x D 2, m1 D 4. The slope of x C 4y D 18, i.e. y D 1 x C m2 D 1 .
28. INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.3 (PAGE 115)SECTION 2.3 (PAGE 115) ADAMS and ESSEX: CALCULUS 9 54 54 Copyright © 2018 Pearson Canada Inc. 54 54 Copyright © 2018 Pearson Canada Inc. ˇ 2 a D .x C 2/2 D .x 2/2 : corres pond to a D 2 and a D 1, so their equati ons are y D 12x 16 and y D 3x C 2. In order to be parallel to y D 4x, the tangent line must have slope equal to 4, i.e., 50. The tangent to y D x2 =.x 1/ at .a; a2=.a 1// has slope .x 1/2x x2 .1/ ˇ a2 2a 2 : 1 .x C 2/2 D 4; or .x C 2/2 D 1 : m .x 1/2 ˇ ˇ ˇ xDa D .a 1/ Hence x C 2 D ˙1 , and x D 3 or 5 . At x D 3 , The equation of the tangent is 2 2 2 2 y D 1, and at x D 5 , y D 3. a2 y D a 1 a2 2a .a 1/2 .x a/: Hence, the tangent is parallel to y D 4x at the points 3 5 This line passes through .2; 0/ provided 2 ; 1 and 2 ; 3 . a2 a2 2a 47. Let the point of tangency be .a; 1 /. The slope of the tan- 0 D .2 a/; a a 1 .a 1/2 1 b 1 2 a 1 1 gent is a2 D 0 . Thus b a D a and a D b . or, upon simplification, 3a2 4a D 0. Thus we can have b2 b2 Tangent has slope 4 so has equation y D b 4 x. 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.
36. 62 62 Copyright © 2018 Pearson Canada Inc. 62 62 Copyright © 2018 Pearson Canada Inc. INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.5 (PAGE 126)SECTION 2.5 (PAGE 126) ADAMS and ESSEX: CALCULUS 9 2t sint 33. v D sec.x2 / tan.x2 / 2 2 2 3 2 f 0 .x/ D .1 C cos x/ sin.x C sin x/ v0 D 2x sec.x / tan .x / C 2x sec .x / sin p x 16. g. / D tan. sin / g0 . / D .sin C cos / sec2 . sin / 34. z D 1 C cos p x p p p p p 3 z0 D .1 C cos p x/.cos x=2 x/ .sin x/. sin x=2 x/ p 17. u D sin3 . x=2/; u0 D cos. x=2/ sin2 . x=2/ .1 C cos x/2 2 1 C cos p x 1 18. y D sec.1=x/; y 0 D .1=x2 / sec.1=x/ tan.1=x/ 1 D 2 p x.1 C cos p x/2 D 2 p x.1 C cos p x/ 19. F .t/ D sin at cos at .D 2 sin 2at/ 35. d dt sin.cos.tan t// D .sec2 t/.sin.tan t// cos.cos.tan t// F 0 .t/ D a cos at cos at a sin at sin at 36. f.s/ D cos.s C cos.s C cos s// . D a cos 2at/ sin a f 0 .s/ D Œsin.s C cos.s C cos s// 20. G. / D cos b a cos b cos a C b sin a sin b Œ1 .sin.s C cos s//.1 sin s/ 37. Differentiate both sides of sin.2x/ D 2 sin x cos x and G0 . / D cos2 b : divide by 2 to get cos.2x/ D cos2 x sin2 x. 2 2 21. d sin.2x/ cos.2x/ 2 cos.2x/ 2 sin.2x/ 38. Differentiate both sides of cos.2x/ D cos x sin x and dx 22. d .cos2 x dx sin2 x/ D d D dx C cos.2x/ 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 D 2 sin.2x/ D 4 sin x cos x equations y D .x / and y D x , respectively.
37. 63 63 Copyright © 2018 Pearson Canada Inc. 63 63 Copyright © 2018 Pearson Canada Inc. INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.5 (PAGE 126)SECTION 2.5 (PAGE 126) ADAMS and ESSEX: CALCULUS 9 D D D D D D ˇ dx 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. 41. The slope of y p 2 cos.x=4/ at . ; 1/ is p . 2=4/ sin. =4/ D 1=4. Therefore the tangent and 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 normal lines to y D p 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 cos2 x at . =3; 1=4/ is p 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 sin.2 =3/ D 3=2. Therefore the tangent and normal x D 5 =6 because dy=dx D 1 2 sin x D 0 at those lines to y tan.2x/ at .0; 0/ have equations p points. y D .1=4/ . 3=2/.x . =3// and y D .1=4/ C .2= p 3/.x . =3//, respectively. 53. lim tan.2x/ sin.2x/ 2 lim 1 2 2 x x x!0 x D x!0 2x cos.2x/ D D 43. Slope of y sin.xı / sin is y0 180 At x D 45 the tangent line has equation 180 cos . 180 54. lim sec.1 x! C cos x/ D sec.1 1/ D sec 0 D 1 1 2 x 2 2 y D p 2 C 180 p 2 .x 45/. x 55. lim x x!0 csc x cot x D lim x!0 2 sin x cos x D 1 1 D 1 2 44. For y D sec .xı/ D sec we have 56. lim cos cos x lim cos sin x cos 1 180 x!0 x2 D x!0 2 x D D 2 dy x tan x : 57. lim 1 cos h 2 sin .h=2/ 1 sin.h=2/ 1 dx D 180 sec 180 180 h 0 h2 D lim h2 D lim 2 h=2 D 2 ! h!0 h!0 At x D 60 the slope is 180 .2 p 3/ D p 3 : 90 58. f will be differentiable at x D 0 if 90 2 sin 0 C 3 cos 0 D b; and Thus, the normal line has slope p 3 and has equation 90 y D 2 p 3 .x 60/. d ˇ .2 sinx C 3 cos x/ ˇ ˇ xD0 D a: 45. The slope of y D tan x at x D a is sec2 a. The tan- gent there is parallel to y D 2x if sec2 a D 2, or cos a D ˙1= p 2. The only solutions in . =2; =2/ are a D ˙ =4. The corresponding points on the graph are . =4; 1/ and . =4; 1/.
40. 66 66 Copyright © 2018 Pearson Canada Inc. 66 66 Copyright © 2018 Pearson Canada Inc. INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.6 (PAGE 131)SECTION 2.6 (PAGE 131) ADAMS and ESSEX: CALCULUS 9 6 y 00 D y 000 D .2 x2 / sin x x 3 .6 x2 / cos x x 3 C 2 c o s x x 2 3 . x 2 2 / s i n x x 4 y 0 D 2x C x2 y 000 D x4 1 1 6 2 13. f.x/ D x D x f 0 .x/ D x 2 3. y D .x 1/2 D 6.x 1/ f 00 .x/ D 2x 3 y 0 D 12.x 1/ 3 y 00 D 36.x 1/ 4 y 000 D 144.x 1/ 5 f 000 .x/ D 3Šx 4 f .4/ .x/ D 4Šx 5 Guess: f .n/.x/ . 1/n .nC1/ D nŠx . / 4. y D p ax C b a y00 D a2 4.ax C b/3=2 Proof: (*) is valid for n D 1 (and 2, 3, 4). Assume f .k/ .x/ D . 1/k kŠx .kC1/ for some k 1 Then f .kC1/.x/ D . 1/kkŠ .k C 1/ x .kC1/ 1 y0 D p 3a3 . 1/kC1 .k 1/Šx ..kC1/C1/ which is (*) for n k 1. 2 ax C b y 000 D 8.ax D C D C b/5=2 C Therefore, (*) holds for n D 1; 2; 3; : : : by induction.
43. 69 69 Copyright © 2018 Pearson Canada Inc. 69 69 Copyright © 2018 Pearson Canada Inc. INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.6 (PAGE 131)SECTION 2.6 (PAGE 131) ADAMS and ESSEX: CALCULUS 9 1 3 2 Differentiating any of these four formulas produces the one for the next higher value of n, so induction confirms the The pattern suggests that .nC1/ overall formula. f .n/ .x/ D nŠjxj sgn x if n is odd 20. f.x/ D x cos x f 0 .x/ D cos x x sinx f 00 .x/ D 2 sin x x cos x f 000 .x/ D 3 cos x C x sin x f .4/ .x/ D 4 sin x C x cosx nŠjxj .nC1/ if n is even 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. 23. f.x/ D p 1 3x D .1 3x/1=2 1 0 1=2 This suggests the formula (for k D 0; 1; 2; : : :) f .x/ D 2 . 3/.1 3x/ 1 1 00 2 3=2 8 n sin x C x cos x if n D 4k ˆ< n cos x x sin x if n D 4k C 1 f .x/ D 2 2 . 3/ .1 3x/ f .n/ .x/ D n sin x x cos x if n D 4k C 2 f 000 .x/ D 1 1 2 2 . 3/3 .1 3x/ 5=2 n cos x C x sin x if n D 4k C 3 Differentiating any of these four formulas produces the one f .4/ .x/ D 1 1 2 2 3 5 2 2 . 3/4 .1 3x/7=2 for the next higher value of n, so induction confirms the overall formula. Guess: f .n/.x/ D 1 3 5 .2n 3/ n 2n 3 21. f.x/ D x sin.ax/ .1 3x/ .2n 1/=2 . / 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 f 0 .x/ D sin.ax/ C ax cos.ax/ 1 3 5 : : : .2k 3/ f 00 .x/ D 2a cos.ax/ a2 x sin.ax/ f 000 .x/ D 3a2 sin.ax/ a3 x cos.ax/ i.e., f .k/.x/ D 3k 2k .1 3x/ .2k 1/=2 f 4/ .x/ D 4a3 cos.ax/ C a4 x sin.ax/ This suggests the formula Then f .kC1/ .x/ D 1 3 5 .2k 3/ 3k 2k 8 nan 1 cos.ax/ C an x sin.ax/ if n D 4k 2.k 1/ 2 .1 3x/ .2k 1/=2 1 . 3/ ˆ n 1 n 1 3 5 2.k 1/ 1 f .n/ .x/ D < na sin.ax/ C a x cos.ax/ if n D 4k C 1 C kC1 nan 1 cos.ax/ an x sin.ax/ if n D 4k C 2 ˆ D 2kC 3 : nan 1 sin.ax/ anx cos.ax/ 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.
46. INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.6 (PAGE 131)SECTION 2.6 (PAGE 131) ADAMS and ESSEX: CALCULUS 9 61 61 . 61 61 .fg/.4/ D .fg/.3/ dx d .3/ 00 0 0 00 .3/ 6%. 12. If V is the volume and x is the edge length of the cube D dx Œf g C 3f g C 3f g C fg  then V D x3. Thus V d V D 3x2 x. If D f .4/ g C f .3/ g0 C 3f .3/ g0 C 3f 00 g00 C 3f 00 g00 C 3f 0 g.3/ C f 0 g.3/ C fg.4/ D f .4/ g C 4f .3/ g0 C 6f 00 g00 C 4f 0 g.3/ C fg.4/ : nŠ V D .6=100/V , then 6x3 =100 3x2 dx, so dx .2=100/x. The edge of the cube decreases by about 2%. 13. Rate change of Area A with respect to side s, where .fg/.n/ D f .n/ g C nf .n 1/g0 C nŠ 2Š.n 2/Š f .n 2/ g00 A D s2 , is dA ds D 2s: When s D 4 ft, the area is changing C 3Š.n 3/Š f .n 3/ g.3/ C C nf 0 g.n 1/ C fg.n/ at rate 8 ft2 /ft. p p n nŠ .n k/ .k/ 14. If A D s2 , then s D 2 A and ds=dA D 1=.2 A/. D X kŠ.n kD0 f g : k/Š If A D 16 m , then the side is changing at rate ds=dA D 1=8 m/m2 .
48. 63 63 . 63 63 INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.7 (PAGE 137)SECTION 2.7 (PAGE 137) ADAMS and ESSEX: CALCULUS 9 ˇ ˇ ˇ D (a) n D 100; d n D 2. Thus, the marginal cost of d V ˇ ˇ production is $2. ˇ dt ˇ ˇ tD15 D 700.20 t/ ˇ ˇ tD15 D 3; 500: (b) dC 82 n D 300; d n D 9 9:11. Thus, the marginal cost Water is draining out at 3,500 L/min at that time. b) Average rate of change between t D 5 and t D 15 is of production is approximately $9.11. 28. x2 Daily profit P D 13x Cx D 13x 10x 20 1000 x2 V.15/ V.5/ 350 .25 225/ 7; 000: D 3x 20 15 5 D 10 D 1000 Graph of P is a parabola opening downward. P will be The average rate of draining is 7,000 L/min over that maximum where the slope is zero: interval. 22. Flow rate F D kr4 , so F 4kr3 r. If F D F =10, then dP 0 dx D 3 2x 1000 so x D 1500 F kr4 r 40k r3 D 40k r3 D 0:025r: Should extract 1500 tonnes of ore per day to maximize profit.
50. 65 65 . 65 65 INSTRUCTOR'S SOLUTIONS MANUAL SECTION 2.7 (PAGE 137)SECTION 2.7 (PAGE 137) ADAMS and ESSEX: CALCULUS 9 1 f 0 .c/ 8. If f .x/ D x3 12x C 1, then f 0.x/ D 3.x2 4/. 2 1 D 2 D 2 D c2 D 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 where c D p 2 lies between 1 and 2. on . 2; 2/ where f 0 3. f.x/ D x3 3x C 1, f 0.x/ D 3x2 3, a D 2, b D 2 9. If f.x/ D x2 .x/ < 0. 4, then f 0.x/ D 2x. The critical point of f .b/ f .a/ b a D D f .2/ f . 2/ 4 8 6 C 1 . 8 C 6 C 1/ 4 4 f is x D 0. f is increasing on .0; 1/ and decreasing on . 1; 0/. 10. If y D 1 x x5 , then y 0 D 1 5x4 < 0 for all x. Thus y has no critical points and is decreasing on the D 4 D 1 f 0.c/ D 3c2 3 2 whole real line. 11. If y D x3 C 6x2 , then y 0 D 3x2 C 12x D 3x .x C 4/. The critical points of y are x D 0 and x D 4. y is increasing 3c2 3 D 1 ) 3c2 D 4 ) c D ˙p 3 (Both points will be in . 2; 2/.) on . 1; 4/ and .0; 1/ where y 0 > 0, and is decreasing on . 4; 0/ where y 0 < 0.
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