Calculus Early Transcendentals 11th Edition Anton Solutions Manual

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Topics in Differentiation Exercise Set 3.1 1. (a) 1 + y + x

(b) y =

dy 6x2 − y − 1 dy − 6x2 = 0, = . dx dx x

2 + 2x3 − x 2 dy 2 = + 2x2 − 1, = − 2 + 4x. x x dx x

dy 1 1 1 1 (c) From part (a), = 6x − − y = 6x − − dx x x x x 2. (a)

2 + 2x2 − 1 x

= 4x −

2 . x2

1 −1/2 dy dy √ y − cos x = 0 or = 2 y cos x.

2 dx dx

(b) y = (2 + sin x)2 = 4 + 4 sin x + sin2 x so

(c) From part (a),

3. 2x + 2y

dy = 4 cos x + 2 sin x cos x. dx

dy √ = 2 y cos x = 2 cos x(2 + sin x) = 4 cos x + 2 sin x cos x. dx

dy dy x = 0 so =− . dx dx y

4. 3x2 + 3y 2

5. x2

dy dy dy 3y 2 − 3x2 y 2 − x2 = 3y 2 + 6xy , = 2 = 2 . dx dx dx 3y − 6xy y − 2xy

dy dy dy dy 1 − 2xy − 3y 3 + 2xy + 3x(3y 2 ) + 3y 3 − 1 = 0, (x2 + 9xy 2 ) = 1 − 2xy − 3y 3 , so = . dx dx dx dx x2 + 9xy 2

6. x3 (2y)

dy dy dy 10xy − 3x2 y 2 − 1 dy + 3x2 y 2 − 5x2 − 10xy + 1 = 0, (2x3 y − 5x2 ) = 10xy − 3x2 y 2 − 1, so = . dx dx dx dx 2x3 y − 5x2 dy

7. −

1 dy y 3/2 − dx = 0, so = − 3/2 .

3/2 3/2 dx 2x 2y x

8. 2x =

(x − y)(1 + dy/dx) − (x + y)(1 − dy/dx) dy dy x(x − y)2 + y 2 , 2x(x − y) = −2y + 2x , so = . (x − y)2 dx dx x

dy dy 1 − 2xy 2 cos(x2 y 2 ) 2 2 9. cos(x y ) x (2y) + 2xy = 1, so = . dx dx 2x2 y cos(x2 y 2 ) 2 2

dy dy dy y 2 sin(xy 2 ) 2 10. − sin(xy ) y + 2xy = , so =− . dx dx dx 2xy sin(xy 2 ) + 1 2

81

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82

Chapter 3

dy dy dy 1 − 3y 2 tan2 (xy 2 + y) sec2 (xy 2 + y) 11. 3 tan2 (xy 2 + y) sec2 (xy 2 + y) 2xy . + y2 + = 1, so = dx dx dx 3(2xy + 1) tan2 (xy 2 + y) sec2 (xy 2 + y) 12.

(1 + sec y)[3xy 2 (dy/dx) + y 3 ] − xy 3 (sec y tan y)(dy/dx) dy = 4y 3 , multiply through by (1 + sec y)2 and solve for 2 (1 + sec y) dx dy dy y(1 + sec y) to get = . dx dx 4y(1 + sec y)2 − 3x(1 + sec y) + xy sec y tan y

13. 4x − 6y 14.

15.

dy 2x dy = 0, = , 4−6 dx dx 3y

dy dx

2

2

2

− 6y

d y d y = 0, so =− dx2 dx2

3

dy dx

2

3y

−2

=

8 2(3y 2 − 2x2 ) = − 3.

9y 3 9y

dy y 2 (2x) − x2 (2ydy/dx) 2xy 2 − 2x2 y(−x2 /y 2 ) 2x(y 3 + x3 ) x2 d2 y = − = − = − , but x3 + y 3 = 1, so = − 2, 2 4 4 dx y dx y y y5 d2 y 2x =− 5. dx2 y dy y d2 y x(dy/dx) − y(1) x(−y/x) − y 2y =− , =− =− = 2.

2 2 2 dx x dx x x x

16. y + x

dy dy dy y dy d2 y + 2y = 0, =− ,2 +x 2 +2 dx dx dx x + 2y dx dx

dy dx

2 + 2y

d2 y 2y(x + y) d2 y = 0, = .

2 dx dx2 (x + 2y)3

17.

dy d2 y dy sin y = (1 + cos y)−1 , = −(1 + cos y)−2 (− sin y) = .

2 dx dx dx (1 + cos y)3

18.

cos y d2 y (1 + x sin y)(− sin y)(dy/dx) − (cos y)[(x cos y)(dy/dx) + sin y] dy = , = = dx 1 + x sin y dx2 (1 + x sin y)2 2 sin y cos y + (x cos y)(2 sin2 y + cos2 y) , but x cos y = y, 2 sin y cos y = sin 2y, and sin2 y + cos2 y = 1, so (1 + x sin y)3 d2 y sin 2y + y(sin2 y + 1) =− .

2 dx (1 + x sin y)3

−

√ √ √ x dy dy = − ; at (1/2, 3/2), = − 3/3; at (1/2, − 3/2), dx y dx √ √ √ dy −x 1/2 dy = + 3/3. Directly, at the upper point y = 1 − x2 , = √ = −p = −1/ 3 and at the lower dx dx 1 − x2 3/4 √ √ dy x 2 point y = − 1 − x , =√ = +1/ 3. dx 1 − x2

19. By implicit differentiation, 2x + 2y(dy/dx) = 0,

√ √ 20. If y 2 − x + 1 = 0, then y = x − 1 goes through the point (10, 3) so dy/dx √ = 1/(2 x − 1). By implicit differentiation dy/dx = 1/(2y). In both cases, dy/dx|(10,3) = 1/6. Similarly y = − x − 1 goes through (10, −3) √ so dy/dx = −1/(2 x − 1) = −1/6 which yields dy/dx = 1/(2y) = −1/6. √ 21. False; x = y 2 defines two functions y = ± x. See Definition 3.1.1.

22. True.

23. False; the equation is equivalent to x2 = y 2 which is satisfied by y = |x|.

24. True.

25. 4x3 + 4y 3

dy dy x3 1 = 0, so = − 3 = − 3/4 ≈ −0.1312. dx dx y 15

Exercise Set 3.1 26. 3y 2

83

dy dy y+1 dy dy + x2 + 2xy + 2x − 6y = 0, so = −2x 2 = 0 at x = 0. dx dx dx dx 3y + x2 − 6y

dy dy dy x[25 − 4(x2 + y 2 )] dy 27. 4(x2 + y 2 ) 2x + 2y = 25 2x − 2y , = ; at (3, 1) = −9/13. dx dx dx y[25 + 4(x2 + y 2 )] dx 2 28.

3

29. 4a3

x

−1/3

+y

−1/3 dy

= 0,

dx

√ √ dy y 1/3 = − 1/3 = 3 at (−1, 3 3). dx x

da da da da 2t3 + 3a2 − 4t3 = 6 a2 + 2at , solve for to get = 3 . dt dt dt dt 2a − 6at

√ 1 −1/2 du 1 −1/2 u du 30. u + v = −√ . = 0, so 2 dv 2 dv v 31. 2a2 ω

b2 λ dω dω + 2b2 λ = 0, so =− 2 . dλ dλ a ω

32. 1 = (cos x)

dx dx 1 , so = = sec x. dy dy cos x

dy dy dy 33. 2x + x + y + 2y = 0. Substitute y = −2x to obtain −3x = 0. Since x = ±1 at the indicated points, dx dx dx dy = 0 there. dx 34. (a) The equation and the point (1, 1) are both symmetric in x and y (if you interchange the two variables you get the same equation and the same point). Therefore the outcome ”horizontal tangent at (1, 1)” could be replaced by ”vertical tangent at (1, 1)”, and these cannot both be the case. dy 2x − y = , which is zero only if y = 2x; coupled with the equation x2 − xy + dx x − 2y √ √ √ √ y 2 = 1 we obtain x2 − 2x2 + 4x2 = 1, or 3x2 = 1, x = ( 3/3, 2 3/3) and (− 3/3, −2 3/3). (b) Implicit differentiation yields

y

2 x

–4

4 –2

35. (a) (b) Implicit differentiation of the curve yields (4y 3 + 2y) so x = 1/2 is impossible.

2

4

2

dy dy = 2x − 1, so = 0 only if x = 1/2 but y 4 + y 2 ≥ 0 dx dx

(c) x − x − (y + y ) = 0, so by the Quadratic Formula, x =

the two parabolas x = −y 2 , x = 1 + y 2 .

−1 ±

p (2y 2 + 1)2 = 1 + y 2 or −y 2 , and we have 2

dy dx 2y(2y 2 + 1) = 2x − 1, = = 0 only if 2y(2y 2 + 1) = 0, which can only dx dy 2x − 1 hold if y = 0. From y 4 + y 2 = x(x − 1), if y = 0 then x = 0 or 1, and so (0, 0) and (1, 0) are the two points where the tangent is vertical.

36. By implicit differentiation, 2y(2y 2 + 1)

84

Chapter 3

37. The point (1,1) is on the graph, so 1 + a = b. The slope of the tangent line at (1,1) is −4/3; use implicit dy 2xy 2 4 differentiation to get =− 2 so at (1,1), − = − , 1 + 2a = 3/2, a = 1/4 and hence b = 1 + 1/4 = dx x + 2ay 1 + 2a 3 5/4.

38. The slope of the line x + 2y − 2 = 0 is m1 = −1/2, so the line perpendicular has slope m = 2 (negative reciprocal). dy dy 4x . Set The slope of the curve y 3 = 2x2 can be obtained by implicit differentiation: 3y 2 = 4x, = dx dx 3y 2 dy 4x = 2; 2 = 2, x = (3/2)y 2 . Use this in the equation of the curve: y 3 = 2x2 = 2((3/2)y 2 )2 = (9/2)y 4 , y = dx 3y 2 3 2 2 2/9, x = = .

2 9 27 39. We shall find when the curves intersect and check that the slopes are negative reciprocals. For the intersection 1 solve the simultaneous equations x2 + (y − c)2 = c2 and (x − k)2 + y 2 = k 2 to obtain cy = kx = (x2 + y 2 ). Thus 2 y−c x−k x2 + y 2 = cy + kx, or y 2 − cy = −x2 + kx, and =− . Differentiating the two families yields (black) x y x dy x−k dy =− , and (gray) =− . But it was proven that these quantities are negative reciprocals of each dx y−c dx y other. dy dy 40. Differentiating, we get the equations (black) x + y = 0 and (gray) 2x − 2y = 0. The first says the (black) dx dx y x slope is − and the second says the (gray) slope is , and these are negative reciprocals of each other. x y y

2 x

–3

–1

2 –1

–3

41. (a) (b) x ≈ 0.84

(c) Use implicit differentiation to get dy/dx = (2y − 3x2 )/(3y 2 − 2x), so dy/dx = 0 if y = (3/2)x2 . Substitute this into x3 − 2xy + y 3 = 0 to obtain 27x6 − 16x3 = 0, x3 = 16/27, x = 24/3 /3 and hence y = 25/3 /3. y

2 x

–3

–1

2 –1

42. (a)

–3

(b) Evidently (by symmetry) the tangent line at the point x = 1, y = 1 has slope −1. (c) Use implicit differentiation to get dy/dx = (2y − 3x2 )/(3y 2 − 2x), so dy/dx = −1 if 2y − 3x2 = −3y 2 + 2x, 2(y − x) + 3(y − x)(y + x) = 0. One solution is y = x; this together with x3 + y 3 = 2xy yields x = y = 1.

Exercise Set 3.2

85

For these values dy/dx = −1, so that (1, 1) is a solution. To prove that there is no other solution, suppose y 6= x.

From dy/dx = −1 it follows that 2(y − x) + 3(y − x)(y + x) = 0. But y 6= x, so x + y = −2/3, which is not true for any point in the first quadrant. dy dt dy 2y 3 + 3t2 y dy , but = . Using implicit differentiation for 2y 3 t + t3 y = 1 we get =− dx dt dx dt 6ty 2 + t3 3 2 dt 1 dy 2y + 3t y = , so =− . dx cos t dx (6ty 2 + t3 ) cos t

43. By the chain rule,

44. Let P (x0 , y0 ) be a point where a line through the origin is tangent to the curve 2x2 − 4x + y 2 + 1 = 0. Implicit differentiation applied to the equation of the curve gives dy/dx = (2−2x)/y. At P the slope of the curve must equal the slope of the line so (2 − 2x0 )/y0 = y0 /x0 , or y02 = 2x0 (1 − x0 ). But 2x20 − 4x0 + y02 + 1 = 0 because (x0 , y0 ) is on the curve, and elimination of y02 in the latter two equations gives 2x0 = 4x0 − 1, x0 = 1/2√which when substituted √ √ 2 2/2.

The slopes of the lines are (± 2/2)/(1/2) = ± 2 and into y02 = 2x0 (1 − x0 ) yields y = 1/2, so y = ± 0 0 √ √ their equations are y = 2x and y = − 2x.

Exercise Set 3.2 1.

1 1 (5) = .

5x x

2.

1 1 1 = . x/3 3 x

3.

1 .

1+x

4.

1 √ 2+ x

5.

2x 1 (2x) = 2 . x2 − 1 x −1

6.

3x2 − 14x . x3 − 7x2 − 3

7.

d d 1 2x 1 − x2 ln x − ln(1 + x2 ) = − = . dx dx x 1 + x2 x(1 + x2 )

8.

d 1 −1 2 (ln |1 + x| − ln |1 − x|) = − = . dx 1+x 1−x 1 − x2

9.

d d 2 (2 ln x) = 2 ln x = . dx dx x 2

10. 3 (ln x)

1 √ 2 x

1 . x

1 1 = √ . x 2x ln x

11.

1 (ln x)−1/2 2

12.

d 1 1 ln x = . dx 2 2x

13. ln x + x

1 √ . = √ 2 x(2 + x)

1 = 1 + ln x. x

86

Chapter 3

14. x3

1 + (3x2 ) ln x = x2 (1 + 3 ln x). x

15. 2x log2 (3 − 2x) +

−2x2 . (ln 2)(3 − 2x)

3

2 16. log2 (x2 − 2x) + 3x log2 (x2 − 2x) 17.

2x − 2 . (x2 − 2x) ln 2

2x(1 + log x) − x/(ln 10) .

(1 + log x)2

18. 1/[x(ln 10)(1 + log x)2 ].

1 1 = . x x ln x

19.

1 ln x

20.

1 1 1 . ln(ln(x)) ln x x

21.

1 (sec2 x) = sec x csc x. tan x

22.

1 (− sin x) = − tan x. cos x

1 23. − sin(ln x) . x 24. 2 sin(ln x) cos(ln x)

1 sin(2 ln x) sin(ln x2 ) = = . x x x

25.

1 cot x .

(2 sin x cos x) = 2 ln 10 ln 10 sin2 x

26.

1 d 1 −2 sin x cos x 2 tan x ln cos2 x = =− .

2 ln 10 dx ln 10 cos x ln 10

27.

3 d 8x 11x2 − 8x + 3 3 ln(x − 1) + 4 ln(x2 + 1) = + 2 = . dx x−1 x +1 (x − 1)(x2 + 1)

d 1 2x3 [2 ln cos x + ln(1 + x4 )] = −2 tan x + . dx 2 1 + x4

d 1 3x 29. ln cos x − ln(4 − 3x2 ) = − tan x + dx 2 4 − 3x2

28.

d 30. dx

1 [ln(x − 1) − ln(x + 1)] 2

1 = 2

1 1 − . x−1 x+1

dy 1 = , so as x = a → 0+ , the slope approaches infinity. dx x √ 32. False, e.g. f (x) = x.

31. True, because

33. True; if x > 0 then

d d ln |x| = 1/x; if x < 0 then ln |x| = 1/x. dx dx

Exercise Set 3.2

34. False;

87

2 d 1 (ln x)2 = 2 ln x 6= . dx x x

p 1 1 2x dy 3 2 2 35. ln |y| = ln |x| + ln |1 + x |, so =x 1+x + .

3 dx x 3(1 + x2 ) dy 1 1 = 36. ln |y| = [ln |x − 1| − ln |x + 1|], so 5 dx 5

r 5

1 1 x−1 − . x+1 x−1 x+1

1 1 ln |x2 − 8| + ln |x3 + 1| − ln |x6 − 7x + 5|, so 3 2 √

2 1/3 3 (x − 8) 2x 3x2 6x5 − 7 x +1 dy = + − . dx x6 − 7x + 5 3(x2 − 8) 2(x3 + 1) x6 − 7x + 5

37. ln |y| =

38. ln |y| = ln | sin x| + ln | cos x| + 3 ln | tan x| − 39. (a) logx e =

(b) logx 2 =

1 d 1 ln e = , so [logx e] = − . ln x ln x dx x(ln x)2 ln 2 d ln 2 , so [logx 2] = − . ln x dx x(ln x)2

40. (a) From loga b =

(b) log(ln x) e = 41. f 0 (x0 ) =

1 dy sin x cos x tan3 x 3 sec2 x 1 √ ln |x|, so = cot x − tan x + − 2 dx tan x 2x x

i ln e 1 d h ln b 1 for a, b > 0 it follows that log(1/x) e = =− , so . log(1/x) e = ln a ln(1/x) ln x dx x(ln x)2

ln e 1 d 1 1 1 = , so log(ln x) e = − =− .

2 ln(ln x) ln(ln x) dx (ln(ln x)) x ln x x(ln x)(ln(ln x))2

1 = e, y − (−1) = e(x − x0 ) = ex − 1, y = ex − 2. x0

42. y = log x =

1 1 ln x , y0 = , y0 = log 10 = 1, y − 1 = (x − 10). ln 10 x ln 10 10 ln 10

1 1 1 43. f (x0 ) = f (−e) = 1, f 0 (x)|x=−e = − , y − 1 = − (x + e), y = − x. e e e 1 1 44. y − ln 2 = − (x + 2), y = − x + ln 2 − 1.

2 2 45. (a) Let

the equation of the tangent line be y = mx and suppose that it meets the curve at (x0 , y0 ). Then 1

1 ln x0 1 1 m=

and y0 = mx0 + b = ln x0 . So m = = and ln x0 = 1, x0 = e, m = and the equation = x x=x0 x0 x0 x0 e 1 of the tangent line is y = x. e (b) Let y = mx + b be a line tangent to the curve at (x0 , y0 ). Then b is the y-intercept and the slope of the 1 1 tangent line is m = . Moreover, at the point of tangency, mx0 + b = ln x0 or x0 + b = ln x0 , b = ln x0 − 1, as x0 x0 required.

46. Let y(x) = u(x)v(x), then ln y = ln u + ln v, so y 0 /y = u0 /u + v 0 /v, or y 0 = uv 0 + vu0 . Let y = u/v, then ln y = ln u − ln v, so y 0 /y = u0 /u − v 0 /v, or y 0 = u0 /v − uv 0 /v 2 = (u0 v − uv 0 )/v 2 . The logarithm of a product (quotient) is the sum (difference) of the logarithms.

88

Chapter 3

47. The area of the triangle P QR is given by the formula |P Q||QR|/2. |P Q| = w, and, by Exercise 45 part (b), |QR| = 1, so the area is w/2. y

1 Q

P (w, ln w)

x

w

2

R –2

48. Since y = 2 ln x, let y = 2z; then z = ln x and we apply the result of Exercise 45 to find that the area is, in the x-z plane, w/2. In the x-y plane, since y = 2z, the vertical dimension gets doubled, so the area is w.

49. If x = 0 then y = ln e = 1, and

dy 1 dy 1 = . But ey = x + e, so = y = e−y . dx x+e dx e

50. If x = 0 then y = − ln e2 = −2, and

dy 1 1 dy = 2 . But ey = 2 , so = ey . dx e −x e −x dx

51. Let y = ln(x + a). Following Exercise 49 we get a = 1, then y = ln(x + 1).

52. Let y = − ln(a − x), then

a = 2, so y = − ln(2 − x).

dy 1 = = e−y , and when x = 0, y = ln(a) = 0 if a = 1, so let dx x+a

dy 1 1 dy = . But ey = , so = ey . If x = 0 then y = − ln(a) = − ln 2 provided dx a−x a−x dx

53. (a) Set f (x) = ln(1 + 3x). Then f 0 (x) =

(b) Set f (x) = ln(1 − 5x). Then f 0 (x) =

f (x) − f (0) ln(1 + 3x) 3 , f 0 (0) = 3. But f 0 (0) = lim = lim . x→0 x→0 1 + 3x x x −5 f (x) − f (0) ln(1 − 5x) , f 0 (0) = −5. But f 0 (0) = lim = lim . x→0 x→0 1 − 5x x x

ln(e2 + ∆x) − 2 d 1

= (ln x)

=

= e−2 . ∆x→0 ∆x dx x 2 2 x=e x=e

54. (a) f (x) = ln x; f 0 (e2 ) = lim

ln w 1

ln w − ln 1 = lim =

(b) f (w) = ln w; f (1) = lim = 1. w→1 w − 1 w→1 w−1 w w=1 0

f (x) − f (0) ln(cos x) = lim , and x→0 x→0 x x

55. (a) Let f (x) = ln(cos x), then f (0) = ln(cos 0) = ln 1 = 0, so f 0 (0) = lim f 0 (0) = − tan 0 = 0.

√ √

(b) Let f (x) = x √ √2−1 0 √ 2x , f (1) = 2.

2

f (1 + h) − f (1) (1 + h) , then f (1) = 1, so f (1) = lim = lim h→0 h→0 h h 0

2

−1

, and f 0 (x) =

Exercise Set 3.3

56.

89

d logb (x + h) − logb (x) [logb x] = lim h→0 dx h

1 x+h = lim logb h→0 h x

h 1 = lim logb 1 + h→0 h x = lim

v→0

1 logb (1 + v) vx

Let v = h/x and note that v → 0 as h → 0

=

1 1 lim logb (1 + v) x v→0 v

=

1 lim log (1 + v)1/v x v→0 b

=

1 logb lim (1 + v)1/v v→0 x

Theorem 1.5.5

=

1 1 ln e 1 logb e = · = . x x ln b x ln b

Formula 7 of Section 1.3

57. Differentiating implicitly gives 0 = claimed.

h and v are variable, whereas x is constant

1 dp dp 1 dp − (−0.0046) − 2.3, from which = 0.0046p(500 − p) as p dt 2.3 − 0.0046p dt dt

1 dy 1 dy dy + = α, from which we obtain that = αy(1 − y). The right side is y dt 1 − y dt dt an inverted parabola, with a maximum value of α/4 at y = 1/2. Thus y is growing most rapidly when half the population has the information.

58. Implicit differentiation yields

Exercise Set 3.3 1. (a)

f 0 (x) = 5x4 + 3x2 + 1 ≥ 1 so f is increasing and one-to-one on −∞ < x < +∞.

(b) f (1) = 3 so 1 = f −1 (3);

d −1 1 1 1 f (x) = 0 −1 , (f −1 )0 (3) = 0 = . dx f (f (x)) f (1) 9

2. (a) f 0 (x) = 3x2 + 2ex ; f 0 (x) > 0 for all x (since 3x2 ≥ 0 and 2ex > 0), so f is increasing and one-to-one on −∞ < x < +∞. (b) f (0) = 2 so 0 = f −1 (2);

d −1 1 1 1 f (x) = 0 −1 , (f −1 )0 (2) = 0 = . dx f (f (x)) f (0) 2

2 d −1 2 −2 1 − 3, so directly f (x) = − 2 . Using Formula (2), f 0 (x) = , so 0 −1 = x dx x (x + 3)2 f (f (x)) 2 d −1 2 2 −(1/2)(f −1 (x) + 3)2 , and f (x) = −(1/2) = − 2. dx x x

3. f −1 (x) =

ex − 1 d −1 ex 2 d −1 , so directly, f (x) = . Next, f 0 (x) = , and using Formula (2), f (x) = 2 dx 2 2x + 1 dx −1 x 2f (x) + 1 e = .

2 2

4. f −1 (x) =

5. (a) f 0 (x) = 2x + 8; f 0 < 0 on (−∞, −4) and f 0 > 0 on (−4, +∞); not enough information. By inspection, f (1) = 10 = f (−9), so not one-to-one. (b) f 0 (x) = 10x4 + 3x2 + 3 ≥ 3 > 0; f 0 (x) is positive for all x, so f is one-to-one.

90

Chapter 3 (c) f 0 (x) = 2 + cos x ≥ 1 > 0 for all x, so f is one-to-one. (d) f 0 (x) = −(ln 2)

1 x 2

< 0 because ln 2 > 0, so f is one-to-one for all x.

6. (a) f 0 (x) = 3x2 + 6x = x(3x + 6) changes sign at x = −2, 0, so not enough information; by observation (of the graph, and using some guesswork), f (0) = −8 = f (−3), so f is not one-to-one. (b) f 0 (x) = 5x4 + 24x2 + 2 ≥ 2 > 0; f 0 is positive for all x, so f is one-to-one. (c) f 0 (x) =

1 ; f is one-to-one because: (x + 1)2

if x1 < x2 < −1 then f 0 > 0 on [x1 , x2 ], so f (x1 ) 6= f (x2 )

if −1 < x1 < x2 then f 0 > 0 on [x1 , x2 ], so f (x1 ) 6= f (x2 ) if x1 < −1 < x2 then f (x1 ) > 1 > f (x2 ) since f (x) > 1 on (−∞, −1) and f (x) < 1 on (−1, +∞)

d 1 (d) Note that f (x) is only defined for x > 0. logb x = , which is always negative (0 < b < 1), so f is dx x ln b one-to-one.

7. y = f −1 (x), x = f (y) = 5y 3 + y − 7, 8. y = f −1 (x), x = f (y) = 1/y 2 ,

dy 1 dy dy dy 1 dx = 15y 2 + 1, = ; check: 1 = 15y 2 + , = . dy dx 15y 2 + 1 dx dx dx 15y 2 + 1

dx dy dy dy = −2y −3 , = −y 3 /2; check: 1 = −2y −3 , = −y 3 /2. dy dx dx dx

9. y = f −1 (x), x = f (y) = 2y 5 + y 3 + 1, dy 1 = .

4 dx 10y + 3y 2 10. y = f −1 (x), x = f (y) = 5y − sin 2y, dy 1 = . dx 5 − 2 cos 2y

dx dy 1 dy dy = 10y 4 + 3y 2 , = ; check: 1 = 10y 4 + 3y 2 , 4 2 dy dx 10y + 3y dx dx

dx dy 1 dy = 5 − 2 cos 2y, = ; check: 1 = (5 − 2 cos 2y) , dy dx 5 − 2 cos 2y dx

11. Let P (a, b) be given, not on the line y = x. Let Q1 be its reflection across the line y = x, yet to be determined.

Let Q have coordinates (b, a). (a) Since P does not lie on y = x, we have a 6= b, i.e. P 6= Q since they have different abscissas. The line P~Q has slope (b − a)/(a − b) = −1 which is the negative reciprocal of m = 1 and so the two lines are perpendicular. (b) Let (c, d) be the midpoint of the segment P Q. Then c = (a + b)/2 and d = (b + a)/2 so c = d and the midpoint is on y = x. (c) Let Q(c, d) be the reflection of P through y = x. By definition this means P and Q lie on a line perpendicular to the line y = x and the midpoint of P and Q lies on y = x. (d) Since the line through P and Q is perpendicular to the line y = x it is parallel to the line through P and Q1 ; since both pass through P they are the same line. Finally, since the midpoints of P and Q1 and of P and Q both lie on y = x, they are the same point, and consequently Q = Q1 .

12. Let (a, b) and (A, B) be points on a line with slope m. Then m = (B − b)/(A − a). Consider the associated points (B, A) and (b, a). The line through these two points has slope (A − a)/(B − b), which is the reciprocal of m. Thus (B, A) and (b, a) define the line with slope 1/m.

Exercise Set 3.3

91

13. If x < y then f (x) ≤ f (y) and g(x) ≤ g(y); thus f (x) + g(x) ≤ f (y) + g(y). Moreover, g(x) ≤ g(y), so f (g(x)) ≤ f (g(y)). Note that f (x)g(x) need not be increasing, e.g. f (x) = g(x) = x, both increasing for all x, yet f (x)g(x) = x2 , not an increasing function.

14. On [0, 1] let f (x) = x−2, g(x) = 2−x, then f and g are one-to-one but f +g is not. If f (x) = x+1, g(x) = 1/(x+1) then f and g are one-to-one but f g is not. Finally, if f and g are one-to-one and if f (g(x)) = f (g(y)) then, because f is one-to-one, g(x) = g(y), and since g is one-to-one, x = y, so f (g(x)) is one-to-one.

15.

dy = 7e7x . dx

16.

2 dy = −10xe−5x . dx

17.

dy = x3 ex + 3x2 ex = x2 ex (x + 3). dx

18.

1 dy = − 2 e1/x . dx x

19.

(ex + e−x )(ex + e−x ) − (ex − e−x )(ex − e−x ) (e2x + 2 + e−2x ) − (e2x − 2 + e−2x ) dy = = = 4/(ex + e−x )2 . x −x 2 dx (e + e ) (ex + e−x )2

20.

dy = ex cos(ex ). dx

21.

dy = (x sec2 x + tan x)ex tan x . dx

22.

dy (ln x)ex − ex (1/x) ex (x ln x − 1) = = .

2 dx (ln x) x(ln x)2

23.

3x dy = (1 − 3e3x )e(x−e ) . dx

24.

p p dy 1 1 15 2 = √ x (1 + 5x3 )−1/2 exp( 1 + 5x3 ).

15x2 exp( 1 + 5x3 ) = dx 2 1 + 5x3 2

25.

dy (x − 1)e−x x−1 = = x . −x dx 1 − xe e −x

26.

1 dy = [− sin(ex )]ex = −ex tan(ex ). dx cos(ex )

27. f 0 (x) = 2x ln 2; y = 2x , ln y = x ln 2,

1 0 y = ln 2, y 0 = y ln 2 = 2x ln 2. y

28. f 0 (x) = −3−x ln 3; y = 3−x , ln y = −x ln 3,

1 0 y = − ln 3, y 0 = −y ln 3 = −3−x ln 3. y

29. f 0 (x) = π sin x (ln π) cos x; y = π sin x , ln y = (sin x) ln π,

1 0 y = (ln π) cos x, y 0 = π sin x (ln π) cos x. y

30. f 0 (x) = π x tan x (ln π)(x sec2 x + tan x); y = π x tan x , ln y = (x tan x) ln π, π x tan x (ln π)(x sec2 x + tan x).

1 0 y = (ln π)(x sec2 x + tan x), y 0 = y

92

Chapter 3

31. ln y = (ln x) ln(x3 − 2x),

2

1 dy 3x2 − 2 1 dy 3x − 2 1 = 3 ln x + ln(x3 − 2x), = (x3 − 2x)ln x 3 ln x + ln(x3 − 2x) . y dx x − 2x x dx x − 2x x

sin x dy 1 dy sin x sin x = + (cos x) ln x, =x + (cos x) ln x .

32. ln y = (sin x) ln x, y dx x dx x 33. ln y = (tan x) ln(ln x),

1 dy 1 dy tan x = tan x + (sec2 x) ln(ln x), = (ln x)tan x + (sec2 x) ln(ln x) . y dx x ln x dx x ln x

34. ln y = (ln x) ln(x2 + 3),

35. ln y = (ln x)(ln(ln x)),

2x 1 1 1 dy dy 2x = 2 ln x + ln(x2 + 3), = (x2 + 3)ln x 2 ln x + ln(x2 + 3) . y dx x +3 x dx x +3 x

dy/dx 1/x 1 = (1/x)(ln(ln x)) + (ln x) = (1/x)(1 + ln(ln x)), dy/dx = (ln x)ln x (1 + ln ln x). y ln x x

36. (a) Because xx is not of the form ax where a is constant. (b) y = xx , ln y = x ln x,

1 0 y = 1 + ln x, y 0 = xx (1 + ln x). y

37.

dy = (3x2 − 4x)ex + (x3 − 2x2 + 1)ex = (x3 + x2 − 4x + 1)ex . dx

38.

dy = (4x − 2)e2x + (2x2 − 2x + 1)2e2x = 4x2 e2x . dx

39.

√ dy 1 = (2x + √ )3x + (x2 + x)3x ln 3. dx 2 x

40.

√ 1 dy = (3x2 + x−2/3 )5x + (x3 + 3 x)5x ln 5. dx 3

41.

x dy = 43 sin x−e ln 4(3 cos x − ex ). dx

42.

dy 1 = 2cos x+ln x ln 2(− sin x + ). dx x

43.

dy 3 3 =p =√ .

2 dx 1 − 9x2 1 − (3x)

44.

dy = −q dx

1/2 1−

x+1 2 2

1 = −p .

4 − (x + 1)2

45.

dy 1 1 =p (−1/x2 ) = − √ . dx |x| x2 − 1 1 − 1/x2

46.

dy sin x sin x =√ = = dx | sin x| 1 − cos2 x

47.

dy 3x2 3x2 = = . dx 1 + (x3 )2 1 + x6

48.

dy 5x4 5 p √ = = . dx |x| x10 − 1 |x5 | (x5 )2 − 1

1, sin x > 0 . −1, sin x < 0

Exercise Set 3.3

93

49. y = 1/ tan x = cot x, dy/dx = − csc2 x. −1

50. y = (tan

−1

x)

−1

, dy/dx = −(tan

−2

x)

51.

dy ex √ + ex sec−1 x. = dx |x| x2 − 1

52.

1 dy √ =− . dx (cos−1 x) 1 − x2

53.

dy = 0. dx

54.

dy 3x2 (sin−1 x)2 √ + 2x(sin−1 x)3 . = dx 1 − x2

55.

dy = 0. dx

1 .

1 + x2

p dy = −1/ e2x − 1. dx

dy 1 1 −1/2 1 √ .

57. =− x =− dx 1+x 2 2(1 + x) x

56.

58.

1 dy =− √ . dx 2 cot−1 x (1 + x2 )

59. False; y = Aex also satisfies

dy = y. dx

60. False; dy/dx = 1/x is rational, but y = ln x is not.

61. True; examine the cases x > 0 and x < 0 separately.

62. True;

d d sin−1 x + cos−1 x = 0. dx dx

63. (a) Let x = f (y) = cot y, 0 < y < π, −∞ < x < +∞. Then f is differentiable and one-to-one and f 0 (f −1 (x)) =

d 1 1 − csc2 (cot−1 x) = −x2 − 1 6= 0, and [cot−1 x]

= lim 0 −1 = − lim 2 = −1. x→0 x + 1 x→0 f (f dx (x)) x=0 d d 1 1 1 1 cot−1 x = tan−1 = − 2 =− 2 . dx dx x x 1 + (1/x)2 x +1 d 1 For x = 0, part (a) shows the same; thus for −∞ < x < +∞, [cot−1 x] = − 2 . dx x +1 (b) If x 6= 0 then, from Exercise 48(a) of Section 0.4,

(c) For −∞ < u < +∞, by the chain rule it follows that 64. (a) By the chain rule,

(b) By the chain rule,

d 1 du [cot−1 u] = − 2 . dx u + 1 dx

d d 1 1 1 −1 √ [csc−1 x] = sin−1 = − 2 p = .

2 dx dx x x |x| x2 − 1 1 − (1/x) d du d −1 du √ [csc−1 u] = [csc−1 u] = . dx dx du |u| u2 − 1 dx

94

Chapter 3 (c) From Section 0.4 equation (11), sec−1 x + csc−1 x = π/2, so (a). (d) By the chain rule,

d du du d 1 √ [sec−1 u] = [sec−1 u] = . dx dx du |u| u2 − 1 dx (3x2 + tan−1 y)(1 + y 2 ) x y 0 + tan−1 y = ey y 0 , y 0 = .

2 1+y (1 + y 2 )ey − x

65. x3 + x tan−1 y = ey , 3x2 +

−1

66. sin

−1

(xy) = cos

d d 1 √ sec−1 x = − csc−1 x = by part dx dx |x| x2 − 1

p p y 1 − (x − y)2 + 1 − x2 y 2 p (1 − y ), y = p (x − y), p (xy + y) = − p .

1 − (x − y)2 1 − x2 y 2 1 − x2 y 2 − x 1 − (x − y)2 1

1

0

0

0

67. (a) f (x) = x3 − 3x2 + 2x = x(x − 1)(x − 2) so f (0) = f (1) = f (2) = 0 thus f is not one-to-one. √

√ √ 36 − 24 = 1± 3/3. f 0 (x) > 0 (f is increasing) if x < 1− 3/3, (b) f (x) = 3x −6x+2, f (x) = 0 when x = 6√ √ √ f 0 (x) < 0 (f√is decreasing) √ if 1 − 3/3 < x < 1 + 3/3, so f (x) takes on values less than f (1 − 3/3) on both sides of 1 − 3/3 thus 1 − 3/3 is the largest value of k.

0

6±

0

2

68. (a) f (x) = x3 (x − 2) so f (0) = f (2) = 0 thus f is not one-to-one. (b) f 0 (x) = 4x3 − 6x2 = 4x2 (x − 3/2), f 0 (x) = 0 when x = 0 or 3/2; f is decreasing on (−∞, 3/2] and increasing on [3/2, +∞) so 3/2 is the smallest value of k.

69. (a) f 0 (x) = 4x3 + 3x2 = (4x + 3)x2 = 0 only at x = 0. But on [0, 2], f 0 has no sign change, so f is one-to-one. (b) F 0 (x) = 2f 0 (2g(x))g 0 (x) so F 0 (3) = 2f 0 (2g(3))g 0 (3). By inspection f (1) = 3, so g(3) = f −1 (3) = 1 and g 0 (3) = (f −1 )0 (3) = 1/f 0 (f −1 (3)) = 1/f 0 (1) = 1/7 because f 0 (x) = 4x3 + 3x2 . Thus F 0 (3) = 2f 0 (2)(1/7) = 2(44)(1/7) = 88/7. F (3) = f (2g(3)) = f (2 · 1) = f (2) = 25, so the line tangent to F (x) at (3, 25) has the equation y − 25 = (88/7)(x − 3), y = (88/7)x − 89/7.

0

70. (a) f (x) = −e

4−x2

1 2 + 2 < 0 for all x > 0, so f is one-to-one. x

1 9 (b) By inspection, f (2) = 1/2, so 2 = f −1 (1/2) = g(1/2). By inspection, f 0 (2) = − 2 + = − , and 4 4

0

(4) 1 f 0 0 2 d 2

0 2 0 0 2

F (1/2) = f ([g(x)] ) [g(x) ]

= f ([g(x)] )2g(x)g (x)

= 4 0 = = f (2 )2 · 2 0

dx f (g(x)) f (2) x=1/2 x=1/2 x=1/2 1 e−12 (2 + 16 ) 33 11 4 = 12 = 12 .

1 9e 3e (2 + 4 ) 71. y = Aekt , dy/dt = kAekt = k(Aekt ) = ky.

72. y = Ae2x + Be−4x , y 0 = 2Ae2x − 4Be−4x , y 00 = 4Ae2x + 16Be−4x so y 00 + 2y 0 − 8y = (4Ae2x + 16Be−4x ) + 2(2Ae2x − 4Be−4x ) − 8(Ae2x + Be−4x ) = 0.

73. (a) y 0 = −xe−x + e−x = e−x (1 − x), xy 0 = xe−x (1 − x) = y(1 − x). (b) y 0 = −x2 e−x

2

/2

+ e−x

2

/2

= e−x

2

/2

2

(1 − x2 ), xy 0 = xe−x

/2

(1 − x2 ) = y(1 − x2 ).

74. (a) Losing 15% of the value means the value after a year’s depreciation is Vnew = 0.85Vprevious . Using induction gives the formula.

Exercise Set 3.3

95

dV dV = 20000 ln 0.85(0.85)t . Evaluating this at t = 5 gives ≈ −1442 dt dt dollars/year. Thus the car is losing value at about $1442 dollars/year when it is 5 years old. (b) Differentiating V = 20000(0.85)t gives

75. (a) (b) The percentage converges to 100%, full coverage of broadband internet access. The limit of the expression in the denominator is clearly 53 as t → ∞.

(c) The rate converges to 0 according to the graph.

12

76. (a)

0

9

0

(b) P tends to 12 as t gets large; lim P (t) = lim t→+∞

t→+∞

60 60 60 = = 12. −t = −t 5 + 7e 5 5 + 7 lim e t→+∞

3.2

(c) The rate of population growth tends to zero.

0

0

f (x) − f (0) = 3e3x x=0 = 3. x→0 x−0

2

2 78. f (x) = ex , f 0 (0) = 2xex

= 0.

77. f (x) = e3x , f 0 (0) = lim

x=0

10h − 1 d x ln 10

d x

79. lim = 10

= e = ln 10.

h→0 h dx dx x=0 x=0

tan−1 (1 + h) − π/4 d 1

1 −1

80. lim = tan x

= = .

2

h→0 h dx 1 + x 2 x=1 x=1

9

96

Chapter 3

81. lim

9[sin−1 (

∆x→0

√

3 2

+ ∆x)]2 − π 2 3 π d 3 −1 −1 2

= 2(3 ) p = 12π. = (3 sin x) √ = 2(3 sin x) √ √

2 ∆x dx 3 1 − x x= 3 1 − (3/4) x= 3 2

2

√

3 sec−1 w − π 3 d 3 −1

82. lim = √ = = 3 sec x

.

2 w→2 w−2 dx 2 |2| 2 − 1 x=2

d 1 − e−kt 1 − e−kt = 9.8 lim = 9.8 (−e−kt ) k=0 = 9.8 t, so if the fluid offers no resistance, then the k k dk k→0+ k→0+ speed will increase at a constant rate of 9.8 m/s2 .

83. lim 9.8

Exercise Set 3.4 1.

2.

dy dx =3 dt dt dy = 3(2) = 6. (a) dt

(b) −1 = 3

dx dx 1 , =− . dt dt 3

dx dy +4 =0 dt dt dy dy 1 (a) 1 + 4 = 0 so = − when x = 2. dt dt 4

3. 8x

dy dx + 18y =0 dt dt

(a)

1 1 dy dy 8 √ · 3 + 18 √ = 0, = −2. dt 2 2 3 2 dt

(b)

dx dx + 4(4) = 0 so = −16 when x = 3. dt dt

√

√ 1 dx dx 5 (b) 8 − 18 · 8 = 0, = 6 5.

3 dt 9 dt

dx dy dx dy + 2y =2 +4 dt dt dt dt dy dy dy = 2(−5) + 4 , = −10. (a) 2 · 3(−5) + 2 · 1 dt dt dt

4. 2x

(b) 2(1 +

√

√ √ √ √ dx dx 3 dx 2) + 2(2 + 3) · 6 = 2 + 4 · 6, = −12 √ = −3 3 2. dt dt dt 2 2

5. (b) A = x2 . dA dx = 2x . dt dt

dA

dx

dA

(d) Find given that = 2.

From part (c), = 2(3)(2) = 12 ft2 /min. dt x=3 dt x=3 dt x=3

(c)

6. (b) A = πr2 . dr dA = 2πr . dt dt

dA

dr

dA

(d) Find given that = 2. From part (c), = 2π(5)(2) = 20π cm2 /s. dt r=5 dt r=5 dt r=5 (c)

dV dr 2 dh 7. (a) V = πr h, so =π r + 2rh . dt dt dt 2

Exercise Set 3.4

(b) Find

97

dh

dr

dV

dV

given that = 1 and = −1.

From part (a), = π[102 (1)+2(10)(6)(−1)] = dt h=6, dt h=6, dt h=6, dt h=6, r=10

r=10

−20π in3 /s; the volume is decreasing. d` 1 8. (a) ` = x + y , so = dt ` 2

2

2

r=10

r=10

dx dy x +y . dt dt

dx d`

1 dy 1 given that (b) Find = and = − . From part (a) and the fact that ` = 5 when x = 3 and y = 4, dt x=3, dt 2 dt 4

y=4

d`

1 1 1 1 = 3 +4 − = ft/s; the diagonal is increasing. dt x=3, 5 2 4 10 y=4

dy dx

x −y dθ dy cos2 θ dx y 2 dθ dt dt , x = = −y .

9. (a) tan θ = , so sec θ x dt x2 dt x2 dt dt

π dθ

dx

dy

1 (b) Find given that = 1 and = − . When x = 2 and y = 2, tan θ = 2/2 = 1 so θ = dt x=2, dt x=2, dt x=2, 4 4 y=2 y=2 √ 2

y=2

1 1 5 π dθ

(1/ 2) 2 − − 2(1) = − rad/s; θ is decreasing. and cos θ = cos = √ . Thus from part (a), = 4 dt x=2, 22 4 16 2 y=2

dz

dx

dy

dz 3 dy 2 2 dx dz

10. Find = 2x y + 3x y , given that = −2 and = 3. = (4)(3) + (12)(−2) =

dt x=1, dt x=1, dt x=1, dt dt dt dt x=1, y=2 y=2 y=2 y=2 −12 units/s; z is decreasing. dA given that 11. Let A be the area swept out, and θ the angle through which the minute hand has rotated. Find dt π 1 dA dθ 4π 2 dθ = rad/min; A = r2 θ = 8θ, so =8 = in /min. dt 30 2 dt dt 15

dA

dr 12. Let r be the radius and A the area enclosed by the ripple. We want = 3. We know that given that

dt t=10 dt dA dr A = πr2 , so = 2πr . Because r is increasing at the constant rate of 3 ft/s, it follows that r = 30 ft after 10 dt

dt dA

seconds so = 2π(30)(3) = 180π ft2 /s. dt t=10

dr

dA dr dr 1 dA dA 13. Find = 6. From A = πr2 we get = 2πr so = . If A = 9 then πr2 = 9, given that dt A=9

dt dt dt dt 2πr dt √ √ dr

1 √ (6) = 1/ π mi/h. r = 3/ π so =

dt A=9 2π(3/ π) 4 D 14. The volume V of a sphere of radius r is given by V = πr3 or, because r = where D is the diameter, 3 2

3 D 1 dD

dV 1 dV 1 dD 4 = πD3 . We want given that = 3. From V = πD3 we get = πD2 , V = π 3 2 6

dt r=1 dt 6 dt 2 dt dD 2 dV dD

2 3 = , so = (3) = ft/min.

2 2 dt πD dt dt r=1 π(2) 2π

dV

dr 4 dV dr dV 15. Find given that = −15. From V = πr3 we get = 4πr2 so

dt r=9 dt 3 dt dt dt Air must be removed at the rate of 4860π cm3 /min.

r=9

= 4π(9)2 (−15) = −4860π.

98

Chapter 3

16. Let x and y be the distances shown in the diagram. We want to find

dx dy

given that = 5. From x2 +y 2 = 172

dt y=8 dt

dy dy x dx dx we get 2x + 2y = 0, so = − . When y = 8, x2 + 82 = 172 , x2 = 289 − 64 = 225, x = 15 so dt dt dt y dt

15 75 dy

= − (5) = − ft/s; the top of the ladder is moving down the wall at a rate of 75/8 ft/s. dt y=8 8 8

17

y

x

dx dx

dy dy dx y dy 17. Find given that = −2. From x2 +y 2 = 132 we get 2x +2y = 0 so =− . Use x2 +y 2 = 169 dt y=5 dt dt dt dt x dt

5 dx

5 = − (−2) = ft/s. to find that x = 12 when y = 5 so dt y=5 12 6

13

y

x

dθ

18. Let θ be the acute angle, and x the distance of the bottom of the plank from the wall. Find given dt x=2

dx

1 x dθ 1 dx that = − ft/s. The variables θ and x are related by the equation cos θ = so − sin θ = , dt x=2 2 10 dt 10 dt √ √ √ dθ 1 dx =− . When x = 2, the top of the plank is 102 − 22 = 96 ft above the ground so sin θ = 96/10 dt 10 sin θ dt

dθ

1 1 1 √ and = − − = √ ≈ 0.051 rad/s. dt x=2 2 96 2 96

ft

dx dy 19. Let x denote the distance from first base and y the distance from home plate. Then x2 +602 = y 2 and 2x = 2y . dt dt √ dy x dx 50 125 When x = 50 then y = 10 61 so = = √ (25) = √ ft/s. dt y dt 10 61 61

60

x y

First

Home

dx

dy

dx dy dx y dy 20. Find given that = 2000. From x2 + 52 = y 2 we get 2x = 2y so = . Use x2 + 25 = y 2 dt x=4 dt x=4 dt dt dt x dt

Exercise Set 3.4

99

to find that y =

√

41 when x = 4 so

√

√ dx

41 = (2000) = 500 41 mi/h.

dt x=4 4

Rocket

y

Radar station

x

5 mi

dx

dy dx dy x dx dy

given that = 880. From y 2 = x2 + 30002 we get 2y = 2x so = . If dt x=4000 dt

x=4000 dt dt dt y dt dy

4000 x = 4000, then y = 5000 so = (880) = 704 ft/s. dt x=4000 5000

21. Find

Rocket

y

x

Camera 3000 ft

dφ

dx dφ dx

π dx

2 given that = 0.2.

But x = 3000 tan φ so = 3000(sec φ) , = 3000 sec2 (0.2) =

dt φ=π/4 dt φ=π/4 dt dt dt φ=π/4 4 1200 ft/s.

22. Find

23. (a) If x denotes the altitude, then r−x = 3960, the radius of the Earth. θ = 0 at perigee, so r = 4995/1.12 ≈ 4460; the altitude is x = 4460 − 3960 = 500 miles. θ = π at apogee, so r = 4995/0.88 ≈ 5676; the altitude is x = 5676 − 3960 = 1716 miles. (b) If θ = 120◦ , then r = 4995/0.94 ≈ 5314; the altitude is 5314 − 3960 = 1354 miles. The rate of change of the dx dr dr dθ 4995(0.12 sin θ) dθ altitude is given by = = = . Use θ = 120◦ and dθ/dt = 2.7◦ /min = (2.7)(π/180) dt dt dθ dt (1 + 0.12 cos θ)2 dt rad/min to get dr/dt ≈ 27.7 mi/min.

24. (a) Let x be the horizontal distance shown in the figure. Then x = 4000 cot θ and

dx dθ = −4000 csc2 θ , so dt dt

dθ sin2 θ dx = − . Use θ = 30◦ and dx/dt = 300 mi/h = 300(5280/3600) ft/s = 440 ft/s to get dθ/dt = dt 4000 dt −0.0275 rad/s ≈ −1.6◦ /s; θ is decreasing at the rate of 1.6◦ /s. (b) Let y be the distance between the observation point and the aircraft. Then y = 4000 csc θ so dy/dt = −4000(csc θ cot θ)(dθ/dt). Use θ = 30◦ and dθ/dt = −0.0275 rad/s to get dy/dt ≈ 381 ft/s.

dh

dV 1 given that = 20. The volume of water in the tank at a depth h is V = πr2 h. Use similar dt h=16 dt 3

2 r 10 5 1 5 25 dV 25 dh dh triangles (see figure) to get = so r = h thus V = π h h = πh3 , = πh2 ; = h 24 12 3 12 432 dt 144 dt dt

144 dV dh

144 9 , = (20) = ft/min.

25πh2 dt dt h=16 25π(16)2 20π

25. Find

100

Chapter 3 10 r 24 h

2 dV 1 1 2 1 1 h dV 1 dh dh

given that h = = 8.

V = πr h, but r = h so V = π πh3 , = πh2 ,

dt h=6 dt 3 2 3 2 12 dt 4 dt

4 8 dh 4 dV dh

= (8) = = , ft/min. dt πh2 dt dt h=6 π(6)2 9π

26. Find

h

r

2 dV

dh 1 2 1 1 h 1 2 dh dV

1 3 dV 27. Find given that = 5. V = πr h, but r = h so V = π πh , = πh , = h= dt h=10 dt 3 2 3 2 12 dt 4 dt dt h=10 1 π(10)2 (5) = 125π ft3 /min.

4

h

r

dC

given dt h=8 1 dC dh 1 1 dV = 10. It is given that r = h, thus C = 2πr = πh so = π . Use V = πr2 h = πh3 to get that dt 2 dt dt 3

12 dV 1 dh dh 4 dV dh dC dC 4 dV dC

4 5 = πh2 , so = . Substitution of into gives = 2 so = (10) =

2 dt 4 dt dt πh dt dt dt dt h dt dt h=8 64 8 ft/min.

28. Let r and h be as shown in the figure. If C is the circumference of the base, then we want to find

h

r

Exercise Set 3.4

101

ds dh 1 given that = 500. From the figure, h = s sin 30◦ = s 29. With s and h as shown in the figure, we want to find dt dt 2 dh 1 ds 1 so = = (500) = 250 mi/h. dt 2 dt 2

s

h

30° Ground

dx

dy dx dy dx y dy given that = −20. From x2 + 102 = y 2 we get 2x = 2y so = . Use x2 + 100 = y 2 dt y=125 dt dt dt dt x dt

√ √ dx

125 500 = √ (−20) = − √ . The boat is approaching to find that x = 15, 525 = 15 69 when y = 125 so

dt y=125 15 69 3 69 500 the dock at the rate of √ ft/min.

3 69

30. Find

Pulley y

10

Boat

x

dy dx

dx dy dy x dx given that = −12. From x2 + 102 = y 2 we get 2x = 2y so = . Use x2 + 100 = y 2 dt dt y=125 dt dt dt y dt √ √ p √ dy 15 69 36 69 to find that x = 15, 525 = 15 69 when y = 125 so = (−12) = − . The rope must be pulled at dt 125 25 √ 36 69 the rate of ft/min.

25

31. Find

Pulley y

10

Boat

x

dx dy 32. (a) Let x and y be as shown in the figure. It is required to find , given that = −3. By similar triangles, dt dt x x+y 1 dx 1 dy 1 3 = , 18x = 6x + 6y, 12x = 6y, x = y, so = = (−3) = − ft/s.

6 18 2 dt 2 dt 2 2 Light

18 Man Shadow x

6 y

(b) The tip of the shadow is z = x + y feet from the street light, thus the rate at which it is moving is given by dz dx dy dx 3 dy dz = + . In part (a) we found that = − when = −3 so = (−3/2) + (−3) = −9/2 ft/s; the tip dt dt dt dt 2 dt dt

102

Chapter 3 of the shadow is moving at the rate of 9/2 ft/s toward the street light.

dθ 2π π dx dx

dx

2 dθ given that = = = rad/s. Then x = 4 tan θ (see figure) so = 4 sec θ , 33. Find dt θ=π/4 dt 10 5 dt dt dt θ=π/4

π π 4 sec2 = 8π/5 km/s.

4 5 x

4

θ

Ship

dx dz

dy

given that = −1200.

34. If x, y, and z are as shown in the figure, then we want = −600 and dt x=2, dt dt x=2, y=4 y=4

dx dy dz 1 dx dy dz But z 2 = x2 + y 2 so 2z = 2x + 2y , = x +y . When x = 2 and y = 4, z 2 = 22 + 42 = 20, dt dt dt dt z dt dt

√ √ √ dz

3000 1 z = 20 = 2 5 so = √ [2(−600) + 4(−1200)] = − √ = −600 5 mi/h; the distance between missile

dt x=2, 2 5 5 y=4 √ and aircraft is decreasing at the rate of 600 5 mi/h.

P

Aircraft

x

y z

Missile

dy

dz

dx = −600 and given = −1200 (see figure). From the law of cosines, z 2 = dt x=2, dt dt x=2, y=4 y=4 dx dy dy dx dz dz = 2x + 2y +x +y , = x2 + y 2 − 2xy cos 120◦ = x2 + y 2 − 2xy(−1/2) = x2 + y 2 + xy, so 2z dt dt dt dt dt dt

√ √ 1 dx dy (2x + y) + (2y + x) . When x = 2 and y = 4, z 2 = 22 + 42 + (2)(4) = 28, so z = 28 = 2 7, thus 2z

dt dt √ 4200 dz

1 √ [(2(2) + 4)(−600) + (2(4) + 2)(−1200)] = − √ = −600 7 mi/h; the distance between missile =

dt x=2, 2(2 7) 7 y=4 √ and aircraft is decreasing at the rate of 600 7 mi/h.

35. We wish to find

P

x

Aircraft

120º y z

Missile

36. (a) Let P be the point on the helicopter’s path that lies directly above the car’s path. Let x, y, and z be the dz

dx dy distances shown in the first figure. Find given that = −75 and = 100. In order to find an equation dt x=2, dt dt y=0

Exercise Set 3.4

103

relating x, y, and z, first draw the line segment that joins the point P p to the car, as shown in the second figure.

Because triangle OP C is a right triangle, it follows that P C has length x2 + (1/2)2 ; but triangle HP C is also a 2 p dz dx dy dz 1 dx dy right triangle so z 2 = x2 + (1/2)2 + y 2 = x2 + y 2 + 1/4 and 2z = 2x + 2y + 0, = x +y . dt dt

dt dt z dt dt √ dz

1 = √ [2(−75) + 0(100)] = Now, when x = 2 and y = 0, z 2 = (2)2 + (0)2 + 1/4 = 17/4, z = 17/2 so

x=2, dt ( 17/2) y=0 √ −300/ 17 mi/h.

North

1 2

West

mi

1 2

P y

x

Car

P y

East

z Helicopter

(b) Decreasing, because

mi

x O

C

z

H

dz < 0. dt

dy

dx

8 8 given that = 6. For convenience, first rewrite the equation as xy 3 = + y 2 then dt x=1, dt x=1, 5 5 y=2 y=2

3 3

dy dx 16 dy dy y dy

dx 2 3xy 2 + y3 = y , = , so (6) = −60/7 units/s. = 16 16 dt dt 5 dt dt dt x=1, 2 dt y − 3xy (2) − 3(1)22 y=2 5 5

37. (a) We want

dy < 0. dt

dy dx 2y dy dx

dy

dx = 2y , = 2 , 38. Find given that = 2. Square and rearrange to get x3 = y 2 − 17, so 3x2 dt (2,5) dt (2,5) dt dt dt 3x dt

dx

5 5 = (2) = units/s.

dt (2,5) 6 3 (b) Falling, because

p 39. The coordinates of P are (x, 2x), so the distance between P and the point (3, 0) is D = (x − 3)2 + (2x − 0)2 =

√ dx

dD 5x − 3 dD

12 dD

dx given that = −2. = √ , so = √ (−2) = −4 5x2 − 6x + 9. Find

2 dt x=3 dt x=3 dt dt dt 36 5x − 6x + 9 x=3 units/s.

p dD

dx

given that = 4. D = (x − 2)2 + y 2 =

dt x=3

dt x=3 p

√ 2x − 3 dD dD dx 3

= √ ; (x − 2)2 + x = x2 − 3x + 4, so = √ 4 = 3 units/s. dt 2 4 2 x2 − 3x + 4 dt dt x=3

40. (a) Let D be the distance between P and (2, 0). Find

√ dθ

dx

y x given that = 4. tan θ = = , so dt x=3 dt x=3 x−2 x − 2 dθ x+2 dx dθ x+2 dx 1 dθ

sec2 θ =− √ , = − cos2 θ √ . When x = 3, D = 2 so cos θ = and = dt 2 dt x=3 2 x(x − 2)2 dt dt 2 x(x − 2)2 dt 1 5 5 − √ (4) = − √ rad/s.

42 3 2 3 (b) Let θ be the angle of inclination. Find

dx dy dy =3 given y = x/(x2 +1). Then y(x2 +1) = x. Differentiating with respect to x, (x2 +1) +y(2x) = 1. dt dt dx dy dy/dt 1 1 But = = so (x2 +1) +2xy = 1, x2 +1+6xy = 3, x2 +1+6x2 /(x2 +1) = 3, (x2 +1)2 +6x2 −3x2 −3 = dx dx/dt 3 3

41. Solve

104

Chapter 3 √ 2 0, x4 + 5x2 − 2 = 0. By the quadratic formula applied to x2 we obtain xq = (−5 ± 25 + 8)/2. The minus sign is √ √ spurious since x2 cannot be negative, so x2 = (−5 + 33)/2, and x = ± (−5 + 33)/2.

dP dv dl = 0.87(3l2 v 2 + 2lv 3 ). Substituting l = 16, v = 4, and dt dt dt dl dv/dt = 0.01 gives 0 = 3 · 162 · 42 · 0.01 + 2 · 16 · 43 . Solving for the rate of change of the blade length, we obtain dt dl 122.88 =− = −0.06 m/s. dt 2048

dS

ds

1 1 1 1 ds 1 dS 43. Find given that = −2.

From + = we get − 2 − 2 = 0, so

dt s=10 dt s=10 s S 6 s dt S dt

225 dS S 2 ds 1 1 1 dS

= − = − 2 . If s = 10, then + = which gives S = 15. So (−2) = 4.5 cm/s. dt s dt 10 S 6 dt s=10 100 The image is moving away from the lens.

42. Since P is constant, differentiation yields 0 =

44. Suppose that the reservoir has height H and that the radius at the top is R. At any instant of time let h and dh r be the corresponding dimensions of the cone of water (see figure). We want to show that is constant and dt dV = −kA where V is the volume of water, A is the area of a circle of radius independent of H and R, given that dt 1 r, and k is a positive constant. The volume of a cone of radius r and height h is V = πr2 h. By similar triangles 3 2 2 R R 1 R R dV r dV dh 3 = ,r= h thus V = π =π = −kA or, because h , so h2 . But it is given that h H H 3 H dt H dt dt 2 2 R dV R dV A = πr2 = π h2 , = −kπ h2 , which when substituted into the previous equation for gives H dt H dt 2 2 R dh R dh −kπ = −k. h2 = π h2 , and H H dt dt R r H h

dr 45. Let r be the radius, V the volume, and A the surface area of a sphere. Show that is a constant given dt dV 4 dV dr that = −kA, where k is a positive constant. Because V = πr3 , = 4πr2 . But it is given that dt 3 dt dt dV dV 2 2 dV = −kA or, because A = 4πr , = −4πr k which when substituted into the previous equation for gives dt dt dt dr dr −4πr2 k = 4πr2 , and = −k. dt dt 46. Let x be the distance between the tips of the minute and hour hands, and α and β the angles shown in the dα 2π figure. Because the minute hand makes one revolution in 60 minutes, = = π/30 rad/min; the hour dt 60 dβ 2π hand makes one revolution in 12 hours (720 minutes), thus = = π/360 rad/min. We want to find dt 720

dx

dα dβ given that = π/30 and = π/360. Using the law of cosines on the triangle shown in the dt α=2π, dt dt β=3π/2

dx dα dβ 2 2 2 figure, x = 3 + 4 − 2(3)(4) cos(α − β) = 25 − 24 cos(α − β), so 2x = 0 + 24 sin(α − β) − , dt dt dt

Exercise Set 3.5

105

dx 12 dα dβ = − sin(α − β). When α = 2π and β = 3π/2, x2 = 25 − 24 cos(2π − 3π/2) = 25, x = 5; so dt x dt dt 12 dx

11π = (π/30 − π/360) sin(2π − 3π/2) = in/min. dt α=2π, 5 150 β=3π/2

4

x

" !

3

47. Extend sides of cup to complete the cone and let V0 be the volume of the portion added, then (see figure) 2 1 r 4 1 1 1 h dV 1 1 dh V = πr2 h − V0 where = = so r = h and V = π πh3 − V0 , = πh2 , h − V0 = 3 h 12 3 3 3 3 27 dt 9 dt

9 dV dh

dh 9 20 = , cm/s. = (20) = dt πh2 dt dt h=9 π(9)2 9π 4 r

6 2

h

6

Exercise Set 3.5 1. (a) f (x) ≈ f (1) + f 0 (1)(x − 1) = 1 + 3(x − 1). (b) f (1 + ∆x) ≈ f (1) + f 0 (1)∆x = 1 + 3∆x. (c) From part (a), (1.02)3 ≈ 1 + 3(0.02) = 1.06. From part (b), (1.02)3 ≈ 1 + 3(0.02) = 1.06.

2. (a) f (x) ≈ f (2) + f 0 (2)(x − 2) = 1/2 + (−1/22 )(x − 2) = (1/2) − (1/4)(x − 2). (b) f (2 + ∆x) ≈ f (2) + f 0 (2)∆x = 1/2 − (1/4)∆x. (c) From part (a), 1/2.05 ≈ 0.5 − 0.25(0.05) = 0.4875, and from part (b), 1/2.05 ≈ 0.5 − 0.25(0.05) = 0.4875.

√ 0 3. (a) √ f (x) ≈ f (x0 ) + f (x0 )(x − x0 ) = 1 + (1/(2 1)(x − 0) = 1 + (1/2)x, so √ with x0 = 0 and x = −0.1, we have 0.9 = f (−0.1) ≈ 1 + (1/2)(−0.1) = 1 − 0.05 = 0.95. With x = 0.1 we have 1.1 = f (0.1) ≈ 1 + (1/2)(0.1) = 1.05. y

∆y

∆y

dy

dy

x

(b)

–0.1

0.1

106

Chapter 3 √ 1 1 √ x0 + √ (x − x0 ), so show that x0 + √ (x − x0 ) ≥ x which is equivalent 2 x 2 x 0 √ 0 √ √ x0 x0 x 1 1 0 . But g(x0 ) = , and g (x) = √ − √ which is negative for x > x0 and to g(x) = x − √ ≤ 2 x0 2 2 2 x0 2 x positive for x < x0 . This shows that g has a maximum value at x = x0 , so the student’s observation is correct.

4. (b) The approximation is

√

x≈

√

5. f (x) = (1 + x)15 and x0 = 0. Thus (1 + x)15 ≈ f (x0 ) + f 0 (x0 )(x − x0 ) = 1 + 15(1)14 (x − 0) = 1 + 15x.

6. f (x) = √

1 1 1 (x − 0) = 1 + x/2. and x0 = 0, so √ ≈ f (x0 ) + f 0 (x0 )(x − x0 ) = 1 + 2(1 − 0)3/2 1−x 1−x

7. tan x ≈ tan(0) + sec2 (0)(x − 0) = x.

8.

1 −1 (x − 0) = 1 − x. ≈1+ 1+x (1 + 0)2

9. x0 = 0, f (x) = ex , f 0 (x) = ex , f 0 (x0 ) = 1, hence ex ≈ 1 + 1 · x = 1 + x.

10. x0 = 0, f (x) = ln(1 + x), f 0 (x) = 1/(1 + x), f 0 (x0 ) = 1, hence ln(1 + x) ≈ 0 + 1 · (x − 0) = x.

11. x4 ≈ (1)4 + 4(1)3 (x − 1). Set ∆x = x − 1; then x = ∆x + 1 and (1 + ∆x)4 = 1 + 4∆x. √

√ 1 1 + √ (x − 1), and x = 1 + ∆x, so 1 + ∆x ≈ 1 + ∆x/2.

2 1

12.

√

13.

1 1 1 1 1 1 ≈ − (x − 1), and 2 + x = 3 + ∆x, so ≈ − ∆x.

2+x 2 + 1 (2 + 1)2 3 + ∆x 3 9

x≈

14. (4 + x)3 ≈ (4 + 1)3 + 3(4 + 1)2 (x − 1) so, with 4 + x = 5 + ∆x we get (5 + ∆x)3 ≈ 125 + 75∆x.

15. Let f (x) = tan−1 x, f (1) = π/4, f 0 (1) = 1/2, tan−1 (1 + ∆x) ≈ −1

16. f (x) = sin √

17. f (x) =

x 2

−1

, sin

√ 1 1 1 π 0 1/2 π 1 −1 0 = , f (x) = p , f (1) = 1/ 3. sin + ∆x ≈ + √ ∆x.

2 2 6 2 2 6 3 1 − x /4

x + 3 and x0 = 0, so

|x| < 1.692.

0

-2

|

π 1 + ∆x.

4 2

√

x+3 ≈

√

√

√ 1 1 1

3 + √ (x − 0) = 3 + √ x, and f (x) − 3 + √ x

< 0.1 if 2 3 2 3 2 3

2

-0.1

f (x) –

( 3 + 2 13 x)|

1 1 1 1 1 1 1 1

18. f (x) = √ so √ ≈√ + (x − 0) = + x, and f (x) − + x

< 0.1 if |x| < 5.5114.

3/2 3 54 3 54 2(9 − 0) 9−x 9−x 9

Exercise Set 3.5

107 0.06

-6

6

0

| f (x) â&#x20AC;&#x201C; (

1 3

+

)|

1 x 54

19. tan 2x â&#x2030;&#x2C6; tan 0 + (sec2 0)(2x â&#x2C6;&#x2019; 0) = 2x, and | tan 2x â&#x2C6;&#x2019; 2x| < 0.1 if |x| < 0.3158.

\ f x n 2x\

20.

1 1 â&#x2C6;&#x2019;5(2) â&#x2030;&#x2C6; + (x â&#x2C6;&#x2019; 0) = 1 â&#x2C6;&#x2019; 10x, and |f (x) â&#x2C6;&#x2019; (1 â&#x2C6;&#x2019; 10x)| < 0.1.

(1 + 2x)5 (1 + 2 · 0)5 (1 + 2 · 0)6 0.12

-0.04

0 | f (x) â&#x20AC;&#x201C; (1 â&#x20AC;&#x201C; 10x)|

0.04

21. (a) The local linear approximation sin x â&#x2030;&#x2C6; x gives sin 1â&#x2014;¦ = sin(Ï&#x20AC;/180) â&#x2030;&#x2C6; Ï&#x20AC;/180 = 0.0174533 and a calculator gives sin 1â&#x2014;¦ = 0.0174524. The relative error | sin(Ï&#x20AC;/180) â&#x2C6;&#x2019; (Ï&#x20AC;/180)|/(sin Ï&#x20AC;/180) = 0.000051 is very small, so for such a small value of x the approximation is very good. (b) Use x0 = 45â&#x2014;¦ (this assumes you know, or can approximate,

â&#x2C6;&#x161;

2/2).

44Ï&#x20AC; 45Ï&#x20AC; Ï&#x20AC; 44Ï&#x20AC; Ï&#x20AC; 44Ï&#x20AC; (c) 44â&#x2014;¦ = radians, and 45â&#x2014;¦ = = radians. With x = and x0 = we obtain sin 44â&#x2014;¦ = sin â&#x2030;&#x2C6; 180 180 4 180 â&#x2C6;&#x161; 4 â&#x2C6;&#x161; 180

Ï&#x20AC; Ï&#x20AC; 44Ï&#x20AC; Ï&#x20AC; 2 2 â&#x2C6;&#x2019;Ï&#x20AC; sin + cos â&#x2C6;&#x2019; = + = 0.694765. With a calculator, sin 44â&#x2014;¦ = 0.694658.

4 4 180 4 2 2 180 22. (a) tan x â&#x2030;&#x2C6; tan 0 + sec2 0(x â&#x2C6;&#x2019; 0) = x, so tan 2â&#x2014;¦ = tan(2Ï&#x20AC;/180) â&#x2030;&#x2C6; 2Ï&#x20AC;/180 = 0.034907, and with a calculator tan 2â&#x2014;¦ = 0.034921. (b) Use x0 = Ï&#x20AC;/3 because we know tan 60â&#x2014;¦ = tan(Ï&#x20AC;/3) =

â&#x2C6;&#x161;

3.

Ï&#x20AC; 60Ï&#x20AC; 61Ï&#x20AC; 61Ï&#x20AC; Ï&#x20AC; Ï&#x20AC; = and x = we have tan 61â&#x2014;¦ = tan â&#x2030;&#x2C6; tan + sec2 3 180 180 180 3 3 1.8019, and with a calculator tan 61â&#x2014;¦ = 1.8040. (c) With x0 =

61Ï&#x20AC; Ï&#x20AC; â&#x2C6;&#x2019; 180 3

23. f (x) = x4 , f 0 (x) = 4x3 , x0 = 3, â&#x2C6;&#x2020;x = 0.02; (3.02)4 â&#x2030;&#x2C6; 34 + (108)(0.02) = 81 + 2.16 = 83.16.

=

â&#x2C6;&#x161;

3+4

Ï&#x20AC; = 180

108

Chapter 3

24. f (x) = x3 , f 0 (x) = 3x2 , x0 = 2, ∆x = −0.03; (1.97)3 ≈ 23 + (12)(−0.03) = 8 − 0.36 = 7.64.

25. f (x) =

√

√ √ 1 1 1 x, f 0 (x) = √ , x0 = 64, ∆x = 1; 65 ≈ 64 + (1) = 8 + = 8.0625.

16 16 2 x

26. f (x) =

√

√ √ 1 1 x, f 0 (x) = √ , x0 = 25, ∆x = −1; 24 ≈ 25 + (−1) = 5 − 0.1 = 4.9.

10 2 x

27. f (x) =

√

√ √ 1 1 x, f 0 (x) = √ , x0 = 81, ∆x = −0.1; 80.9 ≈ 81 + (−0.1) ≈ 8.9944.

18 2 x

28. f (x) =

√

√ √ 1 1 x, f 0 (x) = √ , x0 = 36, ∆x = 0.03; 36.03 ≈ 36 + (0.03) = 6 + 0.0025 = 6.0025.

12 2 x

29. f (x) = sin x, f 0 (x) = cos x, x0 = 0, ∆x = 0.1; sin 0.1 ≈ sin 0 + (cos 0)(0.1) = 0.1.

30. f (x) = tan x, f 0 (x) = sec2 x, x0 = 0, ∆x = 0.2; tan 0.2 ≈ tan 0 + (sec2 0)(0.2) = 0.2. √

1 π π 3 31. f (x) = cos x, f (x) = − sin x, x0 = π/6, ∆x = π/180; cos 31 ≈ cos 30 + − − ≈ 0.8573. = 2 180 2 360 ◦

0

◦

32. f (x) = ln x, x0 = 1, ∆x = 0.01, ln x ≈ ∆x, ln 1.01 ≈ 0.01.

33. tan−1 (1 + ∆x) ≈

π 1 π + ∆x, ∆x = −0.01, tan−1 0.99 ≈ − 0.005 ≈ 0.780398.

4 2 4

34. (a) Let f (x) = (1 + x)k and x0 = 0. Then (1 + x)k ≈ 1k + k(1)k−1 (x − 0) = 1 + kx. Set k = 37 and x = 0.001 to obtain (1.001)37 ≈ 1.037. (b) With a calculator (1.001)37 = 1.03767. (c) 35.

√ 3

It is the linear term of the expansion.

√ 8.24 = 81/3 3 1.03 ≈ 2(1 + 13 0.03) ≈ 2.02, and 4.083/2 = 43/2 1.023/2 = 8(1 + 0.02(3/2)) = 8.24.

36. 6◦ = π/30 radians; h = 500 tan(π/30) ≈ 500[tan 0 + (sec2 0)

π ] = 500π/30 ≈ 52.36 ft.

30

37. (a) dy = (−1/x2 )dx = (−1)(−0.5) = 0.5 and ∆y = 1/(x + ∆x) − 1/x = 1/(1 − 0.5) − 1/1 = 2 − 1 = 1. y 2

∆y = 1 dy = 0.5

1

x

(b)

0.5

1

p √ √ √ √ √ 38. (a) dy = (1/2 x)dx = (1/(2 · 3))(−1) = −1/6 ≈ −0.167 and ∆y = x + ∆x − x = 9 + (−1) − 9 = 8 − 3 ≈ −0.172.

Exercise Set 3.5

109

(b) 39. dy = 3x2 dx; ∆y = (x + ∆x)3 − x3 = x3 + 3x2 ∆x + 3x(∆x)2 + (∆x)3 − x3 = 3x2 ∆x + 3x(∆x)2 + (∆x)3 .

40. dy = 8dx; ∆y = [8(x + ∆x) − 4] − [8x − 4] = 8∆x.

41. dy = (2x−2)dx; ∆y = [(x+∆x)2 −2(x+∆x)+1]−[x2 −2x+1] = x2 +2x ∆x+(∆x)2 −2x−2∆x+1−x2 +2x−1 = 2x ∆x + (∆x)2 − 2∆x.

42. dy = cos x dx; ∆y = sin(x + ∆x) − sin x.

43. (a) dy = (12x2 − 14x)dx. (b) dy = x d(cos x) + cos x dx = x(− sin x)dx + cos xdx = (−x sin x + cos x)dx.

44. (a) dy = (−1/x2 )dx. (b) dy = 5 sec2 x dx.

45. (a) dy =

√

x 1−x− √ 2 1−x

2 − 3x dx = √ dx.

2 1−x

(b) dy = −17(1 + x)−18 dx.

46. (a) dy =

(b) dy =

(x3 − 1)d(1) − (1)d(x3 − 1) (x3 − 1)(0) − (1)3x2 dx 3x2 = =− 3 dx.

3 2 3 2 (x − 1) (x − 1) (x − 1)2 (2 − x)(−3x2 )dx − (1 − x3 )(−1)dx 2x3 − 6x2 + 1 = dx.

2 (2 − x) (2 − x)2

47. False; dy = (dy/dx)dx.

48. True.

49. False; they are equal whenever the function is linear.

50. False; if f 0 (x0 ) = 0 then the approximation is constant.

3 3 51. dy = √ dx, x = 2, dx = 0.03; ∆y ≈ dy = (0.03) = 0.0225.

4 2 3x − 2 52. dy = √ 53. dy =

1 − x2 dx, x = 2, dx = −0.04; ∆y ≈ dy = (x2 + 1)2

54. dy =

x dx, x = 1, dx = −0.03; ∆y ≈ dy = (1/3)(−0.03) = −0.01. x2 + 8

−

3 25

(−0.04) = 0.0048.

√ 4x √ + 8x + 1 dx, x = 3, dx = 0.05; ∆y ≈ dy = (37/5)(0.05) = 0.37.

8x + 1

110

Chapter 3

55. (a) A = x2 where x is the length of a side; dA = 2x dx = 2(10)(±0.1) = ±2 ft2 . dx ±0.1 (b) Relative error in x is within = = ±0.01 so percentage error in x is ±1%; relative error in A is within x 10 dx 2x dx dA =2 = = 2(±0.01) = ±0.02 so percentage error in A is ±2%.

A x2 x 56. (a) V = x3 where x is the length of a side; dV = 3x2 dx = 3(25)2 (±1) = ±1875 cm3 . (b) Relative error in x is within

dx ±1 = = ±0.04 so percentage error in x is ±4%; relative error in V is within x 25

dx 3x2 dx dV =3 = = 3(±0.04) = ±0.12 so percentage error in V is ±12%.

V x3 x

π π 57. (a) x = 10 sin θ, y = 10 cos θ (see figure), dx = 10 cos θdθ = 10 cos ± = 10 6 180

π π 1 π ±0.151 in, dy = −10(sin θ)dθ = −10 sin ± = −10 ± ≈ ±0.087 in.

6 180 2 180

10″

√ !

3 π ± ≈ 2 180

x

θ

y

dx π π √ π = (cot θ)dθ = cot ± = 3 ± ≈ ±0.030, so percentage error x 6 180 180 dy π π 1 π in x is ≈ ±3.0%; relative error in y is within = − tan θdθ = − tan ± = −√ ± ≈ ±0.010, so y 6 180 180 3 percentage error in y is ≈ ±1.0%. (b) Relative error in x is within

π π π 4 58. (a) x = 25 cot θ, y = 25 csc θ (see figure); dx = −25 csc2 θdθ = −25 csc2 ± ± = −25 ≈ 3 360 3 360

π π 2 1 π π √ cot ± = −25 √ ± ≈ ±0.145 cm. ±0.291 cm, dy = −25 csc θ cot θdθ = −25 csc 3 3 360 360 3 3

25 cm

y

θ

x

dx csc2 θ 4/3 π = − dθ = − √ ± ≈ ±0.020, so percentage error in x is x cot θ 360 1/ 3

dy 1 π ≈ ±2.0%; relative error in y is within = − cot θdθ = − √ ± ≈ ±0.005, so percentage error in y is y 360 3 ≈ ±0.5%.

(b) Relative error in x is within

59.

dR (−2k/r3 )dr dr dr dR = = −2 , but = ±0.05 so = −2(±0.05) = ±0.10; percentage error in R is ±10%.

2 R (k/r ) r r R

60. F = 36/L, thus ∆F ≈ dF = −36L−2 dL = −36 · 18−2 · 0.9 = −0.1. Hence the value of F decreases by about 0.1 microtesla.

Exercise Set 3.6

111

1 2 (4) sin 2θ = 4 sin 2θ thus dA = 8 cos 2θdθ so, with θ = 30◦ = π/6 radians and dθ = ±150 = ±1/4◦ = ±π/720 4 radians, dA = 8 cos(π/3)(±π/720) = ±π/180 ≈ ±0.017 cm2 .

61. A =

62. A = x2 where x is the length of a side; percentage error in A is ±2% 63. V = x3 where x is the length of a side; percentage error in V is ±6%.

64.

dx 2x dx dx dA dA = 2 , but = = ±0.01, so = 2(±0.01) = ±0.02; A x2 x x A dx dV 3x2 dx dx dV =3 = , but = ±0.02, so = 3(±0.02) = ±0.06; V x3 x x V

dr dV 4πr2 dr dV dr dr = = 3 , but = ±0.03 so 3 = ±0.03, = ±0.01; maximum permissible percentage error in r V 4πr3 /3 r V r r is ±1%.

65. A =

1 dA (πD/2)dD dD dA dD πD2 where D is the diameter of the circle; = =2 , but = ±0.01 so 2 = ±0.01, 2 4 A πD /4 D A D

dD = ±0.005; maximum permissible percentage error in D is ±0.5%.

D 66. V = x3 where x is the length of a side; approximate ∆V by dV if x = 1 and dx = ∆x = 0.02, dV = 3x2 dx = 3(1)2 (0.02) = 0.06 in3 .

67. V = volume of cylindrical rod = πr2 h = πr2 (15) = 15πr2 ; approximate ∆V by dV if r = 2.5 and dr = ∆r = 0.1. dV = 30πr dr = 30π(2.5)(0.1) ≈ 23.5619 cm3 .

2π √ π dP 2π 1 1 dL 1 68. P = √ L, dP = √ √ dL = √ √ dL, = so the relative error in P ≈ the relative error in L. g g2 L P 2 L 2 g L 1 Thus the percentage error in P is ≈ the percentage error in L. 2 69. Differentiating R = log10 (A/A0 ), we obtain 70. Differentiation gives dT = − dT = −

dR 1 1 dA dA = . Thus dR = , and ∆R ≈ dR ≈ 0.4343 . dA A ln 10 ln 10 A A

10 ln 2 dV1 . Using V1 = 33 and dV1 = ∆V1 = ±0.4, we obtain that ∆T ≈ (ln V1 − ln 20)2 V1

10 ln 2 ±0.4 ≈ ±0.3 days. (ln 33 − ln 20)2 33

Exercise Set 3.6 x2 − 4 (x − 2)(x + 2) x+2 2 = lim = lim = or, using L’Hôpital’s rule, 2 x→2 x + 2x − 8 x→2 (x + 4)(x − 2) x→2 x + 4 3

1. (a) lim lim

x→2 x2

x2 − 4 2x 2 = lim = . + 2x − 8 x→2 2x + 2 3

5 2 − lim 2 2x − 5 2 2 2x − 5 x→+∞ x (b) lim = = or, using L’Hôpital’s rule, lim = lim = .

7 x→+∞ 3x + 7 x→+∞ x→+∞ 3 3x + 7 3 3 3 + lim x→+∞ x 2. (a)

(b)

sin x sin x sin x cos x = cos x so lim = lim cos x = 1 or, using L’Hôpital’s rule, lim = lim = 1. x→0 tan x x→0 x→0 tan x x→0 sec2 x tan x x2 − 1 (x − 1)(x + 1) x+1 x2 − 1 2 = = so lim = or, using L’Hôpital’s rule, x→1 x3 − 1 x3 − 1 (x − 1)(x2 + x + 1) x2 + x + 1 3

112

Chapter 3 x2 − 1 2 2x = lim 2 = . x→1 x3 − 1 x→1 3x 3 lim

3. True; ln x is not defined for negative x.

4. True; apply L’Hôpital’s rule n times, where n = deg p(x).

5. False; apply L’Hôpital’s rule n times.

6. True; the logarithm of the expression approaches −∞. ex = 1. x→0 cos x

7. lim

8. lim

x→0

2 cos 2x 2 = .

5 cos 5x 5

sec2 θ = 1. θ→0 1

9. lim

tet + et = −1. t→0 −et

10. lim

11. lim+

cos x = −1.

1

12. lim+

cos x = +∞.

2x

x→π

x→0

13.

14.

lim

x→+∞

1/x = 0.

1

3e3x 9e3x = lim = +∞. x→+∞ 2x x→+∞ 2 lim

15. lim

−x − csc2 x −1 = lim = −∞. = lim 2 + + 1/x x→0 sin x x→0 2 sin x cos x

16. lim+

−1/x x = lim+ 1/x = 0. x→0 e (−1/x2 )e1/x

x→0+

x→0

17.

100x99 (100)(99)x98 (100)(99)(98) · · · (1) = lim = · · · = lim = 0. x x→+∞ x→+∞ x→+∞ e ex ex lim

18. lim

x→0+

cos x/ sin x = lim cos2 x = 1. sec2 x/ tan x x→0+

√ 2/ 1 − 4x2 = 2. x→0 1

19. lim

20. lim

1−

x→0

21.

1 1 1 1 + x2 = lim = .

2 2 x→0 3x 3(1 + x ) 3 x 1 = lim x = 0. x x→+∞ e x→+∞ e

lim xe−x = lim

x→+∞

Exercise Set 3.6

113

22. lim (x − π) tan(x/2) = lim x→π

23.

x→π

(−π/x2 ) cos(π/x) sin(π/x) = lim π cos(π/x) = π. = lim x→+∞ x→+∞ x→+∞ 1/x −1/x2

lim x sin(π/x) = lim

x→+∞

24. lim+ tan x ln x = lim+ x→0

25.

x−π 1 = lim = −2. cot(x/2) x→π −(1/2) csc2 (x/2)

x→0

lim

x→(π/2)−

ln x 1/x − sin2 x −2 sin x cos x = lim+ = lim = lim+ = 0. cot x x→0 − csc2 x x→0+ x 1 x→0

sec 3x cos 5x =

lim

x→(π/2)−

1 x−π = lim = 1. x→π sec2 x tan x

26. lim (x − π) cot x = lim x→π

cos 5x −5 sin 5x −5(+1) 5 = lim − = =− . cos 3x x→(π/2) −3 sin 3x (−3)(−1) 3

x→π

27. y = (1 − 3/x)x , lim ln y = lim x→+∞

x→+∞

28. y = (1 + 2x)−3/x , lim ln y = lim − x→0

x→0

ln(1 − 3/x) −3 = lim = −3, lim y = e−3 . x→+∞ 1 − 3/x x→+∞ 1/x

6 3 ln(1 + 2x) = lim − = −6, lim y = e−6 . x→0 x→0 x 1 + 2x

ln(ex + x) ex + 1 = lim x = 2, lim y = e2 . x→0 x→0 e + x x→0 x

29. y = (ex + x)1/x , lim ln y = lim x→0

30. y = (1 + a/x)bx , lim ln y = lim x→+∞

x→+∞

b ln(1 + a/x) ab = lim = ab, lim y = eab . x→+∞ 1 + a/x x→+∞ 1/x

ln(2 − x) 2 sin2 (πx/2) = lim = 2/π, lim y = e2/π . x→1 cot(πx/2) x→1 x→1 π(2 − x)

31. y = (2 − x)tan(πx/2) , lim ln y = lim x→1

ln cos(2/x) (−2/x2 )(− tan(2/x)) − tan(2/x) = = lim = lim 2 x→+∞ x→+∞ x→+∞ 1/x −2/x3 1/x

2

32. y = [cos(2/x)]x , lim ln y = lim x→+∞

2

lim

x→+∞

33. lim

x→0

2

(2/x ) sec (2/x) = −2, −1/x2 1 1 − sin x x

= lim

x→0

lim y = e−2 .

x→+∞

x − sin x 1 − cos x sin x = lim = lim = 0. x→0 x cos x + sin x x→0 2 cos x − x sin x x sin x

9 9 1 − cos 3x 3 sin 3x = lim cos 3x = . = lim 2 x→0 x→0 x→0 2 x 2x 2

34. lim

35.

(x2 + x) − x2 x 1 √ = lim √ = lim p = 1/2. x→+∞ x→+∞ x2 + x + x x2 + x + x x→+∞ 1 + 1/x + 1 lim

ex − 1 ex ex − 1 − x = lim x = lim x = 1/2. x x x→0 xe − x x→0 xe + e − 1 x→0 xe + 2ex

36. lim

37.

lim [x − ln(x2 + 1)] = lim [ln ex − ln(x2 + 1)] = lim ln

x→+∞

x→+∞

so lim [x − ln(x2 + 1)] = +∞ x→+∞

38.

lim ln

x→+∞

x 1 = lim ln = ln(1) = 0.

1 + x x→+∞ 1/x + 1

x→+∞

ex ex ex ex , lim 2 = lim = lim = +∞, + 1 x→+∞ x + 1 x→+∞ 2x x→+∞ 2

x2

114

Chapter 3

39. y = xsin x , ln y = sin x ln x, lim+ ln y = lim+ x→0

x→0

lim xsin x = lim+ y = e0 = 1.

x→0+

ln x 1/x = lim+ = lim csc x x→0 − csc x cot x x→0+

sin x x

(− tan x) = 1(−0) = 0, so

x→0

ln(e2x − 1) 2e2x = lim+ 2x (−x2 ) = 1/x −1 x→0 x→0 x→0 e 1 x 1 = lim+ 2x lim (−2xe2x ) = lim+ 2x lim+ (−2xe2x ) = · 0 = 0, lim+ y = e0 = 1. − 1 x→0+ 2 x→0 x→0 e x→0 2e x→0

x

1 ln − ln1x 1 1 x , ln y = x ln − ln x , lim ln y = lim 41. y = − = lim − (−x2 ) = − lim = 0, so ln x 1/x x ln x x→0+ x→0+ x→0+ x→0+ ln x 0 lim+ y = e = 1.

40. y = (e2x − 1)x , ln y = x ln(e2x − 1), lim+ ln y = lim+

x→0

42. y = x1/x , ln y =

ln x 1/x ln x , lim ln y = lim = lim = 0, so lim y = e0 = 1. x→+∞ x→+∞ x→+∞ x→+∞ x x 1

43. y = (ln x)1/x , ln y = (1/x) ln ln x, lim ln y = lim x→+∞

x→+∞

1/(x ln x) ln ln x = lim = 0, so lim y = 1. x→+∞ x→+∞ x 1

44. y = (− ln x)x , ln y = x ln(− ln x), lim+ ln y = lim+ ln(− ln x)/(1/x) = lim+ x→0

x→0

x→0

(1/(x ln x) x = lim+ (− ) = 0, so (−1/x2 ) ln x x→0

lim y = 1.

x→0+

45. y = (tan x)π/2−x , ln y = (π/2 − x) ln tan x, lim

x→(π/2)−

46. (a)

(b)

lim

x→(π/2)

ln y = −

lim

x→(π/2)−

ln tan x = 1/(π/2 − x)

(π/2 − x) (π/2 − x) (π/2 − x) (π/2 − x) = lim − lim = 1 · 0 = 0, so cos x sin x cos x x→(π/2)− sin x x→(π/2)

lim

x→+∞

lim

x→(π/2)−

lim

x→(π/2)−

(sec2 x/ tan x) = 1/(π/2 − x)2

y = 1.

ln x 1/x 1 = lim = lim = 0. n n−1 x→+∞ x→+∞ x nx nxn

xn nxn−1 = lim = lim nxn = +∞. x→+∞ ln x x→+∞ 1/x x→+∞ lim

3x2 − 2x + 1 because it is not an indeterminate form. x→1 3x2 − 2x

47. (a) L’Hôpital’s rule does not apply to the problem lim 3x2 − 2x + 1 = 2. x→1 3x2 − 2x

(b) lim

2

e3x −12x+12 , because it is not an indeterminate form. x→2 x4 − 16

48. (a) L’Hôpital’s rule does not apply to the problem lim (b)

lim and lim+ exist, with values −∞ if x approaches 2 from the left and +∞ if from the right. The general

x→2−

x→2

limit lim does not exist. x→2

0.15

49.

2 1/(x ln x) √ = lim √ = 0. x→+∞ 1/(2 x) x→+∞ x ln x lim

100

0

10000

Exercise Set 3.6

115 1

50. y = xx , lim+ ln y = lim+ x→0

x→0

ln x = lim −x = 0, lim+ y = 1.

1/x x→0+ x→0

0

0.5

0

25

51. y = (sin x)3/ ln x , lim+ ln y = lim+ x→0

x→0

3 ln sin x x = lim+ (3 cos x) = 3, lim+ y = e3 . ln x sin x x→0 x→0

0

19

0.5

4.1

52.

lim −

x→π/2

4 4 sec2 x = lim − = 4. sec x tan x x→π/2 sin x

1.4

1.6 3.3

e−x ln x − 1 ln x ; lim e−x ln x = lim = −x −x x→+∞ x→+∞ ex e e −x e ln x − 1 lim [ln x − ex ] = lim = −∞; no horizontal asymptote. x→+∞ x→+∞ e−x

53. ln x − ex = ln x −

0

1

0

=

1/x = 0 by L’Hôpital’s rule, so x→+∞ ex lim

3

–16

54.

lim [ln ex − ln(1 + 2ex )] = lim ln

x→+∞

ex lim ln = −∞. x→−∞ 1 + 2ex –0.6 0

–1.2

x→+∞

12

ex 1 1 = ln ; horizontal asymptote y = − ln 2. Also, = lim ln −x x→+∞ 1 + 2ex e +2 2

116

Chapter 3 ln(ln x) 1 = lim = 0; lim y = 1, y = 1 is the horizontal asymptote. x→+∞ x→+∞ x ln x x→+∞ x

55. y = (ln x)1/x , lim ln y = lim x→+∞

1.02

100

10000

1

x+1 56. y = x+2 asymptote.

ln

x ,

lim ln y =

x→±∞

lim

x→±∞

x+1 x+2 = 1/x

−x2 = −1; x→±∞ (x + 1)(x + 2) lim

lim y = e−1 is the horizontal

x→±∞

1

0

50 0

57. (a) 0

(b) +∞

(c) 0

(d)

−∞

58. (a) Type 00 ; y = x(ln a)/(1+ln x) ; lim+ ln y = lim+ x→0

x→0

(e)

+∞

(f )

−∞

(ln a) ln x (ln a)/x = lim+ = lim+ ln a = ln a, so we obtain 1 + ln x 1/x x→0 x→0

that lim+ y = eln a = a. x→0

(b) Type ∞0 ; same calculation as part (a) with x → +∞. (c) Type 1∞ ; y = (x + 1)(ln a)/x , lim ln y = lim x→0

x→0

(ln a) ln(x + 1) ln a = lim = ln a, so limx→0 y = eln a = a. x→0 x + 1 x

59.

1 + 2 cos 2x x + sin 2x sin 2x does not exist, nor is it ±∞; lim = lim 1+ = 1. x→+∞ x→+∞ x→+∞ 1 x x

60.

2 − cos x 2x − sin x 2 − (sin x)/x 2 does not exist, nor is it ±∞; lim = lim = . x→+∞ 3 + cos x x→+∞ 3x + sin x x→+∞ 3 + (sin x)/x 3

61.

lim

lim

lim (2 + x cos 2x + sin 2x) does not exist, nor is it ±∞; lim

x→+∞

x→+∞

x(2 + sin 2x) 2 + sin 2x = lim , which does not x→+∞ 1 + 1/x x+1

exist because sin 2x oscillates between −1 and 1 as x → +∞.

62.

lim

x→+∞

63. lim

V t −Rt/L L e

R→0+

64. (a)

1 1 sin x + cos x + x 2 2x

1

=

does not exist, nor is it ±∞; lim

x→+∞

Vt .

L

lim (π/2 − x) tan x = lim

x→π/2

x(2 + sin x) 2 + sin x = lim = 0. x→+∞ x + 1/x x2 + 1

x→π/2

π/2 − x −1 = lim = lim sin2 x = 1. cot x x→π/2 − csc2 x x→π/2

Chapter 3 Review Exercises

(b)

lim

x→π/2

= lim

x→π/2

(c) 65. (b)

117

1 − tan x π/2 − x

= lim

x→π/2

1 sin x − π/2 − x cos x

cos x − (π/2 − x) sin x = (π/2 − x) cos x x→π/2

= lim

−(π/2 − x) cos x (π/2 − x) sin x + cos x = lim = 0 (by applying L’H’s rule twice). −(π/2 − x) sin x − cos x x→π/2 −(π/2 − x) cos x + 2 sin x

1/(π/2 − 1.57) ≈ 1255.765534, tan 1.57 ≈ 1255.765592; 1/(π/2 − 1.57) − tan 1.57 ≈ 0.000058. lim x(k 1/x − 1) = lim+

x→+∞

t→0

(c) ln 0.3 = −1.20397, 1024

(ln k)k t kt − 1 = lim+ = ln k. t 1 t→0 √

1024

0.3 − 1 = −1.20327; ln 2 = 0.69315, 1024

√

1024

2 − 1 = 0.69338.

k + cos `x = ±∞. Hence k = −1, and by the rule x2 √ −1 + cos `x −` sin `x −`2 cos `x `2 lim = lim = lim = − = −4 if ` = ±2 2.

2 x→0 x→0 x→0 x 2x 2 2

66. If k 6= −1 then lim (k + cos `x) = k + 1 6= 0, so lim x→0

x→0

67. (a) No; sin(1/x) oscillates as x → 0.

0.05

–0.35

(b)

0.35

–0.05

(c) For the limit as x → 0+ use the Squeezing Theorem together with the inequalities −x2 ≤ x2 sin(1/x) ≤ x2 .

For x → 0− do the same; thus lim f (x) = 0. x→0

68. (a) Apply the rule to get lim

x→0

− cos(1/x) + 2x sin(1/x) which does not exist (nor is it ±∞). cos x

h x i x 1 [x sin(1/x)], but lim = lim = 1 and lim x sin(1/x) = 0, thus x→0 sin x x→0 sin x x→0 cos x x→0

(b) Rewrite as lim

h x i [x sin(1/x)] = (1)(0) = 0. x→0 sin x lim

sin x sin(1/x) , lim = 1 but lim+ sin(1/x) does not exist because sin(1/x) oscillates between −1 and 1 as (sin x)/x x→0+ x x→0 x sin(1/x) x → +∞, so lim+ does not exist. sin x x→0

69. lim+ x→0

70. Since f (a) = g(a) = 0, then for x 6= a, (f (x) − f (a)/(x − a) f 0 (a) = 0 . x→a (g(x) − g(a))/(x − a) g (a)

f (x) (f (x) − f (a)/(x − a) f (x) = . Now take the limit: lim = x→a g(x) g(x) (g(x) − g(a))/(x − a)

lim

Chapter 3 Review Exercises 1. (a) 3x2 + x

dy dy 2 − y − 3x2 + y − 2 = 0, = . dx dx x

(b) y = (1 + 2x − x3 )/x = 1/x + 2 − x2 , dy/dx = −1/x2 − 2x.

118

Chapter 3 dy 2 − (1/x + 2 − x2 ) − 3x2 = = −1/x2 − 2x. dx x

(c)

2. (a) xy = x − y, x

dy dy 1−y dy +y =1− , = . dx dx dx x+1

(b) y(x + 1) = x, y =

x 1 − x+1 dy 1−y 1 . = = = dx x+1 1+x (x + 1)2

(c)

3. −

dy 1 dy 1 y2 − 2 = 0 so = − 2.

2 y dx x dx x

4. 3x2 − 3y 2

5.

x 1 . , y0 = x+1 (x + 1)2

x

dy dy x2 − 2y dy dy = 6(x + y), −(3y 2 + 6x) = 6y − 3x2 so = 2 . dx dx dx dx y + 2x

dy dy y sec(xy) tan(xy) dy + y sec(xy) tan(xy) = , = . dx dx dx 1 − x sec(xy) tan(xy)

(1 + csc y)(− csc2 y)(dy/dx) − (cot y)(− csc y cot y)(dy/dx) dy , 2x(1+csc y)2 = − csc y(csc y+csc2 y−cot2 y) , (1 + csc y)2 dx 2x(1 + csc y) dy =− . but csc2 y − cot2 y = 1, so dx csc y

6. 2x =

7.

dy 3x d2 y (4y)(3) − (3x)(4dy/dx) 12y − 12x(3x/(4y)) 12y 2 − 9x2 −3(3x2 − 4y 2 ) = , 2 = = = = , but 3x2 − 4y 2 = dx 4y dx 16y 2 16y 2 16y 3 16y 3 d2 y −3(7) 21 7 so = =− . dx2 16y 3 16y 3

8.

y d2 y (y − x)(dy/dx) − y(dy/dx − 1) dy = , = = 2 dx y − x dx (y − x)2 d2 y 3 2xy = −3, so =− .

2 dx (y − x)3

(y − x)

y y −y −1 y 2 − 2xy y−x y−x = , but y 2 − (y − x)2 (y − x)3

dy dy dy

dy

dy

2 2 9. = tan(πy/2) + x(π/2) sec (πy/2), = 1 + (π/4)

(2), = . dx dx dx y=1/2 dx y=1/2 dx y=1/2 2−π 10. Let P (x0 , y0 ) be the required point. The slope of the line 4x − 3y + 1 = 0 is 4/3 so the slope of the tangent to y 2 = 2x3 at P must be −3/4. By implicit differentiation dy/dx = 3x2 /y, so at P , 3x20 /y0 = −3/4, or y0 = −4x20 .

But y02 = 2x30 because P is on the curve y 2 = 2x3 . Elimination of y0 gives 16x40 = 2x30 , x30 (8x0 − 1) = 0, so x0 = 0 or 1/8. From y0 = −4x20 it follows that y0 = 0 when x0 = 0, and y0 = −1/16 when x0 = 1/8. It does not follow, however, that (0, 0) is a solution because dy/dx = 3x2 /y (the slope of the curve as determined by implicit differentiation) is valid only if y 6= 0. Further analysis shows that the curve is tangent to the x-axis at (0, 0), so the point (1/8, −1/16) is the only solution.

11. Substitute y = mx into x2 + xy + y 2 = 4 to get x2 + mx2 + m2 x2 = 4, which has distinct solutions x = √ 2 ±2/ m + m + 1. They are distinct because m2 + m + 1 = (m + 1/2)2 + 3/4 ≥ 3/4, so m2 + m + 1 is never zero.

Note that the points of intersection occur in pairs (x0 , y0 ) and (−x0 , −y0 ). By implicit differentiation, the slope of the tangent line to the ellipse is given by dy/dx = −(2x + y)/(x + 2y). Since the slope is unchanged if we replace (x, y) with (−x, −y), it follows that the slopes are equal at the two point of intersection. Finally we must examine the special case x = 0 which cannot be written in the form y = mx. If x = 0 then y = ±2, and the formula for dy/dx gives dy/dx = −1/2, so the slopes are equal.

Chapter 3 Review Exercises

119

12. By implicit differentiation, 3x2 − y − xy 0 + 3y 2 y 0 = 0, so y 0 = (3x2 − y)/(x − 3y 2 ). This derivative is zero when y = 3x2 . Substituting this into the original equation x3 −xy+y 3 = 0, one has x3 −3x3 +27x6 = 0, x3 (27x3 −2) = 0.

The unique solution in the first quadrant is x = 21/3 /3, y = 3x2 = 22/3 /3.

13. By implicit differentiation, 3x2 −y −xy 0 +3y 2 y 0 = 0, so y 0 = (3x2 −y)/(x−3y 2 ). This derivative exists except when x = 3y 2 . Substituting this into the original equation x3 −xy +y 3 = 0, one has 27y 6 −3y 3 +y 3 = 0, y 3 (27y 3 −2) = 0.

The unique solution in the first quadrant is y = 21/3 /3, x = 3y 2 = 22/3 /3 14. By implicit differentiation, dy/dx = k/(2y) so the slope of the tangent to y 2 = kx at (x0 , y0 ) is k/(2y0 ) if y0 6= 0. k (x − x0 ), or 2y0 y − 2y02 = kx − kx0 . But y02 = kx0 because (x0 , y0 ) The tangent line in this case is y − y0 = 2y0 is on the curve y 2 = kx, so the equation of the tangent line becomes 2y0 y − 2kx0 = kx − kx0 which gives y0 y = k(x + x0 )/2. If y0 = 0, then x0 = 0; the graph of y 2 = kx has a vertical tangent at (0, 0) so its equation is x = 0, but y0 y = k(x + x0 )/2 gives the same result when x0 = y0 = 0.

15. y = ln(x + 1) + 2 ln(x + 2) − 3 ln(x + 3) − 4 ln(x + 4), dy/dx = 16. y =

1 2 3 4 + − − . x+1 x+2 x+3 x+4

1 dy 1 1 cos x sin x 5x + 3 1 ln x + ln(x + 1) − ln sin x + ln cos x, so = + − − = − cot x − tan x.

2 3 dx 2x 3(x + 1) sin x cos x 6x(x + 1)

dy 1 = (2) = 1/x. dx 2x

dy 1 2 ln x 18. = 2(ln x) = . dx x x 17.

19.

dy 1 = . dx 3x(ln x + 1)2/3

20. y =

21.

1 1 ln(x + 1), y 0 = .

3 3(x + 1)

ln ln x 0 1 dy = log10 ln x = ,y = . dx ln 10 (ln 10)(x ln x)

22. y =

1 + ln x/ ln 10 ln 10 + ln x 0 (ln 10 − ln x)/x + (ln 10 + ln x)/x 2 ln 10 = ,y = = .

1 − ln x/ ln 10 ln 10 − ln x (ln 10 − ln x)2 x(ln 10 − ln x)2

23. y =

1 3 2x3 3 ln x + ln(1 + x4 ), y 0 = + .

2 2 2x (1 + x4 )

24. y =

1 sin x 2x 1 − 3x2 1 ln x + ln cos x − ln(1 + x2 ), y 0 = − − = − tan x.

2 2x cos x 1 + x2 2x(1 + x2 )

25. y = x2 + 1 so y 0 = 2x.

26. y = ln

(1 + ex + e2x ) dy ex = − ln(1 − ex ), = . x x 2x (1 − e )(1 + e + e ) dx 1 − ex

27. y 0 = 2e 28. y 0 =

√

x

√

+ 2xe

abe−x .

(1 + be−x )2

x

√ √ √ d √ x = 2e x + xe x . dx

120

Chapter 3

29. y 0 =

2 . π(1 + 4x2 )

30. y = e(sin

−1

x) ln 2

y0 = ex y

31. ln y = ex ln x,

−1 ln 2 2 sin x .

2 1−x

h x i x x 1 dy 1 + ln x , = xe ex + ln x = ex xe −1 + xe ln x . x dx x

(1/x) ln(1 + x) y 0 1 x/(1 + x) − ln(1 + x) ln(1 + x) dy 1 (1/x)−1 (1 + x) = , ln(1+x). , = − = (1+x) − x y x2 x(1 + x) x2 dx x x2

32. ln y = 33. y 0 =

, y0 = √

2 p . |2x + 1| (2x + 1)2 − 1

1 1 d x √ 34. y 0 = √ cos−1 x2 = − √ . −1 2 −1 2 1 − x4 2 cos x dx cos x 35. ln y = 3 ln x −

1 3 x 3x2 x4 ln(x2 + 1), y 0 /y = − 2 , y0 = √ − 2 .

2 x x +1 x2 + 1 (x + 1)3/2

1 y0 1 = 36. ln y = (ln(x2 − 1) − ln(x2 + 1)), 3 y 3

2x 2x − x2 − 1 x2 + 1

4x 4x = so y 0 = 3(x4 − 1) 3(x4 − 1)

r 3

x2 − 1 . x2 + 1

y 6 4 2 x 1

37. (b) (c)

2

3

4

1 1 dy dy dy = − , so < 0 at x = 1 and > 0 at x = e. dx 2 x dx dx

(d) The slope is a continuous function which goes from a negative value to a positive value; therefore it must take the value zero between, by the Intermediate Value Theorem. (e)

dy = 0 when x = 2. dx

dβ 10 38. β = 10 log I − 10 log I0 , = . dI I ln 10

dβ

1 dB/(W/m2 ). (a) =

dI I=10I0 I0 ln 10

dβ

1 (b) = dB/(W/m2 ).

dI I=100I0 10I0 ln 10

dβ

1 (c) = dB/(W/m2 ). dI I=100I0 1000I0 ln 10 39. Solve

dy dx dy dy dx dx =3 given y = x ln x. Then = = (1 + ln x) , so 1 + ln x = 3, ln x = 2, x = e2 . dt dt dt dx dt dt

Chapter 3 Review Exercises

121

40. x = 2, y = 0; y 0 = −2x/(5 − x2 ) = −4 at x = 2, so y − 0 = −4(x − 2) or y = −4x + 8.

1 1 = 1 so x = . The curves intersect when (x, x) lies on the graph x ln b ln b ln x of y = logb x, so x = logb x. From Formula (8), Section 1.6, logb x = from which ln x = 1, x = e, ln b = 1/e, ln b 1/e b=e ≈ 1.4447.

41. Set y = logb x and solve y 0 = 1: y 0 =

y

2 x 2

42. (a) Find the point of intersection: f (x) = √ x = 4. Then ln 4 = 4 + k, k = ln 4 − 2.

√

x + k = ln x. The slopes are equal, so m1 =

1 √ 1 = m2 = √ , x = 2, x 2 x

y

2 x 2

(b) Since the slopes are equal m1 =

√ √ k 1 √ = m2 = , so k x = 2. At the point of intersection k x = ln x, x 2 x

2 = ln x, x = e2 , k = 2/e. y 2 x 0

5

43. As long as f 0 6= 0, g must be differentiable; this can be inferred from the graphs. Note that if f 0 = 0 at a point then g 0 cannot exist (infinite slope). (For example, f (x) = x3 at x = 0).

44. (a) f 0 (x) = −3/(x + 1)2 . If x = f (y) = 3/(y + 1) then y = f −1 (x) = (3/x) − 1, so 1

f 0 (f −1 (x))

=−

(f −1 (x) + 1)2 (3/x)2 3 =− = − 2.

3 3 x

d −1 3 f (x) = − 2 ; and dx x

d −1 2 (b) f (x) = ex/2 , f 0 (x) = 12 ex/2 . If x = f (y) = ey/2 then y = f −1 (x) = 2 ln x, so f (x) = ; and dx x −1 1 2 = 2e−f (x)/2 = 2e− ln x = 2x−1 = .

0 −1 f (f (x)) x 45. Let P (x0 , y0 ) be a point on y = e3x then y0 = e3x0 . dy/dx = 3e3x so mtan = 3e3x0 at P and an equation of the tangent line at P is y − y0 = 3e3x0 (x − x0 ), y − e3x0 = 3e3x0 (x − x0 ). If the line passes through the origin then (0, 0) must satisfy the equation so −e3x0 = −3x0 e3x0 which gives x0 = 1/3 and thus y0 = e. The point is (1/3, e).

46. ln y = ln 5000 + 1.07x;

dy/dx dy = 1.07, or = 1.07y. y dx

47. ln y = 2x ln 3 + 7x ln 5;

dy/dx dy = 2 ln 3 + 7 ln 5, or = (2 ln 3 + 7 ln 5)y. y dx

122

48.

Chapter 3

qk0 q(T − T0 ) dk q(T − T0 ) q = k0 exp − − 2 = − 2 exp − . dT 2T0 T 2T 2T 2T0 T

49. y 0 = aeax sin bx + beax cos bx, and y 00 = (a2 − b2 )eax sin bx + 2abeax cos bx, so

y 00 − 2ay 0 + (a2 + b2 )y = (a2 − b2 )eax sin bx + 2abeax cos bx − 2a(aeax sin bx + beax cos bx) + (a2 + b2 )eax sin bx = 0.

√ √ 50. sin(tan−1 x) = x/ 1 + x2 and cos(tan−1 x) = 1/ 1 + x2 , and y 0 = 2 sin y cos3 y =

−2x x 1 + 2√ = 0.

2 )3/2 2 (1 + x2 )2 (1 + x 1+x

−2x 1 , y 00 = , hence y 00 + 2 1+x (1 + x2 )2

100

51. (a)

0

8

20

(b) As t tends to +∞, the population tends to 19: lim P (t) = lim t→+∞

t→+∞

95 95 95 = = = 19. −t/4 5 5 − 4e−t/4 5 − 4 lim e t→+∞

(c) The rate of population growth tends to zero.

0

0

8

–80

(1 + h)π − 1 d = π. = π(1 + x)π−1

= (1 + x)π

h→0 h dx x=0 x=0

52. (a) y = (1 + x)π , lim

1 − ln x 1 − ln x dy

1/x 1 (b) Let y = . Then y(e) = 0, and lim = =− =− . x→e (x − e) ln x ln x dx x=e (ln x)2 e 53. In the case +∞ − (−∞) the limit is +∞; in the case −∞ − (+∞) the limit is −∞, because large positive (negative) quantities are added to large positive (negative) quantities. The cases +∞ − (+∞) and −∞ − (−∞) are indeterminate; large numbers of opposite sign are subtracted, and more information about the sizes is needed.

54. (a) When the limit takes the form 0/0 or ∞/∞. (b) Not necessarily; only if lim f (x) = 0. Consider g(x) = x; lim g(x) = 0. Then lim x→a x→0 x→0 sin x whereas lim is indeterminate. x→0 x 55.

ex ex ex = lim = lim = +∞, so lim (ex /x2 − 1) = +∞ 2 x→+∞ x x→+∞ 2x x→+∞ 2 x→+∞

lim (ex − x2 ) = lim x2 (ex /x2 − 1), but lim

x→+∞

x→+∞

and thus lim x2 (ex /x2 − 1) = +∞. x→+∞

ln x 1/x 1 56. lim 4 = lim = ; lim 3 x→1 x − 1 x→1 4x 4 x→1

r

cos x is not indeterminate, x

ln x = 4 x −1

r lim

ln x 1 = . −1 2

x→1 x4

Chapter 3 Making Connections

123

2

x2 ex 3x ex 1 57. lim = lim lim = . x→0 sin2 3x x→0 sin 3x x→0 9 9 58. lim ax ln a = ln a. x→0

59. The boom is pulled in at the rate of 5 m/min, so the circumference C = 2rπ is changing at this rate, which means dr dC 1 dr dA dA dr that = · = −5/(2π). A = πr2 and = −5/(2π), so = = 2πr(−5/2π) = −250, so the area dt dt 2π dt dt dr dt 2 is shrinking at a rate of 250 m /min.

√ dz dy dθ

given = a and = −b. From the figure sin θ = y/z; when x = y = 1, z = 2. So θ = sin−1 (y/z)

dt x=1 dt dt y=1

1 dy dθ 1 y dz a p and = − 2 = −b − √ when x = y = 1.

2 2 dt z dt z dt 2 1 − y /z

60. Find

z

y

θ

x

61. (a) ∆x = 1.5 − 2 = −0.5; dy =

−1 1 −1 1 ∆x = (−0.5) = 0.5; and ∆y = − = 2 − 1 = 1.

2 2 (x − 1) (2 − 1) (1.5 − 1) (2 − 1)

(b) ∆x = 0 − (−π/4) = π/4; dy = sec2 (−π/4) (π/4) = π/2; and ∆y = tan 0 − tan(−π/4) = 1. (c) ∆x = 3 − 0 = 3; dy = √

√ √ −x −0 (3) = 0; and ∆y = 25 − 32 − 25 − 02 = 4 − 5 = −1. =p 2 2 25 − x 25 − (0)

46π π 46π ; let x0 = and x = . Then 180 4 180

π 2 π π 46π π ◦ cot 46 = cot x ≈ cot − csc x− =1−2 − = 0.9651; with a calculator, cot 46◦ = 0.9657.

4 4 4 180 4

62. cot 46◦ = cot

π 51 π radians and dφ = ±0.5◦ = ±0.5 radians, 180 180 h ± dh = 115(1.2349) ± 2.5340 = 142.0135 ± 2.5340, so the height lies between 139.48 m and 144.55 m.

63. (a) h = 115 tan φ, dh = 115 sec2 φ dφ; with φ = 51◦ =

(b)

If |dh| ≤ 5 then |dφ| ≤

5 51 cos2 π ≈ 0.017 radian, or |dφ| ≤ 0.98◦ .

115 180

Chapter 3 Making Connections 1. (a) If t > 0 then A(−t) is the amount K there was t time-units ago in order that there be 1 unit now, i.e.

1 1 K · A(t) = 1, so K = . But, as said above, K = A(−t). So A(−t) = .

A(t) A(t) (b) If s and t are positive, then the amount 1 becomes A(s) after s seconds, and that in turn is A(s)A(t) after another t seconds, i.e. 1 becomes A(s)A(t) after s + t seconds. But this amount is also A(s + t), so A(s)A(t) = A(s+t). Now if 0 ≤ −s ≤ t then A(−s)A(s+t) = A(t). From the first case, we get A(s+t) = A(s)A(t).

1 1 If 0 ≤ t ≤ −s then A(s + t) = = = A(s)A(t) by the previous cases. If s and t are both A(−s − t) A(−s)A(−t) 1 1 negative then by the first case, A(s + t) = = = A(s)A(t).

A(−s − t) A(−s)A(−t)

Calculus Early Transcendentals 11th Edition Anton Solutions Manual Full Download: https://alibabadownload.com/product/calculus-early-transcendentals-11th-edition-anton-solutions-manual/ 124

Chapter 3

1 1 1 1 1 A ...A =A n = A(1), so A = A(1)1/n = b1/n from part (b). If n n n n n 1 1 1 = = A(1)1/n = b1/n . n < 0 then by part (a), A = n A(1)−1/n A − n1

(c) If n > 0 then A

(d) Let m, n be integers. Assume n 6= 0 and m > 0. Then A

m n

=A

m 1 = A(1)m/n = bm/n . n

(e) If f, g are continuous functions of t and f and g are equal on the rational numbers

nm

o : n 6= 0 , then

n f (t) = g(t) for all t. Because if x is irrational, then let tn be a sequence of rational numbers which converges to x. Then for all n > 0, f (tn ) = g(tn ) and thus f (x) = lim f (tn ) = lim g(tn ) = g(x) . n→+∞

n→+∞

2. (a) From Figure 1.3.4 it is evident that (1+h)1/h < e < (1−h)−1/h provided h > 0, and (1−h)−1/h < e < (1+h)1/h for h < 0. (b) Suppose h > 0. Then (1 + h)1/h < e < (1 − h)−1/h . Raise to the power h: 1 + h < eh < 1/(1 − h); h < eh − 1 < 1/(1 − h); use the Squeezing Theorem as h → 0+ . Use a similar argument in the eh − 1 < h/(1 − h); 1 < h case h < 0. eh − 1 (c) The quotient is the slope of the secant line through (0, 1) and (h, eh ), and this secant line converges h to the tangent line as h → 0. (d)

d x ex+h − ex eh − 1 e = lim = ex lim = ex from part (b). h→0 h→0 dx h h

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