Chapter 03

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CHAPTER 3

The Derivative EXERCISE SET 3.1 1. (a) mtan = (50 − 10)/(15 − 5) = 40/10 = 4 m/s

v (m/s)

(b) 4

t (s) 10

20

2. (a) mtan ≈ (90 − 0)/(10 − 2) = 90/8 = 11.25 m/s

(b) mtan ≈ (140 − 0)/(10 − 4) = 140/6 ≈ 23.33 m/s

3. (a) mtan = (600 − 0)/(20 − 2.2) = 600/17.8 ≈ 33.71 m/s

(b) mtan ≈ (820 − 600)/(20 − 16) = 220/4 = 55 m/s The speed is increasing with time.

4. (a) (b) (c) (d)

(10 − 10)/(3 − 0) = 0 cm/s t = 0, t = 2, and t = 4.2 (horizontal tangent line) maximum: t = 1 (slope > 0) minimum: t = 3 (slope < 0) (3 − 18)/(4 − 2) = −7.5 cm/s (slope of estimated tangent line to curve at t = 3)

5. From the figure: s

t t0 t1

t2

(a) The particle is moving faster at time t0 because the slope of the tangent to the curve at t0 is greater than that at t2 . (b) The initial velocity is 0 because the slope of a horizontal line is 0. (c) The particle is speeding up because the slope increases as t increases from t0 to t1 . (d) The particle is slowing down because the slope decreases as t increases from t1 to t2 .

6.

s

t t0

t1

7. It is a straight line with slope equal to the velocity.

8. (a) (b) (c) (d)

decreasing (slope of tangent line decreases with increasing time) increasing (slope of tangent line increases with increasing time) increasing (slope of tangent line increases with increasing time) decreasing (slope of tangent line decreases with increasing time) 75

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76

Chapter 3

9. (a) msec =

2 f (1) − f (0) = =2 1−0 1 f (x1 ) − f (0) 2x21 − 0 = lim = lim 2x1 = 0 x1 →0 x1 →0 x1 − 0 x1 →0 x1 − 0

(b)

mtan = lim

(c)

mtan = lim

x1 →x0

f (x1 ) − f (x0 ) x1 − x0

y

(d) 3

2x21 − 2x20 = lim x1 →x0 x1 − x0

Tangent

2 1

= lim (2x1 + 2x0 )

x

x1 →x0

–2

–1

= 4x0

1 –1

2

Secant

f (2) − f (1) 23 − 13 = =7 2−1 1 f (x1 ) − f (1) x3 − 1 (x1 − 1)(x21 + x1 + 1) = lim = lim 1 = lim x1 →1 x1 →1 x1 − 1 x1 →1 x1 − 1 x1 − 1

10. (a) msec = (b)

mtan

= lim (x21 + x1 + 1) = 3 x1 →1

(c)

mtan = lim

x1 →x0

= lim

x1 →x0

f (x1 ) − f (x0 ) x1 − x0

y

(d)

9

x31 − x30 x1 − x0

Tangent

= lim (x21 + x1 x0 + x20 ) x1 →x0

x

= 3x20 Secant

5

f (3) − f (2) 1/3 − 1/2 1 = =− 3−2 1 6 f (x1 ) − f (2) 1/x1 − 1/2 = lim = lim x1 →2 x1 →2 x1 − 2 x1 − 2

11. (a) msec = (b)

mtan

= lim

x1 →2

(c)

mtan = lim

2 − x1 −1 1 = lim =− 2x1 (x1 − 2) x1 →2 2x1 4

x1 →x0

f (x1 ) − f (x0 ) x1 − x0

1/x1 − 1/x0 = lim x1 →x0 x1 − x0 x0 − x1 = lim x1 →x0 x0 x1 (x1 − x0 ) −1 1 = lim =− 2 x1 →x0 x0 x1 x0

y

(d) 4

Secant

x 1

Tangent

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Exercise Set 3.1

77

1/4 − 1 3 f (2) − f (1) = =− 2−1 1 4 f (x1 ) − f (1) 1/x21 − 1 = lim = lim x1 →1 x1 →1 x1 − 1 x1 − 1

12. (a) msec = (b)

mtan

1 − x21 −(x1 + 1) = lim = −2 x1 →1 x2 x →1 (x − 1) x21 1 1 1

= lim (c)

mtan = lim

x1 →x0

= lim

1/x21 − 1/x20 x1 − x0

= lim

x20 − x21 2 x0 x21 (x1 − x0 )

x1 →x0

x1 →x0

= lim

x1 →x0

13. (a)

f (x1 ) − f (x0 ) x1 − x0

mtan = lim

x1 →x0

= lim

x1 →x0

y

(d)

1 x 2

−(x1 + x0 ) 2 =− 3 2 2 x0 x1 x0

Tangent

Secant

f (x1 ) − f (x0 ) (x21 − 1) − (x20 − 1) = lim x1 →x0 x1 − x0 x1 − x0 (x21 − x20 ) = lim (x1 + x0 ) = 2x0 x1 →x0 x1 − x0

(b) mtan = 2(−1) = −2 14. (a)

mtan = lim

f (x1 ) − f (x0 ) (x21 + 3x1 + 2) − (x20 + 3x0 + 2) = lim x1 →x0 x1 − x0 x1 − x0

= lim

(x21 − x20 ) + 3(x1 − x0 ) = lim (x1 + x0 + 3) = 2x0 + 3 x1 →x0 x1 − x0

x1 →x0

x1 →x0

(b) mtan = 2(2) + 3 = 7 15. (a)

mtan

f (x1 ) − f (x0 ) = lim = lim x1 →x0 x1 →x0 x1 − x0 = lim √ x1 →x0

√

√ x1 − x0 x1 − x0

1 1 √ = √ 2 x0 x1 + x0

1 1 (b) mtan = √ = 2 2 1 16. (a)

mtan

√ √ 1/ x1 − 1/ x0 f (x1 ) − f (x0 ) = lim = lim x1 →x0 x1 →x0 x1 − x0 x1 − x0 √ √ x0 − x1 −1 1 = lim √ √ = lim √ √ = − 3/2 √ √ x1 →x0 x →x x0 x1 (x1 − x0 ) x0 x1 ( x1 + x0 ) 1 0 2x0

(b) mtan = −

1 1 =− 3/2 16 2(4)

17. (a) 72◦ F at about 4:30 P.M. (b) about (67 − 43)/6 = 4◦ F/h (c) decreasing most rapidly at about 9 P.M.; rate of change of temperature is about −7◦ F/h (slope of estimated tangent line to curve at 9 P.M.)

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78

Chapter 3

18. For V = 10 the slope of the tangent line is about −0.25 atm/L, for V = 25 the slope is about −0.04 atm/L. 19. (a) (b) (c) (d)

during the first year after birth about 6 cm/year (slope of estimated tangent line at age 5) the growth rate is greatest at about age 14; about 10 cm/year 40

Growth rate (cm/year)

30 20 10 t (yrs) 5

10

15

20

20. (a) The rock will hit the ground when 16t2 = 576, t2 = 36, t = 6 s (only t ≥ 0 is meaningful) (b) vave =

16(6)2 − 16(0)2 = 96 ft/s 6−0

(c) vave =

16(3)2 − 16(0)2 = 48 ft/s 3−0 16t21 − 16(6)2 16(t21 − 36) = lim t1 →6 t1 →6 t1 − 6 t1 − 6

(d) vinst = lim

= lim 16(t1 + 6) = 192 ft/s t1 →6

√ (b) vave = 20,238.6/40 = 505.96 ft/s 21. (a) (40)3 / 10 = 20,238.6 ft √ (c) Solve s = t3 / 10 = 135, t ≈ 7.53 so vave = 135/7.53 = 17.93 ft/s. √ √ t31 / 10 − (40)3 / 10 (t31 − 403 ) √ = lim (d) vinst = lim t1 →40 t1 →40 (t1 − 40) 10 t1 − 40 1 = lim √ (t21 + 40t1 + 1600) = 1517.89 ft/s t1 →40 10

22. (a) vave =

4.5(12)2 − 4.5(0)2 = 54 ft/s 12 − 0 4.5t21 − 4.5(6)2 4.5(t21 − 36) = lim t1 →6 t1 →6 t1 − 6 t1 − 6

(b) vinst = lim

= lim

t1 →6

23. (a) vave =

4.5(t1 + 6)(t1 − 6) = lim 4.5(t1 + 6) = 54 ft/s t1 →6 t1 − 6

6(4)4 − 6(2)4 = 720 ft/min 4−2 6t41 − 6(2)4 6(t41 − 16) = lim t1 →2 t1 →2 t1 − 2 t1 − 2

(b) vinst = lim

6(t21 + 4)(t21 − 4) = lim 6(t21 + 4)(t1 + 2) = 192 ft/min t1 →2 t1 →2 t1 − 2

= lim

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Exercise Set 3.2

79

EXERCISE SET 3.2 1. f (1) = 2, f (3) = 0, f (5) = −2, f (6) = −1 2. f (4) < f (0) < f (2) < 0 < f (−3) 3. (b) m = f (2) = 3

4. m =

(c) the same, f (2) = 3

2 − (−1) 3 3 1 = , y − 2 = m(x − 1), y = x + 1 − (−1) 2 2 2 y

5.

6.

y

x

x

–1

7. y − (−1) = 5(x − 3), y = 5x − 16

8. y − 3 = −4(x + 2), y = −4x − 5

f (1 + h) − f (1) 2(1 + h)2 − 2 · 12 4h + 2h2 = lim = lim = 4; h→0 h→0 h→0 h h h

9. f (1) = lim

y − 2 = 4(x − 1), y = 4x − 2 f ((−1) + h) − f (−1) 1/(h − 1)2 − 1 1 − (h − 1)2 = lim = lim h→0 h→0 h→0 h(h − 1)2 h h

10. f (−1) = lim

2h − h2 2−h = lim = 2; 2 h→0 h(h − 1) h→0 (h − 1)2

= lim

y − 1 = 2(x + 1), y = 2x + 3 f (0 + h) − f (0) h 3 − 03 = lim = lim h2 = 0; h→0 h→0 h→0 h h

11. f (0) = lim

f (0) = 0 so y − 0 = (0)(x − 0), y = 0 f (−1 + h) − f (−1) 2(−1 + h)3 + 1 − (2(−1)3 + 1) = lim = lim (6 − 6h + 2h2 ) = 6; h→0 h→0 h→0 h h

12. f (−1) = lim

f (−1) = 2(−1)3 + 1 = −1, f (−1) = 6 so y + 1 = 6(x + 1), y = 6x + 5

13.

√ f (8 + h) − f (8) 9+h−3 = lim f (8) = lim h→0 h→0 h h √ √ 1 9+h−3 9+h+3 h √ = lim √ = ; = lim h→0 h→0 h 6 9+h+3 h( 9 + h + 3)

f (8) =

√

8 + 1 = 3, f (8) =

1 1 1 5 so y − 3 = (x − 8), y = x + 6 6 6 3

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80

Chapter 3

f (4 + h) − f (4) = lim 14. f (4) = lim h→0 h→0 h

2(4 + h) + 1 − h

√

2·4+1

√ 9 + 2h + 3 √ 9 + 2h + 3

2h 2 1 √ = lim √ = h→0 3 h( 9 + 2h + 3) 9 + 2h + 3

√ 1 5 1 f (4) = 2(4) + 1 = 9 = 3, f (4) = 1/3 so y − 3 = (x − 4), y = x + 3 3 3 = lim

h→0

x − (x + ∆x) 1 1 − x(x + ∆x) 15. f (x) = lim x + ∆x x = lim ∆x→0 ∆x→0 ∆x ∆x = lim

∆x→0

−∆x 1 1 = lim − =− 2 x∆x(x + ∆x) ∆x→0 x(x + ∆x) x

1 (x + 1) − (x + ∆x + 1) 1 − (x + ∆x) + 1 x + 1 (x + 1)(x + ∆x + 1) = lim 16. f (x) = lim ∆x→0 ∆x→0 ∆x ∆x = lim

x + 1 − x − ∆x − 1 −∆x = lim ∆x(x + 1)(x + ∆x + 1) ∆x→0 ∆x(x + 1)(x + ∆x + 1)

= lim

−1 1 =− (x + 1)(x + ∆x + 1) (x + 1)2

∆x→0

∆x→0

(x + ∆x)2 − (x + ∆x) − (x2 − x) 2x∆x + ∆x2 − ∆x = lim ∆x→0 ∆x→0 ∆x ∆x

17. f (x) = lim

= lim (2x − 1 + ∆x) = 2x − 1 ∆x→0

(x + ∆x)4 − x4 4x3 ∆x + 6x2 (∆x)2 + 4x(∆x)3 + (∆x)4 = lim ∆x→0 ∆x→0 ∆x ∆x

18. f (x) = lim

= lim (4x3 + 6x2 ∆x + 4x(∆x)2 + (∆x)3 ) = 4x3 ∆x→0

19. f (x) = lim

√

∆x→0

= lim

∆x→0

20. f (x) = lim

∆x→0

1 1 √ √ −√ x − x + ∆x x x + ∆x = lim √ √ ∆x→0 ∆x x x + ∆x ∆x

1 x − (x + ∆x) −1 √ √ = lim √ √ = − 3/2 √ √ √ √ 2x ∆x x x + ∆x( x + x + ∆x) ∆x→0 x x + ∆x( x + x + ∆x) √

1 1 √ √ √ √ −√ x − 1 − x + ∆x − 1 x − 1 + x + ∆x − 1 x−1 x + ∆x − 1 √ √ √ √ = lim ∆x→0 ∆x x − 1 x + ∆x − 1 ∆x x − 1 + x + ∆x − 1

−∆x √ √ √ √ ∆x x − 1 x + ∆x − 1( x − 1 + x + ∆x − 1) 1 −1 √ √ √ =− = lim √ ∆x→0 2(x − 1)3/2 x − 1 x + ∆x − 1( x − 1 + x + ∆x − 1) = lim

∆x→0

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Exercise Set 3.2

81

f (t + h) − f (t) [4(t + h)2 + (t + h)] − [4t2 + t] = lim h→0 h→0 h h

21. f (t) = lim

4t2 + 8th + 4h2 + t + h − 4t2 − t h→0 h

= lim

8th + 4h2 + h = lim (8t + 4h + 1) = 8t + 1 h→0 h→0 h

= lim

22.

4 4 4 π(r + h)3 − πr3 π(r3 + 3r2 h + 3rh2 + h3 − r3 ) dV 3 3 = lim = lim 3 h→0 h→0 dr h h = lim

h→0

4 π(3r2 + 3rh + h2 ) = 4πr2 3

23. (a) D

(b) F

(c) B

(d) C

(e) A

√ √ 24. The point ( 2/2, 2/2) lies on the line y = x.

(f ) E y m = –1 x 1

y

25. (a)

y

(b)

(c)

x

y

x x –1

y

26. (a)

(b) x

27. (a) f (x) =

√

x and a = 1

28. (a) f (x) = cos x and a = π

29.

1

y

(c)

x

(b) f (x) = x2 and a = 3 (b) f (x) = x7 and a = 1

dy (1 − (x + h)2 ) − (1 − x2 ) −2xh − h2 = lim = lim = lim −2(x + h) = −2x, h→0 h→0 dx h→0 h h

dy = −2 and dx x=1

2

y

x

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Chapter 3

x+2+h x+2 − −2 −2 x+h x = lim x(x + 2 + h) − (x + 2)(x + h) = lim = 2, 30. h→0 h→0 x(x + h) h hx(x + h) x

1 dy =− and thus 2 dx x=−2 dy = lim dx h→0

31. y = −2x + 1

32.

1.5

5

–2

2 0

2.5 0

–3

33. (b)

h (f (1 + h) − f (1))/h

0.5 0.1 1.6569 1.4355

34. (b)

h (f (1 + h) − f (1))/h

0.5 0.50489

0.01 0.001 0.0001 0.00001 1.3911 1.3868 1.3863 1.3863

0.1 0.01 0.67060 0.70356

0.001 0.70675

0.0001 0.00001 0.70707 0.70710

35. (a) dollars/ft (b) As you go deeper the price per foot may increase dramatically, so f (x) is roughly the price per additional foot. (c) If each additional foot costs extra money (this is to be expected) then f (x) remains positive. f (301) − f (300) (d) From the approximation 1000 = f (300) ≈ 301 − 300 we see that f (301) ≈ f (300) + 1000, so the extra foot will cost around $1000.

36. (a) gallons/dollar (b) The increase in the amount of paint that would be sold for one extra dollar. (c) It should be negative since an increase in the price of paint would decrease the amount of paint sold. f (11) − f (10) (d) From −100 = f (10) ≈ we see that f (11) ≈ f (10) − 100, so an increase of one 11 − 10 dollar would decrease the amount of paint sold by around 100 gallons.

37. (a) F ≈ 200 lb, dF/dθ ≈ 50 lb/rad

(b) µ = (dF/dθ)/F ≈ 50/200 = 0.25

38. The derivative at time t = 100 of the velocity with respect to time is equal to the slope of the 10272 − 0 tangent line, which is approximately m ≈ = 128.4 ft/s2 . Thus the mass is approximately 120 − 40 7680982 T = ≈ 59820.73 lb/ft/s2 .

M (100) ≈ dv/dt 128.4 39. (a) T ≈ 115◦ F, dT /dt ≈ −3.35◦ F/min (b) k = (dT /dt)/(T − T0 ) ≈ (−3.35)/(115 − 75) = −0.084

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Exercise Set 3.2

41.

83

lim f (x) = lim

√ 3

2

x→0

x –2

f (0) does not exist.

42.

y

x = 0 = f (0), so f is continuous at x = 0. √ 3 f (0 + h) − f (0) h−0 1 = lim = lim 2/3 = +∞, so lim h→0 h→0 h→0 h h h x→0

lim f (x) = lim (x − 2)2/3 = 0 = f (2) so f is continuous at

x→2

2

y 5

x→2

f (2 + h) − f (2) h2/3 − 0 1 = lim = lim 1/3 h→0 h→0 h→0 h h h

x = 2. lim

which does not exist so f (2) does not exist. x 2

43.

y

lim f (x) = lim f (x) = f (1), so f is continuous at x = 1.

x→1−

x→1+

lim

f (1 + h) − f (1) [(1 + h)2 + 1] − 2 = lim− = lim− (2 + h) = 2; h h h→0 h→0

lim

f (1 + h) − f (1) 2(1 + h) − 2 = lim = lim 2 = 2, so f (1) = 2. h h h→0+ h→0+

h→0−

h→0+

5

x –3

44.

3

y

lim f (x) = lim f (x) = f (1) so f is continuous at x = 1.

x→1−

x→1+

5

f (1 + h) − f (1) [(1 + h)2 + 2] − 3 lim = lim = lim− (2 + h) = 2; h h h→0− h→0− h→0 lim

h→0+

f (1 + h) − f (1) [(1 + h) + 2] − 3 = lim = lim 1 = 1, + h h h→0 h→0+

so f (1) does not exist.

45. Since −|x| ≤ x sin(1/x) ≤ |x| it follows by the Squeezing Theorem

x –3

3

y

(Theorem 2.6.3) that lim x sin(1/x) = 0. The derivative cannot x→0

f (x) − f (0) = sin(1/x). This function oscillates x between −1 and +1 and does not tend to zero as x tends to zero.

x

exist: consider

46. For continuity, compare with ±x2 to establish that the limit is zero. The differential quotient is x sin(1/x) and (see Exercise 45) this has a limit of zero at the origin.

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Chapter 3

47. Let = |f (x0 )/2|. Then there exists δ > 0 such that if 0 < |x − x0 | < δ, then

f (x) − f (x0 )

− f (x0 ) < . Since f (x0 ) > 0 and = f (x0 )/2 it follows that x − x0 f (x) − f (x0 ) > > 0. If x = x1 < x0 then f (x1 ) < f (x0 ) and if x = x2 > x0 then f (x2 ) > f (x0 ). x − x0 48. g (x1 ) = lim

h→0

g(x1 + h) − g(x1 ) f (m(x1 + h) + b) − f (mx1 + b) = lim h→0 h h

f (x0 + mh) − f (x0 ) = mf (x0 ) h→0 mh

= m lim

49. (a) Let = |m|/2. Since m = 0, > 0. Since f (0) = f (0) = 0 we know there exists δ > 0

f (0 + h) − f (0) < whenever |h| < δ. It follows that |f (h)| < 1 |hm| for |h| < δ.

such that 2

h Replace h with x to get the result. (b) For |x| < δ, |f (x)| < 12 |mx|. Moreover |mx| = |mx − f (x) + f (x)| ≤ |f (x) − mx| + |f (x)|, which yields |f (x) − mx| ≥ |mx| − |f (x)| > 12 |mx| > |f (x)|, i.e. |f (x) − mx| > |f (x)|. (c) If any straight line y = mx + b is to approximate the curve y = f (x) for small values of x, then b = 0 since f (0) = 0. The inequality |f (x) − mx| > |f (x)| can also be interpreted as |f (x) − mx| > |f (x) − 0|, i.e. the line y = 0 is a better approximation than is y = mx.

50. If m = f (x0 ) then let = 12 |f (x0 ) − m|. Then there exists δ > 0 such that if |x − x0 | < δ then

f (x) − f (x0 )

1

− f (x0 ) < |f (x0 ) − m|. Then |mx| ≤ |f (x) − mx| + |f (x)|,so |f (x) − mx| > x − x0 2 |mx| − |f (x)| ≥ 12 |mx| > |f (x)|, i.e. |f (x) − mx| > |f (x)|, which shows that the straight line y = 0 is a better approximating straight line than any other choice y = mx + b.

EXERCISE SET 3.3 1. 28x6

2. −36x11

5. 0

6.

9. −3x−4 − 7x−8

10.

√

2

3. 24x7 + 2

4. 2x3

1 7. − (7x6 + 2) 3

8.

√ 11. 24x−9 + 1/ x

1 1 √ − 2 x x2

2 x 5

5 12. −42x−7 − √ 2 x

13. 12x(3x2 + 1)

14.

f (x) = x10 + 4x6 + 4x2 , f (x) = 10x9 + 24x5 + 8x

15. 3ax2 + 2bx + c

16.

1 a

2 1 18. y = √ − 2 , y (1) = −3/2 2 x x

2x +

1 b

19.

17. y = 10x − 3, y (1) = 7

2t − 1

21. dy/dx = 1 + 2x + 3x2 + 4x3 + 5x4 , dy/dx|x=1 = 15 22.

dy −3 2 1 dy = 4 − 3 − 2 + 1 + 2x + 3x2 , =0 dx x x x dx x=1

20.

1 1 − 2 3 3t

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Exercise Set 3.3

23.

85

y = (1 − x2 )(1 + x2 )(1 + x4 ) = (1 − x4 )(1 + x4 ) = 1 − x8 dy = −8x7 , dy/dx|x=1 = −8 dx

24. dy/dx = 24x23 + 24x11 + 24x7 + 24x5 , dy/dx|x=1 = 96 25. f (1) ≈

−0.999699 − (−1) f (1.01) − f (1) = = 0.0301, and by differentiation, f (1) = 3(1)2 −3 = 0 0.01 0.01

26. f (1) ≈

0.980296 − 1 f (1.01) − f (1) ≈ ≈ −1.9704, and by differentiation, f (1) = −2/13 = −2 0.01 0.01

2.0088 − 2.0111 ≈ −0.0116, 1.0981 − 0.9000 and by differentiation, f (1) = 0

27. From the graph, f (1) ≈

2.1

0.9

1.1

1.9

28. f (x) =

√

x + 2x, f (1) ≈

3.2488 − 2.7487 ≈ 2.5006, 1.1 − 0.9

1 5 f (x) = √ + 2, f (1) = 2 2 x

3.3

0

1.1

0

29. 32t 33. (a)

34.

30. 2π dV = 4πr2 dr

32. −2α−2 + 1

31. 3πr2 (b)

dV = 4π(5)2 = 100π dr r=5

d λλ0 + λ6 1 d 1 λ0 + 6λ5 = (λ0 + 6λ5 ) = (λλ0 + λ6 ) = dλ 2 − λ0 2 − λ0 dλ 2 − λ0 2 − λ0

35. y − 2 = 5(x + 3), y = 5x + 17 37. (a) (b) (c) (d)

36. y + 2 = −(x − 2), y = −x

dy/dx = 21x2 − 10x + 1, d2 y/dx2 = 42x − 10 dy/dx = 24x − 2, d2 y/dx2 = 24 dy/dx = −1/x2 , d2 y/dx2 = 2/x3 y = 35x5 − 16x3 − 3x, dy/dx = 175x4 − 48x2 − 3, d2 y/dx2 = 700x3 − 96x

38. (a) y = 28x6 − 15x2 + 2, y = 168x5 − 30x (b) y = 3, y = 0 2 4 (c) y = 2 , y = − 3 5x 5x (d) y = 2x4 + 3x3 − 10x − 15, y = 8x3 + 9x2 − 10, y = 24x2 + 18x

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86

Chapter 3

39. (a) y = −5x−6 + 5x4 , y = 30x−7 + 20x3 , y = −210x−8 + 60x2 (b) y = x−1 , y = −x−2 , y = 2x−3 , y = −6x−4 (c) y = 3ax2 + b, y = 6ax, y = 6a 40. (a) dy/dx = 10x − 4, d2 y/dx2 = 10, d3 y/dx3 = 0 (b) dy/dx = −6x−3 − 4x−2 + 1, d2 y/dx2 = 18x−4 + 8x−3 , d3 y/dx3 = −72x−5 − 24x−4 (c) dy/dx = 4ax3 + 2bx, d2 y/dx2 = 12ax2 + 2b, d3 y/dx3 = 24ax 41. (a) f (x) = 6x, f (x) = 6, f (x) = 0, f (2) = 0

d2 y d2 y dy 4 3 = 30x − 8x, = 120x − 8, = 112 (b) dx dx2 dx2 x=1 (c)

d2 −3 d3 −3 d4 −3 d −3 x = −3x−4 , x = 12x−5 , x = −60x−6 , x = 360x−7 , 2 3 dx dx dx dx4

d4 −3 x = 360 dx4 x=1

42. (a) y = 16x3 + 6x2 , y = 48x2 + 12x, y = 96x + 12, y (0) = 12 3 4 dy d2 y −6 d y −7 d y = 120x , = −720x , = 5040x−8 , (b) y = 6x−4 , = −24x−5 , dx dx2 dx3 dx4

d4 y = 5040 dx4 x=1

43. y = 3x2 + 3, y = 6x, and y = 6 so y + xy − 2y = 6 + x(6x) − 2(3x2 + 3) = 6 + 6x2 − 6x2 − 6 = 0 44. y = x−1 , y = −x−2 , y = 2x−3 so x3 y + x2 y − xy = x3 (2x−3 ) + x2 (−x−2 ) − x(x−1 ) = 2 − 1 − 1 = 0 dy 45. The graph has a horizontal tangent at points where = 0, dx dy = x2 − 3x + 2 = (x − 1)(x − 2) = 0 if x = 1, 2. The but dx corresponding values of y are 5/6 and 2/3 so the tangent line is horizontal at (1, 5/6) and (2, 2/3).

1.5

0

3 0

46. Find where f (x) = 0; f (x) = 1 − 9/x2 = 0, x2 = 9, x = ±3.

14

–7

7

–14

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Exercise Set 3.3

87

47. The y-intercept is −2 so the point (0, −2) is on the graph; −2 = a(0)2 + b(0) + c, c = −2. The x-intercept is 1 so the point (1,0) is on the graph; 0 = a + b − 2. The slope is dy/dx = 2ax + b; at x = 0 the slope is b so b = −1, thus a = 3. The function is y = 3x2 − x − 2. dy 48. Let P (x0 , y0 ) be the point where y = x2 +k is tangent to y = 2x. The slope of the curve is = 2x dx and the slope of the line is 2 thus at P , 2x0 = 2 so x0 = 1. But P is on the line, so y0 = 2x0 = 2.

Because P is also on the curve we get y0 = x20 + k so k = y0 − x20 = 2 − (1)2 = 1.

49. The points (−1, 1) and (2, 4) are on the secant line so its slope is (4 − 1)/(2 + 1) = 1. The slope of the tangent line to y = x2 is y = 2x so 2x = 1, x = 1/2.

50. The points √ (1, 1) and (4, √ 2) are on the √ secant line √ so its slope is 1/3. The slope of the tangent line to y = x is y = 1/(2 x) so 1/(2 x) = 1/3, 2 x = 3, x = 9/4.

51. y = −2x, so at any point (x0 , y0 ) on y = 1 − x2 the tangent line is y − y0 = −2x0 (x − x0 ), or 2 2 y = −2x0 x + x20 + 1. The point (2, 0) is to be √ on the line, so 0 = −4x0 + x0 + 1, x0 − 4x0 + 1 = 0. √ 4 ± 16 − 4 Use the quadratic formula to get x0 = = 2 ± 3.

2 52. Let P1 (x1 , ax21 ) and P2 (x2 , ax22 ) be the points of tangency. y = 2ax so the tangent lines at P1 and P2 are y − ax21 = 2ax1 (x − x1 ) and y − ax22 = 2ax2 (x − x2 ). Solve for x to get x = 12 (x1 + x2 ) which is the x-coordinate of a point on the vertical line halfway between P1 and P2 .

53. y = 3ax2 + b; the tangent line at x = x0 is y − y0 = (3ax20 + b)(x − x0 ) where y0 = ax30 + bx0 .

Solve with y = ax3 + bx to get (ax3 + bx) − (ax30 + bx0 ) = (3ax20 + b)(x − x0 ) ax3 + bx − ax30 − bx0 = 3ax20 x − 3ax30 + bx − bx0 x3 − 3x20 x + 2x30 = 0 (x − x0 )(x2 + xx0 − 2x20 ) = 0 (x − x0 )2 (x + 2x0 ) = 0, so x = −2x0 .

54. Let (x0 , y0 ) be the point of tangency. Note that y0 = 1/x0 . Since y = −1/x2 , the tangent line 1 1 1 1 2 = − 2x + or y = − 2 x + , with has the form y − y0 = (−1/x20 )(x − x0 ), or y − x x x x x 0 0 0 0 0

2 intercepts at 0, and (2x0 , 0). The distance from the y-intercept to the point of tangency x0

is (0 − x0 )2 + (y0 − 2y0 )2 , and the distance from the x-intercept to the point of tangency is

(x0 − 2x0 )2 + y02 so that they are equal (and equal the distance from the point of tangency to the origin).

1 1 x 2 ; the tangent line at x = x0 is y − y0 = − 2 (x − x0 ), or y = − 2 + . The tangent line x2 x0 x0 x0 1 crosses the x-axis at 2x0 , the y-axis at 2/x0 , so that the area of the triangle is (2/x0 )(2x0 ) = 2.

2

55. y = −

56. f (x) = 3ax2 + 2bx + c; there is a horizontal √ tangent where f (x) = 0. Use the quadratic formula 2 on 3ax + 2bx + c = 0 to get x = (−b ± b2 − 3ac)/(3a) which gives two real solutions, one real solution, or none if

(a) b2 − 3ac > 0 57. F = GmM r−2 ,

(b) b2 − 3ac = 0

dF 2GmM = −2GmM r−3 = − dr r3

(c) b2 − 3ac < 0

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88

Chapter 3

58. dR/dT = 0.04124 − 3.558 × 10−5 T which decreases as T increases from 0 to 700. When T = 0, dR/dT = 0.04124 Ω/◦ C; when T = 700, dR/dT = 0.01633 Ω/◦ C. The resistance is most sensitive to temperature changes at T = 0◦ C, least sensitive at T = 700◦ C. 59. f (x) = 1 + 1/x2 > 0 for all x = 0

60. f (x) = 3x2 − 3 = 0 when x = ±1; increasing for −∞ < x < −1 and 1 < x < +∞

6

y

(–1, 2) –6

2

6 x –2

–1

1

2

–6 –2

(1, –2)

61. f is continuous at 1 because lim f (x) = lim f (x) = f (1); also lim f (x) = lim (2x + 1) = 3 x→1−

x→1−

x→1+

x→1−

and lim+ f (x) = lim+ 3 = 3 so f is differentiable at 1. x→1

x→1

62. f is not continuous at x = 9 because lim f (x) = −63 and lim f (x) = 36. x→9−

x→9+

f cannot be differentiable at x = 9, for if it were, then f would also be continuous, which it is not.

63. f is continuous at 1 because lim f (x) = lim f (x) = f (1), also lim f (x) = lim 2x = 2 and x→1−

x→1−

x→1+

x→1−

1 1 lim+ f (x) = lim+ √ = so f is not differentiable at 1.

2 x→1 x→1 2 x 64. f is continuous at 1/2 because

lim f (x) =

x→1/2−

2

lim f (x) =

x→1/2−

lim 3x = 3/4 and

x→1/2−

lim f (x) = f (1/2), also

x→1/2+

lim f (x) =

x→1/2+

lim 3x/2 = 3/4 so f (1/2) = 3/4, and

x→1/2+

f is differentiable at x = 1/2.

65. (a) f (x) = 3x − 2 if x ≥ 2/3, f (x) = −3x + 2 if x < 2/3 so f is differentiable everywhere except perhaps at 2/3. f is continuous at 2/3, also lim f (x) = lim (−3) = −3 and lim + f (x) =

x→2/3

x→2/3−

x→2/3−

lim + (3) = 3 so f is not differentiable at x = 2/3.

x→2/3

(b) f (x) = x2 − 4 if |x| ≥ 2, f (x) = −x2 + 4 if |x| < 2 so f is differentiable everywhere except perhaps at ±2. f is continuous at −2 and 2, also lim f (x) = lim (−2x) = −4 x→2−

x→2−

and lim+ f (x) = lim+ (2x) = 4 so f is not differentiable at x = 2. Similarly, f is not x→2

x→2

differentiable at x = −2.

66. (a) f (x) = −(1)x−2 , f (x) = (2 · 1)x−3 , f (x) = −(3 · 2 · 1)x−4 f (n) (x) = (−1)n

n(n − 1)(n − 2) · · · 1 xn+1

(b) f (x) = −2x−3 , f (x) = (3 · 2)x−4 , f (x) = −(4 · 3 · 2)x−5 f (n) (x) = (−1)n

(n + 1)(n)(n − 1) · · · 2 xn+2

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Exercise Set 3.4

67. (a)

89

d d d d d d2 d2 d [cf (x)] = c [f (x)] = c [f (x)] = c 2 [f (x)] [cf (x)] = 2 dx dx dx dx dx dx dx dx

d d d2 d d d [f (x) + g(x)] = [f (x) + g(x)] = [f (x)] + [g(x)] 2 dx dx dx dx dx dx =

d2 d2 [f (x)] + [g(x)] dx2 dx2

(b) yes, by repeated application of the procedure illustrated in Part (a) 68.

f (2 + h) − f (2) = f (2); f (x) = 8x7 − 2, f (x) = 56x6 , so f (2) = 56(26 ) = 3584. h→0 h lim

69. (a) f (x) = nxn−1 , f (x) = n(n − 1)xn−2 , f (x) = n(n − 1)(n − 2)xn−3 , . . ., f (n) (x) = n(n − 1)(n − 2) · · · 1 (b) from Part (a), f (k) (x) = k(k − 1)(k − 2) · · · 1 so f (k+1) (x) = 0 thus f (n) (x) = 0 if n > k (c) from Parts (a) and (b), f (n) (x) = an n(n − 1)(n − 2) · · · 1 70. (a) If a function is differentiable at a point then it is continuous at that point, thus f is continuous on (a, b) and consequently so is f . (b) f and all its derivatives up to f (n−1) (x) are continuous on (a, b) 71. Let g(x) = xn , f (x) = (mx + b)n . Use Exercise 48 in Section 3.2, but with f and g permuted. If x0 = mx1 + b then Exercise 48 says that f is differentiable at x1 and f (x1 ) = mg (x0 ). Since g (x0 ) = nxn−1 , the result follows.

0 72. f (x) = 2 · 2(2x + 3) = 8x + 12

73. f (x) = 3 · 3(3x − 1)2 = 81x2 − 54x + 9

74. f (x) = 1 · (−1)(x − 1)−2 = −1/(x − 1)2

75. f (x) = 2 · 3 · (−2)(2x + 1)−3 = −12/(2x + 1)3

76. f (x) =

1 x+1−1 =1− , and f (x) = −1(x + 1)−2 = −1/(x + 1)2 x+1 x+1

77. f (x) =

2x2 + 4x + 2 + 1 1 =2+ , so f (x) = −2(x + 1)−3 = −2/(x + 1)3 (x + 1)2 (x + 1)2

EXERCISE SET 3.4 1. (a) f (x) = 2x2 + x − 1, f (x) = 4x + 1 (b) f (x) = (x + 1) · (2) + (2x − 1) · (1) = 4x + 1 2. (a) f (x) = 3x4 + 5x2 − 2, f (x) = 12x3 + 10x (b) f (x) = (3x2 − 1) · (2x) + (x2 + 2) · (6x) = 12x3 + 10x 3. (a) f (x) = x4 − 1, f (x) = 4x3 (b) f (x) = (x2 + 1) · (2x) + (x2 − 1) · (2x) = 4x3 4. (a) f (x) = x3 + 1, f (x) = 3x2 (b) f (x) = (x + 1)(2x − 1) + (x2 − x + 1) · (1) = 3x2

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Chapter 3

5. f (x) = (3x2 + 6)

d dx

2x −

1 4

1 d 1 + 2x − (3x2 + 6) = (3x2 + 6)(2) + 2x − (6x) 4 dx 4

3 = 18x2 − x + 12 2 6. f (x) = (2 − x − 3x3 )

d d (7 + x5 ) + (7 + x5 ) (2 − x − 3x3 ) dx dx

= (2 − x − 3x3 )(5x4 ) + (7 + x5 )(−1 − 9x2 ) = −24x7 − 6x5 + 10x4 − 63x2 − 7 7. f (x) = (x3 + 7x2 − 8)

d d (2x−3 + x−4 ) + (2x−3 + x−4 ) (x3 + 7x2 − 8) dx dx

= (x3 + 7x2 − 8)(−6x−4 − 4x−5 ) + (2x−3 + x−4 )(3x2 + 14x) = −15x−2 − 14x−3 + 48x−4 + 32x−5 8. f (x) = (x−1 + x−2 )

d d (3x3 + 27) + (3x3 + 27) (x−1 + x−2 ) dx dx

= (x−1 + x−2 )(9x2 ) + (3x3 + 27)(−x−2 − 2x−3 ) = 3 + 6x − 27x−2 − 54x−3 9. f (x) = x2 + 2x + 4 + (2x + 2)(x − 2) = 3x2 10. f (x) = x4 − x2 , f (x) = 4x3 − 2x

11.

12.

13.

d d (5x − 3) (1) − (1) (5x − 3) 5 dy dx dx = =− ; y (1) = −5/4 dx (5x − 3)2 (5x − 3)2 √ d √ d ( x + 2) (3) − 3 ( x + 2) √ √ dy dx√ dx = = −3/(2 x( x + 2)2 ); y (1) = −3/18 = −1/6 dx ( x + 2)2 d d (x + 3) (2x − 1) − (2x − 1) (x + 3) dy dx dx = dx (x + 3)2 =

14.

d d (x2 − 5) (4x + 1) − (4x + 1) (x2 − 5) dy dx dx = dx (x2 − 5)2 =

15.

(x + 3)(2) − (2x − 1)(1) 7 dy 7 = ; = (x + 3)2 (x + 3)2 dx x=1 16

(x2 − 5)(4) − (4x + 1)(2x) 4x2 + 2x + 20 =− ; 2 2 (x − 5) (x2 − 5)2

d 3x + 2 d −5 x + 1 + x−5 + 1 dx dx x

−5

x(3) − (3x + 2)(1) 3x + 2 −6 = −5x + x +1 x x2

3x + 2 2 = −5x−6 + x−5 + 1 − 2 ; x x

dy = dx

3x + 2 x

dy 13 = dx x=1 8

dy = 5(−5) + 2(−2) = −29 dx x=1

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Exercise Set 3.4

16.

91

x−1 d d x−1 dy = (2x7 − x2 ) + (2x7 − x2 ) dx dx x + 1 x + 1 dx

x−1 (x + 1)(1) − (x − 1)(1) (14x6 − 2x) = (2x7 − x2 ) + (x + 1)2 x+1

2 x−1 (14x6 − 2x); = (2x7 − x2 ) · + (x + 1)2 x+1

1 dy 2 = (2 − 1) + 0(14 − 2) = dx x=1 4 2 18. f (1) = 1

17. f (1) = 0 19. (a) g (x) = (b) g (x) =

√

1 1 xf (x) + √ f (x), g (4) = (2)(−5) + (3) = −37/4 4 2 x

xf (x) − f (x) (4)(−5) − 3 , g (4) = = −23/16 x2 16

20. (a) g (x) = 6x − 5f (x), g (3) = 6(3) − 5(4) = −2 (b) g (x) = 21. (a) (b) (c) (d)

2f (x) − (2x + 1)f (x) 2(−2) − 7(4) = −8 , g (3) = 2 f (x) (−2)2

F (x) = 5f (x) + 2g (x), F (2) = 5(4) + 2(−5) = 10 F (x) = f (x) − 3g (x), F (2) = 4 − 3(−5) = 19 F (x) = f (x)g (x) + g(x)f (x), F (2) = (−1)(−5) + (1)(4) = 9 F (x) = [g(x)f (x) − f (x)g (x)]/g 2 (x), F (2) = [(1)(4) − (−1)(−5)]/(1)2 = −1

22. (a) F (x) = 6f (x) − 5g (x), F (π) = 6(−1) − 5(2) = −16 (b) F (x) = f (x) + g(x) + x(f (x) + g (x)), F (π) = 10 − 3 + π(−1 + 2) = 7 + π (c) F (x) = 2f (x)g (x) + 2f (x)g(x) = 2(20) + 2(3) = 46 (d) F (x) =

23.

24.

(4 + g(x))f (x) − f (x)g (x) (4 − 3)(−1) − 10(2) = = −21 (4 + g(x))2 (4 − 3)2

2x(x + 2) − (x2 − 1) dy = , dx (x + 2)2 dy = 0 if x2 + 4x + 1 = 0. By the quadratic formula, dx √ √ √ −4 ± 16 − 4 = −2 ± 3. The tangent line is horizontal at x = −2 ± 3. x= 2 x2 − 2x − 1 dy 2x(x − 1) − (x2 + 1) = . The tangent line is horizontal when it has slope 0, i.e. = dx (x − 1)2 (x − 1)2 √ √ 2± 4+4 x2 − 2x − 1 = 0 which, by the quadratic formula, has solutions x = = 1 ± 2, the 2 √ tangent line is horizontal when x = 1 ± 2.

25. The tangent line is parallel to the line y = x when it has slope 1. dy x2 + 2x − 1 2x(x + 1) − (x2 + 1) = = 1 if x2 +2x−1 = (x+1)2 , which reduces to −1 = +1, = 2 dx (x + 1) (x + 1)2 impossible. Thus the tangent line is never parallel to the line y = x.

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Chapter 3

26. The tangent line is perpendicular to the line y = x when the tangent line has slope −1.

1 dy 1 x+2+1 =1+ , hence =− y= = −1 when (x + 2)2 = 1, x2 + 4x + 3 = 0, x+2 x+2 dx (x + 2)2 (x + 1)(x + 3) = 0, x = −1, −3. Thus the tangent line is perpendicular to the line y = x at the points (−1, 2), (−3, 0).

1 at the point (x0 , 1/(x0 + 4)) is 27. Fix x0 . The slope of the tangent line to the curve y = x + 4

−1 dy −1 = given by = . The tangent line to the curve at (x0 , y0 ) thus has the (x + 4)2 x=x0 (x0 + 4)2 dx −1 (x − x0 ), and this line passes through the origin if its constant term (x0 + 4)2 −x0 −1 1 = y0 − x0 is zero. Then , x0 + 4 = −x0 , x0 = −2. (x0 + 4)2 x0 + 4 (x0 + 4)2 equation y − y0 =

2x + 5 2x + 4 + 1 1 dy −1 , thus the tangent line at the = = 2+ , and hence = x+2 x+2 x+2 dx (x + 2)2 −1 1 point (x0 , y0 ) is given by y − y0 = (x − x0 ) , where y0 = 2 + . (x0 + 2)2 x0 + 2 If this line is to pass through (0, 2), then x0 −1 −1 = (−x0 ), , −x0 − 2 = x0 , x0 = −1 2 − y0 = x0 + 2 (x0 + 2)2 (x0 + 2)2

28. y =

1 1 = , x = 2 − x, x = 1, y = 1. The first curve has derivative x 2−x 1 1 so y = − 2 , so the slope when x = 1 is m = −1. Second curve has derivative y = (2 − x)2 x the slope when x = 1 is m = 1. Since the two slopes are negative reciprocals of each other, the tangent lines are perpendicular at the point (1, 1).

29. (b) They intersect when

30. The curves intersect when a/(x − 1) = x2 − 2x + 1, or (x − 1)3 = a, x = 1 + a1/3 . They are −a perpendicular when their slopes are negative reciprocals of each other, i.e.

(2x − 2) = −1, (x − 1)2 which has the solution x = 2a + 1. Solve x = 1 + a1/3 = 2a + 1, 2a2/3 = 1, a = 2−3/2 . Thus the curves intersect and are perpendicular at the point (2a + 1, 1/2) provided a = 2−3/2 .

31. F (x) = xf (x) + f (x), F (x) = xf (x) + f (x) + f (x) = xf (x) + 2f (x) 32. (a) F (x) = xf (x) + 3f (x) (b) Assume that F n) (x) = xf (n) (x) + nf (n−1) (x) for some n (for instance n = 3, as in part (a)).

Then F (n+1) (x) = xf (n+1) (x) + (1 + n)f (n) (x) = xf (n+1) (x) + (n + 1)f (n) (x), which is an inductive proof.

33. (f · g · h) = [(f · g) · h] = (f · g)h + h(f · g) = (f · g)h + h[f g + f g] = f gh + f g h + f gh 34. (f1 f2 · · · fn ) = (f1 f2 · · · fn ) + (f1 f2 · · · fn ) + · · · + (f1 f2 · · · fn )

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35. (a) 2(1 + x−1 )(x−3 + 7) + (2x + 1)(−x−2 )(x−3 + 7) + (2x + 1)(1 + x−1 )(−3x−4 ) (b) (x7 + 2x − 3)3 = (x7 + 2x − 3)(x7 + 2x − 3)(x7 + 2x − 3) so d 7 (x + 2x − 3)3 = (7x6 + 2)(x7 + 2x − 3)(x7 + 2x − 3) dx +(x7 + 2x − 3)(7x6 + 2)(x7 + 2x − 3) +(x7 + 2x − 3)(x7 + 2x − 3)(7x6 + 2) = 3(7x6 + 2)(x7 + 2x − 3)2 36. (a) −5x−6 (x2 + 2x)(4 − 3x)(2x9 + 1) + x−5 (2x + 2)(4 − 3x)(2x9 + 1) +x−5 (x2 + 2x)(−3)(2x9 + 1) + x−5 (x2 + 2x)(4 − 3x)(18x8 ) (b) (x2 + 1)50 = (x2 + 1)(x2 + 1) · · · (x2 + 1), where (x2 + 1) occurs 50 times so d 2 (x + 1)50 = [(2x)(x2 + 1) · · · (x2 + 1)] + [(x2 + 1)(2x) · · · (x2 + 1)] dx + · · · + [(x2 + 1)(x2 + 1) · · · (2x)] = 2x(x2 + 1)49 + 2x(x2 + 1)49 + · · · + 2x(x2 + 1)49 = 100x(x2 + 1)49 because 2x(x2 + 1)49 occurs 50 times.

37. By the product rule, g (x) is the sum of n terms, each containing n factors of the form f (x)f (x)f (x) . . . f (x); the function f (x) occurs n − 1 times, and f (x) occurs once. Each of these n terms is equal to f (x)(f (x))n−1 , and so g (x) = n(f (x))n−1 f (x).

38. g (x) = 10(x2 − 1)9 (2x) = 20x(x2 − 1)9 39. f (x) =

1 xn · (0) − 1 · (nxn−1) n

so f (x) = = − n+1 xn x2n x

40. f (x) = g(x)h(x), f (x) = g (x)h(x) + g(x)h (x), solve for h : h (x) = h = f /g so

f (x) f (x)g (x) f (x)g(x) − f (x)g (x) d f (x) − = = 2 g(x) g(x) g(x)2 dx g(x)

EXERCISE SET 3.5 −10 + cos x x3

1. f (x) = −4 sin x + 2 cos x

2. f (x) =

3. f (x) = 4x2 sin x − 8x cos x

4. f (x) = 4 sin x cos x

5. f (x) =

sin x(5 + sin x) − cos x(5 − cos x) 1 + 5(sin x − cos x) = 2 (5 + sin x) (5 + sin x)2

6. f (x) =

(x2 + sin x) cos x − sin x(2x + cos x) x2 cos x − 2x sin x = (x2 + sin x)2 (x2 + sin x)2

7. f (x) = sec x tan x −

√

2 sec2 x

8. f (x) = (x2 + 1) sec x tan x + (sec x)(2x) = (x2 + 1) sec x tan x + 2x sec x

f (x) − g (x)h(x) , but g(x)

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Chapter 3

9. f (x) = −4 csc x cot x + csc2 x

10. f (x) = − sin x − csc x + x csc x cot x

11. f (x) = sec x(sec2 x) + (tan x)(sec x tan x) = sec3 x + sec x tan2 x 12. f (x) = (csc x)(− csc2 x) + (cot x)(− csc x cot x) = − csc3 x − csc x cot2 x 13. f (x) =

(1 + csc x)(− csc2 x) − cot x(0 − csc x cot x) csc x(− csc x − csc2 x + cot2 x) = but 2 (1 + csc x) (1 + csc x)2

1 + cot2 x = csc2 x (identity) thus cot2 x − csc2 x = −1 so f (x) =

14. f (x) = =

csc x(− csc x − 1) csc x =− 1 + csc x (1 + csc x)2 (1 + tan x)(sec x tan x) − (sec x)(sec2 x) sec x tan x + sec x tan2 x − sec3 x = (1 + tan x)2 (1 + tan x)2 sec x(tan x + tan2 x − sec2 x) sec x(tan x − 1) = (1 + tan x)2 (1 + tan x)2

15. f (x) = sin2 x + cos2 x = 1 (identity) so f (x) = 0 2 sin x sin x −2 3 =0 cos3 x cos x OR f (x) = sec2 x − tan2 x = 1 (identity), f (x) = 0

16. f (x) = 2 sec x tan x sec x − 2 tan x sec2 x =

17.

18.

f (x) =

tan x (because sin x sec x = (sin x)(1/ cos x) = tan x), 1 + x tan x

f (x) =

(1 + x tan x)(sec2 x) − tan x[x(sec2 x) + (tan x)(1)] (1 + x tan x)2

=

sec2 x − tan2 x 1 = (because sec2 x − tan2 x = 1) 2 (1 + x tan x) (1 + x tan x)2

f (x) =

(x2 + 1) cot x (because cos x csc x = (cos x)(1/ sin x) = cot x), 3 − cot x

f (x) =

(3 − cot x)[2x cot x − (x2 + 1) csc2 x] − (x2 + 1) cot x csc2 x (3 − cot x)2

=

6x cot x − 2x cot2 x − 3(x2 + 1) csc2 x (3 − cot x)2

19. dy/dx = −x sin x + cos x, d2 y/dx2 = −x cos x − sin x − sin x = −x cos x − 2 sin x 20. dy/dx = − csc x cot x, d2 y/dx2 = −[(csc x)(− csc2 x) + (cot x)(− csc x cot x)] = csc3 x + csc x cot2 x 21. dy/dx = x(cos x) + (sin x)(1) − 3(− sin x) = x cos x + 4 sin x, d2 y/dx2 = x(− sin x) + (cos x)(1) + 4 cos x = −x sin x + 5 cos x 22. dy/dx = x2 (− sin x) + (cos x)(2x) + 4 cos x = −x2 sin x + 2x cos x + 4 cos x, d2 y/dx2 = −[x2 (cos x) + (sin x)(2x)] + 2[x(− sin x) + cos x] − 4 sin x = (2 − x2 ) cos x − 4(x + 1) sin x 23. dy/dx = (sin x)(− sin x) + (cos x)(cos x) = cos2 x − sin2 x, d2 y/dx2 = (cos x)(− sin x) + (cos x)(− sin x) − [(sin x)(cos x) + (sin x)(cos x)] = −4 sin x cos x

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Exercise Set 3.5

95

24. dy/dx = sec2 x; d2 y/dx2 = 2 sec2 x tan x 25. Let f (x) = tan x, then f (x) = sec2 x. (a) f (0) = 0 and f (0) = 1 so y − 0 = (1)(x − 0), y = x.

π

π

π

π = 1 and f = 2 so y − 1 = 2 x − , y = 2x − + 1. (b) f 4 4 4 2 π

π

π

π

(c) f − = −1 and f − = 2 so y + 1 = 2 x + , y = 2x + − 1.

4 4 4 2 26. Let f (x) = sin x, then f (x) = cos x. (a) f (0) = 0 and f (0) = 1 so y − 0 = (1)(x − 0), y = x (b) f (π) = 0 and f (π) = −1 so y − 0 = (−1)(x − π), y = −x + π π

π

1 1 1 1 1 π

π 1 = √ so y − √ = √ x − , y = √ x− √ + √ = √ and f (c) f 4 4 4 2 2 2 2 2 4 2 2 27. (a) If y = x sin x then y = sin x + x cos x and y = 2 cos x − x sin x so y + y = 2 cos x. (b) If y = x sin x then y = sin x+x cos x and y = 2 cos x−x sin x so y +y = 2 cos x; differentiate twice more to get y (4) + y = −2 cos x.

28. (a) If y = cos x then y = − sin x and y = − cos x so y + y = (− cos x) + (cos x) = 0; if y = sin x then y = cos x and y = − sin x so y + y = (− sin x) + (sin x) = 0. (b) y = A cos x − B sin x, y = −A sin x − B cos x so y + y = (−A sin x − B cos x) + (A sin x + B cos x) = 0.

29. (a) (b) (c) (d)

f (x) = cos x = 0 at x = ±π/2, ±3π/2. f (x) = 1 − sin x = 0 at x = −3π/2, π/2. f (x) = sec2 x ≥ 1 always, so no horizontal tangent line. f (x) = sec x tan x = 0 when sin x = 0, x = ±2π, ±π, 0 0.5

30. (a)

0

2c

-0.5

(b) y = sin x cos x = (1/2) sin 2x and y = cos 2x. So y = 0 when 2x = (2n + 1)π/2 for n = 0, 1, 2, 3 or x = π/4, 3π/4, 5π/4, 7π/4 31. x = 10 sin θ, dx/dθ = 10 cos θ; if θ = 60◦ , then dx/dθ = 10(1/2) = 5 ft/rad = π/36 ft/deg ≈ 0.087 ft/deg ◦ 32. s = 3800 csc θ, ds/dθ √= −3800 csc√θ cot θ; if θ = 30 ,√then ds/dθ = −3800(2)( 3) = −7600 3 ft/rad = −380 3π/9 ft/deg ≈ −230 ft/deg

33. D = 50 tan θ,√ dD/dθ = 50 sec2 θ; if θ = 45◦ , then dD/dθ = 50( 2)2 = 100 m/rad = 5π/9 m/deg ≈ 1.75 m/deg 34. (a) From the right triangle shown, sin θ = r/(r + h) so r + h = r csc θ, h = r(csc θ − 1). (b) dh/dθ = −r csc θ cot√θ; if θ = 30◦ , then dh/dθ = −6378(2)( 3) ≈ −22, 094 km/rad ≈ −386 km/deg

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Chapter 3

35. (a) (b)

36.

d4 d4k d87 d3 d4¡21 d3 sin x = sin x, so sin x = sin x; sin x = 3 4¡21 sin x = 3 sin x = â&#x2C6;&#x2019; cos x 4k 87 4 dx dx dx dx dx dx d4k d100 cos x = cos x = cos x dx100 dx4k

d [x sin x] = x cos x + sin x dx

d2 [x sin x] = â&#x2C6;&#x2019;x sin x + 2 cos x dx2

d3 [x sin x] = â&#x2C6;&#x2019;x cos x â&#x2C6;&#x2019; 3 sin x dx3

d4 [x sin x] = x sin x â&#x2C6;&#x2019; 4 cos x dx4

By mathematical induction one can show d4k [x sin x] = x sin x â&#x2C6;&#x2019; (4k) cos x; dx4k

d4k+1 [x sin x] = x cos x + (4k + 1) sin x; dx4k+1

d4k+2 [x sin x] = â&#x2C6;&#x2019;x sin x + (4k + 2) cos x; dx4k+2

d4k+3 [x sin x] = â&#x2C6;&#x2019;x cos x â&#x2C6;&#x2019; (4k + 3) sin x; dx4k+3

Since 17 = 4 ¡ 4 + 1,

d17 [x sin x] = x cos x + 17 sin x dx17

37. f (x) = â&#x2C6;&#x2019; sin x, f (x) = â&#x2C6;&#x2019; cos x, f (x) = sin x, and f (4) (x) = cos x with higher order derivatives repeating this pattern, so f (n) (x) = sin x for n = 3, 7, 11, . . .

38. f (x) = sin x, f (x) = cos x, f (x) = â&#x2C6;&#x2019; sin x, f (4) (x) = â&#x2C6;&#x2019; cos x, f (5) (x) = sin x, and the right-hand sides continue with a period of 4, so that f (n) (x) = sin x when n = 4k for some k > 0.

39. (a) (c) (e) (g) (i)

40. (a)

(b)

all x x = Ď&#x20AC;/2 + nĎ&#x20AC;, n = 0, Âą1, Âą2, . . . x = Ď&#x20AC;/2 + nĎ&#x20AC;, n = 0, Âą1, Âą2, . . . x = (2n + 1)Ď&#x20AC;, n = 0, Âą1, Âą2, . . . all x

(b) (d) (f ) (h)

all x x = nĎ&#x20AC;, n = 0, Âą1, Âą2, . . . x = nĎ&#x20AC;, n = 0, Âą1, Âą2, . . . x = nĎ&#x20AC;/2, n = 0, Âą1, Âą2, . . .

d cos(x + h) â&#x2C6;&#x2019; cos x cos x cos h â&#x2C6;&#x2019; sin x sin h â&#x2C6;&#x2019; cos x [cos x] = lim = lim hâ&#x2020;&#x2019;0 hâ&#x2020;&#x2019;0 dx h h

sin h cos h â&#x2C6;&#x2019; 1 â&#x2C6;&#x2019; sin x = (cos x)(0) â&#x2C6;&#x2019; (sin x)(1) = â&#x2C6;&#x2019; sin x = lim cos x hâ&#x2020;&#x2019;0 h h d d cos x sin x(â&#x2C6;&#x2019; sin x) â&#x2C6;&#x2019; cos x(cos x) = [cot x] = dx dx sin x sin2 x â&#x2C6;&#x2019;1 â&#x2C6;&#x2019; sin2 x â&#x2C6;&#x2019; cos2 x = = â&#x2C6;&#x2019; csc2 x 2 sin x sin2 x

1 0 ¡ cos x â&#x2C6;&#x2019; (1)(â&#x2C6;&#x2019; sin x) d d sin x = [sec x] = = = sec x tan x dx dx cos x cos2 x cos2 x =

(c)

1 (sin x)(0) â&#x2C6;&#x2019; (1)(cos x) d d cos x = (d) [csc x] = = â&#x2C6;&#x2019; 2 = â&#x2C6;&#x2019; csc x cot x 2 dx dx sin x sin x sin x

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Exercise Set 3.5

41.

42.

97

w+x 2 sin w−x sin w − sin x d 2 cos 2 = lim sin x = lim w→x w→x w−x w−x dx w−x sin w+x = 1 · cos x = cos x = lim w−x2 cos w→x 2 2 w+x −2 sin( w−x d cos w − cos x 2 ) sin( 2 ) [cos x] = lim = lim w→x w→x dx w−x w−x

w−x sin( 2 ) w+x lim = − sin x = − lim sin w−x w→x w→x 2 2

43. (a)

tan h = lim h→0 h→0 h lim

sin h cos h h

= lim

h→0

sin h h cos h

=

1 =1 1

tan x + tan h − tan x tan(x + h) − tan x d 1 = lim − tan x tan h [tan x] = lim (b) h→0 h→0 h h dx 2 tan x + tan h − tan x + tan x tan h tan h(1 + tan2 x) = lim = lim h→0 h→0 h(1 − tan x tan h) h(1 − tan x tan h) tan h tan h sec2 x h = sec2 x lim = lim h→0 1 − tan x tan h h→0 h(1 − tan x tan h) tan h lim h→0 h 2 = sec x = sec2 x lim (1 − tan x tan h) h→0

44.

45.

lim

x→0

d tan(x + y) − tan y tan(y + h) − tan y = lim = (tan y) = sec2 y h→0 x h dy

cos k(x + h) − cos kx cos(kx + kh) − cos kx d [cos kx] = lim = k lim = −k sin kx h→0 kh→0 dx h kh

46. The position function is given by s = −4 cos πt, so by Exercise 45 the velocity is the derivative ds = 4π sin πt. The position function shows that the top of the mass moves from a low point of dt s = −4π to a high point of s = 4π, and passes through the origin when t = π/2, 3π/2, . . .. On the first pass through the origin the velocity is v = 4π sin(π/2) = 4π.

180 t and cos h = cos t, sin h = sin t. Then π π cos h − 1 cos t − 1 cos t − 1 = lim = lim =0 lim t→0 180t/π h→0 h 180 t→0 t sin h sin t sin t π π lim = lim = lim = t→0 180t/π h→0 h 180 t→0 t 180

47. Let t be the radian measure, then h =

sin(x + h) − sin x cos h − 1 sin h d [sin x] = lim = sin x lim + cos x lim h→0 h→0 h→0 h dx h h π cos x = (sin x)(0) + (cos x)(π/180) = 180 d cos(x + h) − cos x cos x cos h − sin x sin h − cos x (b) [cos x] = lim = lim h→0 h→0 dx h h

(a)

= cos x lim

h→0

π π cos h − 1 sin h − sin x lim = 0 · cos x − · sin x = − sin x h→0 h h 180 180

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Chapter 3

EXERCISE SET 3.6 1. (f ◦ g) (x) = f (g(x))g (x) so (f ◦ g) (0) = f (g(0))g (0) = f (0)(3) = (2)(3) = 6 2. (f ◦ g) (2) = f (g(2))g (2) = 5(−3) = −15 3. (a) (f ◦ g)(x) = f (g(x)) = (2x − 3)5 and (f ◦ g) (x) = f (g(x))g (x) = 5(2x − 3)4 (2) = 10(2x − 3)4 (b) (g ◦ f )(x) = g(f (x)) = 2x5 − 3 and (g ◦ f ) (x) = g (f (x))f (x) = 2(5x4 ) = 10x4 √ 5 4. (a) (f ◦ g)(x) = 5 4 + cos x and (f ◦ g) (x) = f (g(x))g (x) = √ (− sin x) 2 4 + cos x √ √ 5 (b) (g ◦ f )(x) = 4 + cos(5 x) and (g ◦ f ) (x) = g (f (x))f (x) = − sin(5 x) √ 2 x 5. (a) F (x) = f (g(x))g (x) = f (g(3))g (3) = −1(7) = −7 (b) G (x) = g (f (x))f (x) = g (f (3))f (3) = 4(−2) = −8 6. (a) F (x) = f (g(x))g (x), F (−1) = f (g(−1))g (−1) = f (2)(−3) = (4)(−3) = −12 (b) G (x) = g (f (x))f (x), G (−1) = g (f (−1))f (−1) = −5(3) = −15 7. f (x) = 37(x3 + 2x)36

d 3 (x + 2x) = 37(x3 + 2x)36 (3x2 + 2) dx

8. f (x) = 6(3x2 + 2x − 1)5

7 9. f (x) = −2 x − x

−3

3

d (3x2 + 2x − 1) = 6(3x2 + 2x − 1)5 (6x + 2) = 12(3x2 + 2x − 1)5 (3x + 1) dx

d dx

10. f (x) = (x5 − x + 1)−9 , f (x) = −9(x5 − x + 1)−10

7 x − x 3

7 = −2 x − x 3

−3

7 3x + 2 x

d 5 9(5x4 − 1) (x − x + 1) = −9(x5 − x + 1)−10 (5x4 − 1) = − 5 dx (x − x + 1)10

11. f (x) = 4(3x2 − 2x + 1)−3 , f (x) = −12(3x2 − 2x + 1)−4

d 24(1 − 3x) (3x2 − 2x + 1) = −12(3x2 − 2x + 1)−4 (6x − 2) = dx (3x2 − 2x + 1)4

1 3x2 − 2 d 3 12. f (x) = √ (x − 2x + 5) = √ 2 x3 − 2x + 5 dx 2 x3 − 2x + 5 √ d 1 3 13. f (x) = (4 + 3 x) = √ √ √ dx 2 4+3 x 4 x 4+3 x d √ 1 1 1 1 14. f (x) = √ ( x) = 1/4 √ = 3/4 2x 2 x 4x x dx 2 15. f (x) = cos(1/x2 )

d 2 (1/x2 ) = − 3 cos(1/x2 ) dx x

√ d √ √ 1 16. f (x) = sec2 x x = sec2 x √ dx 2 x 17. f (x) = 20 cos4 x

2

d (cos x) = 20 cos4 x(− sin x) = −20 cos4 x sin x dx

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Exercise Set 3.6

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18. f (x) = 4 + 20(sin3 x)

d (sin x) = 4 + 20 sin3 x cos x dx

â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; d â&#x2C6;&#x161; â&#x2C6;&#x161; d 3 cos(3 x) sin(3 x) â&#x2C6;&#x161; 19. f (x) = 2 cos(3 x) [cos(3 x)] = â&#x2C6;&#x2019;2 cos(3 x) sin(3 x) (3 x) = â&#x2C6;&#x2019; dx dx x â&#x2C6;&#x161;

20. f (x) = 4 tan3 (x3 )

d d [tan(x3 )] = 4 tan3 (x3 ) sec2 (x3 ) (x3 ) = 12x2 tan3 (x3 ) sec2 (x3 ) dx dx

d d [sec(x7 )] = 4 sec(x7 ) sec(x7 ) tan(x7 ) (x7 ) = 28x6 sec2 (x7 ) tan(x7 ) dx dx

d x (x + 1)(1) â&#x2C6;&#x2019; x(1) x x x 2 2

= 3 cos â&#x2C6;&#x2019; sin 22. f (x) = 3 cos cos x + 1 dx x+1 x+1 x+1 (x + 1)2

x x 3 sin =â&#x2C6;&#x2019; cos2 2 (x + 1) x+1 x+1

21. f (x) = 4 sec(x7 )

1 d 5 sin(5x) 23. f (x) = [cos(5x)] = â&#x2C6;&#x2019; dx 2 cos(5x) 2 cos(5x) d 1 3 â&#x2C6;&#x2019; 8 sin(4x) cos(4x)

[3x â&#x2C6;&#x2019; sin2 (4x)] = 24. f (x) = dx 2 2 3x â&#x2C6;&#x2019; sin (4x) 2 3x â&#x2C6;&#x2019; sin2 (4x)

â&#x2C6;&#x2019;4 d

x + csc(x3 + 3) 25. f (x) = â&#x2C6;&#x2019;3 x + csc(x3 + 3) dx

d â&#x2C6;&#x2019;4 = â&#x2C6;&#x2019;3 x + csc(x3 + 3) 1 â&#x2C6;&#x2019; csc(x3 + 3) cot(x3 + 3) (x3 + 3) dx

â&#x2C6;&#x2019;4 1 â&#x2C6;&#x2019; 3x2 csc(x3 + 3) cot(x3 + 3) = â&#x2C6;&#x2019;3 x + csc(x3 + 3)

â&#x2C6;&#x2019;5 d 4

x â&#x2C6;&#x2019; sec(4x2 â&#x2C6;&#x2019; 2) 26. f (x) = â&#x2C6;&#x2019;4 x4 â&#x2C6;&#x2019; sec(4x2 â&#x2C6;&#x2019; 2) dx

â&#x2C6;&#x2019;5 d = â&#x2C6;&#x2019;4 x4 â&#x2C6;&#x2019; sec(4x2 â&#x2C6;&#x2019; 2) 4x3 â&#x2C6;&#x2019; sec(4x2 â&#x2C6;&#x2019; 2) tan(4x2 â&#x2C6;&#x2019; 2) (4x2 â&#x2C6;&#x2019; 2) dx

4

â&#x2C6;&#x2019;5 x2 â&#x2C6;&#x2019; 2 sec(4x2 â&#x2C6;&#x2019; 2) tan(4x2 â&#x2C6;&#x2019; 2) = â&#x2C6;&#x2019;16x x â&#x2C6;&#x2019; sec(4x2 â&#x2C6;&#x2019; 2)

27.

d dy = x3 (2 sin 5x) (sin 5x) + 3x2 sin2 5x = 10x3 sin 5x cos 5x + 3x2 sin2 5x dx dx

28.

â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 1 1 3 1 dy 2 2 = x 3 tan ( x) sec ( x) â&#x2C6;&#x161; + â&#x2C6;&#x161; tan3 ( x) = tan2 ( x) sec2 ( x) + â&#x2C6;&#x161; tan3 ( x) dx 2 2 x 2 x 2 x

29.

dy = x5 sec dx

1 d 1 1 1 tan + sec (5x4 ) x x dx x x

1 1 1 1 5 4 tan â&#x2C6;&#x2019; 2 + 5x sec = x sec x x x x

1 1 1 = â&#x2C6;&#x2019;x3 sec tan + 5x4 sec x x x

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Chapter 3

30.

sec(3x + 1) cos x − 3 sin x sec(3x + 1) tan(3x + 1) dy = = cos x cos(3x + 1) − 3 sin x sin(3x + 1) dx sec2 (3x + 1)

31.

dy d = − sin(cos x) (cos x) = − sin(cos x)(− sin x) = sin(cos x) sin x dx dx

32.

d dy = cos(tan 3x) (tan 3x) = 3 sec2 3x cos(tan 3x) dx dx

33.

d d dy = 3 cos2 (sin 2x) [cos(sin 2x)] = 3 cos2 (sin 2x)[− sin(sin 2x)] (sin 2x) dx dx dx = −6 cos2 (sin 2x) sin(sin 2x) cos 2x

34.

35.

36.

37.

38.

1 + cot x2 + csc x2 dy (1 − cot x2 )(−2x csc x2 cot x2 ) − (1 + csc x2 )(2x csc2 x2 ) = −2x csc x2 = 2 2 dx (1 − cot x ) (1 − cot x2 )2 √ √ d d dy = (5x + 8)7 (1 − x)6 + (1 − x)6 (5x + 8)7 dx dx dx √ √ −1 = 6(5x + 8)7 (1 − x)5 √ + 7 · 5(1 − x)6 (5x + 8)6 2 x √ √ −3 = √ (5x + 8)7 (1 − x)5 + 35(1 − x)6 (5x + 8)6 x d dy d = (x2 + x)5 sin8 x + (sin8 x) (x2 + x)5 = 8(x2 + x)5 sin7 x cos x + 5(sin8 x)(x2 + x)4 (2x + 1) dx dx dx 2

2

x−5 dy d x−5 11 33(x − 5)2 x−5 =3 · = =3 2 dx 2x + 1 dx 2x + 1 2x + 1 (2x + 1) (2x + 1)4 dy = 17 dx

= 17

39.

1 + x2 1 − x2

16 16

d dx

1 + x2 1 − x2

= 17

1 + x2 1 − x2

16

(1 − x2 )(2x) − (1 + x2 )(−2x) (1 − x2 )2

4x 68x(1 + x2 )16 = 2 2 (1 − x ) (1 − x2 )18

2(2x + 3)2 (4x2 − 1)7 [3(4x2 − 1) − 32x(2x + 3)] 2(2x + 3)2 (52x2 + 96x + 3) =− 2 16 (4x − 1) (4x2 − 1)9

d dy = 12[1 + sin3 (x5 )]11 [1 + sin3 (x5 )] dx dx = 12[1 + sin3 (x5 )]11 3 sin2 (x5 )

41.

(4x2 − 1)8 (3)(2x + 3)2 (2) − (2x + 3)3 (8)(4x2 − 1)7 (8x) dy = dx (4x2 − 1)16 =

40.

1 + x2 1 − x2

d sin(x5 ) = 180x4 [1 + sin3 (x5 )]11 sin2 (x5 ) cos(x5 ) dx

4 d

dy = 5 x sin 2x + tan4 (x7 ) x sin 2x tan4 (x7 ) dx dx

d 4 7 4 3 7 d 7 = 5 x sin 2x + tan (x ) x cos 2x (2x) + sin 2x + 4 tan (x ) tan(x ) dx dx

4 = 5 x sin 2x + tan4 (x7 ) 2x cos 2x + sin 2x + 28x6 tan3 (x7 ) sec2 (x7 )

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Exercise Set 3.6

42.

dy = 4 tan3 dx

101

√ √ (7 − x) 3x2 + 5 (7 − x) 3x2 + 5 2 sec 2 + 2+ x3 + sin x x3 + sin x

√

√ 3x2 + 5 (7 − x) 3x2 + 5 (3x2 + cos x) (7 − x)x × − 3 − + 3√ x + sin x (x3 + sin x)2 3x2 + 5 (x3 + sin x)

43.

44.

45.

dy dy = cos 3x − 3x sin 3x; if x = π then = −1 and y = −π, so the equation of the tangent line dx dx is y + π = −(x − π), y = −x dy dy = 3x2 cos(1 + x3 ); if x = −3 then y = − sin 26, = −27 cos 26, so the equation of the tangent dx dx line is y + sin 26 = −27(cos 26)(x + 3), y = −27(cos 26)x − 81 cos 26 − sin 26 dy dy = −3 sec3 (π/2 − x) tan(π/2 − x); if x = −π/2 then = 0, y = −1 so the equation of the dx dx tangent line is y + 1 = 0, y = −1

2

95 135 1 27 dy dy 1 , = 3 = so the equation of the 1 + 2 ; if x = 2 then y = 46. = 3 x− x x 8 dx 44 16 dx 27 135 x− tangent line is y − 27/8 = (135/16)(x − 2), y = 16 2 47.

√ √ √ dy d dy = sec2 (4x2 ) (4x2 ) = 8x sec2 (4x2 ), √ = 8 π sec2 (4π) = 8 π. When x = π, dx dx dx x= π √ √ √ y = tan(4π) = 0, so the equation of the tangent line is y = 8 π(x − π) = 8 πx − 8π.

48.

dy d dy = 12 cot3 x cot x = −12 cot3 x csc2 x, = −24. When x = π/4, y = 3, so the

dx dx dx x=π/4 equation of the tangent line is y − 3 = −24(x − π/4), or y = −24x + 3 + 6π.

49.

x2 dy dy (−2x), = 4 − 1/2 = 7/2. When x = 1, y = 2, so the equation = 2x 5 − x2 + √

dx dx x=1 2 5 − x2 7 3 of the tangent line is y − 2 = (7/2)(x − 1), or y = x − .

2 2

50.

dy x dy 1 − (1 − x2 )3/2 (−2x), = 1. When x = 0, y = 0, so the equation of the = √

dx 2 dx x=0 1 − x2 tangent line is y = x.

51.

dy d d = x(− sin(5x)) (5x) + cos(5x) − 2 sin x (sin x) dx dx dx = −5x sin(5x) + cos(5x) − 2 sin x cos x = −5x sin(5x) + cos(5x) − sin(2x), 2

d d d d y = −5x cos(5x) (5x) − 5 sin(5x) − sin(5x) (5x) − cos(2x) (2x) 2 dx dx dx dx = −25x cos(5x) − 10 sin(5x) − 2 cos(2x) 52.

dy d = cos(3x2 ) (3x2 ) = 6x cos(3x2 ), dx dx d d2 y = 6x(− sin(3x2 )) (3x2 ) + 6 cos(3x2 ) = −36x2 sin(3x2 ) + 6 cos(3x2 ) 2 dx dx

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102

Chapter 3

53.

(1 − x) + (1 + x) dy 2 d2 y = = = 2(1 − x)−2 and = −2(2)(−1)(1 − x)−3 = 4(1 − x)−3 2 2 dx (1 − x) (1 − x) dx2

54.

dy = x sec2 dx

1 1 1 1 d 1 1 2 + tan = − sec + tan , x dx x x x x x

d2 y 1 d 1 1 2 1 1 1 d 1 1 2 2 2 2 sec + 2 sec + sec = 3 sec tan = − sec x x dx x x x x dx x x x x dx2

55. y = cot3 (π − θ) = − cot3 θ so dy/dx = 3 cot2 θ csc2 θ

56. 6

57.

au + b cu + d

5

ad − bc (cu + d)2

d [a cos2 πω + b sin2 πω] = −2πa cos πω sin πω + 2πb sin πω cos πω dω = π(b − a)(2 sin πω cos πω) = π(b − a) sin 2πω

58. 2 csc2 (π/3 − y) cot(π/3 − y) (c) f (x) = x √

2

59. (a)

−x 4 − 2x2 + 4 − x2 = √ 4 − x2 4 − x2 2

–2

2 –2

2

–2 –6

(d) f (1) =

√

√ 2 2 3 and f (1) = √ so the tangent line has the equation y − 3 = √ (x − 1).

3 3

3

0

2 0

(c) f (x) = 2x cos(x2 ) cos x − sin x sin(x2 )

0.5

60. (a)

1.2

^ ^

0

6

6 –1.2

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Exercise Set 3.6

103

(d) f (1) = sin 1 cos 1 and f (1) = 2 cos2 1 â&#x2C6;&#x2019; sin2 1, so the tangent line has the equation y â&#x2C6;&#x2019; sin 1 cos 1 = (2 cos2 1 â&#x2C6;&#x2019; sin2 1)(x â&#x2C6;&#x2019; 1).

0.8

^

6

0

61. (a) dy/dt = â&#x2C6;&#x2019;AĎ&#x2030; sin Ď&#x2030;t, d2 y/dt2 = â&#x2C6;&#x2019;AĎ&#x2030; 2 cos Ď&#x2030;t = â&#x2C6;&#x2019;Ď&#x2030; 2 y (b) one complete oscillation occurs when Ď&#x2030;t increases over an interval of length 2Ď&#x20AC;, or if t increases over an interval of length 2Ď&#x20AC;/Ď&#x2030; (c) f = 1/T (d) amplitude = 0.6 cm, T = 2Ď&#x20AC;/15 s/oscillation, f = 15/(2Ď&#x20AC;) oscillations/s 62. dy/dt = 3A cos 3t, d2 y/dt2 = â&#x2C6;&#x2019;9A sin 3t, so â&#x2C6;&#x2019;9A sin 3t + 2A sin 3t = 4 sin 3t, â&#x2C6;&#x2019;7A sin 3t = 4 sin 3t, â&#x2C6;&#x2019;7A = 4, A = â&#x2C6;&#x2019;4/7   63. f (x) =

4 3x

+ 5,

x<0



â&#x2C6;&#x2019; 52 x + 5, x > 0  7 â&#x2C6;&#x161;  2]â&#x2C6;&#x161;21x+45 , x < 0 d 7 6 when x = â&#x2C6;&#x2019;1 x + f (x) = =  â&#x2C6;&#x2019; â&#x2C6;&#x161; 3 dx 24 , x > 0 2 â&#x2C6;&#x2019;6x+20 64. 2 sin(Ď&#x20AC;/6) = 1, so we can assume f (x) = â&#x2C6;&#x2019; 52 x + 5. Thus for suďŹ&#x192;ciently small values of |x â&#x2C6;&#x2019; Ď&#x20AC;/6| we have â&#x2C6;&#x161;

5â&#x2C6;&#x161; d 3 d 5 5 [f (2 sin x)] =â&#x2C6;&#x2019; = f (2 sin x) 2 sin x = â&#x2C6;&#x2019; 2 cos x =â&#x2C6;&#x2019; 2 3 dx dx 2 2 2 2 x=Ď&#x20AC;/6 x=Ď&#x20AC;/6 x=Ď&#x20AC;/6 65. (a) p â&#x2030;&#x2C6; 10 lb/in2 , dp/dh â&#x2030;&#x2C6; â&#x2C6;&#x2019;2 lb/in2 /mi (b)

dp dp dh = â&#x2030;&#x2C6; (â&#x2C6;&#x2019;2)(0.3) = â&#x2C6;&#x2019;0.6 lb/in2 /s dt dh dt

66. (a) F =

45 dF 45(â&#x2C6;&#x2019; sin θ + 0.3 cos θ) ; , =â&#x2C6;&#x2019; cos θ + 0.3 sin θ dθ (cos θ + 0.3 sin θ)2

if θ = 30â&#x2014;Ś , then dF/dθ â&#x2030;&#x2C6; 10.5 lb/rad â&#x2030;&#x2C6; 0.18 lb/deg (b)

dF dF dθ = â&#x2030;&#x2C6; (0.18)(â&#x2C6;&#x2019;0.5) = â&#x2C6;&#x2019;0.09 lb/s dt dθ dt

d d du d d cos x, u > 0 (| sin x|) = (|u|) = (|u|) = (|u|) cos x = 67. With u = sin x, â&#x2C6;&#x2019; cos x, u < 0 dx dx du dx du

cos x, sin x > 0 cos x, 0<x<Ď&#x20AC; = = â&#x2C6;&#x2019; cos x, sin x < 0 â&#x2C6;&#x2019; cos x, â&#x2C6;&#x2019;Ď&#x20AC; < x < 0 68.

d d (cos x) = [sin(Ď&#x20AC;/2 â&#x2C6;&#x2019; x)] = â&#x2C6;&#x2019; cos(Ď&#x20AC;/2 â&#x2C6;&#x2019; x) = â&#x2C6;&#x2019; sin x dx dx

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104

Chapter 3

69. (a) For x = 0, |f (x)| ≤ |x|, and lim |x| = 0, so by the Squeezing Theorem, lim f (x) = 0. x→0

x→0

(b) If f (0) were to exist, then the limit doesn’t.

1 (c) For x = 0, f (x) = x cos x

(d) lim

x→0

1 − 2 x

f (x) − f (0) = sin(1/x) would have to exist, but it x−0

+ sin

1 1 1 1 = − cos + sin x x x x

f (x) − f (0) = lim sin(1/x), which does not exist, thus f (0) does not exist. x→0 x−0

70. (a) −x2 ≤ x2 sin(1/x) ≤ x2 , so by the Squeezing Theorem lim f (x) = 0. x→0

(b) f (0) = lim

x→0

f (x) − f (0) = lim x sin(1/x) = 0 by Exercise 69, Part a. x→0 x−0

(c) For x = 0, f (x) = 2x sin(1/x) + x2 cos(1/x)(−1/x2 ) = 2x sin(1/x) − cos(1/x) (d) If f (x) were continuous at x = 0 then so would cos(1/x) = f (x) − 2x sin(1/x) be, since 2x sin(1/x) is continuous there. But cos(1/x) oscillates at x = 0.

71. (a) g (x) = 3[f (x)]2 f (x), g (2) = 3[f (2)]2 f (2) = 3(1)2 (7) = 21 (b) h (x) = f (x3 )(3x2 ), h (2) = f (8)(12) = (−3)(12) = −36

√ 3(x2 − 1) + 4(2x) = 2x 3x2 + 1 √ √ 3 3x − 1 3 1

√ 73. F (x) = f (g(x))g (x) = f ( 3x − 1) √ = = (3x − 1) + 1 2 3x − 1 2x 2 3x − 1 72. F (x) = f (g(x))g (x) =

74.

75.

d [f (x2 )] = f (x2 )(2x), thus f (x2 )(2x) = x2 so f (x2 ) = x/2 if x = 0 dx d d 2 [f (3x)] = f (3x) (3x) = 3f (3x) = 6x, so f (3x) = 2x. Let u = 3x to get f (u) = u; dx dx 3 d 2

[f (x)] = f (x) = x. dx 3

76. (a) If f (−x) = f (x), then f is odd.

d d [f (−x)] = [f (x)], f (−x)(−1) = f (x), f (−x) = −f (x) so dx dx

(b) If f (−x) = −f (x), then f is even.

d d [f (−x)] = − [f (x)], f (−x)(−1) = −f (x), f (−x) = f (x) so dx dx

77. For an even function, the graph is symmetric about the y-axis; the slope of the tangent line at (a, f (a)) is the negative of the slope of the tangent line at (−a, f (−a)). For an odd function, the graph is symmetric about the origin; the slope of the tangent line at (a, f (a)) is the same as the slope of the tangent line at (−a, f (−a)). y

y

f(x)

f(x)

f '(x) x x

f '(x)

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Exercise Set 3.7

105

78.

dy du dv dw dy = dx du dv dw dx

79.

d d [f (g(h(x)))] = [f (g(u))], dx dx =

u = h(x)

du d du [f (g(u))] = f (g(u))g (u) = f (g(h(x)))g (h(x))h (x) du dx dx

EXERCISE SET 3.7 1.

dx dy =3 dt dt (a)

2.

dy = 3(2) = 6 dt

(b) −1 = 3

dx dx 1 , =− dt dt 3

dx dy +4 =0 dt dt (a) 1 + 4

dy dy 1 = 0 so = − when x = 2. dt dt 4

dy dx + 18y =0 dt dt 1 1 dy dy (a) 8 √ · 3 + 18 √ = 0, = −2 dt 2 2 3 2 dt

(b)

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

3. 8x

√

√ 5 1 dx dx (b) 8 − 18 · 8 = 0, =6 5 3 dt 9 dt

dy dx dy dx + 2y =2 +4 dt dt dt dt dy dy dy = 2(−5) + 4 , = −10 (a) 2 · 3(−5) + 2 · 1 dt dt dt √ √ √ dx √ √ dx 3 dx + 2(2 + 3) · 6 = 2 + 4 · 6, = −12 √ = −3 3 2 (b) 2(1 + 2) dt dt dt 2 2

4. 2x

5. (b) A = x2

dA dx (d) Find given that dt x=3 dt x=3 6. (b) A = πr2 (d) Find

dA dx = 2x dt dt

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

dA dr given that dt r=5 dt r=5

dr dA = 2πr dt dt

dA = 2. From Part (c), = 2π(5)(2) = 20π cm2 /s. dt (c)

r=5

dr dV 2 dh =π r + 2rh .

7. (a) V = πr h, so dt dt dt

dV dh dr (b) Find given that = 1 and = −1. From Part (a), dt h=6, dt h=6, dt h=6, r=10 r=10 r=10

dV = π[102 (1) + 2(10)(6)(−1)] = −20π in3 /s; the volume is decreasing. dt h=6, 2

r=10

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106

Chapter 3

8. (a) +2 = x2 + y 2 , so (b) Find

1 d+ = dt +

x

dx dy +y . dt dt

1 dy 1 d+ dx = and =− . given that dt x=3, dt 2 dt 4 y=4

From Part (a) and the fact that + = 5 when x = 3 and 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 2 y 2 dθ dt , dθ = cos θ x dy − y dx dt = 9. (a) tan θ = , so sec θ x dt x2 dt x2 dt dt

dθ dx dy 1 (b) Find given that = 1 and =− .

dt x=2, dt x=2, dt x=2, 4 y=2 y=2 y=2 π π 1 When x = 2 and y = 2, tan θ = 2/2 = 1 so θ = and cos θ = cos = √ . Thus 4 4 2 √

5 1 dθ (1/ 2)2 − 2(1) = − rad/s; θ is decreasing.

2 − from Part (a), = 22 4 16 dt x=2, y=2

dz dx dy given that = −2 and = 3. dt x=1, dt x=1, dt x=1, y=2 y=2 y=2

dz 3 dy 2 2 dx dz = 2x y + 3x y , = (4)(3) + (12)(−2) = −12 units/s; z is decreasing. dt dt dt dt x=1,

10. Find

y=2

11. Let A be the area swept out, and θ the angle through which the minute hand has rotated. dθ π 1 dθ 4π 2 dA dA given that = rad/min; A = r2 θ = 8θ, so =8 = in /min.

Find dt dt 30 2 dt dt 15 12. Let r be the radius and A the area enclosed by the ripple. We want

dr dA given that = 3. dt t=10 dt

dA dr = 2πr . Because r is increasing at the constant rate of 3 ft/s, it dt dt

dA = 2π(30)(3) = 180π ft2 /s. follows that r = 30 ft after 10 seconds so dt We know that A = πr2 , so

t=10

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

dt A=9 dt dt dt dt 2πr dt √ √ dr 1 √ (6) = 1/ π mi/h. = then πr2 = 9, r = 3/ π so

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

3 4 D 1 1 dD dV is the diameter, V = π = 3. From V = πD3 = πD3 . We want given that 3 2 6 dt r=1 dt 6

1 dD dD dV 2 3 2 dV dD = πD2 we get = (3) = , = , so ft/min.

2 2 dt dt πD dt dt r=1 π(2)2 2π dt

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Exercise Set 3.7

107

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

dt r=9 dt 3 dt dt

dV = 4π(9)2 (−15) = −4860π. Air must be removed at the rate of 4860π cm3 /min. dt r=9 16. Let x and y be the distances shown in the diagram.

dx dy given that = 5. From We want to find

dt y=8 dt x2 +y 2 = 172 we get 2x

dy dy x dx dx +2y = 0, so =− . dt dt dt y dt

17

y

When y = 8, x2 +82 = 172 , x2 = 289−64 = 225, x = 15

15 dy 75 = − (5) = − so ft/s; the top of the ladder dt 8 8

x

y=8

is moving down the wall at a rate of 75/8 ft/s.

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

dy dx y dy dx + 2y = 0 so = − . dt dt dt x dt

Use

2

x + y = 169 to find that x = 12 when y = 5 so

dx 5 5 = − (−2) = ft/s. dt 12 6

13

y

x

y=5

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

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

√ 1 dθ 1 1 √ = √ ≈ 0.051 rad/s. = − − above the ground so sin θ = 96/10 and dt x=2 2 96 2 96

x

60

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

ft

− sin θ

y

Home

First

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108

Chapter 3

dy dx given that = 2000.

20. Find dt x=4 dt x=4

From

dy dx y dy dx = 2y so = . dt dt dt x dt √ Use x2 + 25 = y 2 to find that y = 41 when x = 4 so √

√ 41 dx = (2000) = 500 41 mi/h.

dt x=4 4

Rocket

x2 + 52 = y 2 we get 2x

21. Find

y

Radar station

5 mi

dy dx given that = 880. From dt x=4000 dt x=4000

y 2 = x2 + 30002 we get 2y

Rocket

dy dx dy x dx = 2x so = . dt dt dt y dt

If x = 4000, then y = 5000 so

dy 4000 (880) = 704 ft/s. =

dt x=4000 5000

x

y

x

Camera 3000 ft

dφ dx given that = 0.2. But x = 3000 tan φ so 22. Find dt φ=π/4 dt φ=π/4

dx dφ dx 2 2 π = 3000(sec φ) , (0.2) = 1200 ft/s. = 3000 sec dt dt dt φ=π/4 4 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 altitude is given by dx dr dr dθ 4995(0.12 sin θ) dθ = = = . dt dt dθ dt (1 + 0.12 cos θ)2 dt Use θ = 120◦ and dθ/dt = 2.7◦ /min = (2.7)(π/180) 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 dθ dx dθ sin2 θ dx = −4000 csc2 θ , so =− . Use θ = 30◦ and dt dt dt 4000 dt dx/dt = 300 mi/h = 300(5280/3600) ft/s = 440 ft/s to get dθ/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.

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Exercise Set 3.7

109

dV dh given that 25. Find

dt h=16 dt

= 20.

The volume of

1 water in the tank at a depth h is V = πr2 h. Use 3 r 10 5 similar triangles (see figure) to get = so r = h h 24 12

2 dh 25 dV 5 25 1 h h = πh3 , = πh2 ; thus V = π 3 12 432 dt 144 dt

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

2 2 dt 25πh dt dt h=16 25π(16) 20π

dh dV 1 2 26. Find given that = 8.

V = πr h, dt h=6 dt 3 2 1 dh 1 dV 1 h 1 but r = h so V = π h= πh3 , = πh2 , 2 3 2 12 dt 4 dt

4 dV dh 4 8 dh = , ft/min. = (8) =

2 2 dt πh dt dt h=6 π(6) 9π

10 r 24 h

h

r

1 dV dh = 5. V = πr2 h, but 27. Find given that dt h=10 dt 3 2 1 1 dh 1 dV h 1 r = h so V = π h = πh3 , = πh2 , 2 3 2 12 dt 4 dt

1 dV = π(10)2 (5) = 125π ft3 /min. dt h=10 4

h

r

28. Let r and h be as shown in the figure. If C is the circumference

dV dC given that of the base, then we want to find = 10. dt h=8 dt It is given that r =

1 h, thus C = 2πr = πh so 2

dh dC =π dt dt Use

V =

(1)

dh 1 2 1 dV 1 πr h = πh3 to get = πh2 , so 3 12 dt 4 dt 4 dV dh = dt πh2 dt

(2)

dC 4 4 dV dC 5 Substitution of (2) into (1) gives = = 2 so (10) = ft/min. dt h dt dt h=8 64 8

h

r

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110

Chapter 3

dh dt 1 ds given that = 500. From the figure, h = s sin 30◦ = s dt 2

29. With s and h as shown in the figure, we want to find

so

s h

1 ds 1 dh = = (500) = 250 mi/h. dt 2 dt 2

30° Ground

dx dy 30. Find given that = −20. From x2 + 102 = y 2 dt y=125 dt dx dy dx y dy we get 2x = 2y so = . Use x2 + 100 = y 2 dt dt dt x dt √ √ to find that x = 15, 525 = 15 69 when y = 125 so

dx 125 500 = √ (−20) = − √ .

dt y=125 15 69 3 69

Pulley y 10 Boat

x

500 The boat is approaching the dock at the rate of √ ft/min.

3 69

dx dy given that 31. Find = −12. From x2 + 102 = y 2 dt y=125 dt dy dy x dx dx = 2y so = . Use x2 + 100 = y 2 dt dt dt y dt

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

Pulley y 10 Boat

x

32. (a) Let x and y be as shown in the figure. It is required to dy dx , given that = −3. By similar triangles, dt dt x x+y 1 = , 18x = 6x + 6y, 12x = 6y, x = y, so 6 18 2 1 dy 1 3 dx = = (−3) = − ft/s. dt 2 dt 2 2

Light

find

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 = + . In part (a) we found that = − when = −3 so dt dt dt dt 2 dt

dz = (−3/2) + (−3) = −9/2 ft/s; the tip of the shadow is moving at the rate of 9/2 ft/s dt toward the street light.

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Exercise Set 3.7

111

2π π dθ dx = = rad/s. given that 33. Find

dt θ=π/4 dt 10 5

x

dx dθ = 4 sec2 θ , Then x = 4 tan θ (see figure) so dt dt

π dx π = 8π/5 km/s. = 4 sec2 4 5 dt θ=π/4

4

θ

Ship

34. If x, y, and z are as shown in the figure, then we want

dz dx dy given that = −1200. But = −600 and dt dt x=2, dt x=2, y=4

y=4

P

dx dy +y . dt dt √ √ When x = 2 and y = 4, z 2 = 22 + 42 = 20, z = 20 = 2 5 so

√ 1 3000 dz = √ [2(−600) + 4(−1200)] = − √ = −600 5 mi/h;

dt x=2, 2 5 5 y=4 z 2 = x2 + y 2 so 2z

dx dy dz 1 dz = 2x + 2y , = dt dt dt dt z

Aircraft

x

y z

x

Missile

√ the distance between missile and aircraft is decreasing at the rate of 600 5 mi/h.

dz dx = −600 and 35. We wish to find given dt x=2, dt y=4

dy = −1200 (see figure). From the law of cosines, dt x=2,

P

◦

y

z = x + y − 2xy cos 120 = x + y − 2xy(−1/2) 2

2

2

2

dz dx dy dy dx = x2 +y 2 +xy, so 2z = 2x +2y +x +y , dt dt dt dt dt

dx 1 dy dz (2x + y) . = + (2y + x) dt 2z dt dt

Aircraft

120º

y=4

2

x

z

Missile

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

√ dz 1 4200 √ [(2(2) + 4)(−600) + (2(4) + 2)(−1200)] = − √ = −600 7 mi/h; =

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

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

dz dx y, and z be the distances shown in the first figure. Find given that = −75 and dt x=2, dt y=0 dy = 100. In order to find an equation relating x, y, and z, first draw the line segment dt that joins the point 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 right

2

dx dy dz = 2x + 2y + 0, x2 + (1/2)2 + y 2 = x2 + y 2 + 1/4 and 2z triangle so z 2 = dt dt dt

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112

Chapter 3

dx dy +y . Now, when x = 2 and y = 0, z 2 = (2)2 + (0)2 + 1/4 = 17/4, dt dt

√ √ dz 1 z = 17/2 so = √ [2(−75) + 0(100)] = −300/ 17 mi/h

x=2, dt ( 17/2) y=0 1 dz = dt z

x

North 1 2 1 2

mi

mi y

P y

x x

West

Car

C

O

East

z

z Helicopter

(b) decreasing, because

P

H

dz < 0. dt

dy dx 37. (a) We want given that = 6. For convenience, first rewrite the equation as dt x=1, dt x=1, y=2

y=2

16 dy dy y3 dy dx dx 8 8 2 + y then 3xy 2 + y3 = y , = so 16 5 5 dt dt 5 dt dt 2 dt y − 3xy 5

dy 23 = (6) = −60/7 units/s.

16 dt x=1, y=2 (2) − 3(1)22 5

xy 3 =

(b) falling, because

dy <0 dt

dx dy given that = 2. Square and rearrange to get x3 = y 2 − 17 dt (2,5) dt (2,5)

5 5 dy dx 2y dy dx 2 dx (2) = units/s. so 3x = = 2y , = 2 , dt dt dt 3x dt dt (2,5) 6 3

38. Find

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

√ dD dx 2 2 2 given that = −2.

D = (x − 3) + (2x − 0) = 5x − 6x + 9. Find dt x=3 dt x=3

5x − 3 dD dD 12 dx =√ , so = √ (−2) = −4 units/s. dt dt x=3 36 5x2 − 6x + 9 dt

dx dD given that = 4. dt x=3 dt x=3

√ 2x − 3 dD = √ ; D = (x − 2)2 + y 2 = (x − 2)2 + x = x2 − 3x + 4 so dt 2 x2 − 3x + 4

3 dD 3 = √ = units/s. dt x=3 4 2 4

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

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Exercise Set 3.7

113

dx dθ given that = 4. (b) Let θ be the angle of inclination. Find dt x=3 dt x=3 √ x+2 y x dx dθ dx dθ x+2 = so sec2 θ =− √ , = − cos2 θ √ . tan θ = 2 2 x−2 x−2 dt 2 x(x − 2) dt dt 2 x(x − 2) dt

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

2 dt x=3 42 3 2 3 dx dy =3 given y = x/(x2 + 1). Then y(x2 + 1) = x. Differentiating with respect to x, dt dt dy 1 dy dy/dt 1 (x2 + 1) + y(2x) = 1. But = = so (x2 + 1) + 2xy = 1, x2 + 1 + 6xy = 3, dx dx dx/dt 3 3 2 2 2 2 2 2 2 4 6x − 3x − 3 = 0, x + 5x2 − 3 = 0. By the binomial theorem x + 1 + 6x /(x + 1) = 3, (x + 1) + √ 2 2 x = (−5 ± 25 + 12)/2. The minus sign is spurious since x2 cannot be applied to x we obtain √ 2 negative, so x = ( 37 − 5)/2, x ≈ ±0.7357861545, y = ±0.4773550654.

41. Solve

dy dy dx dx 16 dx + 18y = 0; if =

= 0, then (32x + 18y) = 0, 32x + 18y = 0, y = − x so dt dt dt dt dt 9 9 9 16 9 16 256 2 400 2 81 16x2 + 9 x = 144, x = 144, x2 = , x = ± . If x = , then y = − = − .

81 9 25 5 5 9 5 5

9 16 16 9 9 16 and − , .

Similarly, if x = − , then y = . The points are ,− 5 5 5 5 5 5

42. 32x

1 1 1 ds 1 dS dS ds 1 = we get − 2 − 2 = 0, so 43. Find given that = −2. From +

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

dS 225 S 2 ds 1 1 1 dS =− =− 2 . If s = 10, then + = which gives S = 15. So (−2) = 4.5

dt s dt 10 S 6 dt s=10 100 cm/s. The image is moving away from the lens.

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 r be the corresponding dimensions of the cone of water (see figure). We want to show dh dV that is constant and independent of H and R, given that = −kA where V is the volume dt dt of water, A is the area of a circle of radius r, and k is a positive constant. The volume of a cone of 2 1 2 r R R 1 R radius r and height h is V = πr h. By similar triangles = , r = h thus V = π h3 3 h H H 3 H so dV =π dt

R H

2 h2

dh dt

(1)

dV = −kA or, because dt 2 2 R R 2 2 dV h , h2 , = −kπ A = πr = π H dt H

But it is given that

which when substituted into equation (1) gives 2 2 R dh dh R −kπ h2 = π h2 , = −k.

H H dt dt

R r H h

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114

Chapter 3

dr is a constant 45. Let r be the radius, V the volume, and A the surface area of a sphere. Show that dt dV 4 given that = −kA, where k is a positive constant. Because V = πr3 , dt 3 dr dV = 4πr2 dt dt

(1)

dV dV = −kA or, because A = 4πr2 , = −4πr2 k which when substituted into dt dt dr dr = −k. equation (1) gives −4πr2 k = 4πr2 , dt dt

But it is given that

46. Let x be the distance between the tips of the minute and hour hands, and α and β the angles shown in the figure. Because the minute hand makes one revolution in 60 minutes, dα 2π = = π/30 rad/min; the hour hand makes one revolution in 12 hours (720 minutes), thus dt 60

2π dx dβ dβ dα = = π/360 rad/min. We want to find = π/30 and = π/360. given that

dt 720 dt α=2π, dt dt β=3π/2

Using the law of cosines on the triangle shown in the figure, x2 = 32 + 42 − 2(3)(4) cos(α − β) = 25 − 24 cos(α − β), so

dα dβ dx = 0 + 24 sin(α − β) , 2x − dt dt dt

dx 12 dα dβ = − sin(α − β). When α = 2π and β = 3π/2, dt x dt dt

4 x

x2 = 25 − 24 cos(2π − 3π/2) = 25, x = 5; so

11π dx 12 (π/30 − π/360) sin(2π − 3π/2) = in/min. = dt α=2π, 5 150

␤ ␣

3

β=3π/2

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

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

4 r

6 2 6

EXERCISE SET 3.8 1. (a) f (x) ≈ f (1) + f (1)(x − 1) = 1 + 3(x − 1) (b) f (1 + ∆x) ≈ f (1) + f (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 (2)(x − 2) = 1/2 + (−1/22 )(x − 2) = (1/2) − (1/4)(x − 2) (b) f (2 + ∆x) ≈ f (2) + f (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.

h

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Exercise Set 3.8

115

√ 3. (a) f (x) ≈ f (x0 ) + f (x √0 )(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 (b) ∆y

∆y

dy

dy

x 0.1

– 0.1

4. (a) f (x) ≈ f (x0 ) + f (x0 )(x − x0 ) = 1/2 − 1/(2 · 43/2 ) (x − 4) = 1/2 − (x − 4)/16, so with x0 = 4 √ and x =√3.9 we have 1/ 3.9 = f (3.9) ≈ 0.5 − (−0.1)/16 = 0.50625. If x0 = 4 and x = 4.1 then 1/ 4.1 = f (4.1) ≈ 0.5 − (0.1)/16 = 0.49375 (b)

y ∆y dy

∆y

dy 3.9

4

x

4.1

5. f (x) = (1+x)15 and x0 = 0. Thus (1+x)15 ≈ f (x0 )+f (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 (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. x4 ≈ (1)4 + 4(1)3 (x − 1). Set ∆x = x − 1; then x = ∆x + 1 and (1 + ∆x)4 = 1 + 4∆x.

10.

11.

√

x≈

√

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

1 1 1 1 1 1 ≈ − ≈ − ∆x (x − 1), and 2 + x = 3 + ∆x, so 2+x 2 + 1 (2 + 1)2 3 + ∆x 3 9

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

√ 1 f (x) − 3 + √ x < 0.1 if |x| < 1.692.

2 3

0

–2

|

–0.1

f (x) –

( 3 + 2 13 x)|

2

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116

Chapter 3

1 so 9−x 1 1 1 1 1 √ (x − 0) = + x, ≈√ + 3 54 9−x 9 2(9 − 0)3/2

1 1 and f (x) − + x < 0.1 if |x| < 5.5114 3 54

14. f (x) = √

0.06

–6

6 0

| f (x) – ( 15. tan 2x ≈ tan 0 + (sec2 0)(2x − 0) = 2x, and | tan 2x − 2x| < 0.1 if |x| < 0.3158

1 3

+

)|

1 x 54

0.06

–0.8

0.8 0 | f (x) – 2x|

16.

1 1 −5(2) ≈ + (x − 0) = 1 − 10x, (1 + 2x)5 (1 + 2 · 0)5 (1 + 2 · 0)6

0.12

and |f (x) − (1 − 10x)| < 0.1 if |x| < 0.0372

–0.04

0.04 0 | f (x) – (1 – 10x)|

17. (a) The local linear approximation sin x ≈ x gives sin 1◦ = sin(π/180) ≈ π/180 = 0.0174533 and a calculator gives sin 1◦ = 0.0174524. The relative error | sin(π/180)−(π/180)|/(sin π/180) = 0.000051 is very small, so for such a small value of x the approximation is very good. √ (b) Use x0 = 45◦ (this assumes you know, or can approximate, 2/2). π 44π 44π 45π π radians, and 45◦ = = radians. With x = and x0 = we obtain (c) 44◦ = 180 180 4 180 4 √ √

44π π 2 2 −π 44π π π = = 0.694765. With a sin 44◦ = sin ≈ sin + cos − + 180 4 4 180 4 2 2 180 calculator, sin 44◦ = 0.694658.

18. (a) tan x ≈ tan 0 + sec2 0(x − 0) = x, so tan 2◦ = tan(2π/180) ≈ 2π/180 = 0.034907, and with a calculator tan 2◦ = 0.034921 √ (b) use x0 = π/3 because we know tan 60◦ = tan(π/3) = 3 π 60π 61π (c) with x0 = = and x = we have 3 180 180

√ π π 61π π 61π π ≈ tan + sec2 − = 3+4 = 1.8019, tan 61◦ = tan 180 3 3 180 3 180 and with a calculator tan 61◦ = 1.8040

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Exercise Set 3.8

117

19. f (x) = x4 , f (x) = 4x3 , x0 = 3, ∆x = 0.02; (3.02)4 ≈ 34 + (108)(0.02) = 81 + 2.16 = 83.16 20. f (x) = x3 , f (x) = 3x2 , x0 = 2, ∆x = −0.03; (1.97)3 ≈ 23 + (12)(−0.03) = 8 − 0.36 = 7.64 21. f (x) =

22. f (x) =

23. f (x) =

24. f (x) =

√

√ √ 1 1 1 = 8.0625 x, f (x) = √ , x0 = 64, ∆x = 1; 65 ≈ 64 + (1) = 8 + 16 16 2 x

√

√ √ 1 1 x, f (x) = √ , x0 = 25, ∆x = −1; 24 ≈ 25 + (−1) = 5 − 0.1 = 4.9 10 2 x

√

√ √ 1 1 x, f (x) = √ , x0 = 81, ∆x = −0.1; 80.9 ≈ 81 + (−0.1) ≈ 8.9944 18 2 x

√

√ √ 1 1 x, f (x) = √ , x0 = 36, ∆x = 0.03; 36.03 ≈ 36 + (0.03) = 6 + 0.0025 = 6.0025 12 2 x

25. f (x) = sin x, f (x) = cos x, x0 = 0, ∆x = 0.1; sin 0.1 ≈ sin 0 + (cos 0)(0.1) = 0.1 26. f (x) = tan x, f (x) = sec2 x, x0 = 0, ∆x = 0.2; tan 0.2 ≈ tan 0 + (sec2 0)(0.2) = 0.2 27. f (x) = cos x, f (x) = − sin x, x0 = π/6, ∆x = π/180; √

π 3 1 π

◦ ◦ = − ≈ 0.8573 cos 31 ≈ cos 30 + − 2 180 2 360 28. (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) The approximation is (1.1)37 ≈ 1 + 37(0.1) = 4.7, and the calculator value is 34.004. The error is due to the relative largeness of f (1)∆x = 37(0.1) = 3.7.

29. (a) dy = f (x)dx = 2xdx = 4(1) = 4 and ∆y = (x + ∆x)2 − x2 = (2 + 1)2 − 22 = 5

y

(b)

∆y

dy 4

x 2

30. (a) dy = 3x2 dx = 3(1)2 (1) = 3 and ∆y = (x + ∆x)3 − x3 = (1 + 1)3 − 13 = 7

3

y

(b)

8 ∆y dy 1

(1,1)

x

dx 1

2

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118

Chapter 3

31. (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

(b) 2

∆y = 1 dy = 0.5

1

x 0.5

1

√ 32. (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 y

(b) 3

dy = –0.167 ∆ y = –0.171

x 8

9

33. dy = 3x2 dx; ∆y = (x + ∆x)3 − x3 = x3 + 3x2 ∆x + 3x(∆x)2 + (∆x)3 − x3 = 3x2 ∆x + 3x(∆x)2 + (∆x)3 34. dy = 8dx; ∆y = [8(x + ∆x) − 4] − [8x − 4] = 8∆x 35. 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 36. dy = cos x dx; ∆y = sin(x + ∆x) − sin x 37. (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 38. (a) dy = (−1/x2 )dx

39. (a) dy =

√

x 1−x− √ 2 1−x

(b) dy = 5 sec2 x dx

2 − 3x dx = √ dx 2 1−x

(b) dy = −17(1 + x)−18 dx 40. (a) dy = (b) dy =

(x3 − 1)d(1) − (1)d(x3 − 1) (x3 − 1)(0) − (1)3x2 dx 3x2 = =− 3 dx 3 3 2 2 (x − 1) (x − 1) (x − 1)2 2x3 − 6x2 + 1 (2 − x)(−3x2 )dx − (1 − x3 )(−1)dx = dx 2 (2 − x) (2 − x)2

3 3 41. dy = √ dx, x = 2, dx = 0.03; ∆y ≈ dy = (0.03) = 0.0225 4 2 3x − 2 42. dy = √

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

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Exercise Set 3.8

43. dy =

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

44. dy =

119

√

−

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

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

(b) relative error in x is ≈

46. (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 ≈ in V is ≈

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

3x2 dx dx dV = = 3(±0.04) = ±0.12 so percentage error in V is ≈ ±12% =3 V x3 x

47. (a) x = 10 sin θ, y = 10 cos θ (see figure), √

π π

3 π

± = 10 ± dx = 10 cos θdθ = 10 cos 10″ 6 180 2 180 ≈ ±0.151 in,

θ

π π

1 π

y dy = −10(sin θ)dθ = −10 sin ± = −10 ± 6 180 2 180 ≈ ±0.087 in

π π √ π

dx = (cot θ)dθ = cot ± = 3 ± ≈ ±0.030 (b) relative error in x is ≈ x 6 180 180 so percentage error in x is ≈ ±3.0%;

π π

1 π

dy = − tan θdθ = − tan ± = −√ ± ≈ ±0.010 relative error in y is ≈ y 6 180 180 3

x

so percentage error in y is ≈ ±1.0% 48. (a) x = 25 cot θ, y = 25 csc θ (see figure);

π π

dx = −25 csc2 θdθ = −25 csc2 ± 3 360

4 π

= −25 ± ≈ ±0.291 cm, 3 360

π

π π

dy = −25 csc θ cot θdθ = −25 csc cot ± 3 3 360

2 1 π √ = −25 √ ≈ ±0.145 cm ± 360 3 3

25 cm

y

θ

x

csc2 θ dx 4/3 π

=− ≈ ±0.020 so percentage dθ = − √ ± x cot θ 360 1/ 3 dy 1 π

≈ ±0.005 error in x is ≈ ±2.0%; relative error in y is ≈ = − cot θdθ = − √ ± y 360 3 so percentage error in y is ≈ ±0.5%

(b) relative error in x is ≈

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120

49.

Chapter 3

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

50. h = 12 sin θ thus dh = 12 cos θdθ so, with θ = 60◦ = π/3 radians and dθ = −1◦ = −π/180 radians, dh = 12 cos(π/3)(−π/180) = −π/30 ≈ −0.105 ft 1 2 (4) sin 2θ = 4 sin 2θ thus dA = 8 cos 2θdθ so, with θ = 30◦ = π/6 radians and 4 dθ = ±15 = ±1/4◦ = ±π/720 radians, dA = 8 cos(π/3)(±π/720) = ±π/180 ≈ ±0.017 cm2

51. A =

dA dx 2x dx dx 52. A = x2 where x is the length of a side; = 2 , but = ≈ ±0.01 so A x2 x x dA ≈ 2(±0.01) = ±0.02; percentage error in A is ≈ ±2% A dV dx 3x2 dx dx =3 53. V = x3 where x is the length of a side; = , but ≈ ±0.02 3 V x x x dV ≈ 3(±0.02) = ±0.06; percentage error in V is ≈ ±6%. so V

54.

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

1 (πD/2)dD dD dA dA πD2 where D is the diameter of the circle; = =2 , but ≈ ±0.01 so 4 A πD2 /4 D A dD dD 2 ≈ ±0.005; maximum permissible percentage error in D is ≈ ±0.5%. ≈ ±0.01, D D

55. A =

56. 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 .

57. V = volume of cylindrical rod = πr2 h = πr2 (15) = 15πr2 ; approximate ∆V by dV if r = 2.5 and dr = ∆r = 0.001. dV = 30πr dr = 30π(2.5)(0.001) ≈ 235.62 cm3 . π dP 2π 1 2π √ 1 dL 1 58. P = √ L, dP = √ √ dL = √ √ dL, = so the relative error in P ≈ the g g2 L P 2 L 2 g L 1 relative error in L. Thus the percentage error in P is ≈ the percentage error in L. 2 59. (a) α = ∆L/(L∆T ) = 0.006/(40 × 10) = 1.5 × 10−5/◦ C (b) ∆L = 2.3 × 10−5 (180)(25) ≈ 0.1 cm, so the pole is about 180.1 cm long.

60. ∆V = 7.5 × 10−4 (4000)(−20) = −60 gallons; the truck delivers 4000 − 60 = 3940 gallons.

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Review Exercises, Chapter 3

121

REVIEW EXERCISES, CHAPTER 3 2. (a) msec = (b)

(4)2 /2 − (3)2 /2 7 f (4) − f (3) = = 4−3 1 2 f (x1 ) − f (3) x2 /2 − 9/2 = lim 1 x1 →3 x1 →3 x1 − 3 x1 − 3

mtan = lim

x21 − 9 (x1 + 3)(x1 − 3) x1 + 3 = lim = lim =3 x1 →3 2(x1 − 3) x1 →3 x1 →3 2(x1 − 3) 2

= lim

(c)

mtan = lim

x1 →x0

f (x1 ) − f (x0 ) x1 − x0

(d)

x2 /2 − x20 /2 = lim 1 x1 →x0 x1 − x0 x21 − x20 x1 →x0 2(x1 − x0 ) x1 + x0 = x0 = lim x1 →x0 2

10

y Tangent

= lim

3. (a)

mtan = lim

x1 →x0

= lim

x1 →x0

Secant

x

5

f (x1 ) − f (x0 ) (x21 + 1) − (x20 + 1) = lim x1 →x0 x1 − x0 x1 − x0 x21 − x20 = lim (x1 + x0 ) = 2x0 x1 →x0 x1 − x0

(b) mtan = 2(2) = 4 4. To average 60 mi/h one would have to complete the trip in two hours. At 50 mi/h, 100 miles are completed after two hours. Thus time is up, and the speed for the remaining 20 miles would have to be infinite.

3(h + 1)2.5 + 580h − 3 1 1 d 2.5 5. vinst = lim = 58 + (2.5)(3)(1)1.5 = 58.75 ft/s = 58 + 3x h→0 10h 10 dx 10 x=1 6. 164 ft/s 2500

1

20 0

7. (a) vave =

[3(3)2 + 3] − [3(1)2 + 1] = 13 mi/h 3−1 (3t21 + t1 ) − 4 (3t1 + 4)(t1 − 1) = lim = lim (3t1 + 4) = 7 mi/h t1 →1 t1 →1 t1 →1 t1 − 1 t1 − 1

(b) vinst = lim

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122

Chapter 3

dy = lim dx h→0

9. (a)

= lim

h→0

√ 9 − 4(x + h) − 9 − 4x 9 − 4(x + h) − (9 − 4x) = lim √ h→0 h( 9 − 4(x + h) + h 9 − 4x) −4h −4 −2

= √ =√ √ 2 9 − 4x 9 − 4x h( 9 − 4(x + h) + 9 − 4x)

x x+h − dy x + h + 1 x + 1 = lim (x + h)(x + 1) − x(x + h + 1) = lim h→0 dx h→0 h(x + h + 1)(x + 1) h h 1 = lim = h→0 h(x + h + 1)(x + 1) (x + 1)2

(b)

10. f (x) is continuous and differentiable at any x = 1, so we consider x = 1. (a) (b)

lim (x2 − 1) = lim k(x − 1) = 0 = f (1), so any value of k gives continuity at x = 1.

x→1−

x→1+

lim f (x) = lim 2x = 2, and lim f (x) = lim k = k, so only if k = 2 is f (x) differen-

x→1−

x→1−

x→1+

x→1+

tiable at x = 1.

11. (a) (b) (c) (d)

x = −2, −1, 1, 3 (−∞, −2), (−1, 1), (3, +∞) (−2, −1), (1, 3) g (x) = f (x) sin x + 2f (x) cos x − f (x) sin x; g (0) = 2f (0) cos 0 = 2(2)(1) = 4 y

12.

1

x

10 − 2.2 = 0.078 billion, or in 2050 the world population 2050 − 1950 was increasing at the rate of about 78 million per year.

13. (a) The slope of the tangent line ≈

(b)

dN/dt 0.078 ≈ = 0.013 = 1.3 %/year N 6

14. When x4 − x − 1 > 0, f (x) = x4 − 2x − 1; when x4 − x − 1 < 0, f (x) = −x4 + 1, and f is differentiable in both cases. The roots of x4 − x − 1 = 0 are x1 = −0.724492, x2 = 1.220744. So x4 − x − 1 > 0 on (−∞, x1 ) and (x2 , +∞), and x4 − x − 1 < 0 on (x1 , x2 ).

Then

lim f (x) =

x→x− 1

lim (4x − 2) = 3

x→x− 1 3

4x31

1.5

–1.5

− 2 and

lim f (x) = lim+ −4x = −4x31 which is not equal to

x→x+ 1 4x31 −

–1.5

x→x1

2, so f is not differentiable at x = x1 ; similarly f is not differentiable at x = x2 .

15. f (x) = 2x sin x + x2 cos x

2

16. f (x) =

√ 1 − 2 x sin 2x √ 2 x

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Review Exercises, Chapter 3

17. f (x) = 19. (a)

(b)

20. (a)

6x2 + 8x − 17 (3x + 2)2

123

18. f (x) =

(1 + x2 ) sec2 x − 2x tan x (1 + x2 )2

dW dW = 200(t − 15); at t = 5, = −2000; the water is running out at the rate of 2000 dt dt gal/min.

W (5) − W (0) 10000 − 22500 = = −2500; the average rate of flow out is 2500 gal/min.

5−0 5 43 − 23 56 = = 28 4−2 2

(b)

(dV /d+)| =5 = 3+2 =5 = 3(5)2 = 75

21. (a) f (x) = 2x, f (1.8) = 3.6 (b) f (x) = (x2 − 4x)/(x − 2)2 , f (3.5) ≈ −0.7777778 22. (a) f (x) = 3x2 − 2x, f (2.3) = 11.27 (b) f (x) = (1 − x2 )/(x2 + 1)2 , f (−0.5) = 0.48 23. f is continuous at x = 1 because it is differentiable there, thus lim f (1 + h) = f (1) and so f (1) = 0 h→0

f (1 + h) f (1 + h) − f (1) f (1 + h) exists; f (1) = lim = lim = 5. because lim h→0 h→0 h→0 h h h 24.

f (x) = lim

h→0

f (x + h) − f (x) f (x)f (h) − f (x) f (x)[f (h) − 1] = lim = lim h→0 h→0 h h h

= f (x) lim

h→0

f (h) − f (0) = f (x)f (0) = f (x) h

25. The equation of such a line has the form y = mx. The points (x0 , y0 ) which lie on both the line and the parabola and for which the slopes of both curves are equal satisfy y0 = mx0 = x30 − 9x20 − 16x0 , so that m = x20 − 9x0 − 16. By differentiating, the slope is also given by m = 3x20 − 18x0 − 16.

Equating, we have x20 − 9x0 − 16 = 3x20 − 18x0 − 16, or 2x20 − 9x0 = 0. The root x0 = 0 corresponds to m = −16, y0 = 0 and the root x0 = 9/2 corresponds to m = −145/4, y0 = −1305/8. So the line y = −16x is tangent to the curve at the point (0, 0), and the line y = −145x/4 is tangent to the curve at the point (9/2, −1305/8).

26. The slope of the line x + 4y = 10 is m1 = −1/4, so we set the negative reciprocal d 4 = m2 = (2x3 − x2 ) = 6x2 − 2x and obtain 6x2 − 2x − 4 = 0 with roots dx √ 1 ± 1 + 24 = 1, −2/3. x= 6 27. The slope of the tangent line is the derivative

y = 2x 1 = a + b. The slope of the secant is

y

x= 2 (a+b)

a2 − b2 = a + b, so they are equal. a−b

(b, b 2)

(a, a 2) x a a+b b 2

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124

Chapter 3

28. (a) f (1)g(1) + f (1)g (1) = 3(−2) + 1(−1) = −7 (b) (c)

−2(3) − 1(−1) 5 g(1)f (1) − f (1)g (1) = =− 2 2 g(1) (−2) 4 1 1 3

f (1) = √ (3) = 2 2 1 2 f (1)

(d) 0 (because f (1)g (1) is constant) 3 29. (a) 8x7 − √ − 15x−4 2 x (b) 2 · 101(2x + 1)100 (5x2 − 7) + 10x(2x + 1)101 √ 3 (c) 2(x − 1) 3x + 1 + √ (x − 1)2 2 3x + 1 2 2

x (3) − (3x + 1)(2x) 3x + 1 3(3x + 1)2 (3x + 2) (d) 3 = − x2 x4 x7 30. (a) cos x − 6 cos2 x sin x (b) (1 + sec x)(2x − sec2 x) + (x2 − tan x) sec x tan x

csc 2x −2(x3 + 5) csc 2x cot 2x − 3x2 csc 2x (c) − csc2 x3 + 5 (x3 + 5)2 2 + 3 sin2 x cos x (d) − (2x + sin3 x)2 31. Set f (x) = 0: f (x) = 6(2)(2x + 7)5 (x − 2)5 + 5(2x + 7)6 (x − 2)4 = 0, so 2x + 7 = 0 or x − 2 = 0 or, factoring out (2x + 7)5 (x − 2)4 , 12(x − 2) + 5(2x + 7) = 0. This reduces to x = −7/2, x = 2, or 22x + 11 = 0, so the tangent line is horizontal at x = −7/2, 2, −1/2.

4(x2 + 2x)(x − 3)3 − (2x + 2)(x − 3)4 , and a fraction can equal zero (x2 + 2x)2 only if its numerator equals zero. So either x − 3 = 0 or, after factoring out (x − 3)3 , 4(x2 + 2x) − (2x + 2)(x − 3) = 0, 2x2 + 12x + 6 = 0, whose roots are (by the quadratic formula) √ √ √ −6 ± 36 − 4 · 3 = −3 ± 6. So the tangent line is horizontal at x = 3, −3 ± 6. x= 2

32. Set f (x) = 0: f (x) =

33. Let y = mx + b be a line tangent to y = x2 + 1 at the point (x0 , x20 + 1) with slope m = 2x0 .

By inspection if x1 = −x0 then the same line is also tangent to the curve y = −x2 −1 at (−x1 , −y1 ), since y1 = −y0 = −x20 − 1 = −(x20 + 1) = −x21 − 1.

Thus the tangent line passes through the points (x0 , y0 ) and (x1 , y1 ) = (−x0 , −y0 ), so its slope y0 − y 1 2y0 x2 + 1 x2 + 1 m= = = 0 . But, from the above, m = 2x0 ; equate and get 0 = 2x0 , with x0 − x1 2x0 x0 x0 solution x0 = ±1. Thus the only possible such tangent lines are y = 2x and y = −2x.

34. (a) Suppose y = mx + b is tangent to y = xn + n − 1 at (x0 , y0 ) and to y = −xn − n + 1 at (x1 , y1 ).

Then m = nxn−1 = −nxn−1 , and hence x1 = −x0 . Since n is even, y1 = −xn1 − n + 1 = 0 1 n n −x0 − n + 1 = −(x0 + n − 1) = −y0 . Thus the points (x0 , y0 ) and (x1 , y1 ) are symmetric with respect to the origin and both lie on the tangent line and thus b = 0.

The slope m is given by m = nxn−1 and by m = y0 /x0 = (xn0 + n − 1)/x0 , hence nxn0 = 0 xn0 + n − 1, (n − 1)xn0 = n − 1, xn0 = 1. Since n is even, x0 = ±1. One easily checks that y = nx is tangent to y = xn + n − 1 at (1, n) and to y = −xn − n + 1 at (−1, −n).

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Review Exercises, Chapter 3

125

(b) Suppose there is such a common tangent line with slope m. The function y = xn + n − 1 is always increasing, so m ≥ 0. Moreover the function y = −xn − n + 1 is always decreasing, so m ≤ 0. Thus the tangent line has slope 0, which only occurs on the curves for x = 0. This would require the common tangent line to pass through (0, n − 1) and (0, −n + 1) and do so with slope m = 0, which is impossible. d (3x − tan x) = 3 − sec2 x = 1, or sec2 x = 2, 35. The line y − x = 2 has slope m1 = 1 so we set m2 = dx √ sec x = ± 2 so x = nπ ± π/4 where n = 0, ±1, ±2, . . . .

36. Solve 3x2 − cos x = 0 to get x = ±0.535428. √ √ 37. 3 = √ f (π/4) = (M + N ) 2/2 and 1 = f (π/4)√= (M − N√) 2/2. Add these√two equations √ to get 4 = 2M, M = 23/2 . Subtract to obtain 2 = 2N, N = 2. Thus f (x) = 2 2 sin x + 2 cos x.

3π 3π = −3 so tangent line is y − 1 = −3 x − . f 4 4 √ √ 38. f (x) = M tan x+N sec x, f (x) = M sec2 x+N √ x. At √ x = π/4, 2M + 2N, 0√= 2M + 2N . √ sec x tan Add to get M = −2, subtract to get √ N = 2 + M/ 2 = 2 2, f (x) = −2 tan x + 2 2 sec x. f (0) = −2 so tangent line is y − 2 2 = −2x.

39. f (x) = 2xf (x), f (2) = 5 (a) g(x) = f (sec x), g (x) = f (sec x) sec x tan x = 2 · 2f (2) · 2 · 3

(x − 1)f (x) − f (x) f (x)

(b) h (x) = 4 , x−1 (x − 1)2 h (2) = 4 40. (a)

√

√ 3 = 40 3.

2 · 2f (2) − f (2) 53 f (2) − f (2) = 4 · 53 = 4 · 53 · 3 · 5 = 7500 1 1 1

1 dT 2 1 =√ √ =√ dL g2 L gL

(b) s/m

dT > 0 an increase in L gives an increase in T , which is the period. To speed up a (c) Since dL clock, decrease the period; to decrease T , decrease L. √ dT L (d) = − 3/2 < 0; a decrease in g will increase T and the clock runs slower dg g √

√ L dT −1 −3/2 =2 L g (e) = − 3/2 (f ) s3 /m dg 2 g 41. A = πr2 and 500π m2 /min.

dr dA dA dr = −5, so = = 2πr(−5) = −500π, so the area is shrinking at a rate of dt dt dr dt

dy dθ dz = a and = −b. From the figure 42. Find given dt x=1 dt dt y=1 √ sin θ = y/z; when x = y = 1, z = 2. So θ = sin−1 (y/z)

a dθ 1 dy 1 y dz = −b − √ and = − 2 2 2 dt z dt z dt 2 1 − y /z when x = y = 1.

z

y

θ

x

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126

Chapter 3

43. (a) ∆x = 1.5 − 2 = −0.5; dy =

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

1 1 = 2 − 1 = 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. ∆y =

(c) ∆x = 3 − 0 = 3; dy = √ ∆y =

√

25 − 32 −

√

−0 −x = (3) = 0; and 2 25 − x 25 − (0)2

25 − 02 = 4 − 5 = −1.

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

π 46π π π π x− =1−2 − = 0.9651; cot 46◦ = cot x ≈ cot − csc2 4 4 4 180 4 with a calculator, cot 46◦ = 0.9657.

44. cot 46◦ = cot

51 π radians and 45. (a) h = 115 tan φ, dh = 115 sec2 φ dφ; with φ = 51◦ = 180 π

radians, h ± dh = 115(1.2349) ± 2.5340 = 142.0135 ± 2.5340, so dφ = ±0.5◦ = ±0.5 180 the height lies between 139.48 m and 144.55 m. (b) If |dh| ≤ 5 then |dφ| ≤

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

115 180