The domain is () 0, ∞ .

41. The domain is () 0, ∞ .

∫ –1/ 2 dt < ∫ dt = ∫ t dt

() 1 = if ln = , or =.

f ' () x < for 0 x < and 1 f ' () x > for 0 x > 1 = ⎡ 2 t ⎣ ⎤ ⎦ = 2( x –1) 1

so f(1) = –1 is a minimum.

so ln x < 2( x – 1)

Section 6.1 Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No

51. From Ex 10,

ln 3 x 2( x – 1) 3

so 0 <

. ∫ π sec csc x x dx = ⎡− ⎣ ln cos x + ln sin x ⎤ ⎦

ln x

2( x + 1)

Hence 0 ≤ lim

= ln( 3) ln1 0.5493 0 − = −= 0.5493

52. Let u =+ 1 sin x ; then du = cos x dx so that

dx = ∫ du =

→∞ n 1 n 2 2 n

1 sin x

= ln 1 sin + x += C ln(1 sin ) + x + C

(since 1 sin

+ x ≥ for all x ). 0

= lim ∑ ⎜ i ⎟ ⋅ = ∫ dx = ln 2 ≈ 0.693 53. V =π 2 ∫ xf x dx = 1 () ∫ 1 2 dx

48. ≈ 72,382 ∫ 2 dx =π ∫ du =π ln u + C

2 ⎝ 4 ax + b ⎠ ⎝ ax + b ⎠ ⎡ ⎤ ∫

1 x 2 dx =π ⎢ ln x + 4 ⎥

[ln( ax – ) – ln( b ax + b )] = π ln 20 − π ln 5 = π ln 4 ≈ 4.355

2 2 = 1 L = ∫ 1 + ⎜ ⎟ dx = ∫ 1 + ⎜ – ⎟ a dx − b 1 ⎝ dx ⎠ 1 ⎝ 2 2 x ⎠

b. fx ′ () = cos u ⋅

2 du

= ∫ 1 ⎜ + ⎟ dx = ⎜ + ⎟ dx

2 2 2 x + 1 2 = 2 cos [ln( x + x –1)] ⋅ 1 ⎡ x ⎤ 1⎡

f ′ (1) = cos [ln(1 2 2 211 + ⋅+ 1 –1)] ⋅ 2 3 1

50. From Ex 9,

0 tan x dx = ⎡− ⎣ ln cos x ⎦ ⎤ 0

= ln cos 0 − ln cos π

ln(1) ln(0.5) =⎜ ln

= ln 2 ≈ 0.69315

Instructor’s Resource Manual

Section 6.1 351

+ 1 1.5sin x

b. f ′′ () x =−

(1.5 sin ) + x 2

On [0,3 ], π f ′′ () x = 0 when x ≈ 3.871,

5.553. Inflection points are (3.871, –0.182), (5.553, –0.182).

c. ∫

0 ln(1.5 sin ) x dx ≈ 4.042

1 1 1 + + ⋅⋅⋅ + = the lower approximate area

sin(ln ) x

= the upper approximate area

On [0.1, 20], fx ′ () = 0 when x = 1.

2 n –1

ln n = the exact area under the curve

Critical points: 0.1, 1, 20

f (0.1) ≈ –0.668, f(1) = 1, f(20) ≈ –0.989 Thus,

On [0.1, 20], the maximum value point is

1 1 1 1 1 1 (1, 1) and minimum value point is + + ⋅⋅⋅ + < ln n < + + + ⋅⋅⋅ + 1 . (20, –0.989).

b. On [0.01, 0.1], fx ′ () = 0 when x ≈ 0.043.

dt – ∫ dt ∫ dt

f (0.01) ≈ –0.107, f(0.043) ≈ –1

ln – ln y x

1 1 56. x = t t = t On [0.01, 20], the maximum value point is y – x

(1, 1) and the minimum value point is

= the average value of

on [x, y].

c. ∫ cos(ln ) x dx ≈− 8.37

Since is decreasing on the interval [x, y], the

t average value is between the minimum value of

and the maximum value of . y

fx ′ () = 0 when cos x = 0.

πππ 3 5 a. ∫ x ln − x ln dx = ≈ 0.139

Critical points:

b. Maximum of ≈ 0.260 at x ≈ 0.236

f ⎜⎟ ≈ 0.916, 0.693, f ⎜ ⎟ ≈−

⎝ 2 ⎠ On [0,3 ], π the maximum value points are

⎜ , 0.916 , , 0.916 2 ⎟⎜ ⎟ and the minimum

a. ∫ [ ln x x − x ln ] x dx = ≈ 0.194

value point is ⎜ , 0.693 . − 0 36

b. Maximum of ≈ 0.521 at x ≈ 0.0555

Section 6.1 Instructor’s Resource Manual

6.2 Concepts Review

11. fz ′ () = 2( – 1) z > 0 for z > 1

f (z) is increasing at z = 1 because f(1) = 0 and

fx () 1 ≠ fx ( 2 )

f (z) > 0 for z > 1. Therefore, f(z) is strictly

increasing on z ≥

1 and so it has an inverse.

2. x; f () y

12. fx ′ () = 2 x +> 10 for x ≥ 2 . f(x) is strictly

3. monotonic; strictly increasing; strictly decreasing

increasing on x ≥ 2 and so it has an inverse.

13. fx ′ () = x 4 + x 2 + 10 > 0 for all real x. f(x) is strictly increasing and so it has an inverse.

Problem Set 6.2

cos

14. fr () = cos 4 ∫ r tdt = – 4 ∫ tdt 1

1. f(x) is one-to-one, so it has an inverse.

Since f (4) = f 2, (2) − 1 = 4 .

π r < 0 for all r ≠π+ k , k any

fr ′ () = – cos 4

2. f(x) is one-to-one, so it has an inverse.

integer.

Since f(1) = 2, f − 1 (2) 1 = .

f (r) is decreasing at r =π+ k since fr ′ () < 0 2

3. f(x) is not one-to-one, so it does not have an

on the deleted neighborhood

inverse.

⎜ k π+− ,. k π++ ⎞ ε ε Therefore, f(r) is 2 2 ⎟

4. f(x) is not one-to-one, so it does not have an

strictly decreasing for all r and so it has an

inverse.

5. f(x) is one-to-one, so it has an inverse.

15. Step 1:

Since f(–1.3)

y =x+1 x =y–1

6. f(x) is one-to-one, so it has an inverse. Since

Step 2: f –1 () y = y –1

f ⎛⎞= f − ⎜⎟ 1 2, (2) = .

⎝⎠ 2 2 Step 3: f –1 () x = x –1 Check:

x ≠ 0. f(x) is strictly decreasing at x = 0 because

( –1 ( )) x = ( – 1) 1 x += x

(x) > 0 for x < 0 and f(x) < 0 for x > 0. Therefore

f (x) is strictly decreasing for x and so it has an inverse.

16. Step 1:

8. fx ′ () = 7 x 6 + 5 x 4 > 0 for all x ≠ 0.

f (x) is strictly increasing at x = 0 because f(x) > 0

– = y –1

for x > 0 and f(x) < 0 for x < 0. Therefore f(x) is

strictly increasing for all x and so it has an

x = –3(y – 1) = 3 – 3y

inverse.

Step 2: f –1 () y = 3–3 y

9. f ′ () θ = – sin θ < 0 for 0 < θ < π

Step 3: f –1 () x = 3–3 x

f (θ) is decreasing at θ = 0 because f(0) = 1 and

Check:

f (θ) < 1 for 0 < θ < π . f(θ) is decreasing at

f –1

( ( )) fx = 3–3– ⎜ + 1 =+ 3 ( – 3) x = x

θ = π because f(π ) = –1 and f(θ) > –1 for

0 < θ < π . Therefore f(θ) is strictly decreasing

on 0 ≤ θ ≤ π and so it has an inverse.

f (x) is decreasing on 0 << x and so it has an 2

inverse. Instructor’s Resource Manual

Section 6.2 353

17. Step 1:

x –2 =

y = x + 1 (note that y ≥ 0 )

x += 1 y 2 1 x =+ 2 2 , y > 0

x = y 2 – 1, 0 y ≥ y

Step 2: f –1 () y = y 2 – 1, 0 y ≥ –1

Step 2: f () y =+ 2 2 , 0 y >

Step 3: f –1

() x = x 2 – 1, 0 x ≥

Check:

Step 3: f –1 () x =+ 2 , 0 x >

() x –2

18. Step 1:

x –2

y = –1– x (note that y ≤ 0 )

x = 1– y 2 , 0 y ≤ = x = x Step 2: f –1 ()1– y = y 2 , 0 y ≤

2 21. Step 3: Step 1: f ()1– x = x , 0 x ≤

y = 4 x 2 , x ≤ 0 (note that y ≥ 0 )

, negative since x = –(–x) = x ≤ 0

19. Step 1:

Step 2: f –1 () y =−

y = – x –3

Step 3: f –1 () x =−

Step 2: x

1 Step 3: f –1 () x = 3–

22. Step 1:

Check:

y = ( – 3) , 3 x 2 x ≥ (note that y ≥ 0 )

= x Step 2: f – –1 () y =+ y

3– x –3

Step 3: f –1 () x =+ 3 x

(note that y > 0)

Section 6.2 Instructor’s Resource Manual

23. Step 1:

1 + y 1/ 3

Step 2: f –1 () y =

Step 2: f –1 ()1 y =+ 3 y Check:

Step 3: f –1 ()1 x =+ 3 x 1 ⎡ x –1 3 () ⎤

Check: f ( ( )) 1 fx =+ ( –1) x =+ 1 ( –1) x

= ( ( )) x

1– ⎡ x –1 ⎤

⎢ ⎣ () x + 1 ⎥ ⎦

Step 2: f () y = y

Step 3: f () x = x

⎛ 2 x 1/ 3 Check: 3 ⎞

ff ( –1 ( )) x = ( x 2/55/2 ) = x

27. Step 1:

x 3 25. Step 1: + 2

Step 2: f () y =

Step 2: f () y

Step 3: f () x =⎜

1– x + 1 x + 1– x + 1 2 Check:

⎝ 26. Step 1: x 31 + ⎠

⎣ ⎢ () x –1 ⎦ ⎥ ⎤+

⎢ ⎣ () x ⎥ ⎦

Instructor’s Resource Manual

Section 6.2 355

28. Step 1: r 4 2 h

5 29. By similar triangles, = . Thus, r =

h 6 3 y =⎜

⎜ 3 ⎟ This gives

⎟ v = 0 when v 0 = 32 t ⎜ , that is, when y ⎝ 1/ 5 –1 ⎟ ⎠

t = 0 . The position function is

⎛ 2– y 1/ 5 1/ 3 ⎞

Step 2: f –1 () y =⎜ ⎜

y 1/ 5 –1

0 16 t 2 ⎠ . The ball then reaches a height

Step 3: f () x =⎜

31. fx () = 4 x + 1; ( ) fx > 0 when x >− and

1 fx () < 0 when x <− .

The function is decreasing on ⎜ −∞ − ,

. Restrict the domain to ⎛ 3 1/ 3

increasing on − ,

or restrict it to − , ∞⎟ .

Then f − 1 ⎢ 1 ⎜ ⎟ ⎥ + 2 () x = (1 −− 8 x + 33) or

f () x = (1 −+ 8 x + 33).

Section 6.2 Instructor’s Resource Manual

() < 0 when x < .

The function is decreasing on ⎜ −∞ ,

and

increasing on

, ∞⎟ ⎢⎣ . Restrict the domain to

or restrict it to

2 37. fx ′ ( ) 15 = x 4 + 1 and y = 2 corresponds to x = 1, − 1 f 1 () x = (3 + 4 x + 5). 1 1 1

38. fx ′ () = 5 x 4 + 5 and y = 2 corresponds to x = 1, 1 1 1

so ( f − 1 ) (2) ′ =

= f ′ (1) 55 + 10

39. fx ′ () = 2sec 2 x and y = 2 corresponds to x = , 4

so ( f ) (2) =

2 π = cos 2 ⎜⎟ f ′ ()

4 2sec () 4 ⎝⎠

and y = 2 corresponds to x = 3,

41. ( g –1 f –1 )( ( )) hx = ( g –1 f –1 )( ( ( ))) fgx = g –1 [ f –1 ( ( ( )))] fgx = g –1 [ ( )] gx = x

Similarly,

hg (( –1 f –1 )( )) x = fgg ( (( –1 f –1 )( ))) x = fgg (( –1 ( f –1 ( )))) x = ff ( –1 ( )) x = x

Thus h –1 = g –1 f –1

( f − 1 ) (3) 1 ′ ≈− 3

Instructor’s Resource Manual

Section 6.2 357

42. Find f − 1 (): x ax + 44. a. b y =

Find g − 1 (): x 1 dx b −

b. If bc – ad = 0, then f(x) is either a constant y − 2

function or undefined.

g − 1 () y =

c. If f = f − 1 , then for all x in the domain we

− 1 x − g 2 () x = have:

3 x + 2 (ax + b)(cx – a) + (dx – b)(cx + d) = 0

() x − 2 acx h 2 − 1 x = g − 1 f − 1 x = g − 1 = + ( bc − a 2 ) x − ab + dcx 2

Setting the coefficients equal to 0 gives three

⎠⎣ () x 2 ⎦ + 2 () 1 x

(2) a = ±d (3) a = –d or b = 0

43. f has an inverse because it is monotonic If a = d, then f = f − 1 requires b = 0 and (increasing):

= . If a = –d, there are x

no requirements on b and c (other than

− 1 1 1 a. ( f )() ′ A = =

= 1 bc − ad ≠ 0 ). Therefore, f = f − 1 if a = –d

f ′ π () + 2 2 1 cos π ()

2 or if f is the identity function.

b. ( f − 1 1 )() 1 ′ B = =

f ′ 5 () π + 1 cos 2 6 5 π () 7

f − ∫ 1 () y dy = 0 (Area of region B)

= 1 – (Area of region A)

1 ∫ () f x dx =−= 1

Section 6.2 Instructor’s Resource Manual

46. a

1 1 p ∫ –1 () f x dx = the area bounded by y = f(x), y = 0,

0 47. Given p > 1, q > 1, += 1, and fx () = x , p q

and x = a [the area under the curve].

, ∫ so

+= 1 for p gives p =

0 ()

the area bounded by

ab = the area of the rectangle bounded by x = 0,

Case 1: b > f(a)

q –1

⎢ ⎣ q –1 ⎥ ⎦

Thus, if y = x p –1 then x = y p –1 = y q –1 , so

f –1 () y = y q –1 .

By Problem 44, since fx () = x p − 1 is strictly increasing for p > 1, a

ab ≤

y q ∫ –1

0 ⎢ ⎣ ⎥ The area above the curve is greater than the area ⎦ 0

of the part of the rectangle above the curve, so q a b

ab ≤

the total area represented by the sum of the two

integrals is greater than the area ab of the rectangle. Case 2: b = f(a)

6.3 Concepts Review

1. increasing; exp

2. ln e = 1; 2.72

3. x; x x

The area represented by the sum of the two

4. e x ; e + C

integrals = the area ab of the rectangle. Case 3: b < f(a)

Problem Set 6.3

1. a. 20.086

b. 8.1662

c. e 2 ≈ e 1.41 ≈ 4.1

d. e cos(ln 4) ≈ e 0.18 ≈ 1.20

2. a. e 3ln 2 = e ln(2 ) = e ln 8 = 8

The area below the curve is greater than the area

ln 64

1/ 2

of the part of the rectangle which is below the

b. e 2 = e ln(64 ) = e ln 8 = 8

curve, so the total area represented by the sum of the two integrals is greater than the area ab of the

3. e 3ln x = e ln x rectangle. 3 = x 3

a ab b ≤ () f x dx + − 1 ∫

0 ∫ f () 0 y dy with equality

e –2 ln x

4. = e ln x = x − 2 = 1

holding when b = f(a).

Instructor’s Resource Manual

Section 6.3 359

The graph of y = e x is reflected across the

= e x 3 ln x ( x 2 + 3 x 2 ln ) x x -axis.

2 x = 3 xe ln x (1 3ln ) + x

Section 6.3 Instructor’s Resource Manual

26. fx () = e 2 Domain = ( −∞ ∞ ,)

fx ′

() =− e 2 1 , − f ′′ () x = e 2

2 4 Since fx ′ () < for all x, f is decreasing on 0

Since f ′′ () x > for all x, f is concave upward on 0 ( −∞ ∞ . ,)

Since f and f′ are both monotonic, there are no

The graph of

is reflected across the

extreme values or points of inflection.

-axis.

24. a <⇒ b – a > – b ⇒ e – a > e – b , since e x is an increasing function.

25. fx = x () e 2 Domain = ( −∞ ∞ ,) fx ′ = e 2 () x 2 , f ′′ () x = 4 e 2 x 4 Since fx ′ () > for all x, f is increasing on 0 ( −∞ ∞ . ,)

Since f ′′ () x > for all x, f is concave upward on 0 −5

Since f and f′ are both monotonic, there are no

27. fx = xe − x ()

Domain = ( −∞ ∞ ,)

extreme values or points of inflection.

f is increasing on ( −∞ and decreasing on ,1]

[1, ) ∞ . f has a maximum at (1, 1 e )

f is concave up on (2, ) ∞ and concave down on

, 2) . f has a point of inflection at (2, 2 2 x ( −∞ 2 )

Instructor’s Resource Manual

Section 6.3 361

28. fx () = e x + Domain = ( x −∞ ∞ ,)

30. fx () = ln(2 x − . Since 2 1) x − > if and only if 10

1 fx 1 ′ () = e x + 1, f ′′ () x = e x x > , domain =

Since fx ′ () > for all x, f is increasing on 0 −

Since f ′′ () x > for all x, f is concave upward on 0 Since fx ′ () > for all domain values, f is 0 ( −∞ ∞ . ,)

2 are both monotonic, there are no ∞.

increasing on (,) 1

Since f and f′

extreme values or points of inflection. Since f ′′ () x < for all domain values, f is 0

concave downward on (,) y 1 2 ∞.

Since f and f′ are both monotonic, there are no

extreme values or points of inflection.

29. fx () = ln( x 2 + Since 1) x 2 + > for all x, 10

x x + 1 ( x 2 + 1) 2 31. fx () = ln(1 + e ) Since 1 + e > for all x, 0 x ( −∞ − , 1) − 1 ( 1,0) 0 (0,1) 1 (1, ) − ∞

f is increasing on (0, ) ∞ and decreasing on Since fx ′ () > for all x, f is increasing on 0 ( −∞ , 0) . f has a minimum at (0, 0)

f is concave up on ( 1,1) − and concave down on Since f ′′ () x > for all x, f is concave upward on 0 ( −∞ − ∪ ∞ . f has points of inflection at , 1) (1, )

( 1, ln 2) − and (1, ln 2) Since f and f′ are both monotonic, there are no

extreme values or points of inflection.

Section 6.3 Instructor’s Resource Manual

[0, ) ∞ . f has a maximum at (0, ) e −3

f is concave up on ( −∞ − , 2 ) ∪ ( 2 ,) ∞ and

34. fx () = e x − e x

Domain = ( −∞ ∞ ,) fx ′ () = e x + e − x , f ′′ () x = e x − e − 2 x 2

concave down on ( − 2 , 2 ) . f has points of x ( −∞ , 0) 0 (0, ) ∞

2 inflection at 2 ( −

2 , e ) and ( 2 , e ) f ′

f is increasing on ( −∞ ∞ and so has no extreme ,) values. f is concave up on (0, ) ∞ and concave down on ( −∞ , 0) . f has a point of inflection at

33. fx () = e −− ( x 2) 2 Domain = ( −∞ ∞ ,)

Note that 4 x 2 − 16 x + 14 = when 0

f is increasing on ( −∞ , 2] and decreasing on [2, ) ∞ . f has a maximum at (2,1)

f is concave up on ( −∞ , 2 ) ∪ (

concave down on (

2 2 . f has points

of inflection at (

2 , ) e and (

Instructor’s Resource Manual

Section 6.3 363

37. Let u = 3x + 1, so du = 3dx.

f is increasing on ( −∞ ∞ and so has no extreme ,)

∫ xe dx = e − 3 ∫ u 2 x dx 2 = 2 ∫ e du

values. f is concave up on ( −∞ , 0) and concave

down on (0, ) ∞ . f has a point of inflection at

e u += C e x 23 − + C

39. Let u = x 2 + 6 x , so du = (2x + 6)dx.

∫ x ( x + 3) e x 26 + 1 dx 1 = u u

2 ∫ e du = e + C 2

dx = ∫ du = u += C e x −+ C

36. x fx () = te dt − t ∫

e − 1/ x x ( −∞ ,0) 0 (0,1) 1 (1, ) ∞

dx = u e du = e u += C e − 1/ x

f ′′ +

f x is increasing on [0, ) ∞ and decreasing on

e xe + dx = e x ⋅ e e 42. x ∫ ∫ dx

( −∞ , 0] . f has a minimum at (0, 0)

Let x u = e , so du = e dx .

f is concave up on ( −∞ and concave down on ,1)

∫ e ⋅ e dx = ∫ e du = e u += C e e + C (1, ) ∞ . f has a point of inflection at (1, ∫ te dt ) .

Note: It can be shown with techniques in

43. Let u = 2x + 3, so du = 2dx

∫ te dt =−≈ 1 0.264 ∫ e dx = ∫ e du = e u += C e 2 x + 3 + C 0 e 2 2 2

Chapter 7 that

∫ e dx = e 2 x + ⎢ 3 ⎥ = e 5 – e 3 0 ⎣ 2 ⎦

∫ u e du = – e u + C

– e 3/ x + C

Section 6.3 Instructor’s Resource Manual

45. ln 3

V =π

e 2 x dx ⎧ ⎡ ⎛ 0.3 ⎞ 0.3 ⎤ 0.3 ∫ ⎫

ln 3

( e x ) 2 dx =π

0 ∫ 0 50. e 0.3 ≈ ⎨ ⎢ ⎜ + 1 ⎟ + 1 ⎥ + ⎬ 1 0.3 () + 1

ln 3

=π e e 2 ln 3 =π 1 ⎜ − e 0 ⎟ = 4 π ≈ 12.57 = 1.3498375

e 0.3 ≈ 1.3498588 by direct calculation

46. t V = ∫ 2 π xe − x

ds = dx 2 + dy ∫ 2 2 π xe dx = −π ∫ e (2) − x dx = −π ∫ e du

2 = e u t − x (sin t + cos ) t 2 + (cos t − sin ) t 2 dt

= −π + = −π e C e + C

1 = e t 2sin 2 t +

2 cos tdt = 2 t e dt

2 π xe x dx = −π e − x

=π –( e − ∫ 1 ⎢ ⎥ − e 0 ) The length of the curve is

− 1 e dt

⎡⎤ ∫ t 2 = 2 e = e π −≈

52. Use x = 30, n = 8, and k = 0.25.

47. The line through (0, 1) and ⎜ 1, ⎟ has slope

n − kx

() kx e (0.25 30) ⋅ e

2 is of the form .

– 1)(1) – ( x xe )

48. fx ′ () =

x →∞ + 1 (ln ) x 2 x →∞ 2 ln x

x (– e )(–1)

0 when ln x =± 1 so x = e 1 = e

When x > 0, fx ′ () < so f(x) is decreasing for 0,

49. a. Exact:

⎝⎠ e 2 1 2 + ln 1 + 1 (–1)

Approximate: e

Maximum value of

at x = e; minimum

value of

Instructor’s Resource Manual

Section 6.3 365

y -axis so the area is

c. Fx () = ∫

1 (ln ) 3 t 2 ⎧⎪ x 2 x 2

2 2 [ e − − 2 e (2 x ⎨ 2 ∫ − 0 1)] dx

ln( e ) 2 1 ∫

3 [2 e (2 x −− 1) e ] dx ⎬

⋅ 2 e 2 ⎪⎭

+ 1 [ln( e )] 22 11 + 2 ≈ 4.2614 = e ≈ 1.65

be the point of tangency. Then

x 0 b. fx ′ () = xe p – x (–1) + e – x ⋅ px p –1

so the line is y = e x 0 x or y = ex.

fx ′ () = when x = p 0

ex 2 ⎤

59. lim ln( x 2 + e – x )

= ∞ (behaves like x −)

⎥ x →∞ ∫ – 0

a. A = ( e – ex dx ) =⎢ e –

e e lim ln( x + e – x ) = ∞ (behaves like 2ln x )

= e −− ( e 0 − 0) = – 1 0.36 ≈ x →∞

60. fx ′ () = –(1 + e x –1 ) –2 1 –1 x 2 2 ⋅ e x (– x –2 )

b. V =π ∫ [( e )–()] ex dx

∫ ( e – ex ) dx =π⎢ e –

=π ⎢ e 2 − −

0 ⎜ 2 e ⎟ ⎥ = ( e – 3) ≈ 2.30

55. a. ∫ exp ⎜ − 2 ⎟ dx = 2 ∫ exp ⎜ − 2 ⎟ dx ≈ 3.11

e 0.1 ∫ x sin x dx

8 b. π−

56. a. lim (1 + x ) 1 x = e ≈ 2.72

b. lim (1 + x ) − 1 x 1 = ≈ 0.368

y = f(x) and y = f ′′ () x intersect when

2 e. f has no minimum or maximum values. Both graphs are symmetric with respect to the

Section 6.3 Instructor’s Resource Manual

6.4 Concepts Review

2. e ln12

Problem Set 6.4

12. 7 log (8.57) ln 8.57 10 = 7 ≈ 6.5309

2. x = 5 2 = 25 ln17

4. x 4 = 64 x ln13 = ≈ 1.5937 ln 5 x = 4 64 = 22

16. ln12 = ln 4 =

Alternate method:

D x log 3 e = D x ( log x 3 e ) = log e

ln e = 1 = ≈

0.9102 ln 3 ln 3

Instructor’s Resource Manual

Section 6.4 367

20. D x log ( 10 x + 9) =

–3 x

⋅ D x ( x + 9) 26. ∫ (10 + 10 ) dx = 10 3 ∫ x dx + 10 –3 x dx

x 3 + 9) ln10

Let u = 3x, so du = 3dx.

1 10 = u ( x 3 + 9) ln10

∫ 10 dx = 10 du =⋅ + 3 C ∫ 3 ln10

Now let u = –3x, so du = –3dx.

+ ln( z + 5) ln 3 ∫

22. D θ 2– log (3 10 ) = D θ ( θ 2 – ) log 3 θ 10 1 10 Thus, –10 (10 3 ∫ x + 10 –3 x ) dx

–3 x 1 ⎤

⎣ ⎢ 3ln10 ⎦ ⎥ ( θ – ) ln 3 θ

27. 10 ( ) = 10 ( ) ln10 x 2 = 10 ( ) x 2 ln10

2 dx θ 2 – θ ln10

( x ) 2 10 d = x 20 = 20 x 19

23. Let u = x 2 so du = 2xdx.

+ ( x )] ∫

2 ∫ 2 du =⋅ + C dx dx

= 10 ) x 2 ln10 20 + x 19

2 ln 2

ln 2

28. sin 2 x d = 2sin x sin x = 2sin cos x x

x = sin ∫ x 10 dx = ∫ 10 du =⋅ + C 2 2 ln 2 sin 2 ln 2 cos x

x + 2 sin = x C ) + 5ln10

10 x 5 –1

dy d = (sin 2

dx dx

= 2sin cos x x + 2 sin x cos ln 2 x

dx = 25 ∫ u du =⋅ 2 + C d x

Section 6.4 Instructor’s Resource Manual

35. fx () = 2 x

(ln 2)( − x )

30. 2 = 2 ln 2 e = 2 e ln 2 = e Domain = ( −∞ ∞ ,)

e ln 2 Since fx ′ () < 0 for all x, f is decreasing on

[2 ( e ) + (2 ) ] ex Since f ′′ () x > 0 for all x, f is concave upward on

dx dx

= 2 ( ) e x ln 2 (2 ) + ex e ln 2 Since f and f′ are both monotonic, there are no extreme values or points of inflection.

e (ln ) ln( x x 2 + 1) d [(ln ) ln( x x 2 + 1)] 4

32. y = (ln x 22 ) x + 3 = 2 e (2 x + 3) ln(ln x ) −2

dy =

e (2 x + 3) ln(ln x 2 ) d [(2 x +

3) ln(ln x 2 )] − 36. x fx () = x 2 Domain = ( −∞ ∞ ,)

3) ln(ln x 2 ) ⎡

fx ′ ( ) [1 (ln 2) ]2 =− x ,

(2 x +

= e ⎢ 2 ln(ln x ) (2 + x + 3)

(2 ) x ⎥

ln xx 2 2 ⎦

f ′′ () x = (ln 2)[(ln 2) x − 2]2

fx ′ () = e (sin ln ) x x f is increasing on ⎜ −∞ ,

⎥ and decreasing on

⎢ (sin ) x ⎜⎟ + (cos )(ln ) x x ⎥ ⎢

, ∞⎟ . f has a maximum at ( , 1

+ cos ln x x ⎟ f 2 is concave up on ( ,) ∞ and concave down on

ln 2

f (1) 1 = sin1

⎛ sin1

+ cos1ln1 = sin1 0.8415 ⎜ 2 ≈ ( −∞ , ) . f has a point of inflection at ⎝ 1

ge () = e π ≈ 23.14

g (e) is larger than f(e).

() e =π ln π≈ 25.71

ge ′ () =π e π− 1 ≈ 26.74

ge ′ () is larger than fe ′ () .

Instructor’s Resource Manual

Section 6.4 369

x ( −∞ − , 1) − 1 ( 1, 0) 0 (0,1) 1 (1, ) − ∞ f ′

f ′′ −

f is increasing on [0, ) ∞ and decreasing on

39. fx () = ∫ 1 2 − t dt

Domain = ( −∞ ∞ ,) − x 2 ( 2 −∞ , 0] . f has a minimum at (0, 0) fx ′ () = 2 , f ′′ () x =− 2(ln 2) 2 x − x

f is concave up on ( 1,1) − and concave down on

x ( −∞ , 0) 0 (0, ) ∞

( −∞ − ∪ ∞ , 1) (1, ) . f has points of inflection at

( 1,1) − and (1,1) f ′′

f is increasing on ( −∞ ∞ ,) and so has no extreme values.

f is concave up on ( −∞ , 0) and concave down on (0, ) ∞ . f has a point of inflection at

0 − t (0, 2 ∫

x 2 +> 10 for all x, domain = ( −∞ ∞ ,)

f is increasing on ( −∞ ∞ ,) and so has no extreme values. f is concave up on (0, ) ∞ and concave down on ( −∞ , 0) . f has a point of inflection at

Section 6.4 Instructor’s Resource Manual

40. x fx () =

2 ∫ 2 log (

10 t + 1) dt . Since log ( 10 t + 1) has

20 log (121.3 ) 10 P

0 log (121.3 )

10 P = 5.75

domain = ( −∞ ∞ ,) , f also has domain = ( −∞ ∞ ,)

2 ln( x 2 + 1)

≈ 4636 lb/in.

45. If r is the ratio between the frequencies of

ln10 ⎟⎜ 2 ⎝ ⎟ ⎠ ⎝ x + 1 ⎠

successive notes, then the frequency of C = r 12

x ( −∞ , 0) 0 (0, ) ∞

(the frequency of C). Since C has twice the

f 1/12 + 0 + frequency of C, r = 2 ≈ 1.0595

f ′′

Frequency of C = 440(2 1/12 3 ) = 440 2 4 ≈ 523.25

f is increasing on ( −∞ ∞ ,) and so has no extreme

values. 46. Assume log 3 2 =

where p and q are integers,

f is concave up on (0, ) ∞ and concave down on

( −∞ , 0) . f has a point of inflection at (0, 0) q ≠ 0 . Then 2

2 =⋅… 22 2 (p times) and has only powers of 2 as factors and 3 q = ⋅ … (q times) and has 33 3

only powers of 3 as factors.

2 p = 3 q only for p = q = 0 which contradicts our assumption, so log 3 2 cannot be rational.

47. If y =⋅ Ab x , then ln y = ln A + x ln b, so the ln y vs. x plot will be linear.

y =⋅ Cx If d , then ln y = ln C + d ln x, so the

ln y vs. ln x plot will be linear.

y = fx () () y ′ = gxfx gx ()() ()1 − fx ′ ()

WRONG 2: y = fx () gx ()

y ′ = fx () gx () (ln ( )) fx ⋅ gx ′ () = fx () gx () gx ′ ( ) ln ( ) fx

RIGHT: y = fx gx () e () gx = ( ) ln ( ) fx

log (0.37 ) E =

fx + fx gx ()1 − gxfx

Note that RIGHT = WRONG 2 + WRONG 1.

0.37 Evaluating this expression for M = 7 and M = 8 gives E ≈ 5.017 10 × 8 kW-h and

E ≈ 1.560 10 × 10 kW-h, respectively.

Instructor’s Resource Manual

Section 6.4 371

49. fx () = ( x ) = x

≠ x ) = gx () 51. a. Let x g (x) = ln f(x) = ln ⎜

gx ′ () < 0 when x >

⎜ x 2 ln x + x 2 ⋅ ⎟

, so as x → ∞ g(x)

ln a

is decreasing. gx ′′ () =− , so g(x) is

concave down. Thus, lim gx () = −∞ , so

= e x x x →∞

Using the result from Example 5

b. Again let g (x) = ln f(x) = a ln x – x ln a.

Since y = ln x is an increasing function, f(x) ()

gx ′ = e x ln x

is maximized when g(x) is maximized. = x x e ln x ⎡ x

Therefore, g(x) (and hence f(x)) is

ln x + (ln ) x 2 + ⎤ a

maximized at x 0 =

ln a

50. fx () = x

c. Note that x a = a x is equivalent to g(x) = 0. a + 1

a ( a + 1) a x ln a − ( a x − 1) a x ln a 2 a x ln a By part b., g(x) is maximized at x 0 = . fx ′ () =

ln a

+ 1) 2 ( a x + 1) 2 If a = e, then

⎟ = ge () = e ln ee − ln e = 0. x

Since a is positive, x a is always positive.

⎛ gx e = g ⎞

( a + 1) 2 fx ′

is also always positive, thus e () > 0 ⎝ ln ⎠ gx < gx =

x ≠ x if ln a > 0 and fx ′

() < 0 if ln a < 0. f(x) is either

always increasing or always decreasing, a equation g(x) = 0 (and hence x = a x ) has depending on a, so f(x) has an inverse.

just one positive solution. If a ≠ e , then a x − 1

a ( y −=−− 1) 1 y

Now

> e (justified below), so

−> 1 ⎥ a (ln e −= 1) 0. Since

x = = log a lim ( ) gx = −∞ , g(x) = 0 has exactly one

ln a 1 − y

f − 1 () y = log a

solution on (0, ). x 0

Since gx ′ () <

0 on ( x 0 , ) ∞ ,

gx ( 0 ) >

0, and lim gx () = −∞ , g(x) = 0 has

f () x = log a x →∞

exactly one solution on ( x 0 , ). ∞ Therefore,

Section 6.4 Instructor’s Resource Manual

Since u lim

e = e , this implies that

has exactly two positive solutions.

u →∞ u + 1

To show that

> e when a ≠ e :

⎟ = e , i.e., lim 1 ⎜ + ⎟ = e u .

Consider the function hx () =

ln( )(1) x − x 1 () x ln x − 1 Let gx () = x ln . x

hx ′ () =

2 (ln ) x (ln ) x

Using L’Hôpital’s Rule,

Note that hx ′ () < 0 on (1, e) and hx ′ () > 0 ln lim x gx () = lim

on (e, ∞ ), so h(x) has its minimum at (e, e).

d. For the case a = e, part c. shows that

Therefore, lim

Therefore, when x ≠ e e , x ln x < ln e , which

gx ′ ( ) 1 ln =+ x

implies e x < e . In particular, π< e π . Since gx ′ () < 0 on ( 0,1/ e ) and gx ′ () > 0 on , g(x) has its minimum at x ) 1

( 1/ , e ∞

Therefore, f(x) has its minimum at ( e , e ) .

Since f u ′ () x > 0 on (0, u) and f u ′ () x < 0 on

Note: this point could also be written as

⎜ 1 , 1 ⎟ . ⎜ e () e ⎟

(u, ∞ ), f u () x attains its maximum at x 0 = u.

b. f u () u > f u ( u + 1) means

ue u ( u 1) e −+ ( 1) .

e ⎛ u + 1 Multiplying by ⎞ u gives >⎜ ⎟ .

f u + 1 ( u +> 1) f u + 1 ( ) means u

u + ( u 1) +−+ 1 e ( u 1) > u u +− 1 e u . (2.4781, 15.2171), (3, 27) e u + 1 u + ⎛ 1 u + 1 ⎞

Multiplying by u + 1 gives ⎜

55. x sin ∫ x dx ≈ 20.2259

Combining the two inequalities, 0

⎟ << e ⎜

c. From part b., e <⎜

Multiplying by

gives

e ⎟ . u + 1 <⎜ ⎝ u ⎠

⎞< ⎟ e in part b., so

We showed ⎜

Instructor’s Resource Manual

Section 6.4 373

57. a. In order of increasing slope, the graphs

6.5 Concepts Review

= represent the curves x 2, 3, = x y y and

b. ln y is linear with respect to x, and at x = 0,

3. half-life

4. ( 1 + h )

y = 1 since C = 1.

c. The graph passes through the points (0.2, 4)

and (0.6, 8). Thus, 4 = Cb 0.2 and 8 = Cb 0.6 . Problem Set 6.5

Dividing the second equation by the first,

gets 2 = b so b = 2 52 . 1. k =− 6 , y 0 =

4, so y = 4 e

Therefore C = 2 32 .

2. k = y 6, 1, = y = e 0 6 so t

58. The graph of the equation whose log-log plot has negative slope contains the points (2, 7) and (7,

3. k = 0.005, so y = ye 0.005 t 0

y (10) = ye 0.005(10)

0 = ye 0

Thus, 7 = C r 2 and 2 = C 7, r so

2 =⎜⎟ . ⎝⎠ 2 7 y (10) =⇒ 2 y 0 =

y = 0.05 e 0.005 t = 2 e 0.005 0.05 t − = 2 e 0.005( 10) t −

Hence, one equation is y = 14 x − 1 . e

The graph of one equation contains the points

4. k = –0.003, so y = ye –0.003 0

(7, 30) and (10, 70). Thus, 30 = C 7 and

y (–2) = ye 0 = ye 0.006 0

e 0.006 3 7 ln 3 ln 7 −

ln = r ln ⇒= r

≈ 2.38 and

7 10 ln 7 ln10 −

e –0.003 t

⋅ 30 7 − 2.38 ≈ 0.29 . Hence, another equation is

y = 0.29 x 2.38 . 5. y 0 =

10, 000, y(10) = 20,000

The graph of another equation contains the points

20, 000 10, 000 = e k (10)

(1, 2) and (7, 5). Thus, 2 = C 1 r and 5 = C 7, r so

ln 2 = 10k;

10, 000 e ((ln 2) /10) t =

10, 000 2 ⋅ t /10

Hence, the last equation is y = 2 x

≈ 56,568. Student answers may also vary.

After 25 days, y =

The given answers are only approximate.

6. Since the growth is exponential and it doubles in

10 days (from t = 0 to t = 10), it will always double in 10 days.

7. 3 y 0 = ye ((ln 2) /10) t 0

3 = e ((ln 2) /10) t

Section 6.5 Instructor’s Resource Manual

8. Let P(t) = population (in millions) in

13. = year 1790 + t. k e (700) and 10 y 0 =

In 1960, t = 170.

–ln 2 = 700k

45.64 = e 170 k y = 10 e − 0.00099 t

ln 45.64

≈ 0.02248 At t = 300, y = 10 e ≈ 7.43.

After 300 years there will be about 7.43 g.

In 2000, t = 210 P (210) ≈ 3.9 e 0.02248 210 ⋅ ≈ 438 14. 0.85 = e k (2)

The model predicts that the population will be about

ln 0.85 = 2k

438 million. The actual number, 275 million, is

ln 0.85

quite a bit smaller because the rate of growth has

declined in recent decades.

e − 0.0813 t

9. 1 year: (4.5 million) (1.032) ≈ 4.64 million

2 2 years: (4.5 million) =− (1.032) ≈ 4.79 million – ln 2 0.0813t

ln 2

10 years: (4.5 million)

100 years: (4.5 million) (1.032) 100 ≈ 105 million

The half-life is about 8.53 days.

10. kt y = ye kt

0 15. The basic formula is y = ye 0 . If t * denotes the

1.032 A = Ae k (1)

half-life of the material, then (see Example 3)

k = ln1.032 ≈ 0.03150 = e kt *

At t = 100,

After 100 years, the population will be about

=− 0.0229 and k S = =− 0.0241

105 million.

30.22 28.8 To find when 1% of each material will remain, we

11. The formula to use is y = ye kt 0 , where y =

ln(0.01)

use

0.01 y 0 = ye kt 0 or t =

. Thus

population after t years, y 0 =population at time t =

0, and k is the rate of growth. We are given

≈ 201 years (2187) and

k (12)

235, 000 = ye 0 and − 0.0229

164, 000 = ye k (5)

≈ 191 years (2177)

− 0.0241 Dividing one equation by the other yields

1.43293 − = e 12 k 5 k = e 7 k or

16. The basic formula is y = ye kt 0 . We are given

ln(1.43293)

≈ 0.0513888 15.231 = ye k (2) 0 and 9.086 = ye k (8)

Thus y 0 =

Dividing one equation by the other gives

e 12(0.0513888)

15.231 = k − k

12. The formula to use is y = ye 0 , where y = mass t

e (2) (8) = e ( 6) so k =− 0.0861

months after initial measurement, y 0 = mass at time

Thus y 0 = ( .0861)(2) − ≈ 18.093 grams.

of initial measurement, and k is the rate of growth.

We are given

6.76 = 4 e k (4) so that To find the half-life: 1 ⎛ 6.76 ⎞ 0.5247

Thus, 6 months before the initial measurement, the

mass was − 4 e (0.1312)( 6) ≈ 1.82 grams. The

tumor would have been detectable at that time. Instructor’s Resource Manual

Section 6.5 375

17. = e 5730 k 22. From Example 4, Tt () =+ T

1 ( T 0 − Te 1 ) . In this 2

ln 1 ()

problem,

250 (15) T (15) = + 40 (350 40) − e k so

− 4 k = ⎝ 310 ⎠ =− 0.026 ( 1.210 10 ; the brownies will be −

F when 110 = + 40 (310) e or

The fort burned down about 2950 years ago.

t = ⎝ 310 ⎠ = 57.2 min.

18. = 5730 k e

23. From Example 4, Tt () =+ T 1 − Te ( kt T 0 1 ) .

ln 1 ()

2 ≈− 1.210 10 × − 4 Let w = the time of death; then

82 = T − = +

− (10 w w ) 70 (98.6 70) − e (10 )

k (11 − 76 w T (11 w ) 70 (98.6 70) e )

0.51 y = ye ( 1.210 10 − × ) t 0 0

− or w 12 28.6 e (10 )

ln 0.51 t =

≈ 5565 k (11 − w )

6 28.6 e

The body was buried about 5565 years ago. k Dividing: − 2 = e ( 1) or k = ln (0.5) =− 0.693

19. From Example 4, Tt () =+ T 1 ( T 0 − Te kt 1 ) . In this

To find w :

problem,

k 200 (0.5) T (0.5) = + 75 (300 75) e so

12 = 28.6 e so 10 −= w

− 0.693 k = ⎝ 225 ⎠ =− 1.1756 and

Therefore w = − 10 1.25 8.75 8 : 45 pm = = .

0.5 T (3) =

+ e ( 1.1756)(3) 75 225 − =

81.6 F 24. a. From example 4 of this section,

20. From Example 4, Tt =+ T T − Te kt . In this

problem, 0 = T = +−−

24 ( 20 24) k e (5) (5) so

∫ dT =

k dt

or ln T(t)-T 1 =+ kt C

This gives

Tt () − T 1 = ee kt C . Now, if T is

0.1212 ; the thermometer will

the temperature at t = 0, T −

0 T 1 = e and the

⎛ − 4 ⎞ ln kt Tt − T = T − Te ⎜ Tt ⎟ () 1 0 1 . Note that () is

t = ⎝ 44 ⎠ = 19.78 min.

always between T 0 and T 1 so that

− 0.1212 Tt () − T 1 and T 0 − T 1 always have the same

sign; this simplifies the Law of Cooling to problem,

21. From Example 4, Tt =+ T T − Te kt () 1 ( 0 1 ) . In this

Since Tt () is always between T 0 and T 1 , it

T (10) = − 90 64 e = − 90 64(0.0977) =

follows that e =

1 < 1 so that k < 0 .

T 0 − T 1 Hence

Section 6.5 Instructor’s Resource Manual

25. a. ($375)(1.035) 2 ≈ $401.71 34. dy = ky L (–) y 24 dt

d. ($375) e 0.035 2 ⋅ ≈ $402.19 1 ⎛ 1 1 ⎜ ⎞ + ⎟ dy = kdt L ∫ ⎝ y L – y

26. a. ($375)(1.046) 2 = $410.29 1 [ln y – ln L − y ] =+ kt C

d. ($375) e 0.046 2 ⋅ ≈ $411.14 ⎛ Note that: C = Ce 0 = Ce Lk ⋅ 0 ⎞

1.005 12 t = 2 y + yCe Lkt = LCe Lkt

1 + C + – Lkt It will take about 11.58 years or 1 Ce

11 years, 6 months, 29 days. e

Lkt

b. e 0.06 t

It will take about 11.55 years

or 11 years, 6 months, and 18 days.

5 35. y =

− 16(0.00186) t 28. $20, 000(1.025) ≈ $22, 628.16 + 6.4 (16 6.4) − e

29. 1626 to 2000 is 374 years.

0.02976 6.4 9.6 t e −

y = 24 e ⋅ 0.06 374 ≈ $133.6 billion

A 0 = 1000 e − 0.05 ≈ $951.23

33. If t is the doubling time, then

ln 1 +

Instructor’s Resource Manual

Section 6.5 377

36. a. lim (1 + x ) 1000 = 1 1000 = 1 39. Let y = population in millions, t = 0 in 1985,

x → 0 a = 0.012, b = 0.06 , y 0 = 10

b. lim 1 x 1/ = lim 1 1 = dy = 0.012 y + 0.06 x → 0 x → 0 dt

⎟ e 0.012 t –

= 15 e 0.012 t –5

From 1985 to 2010 is 25 years. At t = 25,

d. lim (1 + ε x ) = lim

15 →∞ −≈ 5 15.25. The population in 2010

y = e 0.012 25 ⋅

x → 0 (1 + ε ) n

will be about 15.25 million.

e. lim (1 + x ) 1/ x = e

x → 0 40. Let N(t) be the number of people who have heard

the news after t days. Then

∫ k dt

1 –ln(L – N) = kt + C

Nt () = L (1 − e − 0.1386 t )

= lim (1 − x ) 2/ x

0.99 = L (1 − e − 0.1386 x t → 0 + L )

1 0.01 = e − 0.1386 t

= lim ⎢ (1 − x ) − x ⎥ = 2

e ⎦ ln 0.01 ⎥ t = ≈ 33

dy 99% of the people will have heard about the scandal

38. = ay + b after 33 days.

dt dy

() kt ∫ ke

= ∫ a dt 41. If f(t) = e , then

y 0 =−⇒= A A y 0 +

Section 6.5 Instructor’s Resource Manual

e. The maximum population will occur when

43. => k 0 can be written as

0.0132 t − 0.0001 t 2 ) = 0

= k where y = f(x).

k dx has the solution y = Ce kx . t = 0.0132 / 0.0002 = 66

t = 66 , which is year 2070.

Thus, the equation fx () = Ce kx represents

The population will equal the 2004 value of

exponential growth since k > 0.

6.4 billion when 0.0132 t − 0.0001 t 2 = 0

fx ′ ()

44. =< k 0 can be written as

1 dy

= k where

The model predicts that the population will

y = f(x).

= k dx has the solution y = Ce kx . return to the 2004 level in year 2136.

47. a.

k 0.0132 0.0001 t

Thus, ()

fx = Ce kx

which represents exponential

b. y ' = ( 0.0132 0.0001 − ty )

decay since k < 0.

45. Maximum population:

c. = ( 0.0132 0.0001 − ty )

13,500, 000 mi 2 640 acres 1 person

1 mi 2 1 2 acre dy

10 = ( 0.0132 0.0001 − t dt ) = 1.728 10 people × y

Let t = 0 be in 2004.

ln y = 0.0132 t − 0.00005 t 2 + C

(6.4 10 ) × 9 e 0.0132 t

10 y =

Ce t 0.0132 0.00005 − t

ln ⎜ ⎜

The initial condition y (0) = 6.4 implies that

75.2 years from 2004, or

C = 6.4 . Thus y = 6.4 e 0.0132 0.00005 t − t 0.0132 2 1

sometime in the year 2079.

d. y

46. a. k = 0.0132 0.0002 − t

b. y ' = ( 0.0132 0.0002 − ty ) dy

c. = ( 0.0132 0.0002 − ty ) 10

= ( 0.0132 0.0002 − t dt )

ln y = 0.0132 t − 0.0001 t 2 + C 0

0.0132 0.0001 t − t 2 e. The maximum population will occur when y = Ce

( 0.0132 t − 0.00005 t ) = 0 dt

The initial condition y (0) = 6.4 implies that

6.4 t . Thus y = 6.4 e 0.0132 0.0001 − 0.0132 = 0.0001t

t = 0.0132 / 0.0001 132 =

t = 132 , which is year 2136 The population will equal the 2004 value of

6.4 billion when 0.0132 t − 0.00005 t 2 = 0

t = 0 or t = 264 .

5 The model predicts that the population will return to the 2004 level in year 2268.

Instructor’s Resource Manual

Section 6.5 379

48. Ex ′

Ex ( + )–() Ex

() = lim

6.6 Concepts Review

1. exp ( P x dx () )

= lim

ExEh ()()–() Ex

h → 0 h Eh ()–1

( ∫ P x dx () )

Thus, Ex () = Ex ( ) lim

= kE(x) where k = E′ (0) .

Problem Set 6.6

Hence, Ex = Ee () kx 0 = E (0) e kx =⋅ kx

1 kx e = e .

1. Integrating factor is x e . Check: Eu ( + v ) = e kuv ( + ) = e ku kv + ( x D ye )1 =

= e ku ⋅ e kv = Eu () ⋅ Ev () y = e – x ( x + C )

2. The left-hand side is already an exact derivative.

Integrating factor:

Exponential growth:

exp

∫ 2 dx = exp ln(1 – x ⎣ ) ⎦

In 2010 (t = 6): 6.93 billion

In 2040 (t = 36): 10.29 billion

1– x

In 2090 (t = 86): 19.92 billion 2 –1/ 2 (1 – x ) Logistic growth:

Dy [ (1 – x ) 2 –1/ 2 ] = ax (1 – x ) 2 –3 / 2

In 2010 (t = 6): 7.13 billion In 2040 (t = 36): 10.90 billion In 2090 (t = 86): 15.15 billion

Then y (1 – x ) 2 –1/ 2 = a (1 – x ) 2 –1/ 2 + C , so

4. Integrating factor is sec x.

5. Integrating factor is . x

Section 6.6 Instructor’s Resource Manual

6. y ′ – ay = fx () 14. Integrating factor is sin 2 x . Integrating factor: e ∫ – adx = e – ax Dy [ sin 2 x ] = 2sin 2 x cos x [ D ye – ax ] = e – ax fx () y sin 2 x 2 = sin 3 xC

Then ye – ax = – ∫ e ax (), f x dx so

x ∫ + f x dx sin

7. Integrating factor is x. D[yx] = 1; y =+ 1 Cx 3 12

8. Integrating factor is ( x + 2 1) . goes through ⎜ ,2. ⎟ ⎝ 6 ⎠

Dyx [( + 2 1) ] ( = x + 1) 5 15. Let y denote the number of pounds of chemical A

after t minutes.

lb/min

Integrating factor: e∫

Integrating factor: ∫ ( 3 / 20 ) dt e = e t Then 3 / 20 ye = e + C , so

Then ye t 3 / 20 = 40 e t 3 / 20 + C . t = 0, y = 10

C = –30. [ D ye 2 x ] = xe 2 x Therefore, yt

10. Integrating factor is e 2 x . ⇒

() t = 40 – 30 e –3 / 20 , so

y = ⎜⎟ x – ⎜⎟ + Ce –2 x y (20) =

40 – 30 e ≈ 38.506 lb.

11. Integrating factor is . D ⎡⎤= ⎢⎥ 3 x ; y = x + Cx dt

e Integrating factor is t / 50 .

y = x 4 + 2 x goes through (1, 3).

3 Integrating factor: dx e ∫ =

e 3 x yt () =

e – / 50 400 – 350 t goes through (0, 50). [ D ye 3 x ] = e 5 x y (40) = 400 – 350 e –0.8 ≈ 242.735 lb of salt

Then ye 3 x =

5 5 Integrating factor is (60 – ) . t

e x 2 + 4 e –3 x

Dy [ (60 – ) ] t –3 = 4(60 – ) t –3

Therefore, y =

is the particular

5 yt () = 2(60 – ) t + C (60 – ) t 3

solution through (0, 1).

⎛ yt 1 = ⎞

() 2(60 – ) – t ⎜

⎟ (60 − t ) 3 goes through

13. Integrating factor: xe ⎝ 1800 ⎠

goes through (1, 0).

Instructor’s Resource Manual

Section 6.6 381

2 23. Let y be the number of gallons of pure alcohol in the

dt 50 + t

50 + t

tank at time t.

Integrating factor:

exp ⎜ ∫

2 2 ln(50 ) + t

Integrating factor is e 0.05t . Dy [ (50 + t )]0 2 = t

yt () = 25 + Ce –0.05 ; 100, 0, 75 y = t = C =

Then y (50 + t ) = C . t = 0, y = 30 ⇒

C = 75000

yt =

+ 25 75 e –0.05 () t ; y = 50, t = T,

Thus, y (50 + t ) 2 =

T = 20(ln 3) ≈ 21.97 min

If y = 25, 25(50 + t ) 2 =

75, 000, so

b. Let A be the number of gallons of pure alcohol t = 3000 – 50 ≈ 4.772 min.

drained away.

19. I ′+ 10 I = 1 (100 – A) + 0.25A = 50 ⇒= A

Integrating factor = exp(10 ) 6

t 200

6 6 It took DI 3 [ exp(10 )] exp(10 ) t = t minutes for the draining and the

5 It ( ) 10 = –6 + C exp(–10 ) 6 t same amount of time to refill, so

It ( ) 10 [1 – exp(–10 )] = –6 6 t goes through (0, 0).

≈ 26.67 min.

20. 3.5 I ′= 120sin 377 t

c. c would need to satisfy

⎟ (1 – cos 377 ) t through (0, 0).

(3ln 3 – 2)

d. y ′= 4(0.25) – 0.05 y = 1 – 0.05 y

21. 1000 I = 120 sin 377t

I (t) = 0.12 sin 377t

Solving for y, as in part a, yields

+ 20 80 e –0.05 t . The drain is closed when dx

t = T 0.8 . We require that

dt 100

(20 80 + e 0.05 0.8 ) 4 0.25 0.2 +⋅ ⋅ T = 50, ⎛ 1 ⎞

x ′+ ⎜

or 400 e –0.04 T += T 150.

Integrating factor is t e / 50 . 24. a. v ′+ av = – g [ D xe t / 50 ]0 = Integrating factor: e at

xt () = 50 e satisfies t = 0, x = 50.

Integrating factor is t e /100 . – g ⎛

e – t at /100 – /100 t

Therefore, v =

a ⎜ 0 , [ so D ye ] = e ⎝ a ⎟ ⎠

– at

yt () = e – /100 t ( – 100 C e – /100 t ) vt () = v ∞ + ( v 0 – v ∞ ) e .

yt () = e – /100 t (250 – 100 e – /100 t ) satisfies t = 0,

y = 150.

Section 6.6 Instructor’s Resource Manual

b. = v ∞ + ( v 0 – v ) e – ∞ at , so

26. For t in [0, 15],

e –0.1 a t () – 320 = 320( –1);

e –0.1 (0 320) t

() t = 8000 – 320 t + 10(320)(1 – e –0.1 );

Let t be the number of seconds after the parachute

y = vt

⎜ opens that it takes Megan to reach the ground. y 0 +

For t in [15, 15+T], v = –

= y 0 + vt + 0 ∞ (1 – – ∞ at e ) 1.6

a 0 = yT ( + 15) = [3200(2 – e –1.5 )]

0.05 –20 T + (0.625)[320( e –1.5

+ e – 1) 20](1 – T –1.6 )

vt ( ) [120 ( 640)] = −− e − 0.05 t +− ( 640) = 0 if

≈ 5543 – 20 –142.9 T e − 1.6 T ≈ 5543 – 20T [since

e –1.6 > 50, so T < 10 ⎛ –35 19 ⎞ (very small)]

20 ln ⎜

Therefore, T ≈ 277, so it takes Megan about

yt () =+ 0 (–640) t

292 s (4 min, 52 s) to reach the ground.

= xe 2 − ln xC ⎝ + 0.05 ⎠ ⎜ ⎟

–0.05 +⎜ t ⎟ [120 – (–640)](1 – e )

15, 200(1 – e –0.05 t ) ⎛ dy y ⎞

e − ln xC e ⎜

− ⎟ = xee 2 C − ln x

Therefore, the maximum altitude is

⎝ dx x ⎠

⎜ 20 ln ⎜ ⎟ ⎟ =− 12,800 ln ⎜ ⎟ +

b. –640 T +

= Qxe () ∫ P x dx C ∫ e dx C + 1

− ∫ P x dx ()

∫ P x dx y () = e ∫ Qxe

Instructor’s Resource Manual

Section 6.6 383

6.7 Concepts Review

1. slope field

2. tangent line

y n − 1 ( hf x n − 1 , y n − 1 )

The oblique asymptote is y = x .

4. underestimate

Problem Set 6.7

The oblique asymptote is y =+ 3 x /2 .

lim ( ) 12 yx = and y (2) 10.5 ≈ x →∞

2 y = Ce x 1 /2 To find C 1 , apply the initial condition:

Section 6.7 Instructor’s Resource Manual

ey ' + ye = ⎜ 2 x + ⎟ e x

ey = ⎜ 2 x + ⎟ e x

ey = ∫ ⎜ 2 x + x ⎟ e dx

Integrate by parts: let u = 2 x + , =− dx 2

dv = x e dx . Then du = 2 dx and v = e x .

ln y =−+ x C Thus,

To find C 1 , apply the initial condition:

2 To find C, apply the initial condition:

3 = y (0) =−+ 0 Ce − 0 =− C 2 2

7 Thus C = , so the solution is

y = 2 x −+ e − x

y ' +=+ y x 2 2 2

∫ The integrating factor is 1dx e = x e .

x + ey x ' ye = x e ( x +

Note: Solutions to Problems 22-28 are given along with

2) the corresponding solutions to 11-16.

dx ()

x n Euler's

Improved Euler

ey =

( x + 2) x e dx n

Method y Method y

Integrate by parts: let x u =+ x 2, dv = e dx .

Then du =

dx x and v = e . Thus

ey x = ( x + 2) e x − x ∫ e dx 0.6 8.232

ey x = ( x + x 2) e − e + C 0.8 11.5248

y x =+−+ x 21 Ce − 1.0 16.1347

To find C , apply the initial condition:

4 = y (0) =++ 01 Ce − 0 =+ 1 C → C = 3 12., 23.

x n Euler's

Improved Euler

Thus, y =++ x 13 e − x .

Method y n Method y n

Instructor’s Resource Manual

Section 6.7 385

13., 24. x n Euler's

Improved Euler

17. a. y 0 = 1

Method y n Method y n y 1 = y 0 + (, hf x 0 y 0 )

0.0 0.0 0.0 = y

0 + hy 0 =+ (1 hy ) 0

0.2 0.0 0.02 y 2 = y 1 + (, hf x y 1 1 ) = y 1 + hy 1

0.4 0.04 0.08 =+ (1 hy )

h 2 1 =+ (1 ) y 0

0.6 0.12 0.18 y 3 = y 2 + ( hf x 2 , y 2 ) = y 2 + hy 2

0.8 0.24 0.32 =+ (1 hy ) =+ (1 h ) 3 2 y 0

y n = y n − 1 + ( hf x n − 1 , y n − 1 ) = y n − 1 + hy n − 1

14., 25. x Euler's

Improved Euler

n − 1 =+ (1 h ) y 0 =+ ( 1 h )

Method y Method y

0.0 0.0 0.0 b. Let 1/ N = h . Then y N is an approximation

to the solution at x = Nh = (1/ ) hh = 1 . The

exact solution is y (1) = e . Thus,

N 0.6 0.040 0.076 ( + 1 1/

N ) ≈ e for large N. From Chapter 7,

we know that N lim 1 1/ ( +

15., 26. x Euler's

Improved Euler

y 1 = y 0 + ( hf x 0 ) =+ 0 ( hf x 0 ) = ( hf x

Method y n Method y n y 2 = y 1 + () hf x 1 = ( hf x 0 ) + () hf x 1

hfx ( ( 0 ) + fx () 1 )

= hfx [ ( 0 ) + fx () 1 ] + ( hf x 2 )

= hfx [ ( 0 ) + fx () 1 + fx ( 2 ) ] = h ∑ fx () i

At the nth step of Euler's method,

n − 1 y n = y n − 1 + ( hf x n − 1 ) = h ∑ fx () i

16., 27. x n Euler's

Improved Euler

Method y n Method y n x 1 19. a. x ∫

x y x dx '( ) = 1 ∫ x sin x dx 2 1.0 2.0 2.0 0 0

yx () − yx ( ) ≈ x − x sin x 2 1 0 ( 1 0 ) 0

yx ( 2 ) − yx ( 0 ) ≈ x − x sin x 2 ( 1 0 ) 0

+ ( x − x ) sin x 2 2 1 1 yx ( ) − y (0) = h sin x 2 2 0 + h sin x 2 1 yx ( 2 )0 −≈ 0.1sin 0 2 + 0.1sin 0.1 2

yx ( 2 ) ≈ 0.00099998

Section 6.7 Instructor’s Resource Manual

y x dx '( ) =

sin x dx c. 3 x ∫ 3 x ∫ x ∫ x y x dx '( ) = ∫ x x + 1 dx

yx () 3 − yx ( 0 ) ≈ x

1 − x 0 ) sin x 0 yx () 3 − yx ( 0 ) ≈ ( x 1 − x 0 ) x 0 + 1

yx ()0 3 −≈ 0.1sin 0 2 + 0.1sin 0.1 2 +

+ 0.1sin 0.2 2 yx () 3 ≈ 0.314425 Continuing in this fashion, we have

yx () 3 ≈ 0.004999 x

Continuing in this fashion, we have

∫ y x dx '( ) = ∫ x + x x 1 dx

x n y x dx '( ) = sin x dx 2

0 yx ( n ) − yx ( 0 ) ≈ ∑ ( x i + 1 − x i ) x i − 1 + 1

yx ( n ) − yx ( 0 ) ≈ ∑ ( x

i + 1 − x i ) sin x i n − 1

i = 0 yx ( n ) ≈ h ∑ x i − 1 + 1

yx ( n ) ≈ h ∑ fx ( i − 1 ) When 10 n = , this becomes

i = 0 yx ( ) = y (1) 1.198119 ≈

When 10 n = , this becomes

yx ( 10 ) = y (1) ≈ 0.269097 Δ y 21. a. 1 = [(, fx 0 y 0 ) + fx ( 1 + y ˆ 1 )]

d. The result yx ( n ) ≈ h ∑ fx ( i − 1 ) is the same as

b. 1

= [( fx , y ) + fx ( + y ˆ )] ⇒

0 0 1 x 1 2 that given in Problem 18. Thus, when ( , ) fxy 2( y 1 − y 0 ) = hfx [( 0 , y 0 ) + fx ( 1 + y ˆ 1 )] ⇒

depends only on x , then the two methods (1)

Euler's method for approximating the solution

y 1 − y 0 = [(, fx y ) + fx ( + y ˆ )] ⇒ 2 0 0 1 1

to ' y = fx () at x n , and (2) the left-endpoint

y 1 = y 0 + [( fx 0 , y 0 ) + fx ( 1 + y ˆ 1 )]

Riemann sum for approximating ∫ 0 () f x dx ,

c. 1. x

are equivalent.

x 1 x 1 2. y n − 1 + ( hf x n − 1 , y n − 1 )

20. a. ∫ x y x dx '( ) = ∫ x x + 1 dx

0 0 3. h y

n − 1 + [( fx n − 1 , y n − 1 ) + fx ( n , y ˆ n )]

yx () 1 − yx ( 0 ) ≈ ( x 1 − x 0 ) x 0 + 1 2

yx () 1 − y (0) = hx 0 + 1 22-27. See problems 11-16

28. Error from

Error from

∫ Euler Method x 0 ∫ x 0 0.2 0.229962 0.015574

y x dx '( ) = 2 x + 1 dx h Method

yx ( 2 ) − yx ( 0 ) ≈ ( x 1 − x 0 ) x 0 + 1 0.1 0.124539 0.004201

+ ( x 2 − x 1 ) x 1 + 1 0.05 0.064984 0.001091 yx ( 2 ) − y (0) = hx 0 ++ 1 hx 1 + 1 0.01 0.013468 0.000045

For Euler's method, the error is halved as the step size h is halved. Thus, the error is proportional to h. For the improved Euler method, when h is halved, the error decreases to approximately one-fourth of what is was. Hence, for the improved Euler

method, the error is proportional to h 2

Instructor’s Resource Manual

Section 6.7 387

6.8 Concepts Review

13. cos(arccot 3.212) = cos arctan ⎜

14. sec(arccos 0.5111) =

⎛ cos(arccos 0.5111) ππ ⎞

15. sec (–2.222) –1 = cos –1 ⎜

Problem Set 6.8

since cos =

17. cos(sin(tan − 1 2.001)) ≈ 0.6259

18. sin (ln(cos 0.5555)) ≈ 2 0.02632

2. arcsin – ⎜ ⎜

⎟ ⎟ = – 2 since sin – 3 ⎜ 3 ⎟ = –

19. θ = sin − 1

3. sin –1 ⎜ ⎜ –

⎟ = – since sin –

⎟ ⎟ = – since sin – 4 ⎜

since tan

⎛⎞ π 1 23. Let θ be the angle opposite the side of length 3, 1

6. arcsec(2) = arccos ⎜⎟ =

since cos ⎜⎟ = , so

3 and θ 2 = θθ 1 –, so θθθ = 1 – 2 . Then tan θ= 1 π

7. arcsin – ⎜ ⎟ = – since sin – ⎜ ⎟ = –

6 6 24. Let ⎝ θ be the angle opposite the side of length 5, ⎠ 2 1 and θ 2 = θθ 1 − , and y the length of the unlabeled

8. tan –1 ⎜ ⎜ –

⎟ ⎟ = – since tan –

side. Then θθθ = − and y = x − 25.

9. sin(sin –1 0.4567) = 0.4567 by definition

θ = tan − 1 5 − tan − 1 2

10. cos(sin –1 0.56) = − 1 sin (sin 2 − 1 0.56) 2 x − 25 x 2 − 25

2 ⎡ –1 ⎛ 2 ⎞ 25. ⎤ cos 2sin

⎜ – ⎟ ⎥ = 1 – 2sin ⎢ sin ⎜ – ⎟ ⎥

Section 6.8 Instructor’s Resource Manual

2 tan tan ⎡ –1 1 ⎤

1 – tan 2 ⎡ ⎣ tan –1 1 ⎤ () 3 ⎦

sin(sin –1 x )

29. tan(sin x ) =

cos(sin –1 x )

30. sin(tan –1

csc(tan –1 x )

+ 1 cot (tan 2 –1 x )

tan (tan

= lim cos –1 z =

31. cos(2sin –1 x ) 1 – 2sin (sin = 2 –1 x )1–2 = x 2

35. a. Let L = lim sin − 1 x . Since

32. tan(2 tan x ) =

2 tan(tan x )

1 – tan (tan –1 x ) 1– x 2 sin(sin − 1 x ) = x , lim sin(sin − 1 x ) = lim x = 1 .

since lim tan θ =∞

x →∞

Thus, since sin is continuous, the Composite

Limit Theorem gives us

2 lim sin(sin − 1 x ) = lim sin( ) L ; hence

sin L = 1 and since the range of sin − 1 is

Instructor’s Resource Manual

Section 6.8 389

39. y = ln(2 sin ) + x . Let u =+ 2 sin x ; then

x →− 1 +

y = ln u so by the Chain Rule

sin(sin − 1 x ) = x ,

lim sin(sin x ) = lim x =− 1 .

Thus, since sin is continuous, the

2 sin x

Composite Limit Theorem gives us

− 1 40. d e tan x = tan x d e tan x = e tan lim sin(sin x x ) = lim sin( ) L ; sec 2 x

sin L = − and since the range of 1 sin − 1 is

d sec tan x x + sec 2 x

41. ln(sec x + tan ) x =

(sec )(tan x x + sec ) x

36. No. Since sin − 1 x is not defined on (1, ) ∞ ,

d – csc cot – csc x x 2 x

42. [– ln(csc x + cot )] x = –

csc x + cot x lim sin x does not exist so neither can the

csc (cot x x + csc ) x

= csc x

cot x + csc x two-sided limit lim sin − 1 x .

37. Let fx () == y sin x ; then the slope of the d 1 e x

tangent line to the graph of y at c is

44. arccos( e x ) = –

. Hence, lim fc ′ () =∞ so

that the tangent lines approach the vertical.

46. ( e x arcsin x 2 ) = e x ⋅

+ e x arcsin x 2

1–( x 22 )

⎞ = e + arcsin x 2 ⎟

⎝ 1– x

d –1 3 –1 2 1 3(tan –1 x ) 2

47. (tan x ) = 3(tan x ) ⋅

d sin(cos − 1 x )

d –1

d 1– x 2

48. tan(cos x ) =

dx cos(cos − 1

Section 6.8 Instructor’s Resource Manual

49. sec ( x ) =

d –1 3 1 2 3 54. y = x arc sec( x 2

2 ⎤⎛ d ⎞ ⎢ arcsec( x + 1) ⎥ + ⎜

x ⎟ ⋅ arcsec( x 2 + 1)

50. (sec –1 x ) 3 = 3(sec –1 x ) 2 ⋅

⎥ +⋅ 1 arcsec( x 2 + 2 1) 2 2

x x –1

( x + 1 ) ( x + 1) − 1 ⎥

d –1 3 –1 2 1 ⎣ ⎢ ( )

⎥ + arcsec( x 2 + 1)

⎥ + arcsec( x 2 + 3(1 sin 1) )

+ –1 x 2 = ⎢

x 2 +⋅ 1 x x ⎢ 2 ( ) + 2 ⎣

1– x 2

⎥ + arcsec( x 2 + 1 52. ) y = sin ⎜ ⎟ 2 2

Let u =

; then y = sin − 2 1 ( ux () ) so by the

x + 4 55. ∫ cos 3x dx

Chain Rule:

dy dy du

1 du Let u = = 3, x du = = 3 ⋅ dx = ; then

dx du dx

1 − u 2 dx

cos 3 x dx = ∫ cos 3 (3 ) x dx =

⎜ 2 ⎜ ( x 2 + 4) 2 ⎟ ⎟

∫ cos u du ⎛ = sin u += C sin 3 x + 1 C

+ 8 x 2 + 15 ⎠ ⎝ ( x + 4) 2 ⎟ x sin( x 2 ⎠ 2 ∫ ) dx = ∫ sin( x )2 ⋅ x dx

4) x 4 + 8 x 2 + 15 2 ∫

sin u du =− cos u + C

cos( x 2 + − C 1 2 )

53. y = tan ln x

Let u = x 2 , v = ln u ; then y = tan − 1 ( vux ( ( )) ) so

57. Let u = sin 2x, so du = 2 cos 2x dx.

by the Chain Rule:

2 ∫ sin 2 (2 cos 2 ) x x dx dy dy dv du 1 1

2 ∫ u du

∫ 1 tan x dx = ∫ dx =− ∫ ( sin ) − x dx

du =− ln u +=− C ln cos x + C

u = ln sec x + C

Instructor’s Resource Manual

Section 6.8 391

59. Let u = e 2 x , so du = 2 e 2 x dx .

65. Let u = 2x, so du = 2 dx.

e 2 x cos( e 2 x

) dx = ∫ cos( e 2 x )(2 e 2 x ) dx ∫ 14 + x 2 dx = ∫

2 2 dx

2 + 1 (2 ) x

cos u du = ∫ 2 du = arctan u + C

sin u += C sin( e 2 x )

arctan 2 x + C

e x cos( e 2 x

sin( e 2 x ∫ ⎤ ) x

= sin( e ) − sin( ) e ∫ 2 x dx = ∫

dx =

1 + e 1( + e x ) 2 ∫ 1 + u 2

= arctan u + C = arctan e sin x e 2 − sin1 +C =

dx = ∫

∫ sin x x dx = 0 cos ⎢ ⎥ =−= 0 ⎝ ⎠

Let u =

x du , =

dx ; then

dx = [arcsin ] x 2/2

2 dx 1– = x

2 sec − 1 2 ∫

tan 1 1 − ∫ 1 x ⎤ = −

=− ∫ du =− (2 u ) + C

2 d θ =− ∫

du =− tan − ∫ 1 u + C

=− tan (cos ) C − 1 θ +

π /2 sin θ

− 1 π ∫ /2 d θ =− ⎡ tan (cos ) θ ⎤

0 + 1 cos 2 θ ⎣ ⎦ 0

tan − 1 + 0 tan 1 − 1 =− ππ =−+= 0

Section 6.8 Instructor’s Resource Manual

69. ∫ 2 dx =

1 1 73. The top of the picture is 7.6 ft above eye level,

13 ( x 2 − 6 x ++ 9) 4

and the bottom of the picture is 2.6 ft above eye

level. Let θ be the angle between the viewer’s 1

line of sight to the top of the picture and the

( x − 3) 2 + 4

horizontal. Then call θ 2 = θθ 1 − , so θθθ = 1 − 2 .

Let u =− x 3, du = dx a , = 2; then

2 dx = ∫ 2 2 du = tan

If b = 12.9, θ≈ 0.3335 or 19.1 °. ⎡ π ⎤

74. a. Restrict 2x to [0, ] π , i.e., restrict x to 0, . ⎢

70. ∫ 2 dx =

2 x + 8 x + 25 ∫ 2( x 2 + x

17 dx =

Then y = 3 cos 2x

2 x = arccos

x = f –1 () y = arccos

dx =

2 ∫ ( x + 2) 2 17 ∫

du =

f –1

() x = arccos

b. Restrict 3x to – , , ⎢ ⎥ i.e., restrict x to

Then y = 2 sin 3x

y = sin 3 x

dx . Let u = 2, x du = 2 dx a , = 3 ;

3 x = arcsin

then ∫

dx =

(2 dx ) = 1 y

− 9 2 x 4 x 2 − 9 x = f –1 () y = arcsin 3 2

du = 1 sec − 1 ⎜ 1 ⎟ += C f –1 () x = arcsin

3 3 c. Restrict x ⎝ to – , ⎠ ⎜ ⎟ ⎝ 22 ⎠

1 y = tan x

dx = ∫

dx + ∫

dx 2

49 − x 2 49 − x 2 49 − x 2 2y = tan x

These integrals are evaluated the same as those in

x = f –1 () y = arctan 2 y

problems 67 and 68 (with a constant of 4 rather

f –1 () x =

than 12). Thus

arctan 2 x

dx =− 49 − x + sin ⎜ ⎟ + C

49 − x 2 9 3 ⎝ 2 ⎠

Instructor’s Resource Manual

Section 6.8 393

d. Restrict x to –,– ∞

5 1– 2 restricted to – , 0 119

then y = sin ⎡ –1 ⎛⎞ 1 –1 ⎛ 1 ⎞ tan 4 tan ⎤ – tan

arcsin y ⎡

arcsin y

arcsin x

Thus, 4 tan ⎜⎟ – tan ⎜

4 ⎣ + ⎦ tan tan ⎣ () 4

tan 2 tan ⎡ –1 1 ⎤

1 – tan 2 tan ⎡ − 1 1 –1 ⎣ 1 () ⎤ 4 tan tan ⎡ ⎦ ⎤ ⎣ () 4 ⎦

Let θ represent ∠ DAB , then CAB ∠ is . Since = 15

b b θ b Δ ABC is isosceles, AE = , cos == 2 and ⎡

. Thus sector ADB has area

tan 3 tan ⎡ –1 1 () ⎤ + tan tan ⎡ –1 5

b 2 = b 2 cos –1 . Let φ represent 1 – tan 3 tan ⎡

2 cos –1

2 ⎣ a ⎦ ⎣ ⎦ 47 + 5

∠ DCB , then ACB ∠ is

sin == 2 and φ = 4sin –1 . Thus sector

4 a 2 a 2 ⎜⎟ a ⎜ ⎟ = tan (1) = .

Thus, 3 tan –1 ⎛⎞ 1 tan ⎛ –1 5 + ⎞

4sin –1 b ⎞ a 2 = 2 a 2 –1 ⎜ b sin . 2 ⎝

DCB

has area

These sectors overlap on the triangles ΔDAC and

76. tan 2 tan –1

ΔCAB, each of which has area

= 2 The large circle has area π b , hence the shaded

region has area

Section 6.8 Instructor’s Resource Manual

y = sin(arcsin x) is the line y = x, but only

defined for −≤≤ 1 x 1 .

y = arcsin(sin x) is defined for all x, but only the

portion for – ≤≤ x

is the line y = x.

They have the same graph.

Conjecture: arcsin x = arctan

1– 2 a 2 – x x 2

for

a 2 ⎡ 1– x 2 ⎢ ⎤ () a ⎥

–1 < x < 1

Proof: Let θ = arcsin x, so x = sin θ.

since a > 0

() a 1–

Then

= tan θ 1–

() a

1– x 2

1 – sin 2 θ cos θ

so θ = arctan

∫ du ⋅ = = − ∫

1 dx

sin 1 u + C

1– x 2 1 − u () 2 a

1– x () a

It is the same graph as y = arccos x.

Conjecture: – arcsin x = arccos x =

Proof: Let θ = – arcsin x

, since a > 0

a – x 2 sin a

Then x

θ ⎟ = cos θ

so θ = arccos . x =

a 1 + x 2 () a

tan − ∫ 1 du u + C

Instructor’s Resource Manual

Section 6.8 395

86. a

∫ a − x dx − 0

a 2 2 84. Let a u = , so du = () 1/ a dx . Since a > 0, ∫ a − x dx = 2 2 2

()() a a − 1 ⎢ ⎣ 2 2 a ⎥ ⎦ 0

This result is expected because the integral

85. Note that sin ⎛⎞= ⎜⎟

1 should be half the area of a circle with radius a.

(See

dx ⎝⎠ a a 2 − x 2 87. Let θ be the angle subtended by viewer’s eye. Problem 67).

2 2 2 () b () 22 b a − x

2 2 b 2 + 4 b 2 + 144 ( b 2 + 4)( b 2 + 2 144) a − x

> 0 for in 0, 2 6 b ⎣ )

1 2 1 − x 2 + a 2 Since

= a 2 − x 2 db

and 0 < for b in ( 2 6, ∞ the angle is ) ,

db maximized for b = 26 ≈ 4.899 .

The ideal distance is about 4.9 ft from the wall.

x 2 ⎟ ⎝ ⎠⎝ b dt ⎠ ⎜

() b ⎠

⎣ ⎝ b − x 2 + ( a + x ) 2 ⎟⎜ ( b ⎠⎝ 2 − x 23/2 ) ⎟ ⎠

Section 6.8 Instructor’s Resource Manual

89. Let h(t) represent the height of the elevator (the

92 Let x represent the length of the rope and let θ

number of feet above the spectator’s line of sight) t

represent the angle of depression of the rope.

seconds after the line of sight passes horizontal, and Then θ = sin − 1 ⎛⎞ let 8 θ () t denote the angle of elevation. ⎜⎟ , so

Then h(t) = 15t, so θ () t = tan − 1 = tan − 1 .

xx − () 64

8 2 x 2 dt

2 dt

() dx

dt 1 + t 2 ⎜⎟ ⎝⎠ 4 16 + t 2

When x = 17 and

=− 5 , we obtain

At t = 6,

radians per second or

17 17 2 about 4.41° per second. 51 − 64

The angle of depression is increasing at a rate of

90. Let x(t) be the horizontal distance from the observer

8 / 51 0.16 ≈ radians per second.

to the plane, in miles, at time t., in minutes. Let t = 0 when the distance to the plane is 3 miles. Then

93. Let x represent the distance to the center of the earth

2 2 x and let θ represent the angle subtended by the (0) = 3 − 2 = 5 . The speed of the plane is 10 − 1 ⎛ 6376 ⎞

miles per minute, so xt () =

− 5 10 . t The angle of

earth. Then θ = 2sin ⎜

⎟ , so ⎠

elevation is θ () t = tan

() x

x 6376 2 ⎠ dt

1 + ( 2/ ( − 5 10 t ⎝ ( 5 10 ) − t ) ⎠

xx 2 − 6376 2 dt

. When she is 3000 km from the surface

( 5 10 ) − t 2 + 4

x = 3000 + 6376 = 9376 and

When t = 0,

≈ 2.22 radians per minute.

these values, we obtain

× 3.96 10 − 4 radians

91. Let x represent the position on the shoreline and let θ represent the angle of the beam (x = 0 and per second. θ =

when the light is pointed at P). Then

1 + x 2 dt 4 + x () dt 2

When x = 1,

(5 ) π=π 2 The beacon

revolves at a rate of 2 π radians per minute or 1 revolution per minute.

Instructor’s Resource Manual

Section 6.8 397

6.9 Concepts Review

2. cosh 2 − sinh x 2 x = 1

3. the graph of 2 x 2 − y = 1 , a hyperbola

4. catenary; a hanging cable or chain

Problem Set 6.9

1. cosh x + sinh x =

e 2 x + e –2 x e 2 x – e –2 x

2. cosh 2 x + sinh 2 x =

3. cosh – sinh x x =

e 2 x + e –2 x e 2 x – e –2 x

2 e –2 x

4. cosh 2 – sinh 2 x x =

= –2 = e x

5. sinh cosh x y + cosh sinh x y =

2 2 2 2 e xy + + e xy – – e – xy + – e –– xy e xy + – e xy – + e – xy + – e –– xy

4 4 2 e xy + –2 e –( xy + ) e xy + – e –( xy + )

= sinh( x + y )

6. sinh cosh x y – cosh sinh x y =

= sinh( – ) x y

Section 6.9 Instructor’s Resource Manual

7. cosh cosh x y + sinh sinh x y =

= cosh( x + y )

8. cosh cosh x y – sinh sinh x y =

= cosh( – ) x y

tanh x + tanh y

cosh x + cosh y

sinh x

sinh y

16. D x cosh x = 3cosh x sinh x

1 tanh tanh x y

+ sinh x sinh 1 y cosh x ⋅ cosh y

17. D x cosh(3 x += 1) sinh(3 x +⋅= 1) 3 3sinh(3 x + 1)

sinh cosh x y + cosh sinh x y sinh( x + y )

cosh cosh x y + sinh sinh x y cosh( x + y )

18. D x sinh( x + x ) = cosh( x +⋅ x ) (2 x + 1)

= tanh (x + y)

= (2 x + 1) cosh( x 2 + x )

sinh x sinh y

tanh – tanh x y

cosh x – cosh y

19. D x ln(sinh ) x =

⋅ cosh x =

1 cosh x

1 – tanh tanh x y 1– sinh x

cosh x ⋅ sinh y

sinh x

cosh y

= coth x

sinh cosh x y – cosh sinh x y sinh( – ) x = y =

cosh cosh x y – sinh sinh x y cosh( – ) x y

20. D x ln(coth ) x =

(–csch 2 x )

= tanh(x – y)

coth x

sinh x

11. 2 sinh x cosh x = sinh x cosh x + cosh x sinh x

cosh x sinh 2 x

sinh cosh x x

= sinh (x + x) = sinh 2x

=− csch x sech x

12. 2 cosh 2 x + sinh x = cosh cosh x x + sinh sinh x x 2 2

= cosh( x + x ) = cosh 2 x 21. D x ( x cosh ) x = x ⋅ sinh x + cosh x ⋅ 2 x

= x 2 sinh x + x 2 cosh x

13. sinh D 2 x x = 2 sinh cosh x x = sinh 2 x

22. D ( x –2 sinh ) x = x –2 ⋅ cosh x + sinh x ⋅ (–2 x x –3 )

14. D 2 cosh x = 2 cosh sinh x x = sinh 2 x = x − x 2 cosh x − 2 x − 3 sinh x

15. D 2 x (5sinh x ) = 10 sinh x ⋅ cosh x = 5sinh 2 x

23. D x (cosh 3 sinh ) x x = cosh 3 x ⋅ cosh x + sinh x ⋅ sinh 3 x ⋅ 3 = cosh 3 cosh x x + 3sinh 3 sinh x x

24. D x (sinh cosh 4 ) x x = sinh x ⋅ sinh 4 x ⋅+ 4 cosh 4 x ⋅ cosh x = 4 sinh x sinh 4x + cosh x cosh 4x

25. D x (tanh sinh 2 ) x x = tanh ⋅ cosh 2

2 x 2 x ⋅+ 2 sinh 2 x ⋅ sech x = 2 tanh cosh 2 x x + sinh 2 sech x x

x (coth 4 sinh ) x x = coth 4 x ⋅ cosh x + sinh (–csch 4 ) 4 x x ⋅ = cosh coth 4 – 4sinh csch 4 x x x x

Instructor’s Resource Manual

Section 6.9 399

27. D x sinh ( x ) =

–1 2 1 2 x

( 22 x ) + 1 x 4 + 1

28. D x cosh –1

( 32 x )–1

x 6 –1

29. D x tanh (2 – 3) x =

30. D x coth ( x ) = D x tanh ⎜ 5 ⎟ =

31. Dx x [ cosh (3 )] x =⋅ x

⋅+ 3 cosh (3 ) 1 x ⋅ =

+ cosh –1 3 x

32. D x ( x sinh x ) = x ⋅

⋅ 5 x + sinh x ⋅ 2 x =

+ x 2 sinh –1 5 x

( 52 x ) + 1 x 10 + 1

33. D x ln(cosh –1 x ) =

1 1 38. Let u = 3x + 2, so du = 3 dx.

cosh x

x –1

∫ sinh(3 x + 2) dx = ∫ sinh u du = cosh u + C 3 3

x 2 – 1 cosh –1 x

= cosh(3 x ++ 2) C

34. cosh –1 (cos ) x does not have a derivative, since

39. Let u =π+ x 2 5, so du =π 2 xdx .

D cosh − u 1 u is only defined for u > 1 while

2 ∫ 1 x cosh( π+ x 5) dx =

2 π ∫ cosh u du

cos x ≤ 1 for all x.

35. tanh(cot ) sech (cot ) (– csc 2 2

sinh u += C sinh( π++ x 5) C

= – csc 2 x sech (cot ) 2 x

40. Let u = z , so du =

dz .

D x coth (tanh )

D x tanh ⎜

⎝ tanh x ⎠

cosh z

dz = ∫ 2 cosh u du = 2sinh u + C

= D x tanh (coth ) x z

1 –csch 2 x

= 2 sinh z + C

2 = 1 1 – (coth ) x –csch x

2 (–csch 2 x ) =

37. Area = ∫ cosh 2

ln 3

0 xdx =⎢ sinh 2 x ⎥

⎡ 1 ⎤ ln 3

⎣ 2 ⎦ 0 sinh(2 z ∫ ) dz = ∫ 2 sinh u du = 2 cosh u + C

1 ⎛ e 2 ln 3 – e –2 ln 3 e 0 – 43 e –0 ⎞ z

⎠ 1/ 4 = 2 cosh(2 z ) + C

Section 6.9 Instructor's Resource Manual

42. Let u = e x , so du = x e dx .

48. tanh x = 0 when sinh x = 0, which is when x = 0.

e sinh e dx = ∫ sinh u du = cosh u + C Area = ∫ 8 ( tanh ) − x dx + ∫ tanh x dx − 0

= cosh e x + C

8 8 sinh = x

2 ∫ tanh

0 x dx = ∫ 0 dx cosh x

43. Let u = sin x, so du = cos x dx

Let u = cosh x, so du = sinh xdx.

∫ cos sinh(sin ) x x dx = ∫ sinh u du = cosh u + C sinh x

dx = 2 ∫ du = 2 ln u + C

= cosh(sin x) + C

cosh x

8 sinh x

44. Let u = ln(cosh ), x so

2 ∫ 0 dx =⎡ ⎣ 2 ln cosh x ⎤

cosh x

du =

⋅ sinh x = tanh x dx .

= 2(ln cosh 8 − ln1) = 2 ln(cosh 8) ≈ 14.61

cosh x

tanh ln(cosh ) x x dx = ∫ u du = + C 49. Volume =

2 ∫ 0 π cosh x dx = 0 (1 cosh 2 ) + x dx 2 ∫

[ln(cosh )] x + C π⎡ sinh 2 x

45. Let u = ln(sinh x 2 ) , so

sinh 2

du =

2 cosh x ⋅ 2 xdx = x 2 coth x dx 2 .

ππ sinh 2

sinh x

x coth x 2 ln(sinh x 2 ) dx = ∫ u du =⋅ + C

0 π sinh xdx 1 2 2

2 2 2 50. Volume = ln10

= [ln(sinh x )] + C

2 ln10 ⎛ e x − e − x ⎞

46. Area = ∫ – ln 5 cosh 2 x dx = 2 ∫ 0 cosh 2 x dx

π ∫ ln10 dx = ∫ 0 ( e 2 ln 5 x =π 0 –2 + e –2 x ) dx

ln10 e 2 x –2 + e –2 x

2 sinh 2 x

sinh(2 ln 5)

ln 1 = [ e 2 x –4– x e –2 x ] ln10 1 ln 25 25 1 ⎛ 1 ⎞ 0

51. Note that + 1 sinh 2 x = cosh 2 x and

47. Note that the graphs of y = sinh x and y = 0

2 cosh x x =

+ 1 cosh 2

intersect at the origin.

ln 2

0 0 Surface area = ∫ 0 2 π y 1 + () dx dx

Area = ∫ 0 sinh x dx = [cosh ] x 0 1 dy 2

0 π 2 cosh x + 1 sinh 2 x dx

0 π 2 cosh cosh x x dx

0 π+ (1 cosh 2 ) x dx

=π+ ⎢ x sinh 2 x ⎥ =π+ sinh 2 ≈ 8.84

Instructor’s Resource Manual

Section 6.9 401

1 ⎛ 2 dy ⎞

52. Surface area = ∫ 0 2 π y 1 +⎜⎟ dx = ∫ 0 π 2 sinh x + 1 cosh 2 xdx

⎝ dx ⎠

Let u = cosh x, so du = sinh x dx

2 sinh x + 1 cosh 2 xdx =π 2 ∫ 1 + u du 2 =π 2 ⎡ 1 + u ⎢ 2 + ln u + 1 + u 2 + C ⎤

=π cosh x + 1 cosh 2 x +π ln cosh x + + 1 cosh 2 x + (The integration of C ∫ + 1 u du 2 is shown in Formula 44 of

the Tables in the back of the text, which is covered in Chapter 8.)

π 2 sinh x + 1 cosh 2 xdx =π ⎡ cosh x + 1 cosh 2 x + ln cosh x + + 1 cosh 2 x ∫ ⎤ 0

= π⎡ cosh1 1 cosh 1 ln cosh1 + 2 + + + 1 cos 1 2 − 2 + ln 1 + 2 ⎢ ⎤ ⎣

⎛⎞ x

53. y = a cosh ⎜⎟ + C

⎝⎠ a dy

⎛⎞ x = sinh ⎜⎟

dx ⎝⎠ a

dy 2 1 ⎛⎞ x

2 = cosh a ⎜⎟

We need to show that

Note that + 1 sinh 2 ⎜⎟ = cosh 2 ⎜⎟ and cosh ⎛⎞> ⎜⎟ 0. Therefore,

cosh ⎜⎟ = cosh ⎜⎟ =

54. a. The graph of y =− b a cosh ⎜⎟ is symmetric about the y-axis, so if its width along the ⎝⎠ a

⎛⎞ a

x -axis is 2a, its x-intercepts are (±a, 0). Therefore, ( ) ya =− b a cosh ⎜⎟ = 0, so b = a cosh1 1.54308 . ≈ a

⎝⎠ a

b. The height is (0) y ≈ 1.54308 a − a cosh 0 = 0.54308 a .

c. If 2a = 48, the height is about 0.54308a = (0.54308)(24) 13 ≈ .

55. a.

b. Area under the curve is 24 ⎡ 24 ⎛ x ⎞ ⎤ ⎡ ⎛ x ⎞ ⎤

∫ 24 ⎢ − 37 24 cosh ⎜ ⎟ ⎥ dx = ⎢ 37 x − 576 sinh ⎜ ⎟ ⎥ ≈ 422 −

Volume is about (422)(100) = 42,200 ft 3 .

Section 6.9 Instructor’s Resource Manual

⎟ dx = ∫ cosh ⎜ ⎟ dx = 24 sinh ⎛ ⎞ = 48sinh1 56.4 ≈ − 24

⎛ 24 dy ⎞ 1 24 + dx =

+ 1 sinh 2 ⎛ x ⎞

Surface area ≈ (56.4)(100) = 5640 ft 2

1 cosh t

2 1 1 56. Area = 1 cosh sinh t t − x − 1 dx = cosh sinh t t − ⎡ xx 2 −− 1 ln x + x 2 − 1 ⎤

cosh t

cosh sinh t t − ⎡ ⎢ cosh t cosh 2 t −− 1 ln cosh t + cosh 2 t −− 1 0 ⎤

cosh sinh t t − cosh sinh t t + ln cosh t + sinh t = ln e t =

57. a. (sinh x + cosh ) x r = ⎜

sinh rx + cosh rx =

= e rx

b. (cosh – sinh ) r

cosh rx – sinh rx =

c. ( cos x + i sin x ) = ⎜

d. ( cos xi − sin x ) = ⎜

58. a. gd (– ) t = tan [sinh(– )] –1 t b. If y = tan –1 (sinh ) t then tan y = sinh t so

sin y =

= tan –1 (– sinh ) t = – tan –1 (sinh ) t =− () gd t tan y

sinh t

so gd is odd.

2 tan 2 y + 1 sinh t + 1

t [ D gd t ( )] =

⋅ cosh t =

1 cosh t

sinh t

+ 1 sinh 2 t

cosh 2

= tanh t so y = sin –1 (tanh ) t

cosh t

= sech t > 0 for all t, so gd is increasing.

D t 2 [ ( )] gd t = D t (sech ) t =− sech tanh t t Also, Dy t =

⋅ cosh t

+ 1 sinh 2

D t 2 [ ( )] gd t = 0 when tanh t = 0, since

cosh t

sech , t

sech t > 0 for all t. tanh t = 0 at t = 0 and

cosh 2 t cosh t

tanh t < 0 for t < 0, thus D t 2 [ ( )] gd t > 0 for

so y = ∫ 0 sech u du by the Fundamental

t < 0 and D t 2 [ ( )] gd t < 0 for t > 0. Hence Theorem of Calculus.

gd (t) has an inflection point at (0, gd(0)) = (0, tan − 1 0) = (0, 0).

Instructor’s Resource Manual

Section 6.9 403

59. Area = ∫ 0 cosh t dt = [sinh ] t 0 = sinh x

Arc length =

∫ x 1[ + D cosh ] t 2 dt = + 1 sinh 2 tdt

0 cosh t dt = [ sinh t ] 0 = sinh x

60. From Problem 54, the equation of an inverted

catenary is y =− b a cosh . Given the

information about the Gateway Arch, the curve The functions y = sinh x and y = ln( x + x 2 + 1)

passes through the points (±315, 0) and (0, 630).

are inverse functions.

Thus, b = a cosh

and 630 = b – a, so

62. y = () gd x = tan –1 (sinh ) x

b = a + 630.

tan y = sinh x

315 –1 x = gd () y = sinh (tan ) y

a + 630 = a cosh

⇒≈ a 128, so b ≈ 758 .

a Thus, y = gd –1 () x = sinh –1 (tan ) x

The equation is y = 758 128cosh −

6.10 Chapter Review

ln( x 9. 4) True: fgx ( ( )) =+ 4 e −

=+− 4( x 4) =

Concepts Test

and

1. False: ln 0 is undefined. gfx ( ( )) = ln(4 + e x − 4) = ln e x = x

exp( x + y ) = exp exp x y

dx 2 x 2

11. True:

ln x is an increasing function.

e 3 1 e 12. False:

Only true for x > 1, or ln x > 0.

4. False: The graph is intersected at most once by every horizontal line.

14. True:

e x is an increasing function.

5. True: The range of y = ln x is the set of all

15. True:

lim (ln sin x − ln ) x

real numbers.

2 e x + 1 17. False:

ln π is a constant so

ln π= 0.

8. True:

ln(2 e x + 1 ) – ln(2 e x ) = ln

2 e x = ln e = 1 18. d True: (ln 3 x + C )

Section 6.10 Instructor’s Resource Manual

19. True:

e is a number.

e x > 1 and e − x << 1 e x , thus

20. True:

exp[ ( )] gx ≠ because 0 is not in the 0 e x − e − x

range of the function y = e x .

22. True: 2 ( 2 tan x + sec x )( ' − tan x + sec x

( 2 sec x + sec tan x x )

= sec 2 x − tan 2 x = 1 sin () 2 1

but

23. True:

The integrating factor is

cos − 11 () 2 2

∫ 4/ x dx

ln x 4 e = e = e = () 4 x

cosh(ln 3) =

e () 2x . Thus,

24. True:

The solution is yx = e − 4

slope = 2 e ⋅ e and at x = 2 the

slope is 2.

⎟ ⎞== ln1 () 0 ⎜

⎛ sin x

25. False:

The solution is yx = e 2x

yx ' () = 2 e 2 x . In general, Euler’s

1 method will underestimate the π 35. True: lim tan − x = − , since solution if the slope of the solution is

x →−∞

increasing as it is in this case.

lim tan x = −∞ .

x →− π + 2

sin arcsin(2) ( ) is undefined

26. False:

36. False:

cosh x > 1 for x ≠ 0 , while sin − 1 u is

27. False:

arcsin(sin 2 ) π= arcsin 0 = 0 only defined for −≤≤ 1 u 1.

28. True:

sinh x is increasing.

sinh x

37. True:

tanh x =

; sinh x is an odd

29. False:

cosh x is not increasing.

cosh x

function and cosh x is an even

30. True:

cosh(0) == 1 e 0 function.

If x > 0, e x > 1 while e − x << 1 e x so

38. False:

Both functions satisfy y ′′ − = y 0 .

cosh x = ( e x + e − x ) < (2 e x )

2 2 39. True:

ln 3 100 = 100 ln 3 100 1 > ⋅ since

= x e x = e . If x < 0, –x > 0 and

ln 3 > 1.

e − x > while 1 e x << 1 e − x so

40. False:

ln(x – 3) is not defined for x < 3.

cosh x = ( e x + e − x 1 ) < (2 e − x ) 41. True:

y triples every time t increases by t 1 . 2 2

42. False:

x (0) = C; C = Ce − kt when

31. True: x sinh x ≤ e is equivalent to

= e − kt , so ln =− kt or

e − e − x ≤ e . When x = 0,

ln 1 t = 2

− ln 2 ln 2

sinh x =< 0 e = . If x > 0,

Instructor’s Resource Manual

Section 6.10 405

43. True: (() yt + zt ( )) ′ = yt ′ () + zt ′ () d 1 d cos x

8. tanh − 1 (sin ) x =

44. False: Only true if C = 0;

( yt 1 () + yt 2 ( )) ′ = yt 1 ′ () + yt 2 ′ () x = ky t 1 () ++ C ky t 2 () + C d − 1 1 d

tan x =

9. sinh (tan ) x =

45. False: sec Use the substitution u = –h. 2 x

by Theorem 6.5.A.

10. 2 sin

12 dx

e 0.05 ≈ 1.051 <+ ⎛ 1 ⎜ ⎞ ≈ 1.062 ()

1323 − x x

47. True:

If Da x ) = a x x ( ln a = a x , then

ln a = 1, so a = e.

11. d sec − 1 1 e d x = e x

Sample Test Problems

2. sin ( 2 3 x d ) = 2 sin( 3 x ) sin( 3 x ) 1 ⎛⎞ x d ⎛⎞ x

x ) x = 6 x sin( x ) cos( x )

= 2sin( 3 ) cos( 3 d 3 2 3 3

sin () 2 ⎣ ⎝⎠ ⎦ ⎝⎠ ⎝⎠ 2

= (2 x − 4) x − x e 24 13. 3ln( e += 1)

5 ( x − 1) 14. ln(2 x − 4 x + 5) ( x − 1) ln10 dx

5. tan(ln e ) = tan x = sec x =− sin e e

d d sech 2 x

7. 2 tanh

2sech x 2 = x

Section 6.10 Instructor’s Resource Manual

16. ln(tanh ) x =

tanh x 23. x 1 + x d = e (1 + x ) ln x

tanh x dx

sech 2 x = csch sech x x = e (1 + x ) ln x [(1 + x ) ln ] x

tanh x

d − 1 − 2 d ⎢ (1)(ln ) (1 x ++⎜⎟ x )

d ∫ e − dx = ∫ e − 3 dx = ∫ e du

20. (log 10 2) x

6 cot 3 x dx = 2 ∫

3cos 3 x dx = 2 ∫ du

(log 10 + 2 log 10 x )

= (log 10 2) x

3 x ln10

3 ln10 log

20 tan 5 sec 5 tan 5 x x x ∫

e x sin 2 x e dx =

∫ sin u du =− cos u + C

20 sec 5 sec 5 x x +

20 sec 5 (sec 5 x x + tan 5 ) x =− cos e + C

2 Check:

20 sec 5 (2sec 5 x x − 1) d d ( cos − e x + C ) = (sin e x ) e x = e x sin e x

2 dx = 3 (2 x + 1) dx x 4 x x +− x 5 ∫ 2

x +− x 5

∫ du

dx (

) = 2 ( x +− x x 5) +− x 5 dx

Instructor’s Resource Manual

Section 6.10 407

+ 1 eu ∫ −

dx =− ∫

e e =− ∫ 2 du =− tan − 1 u +=− C tan − 1 (ln ) x + C

∫ 2 x 4 cos x dx = 2 (cos x 2 )2 x dx =

u du

∫ 2 ∫ 2 cos ∫ sech ( x − 3) dx = sech 2 ∫ u du = tanh u + C

= 2 sin u += C 2 sin x 2 + C = tanh( x −+ 3) C

x 2 + C ) = 2 cos x x = x 4 cos x [tanh( x − 3)] = sech ( 2 x − 3) ( x − 3)

= sech ( 2 x − 3)

31. Let u = 2x, so du = 2 dx.

dx = 2 ∫

2 dx 35. fx ′ () = cos – sin ; ( ) x xfx ′ = when tan x = 1, 0

14 − x 2 − 1 (2 ) x 2 π x =

du 4

1 − u 2 fx ′ () > when cos x > sin x which occurs when 0

= 2sin − 1 u += C 2sin − 1 2 xC +

π – ≤< x .

Check:

f ′′ () x = – sin – cos ; ( ) x xf ′′ x = when 0

f ′′ () x > when cos x < –sin x which occurs 0 = − 1 4x 2

when – ≤< x – .

32. Let u = sin x, so du = cos x dx.

Increasing on – ⎡ , ⎤

dx ∫

du = tan u + C

+ 1 sin 2

1 + u 2 Decreasing on ⎡ ππ , ⎤

= tan − 1

⎣ ⎢ 42 (sin ) ⎥ x + C ⎦

Check:

Concave up on – ⎛ π π ⎜ , – ⎞

⎣ tan (sin ) x + C ⎦ dx = + 1 sin 2 sin x x dx

Concave down on – ⎛ , ⎞

Inflection point at – ⎛ ⎜ ,0 ⎞ ⎟

⎝ 4 ⎠ Global maximum at ⎛ π ⎜ ,2 ⎞

Section 6.10 Instructor’s Resource Manual

37. a.

fx ′ () = 5 x 4 Global minimum at – ,–1 + 6 x 2 + ≥ > for all x, so 4 4 0

f (x) is increasing.

b. f(1) = 7, so g(7) = f − 1 (7) = 1.

= 1 100 e − 0.06931 t

f is increasing on [0, 2] because fx ′ () > on 0 t () = 100 ≈ 66.44

f is decreasing on ( −∞ , 0] ∪ [2, ) ∞ because

It will take about 66.44 years.

Inflection points are at

The graph of f is concave up on

∞ because 2, ) f ′′ () x > 0

on these intervals.

40. Let x be the horizontal distance from the airplane

The graph of f is concave down on

2, 2 + 2) because f ′′ () x < on this 0 to the searchlight,

The absolute minimum value is f(0) = 0.

The relative maximum value is f (2) = . d θ

1 ⎛ 500 ⎞ dx

⎝ x 2 ⎠ () dt

dt 1 + 500 2 ⎜

The inflection points are

x 2 + 250, 000 dt

When θ = 30°, x =

. The angle is decreasing at the

rate of 0.15 rad/s ≈ 8.59°/s.

Instructor’s Resource Manual

Section 6.10 409

41. y = (cos ) x sin x = e sin x ln(cos ) x 45. (Linear first-order) y ′+ 2 xy = 2 x

dy sin x ln(cos ) x d

∫ 2 xdx

= e [sin ln(cos )] x x

Integrating factor: e x = e

sin x ln(cos ) x ⎡

cos ln(cos ) (sin ) ⎛

= If x = 0, y = 3, then 3 = 1 + C, so C = 2. (cos ) x

sin x ⎡

sin 2 x ⎤

⎢ cos ln(cos ) x x −

cos x ⎥ ⎦

Therefore, y =+ 12 e – x 2 .

46. Integrating factor is e – ax .

The tangent line has slope 0, so it is horizontal:

[ D ye – ax ] 1; = y = e ax ( x + C )

y = 1.

47. Integrating factor is e –2 42. Let t represent the number of years since 1990. x .

14, 000 10, 000 = e 10 k [ D ye –2 x ] = e – x ; y = – e x + Ce 2 x

ln(1.4) k =

48. a. Qt ′ () = 3 – 0.02 Q

10, 000 0.03365 t e b. Qt ′+ ( ) 0.02 Q = 3

10, 000 y (0.03365)(20) = e ≈

Integrating factor is e 0.02t

The population will be about 19,600.

[ 0.02 t ] = 3 0.02 D Qe t e

The domain is () 0, ∞ .

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