Calculus for the Life Sciences Marvin L. Bittinger Test Bank.pdf (49)

FINAL EXAM ANSWERS TO CHAPTER TEST FORMS

29. 32 in.by 32 in. by 8 in.; 8192 in.3 31. 0.912871 32. (a) P (t ) = 40, 000e0.12t (b) 82,177 (c) 5.8 hr

55. (a)

1 35. 1 sin(2x )+ C ) − xcos(2x 4 2 1 − 36. 3 cos3 x + C 2

⎞

ln x −

2 − 6 x⎟⎟ x ⎠ x + C 38. 10ln ⎜⎜ ⎝ 39. 76 3

(

(b) Asymptotically stable, unstable (c)

y =1

(d)

56. 1774.10 lb 57. y=

)

41. π e36− 1

42. Divergent 43. Convergent12,

⎡−29 10⎤ 44. ⎢ ⎣ 8 −4⎥⎦ 45. 1, invertible 46. r =2, t

y = 0,2

x2 + 6x + C 4

40. 37 6

6

5

Ce3x5 5 , or y = eCe3x 5

54. y = − c o s x − 2 x2 − 2

34. 2e2 + 2

⎛

⎡1⎤ ⎡5⎤ r = 3, t⎢⎥ ⎢⎥; ⎣6⎦ ⎣1⎦

5 x +C e4x Ce−1 2

58. xt ( ) = C1 e4t+ 2Ce−2t ; y (t ) = C1 e4t+ 5Ce−t 59. 2 13 60. 0.952 61. (a)

8 5

(b)

2 5

2 , or 2 6, or 0.3266 3 15

62. 0.3721 63. 0.046875

770 ⎤ 47. (a) ⎡⎢ 203.5 ⎥ ⎣ ⎦

64. − 78

⎡ 1331 ⎤ ⎢ 446.05⎥ ⎣ ⎦ 2 48. −3 x y + 2 y

66. 63 4

(b)

sin( ) ⎥⎦

52. ln2 + 1 3 53. y=

x5 33. 8 + C 5

37.

3y4 ⎡ ⎤ 12 xy3 50. ⎢ ⎥ ye3x −e3 x ⎢ 3ye3x ye3x ⎣− .

51. Relative maximum8= at (0,0)

sin( )

30. 9.055556

367

65. 0.292316

49. −3 y2

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ANSWERS TO EVEN-NUMBERED EXERCISES 369

Chapter 1 Exercise Set 1.1

10. slope: m = −1; y-intercept: (0,4)

2. Horizontal line:y = −3.5

12. slope:m = −3; y-intercept: (0,2)

4. Vertical line: x = 10

14. slope m = 2; y-intercept(:0,3 ) m = 1; y-intercept(:0,2 )

16. : 6. slope:m = −0.5; y-intercept(0, : 0)

slope 3 1 18. slope: m = − ; y-intercept⎛⎜:0, 4 ⎞⎟ 4 ⎝ ⎠ : 20. y = 7 x 22. y = −3 x+ 13 24. y = −5 x+ 25 26. y =

8.slope: m = 3; y-intercept: (0,0)

4 3

x+ 7

28. y = 8 30.

1 4

32. −

3

4 This line has no slope.

34.

36. 0 38. 4 40. 3 42.

y=1x+3 4 2

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370

ANSWERS

44. y = −

29 13 x − 16 4

c) Since cars cannot have negative speed, and since the car will not need to stop if it has speed of 0 then the domain is an positive real number.

NOTE: The domain will have an upper bound since cars have a top speed limit, depending on the make and model of the car.

46. x = −4 48. y =

1 2

50. y = 4 x

66. a) N =1.02 P

52. y = 3 x − 1 54.

56.

2.6 6.2

b) c)

P=360,000

≈0.3171, or 31.71%

8.25 9

68. The use of the slope-intercept equation or the point-slope equation depends on the problem.

If the problem gives the slope and the yintercept then one should use the slope-intercept equation. If the problem gives the slope and a point that falls on the line, or two points that fall on the line then the point-slope equation should be used.

=0.916≈0.9167=91.67%

58. a) F (−10 ) = 14 F

F 0( ) = 32 F F10 (

) = 50 F

F40 (

) = 104 F

Exercise 1.2 2.

b) 86 F c) − 40

60. t ≈ 0.0205 seconds 62. a) M=0.4W b)

40%; This means that 40% of the body weight is the muscles. c) 48 lb

64. a) D5( ) = 0.5ft

D10 ( ) = 11.5ft D20 (

) = 22.5ft

D50 (

) = 55.5ft

D65 (

) = 72ft

4.

b) 6 0 5 0 4 0

r

3 0 2 0 1 0

10

20

N=204,000

30

40

50

60

70

ANSWERS TO EVEN-NUMBERED EXERCISES 371

6.

16. y = − x2 − x + 6

18. y = 3 x2 − 9 x + 2

8.

10

The graph of the function is not a parabola.

.

The graph of the function is a parabola; the

12

vertex is (1,−3).

20. y =

1 2 x + 4x− 2 3

22. 1+

5 and 1 − 5

.14. y = x 2 − 6 x +5

24.

5 + 33 5 − 33 and 4 4

26. −3 + iand − 3 − i 28. −2 +

7 and −2 − 7

30. −5 + i 2 and −5 − i 2

1+ 2 1 2 and − − 32. − 2 2

372

ANSWERS

34. a) $114.4333 million b) 2007– 2008 season

Exercise Set 1.3

36. a) 163.4032 pounds b) Approximately 73.8 inches

2.

38. a) For large values of x,x4will be larger than −10x 3 +3x 2 −2x+7.

b) For very large values ofx , the function will look like x4. c)

4.

40. a) For x values very close to 0, 2xis larger than x 3 since for values of xvalues less than 1 the higher the degree the smaller the values of the term.

b) For xvalues very close to 0, the function will look like 2xsince the x3term may be “ignored.”

6.

c)

8.

42

=x 2.359

.

=x 2.039 and x =3.594

44

y = 8.254x − 5.457

.

y = 1.004 x 2 + 1.904x − 0.601

46 .

48 .

50 .

y = 0.218 x 3+ 0.188x2 − 29.643x + 57.067

ANSWERS TO EVEN-NUMBERED EXERCISES 373

10.

16. y 5

−5

x

5

10

−5

−10

12.

18. x 5 2 20. b 1 2 22. c 1 8 24. t − 6 5 26. m − 1 2

14.

28.

( x +34 )

30.

7

t

32.

5

t2

34.

−12

1 3

6

y4

36.

5

2

38. −6

−4

−2

2 −2 −4 −6

x 4

6

40.

42.

y2

1 b2 1

6

m19 1

4

y+7

5

w4

44. 1024 46. 4 48. 3 1 2 5

50. The domain is (−∞,−2)∪(−2,∞).

52. The domain consists of all real numbers such that x≠−3 and x≠−2.

54. x ≥ 3

374

ANSWERS

56. 84 58. a) b) c) ≈

≈ 26.864 kg ≈ 853.156 kg 11.393 m

6. −5π rad 2

60. 3136m2 62.

1± 5 2

64. At most a function of degree ncan have n y-intercepts. A polynomial of degree ncan be factored into at most nlinear terms and each of those linear terms leads to a y-intercepts. This is sometimes called the Fundamental Theorem of Algebra.

66

x = 1.5 and x= 9.5

.

x = −2 and x = 3

68

x = 1 and x= 2

.

Exercise Set 1.4 70 . 2.

8. 210

10. 540

5π rad 6

12. −132

4. 5π rad 3

14. Coterminal 16. Not 18. coterminal 20. Coterminal 22. Not 24. coterminal 26. 0.9903

0.7547

28. 2.4751

0.6745

30. 1.0515

ANSWERS TO EVEN-NUMBERED EXERCISES 375

y r and csct= y , y ≠ 0 r r x cos = and sect= x (x≠ 0)and r t = y (x≠ 0 )and cott = x (y≠0). y x tan From the above definitions and recalling that the treciprocal of a nonzero number xis given by 1 we show t that : x

32. 0.3090

sint =

34. 0.8665 36. 1.2151 38. 0.6895 40. 60.4586 42. 43.9455 44. 40.0302 46. 0.2630 48. 0.6761

sint =

50. 0.8999 52. 41.2122 54. 3.3127 56. 38.6598 58. 36.8699 60. The coordinate is 187.939. − x − coordinate is 68.404.

The ft 62. 1582.02 y 64. 6.16 cm sec

66. a) x =

h

( )

tan 67

b) 555.567 ft

68. a) Since the triangle has two angles equal in magnitude it should have two sides that are equal as well. (It is an isosceles triangle.) b) h= 2 c) Since the hypotenuse is known then we can use the figure to find the trigonometric

functions ofπ =45 using the ratios of the 4 sides of the triangle.

csct

, cost=

1 and tant= 1 . cott sect

72. a) sin t =u =u 1 b) Consider the triangle made by the sides v, w, and y. The angle vwhas a value of 90−r (completes a straight angle). The sum of angles in any triangle is180. Therefore s +90 + (90 − r ) = 180 s + 180 − r =180 s − r =0 s=r

c ) sins =

w which means that w =( sins ) v , v

v = v. Thus, w = (sins)v=sinscost.

1 u x d) sint = = uand sinr = which means that 1 u x=ucosr. In part b) we showed that r=s, therefore cosr=coss. So, x=ucosr=sintcoss. cost =

=w+x. Using the results s t e) sin(+)=w+x 1 we have obtained from previous parts we conclude that:

sin(s+t)=w+x=sinscost+cosssint.

70. Let (x,y)be a non-origin point that defines the ⎛π ⎞ u − = = u = sint 74. a) cos⎜ ⎝2 t⎟ 1 ⎠ r= x2+y2 be the distance form the origin to v ⎛ π ⎞ the point (x,y). Then the trigonometric b) sin 2 − t ⎟⎠ = 1 = v = cost ⎜ ⎝ functions are defined as follows: 76. cost 2 + sin2 t = 1 t + sin2 t = 1 cos2 sin 2 t sin2t t sin2 cot2t +1 = csc2 t

terminal side of an angle, t, and let

376

ANSWERS

(2t ) = cost(+ t )

78. a) cos

4. 2π

= costcost− sintsint cos2t− sin2t = 2 b) cos2(t ) = cos2 t − sin t

(

= cos2t − 1− cos2t = 2 cos t − 1 c) cos (2t ) = cos2t− sin2t

)

2

=

(1− sin2 t)− sin2t

= − sin t 2

80.

cos (2t ) = 1 2 cos t − 1

cos(2+t 1 = 2 t ) cos (2+t1) 2cos 2

6. −7π 4

=2cos2t

82. a) 0.8208 b)

8. −

c) The result from part b) agrees with definition

10. −

of V(h)since the values of V(h)between 12. 0 0 and 1.

Exercise Set 1.5 2. −5π 6

1 2 1 2

14. − 3 16. 0.2756 18. −0.2924 20. −1.34563 22. −1.37638 24. −0.891929 26.

3π + 2 nπ 2

28. π + 2 nπ 30.

3π + nπ 4

32.

7π + nπ 4 3

34. 2 nπ 36.

3π + 2 nπ 2

ANSWERS TO EVEN-NUMBERED EXERCISES 377

c) Half the difference between the maximum and minimum, using the results from part a), implies that the amplitude is

38. amplitude = 3, period=π , midline is y=−3 maxi mum = 0, minimum=−6

( k + a ) − ( k −=2aa) =2.

40. amplitude= 3, period= 6, midline is y= 2 maximum= 5, minimum=−1

2

a

=1 42. amplitude 2 , period= π , midline is y= 2 2 =5 3 maxi mum 2 , minimum =2

70. Since the apex, Lis large, Tis small, and d is small, then the basilar membrane is affected mostly by low frequency sounds.

72

440

2 44. amplitude = 3, period= , midline is y=−2 3 maximum= 1, minimum= −5

.

There are 12 notes above A above middle C.

.

a) Left to the student.

5 3 46. y = 2 cos⎛⎜ t ⎟⎞ + 2 ⎝2⎠ 1 = 48. y 2 sin(2πt)−1

76

1 b) y =1 cos(2t)− 2 2

74 19.01955

.

78 .

50. 1.12788 megajoules m2 2

52. 0.858372 megajoules m 54. V ( t) = 1700 cost )(π +2800 56. T (t) 0.77 = cos⎛⎜ π12 ⎞ 36.1 ⎝ ⎟+ ⎠ 58. 220 Hz

6.7cos(0.496372t)+137 p(t)= 62. (−0.17365,−0.984 60.

64. (−81)

0.22252,−0.974

1 3 66. − + 2 2 93)

68. a) The amplitude could be thought of as half the difference between the maximum and max−min minimum, a = , which implies that 2 2a=max+min. Solving the system of equations above for max and min gives the desired results.

b) The average mean of the maximum and minimum, using the results from part a), implies that the mid-line equation is ( k + a2 ) + ( k −=2ak) =k. y = k

c) We use the double angle identity obtained in Exercise 79 of Section 1.4 and solve for −sin2t to obtain the model in part b).

80. a) Left to the student. b) y = sin (2 t ) c) We use the double angle identity for

sin(2t)obtained in Exercise 77 of Sectio 1.4 to obtain the model in part b).

82. a) Left to the student. b) Left to the student. c) The horizontal shift moves every point of

π units to the left. the original graph 3

84. Left to the student.

378

ANSWERS

e) False

Chapter 2 Exercise Set 2.1

14. a) False b)

False

c)

False

d)

False

−2

e)

True

c) Yes

f)

False

d) Limit does not exist.

g)

False

e ) −3

h)

True

The function is notx = −2.

2. 4.

6. continuous at The function isx a) − = −2. not continuous at 2; − 2; − 2 b)

f) The function is not continuousxat = −2.

i) True j) False

8. a) ≈ 0.25; ≈ 0.25; ≈ 0.25 b)

≈ 0.25

16. a) True

c)

The function is continuousxat= 1.

b)

True

d)

The limit does not exist.

c)

True

d)

True

e)

True

− 2). at t( e) The function is undefined f) The function is not continuous at = x−2.

b)

0

18. pis continuous at x=2.99; pis not continuous at x=3; pis continuous at x=3.04; and pis not continuous at x=4.

c)

0

20.

d)

3

10. a) 0

e) f (x)is not continuous at=x −2. f) f (x)is continuous at x=1.

12. a) True

x→ 2 −

d) False

x→2+

limp(x) does not exist. x→ 2

22. Limit does not exist.

24. Cis continuous at x=0.1; Cis not continuous at x=0.2; Cis continuous at x=0.25; Cis continuous at x=0.267.

26. lim C (x) = $2.60; lim C(x)= $2.60; 1− x→ 4

1+ x→ 4

lim C (x ) = $2.60

b) False c) False

limp(x)=0.6; limp(x)=0.8;

x →1 4

28.

lim C (x) = $2.90; lim C(x)= $3.20;

x→0.6−

x→0.6+

)does not exist. lim C ( x 0.6

x→

ANSWERS TO EVEN-NUMBERED EXERCISES 379

30.

limC(x)= $2.60;limC(x)=$2.90; x→0.4+ limC(x)does not exist.

14. 3

x→0.4− x → 0.4

32. There was a jump in the population at t=0.5, t=0.75, t = 1.5, and t=1.75 due to births.

The decline in the population at t=1.25was due to deaths.

16.

3 1 + 2 3

18.

2

20. Does not exist.

34

11

22.

.

24.

7

.

There was a jump in the population at t=0.1, t=0.4, t=0.5, and t=0.6 due to births.

The declines in the population at t=0.3and t=0.8were due to deaths.

4 − 83

38

33

.

Once the task is mastered one should get 100 correct trials out of 100 trials.

36

40 42. .

44.

26. 10 x

5 28. − x2 30. 1

limN(t)=100;limN(t)=100; x→ 30+ limN(t)=100. x→ 30 −

32. 0

x→ 30

34. 0

N(t)is discontinuous at t=10. N(t)is continuous at t=26

46. Continuous by C1 and C2; 19.

1 48. Continuous by C4; 2 .

58. 0.92857 60. 0 62. 0

Exercise Set 2.2 2. 5 4. Does not exist.

6. −7 8.

−8

10.

2 5

12. −9

1 2

40. 1

44.

2 3

.

54. Continuous by C5; 19 .

56. Continuous by C3;

38. − 2

.

1

42.

50. Continuous by C5; 1 52. Continuous by C5;

36. Does not exist.

2 1 2 1

46.

7

48. 0

Exercise Set 2.3 2. a) 10 x +5 h b) 50; 45; 40.5; 40.05

4. a) −10x − 5h b)

−50;

− 45; − 40.5;−40.05

6. a) 15x2 + 15xh + 5h2 b) 380;305;246.05;240.6005

8. a)

−4

x ( x+ h)

b) ≈ −0.167;≈ − 0.2;≈ −0.244; ≈ −0.249

380

ANSWERS

10. a)

2

32.

b) 2 for all values of x.

12. a) 2 x + h − 1

34.

b) 11;10;9.1;9.01

−2

3 − 2 ( x + h ) +3 − 2 x −2 x − 1)( x+h−1) (

14. a) 1.4 pounds per month b) ≈1.583pounds per month

Exercise Set 2.3

c) ≈1.767pounds per month d) The average growth rate for the first few weeks of a typical boy’s life will be larger than 1.767.

2. a-b)

16. a) 0 degrees per day. Using this rate of change, we would not be able to conclude the person was sick.

b) 0.9 degrees per day; 1.5 degrees per day; 1 degree per day; 0.5 degrees per day; 0 degrees per day; −0.1degrees per day; −0.4degrees per day; −2 degrees per day; −1.4degrees per day; 0 degrees per day c) The temperature began to rise on day 1, reached the peak on day 5, began to subside on day6, and was back to normal on day 10. d) By examining the graph, we notice an increase in temperature after day 1 which reaches a maximum value of 102.5 on day 5 and continues at the max temp through day 6 at which time the temperature begins to drop until it reaches the normal level of 98.6 on day 10.

c) 14 x d) −28;0,14. These slopes are in agreement with the slopes of the tangent lines drawn in part b).

4. a-b)

18. a) 40 miles; 250 miles b) 210 miles. This represents the distance traveled between t=2and t=5seconds. c) 70 miles per hour

20. a) 14.5 miles per gallon b) 14.5 miles per gallon

22. In the period between 1850 and 1860 the deer population is decreasing which corresponds to a negative rage of change. In the period between 1890 and 1960 the deer population is increasing which corresponds to a positive rate of change

c) −14x d) 28;0,−14. These slopes are in agreement with the slopes of the tangent lines drawn in part b).

6. a-b)

24. m 26.

28.

(3ax2+2 bx )+ h (3b)ax + 4x3+6

4xh2 + h 3 x h+ 30.

1 1−(x + h)) (1−x)( 2

c) −3 x2

ANSWERS TO EVEN-NUMBERED EXERCISES 381

d) −12;0;−3. These slopes are in agreement with the slopes of the tangent lines drawn in

d) −5;−1;1. These slopes are in agreement with the slopes of the tangent lines drawn in part b).

part b).

14. a-b)

8. a-b)

c) 10x−2

c) −2

d) −22;−2;8. These slopes are in agreement with the slopes of the tangent lines drawn in part b).

d) −2;−2;−2. These slopes are in agreement with the slopes of the tangent lines drawn in part b).

10. a-b)

16. a-b)

1 c) 2

c) − 5 x2

1 1 1 ; ; . These slopes are in agreement 2 2 2 with the slopes of the tangent lines drawn in part b).

5 d) − 4 ;does not exist; −5. These slopes are

d)

12. a-b)

in agreement with the slopes of the tangent lines drawn in part b).

18

2ax + b

.

y=12x+16; y=0; y=48x−128

20 22. y=−2x−4; .

x y = − 2 +2; y=−

24. y=−6x−4; y=−1; y=6x−16

x 2 50 + 5

26. The function is not differentiable at 2, x4, x5,

x7and x8.

c) 2 x − 1

28. The function is not differentiable at values of xof the form x=0.2n where n=1,2,3,….

382

ANSWERS

30. The function is not differentiable in the year 1850, 1860, 1865, 1880, 1910, 1960, 1975 and 1980.

32. 4Land L6are the only lines that appear to be tangent since the others intersect the curve and are not “tangent” to it.

34. 4 x 3 36.

38.

40.

1

(1 − x)2 1

8. − x−12

4 10. 7 x− 37 12. −3sin x

1 x− 12+1 − 32 1 1 14. 2 2 x or + 32 2 x 2 x 16.

6(3 x− 2)

18. 4 x11.5

5cos x

20.

2 x 15

( x+5) 2

2 1 22. −2 − 2 x

42. 6; − 2;20

24. −2 x− 53 +

1 44. −5;− 5; − 2000

26.

46. 2;0; − 10

28. − x−2

48. a)

7 2

3 − 34 84 x 15 8 −4 x + − x 5 3 4

x5 2

30. 1 − 3 x−2 − 8 x−3 32. −2 x−2 + 8 x−3 − 18 x −4 34. −12si x + 8 cosx

b) y = 0.41x + 16.067 c) Rate of change is 0.41.

36.

n3 5 cosx + sin x 2 8

38.

1 −4 sinx x−2 3+ 3

40. y = x

42. y = 3.25x − 3

d) y = 0.472416x+15.318, rate of change is 0.47241; y = 0.5685x+14.165, rate of change is 0.05685; y = 0.6328x+13,395, rate of

1 44. 12 x−1112

e) The slope of the tangent line at

48.

change is 0.6326.

46. 4 cosx

1 ( cos x − sin x) 2

(12,w(12))appears to be approaching 0.71.

50. y =

Exercise 2.5 2. 8 x7 3

4. 2 0x

6. 10 cosx

25 25 x− 4 4

52. f ′(x )− g′(x ) 54. The value of the slope of the tangent lines of the sine function correspond to whether the cosine function has positive or negative values. A positive slope of the tangent line means the cosine has positive values at the particular xvalue, and vice versa.

ANSWERS TO EVEN-NUMBERED EXERCISES 383

56. Left to the student.

58. The

tangent

line

isx = 0.

14. a) 0.1x− 0.9x2 b) The sensitivity is affected by the following rule: for every increase in the dosage of xcubic centimeters the blood pressure is changed by an amount equal to 0.1x−0.9x2.

60. horizontal at The tangentx = 1.9384, x=3.6613, and at x=5.6498. line is horizontal at

Section 2.6

16. a) 1.41W0.41 b) For every increase of Win body weight the home range of an animal increases by an amount equal to 1.41W0.41.

2. a ) 3

b) 0 c ) 3 miles per hour ; 0 miles per hours squared

18. a) 4000t d ) Uniform motion means that the velocity and acceleration are constants namely, the velocity b) 300,000 people is the slope of the linear distance function and the acceleration is zero. c) 40,000 people per year

2t− 1 2 b) 2

4. a)

d)

c) 1.5 feet per second; 2 feet per second squared.

6. a) 3+ sint

The population of the city will reach 300,000 people after 10 years. The population growth rate at 10 years will be 40,000 people per year, that is, the population will increase by 40,000 for the next year.

20. a)

0.61 h

b) cost

b) 244 miles

c) ≈3.866msec; ≈0.5msec2

c) 0.00305 miles per feet

d)

d) At a height of 40,000 feet the view is 244 miles. At a height of 40,000 feet, for every one foot increase in height corresponds to an increase in the view of 0.00305 miles.

3π +2nπ 2

8. a) 1.82 − 0.1192t + 0.002274t2 b)

8.15 pounds

c)

1.82 pounds per month

d)

22.72 pounds

e)

0.717 pounds per month

f)

1.21 pounds per month

g)

5.747

10. a) 2 feet per degree b) For every increase of one degree there will be an increase of 2 feet in the stopping distance of an object on glare ice.

12. a) 2πr b) For every increase of one centimeter of the radius the circular area of a healing wound

increases by 2πrsquared centimeters.

22. The derivative of a function at a point xcan be thought of as the slope of the tangent line at that point. It could also be thought of as the instantaneous rate of change of the function at that point.

24.

384

26.

ANSWERS

26. 16 x 3 − 3 x 2 40 − x +5

4 5 28. 4x2 + 3 x − 3 30.

32.

28. a)

34.

b) The acceleration function is the same as the product of the position function times −1.

That is, it is a reflection of the position function about the taxis.

(x2+ )21

cosx xcscx − − 2 2x3 2

40.

−150 x 3 − 60 x 2 − 54 x +18 2

(15x + 9)2

+ 42. −t − t csc t cot t csct− t cot2t + cott

b)

10.

(42x+12x )(9 x10− 7 x )

12.

tanx ⎜ x +2 +2⎟ ( x + 2 x)sec ⎜ 33 x 2 ⎟ ⎛

3

x secx tanx +

1

⎠

33 x2

(6.24+x10.2)(2.9 x 2 +4.3x− 2.1)

csc x(cot + csc22 xx)

1 1 + cosx 2sec2 t tant 24.

(x −1)2 2

−2 x x2 −1

(

)2

⎞

secx

16. 50t − 40

−2 x

c) Since the graphs of both functions are similar then the average rate of change for the functions will be the same (that is why the answers in part a) and part b) are equal).

2

⎝

22.

−2 x

38.

8. 1

20. −

(2t2 − 3t − 7)2

−1 sint+ 1

6. −3 x−4

18.

11t2 − 38t − 52

36.

8 53 x 4.

3

3

2

86. a)

2. 13 x12

14.

( x + 1)

44. - 84. Left to the student.

Section 2.7

5

4

88. a) 2 ta xsec2x b) c)

n

xsec2x

The 2 ta answers in parts a) and b) are the same.

The reason is that tan2x+1=sec2x and n therefore when we take the derivative we get the same result since the derivative of 1 is zero.

90. y =4; y=

x −1

92. y =−x+9;

y=6

ANSWERS TO EVEN-NUMBERED EXERCISES 385

+ π8 94 . y =⎛⎜ ⎝ 8 96 . a)

⎞ π2 x ⎟ − 32 ⎠ + −100t2 450

(2t + 9) 2

2

b) The population will reach approximately 12920 people after 8 years. c) ≈ −163people per year

3sint)− (3 cost )⎛⎜ ⎞⎟ ( t +1)(− ⎝2 t ⎠ 98. a) ( t +1)2 1

2 4 −4 x seems to be the correct derivative 112. y = 2 + 4x of y = . x2 + 1

(x 1)2

b) −1.465 in sec

100.

102.

( ) (t 3+ 1)2

3 t 2 t 3 − 1 csc x

2 xsin 2 x+ xcos2 x

( xsinx + cosx)2

(2 x +) 2 ) ( x sinx − x cosx 104.

32

( x +22x+3)2

106.

Exercise Set 2.8 2. −12 + 8 x 2tan xsec 2x 4.

6.

8.

x + 3 )9

2x

(

)2

3 3 x2 + 1 2 −3 10. ( x + 4 )3 12.

14.

16.

108.

20(2

18.

t2 + 2t 3 3t + 4 3 3t + 4

−3sin(3− t4) sec x tan x

2 1 − secx 1 2 x

+ 3 ( x − 3 )2

20. −x 2sin x +2 sin x + 2 csc 2 x

110.

2

22.

(2 x + 3 )32+ x +

24.

csc (1x)

26.

x2

cos t 2 t

3x + 1

2x + 3

386

ANSWERS

28.

30.

32.

34.

36.

3(5 x− 2)

1 66. − 2

+ 5 3x + 5

2 3x + 4

( cosx ) sec 2( sinx )

(

8x x − 1 3

68. ≈ −42.8513

+ 32) (x − 1 ) )5(x + 3 )3− 15x2(x (x3− 1)10 4

2

3

4

2 x −1

40. −csc +x

+ x−

1sec ( 1) x− x 2 )2

2sin xcos xcos x) (cos2

5

c)

2.199

d)

10.995 dD

represents the rate of change of the dw dosage with respect to the patient’s weight. For each additional kilogram of weight, the dosage is increase by about 11 mg.

e)

74. −0.1859degreeshr

2) sec 2( tant+( t+2)

76.

+1) +14sec2 x tan x 3(x5 + x(5+x4)21 2 (x5+ x + 1)3+7sec2x

78.

− 2 cos 2 xsec ( sin (2 x )) ⋅

t +2

cos t sint 48. − 2 + cos2t

52.

b)

2x + 3 +1 2x + 3

2

2 x2 sin x)(1 + xcosx + sin x) ( x + sinx)2+(−2x2− 4 ( x +sinx)

50.

72. a) ≈ 2.199 w

1

44.

46.

represents the rate at which the amount dt of investment is changing with respect to a quarterly compounded interest rate i.

33(x − 5 )433( 3x − 5 )2

xsecx2−1) (x

dA

b)

−19

− 5.45)4(3.21x − 5.87 ( 60.6−x121.75) 38. (2.36x

42.

10

70. a) 5000⎜⎛1 +i⎞4⎟ ⎝ ⎠

1 − cscxcot x 2 x + cscx −90

(2x3+2 )41

54. 56. − x− 58. 6x2

(x3 )+1

60. y = 0 = −735x 62. y + 1813 64. a)y= 2π r − 2π 2 6 x 5+ 40 x3+ 50 x

b) 6 x 5+ 40 x3+ 50 x c) The answers are the same.

80.

tan ( sin (2 x )) sin ( sec ( sin (2 x )))

2x 2 x2 +1

82. The derivative of a constant is 0.

84.

sin ( a + x )

86. One might use the following: Composition of

functions could be expressed as a process through which one function is evaluated using another function that falls in the original function’s domain.

88. (−1.475,9.488)

ANSWERS TO EVEN-NUMBERED EXERCISES 387

t 1) tan ( −2.6 1) t 36. 91.936 sec2( 2.6− 38. 12t2 + 2 40. 0.0036t − 0.068 42. 200,000 44. 2 ⎡ ⎢ ⎣ 90.

2x − 40

x( x +)21⎤⎥ ⎡⎣ x ( x+1)⎤⎦− ⎦ ⎡ x( x +1 ⎤ ) −3

⎢ ⎣

( x + 10)32 46.

Exercise Set 2.9

−8 x 2

+

3

48. a) − sin x b)

−cosx

−18

c)

sinx

( x −4)3 3 108 6. ( 3 x + 2)5

d)

cosx

e)

cosx

f)

sinx

g)

sinx

2. 0 4.

− 8.

1

( x −1) 10. 810(3x +2)8 12.

3

4

50.

8 sec2 ( 2x ) tan ) ( 2x

14. 26 − csc3 x −cscx

cot2 x

52.

16. 0 18. −

8 9 3 (2x+ 4) 5

20. 91.52 x1.2+ 2.88 x−0.8 + 1.5625 x2.25 22. −a2 sin )(at 24.

26.

15 x 12+3 12+1 x− 32 4 x− 4 4

2sec(2πttanπt ) ( )+ 4sec3( πt (πt)2 ) +4sec(πt)tan

28. 120x 30. 5040x 32.

k ( k−)(1k − 2 )(k − 3)( k−)4xk−5

34. −54csc4(3x − 1 ) −108csc2 (3 x − 1) cot (3 x2−1)

54.

4

(1 + x ) (1+ x) 2 2

360

(x + 2 ) 6

2

2 −2

1

⎥ ⎦

2 x

388

ANSWERS

56.

ANSWERS TO EVEN-NUMBERED EXERCISES 389

Chapter 3 Exercise Set 3.1

at ( 3, − 12).

2. Relative minimum

⎛ 3 , 25 ⎞ .

4. Relative maximum at ⎜− 4 8 ⎟ ⎝ ⎠

⎛ 1, 13 ⎞ ; relative 8. Relative maximum at ⎜− 2 ⎟ ⎝ ⎠ 2 113 ⎞ minimum at ⎛⎜ , ⎟. ⎝ 3 27 ⎠

10. Relative maximum at (0,0); relative min

at (2,−4).

12. This function has no relative extrema.

at (2, − 13).

6. Relative minimum

390

ANSWERS

14. Relative maximum at (30,23.55).

at ( −3, − 5).

20. Relative minimum 2

−10

−5

5

10

x −2

−4

−6

16.

Relative minima at)(−2,−13−and 13); (2, relative maximum) at (0,3.

22. Relative maximum at (0,5).

1 y0 5

−4

−2

2

4

−2

−1

24. Relative minimum.at (0,0)

Relative minimum at ( − 8, − 8 ) ; relative maximum at (8,8).

18.

10

y 5

−4

−2

2

−5

−10

4

26. There are no relative extrema.

ANSWERS TO EVEN-NUMBERED EXERCISES 391

at ( − 3,1).

28. Relative minimum

30. Relative maximum at (0,3).

,1); andrelative (2π 32. Relative maxima at (0,1) minimum (π,−1).

⎞ ⎛ 5π ⎞ 34. Relative maxima at⎛⎜ π 6 ,1,⎟ ⎜ ⎟ ,1and ⎝ ⎠ ⎝ 6⎠ ⎛ π ,−1⎟⎞, ⎜ 6 ⎟ tive minimum at ⎝2⎜ ⎝ ⎠ ⎛ 11 π 7π ⎜ ⎞ ⎛ ,−1.⎞ ,− 1 ⎟and ⎜ ⎝ ⎟ 6 6 ⎝ ⎠ ⎠

⎛9π ,1 ⎞ ; rela

2π 36. Relative maxima at⎛⎜ 3 , 3.826 ⎞⎟ and ⎠ ⎝ 4π 3 ⎛ , 2.457 ⎞⎟ . ; relative minimum at (2π, 2π) ⎠ ⎜ ⎝

38. Relative maximum at⎛⎜ 4π 3 , 0.1812 ⎞⎟ . ⎝ ⎠

392

ANSWERS

2π 40. Relative maximum at⎛⎜ 3 ,0.5774 ⎞⎟ ; relative ⎝ ⎠

⎛ π, 46. Relative maxima at ⎜4 ⎝

minimum at⎛⎜ 4π ,−0.5774.⎞⎟ ⎝ 3 ⎠

⎞ 2,( 1)and ⎟ π, − ⎠

⎛7π , 2 ⎞ ; relative minima ⎛at3π , 2 ⎞ − ⎜ 4 ⎟ ⎝ ⎛5π ⎠

and ⎜ ,− ⎝ 4

2 ⎞. ⎟ ⎠

⎜ 4 ⎝

⎟⎠

⎛5π 1⎞

⎛π 1⎞ ⎟ and ⎜ , ; 42. Relative maxima at ⎜ ⎝ 4, 2 ⎠ ⎝ 4 2 ⎟⎠ ,− 1 3π 1⎞ 7π ⎛ ⎞ ⎛ , − ⎟ and ⎜ retive la minima at ⎜ .

2 2 ⎟⎠ ⎝4 ⎠ ⎝ 4

48. – 92. Left to the student.

94. Answers vary.

96. Relative minimum at latitude of ~ 8.655 degrees east, longitude of ~ 10.786 degrees south.

ππ(2,2); ,0) and 44. Relative maxima at (0,2),( ⎛ 2π 1 ⎞ and ⎛ 4π , 1 ⎞ . relative minima at ⎜ 3 ,− 4 ⎟ ⎜ 3 −4⎟ ⎝ ⎠ ⎝ ⎠ 98.

100.

(1,5 )is a relative maximum; (3,1 )is a relative minimum; (4,5 is) a relative maximum; (5,2 is) a relative minimum. (

1,1)is a relative minimum;

(5,4 is) a relative maximum; (6,2 is) a relative minimum.

ANSWERS TO EVEN-NUMBERED EXERCISES 393

102. Relative maxima at (−6.262,3213.80,)

and (5.054,6674.12); (−0.559,1440.06, ) relative minima at ) (−3.683,−2288.03and 2.116, − 1083.08 . ( )

at 0,80); relative 6. Relative maximum (

minimum at (−)6,−28; inflection point at (−3,26).

8. Relative minima ( at −2,−27)and (1,0);

relative maxima ( 0,5;) inflection points at

(−1.22,−13.48 ) and (0.55,2.31).

Exercise Set 3.2 5 1 2.Relative minimum at⎛⎜ − 2 , − ⎞⎟ ; there are no 4⎠ ⎝ inflection points.

at −2,45); relative 4. Relative maximum (

10. Relative minimum at (1,0); no inflection points.

minimum at (2, ) −51; inflection point at (0,−3).

394

ANSWERS

at maximum at (0,3); inflection points 12. Relative minimum at (0,0); relative maxima 18. Relative

(−1,1)and (1,1); inflection points at

at ⎛⎜− 1 , 9 ⎞⎟ and⎛⎜ 1 , 9 ⎟⎞.

3 4⎠ ⎝ ⎝ 3 4⎠

⎛ 1 ,5⎞ ⎛ 1 , 5 ⎟⎞. ⎜− 3 9 ⎠⎟ and⎝⎜ 3 9 ⎠ ⎝

−2,0).point at ( 20. No relative extrema; inflection ⎛ 3 , −27 ⎞ ; inflection 14. Relative minimum at 16 ⎟ ⎜2 ⎝ ⎠ (1, points at(0,0) and − 1 ) .

22. Relative minima at (0,0)and (3,0); relative

⎛ 3 81 ⎞ maxima at ⎜ , ; inflection points at ⎝ 2 16⎟ ⎠

at −6,400); relative 16. Relative maximum (

minimum at (8,) −972; inflection point at (1,−286).

(0.63,2.23)and (2.37,2.23).

ANSWERS TO EVEN-NUMBERED EXERCISES 395

24. Relative minimum at(−1,−2;)relative

maximum) (1,2; inflection points at (−0.71,−1.25 ) , (0,0 ) and (0.71,1.25).

⎞ 32. Relative maximum at⎛⎜ π 3 ,1.1278;⎟ relative ⎝ ⎠

minimum at⎛⎜ 5π ,0.44297 ⎞⎟ ; inflection point at ⎝ 3 ⎠ ⎛ 2 1, ⎞ 26. Relative maximum at⎜ − 2 ⎟ ; relative ⎜ 2 ⎟ ⎝ ⎠ ⎛ 2 1⎞ minimum at⎜ ,− ; inflection point at ⎜ 2 2 ⎟⎟ ⎝ ⎠

(π,0.785).

(0,0).

34. Relative maximum at⎛⎜ 3π 4 ,1.414;⎞⎟ relative ⎝ ⎠

⎛7π minimum at ⎜ ,−1.414⎟⎞; inflection points at ⎝ 4 ⎠ ⎛ π , 0⎞ and ⎛ 5π ,0 ⎞ . ⎜4 ⎟ ⎜ 4 ⎟ 28. No relative extrema; inflection. point at (0,2) ⎝ ⎠ ⎝ ⎠

30. Relative maximum at⎛⎜ 5π 3 ,6.968⎞⎟ ; relative ⎝ ⎠

⎞ minimum at⎛⎜ π 3 , − 0.6849 ⎟ ; inflection point at ⎝ ⎠

(π,0.785).

396

ANSWERS

36. Relative maximum at⎛⎜ 5π ,2;⎞⎟relative ⎝ 6 ⎠

⎛11π

, −⎞2⎟; inflection points at minimum at ⎜ 6 ⎠ ⎝ ⎛ π , 1 ⎞and ⎛ 7π ,1.⎞ ⎜6 − ⎟ ⎜ 6 ⎟ ⎝ ⎝ ⎠ ⎠

⎞ 42. Relative maximum at⎛⎜ π 2 ,2;⎟relative ⎝ ⎠

minimum at⎛⎜ 3π,−2 ⎞⎟ ; inflection point at ⎝ 2 ⎠

(π,0).

11π , 0.577 ⎞⎟ ; relative 38. Relative maximum at⎛⎜ 6 ⎝ ⎠ ⎛ 7π , −0.577 ⎞ minimum at⎜ ⎟ ; inflection points ⎠ ⎝ 6 ⎛ 3 at ⎛⎜ π,0 ⎞⎟ and ⎜⎝ π ,0 ⎞⎟ .

2 ⎝ 2 ⎠ ⎠ 44.

(2,2 )

46.

(0, 10 ), (2,− 6)

n 48. π + π 4 2 50.

⎞ ⎛ 3π ⎞ 40. Relative maximum at⎛⎜ π 2 ,1 ⎟ and ⎜ 2 ,1 ⎟ ; ⎝ ⎠ ⎝ ⎠ 0,0 ,0, (2π,0);

relative minimum),at(π) (

⎛π inflection points at ⎜ ,0.5625⎟⎞, ⎝3 ⎠ ⎛2π , 0.5625⎞ ⎛4π ⎜ 3 ⎝

⎠⎟, ⎜ ⎝ 3 ⎛5π , 0.5625⎞ ⎟. ⎜ 3 ⎝ ⎠

,0.5625⎟⎞and ⎠

π n 2+ π

52. No inflection points.

54.– 104. Left to the student.

106. The minimum occurs about 3.4 hours after midnight and the minimum value is 0.442.

108. a) Relative minimum after 6.5 years, value is 7977394.5.

b) Inflection point at 3.039 years, value is 857232.4

ANSWERS TO EVEN-NUMBERED EXERCISES 397

c)

120. No relative extrema.

110. g is the graph of the derivative of h since the maximums and minimums occur where g is zero.

122. Relative maximum at (−1,1.414); relativ

112. a)

Relative maximum at (1,5)and (5,5).

b)

Relative minimum at (3,1).

c)

Inflection points at (2,3)and (4,3).

d)

Increasing on (0,1)and (3,5).

e)

Decreasing on (1,3) and (5,6).

f)

Concave up on (2,4).

g)

Concave down on (0,2)and (4,6).

114. Left to the student.

1 1 116. a) y =− 2 cos (2 x )+ 2 b)

Left to the student.

c) sin(2x ) =2sinxcosx

minimum at (−1,−1.414).

Exercise Set 3.3 5 2. −2 x 4.

3 4

6. 7

6 8.

118. Relative maximum at (0,0); relative minimum 5

at (1,−)2.

10. −4 12. 0 14. −∞ 16. ∞

398

ANSWERS

18.

20.

26.

28.

22.

30.

24.

32.

ANSWERS TO EVEN-NUMBERED EXERCISES 399

34.

36.

38.

40.

42.

44.

46.

48. a) 100 cc; 50 cc; 20 cc; 2 cc; ≈0.99 cc b) 0 c) The maximum value of the injection is 100 cc at t=0. d)

400

ANSWERS

100 0has no solution, so e) The equation t 2+1 = according t the given function the medication never completely leaves the bloodstream.

78.

50. a)500; 550; 562.5; 543.1; 524.04 b) 500 c) 562.5 d)

80.

52. Vertical asymptotes occur at values of the variable for which a function is undefined. Thus, they cannot be a part of the graph.

54. −∞ 56. −3 58.

82.

∞ ∞

60.

62. Undefined 64. Undefined 66. −∞

84.

68. ∞ 70. 1 72. ∞ 74. ∞ 76.

0

a) x = −1.732 and x = 1.732 b) y = 0.75 c) Vertical asymptote at=x 2; oblique asymptote at y= 1 x+1.

2

ANSWERS TO EVEN-NUMBERED EXERCISES 401

Exercise Set 3.4

30. Absolute maximum at ( 4,5.6569;) absolute

minimum ( at) 0,0 (and 6,0).

32. Absolute maximum(at 2π,5.2832; ) absolute

2. Absolute maximum at (2,10); absolute

minimum at (0,4).

minimum at( 0,−1.)

4. Absolute maximum at (0,5); absolute34. Absolute maximum at⎛π1⎞ ⎜ , ⎟; absolute ⎛ 2 113 ⎞ ⎟. minimum at ⎜ , ⎝3 27 ⎠

⎛3π

⎝2 2⎠

minimum at ⎜ ,− 1⎞ 4⎟ . ⎝ 2 ⎠

6. Absolute maximum at (1,6); absolute 36. Absolute maximum at ( π ,1); absolute minimum at (−1,2).

8. Absolute maximum at (1,4); absolute minimum at (4,−23).

10. Absolute maximum at ( 10,2000;) absolute

minimum at (−10,) −2000.

12. Absolute maximum at ( 0,10;) absolute

⎛π ⎜ π ,0⎟⎞. minimum at ⎜ ,0⎞⎟ and ⎛3 ⎝2 ⎠ ⎠ ⎝ 2

⎛ π ,2.2556;⎞absolute 38. Absolute maximum at ⎜6 ⎟ ⎝ ⎠

⎛5π minimum at ⎜ ,0.8859⎟⎞. ⎝ 6 ⎠

minimum at (4, ) −22.

⎛3π 40. Absolute maximum at ⎜ ,1.4142 ⎞ ⎝4 ⎟and ⎠ ⎛3 27 ⎞ ⎛5π ⎞ ,1.4142⎟; absolute minimum at ⎟; absolute 14. Absolute maximum at ⎜ , ⎝4 256⎠ ⎜ 4 ⎠ ⎝ minimum at (− 1, − 2. ⎛7π ,−1.4142⎟⎞. ) ⎠ 16. Absolute maximum at −3,12 and (3,12); ⎜⎝ 4

( ) absolute minimum at ⎛ ⎜4,−6 4 ⎞ ; no absolute 42. Absolute maximum at ( ) −2,−13and (2,−13).

18. Absolute maximum at ( 0,1;) absolute

minimum at (− ) 8,−(3 and ) 8,−3.

20. Absolute maximum at ( −2,−4;) absolute

minimum at (−8, ) −8.5.

22. Absolute maximum at ( 1,2;) absolute

minimum at (− ) 1,−2.

24. Absolute maximum at ( 64,4;) absolute

minimum at (8,2).

26. Absolute maximum at ( −1,1; ) absolute

⎛ ⎜−3,−1 ⎞ minimum at 3⎟. ⎝ ⎠ 28. Absolute maximum at (6, 20.785 ) ; absolute

minimum0,0at) and ( (8,0 ).

minimum.

⎝

3⎟ ⎠

44. Absolute maximum at (70,19); absolut minimum.

46. Absolute minimum at (60,120); no abso maximum.

48. Absolute minimum at (5,75); no absolu maximum.

50. The function has no absolute extrema.

52. The function has no absolute extrema.

54. Absolute minimum at (0,0); no absolute maximum.

56. The function has no absolute extrema.

58. Absolute minimum at (±1,0); no absolu maximum.

402

ANSWERS

⎛π⎞

60. Absolute maximum at ⎜,2 2⎟⎝⎠ minimum.

; no absolute

82. Absolute minima at x = ±1; no absolute maximum.

⎛ 5π ,1.1287 ⎞ ; absolute 62. Absolute maximum at ⎜ 6 ⎟ ⎝ ⎠ minimum at⎛⎜ π, 0.4433 ⎞ ⎝ 6 ⎟. ⎠

⎛ 2π ⎞ 64. Absolute minimum at ⎜ 3 ,1.7321 ⎟ ; no ⎠ ⎝ absolute maxima. ⎛ π , −1.7321 ⎞ ; no 66. Absolute maximum at ⎜ ⎟ ⎝6 ⎠ absolute minimum. ⎛ 0, 1 ⎞ ; no absolute 68. Absolute minimum at ⎜ 2⎟ ⎝ ⎠ maximum.

70. The maximum blood pressure of 0.00021 occurs at a dosage of 0.1111.

Exercise Set 3.5 2. The two numbers are x=35and y=35.

4. There cannot be a minimum product since there is only one critical value for the function and the second derivative is positive for values of x.

6. The two numbers arex=−3 and y=3.

72. The maximum temperature value is 102.2 and it occurs after 6 days.

8. 4 when x = 2 andy = 2

74. a) D(h)= 0.139443 h− 0.043705 h 0.725

10 9 when x = 1 andy = 2.

b) The function has an absolute maximum of 0.2497 and an absolute minimum of −0.1010. b 76. x =a − 4 78. Left to the student (answers vary).

80. Absolute maximum atx = 0and absolute minimum at x=3.

. 25 when x = 5 and y = 5.

12 1 2 1 14. 3 when x= 3 and y = 2 . .

16. 3 when x = 1 and y = 4.

18. The maximum area that can be enclosed is 480 square yards.

20. The rectangle is 8.5 feet by 8.5 feet with a maximum area of 72.25 squared feet.

22. The box has a length and width of 20 inches and a height of 5 inches with a maximum volume of 2000 squared inches.

24. The square based box has dimensions 4×4×2 inches and a minimum surface area of 48 squared inches.

26. The maximum area possible is 0.6495 squared units of length.

28. The maximum profit occurs when the price of the room is $48.

30. The maximum revenue occurs when the price of the ticket is $6.50.

ANSWERS TO EVEN-NUMBERED EXERCISES 403

32. The area of the 7×13 sign does not have as large an area as the 10×10sign. The customer could refuse this offer if they are concerned with maximizing the size of the sign. (Answers may vary.)

34. The outside dimensions should be 10.7432×6.9821inches. The actual page measured 10×7.5inches.

36. The boy’s growing rate is slowest after 26.2093 months.

34. Initial guess x = −3leads to solution x=−2.95728;

Initial guess x= −0.25leads to solution x=−0.15237; Initial guess x= 1leads to solution x=1.10964.

36. Initial guess x= −1.5leads to solution x=−1.4613;

Initial guess x= 0.6leads to solution x=0.57437; Initial guess x= 1.25leads to solution

38. The fence should be 6×8yards.

40. Left to the student.

42.

x=1.24907.

1

2 44. The power line should be constructed so that it is 1.25 miles up shore from point B. 46.

pb2± pb 2b2 − a2

(a2 − b2 )

48. Q reaches a minimum when x = −1 andy =−1 .

Exercise Set 3.6 2. L ( x) = 12 x − 16 4. L ( x) = −16 x + 22

1

6.

L(x)=

8.

L (x) = x

20

10. L ( x ) = x 12. 4.9 14. 10.17

x+5

38. The ages for which there are 3 million hearing impaired Americans are x=33.2years and 74.2years.

40. The age at which the median weight of boys is 25 pounds is t=15.77pounds.

42.

The average temperature in NY is 60 on middle of May and middle of October.

44

x=0.94019

.

After 36 iterations of Newton’s method, we reach r=1.19989.

46 48. a) 4.62cms . b) 0.08cms

50. a) Newton’s method does not converge to a solution. b) Newton’s method does not converge to a

solution for f(x)=x13since the derivative x1=0is not defined.

52. a) b)

16. 0.97 18. 12.0286 20. −0.05 22.

0.02 24. 2.61804 26. 0.25099 28. 3.43238 30. 0.51097 32. 0.76317

c) x = 0.14065, x = 0.21719 and x = 0.47752 d) For x = 0.14065, f = 792.43 Hz For x = 0.21719 f = 1223.67 Hz

For x = ,

0.47752 ,

f = 2690.39 Hz

404

ANSWERS

54. For x = 0.15381, f = 866.58 Hz For x = 0.27909, f = 1572.42 H

For x = 0.52360, f = 2950.01 z

H z

Exercise Set 3.7

3 tan xsec 2 x8 ; =− sin y 3

d

d

20.

22.

24.

26.

28.

dR dt

40. 2.083 ft sec 42. a) 1.06667 ft sec b)

0.1333 rad sec

44. 3.75 rad hour 46. −

x6 2 + 6 y2 (2y − x)3 4

2 xy3 −2 x 48. y5

52.

5 x 4 y 3− 3 x2 y2 2 x 3 y −3 x5 y2

y 3 y x d 2x d x = 3 2 y y x+ 1 d d − y d xy 6 2 5 4 + 6y2 = 54 d dx − 12xy x xy x d dy = 2−y y dx 2 d = x− dy tan2y x dx 2y2 − 2 xtan y sec2y =−

(

mmday 9 4

50. Left to the student (answers vary).

14. y = − x y d x 5x4 16. d = = 3y2

y 18. d x

3 dH ; =− dt 24

38. −0.0408 m2 month

sin(π 3 ) ; − 1 = 3 s i n( 4π 3 ) 3 y = x+y 12. d x 3y−x d 10.

=1 ,

b) 0.05 mm3min

dy = 12 x 2 +5 ; − 17 6. dx 4y2 + 3 29 dy 8.

34. −50

36. a) 1000 R

d 2− y 4 2.

2 y; 5 y = x− d dy x 3 4. x = ; dx y 2

dx

dH 32. G = 0, dt = 0 ; dH G = 3 , dt = − G

)

)

( y cosxy dy 30. =− cos sin y + x ( xy) dx

54.

ANSWERS TO EVEN-NUMBERED EXERCISES 405

Chapter 4

22. x 4 ex(x+ 5 )

Exercise Set 4.1

24. x 2 ex 26.

2.

28.

30.

(cosx)ex− (sinx ) ex ex(x− 5)

x

6

(−2x+8)e− x +8 x 2

32. xex2 2 34.

36.

4.

e x−4 2 x−4 ex 2 ex+1

(e 2 x )

38. 2e2x cos

40. −sinxecosx

(

)

42. ex secx+ tanx+ secx tanx + sec2x

(

)

44. x 3 x − ex 46. 3e−3x 48. me− mx 6.

(

50. 8 xex ex − 2 2

2

)

3

e(3 t + t )(3e +1)

1 52. 3 54.

− 23

3t

ex

(1−ex) 2

56. −e−x − x− 2xe 1 8. 1050; 1103; 1276 10. 50950000; 51918050; 54933962 12. 2 e2 x 14. −12−e3x 16. e − x 18. −4ex 20. e4x

58.

x2− x + − e−x (−x3 1) (1+x2)2

+ 60. exe x

406

ANSWERS

c)

62.

No extrema; increasing over the entire real line; no inflection points; concave up over the entire real line.

d) −42e−0.7 t e) The derivative represents the rate of change of the amount of knowledge retained by elephants with respect to time.

80. a)

64.

r(T)= 0.124(e0.129T−1.2255−e0.144T−1

b) r′(T)=

0.017856(e0.144T−1.848)

1 x +1) x⎛ x2 + 5x−1 ⎞ 82. 3 e( ⎜ ⎟ 3 x ⎝ ⎠ 84. −6 x−3 +

5 2 + e2 x 2 x

( x)

86. excos e e

sin(ex )

No extrema; decreasing over the entire real line; no inflection points; concave up over the entire 88. 0 real line.

90. 4; 2.86797; 2.732; 2.71964; 2.71855

92. The minimum occurs at (−1,e).

66.

94. a) 0 b) 0.56978 c) 0.39967

No extrema; increasing over the entire real line; no inflection points; concave down over the entire real line.

68. – 72. Left to the student.

96. a) The rate of change in the annual death rate per 1000 population per year is given by 64.608e−0.072t .

2

(1+29.44e−0.072t)

b) Between 2007 and 2008 the annual death rate will decline by 0.00192 deaths per 1000 population.

74. y = − 6 x +2

98. 0.321570

76. Left to the student.

100. The probability is declining by 0.24705 per week.

78. a) 100%; 69.795%; 54.796%; 40.9%; 40.055% b) 40%

ANSWERS TO EVEN-NUMBERED EXERCISES 407

102. 3xonly have positive values in its range while x3has all real numbers as its range. When x<0, 3xincreases at a slower rate than x3but when x>0, 3xincreases at a higher rate than that of x3.

104.

106. Relative maximum at (0,1).

Exercise Set 4.2 2. 3481 = 4. 271 3=3 6. a k = J 8. w b− = V 10. lnp= t 12. log101000 = 3 14. lo 10 0.01 = −2 16.

QT=n g 18. −0.51 lo 3 20. g2 22. 4.317 24. 0.2231 26. 2.3863 28. 4.6051 30. 9.392662 32. −7.047017 34. −4.509860 36. ≈ 6.9

108.

38. ≈ 4.5 40. ≈ 4.6 42. ≈ 9.9

4 44. − x

110.

46.

x 4(1+5 lnx )3 −x

48.

1 − 5 lnx x6

50.

1

x sin3 x 52. cos3 x 54.

1

xln (3)x

408

ANSWERS

56.

x2− 5

90.

x ( x+25 )

y = 2.997 x−0.4998

2 ln

1 58. e2 x ⎛⎜ + ⎝x

x ⎞⎟ ⎠

60. ex(sin ( 2) +x2 cos (2))x 62.

ex ex − 2

64.

3ln ( x)

66.

n(ln x) −

68.

3(2t + 1)

2

92. y = 11.326x 0.40533

x

n 1

70.

x

t2 + t

1

xln (3)xln ( ln (3 )) x

72.

−2 1 −t 2

74.

x 5 + x2

94. y = 1920497 x −0.993

76. x 4ln x 78.

−x 2ln x + lnx + 2 x2

( x +2)21 or

80.

1 x −x x

x x − x2

2cot(slnx)csc(slnx) 82. − x 84. − sinxln (cot x )− cscx 86. −3 csc (3) x

42 x 88. −4 x e e −1

⎛ A , A lnA−AlnB−A. ⎞ 96. Maximum value is ⎜B ⎟ ⎝ ⎠ 98. a) 78% b)

≈ 54%

c)

≈ 30%

d)

≈ 38.5%

15 e) S′ ( t ) = − t+1 f) S(t)has a maximum value of 78% when t = 0.

g) limS(t)=−∞. This limit indicates that x→∞

eventually everything will be forgotten.

100. a) ≈ 26 words per minute; ≈91 words per minute.

b) W ′ (t=30 ) e−0.3t

ANSWERS TO EVEN-NUMBERED EXERCISES 409

116. The graph of y=ln(x)can be obtained by d) limW(t)=100. This indicates that the taking a reflection around the line y=xof the

c) ≈10 weeks x→∞

typist’s speed will eventually level off at 100 words per minute.

1− (ln x)

2

102.

(ln x)2) x(1+

2

graph of y=ex.

118

πeis larger

.

0

120 .

⎛ P⎞ ln ⎟ ⎜ 0 104. t = − ⎝ P ⎠ k

122 .

106. Left to the student.

108. a) b)

110.

r ( T )=0.153( e − e 0.141T −1.3395

0.152T −1.8123

)

dr 0.141T −1.3395 0.153T −1.823 = 0.021573e − 0.023409e dT dr = dT

124.

b(T −TL ) ceb(T −TL) (L ) A(be − ) U

−cT −T

112. a) – b)

126. The function has a minimum , − 1 e) . at (1e 128. y = 16.485x0.136 130. y = 337219504.1x−1.237 c) The line in the graph in part b) is 2−x1since

log (102 x− 1)=2 x− 1.

114. a)

132. x 4063.76 and x 20000 = = 134. x 40.88 and x 1000 = = 136. x 143.32 and x 10993.19 = = 138. x 143.32 and x 10993.19 = =

Exercise Set 4.3 2. Rt( ) = R0 ekt t 4. H = 10e0.5

b) m = 0.1073 ; b = 2.19589 N = 0.10730t +2.19589 c) 10

6. a) P = 1000e 0.000641t b)

1019

c)

1039

d)

2517

e)

1081

410

ANSWERS

b)

8. a) 0.3466 b)

1414

c)

2049299

d)

3.1697

10. 2712 12. a) P=59.8e0.0021t b) 61.455 million c) 330.07 years

14. a) P ( t) = P0 e0.08t b)

≈ $21,665.74; ≈$23,470.22

c)

≈ 8.7 years

24944.7753 c) A′(x)= − x is close to −2000 when d) The value of A′(x) x is close to 14.

32. a) 400; ≈ 520; ≈ 1214 ; ≈ 2059 ; ≈ 2396 ; ≈ 2478

16. ≈ 5.8% 18. ≈ 1%

b) P′

20. ≈ 4% 22. ≈ 3.5%

c)

(t)

=

4200 e−0.32 t

(1+ 5.25e

−0.32 t

)

2

24. ≈ 0.21%

26. a) y = (45811.9)1.0587t

y

b)

= (4511.9) e 0.057 t

c) When t=50,y=795268

d) t = 54; corresponds to the year 2014. e) 12.16

34. a) 0.74164;9.2046;38.884 28. a) y = (85.886796)1.061898 t ; ln ( b )=0.0601; b) 74.923 This number represents the growth rate. −0.092 t c ) C′ (t)= 1 238.964 e b) y = 85.886796e0.0601t e−0.092 t c) 116 d) Inflection point at t = 70.731. d) 40.87

(1 + 335 )2

e) 11.53

30. a)

A ( x )=67367.603 − 24944.7753 ln ( x )

36. a) ≈ 0.244; ≈ 0.429 ; ≈ 0.753 ; ≈ 0.954 ; ≈ 989; ≈ 996 b) p ′ (t ) = 0.28e−0.28t

ANSWERS TO EVEN-NUMBERED EXERCISES 411

c)

6. ≈ 23.1%per min 8. ≈ 7.2 days 10. ≈ 0.015%per year

12. ≈ 40.8 g 14. The tusk is about 4239 years old.

16. The sample was on the shelf for about 25 days.

18. a) 42.2% 47.3% 38. a) Nt ( ) =

b) 24.94 days

29.47232081 − t 1 + 79.56767122e 0.809743969

20. a)

b) We round up since the function deals with people. The limiting value appears to be about 30 students. c)

b) 0.5749 N0; After 48 days 57.49% of the

iodine 125 will remain 22

2.25×109 years

.

There will be 1 g left after about 11 hr.

24

a) The animal loses 0.9% of its weight each day. b) After 30 days, 78.8% of the initial weight remains.

.

26

.

28. 0.37I 0 or 37%I

d) N′(t)=

30. a) a =

1898.885181 e −0.809743969 t

b) T

t) c) ≈ (1+79.56767122e−0.809743969

0

60

(t) = 60e−0.04t+40 80.2

d) ≈ e) limN′(t)=0; This indicates that eventually 102.4 min t→∞

the population does not change. That is, it reaches and remains at a limiting value.

34. a) N=150e−0.005832t b) 137.44

40. 8.33% 42. 6.4% 44. t =

ln (4) k

46. After the second year.

⎛ y2 ⎞ ln⎜ ⎟ y1 48. k = (t2−⎝ t1 ⎠

)

50. An exponential function grows indefinitely while a logistic function has a limiting growth value.

Exercise Set 4.4 2. (f) 4. (e)

32. The coroner arrived at 2 am, so the murder was committed at about 8 pm.

c) The population of Russia will be 100 million in the year 2065. ≈14.0lbin2 36. a) ≈5.4lbin2 b) c) d)

≈46,052ft The rate of change of pressure is −0.000735e−0.00005a lbin2per unit of altitude.

38. Thus, 62% of the carbon-14 remains, so 38% has been lost.

40. a) N=72r−0.005889t

b) 125.71 c) The population of Russia will be 100 million in the year 2198.

412

ANSWERS

42. Answers vary.

Exercise Set 4.5 2. ≈ e2.0794

e) The intensity is changing at a rate of 0.1⋅ I 0⋅ln10⋅100.1L. 40. a)

6. eRln 5 8. e ln10

16.

b) 0.001t +10 − 7

e) The pH is changing most rapidly at = 0t which corresponds to a pH of 7.

44. 4x 3 (ln2 ) 2

5 ⋅1 x ln10 ⋅ 1 x 1 22. ln10 20.

28.

d) The pH is −log(10−7)=7.

(

1 ⋅1 18. ln5 x

26.

0.001 c) −1 ⋅ ln10 0.001t +10−7

(5.4 x (xln(5.4)+1) ) (5 xx 3( xln (5)+4)

46.

1 + log6 x ln 6

24.

cot ( x ) −xcsc 2( x ) ln ( x ) xln 5

1 ln 2

2x + lnx 32. x ln 2

′ ( t )=$5200( ln 0.80)(0.80) t

at a rate of $5200(ln0.80)(0.80)tper year.

6.9 36. 38. a)

10 10⋅I0

b)

10 ⋅I0 c) The intensity in (a) is 109times that in (b). d)

dI 0.1⋅ 0I ⋅ ln10 ⋅100.1L dL =

x

(ln x +1) 3)( 2

dy f′ x ⎤ g x ⎡ 50. dx = f (x) (⎢ g(x) ) ⋅ f (x ) + g′ ( x) lnf( x)⎥ ( ) ⎥⎦ ⎢ ⎣ logaxwe have 52. If we first write loga x as a

⋅tanx

b) The value of the office machine is changing

2

x4

( x) 48. dy dx = f ′(x)(ln a ) af

n 1 0 ⎛1 −ln1 ⎞ 30. llnx⎜ x⎠⎟ ⎝

34. a) V

10 ⋅1 ln10 I

42. a)10 −7 x

12. Rln 100)100 x 14.

dI

=

b) The loudness is changing at a rate of 10 ⋅1. ln10 I

4. ≈ e

0.941

10. kln 7) 7

dL

alogax = x. Then we can take the natural logarithm on both sides of this expression, use 1 ⋅lnx. This allows P3, and write log ax = lna us to find

d dx

loga xusing the rules for

differentiating f(x) . = c⋅lnx

ANSWERS TO EVEN-NUMBERED EXERCISES 413

Chapter 5

3

42. f ( x) = x + x + 8 3 cos(8x

Exercise Set 5.1

44.

8

2. x + C 8

46.

48.

4. 4 x + C

50.

6. 3 x 4 3+ C 4 8.

x3 3

−

x2

52.

+ 2x + C

54.

2

56.

10. t3 − 2t 2 + 7t + C C 12. 35 e5 x +

58.

f(x) −= 3 8

f( x) =−3e

s (t ) =t2 + 10

v(t) = 3t2 +30

s(t)= −t3+ 7 t2+10t+20 2

− A( t) = 43.4t−1 3.7

b) 2.5

b) 516

18. − 1 x−2 + C or − 1 + C 2 2x2 3 20. 5 x 5 3+ C

c) 1111 d) 595

64. f (t ) = 2 t 3 2+ 2t12− 28 3 3

22. 100 x1 5 + C 24. −28e−0.25 x + C x5 26. 5 +41 x1 2 − 4 x3 5 + C cos ( 2θ3

2π

(t)= 32.307e0.1049 t − 32.307

62. a) I

16. 500 lnx + C

)+C

3 sin(2

+9

455036

5

30.

−x

≈ 5.1 sec 60. aThe ) car travels almost 77 feet during that time.

14. t − 5 t115+ C 5 11

28. 1 8

11

)+ 8

66. 25 t3 + 20t2+16t + C 3 68. 2 t 3 2− 2 t 5 2+ C 3 5 70.

x2 2 −6lnx 27+ x−2+C

72.

ln(x) +C ln(10)

8

πx)+ π cos( πx)+ C

32. −5 2 cot(2x )+ C C 2sec(2 34. x) + 36. − cot ⎛⎜ x ⎞ C ⎝ 2⎟+ ⎠ 7 101 38. 4 e7 w − 5 w35+101 w +C 7 32 40. f ( x) = x − 5x + 21 2 2

74. 2x3 3+ x2−5x+C 2 76.

x2

−x+C 9 78. − c o s x + C

80.

tan(3x) 3

−x+C

414

ANSWERS

82. Answers could vary. The integral is not unique since the student forgot to mention the constant of integration.

Exercise Set 5.2

4. 9 6. 4 4 3 or 11 8.

3 10. e2−1≈6.3891

2. a) 35 b)

12. 2ln4≈2.773

40.625

⎞ 14. −cos ⎛⎜ π 4 ⎟+ 3 ≈ 0.8787 4. The shaded region represents an antiderivative. ⎝ ⎠ It also represents distance, the antiderivative of velocity.

16. The area above the x-axis is greater than the area below the x-axis.

6. The shaded region represents an antiderivative.

It also represents the total number of divorces 18. − 2 e3b+23 or − 2 in time t. The shaded region represents an 3 3 thex-axis.

8. antiderivative.

(e3b−1). The area is below

It also represents total sales.

10. The shaded region represents

an

antiderivative.

It alsoinrepresents the number of pages 12. 14. typed t hours. 0.75 0.58333 1.11072 16. 18. a) The integral is positive because there is more area above the x-axis than below. b) The integral is negative because there is more area below the x-axis than above.

20. ≈ 243 22. −22000 24. 120.75 26. 1 28. 9 30. 1 32. 255 4 34. – 42. Left to the student.

44. 0 46. 4 48. 1

20. – 36. Left to the student.

38.

a3

6 40. −609 42. e2 3 − 44.

2 2

46. 4 3− 4 4 13

48. 15 50. 4 1 2 52. e3 − e −2 or e3 − 1 2 e 54. 153 4 56. 3888 25 58. 81 4 60. 10

Exercise Set 5.3 2. 10

3 62. 30

ANSWERS TO EVEN-NUMBERED EXERCISES 415

64. 16 3

1 12 14. 2

12.

66. 40

16.

5 6 8 3 70. 1 e

68.

72.

18. 1π− 4 20. 22 3

−1

22. 3 2

74. 16 3

24. 625 = 6212 10

76. 21 78.

26. 5

2n+1 −1 n +1

28. 608 15

80. 90 82.

84.

86.

88.

10

(e3+1)

−3600

30. 2276 15 32. 6

3157.72 N⋅mm

34. 4.43

54

36. 2; Subject B memorizes 2 more words then subject A during the first 10 minutes.

90. a) 4 πb3 3

38. 3

b) The volume could be thought of as a stacking of surface areas.

92.

2−1

1

∫ln xdx ≠ x

Exercise 5.4

40. The area bounded in the graph is 4.

42. The area bounded in the graph is 5.886.

44. The area bounded in the graph is 0.238.

Exercise Set 5.5

ln(1 C +x3)+

2. 3 10

2.

4. 9 2

4. 1 e3x + C 3 6. 3ex 3 + C

6. 125 6 8. 13 9

10. 2

8. 1 e x5 + C 5 10. − 1 e −t2 + C 2 12.

(ln 5x )2 2

+C

416

ANSWERS

14. 12 sin(2x + 3)+ C

56. 6561 16

16. tan(2x + 3) + C

1 4 58. 1 e 4 − 1 e1 or +e −e 2 2 2

18. cot (2 − x − x + K )

60. ln ⎛⎜ 92 ⎞⎟ ⎝ ⎠

20. −ln( 1−x +C ) 22.

t (2−1)6 +C

62. −e −2b +1, or 1− 1 2b e

12

64. −e − kb +1, or 1− 1kb e

24. 1 sinx3+C 3

(

)

26. − 13 csc x3 + 3x2 + 6 x − 7 + C 28.

30.

ln(3 +et)+ C +C

(lnx 2 ) + C + b) 2 + 1 34. 3a( ax C 32. 1 ln 2

36. P0 k ekt+ C

6+C

1

24(2 −x ) 4

)

65 42. 5 1 + 6 x2 +C

44. −coset + C 46. − 1 3ex3 48. − 13 cos3x + C 50.

52.

1 e

cost

+C

tan(ln x)+ C

54. e − 1

b) ≈ 17, 970, 604

76. 128 3 78. x +

38. 1b sin( bx + c ) + D

6(

74. a) ≈ 43,526,828

3

2

40.

68. 1 3 70. − 21 512 72. 5

(lnx)3 3

66. 39

1

+C

( x +3)

80. t − ln t − 4 + C 82.

ln(1 +e−x) +C

84. − 1 ln1+ ae−mx + C am 86. − ln csc x + coyx + C

Exercise 5.6 2. xe2x − 12 e2x+ C 4. x sin

x−

cos x + C

6. 1 xe3x − 1 e3 x 3 9 8. −xe−x − e − x + C 4

4

10. x ln x − x + C 4 16

ANSWERS TO EVEN-NUMBERED EXERCISES 417 3 2x3 2x3 48. x (lnx)2 − +C ln x + 9 27 3

3

12. x3 ln x − x + C 3 14.

− x+C ( x+1)ln(x+1)

50.

2 ⎛x2 ⎞ x− −x+ C 16. ⎜ 4 ⎝ 2 +x⎟lnx ⎠

18.

(x−2)sinx+cosx+C

xn+1 (ln x +1ln x 2xn=1 )2 2xn − +C − n +1 (n + 1)3 ( n1)2 +

(

+ 8) )(3x

52. 2 27x3+ 8 3 x − 2 7 −

20.

22.

2 x(x 4)32 4 + − 15 (x+4)52 +C 3

x3ln(5x)x3 −

3

9

4 91( 8 )136 81x2+83) (3x + − 1296 x (3x + 8)196 1729

2592 − 43225 (3x + 8 )256 +C

54. Left to the student.

+C

56. Left to the student.

24. x (lnx)2 − 2 xln x + 2 x +K

58. Left to the student.

x−4 x−4 26. − 4 ln x − +C 16 5 1 16 4 5− x3sin(4x 28. x5sin (4x ) + cos(4x ) )− 16

15 2 15 15 x cos(4x ) + xsin (4x)+ 512 cos4x ( ) 128 64 30. −x cot x + ln sin x + C 32.

76

Exercise 5.7 2.

1 4x e (4x − 1 ) + C 16

4.

ln( x + x2−9)+C 1 2

⎛ ⎞ 6. − ln 2 + + x4x 2 ⎟+ C ⎜ ⎟ ⎜ ⎠ ⎝ 8. 1 + ln(1−x) +C 1−x

15 4ln2 − 16

34. 6 l n 6 − 5

9ln(

36. 1

C +⎦ ) ⎤⎥

40. 251869

2 10. 1 ⎡ x x2 + 9 + x+ x + 9 2 ⎢⎣ ln 12. x ⎛⎜ 45 x⎞⎟−x+ C ⎝ ⎠

⎛t 1 ⎞ 1 42. a) − e−kT ⎜ + 2 ⎟+ 2 ⎝k k ⎠ k

3 3 3 − 2x 2x 14. − 1 x3e− − x2e− − xe−2 x − 8 e 2 x + C 5 4 2 4

38. 10π − 5 3 3 2

b) ≈

44. a)

mg

14.850

3

(x + 2)− 2 ( x+2) + C 73

43

3

b) 37 x ( x + 24 3) − 9 ( x + 2) 7 3C+ 4 28 2 ⎛x ⎞ c) 3 ( x + 2 )43 ⎜ − ⎟+ C ⎝ 7 14 ⎠

46. n 1+ 1xn+1 ln x −

16. x5 ln x − x + C 5

1

n+1

x (n +1)2

+C

18. − 1 ) + cot (3x ) +C 3 ln csc(3x 4 ln sin (3x − 2) 3 20.

+C

⎛ 1−x2 ⎞ 22. −3l n ⎜ 1 + C ⎜ ⎟+ x ⎝ ⎟ ⎠

418

ANSWERS

46. a) Trapezoid: 0.784240767

24. 15 ln ⎛⎜ x ⎞⎟ + C ⎝ 7x + 2 ⎠

b) Simpson’s: 0.785397945

⎡ 1− ⎞⎤ 26. 3 t t2− 1 − 1 ln⎛⎜ ⎝ t2 9 ⎟⎥+C ⎢ 9 9 ⎠⎥⎦ 2 ⎢⎣

48. a) Trapezoid: 0.754348974

, or+ 3( 28. 3x−1 (−ln x −1) C−lnx − 1) + C

50. a) Trapezoid: 0.127255793

x

30.

2 x(lnx)4−4x(ln x)3 +12(ln x) −24x(lnx)+24x + C

32. 32 ln

b) Simpson’s: 0.127440729 52. 15

54. −4ln

+C (x+x2 + 25)

56.

2 (9x 34. 135 − 4 )(2 + 3x)32 +C

58.

36. 12 x2lnx + 1x2 4 +C 38.

b) Simpson’s: 0.748092562

ln

+ x − 1 − ln x + 1 33x −44 x 6 x+1 − + 2 x + ln 3 x − ln3x + 6x + 1 − 2ln6x + 1 +

3 y− 3

− ln3 − y

2ah3 3

+C

+ 2ch

Exercise Set 5.8 +

12 12 5 x +C 5 3 40. 1 x6 sin (x) 2 + 2 x5 cos( 2x)− 2 15 4 (x) 15 x sin 2 − x3 cos( 2x)+ 2 4 ( 45 2 45 x sin (x) 2 − sin 2 x)+ 8 4 45 2 +C x cos (x) 4 −x 42. −1 e − x cos(2x ) + 2 )+ C 5 e sin(2x 5

b) Simpson’s: 17.75228722

x

c) h ⎡⎣ f (−h)+4f) +(0 f (h)⎤⎦ 3

3 3 2 + 2ln 6 x −1 + x lnx − 1 2 −2ln 4 x −1 + 2ln 4 +x− 1

44. a) Trapezoid: 18.21055563

1

b) ah2−bh+C

2ln 6 x − 12 x+ 1 + 4 l n 1 2+x1 + x ln3 x2 + 3 x + 1+ 6 6 −x1212

1

2(2− x)+ 4 ln x − 2 + C

60. a)

1212x7 2ln3x − 1 − − 7 2ln6x + 12x − 4ln12x − 1 −

x +C 3x − 2

2. 2π 4.

2π 2

6. π 2

(e4− e ) −6

8. 3π 4 10. πln4 12. 50π 14. 9π 16. π 18. π 20. 6π 22.

4 3

πr3

ANSWERS TO EVEN-NUMBERED EXERCISES 419

24.

4 3

34. The limit does not exist, the improper integral is convergent.

3

36. The limit does not exist, the area is infinite.

26.

4 972 5 28. 64 3

38. 1 6 40. a) ≈14.891 or 1489.1% per year. b) ≈ 0.47739 rems c) ≈ 0.67155 rems

30. 100 32.

42. ≈ 43.5 lbs

34. 62.832

44. The integral is convergent for r<−1 and divergent otherwise.

36. π ⎛ −3 +2e ⎞ ⎜ ⎝

e

2

⎟ ⎠

46. 1 10 48.

38. a) Trapezoid: 53 b) Simpson’s: 52.06667

Exercise Set 5.9 2. Limit does not converge, the improper integral is divergent.

4. Limit does not converge, the improper 6.

integral is divergent.

The limit does

50. 1.252310134 converge,

the

not

converge,

the

14. improper integral is divergent.

16.

The limit does not

converge,

the

8.

not

improper 10. integral is divergent.

12.

The

limit

does

18. improper 20. integral is divergent.

22. The integral is convergent.

24. The integral is convergent.

The limit does not exist, the improper

26.

integral is divergent.

30.

The integral is convergent.

28.

32.

The integral is divergent.

The integral is convergent.

The limit does not exist, the integral is divergent.

The integral is convergent.

The integral is convergent.

The limit does not exist, the integral is divergent.

420

ANSWERS

Chapter 6 Exercise Set 6.1 ⎡12 2. ⎢ ⎣7

30. (a) (b)

⎡12 −3 ⎤ 6. ⎢ ⎣ 21 −27 ⎥ ⎦ 3 ⎡ − −9⎤ 8. ⎥ ⎢ −2 2⎦ ⎣ −5 6⎤ 10. ⎡ ⎥ 1⎦ ⎢ ⎣2 3 ⎤ 12. ⎡ ⎣ −292⎦ ⎥ ⎢

(c) 32. (a) (b) (c)

⎡27⎤ 14. ⎢ ⎥ ⎣ −62⎦ 16. Undefined

34. (a)

−48 ⎤ ⎥ −342 ⎦

⎡4 −1⎤ 20. ⎢ −9⎦⎥ ⎣7 175 22. ⎡⎢ − ⎣ 40 ⎡ 8 ⎢ 3 ⎢ 24. ⎢ ⎣ −26

⎡−2 28. ⎢ 4 ⎢ ⎢ ⎣0

2 ⎤ ⎥ −12 ⎦

4.Undefined

⎡ −11 18. ⎢ ⎣5

(d) The matrix multiplication is only defined in the direction stated in parts a), b) and c).

−30 −27

28 27 14

⎡ 18 ⎤ 26. (a) ⎢ 16 ⎥ ⎥ ⎢−15 ⎥ ⎢⎣− ⎦ ⎡ 0 ⎤ (b) ⎢ 14 ⎥ ⎢ 10 ⎥ ⎢⎣− ⎥⎦ ⎡−22 ⎤ (c) ⎢ 3 ⎥ ⎢− ⎥ ⎢⎣− 14⎥ ⎦

(b)

−1⎤ 7 ⎥⎦ 14 ⎤ ⎥ ⎥ −25 ⎥⎦ −20

(c)

36. (a)

6 −2⎤ −⎥1⎥ −5 3 −8⎥⎦ Left to the student. ⎡598⎤ ⎥ ⎣⎢ 291⎦ ⎡1446 ⎤ ⎢ 711 ⎥ ⎦ ⎣ ⎡ ⎢ 0.5 1.25 ⎤ ⎣ 0.75 0.25 ⎥⎦ ⎡110 ⎤ ⎢ 87 ⎥ ⎣ ⎦ ⎤ ⎡164 ⎢102 ⎥ ⎣ ⎦ ⎡ ⎢ 0 0.76 0.95⎤ ⎢0.2 0 0 ⎥ ⎥ ⎢ ⎣ 0 0.91 0.91⎥ ⎦ ⎡

639⎤ ⎢116⎥ ⎥ ⎢ ⎢627⎥ ⎣ ⎡⎦ ⎢684⎤ ⎥ ⎢ 676⎥⎦ ⎥ ⎢⎣128 ⎡ ⎢3.6 0.98 0.65⎤ ⎢0.5 0 0 ⎥ ⎢ ⎥ ⎣0 0.5 0.49⎥ ⎡ ⎦

(b)

72⎤ ⎢10 ⎥ ⎢ ⎥ ⎢ 0⎥⎦ ⎣

(c)

⎡ 269 ⎤ ⎢ 36 ⎥ ⎢ ⎥ ⎢ 5 ⎥ ⎣ ⎦

ANSWERS TO EVEN-NUMBERED EXERCISES 421

( d)

38. (a) (b)

⎡1007 ⎤ ⎢ 135 ⎥ ⎥ ⎢ ⎢ 20 ⎥ ⎦ ⎣

24 - 28. Left to the student.

⎡−2 −117⎤ ⎥ ⎢ 39 37 ⎣

.

34.

32

⎡28 ⎢ 33 ⎣

36 .

⎞ 5 z+ , 5 z + 5 , z⎟ ( x, y, z)=⎛⎜ ⎝ 5 ⎠ − ( ( x, y, z)= 40, 3 0,0 )

.

( x, y, z, w) = (−3,2, −1,4)

38

( x, y ,z, w) = (15,31, − 8,− 3)

.

42.

40 44. .

⎡ 598 ⎤ ⎥ ⎢ ⎣ 291 ⎦ 3DP= 59951;2DP=0;DP=4634

−105⎤ ⎥ 7

(c) Matrix multiplication is not commutative. (d)

A2− AB − BA + B2

40. (a) No (b)

Yes

42. True; due to the commutative property of addition for real numbers.

44. Since the matrices Aand Bare square, both ATand BTwill have the same dimensions as AandB. The entries on the diagonal are the same. Using the commutative property of matrix addition, the off-diagonal entries can be shown to be equal.

46. - 48. Left to the student. ⎡1, 748, 798⎤ 50. ⎢ ⎥ ⎣ 859,210 ⎦

No Solution 8

⎡4

11 7

1

5 6⎤ 28 3⎥⎥ 9⎥⎦

46. (⎢⎢1 a) ⎢ ⎣7

⎡1 (b) ⎢⎢4 ⎣− ⎢ 4

1

−3⎤

−2 3

7⎥⎥ 6⎥⎦

2 3⎤ 48. (a) ⎢ 1 81 15⎥⎥ ⎢7 ⎢ 9⎥⎦ ⎣7

⎡ 4 (b) ⎢ ⎢13 ⎢⎣−4

Exercise 6.2 2. ( x,y ) = (10,5 ) 4. (r,s ) = (−1,−1 )

a( ,4a−2)

6. Identity,(a,b ) 8. No Solution

=

10. - 16. Left to the student.

( x,y,z ) = (46,18,−32) x,y,z 20. ( ) = (16,20,−6) 18.

(

30

⎡1

⎡1270 ⎤ 54. ⎢ 235 ⎥ ⎥ ⎢ ⎢⎣1256 ⎥ ⎦

x,y,z

(x,y,z)=(−1,3,5)

(c) Matrix Ais the identity matrix with first and second rows swapped. Therefore, the effect of multiplying A on the left yields a swap of the first and second rows of the matrix being multiplied byA.

⎡ 968, 803 ⎤ 52. ⎢ ⎥ ⎣659,436⎦

22.

.

) = (1,5,5)

−2 7⎤ −5 3

18⎥⎥ 6⎥⎦

(c) Matrix Ais the identity matrix with the second row replaced by 3R1+R2.

Therefore, the effect of multiplying Aon the left with another 3×3matrix will give a matrix with the second row replaced by three times the first row plus the second row.

50. - 54. Left to the student.

422

ANSWERS

30 0, not invertible.

Exercise Set 6.3 5 2 ⎤⎥ 2. ⎡⎢ −3 −3 ⎣ ⎦ ⎡ −1 ⎢ 0⎤ 4. ⎣⎡ 0 ⎥ 1 2⎦ ⎢ ⎣ ⎡7 18 − 118 ⎤ 6. ⎢ ⎥ 29 19 ⎦ 512 ⎤ 2⎥⎦3 8. ⎣112 1 3 ⎡ ⎢ −2 21 0 0 ⎤ ⎢ ⎥ 10. ⎢ 1 2 ⎥ ⎤ ⎣ ⎥ 0 −1 ⎦ ⎡ ⎥ ⎥ 2 ⎢ 0 ⎥ ⎦1 ⎢ 0 12. ⎢ −1 0 −75 ⎣6 26 ⎡ 1 2 ⎢ 38 ⎢ −81⎤ ⎢ −61 ⎣ 21 14. ⎡ 28 ⎥ ⎢ 31 41⎥⎦ ⎢ ⎢ ⎣15 2 17 2 8⎤ 16. ⎡ −1 −2 −2⎥⎥ 4⎥⎦ ⎢ −9 2 −9 2 − ⎢ ⎢⎣ 2 1 −3 2 ⎤ ⎡ −12 18.

12 −12⎥⎥ ⎢3 2 12 −1 ⎥⎦ ⎢ ⎢⎣ 9 10 11⎤ ⎡ ⎢ −5 5⎥⎥ 20. ⎢ −7 − − 1 ⎢ 0 −2⎥⎦ ⎣ ⎡ ⎢ 3115 4 3 −6 5 ⎤ ⎢ 22. ⎢ 2 6 1 5 5 3 −6 5 ⎥ ⎥ ⎢ ⎣ 6 8 1 5 8 3 − 1 3 5⎥⎦

24.

−2 −3

1 0

5 7

5 7

8⎤ 10 ⎥ ⎥ −2 −2 −3⎥ 2 3 4 ⎥⎦

26. 4, invertible.

28. −79, invertible.

.

−157, invertible.

32

−77, invertible.

.

504, invertible.

34

0, not invertible.

.

218, invertible.

36

50, invertible.

.

⎡ 0 ⎢⎢−203

38 (a) 44. .

10 3

⎢ ⎣120 19

40

0 0

0 ⎤ 203 ⎥ ⎥ −100 19⎥⎦

⎡ 242 ⎤

. (b) 42

⎢1⎥ ⎥ ⎢ ⎢ ⎣250⎥⎦

.

⎡ 184 ⎤ (c)

46. (a)

(b)

(c)

⎢⎥13 89⎥ ⎢ 0 ⎢ ⎥⎦ ⎣ ⎡ 2 ⎢ 0 ⎢ 3.157 −22.732 ⎢ ⎣ −3.222 23.196 ⎡ ⎢ 2190 ⎤ ⎢ 220 ⎥ ⎥ ⎢⎣ ⎡ 408 ⎥⎦ ⎢ 441 ⎤ 198 ⎥ ⎥ ⎢ ⎢630 ⎥ ⎦ ⎣

48. ABB−1A−1 =

0 ⎤ −4.188⎥ ⎥ 6.314⎥ ⎦

A(BB−1) A−1

= A ( I ) A− 1 = AA−1 =I

Therefore, B−1A−1is theAB inverse of ( ).

50. – 58. Left to the student.

Section 6.4

2. The vector is an eigenvector with eigenvalues of 3.

4. The vector is not an eigenvector.

6. The vector is an eigenvector with eigenvalues of 10.

ANSWERS TO EVEN-NUMBERED EXERCISES 423

8. The vector is an eigenvector with eigenvalues of 5.

10. The vector is an eigenvector with eigenvalues of−2 .

.

12. The vector is not an eigenvector.

4 14. v = −3w + u 3 16. v = 5 w − 2 u 18. v = −2 w + 4 u 20. Eigenvaluer = 2and

⎡ −t⎤ eigenvector⎢ ⎥ , t ≠ 0; ⎣t⎦ Eigenvaluer=3 and 0 eigenvector⎡ ⎤ ,t≠0 ⎢ ⎥ t ⎣ ⎦ 3 5 97 , 22. Eigenvaluer = 2 + 2

⎡⎛ −7 97 ⎞ ⎤ ⎟ ⎥ + ⎢ eigenvector 6 6 ⎟ , t ≠ 0; ⎜ ⎠t ⎥ ⎢ t ⎥ ⎜ ⎦ ⎝ 3⎢ 5 97 Eigenvaluer = ⎣− , 2 2 ⎡ −7 97 ⎞ ⎤ ⎛ ⎟ ⎥ − eigenvector⎢ 6 6 ⎟ , t≠ 0 ⎠t ⎜ ⎥ t ⎢ ⎥ ⎦ ⎜ 24. Eigenvalue r=3 and ⎝ ⎢ ⎡ 5⎣ t⎤ ⎢ 3 ⎥ ,t ≠ 0 eigenvector ⎢ ⎥ ⎢t ⎥ ⎣ ⎦ Eigenvalue r= 7 and ⎡ t⎤ , t 0 eigenvector⎢ −2⎥ ≠ ⎣

26. Eigenvaluer = 0.5 and ⎡−3t⎤ eigenvector ⎢ ⎥, t ≠ 0 ⎣ t ⎦ Eigenvalue r= 3

5 eigenvector⎡⎢ t2 ⎤⎥ , t ≠ 0 ⎣t ⎦

28. The characteristic equation is

= 0, (r+3)(r2−r)

The eigenvalues are=r0, r = 1 and r = 3.

t ], The eigenvectors [ 0, 5t,are [0, t, 0], respectively. −6t, t], and −t, [

30. The characteristic equation is r3 + 6r2 + 11r + 6 = 0,

The eigenvalues are=r−1, r = −2 and r = −3.

The eigenvectors are [ 0,0, t],

[t−,3t, −9t], and [2t, −5t, −15t ], respectively.

32. The characteristic equation is r−36r2 + 8r = 0,

The eigenvalues are r=−2, r = 2and r = 3.

The eigenvectors are [ 0,t, 0], 2t, [ t, −2t], [−t, −t, t], and respectively.

34. The characteristic equation is r+36r2 +8r = 0,

The eigenvalues are r =0, r = 2and r = 4.

The eigenvectors are [ 0,t, 0,]

[−22tt,, t], and [ −33tt,, t],

respectively.

36. The characteristic equation is r3+r2−4r−4 = 0,

The eigenvalues are=r−2, r = −1and r=2.

The eigenvectors−tare , −2t[0, ], [t2,t, −2t], andt, [ −t]t,, respectively.

2 38. ⎡⎢ ⎤⎥ ⎣5 ⎦

⎡ 162⎤ 40. ⎢ −162⎥ ⎦ ⎣ 42. 1.01; 1%

424

ANSWERS

44

4; 300%

.

≈1.14

46

≈3.74

30. xn = (−1)n−(2)n+1 32. The population grows exponentially to∞.

. ⎡3⎤ 50. (a) 4 ⎢5⎥ 48 ⎣ . (b) If the long-term rate is not 1.4,

1 n⎞ lim the ⎟ Gnpmay not converge. n→∞ 1.4⎟ ⎝ ⎠ ⎛⎜ ⎜ ⎡13 −1 ⎤ 52. (a) 3⎥ ⎢⎣559 17

34. The population grows exponentially to∞.

36. The population decreases exponentially to 0.

38. (a) x = − 46.204 98

(b) The population will die.

40. (a) x n=c1 (0.01009)nc+2(1.13991) +83.03249 n

(b) 0.53845(0.01009)n+

(b) r2−3r+2

182.49404(1.13991)n−83.03249

(c) Since the two matrices have the same characteristic equation, they will have the same eigenvalues.

54. Yes.

(c) The limit grows to ∞.

42. (a) x

. Linear, but not homogeneous. .

n

n

xn = 1 + c2 (5 ) 6 c1(2)n+ c2 (3 )n 10. . xn =

1)n

8 xn = c1(− 12.

28. xn

n

=

n

5 ⎛1− 5 ⎞ 5 ⎛ 1+ 5 ⎞ ⎜ ⎜ ⎟ ⎟ + 5 ⎜ 2 ⎟ 5 ⎜ 2 ⎟ ⎝ ⎝ ⎠ ⎠

(b)

xn = −

(c)

2; 6755

n

n (d) Approximated to .

5

n

46. xn = c1 + c2 n

= 11 )n1 17 2) n 4(− + 4 ( = − n 18. xn 3( 1) + 4 (5 )n = n 20. xn 19(2) − 2 (3 )n = n n 22. xn c1(−1) + c2 (2 ) + 4 = =

⎛1 ⎛ 1+ 5 ⎞ 5⎞ xn=c1 ⎜ − ⎟ +c2⎜ ⎟ ⎜ 2 ⎟ ⎜ 2 ⎟⎠ ⎝ ⎠ ⎝ n

+ c2 4( )

16. xn

26. xn

44. (a)

c(3)

. x = )n4 14. ( n

24. xn

n

Not linear

4

n+1 =1−xn

(b) xn =1−1(−1)n 2 2

Exercise Set 6.5 2 Linear, but not homogeneous.

(−0.13724 )n+ 46.10498(0.94724)n

n

n

c1(2)+ c2 3( ) + 9

(−n2)+ (3 )n+ 6

10(−4)n − 2 (7 )n+ 5

n

48. xn = 7(3) 50. - 54. Left to the student.

56.

(a) 1.425667816 (b) n=1,M 1=296.14 n = 2, M2 = 297.02 n = 3, M3 = 297.7 n = 4, M4 = 298.23

n = 5, M5 = 298.64

(c) The answers in part (b) are expected since they vary by a value close to the equilibrium value found in part (a).

ANSWERS TO EVEN-NUMBERED EXERCISES 425

26.

Chapter 7 Exercise Set 7.1 2. f (−2,0 ) = 0 , f (3,2 ) = 10648 , f −5, 10) = 125000

4. f (3,7 ) = 1 , f (1,99 ) = 2 , f (2,− 1) = 0

( 6. fπ,π

3 ) =1, f (π 3 , 0 )=

f (7π 6, π ) =

1 2

3 , 2

8. f (0,1, − 3) = −14, f(1,0,−3)=1 10.

1.939m2

12. (a)

0.292

(b) 0.738 14. (a) 105 (b) 95 16. A function of one variable has only independent variable, while a function of two variables has two independent variables.

18. −10 20. −64 22.

Exercise Set 7.2 d 2. d = 3 ( x − y)2 =−3 ( x − y , )2 , z z d ddz 5)=− 75 −2, − 3)= 3, ( x ydy (0, − d 4. z = 6 x2 + 3 y−1, dz = 3x, dy d x −2, 3)= 14, dz (0, = 0 ( − dy − 5) d z 6. fx = 5 , f y = 7 , fx (−2, 1) = 5, d xf −3, − 2 ) =7 y( d ⎞ ⎛ x y ⎟, 8. zfx = , f y=−⎜ ⎟ x2 + y2 x2 y2 − ⎜ ⎠ d ⎝ x = 2,1) −23 , fy (−3, −2 fx ( − )= 2 5

10. fx = 2e 2 x− y , f y = − e 2 x− y = 12. fx 2ye2xy , f y = 2xe2xy

24.

14. fx = 1 + y2 , f y = − 1 − x y x x y2 = 5x4 − 8xy2+ 5 16. fx y3, f y =−2− 8x2y+15xy2 = y 2xy 18. fx = + , fy 2 x2+y x +y

ln(x + y)

20. fx =

2

3xy2 + 2 x + y3 , f y = 2x + y3 2 2 x+y3 x

426

ANSWERS

(y − xy) sinxy)(− co xy), ( x − y)2 s( xy) ( xy)(+ co y − xy) sin

54 15.6

2

22. fx =

2

fy =

24. fx =

26. fx =

sin (2 x + 3 y )

e

, fy =

64. ∂Th =1.98−1.09(1−H). This means that for ev ∂H change in temperature of 1 degree at a specific humidity H, the Temperature-Humidity Heat index

+2y

−3tx2 +2tx

(x +t) 2

3

−202 tx

(x +t) 3

4

⎛x 2 +t 2 ⎞ 2⎝ ⎜ 2 −2⎟ , zt= x t ⎠

(x2−t)

(x2−t2)

⎡y2 2xy ⎤ 36. ⎢ ⎥ ⎣1 y −x2 y2⎦ ⎡ 38. ⎢

4

68. ∂E =−1.015. This means that for every increase of ∂s one word for the average words per sentence there is a decrease of 1.015 in the reading ease of a 100-word section.

70.

fy = −10 xy fxy = f yx = −10 y

⎤ −2siny(x+cosy)⎥⎦ ⎥

72. The partial derivatives of a function of two variables represent the slope of the tangent line for each

variable. For example, ∂f∂xat a point (x x,aty) in represents the slope of the tangent line

⎢⎣2(x+cosy)

40.

, ,f xy =−e2x−y fxx=4e2x−y 2 x−y x−y fyx=−2e2 fyy=e2

the x direction.

xx=4y2e2xy ), ,fxy=2e2 xy (2xy+1 42. f

fyx=2e2xy(2xy+1) f = 4 x2e2x yy

Exercise 7.3

y

44. fxx = −23y , f x y= − 1 + 1 , y2 x2 x = 2x 1 1 f yx = − 2 + 2 f yy y x y3 46. fxx = 20 − 8 y , f x y = − 6 2, xy1+15y x xy + 15y2 fyy= −8 x 2 + 30 xy f = −16

2 f .

4

yx

xy3 y3 3xy2 48. fx = , fy = 4 , fz = −4 5 z4 z z x= 1 3 5 50. f , f y = , fz = z y x

( x,yhas ) a relative minimum at −15, −25 ). f ( x,y ) has a saddle point atx(0,0). , y ) hasfa relative maximum at (83,163).

) hasfa has a saddle point atx,y (0,0). .6. f ( x,y ) relative minimum at (2,2).

3

52. 8.2

f(x,y)=x3−5xy2

xfxy−f y = −10 xy −(−10 xy)= 0 3x2y

2xy3

66. 48.465

2

20t2x ⎛ x + 2 ⎞ 2 ⎟ 2 ⎜ 2 ⎝2 t ⎠ x − 34. zx=4x−13−2x−34t12+6x−32t32, 2 t 14−12 −1212 t=−4x −18x z t t

32. zx=

changes by 1.98−1.09(1−H).

, zt= −x −x2 3

−3.32223

62. 97.5

∂f =28m+12b− 28. ∂f 98 ∂b =12m+6b−42 , ∂m 4

(b) 17.27651

60. 117.8

2 sin 2( x + 3 y )

x 2e (x + 2 y − 1) , (x+ 2y − 1) fy = (x+ 2y)2 ( x + 2 y)2

30. zx= − x

56

(b) −5.653761

3cos (2 x + 3 y )

x+ 2 y

(a) 16.99436

. (a) 58.

( x − y )2 s( cos (2 x + 3 y )

.

8. f ( x,yhas ) a relative minimum at 0,0 ) .

10.

12.

f ( x,yhas ) a relative maximum at0,0 ) .

f(x,y)has a saddle point at (0,0).

14. f ( x,yhas ) a relative minimum at1,2 ).

ANSWERS TO EVEN-NUMBERED EXERCISES 427

16. P(a,p)has a relative maximum at (10,2.6983).

The value of the maximum is 33.635.

18. The dimensions of the container are 4.9866ft×4.9866ft×12.8689ft.

20. T(x,y)has a minimum at (12,−1). The value of the minimum is −72.

22

f(x,y)has a relative minimum at (1,2).

.

S(b,m)has a relative minimum at (2113,32).

24 .

26

The relative minimum of a function of two variables is only the minimum in a local neighborhood around the minimum point while the absolute minimum is the smallest value the function has in its entire domain.

.28. (a) ∂R∂s=e−1.236+1.35coslcoss−1.707sinlsins×

(−1.35coslsins−1.797sinlcoss)

(b) The sign of ∂R∂sis determined by the sign of the term in parenthesis in pat a) since eanythingis positive for all values permissible. Since the trigonometric functions are positive in the first quadrant (the permissible values for l and s fall in the first quadrant) then the term in parenthesis will always be negative. Thus, ∂R∂sis always negative for the permissible values of l and s.

(c) Since ∂R∂sis negative for 0≤s≤60then the function is decreasing in the s-direction which means that the maximum occurs when s=0 (the beginning of the interval).

30. Relative minimum of −55 4 4 at (4, −).

Relative maximum of1 at (2, −).

2 2

Exercise Set 7.4 2. (a) y = 0.000466x+0.535 (b) When x = 4000, y = 2.399.

428

ANSWERS

4. (a) 0.174 x + 64.843 (b) When x = 60, y = 75.283.

16. 64 2 27

When x = 65, y =

18. 1.806

76.153.

6. (a) − 1.6x + 33.7

(b) When x = 14,y = 11.3.

8. (a) 0.227x − 357.556 (b) When = 2010,y = 98.7. x = 2050,y = 107.8.

When (c) The world record for x=2050 is a little x unrealistic since it is close to 9 feet. This

10.

unrealistic answer occurs because the regression line is calculated for a small range of years and therefore accurate for years relatively close to that range. When we consider the year 2050 the regression equation becomes less accurate in predicting the world record since 2050 is 61 years form the highest year used to calculate the regression line.

Left to the student.

12. (a)

(b) Y = 0.36427 X + 1.33226 (c) y = 21.49117x0.36427 (d) 38

Section 7.5 2. 1 4. 332 6. 1 8. 1 e 4 − e 2 + 12 2 10. 1 12.

e9 − e

14.

1 3

20. 0.1 22. 760 24. The integral of Example 1 could not be solved if we switch the order of integration because then the volume the solid generated would depend on the value of x and for values of x greater than 1 we would get a negative volume.

26. 0 28.

14 3

30. Left to the student.

32. 11.61953492 34. −0.1202884987

ANSWERS TO EVEN-NUMBERED EXERCISES 429

(b)

Chapter 8 Exercise Set 8.1 2. 13 (2x− 3)32+ C 4. −5cosx − 4 x +C 6. 4x1 2+ tanx +C 8. 1 (lnx)2 +C 2

38. (a) Slope at(−2,1) at − 2.

10. cos x + xsin x + C ⎡ ⎢ ⎢⎣2x ⎤ 12. 1 + ln3x + 2 ⎥+ C 9 =x2+6 (3x+2) ⎥⎦

(b)

14. y 16.

1x2 +4 2

y = −cosx−

18. f (x)= ln x−

x2 2 x3 2 38 + + − ln4 3 3

20. y 1= ex2+ 12 2 22. y =ln2 + cosx + 3

40. Thus,y is a solution to the differential equation.

42. Thus,y is a solution to the differential 44. Thus,y equation. is a solution to the

24. y =sinx − xcosx + 3 26. y = 1 x2lnx 1x2 4− 2

)32

46. Thus, y differential equation. is a solution to the differential equation.

L 48. T ( x )= − 1 x2 + D x 2D

1 +4

(ex

13 28. y = 2 +3 − 3 3 − 30. f (x)=2x38x + 8

L2 50. 3D

52. (a) y = 0 (b) The slope of the tangent line at the point where y′ = 0is 0.

32. f (x) 1= e2x + 4x + 15 4 4 34. f (x) = 1 x4 + 1x2 24 4

−

2,1 is 1 36. (a) Slope at − ( )

3x + 2

2

.

(c) Left to the student.

54. (a)

y=0

(b) The slope of the tangent line at the point where y′=0is 0.

(c) Left to the student.

430

ANSWERS

40 201.16 lb

Exercise Set 8.2 2. The unique solution will exist (on −∞ , 2 ) .

exist ). on ( 4. The unique solution will −1,2 6. The unique solution will exist on 0,6 ) .

(

8. y = 12 e7 x + Ce3x

t4 + t

(c) −0.01431 (d) 18.11 (e) 64.03 pounds per acre

. The solution exists 0, ∞ ) .on (

24. y = 1+ Ce−cos x . The solution exists on all real numbers.

+

1 x3+Cx−3 4

52. (a) Left to the student

3 20. y = 12 tan ()2t − + C sec ( )2t 2 3

t

y=

50. P=0.59016sint−0.4918cost+0.4918e−1.2t

18. y = x − 1 + Ce−x

26. y =

.

48. P=1−e−1.2t

4

t4

(b) Y = 60 − 45ekx

(b) 2

C

+ +1) t2+1 4 (t2

x

42

46. (a)

14. y = 4 t5e−t + Ce−t4 5 sin 16. y = x x − cos x

22. y =

(a) Y′ + kY = 60k

44. This value agrees with the result obtained in Example 7.

10. y = 2x2 − 4 x − 8 +Ce−3 x 9 3 12. y =

.

+ 3

2

+ 1) t +1 4t( 31)+ (t3

. The solution

exists on ( ). −1,∞

(b) Q =

a a (b+k )t − b + ke− b+k

(c) Q =

a b+k

Exercise Set 8.3 2. (a)

4 y= 5

(b) Unstable (c) No inflection points. (d)

28. y= 2 −2e2x e 2 . The solution exists for all real numbers.

30. y =

ln csc x csc x

− sin x . The solution exists

on[0,π].

32. y=1+3t−1 e −1 . The solution exists for all real numbers.

34. y = t cos t + π cos t. The solution exists on

[0,π].

2

(b) (c) y = 1 is a semi-stable equilibrium point.

36

T = 4 +Ce−0.5t

.

y = 15 +Ce−0.2

38 .

4. (a) y = 1

No inflection points. t

ANSWERS TO EVEN-NUMBERED EXERCISES 431

(d)

6. (a) y = 2 (b) y = 0 is an asymptotically stable equilibrium point; y=2is an asymptotically stable equilibrium point.

No (c) inflection points.

(d)

(d)

12. (a) 3.75 million (b) 15.58 million 14. (a) k ln(Ly) (b) y = L is an asymptotically stable equilibrium point. (c)

8. (a) y = ln5 (b) y = ln5 is an asymptotically stable equilibrium point. (c) No inflection points. (d)

16. (a) – (b) y = (kk − s)L +C

(c) If s<k thenk − s > 0→ c> 0. (d) The inequality states that the rate of harvesting is less than the growth rate.

10. (a) y = ln3 (b) y = ln2 is an asymptotically stable equilibrium point; y=ln3is an unstable equilibrium point. (c) Inflection point at y= ln5 2.

18. (aThe ) equation follows the logistic growth model with a threshold with k=2, L=20, 5 T = 5 and a harvesting rage of .

25 (b) y=0is an asymptotically stable equilibrium point; y=9is an unstable equilibrium point; y=16is an asymptotically stable equilibrium point.

(c) It is less than 20 to ensure the survival of the population. (d) It is greater than 5 to ensure the survival of the population.

432

ANSWERS

20. −9

32. Left to the student.

22. (a) By Theorem 4 it follows that c is semistable.

34. (a)

(b) The result follows from Part (a) and the definition on page 564.

3ky2 24. y′′= 2ky − k − LT + 2ky L T y′′(T) = 2k − k − k−

3kT 2kT + = L L

kT >0, unstable

(b) 0.0082

L

(c) t = 0.48

k2L2− 4ksL 26. (a) y = kL± 2k

2 2 kL −4 ksL, which is positive.

Exercise Set 8.4 2kL Thus, the smaller equilibrium point is unstable.

2. y = 3 5 x2 + C 2 2 k L ksL −4 (c) y′′=− , which is negative.

2 2kL Thus, the larger equilibrium point is −3 4. y = asymptotically stable. x3+ C (d) The population grows indefinitely. −1 6. y = ln Cex

(b) y′′=

2(

28. k L y2 − ky + s = 0 4ks k ± k2− L

=y

2k L k ± k 2 −k 2 =y 2k L L =y 2 2k − L y= y ′ ′ 0=y′′⎛⎜ L ⎞ ⎟ ⎝2⎠

Therefore,y L= is a semi-stable equilibrium 2 point.

30. y′′= nkyn−1 − k

y′′(0 ) = − k < 0 y= 0is an asymptotically stable equilibrium point.

)

=C 3 ( )32

8. ln y − 2 e x + 4

10. 12 − 13 et3= C y 2 1 y 6 1 6 1x2 12. 2 y2 + −2 =C 14. e−y+ 1e−2y+ ex = C 2 16. coty + lnt = C 18. y − 1 = Ce2x y+1 20. y

= ex22 −2 2

22. 1 2

ln(y2 + 4)= t + C

24.

1

1 y2 + ln y = sinx − 2

2 e−2y =e−8−x2 26.

ANSWERS TO EVEN-NUMBERED EXERCISES 433

2(

28. y = − ln1 e 2t + 3e 2

)

30. sin y = 1 et 2

2. (a) ≈ 0.429

x2+4, which has a domain of . ( −∞, ∞)

32

y=

.

(a) Left to the student.

34

r ⎞ (b) P′ = −⎛⎜ ⎝V ⎟ P ⎠ k =⎛⎜ ⎝ r ⎞ (c) k V ⎟P ⎠

.

36. (a)

Exercise Set 8.5

(b)

≈ 0.541

(c)

= y4e

(d)

y(2) = 0.541

4. (a) ≈ 7.850 (b) ≈ 8.9999 2

⎛ 2 +3⎞ (c) y =⎜ x ⎟⎟ ⎜ 4⎠ ⎝

T = (T0 −C)e + C −kt

(b) Left to the student.

(d)

38. Left to the student.

1 ; domain of the solution is t .

40. (a) y = 1− ≠1 t

y=

(b) ≈3.354 (c) y = (d)

k ; domain of the solution is 1 − kt

t≠1k.

(d) There is no solution for the initial condition

y(3)=9

6. (a) ≈3.004

(b) y = 2 ; domain of the solution is 1 − 2t t ≠1 2 . (c)

−x

(x2+6) −1 4

y(1)=2.5

8. (a) ≈16.909

≈17.056 (b) y(0)=0since we get a division by zero when

we try to find the constant of integration.

42. Left to the student.

44. (a)

(c)

y=2x2

(d)

y(3)= 18

10. (a)

Q (30)≈ 0.6596

Q(90) Q ≈ 0.9993

(b) Q (90) (30)≈ 0.6574 ≈

0.9990

(c) Q (30)≈ 0.657

Q (90)≈ 0.999

(b) Left to the student.

h(1)≈ 0.8757 h 2 ) ≈ 0.00509 ( ≈ 0.4079 (b) h(1) h 2 ) ≈ 0.00289 (

12. (a)

434

ANSWERS

(c) h (1)≈ 0.3647

h (2 ) ≈ 0.0025

14. ( a ) Q (20)≈ 34.6821

Q (40 ≈ 158.5996 Q ) ≈ 206.5281

(60) (80)≈ 215.0765 Q (100 ) ≈ 216.4213 Q

(b) Q = 0 is asymptotically stable; = Q 216.6667is asymptotically stable. (c) About 60 days 16. (a) 2.4522 (b) 2.4290 (c) 2.1475 (d) 0.3047;0.2815 (e) 1.0824

ANSWERS TO EVEN-NUMBERED EXERCISES 435

r 2 + (c + m+ 2ar) + a ( c + m + a )= 0

44.

− ( c + m a+)2± (c + m+2a )2− 4 a ( c+ m+ a) r

Chapter 9 Exercise Set 9.1

( c+mthe Since always than.

2 )radicand the twoisroots willsmaller be negative +a

5x 2. y = C1 e 3 x+ Ce 2 4.

y = C1e− 3 x+

Cex

46. Nt( Ce− )= 1

2

6. y = C1−e3x + C2 e−2

x

Ce2 e 8. y = C1 e x +22C C 2x e −4x x − 10. y = C1e +

x 4x

12. y = C14ex +

16.

−4x

+ C3xe

x y = C 1−e2x + C 2 ex sin 3 +x C 3 e cos 3 x

+ 2 −x −x 18. y = C1 C C 22 e x + C 3 xe−x +C 4 x e + 4x 20. y = C1 + C3x2 +C4e 22.

y

0.0576 t

3.8976 t

+ C 2e −

3x + 2 x + Ce 48. y (x ) =C e1 x2 +Cxe 2 3

x

14. y = C1 + C2e

−

⎛ 3x ⎞ =C1e x + C 2 ex +Ce−3x 2 sin⎜ ⎟+ 4e ⎜ 2 ⎟ ⎝ ⎠ ⎛ 3x ⎞ C − x 2 cos⎜ ⎟ ⎜ 2 ⎟ ⎝ ⎠

C 4xe 3x + C5x 2 e 3 x+ C6 x ⎛ ⎜ ⎛πkx 2 ⎞ C 2cos⎝ 2πkx⎞ 50. (a) p( x) = C1 sin⎜ ⎟+ c ⎟⎠ ⎝ c ⎠ ⎞ ⎛ ⎜ 2πkx ⎟ (b) p( x) = Csin⎝ ⎠ c (c) k =

(2 n− 1) c 4L

52. Left to the student.

54. (a) Let r=p+iq, the

a ( p+ iq2) + b (p + iq )+ c = 0

a( p+ 22ipg− q2)+bp+ ibq+ c = 0 (ap− 2aq2+ bp+ c)+i (2apq+ bq)= 0 →

a ( p 2 )+bp+c= −q 0 2

24. y = e (C1 sin (3)x + C2 cos (3 ))x 4x

⎛ 2cos ⎜ ⎛ ⎞ ⎛ 3x ⎞ ⎞ 3 x ⎜ 26. y = e− x 2 ⎝ C1 sin⎜ ⎟⎟ ⎟+C ⎜ ⎜ 2 ⎟ ⎜ 2 ⎟⎠ ⎟⎠ ⎝ ⎠ ⎝ = 2x +C 2x x +C 28. y Ce 1

and

=

2 sin 2( ) 3 cos )(

30. y

) x C1 sin(2 x) + C2 cos (2 + C3 xsin 2 x) + C 4 xcos 2x) ( (

32. y = C 1e

(5 − 17) x 4C e(5+17)x 2

3x 2

34. y =e−

=

2

+

2

(C1sinx + C2 cosx)

x − 3cosx 36. y = −4 sin 38. y = e x +xex

40. y = e x(sin(2 x)−cos(2 x)) 42. y = 3 − e−3x−2 xe−3x

(b) y = epxsin(qx)

2apq+ bq = 0

y ′= qepx cos ( qx) + pepxsin ( qx )

y′′=− q2epxsin ( qx) + 2qpepxcos+(qx) p 2 epx sin (qx)

ay′′+ by′+ cy= a ( 2 px = −q e sin(qx+) 2 qpepx cos + ( qx ) p 2 epx sin(qx ) + b ( qepx cos + ( qx ) ) pepx sin ( qx)) +

(

c ( epxsin ) ( qx ) )

epxsin qx ap2 aq2 +bp +c + − ( ) epxcos qx 2 apq+ bq ) = + 0 ( )( =0 0 =

(c) Follow similar steps to part (c).

436

ANSWERS 7t + C 2 e 7t 4. x = Ce− 1

1 − 7 )C2e (3 + 7 )Ce−+(3 y=

Exercise Set 9.2

4

2. y = C1ex + C2e2 x + 4 4. y = C12ex + C2e5 x + x − 2 C + C2 cosx 6. y = e 2 x 1(sinx + x) + 1 8.

y= =

10. y

C1 sin (2+xC)2 cos (2 x + 2) x 2 + 3 x + 1 C1e−x

=

+

C2xe−

x

x + − 6x2 + 20x − 29 3

12. y C1 + C 2 sin x+ C 3 cosx− 2 x x = C1 + C2 2 x + x2 + 14. e y C 3 e2 = x 2 16. x C1 + − −x +7x y = 1 sin 3 18. y ( x) + 2 cos (3+xx) − 1 3 + 20. y = −e2x− e3x 3x2 + 7x + 5 22. y =

13 11 14 e−3x x3 2 + x+ + − 2x 9 3 9

( (3x−)cos(3 2sin x))+ 2 24. y = e− 2−x 3x 26. y = 2x + 2x2 + x3 + 5 − 3e3x+ 4 xe

28. x =

=

1 − cos (2 t ) 1

2

1te−5t 1 11 30. x 5 + t − e−5t− 6 2 30

= 29

15 e−7t2 2e−3t5 34. F − − 7 7 36. Left to the student.

Exercise Set 9.3

et1(sin C 2t()+ C 2 cos ( 2t)) et2(C 1 cos 2 t() − 22Csin2 t()) =

x= y

6. x = C1e−t y=0

t2 8. x = 2 + C1t+ C2 y = t 2 + 2 ( C1 − 1)t +(2 C −2C 1 ) C t 10. x = 1 sin ( 2t )+ C2 cos (2) t + C(−)22C1 y = (C1+ 2 C 2 )sin(2

t

y=

2

) cos ()2t

( ( 1 sint +(C 32 )cos 3t)) e− ((2 C +23C 1) tcos )

12. x = e−t C

5

t 2 t + 2C 2te2+5 14. x = 2Ce 1 2 t − 11 y = Ce 2 + 16. x = C1e−2t C 2 et − t + 1 y = 2 Cet − Ce2−t − 3t 2 − 2 1

18. x = 1 + e 2t y = e2t− 1 5 2 3 20. x = e5t − te5t 2 5t 5t 3 6 e y= − te 22. x = e−t (5 sin t + 10 cos t)

y = e−t

2 4 32. F = e−3t4− 2e−t+ 3 3

2.

7t

7t

(4sin

t − 3 cost )

24. x =et+ 2tet t t y = 3e + 2 te − 5 26. x = C1 + C 2 e−t5

y = 1 C C2e−5t 4 1− t2 −t 2 3 − = 3C1 − C 2e 4t − 2 t 2 − 4 t − 1

28. x = C1 + C 2e −4t −

ANSWERS TO EVEN-NUMBERED EXERCISES 437

30. (a)Left to the student. (b) It follows from the statement of the problem. (c) A (t ) = 38.8235e−0.02 −t38.8235e−0.35t

(c) B′ = −0.06B + 0.03L B′ = −0.06 B+ 0.03(3B−) B′ + 0.09B = 0.09 t Bh = Ce−0.09

t B t( ) = 15.5294e−0.02 +t48.5294e−0.35

(d)

Bp = A 0

′ = Bp 0.09 0+0.09 A = A =1 Bp = 1

B = 1 + Ce−0.09t

(d) C = 2, B=1+2e−0.09t 32. (a) Left to the student. (b) It follows from the statement of the problem. t − + (c) F = − 0.5e−0.01t0.5 0.5 e 0.02 B = 1 − 0.5e−0.01t (d) B = 10;F = 5

( q−p ) t C1 + Ce 42. B = 2 ( q−p ) t L =pC1 + qCe 2

34. Left to the student.

46 . VL

L ′′=− (a + mL ) ′+ cmL− (a + cL) ′ − (a + c)(a + ) − (a + c ) p 2

aL − amL− xaL − cmL− (a + c )p

− (2a+ c + mL ) ′ − aLa( + m+ c) − (a + cp) = L ′′+ (2a + c + m) L′ + a ( a + c + m) L = (a + c ) p

38. L =

(c + a ) p a ( c + m + a)

40. (a) B′ = −0.06B + 0.03L L ′ = 0.06B − 0.03L

B′ = −0.06B + (0.06B − L′ ) B′ = − L ′

B (t )= −L (t) + C B (t )+ L (t ) =C (b) 3 mg

− =

36. L ′′=− ( a + m ′)+Lc M ′

m) L′ + cmL− L′′=− (a + m+ a + cL

44. C1 =

qD pD ; C2 = rVb rVB C2 D

2) C1(1 C +C

Exercise− Set 9.4 ⎡ x ⎤ ′ ⎡0 1⎤⎡x⎤ 2. ⎥⎢ =⎢ ⎥ ⎢ y⎥ ⎣2 ⎣y⎦ 0⎦⎣ ⎡ x ⎤ ′ ⎡− 1 3⎤⎡x⎤ 4. ⎢ ⎥ = ⎢ ⎥ ⎢ y⎥ ⎣y⎦ ⎣3 1⎦⎣ 6. x′=2x−3y and y′=−x+5y 8. x′=4x and y′=2x−3y

⎢ (−e2t + 2e−t )′ ⎤⎥

⎡x⎤′ ⎡ ⎡−2e2t −2 e− ⎤ 10. ⎢ =⎢ ⎥=⎢ e ⎥ y⎥ ⎣−4e2t+ 2 t ⎦ e3t+2 e2 ⎣ ⎦ ⎢ − ⎡−4 −3⎤ ⎡ −e 2 t + 2 e−t ⎤ t t2 e−t ⎡−2e2 − 2⎤⎥ ⎥=⎢ ⎥=⎢ ⎢ t ⎣− 2 e 2 t− 2 e−t⎦ ⎣6 e 2 ⎣−4 + ⎦e−t 5⎦ sint

⎣(−2

t)′⎥

x′ ⎡⎢(−3 sint + cost)′⎤ ⎡−3 cost − ⎤ ⎥ =⎢ 12. ⎡⎢ y⎥ = ⎥ − ⎥ ⎢ 2 cost sint ⎣ ⎣ ⎦ ⎦ 2 sint + cost)′ ⎥ ( ⎡ ⎢⎣ ⎦ ⎢ − 1 −2 ⎤ ⎡ −3 sint cost⎤ ⎡− sint− 3 cost ⎤ ⎣ + ⎥=⎢ ⎥=⎢ 1 2 sint+ cost ⎦ ⎣− sint+ 2 cost⎥⎦ 1⎦

438

ANSWERS

2t 3t ⎤ ⎡C1 2t + 2Ce 2 14. ⎢ e 1e ⎥ 3t ⎥ − 2Ce 2 ⎢C ⎣ ⎦ 3t 3t ⎡C1e− + 2C 2 e ⎤ 16. ⎢ ⎥ −3t 2 ⎥⎦ ⎢⎣ C1e + + Ce 3 t + − 8t 2C1 −4t 2Ce− ⎤ 2 18. ⎡ e ⎢ 2C1e −4 t C 2 e−8t ⎥ ⎥⎦ ⎢⎣ sin t+C2 cost ) 20. x = e−t (C1

y=

28. The origin is an asymptotically stable node.

(−C1+ 2C 2) e−tsint + 2

(C1 − C2 ) e−tcost 2

30. The origin is an unstable node.

22. x=Ce−12t+Cte−22t

et y =(−3C 2+C ) −2C 2tet 1

24. The origin is an unstable saddle points.

32. The origin is an unstable saddle point.

26. The origin is an unstable node.

34. The origin is an asymptotically stable node.

ANSWERS TO EVEN-NUMBERED EXERCISES 439

Let r = aand r =b, a,b ≠ 0be the two 42. eigenvalues , then

36. The origin is an asymptotically stable spiral point.

(r − a)(r − b)= 0 r 2 − (a + b) r + ab = 0 x′′− (a + b) x′ + abx= 0

We have a matrix with a trace of a+band a determinant of abwhich means we can write the matrix as follows:

⎡a 0⎤ ⎢ 0 b⎥ ⎣ y′=bymeans the origin is Then, x′=ax and ⎦ the only equilibrium point.

38. The origin is an asymptotically stable improper node.

⎡C1e−2t− C 2e2t − C3e4t ⎤ ⎢ 1 ⎥ 2t− C2e2t + C3e4t ⎥ 44. ⎢ −Ce− ⎢⎣C1 e− 2 t C 2 e2t C 3 e4t ⎥ + + −3t ⎦ ⎡− 3 1e−9+t 3C 2e

⎤ 1 Ce− 9t+ C2e−3t + C3e3t ⎥⎥ C t 2 C 2 e2t − 2 C 3 e4t ⎥ ⎣C1 e− 9− ⎦ ⎢ ⎢⎢ 5 ⎡ −C1e−t− C 2t ⎤ ⎢ ⎥ + 2t 48. ⎢C 2et C3e ⎥ ⎢C1e−5t C 2et+C3e 2t ⎥ − ⎣ ⎦ 46.

40. (a) (r−λ)(r−λ)−0=0

λ=r repeated

⎡0 0⎤⎡x⎤ ⎡0⎤ ⎥ ⎢ y⎥=⎢ ⎥ ⎢ ⎣0 0⎦⎣ ⎦ The above is true for any vector.

(b) x=Cert , y=Cert 1 2 (c) Note that x = 0is a constant solution y ert= x = C1 C 2 y C2 x x = mx Thus, = ⇒y= C1 C 2 C1 (d) Left to the student. (e)

50. Left to the student.

1 52. r = −8,r = −5, and r = − .

2 54. As time goes on, the amount of carbon-14 will continue to decrease until it eventually vanishes

Exercise Set 9.5 2. (0,0is) an asymptotically stable node; (3,2 ) is an unstable saddle point.

4. (1,2 ) is an unstable spiral point;

(2,1is) an unstable saddle point.

6. (2,1is ) an unstable spiral point;

⎛5 1 ⎞ ⎜ 2 , 2 ⎟ is an unstable spiral point. ⎝ −⎠ 8. (−2, 8 ) is an unstable saddle point; 2, − 8is ) an asymptotically stable spiral point.

(

440

ANSWERS

10.

(−3,e ) is an unstable spiral point; (2,e2 ) is an unstable spiral point. −3

26. (0,0 ) gives an unstable saddle point; (6,3)

gives an asymptotically stable spiral point; (24,0 ) gives an unstable saddle point.

(0,0)is an unstable node; (0,50)is an28. The coexistence equilibrium points are close for asymptotically stable node; (50,0)is an the two sets of exercises. Most of the equilibrium points in the inhibited asymptotically stable node; (12.5,12.5)iscoexistence an

12. (a)

unstable saddle point.

(b) The stable equilibrium point (0,50) indicates that only the second species survives, while the stable equilibrium point

prey growth model are stable and while in the uninhibited model they were centers.

30. (a) Whenx=0and y=0, then ax−by=0and

cx =0 −dy+ c 2 (50,0)indicates that only the first species x= x 2 +1 65 ,y= 65 survives. (b) 9 12 14. (a) (0,0)is an unstable node; (0,100)is an (c) Left to the student.

asymptotically stable node; (30,10)is an 32. (a) Exercises 13 and 16 unstable saddle point; (100,0)is an (b) Exercises 12, 14, and 15 asymptotically stable node.

34. The term represents the amount for the prey that (b) For the equilibrium point (0,100)only the

second species survives while for the

is over the “normal” or expected value for that species

equilibrium point (100,0)only the first species 36. The coexistence equilibrium point is (2,4.8) survives.

100 ⎞ ⎟ is an unstable saddle point; 16. (a) ⎛⎜0, 3 ⎠ ⎝ ⎛50 ⎞ ,0⎟ is an unstable saddle point; ⎜ ⎠ ⎝ 3

which is an unstable saddle point. ⎛r 38. (a) (0,0 ) and ⎜ , p⎞⎟ ⎝s q⎠

(b) (0,0)gives an unstable saddle point;

⎛ r .p ⎞ (12.5,31.25)is an asymptotically stable node ⎜ , ⎟ gives a center.

(b) The stable equilibrium point indicates there is coexistence.

⎝ s q⎠

40.

18. (a) x = 6.4, y = 5.2 (b) The new equilibrium point is (6.4,5.2)and is an unstable node. It is different from

(8,4)due to the addition of the new predation

rate0.01xwhich is the increase of fishing for the first species.

20. (0,0) gives an unstable saddle point; (2,1.25 ) gives the center.

y (−r + sx) d = x ( p − qy) y p − qyd = −r + sx dy dx y x x

p

r

∫y − qdy=∫ − x + sdx

p ln y − qy = rlnx + sx + C pln y− qy+ ln r x −sx = C 42. x=0.57990238 and y = 6.6958675

(0,0) gives an unstable saddle point; 44. x=0.74988795 and y = 5.9118767 (6,4) gives the center.

46. Left to the student. gives an unstable saddle point ; 24. (0,0) (2,1)gives an asymptotically stable Exercise spiral Set 9.6 x 2. (1 point( ; 10,0)gives an unstable saddle point . ≈ 12.41175 y ) ≈ 12.27048 (1 ) 22.

ANSWERS TO EVEN-NUMBERED EXERCISES 441

4. x (2)≈ 14.337

12. (a)

y (2)≈ 14.20849

6. x (3)≈ −128.29883

y (3)≈ −239.83185

8. (a)

(b)

(b)

14. (a)

10. (a)

(b)

(b)

(c) About 6.9 years.

442

ANSWERS

16. (a)

(c) The equilibrium points are approximately 260.6778, 35.4579, 1768.2844 respectively.

22.

(b)

(0,0)is an unstable node; (0,50)is an asymptotically stable node; (12.5,12.5)is an unstable node; (40,0)is an asymptotically stable node.

24.

(0,0)is an unstable saddle point; (5,1)is a center.

26.

(0,0)is an unstable saddle point; (2.5,1.75)is a stable spiral point.

(c) About 9.5 years.

18. S (10 ) ≈ 2.13429, X10( 20. (a)

) ≈ 0.04225 T (35) ≈ 822.87647,

I (35 ) ≈ 100.10859,V 35(

(b)

) ≈ 4187.11131

ANSWERS TO EVEN-NUMBERED EXERCISES 443

Chapter 10 Exercise Set 10.1

54. Exercise 52 asked for the probability of no flips being heads, while Exercise 53 asked for the probability of at most 4 flips begin heads.

56.

1 216

2. 0.20

58. 0

4. 0.0

3 60. 4 29

6. 0.26 8. 0.94 10. 0.48 12. (a) Events A and B are not independent. (b) Events A and B are disjoint.

14. (a) Events A and B are independent. (b) Events A and B are not disjoint.

16. (a) Events A and B are independent. (b) Events A and B are not disjoint.

18. (a) Events A and B are not independent. (b) Events A and B are not disjoint.

62. 36

Exercise Set 10.2 1 2. 3 4.

6.

20. 0.2 22. 0.8

8.

24. 0.6 26. 0.05 28. 0.15 30. 0.04

10.

0 2 9 1 9 1 8 3

12. 8

32. 0.16

1 221

34. 0.3

14.

36. 0.8

16. 0.144796

38. 0.7 2 40. 15

7 42.

30

44. 0.12 46. 0.21 48.

125 216

50. Exercise 48 asked for the probability of no sixes rolled, while Exercise 49 asked for the probability of at most 2 sixes rolled.

52.

1 32

9 16 1 (b) 8 1 (c) 16

18. (a)

20. (a) 0.824 0.818 (b) P ( D + andT + ) = 0.3296 , (c)

P (T + ) =0.4388 ,

P (D + T + ) = 0.7511

444

ANSWERS

22. (a) 0.414 (b) 1.0 (c) P( D +and T + ) = 0.00497 ,

28.

PT + ) = 0.00497, PD ( + T +) = 1

24.- 26. Left to the student.

0.5, 28. PDD = ) Pdd ( ) = 0.25, P(Dd ( )= 0.25 30. 0.875 0.5

Exercise Set 10.3 2. 10

1

Not

4. possible. 1 6. 792

Not

8. possible.

10. 0.29663 12. 0.33554 14. 0.09842 16. 0.08113 0 18.

20.

22.

24.

32. 0.75 34. 0.736776 36. 0.31146 38. 0.15217 40. 0.216 42.

44. (a) 0.00115 (b) 0.31494

46. P ( X = CC ) = 0.1681, P ( X=CG) = 0.4838 ,

P ( X = GG ) = 0.3481

48. 0.0328 50. 0.095 52. (a) X = 1, 2, 3, 4, 5, 6 (b)

P ( X=i)=1, for i = 1,2,3,4,5,6 6

(c)

26.

54. (a) X = 0,1 (b)

P ( X= 0)= 13 , P ( X= 1) = 23

ANSWERS TO EVEN-NUMBERED EXERCISES 445

(c)

28

aE ( X)

a S D( X )

.

30

Exercise Set 10.5 .

2.

1

4.

0.3125

the

6.

0.14645

58. student. Left to

8.

0.31662

56. Left

to

1

10. 0.63212

60. the student. 35 1 62.

12.

⎛ n⎞ 64. ⎜ ⎟ ⎝k⎠

14.

66. (a) SS, SF , FS , FF (b) 2 (c) 2

1 400

16. 0.02695 18. 0.13534 20. 4.64159 22. 2.14597

68. 1

24. 3.52636

70. 0.01281096

26. Since f ( xis ) negative at x=−1, we cannot

make a probability density function of

f(x) on[−1,4].

Exercise Set 10.4 2. 2.6

; SD ( X=) 1.496663 4. Var ( X ) = 2.24

28. Since f ( x) is negative at x= −2 and x = −1, we cannot make a probability density function of f ( x) on− [ 2,2].

6. (a) 2 3 (b)2 3

30.

8. (a) 0.75 (b)0.75

32. (a)

12. 0.75

E ( X ) = 5, SDX (

) = 1.58114

E ( X ) = 9, SDX (

) = 2.4

18

E ( X ) = 70, SDX (

) = 4.582576

.

−0.4

20

0

.

−2.75

16.

22 .

24 .

26 .

512

P( = )

(b) X

10. 0.61237 14.

15

E ( X ) =4.152, SD ( X=) 1.64785

34. (a)

x

1 ln(4 ) x

= 0.79248

P P (( X≤ ) 2 cos2 )(x X 3 = π

(b) P ⎛⎜ X ≥ π ⎞ 0.590845 4 ⎟⎠ = ⎝ = ) x 2( x−sin 36. (a) P ( X= x) = − x)e− (b) P ( X≤x 3) =0.22028

=

38.

3 7

446

ANSWERS

72

1 40. 4 42.

) SD ( X )=2.64575 E ( X =) 3.5, Var (=X7,

) 44. E ( )X =− 1.25, Var ( X = 0.6377, SDX ) = 0.79844

(

) 46. E () X = 4.3125, Var(=X1.727343750, SDX( ) = 1.314284501

48. )E ( X 0.5848739496,

Var X ( = 0.1011044418, SDX ( ) = 0.3179692466

50. E ( X ) = π 2 − 1, Var(X=) 0.141592654,

SDX (

) = 0.3762879934 ) 52. E ( X ) = 2.09726, Var(≈X0.17605, ) = 0.41958

2 4 , Var ( X )=2, SD ( X )= 9 3 3

= 56. E ( X ) 1.5, Var ( X )=49, 12 SD ( X=) 2.02073 58.

74 0.013754 . 0.171797 76 13.277 . 16.812 78 80 2. .

)

54. E ( X ) =

. 0.013262

.

Exercise Set 10.6

=

SDX (

0.069663

E ( X )= 0.0015, Var ( X≈) 0.00000008333333, SD ( X) = 0.000288675

0.2158

4.

82 6. .

0.1014

8.

0.9810

0.1515

10. 0.7762 12. 0.57744 14. 4 2.3875 16. 0.0856 18. 0.0024 20. 0.0025 22. 0.9179 24.

26.

28.

1 10 1 10

3 60. 4

30.– 48. Left to the student.

62. 18

52.

50.

1 64. 2

54.

66. Var ( x =) ∫ (x −µ)2x fdx( ) b

a

(

x −f(x)−2µxfx =∫ ( ) a b

2

dx +µ2f(x))

b b b ⎛ x 2 f (x)dx−2µ∫ xf f ( x ⎞⎟)dx =∫ ( x)+µ2∫ a a a ⎠ ⎜ ⎝ X 2 E = − 2µ(µ) + (µ2 ) 1

( ) = E (X 2 )−µ2 68. (a) 0.217098 (b) 0.24449

70. Answers vary

E(X )=3.5, Var ( X )=3.5, SD (X) = 1.8708

EX)=1 (

5

,Var(X)= 1 ,SD(X)= 1 25

5

56. 0.5654 (a) 0.1465 (b) 0.0993 (c) 0.01375 58. (a) (b) (c) (d) (e)

0.3935 0.4493 0.1353 0.2052 0.4422

60. (a) E(X)=13.1,SD(X)=3.6194 (b) 0.08389

ANSWERS TO EVEN-NUMBERED EXERCISES 447

14. (a) µ= −3 (b) σ= 3 (c) 0.7486 (d) 0.9901 (e) 0.6255

62. (a) e−λc (b) P ( X> a + bX > a )

=

P(X > a + b and X > a) P ( X > a)

( ) = e−λa+b

= =

e−λa e−λb

P ( X> b)

(c) The events are independent.

64. 1 66. The mean and median are equal.

68. (a) m =

ln 2

λ

(b) The median is larger than the mean 1 EX( ) = . This makes sense since at

λ

ln 2

m= , for exponential given, the value of λ f ( x) =0.5.

70. Binomial: P( X = 3 ) = 0.180628

Poisson:P (X = 3 ) = 0.180447

16. (a) 0.0287 0.8159 (b) 0.3056 (c) 18. – 32. Left to the student.

34

0.7049

.

0.9525

36

0.1292

.

(a) 0.9901 (b) 0.0465 (c) 0.0038

38

.

42. 0.7527 40 44. 0.0150 .

46. 0.2327

48. (a) − 0.5 (b) 0 (c) 1.67 50. 0.0166

52. P ( X ≤− 1.96) 54.

56.

72. Binomial: P( X = 5 ) = 0.175555

58.

74. Binomial: P( X = 3 ) = 0.140367

62.

Poisson:P (X = 5 ) = 0.175467 Poisson:P (X = 3 ) = 0.140374

60.

64.

66.

68.

Exercise Set 10.7 2.

0.8529

4.

0.8808

6.

0.0245

8.

0.0906

10. 0.5 12. 0.9989

70.

0.4247 Critical value is 1.64.

Critical value is 2.58.

Critical value is −2.33.

1.3333 0.8474576271 −1.39

1.037344398

E ( X )=9.28476812 SD ( X )=2.649006623

72. The reason is the symmetric nature of the

Normal(µ,σ) distribution about 0.

74. (a) 0.46517 (b) 0.3944 (c) 0.07077 76. (a) 0.39508 0.3944 (b) 0.000677 (c)

448

ANSWERS

78. n = 2 n→ n= 4 n→ =6

→ =8 →

0.842881 0.852459 0.854219 0.854834

449

EXTENDED LIFE SCIENCE CONNECTION

Chapter 1

2 3 . a) y = −0.000307x3 + 0.031536x + 0.50938x+315.8660781

1. a) y = 1.343450619 x+ 311.3019556

b) b)

370 360

360

350

350

340

340

330

330

320 0

320

0

10

20 x

30

40

c) 352.95; 366.38. The estimates seem to be reasonable when compared to the data. d) 379.82; 433.56. The estimates seem to be reasonable when compared to the data. e) 140.5; The carbon dioxide concentration will reach 500 parts per million sometime in the year 2099.

10

20x

30

40

c) 352.82;368.60. The estimates seem to be reasonable when compared to the data. d) 383.15; 392.02. e) Under this model, the carbon dioxide concentration will never reach 500 parts per million.

4. a) 370 360 350

2. a) y = 0.0122244281x2 + 0.8300246407 x+

340

314.8103665

330

b)

320 0

370

10

20x

30

40

b) The graph represents a steady increase in the concentration of carbon dioxide. c)

360 350 340

0.4 330 0.2

320 0

10

20x

30

40

c) 352.29; 369.39. The estimates seem to be reasonable when compared to the data.

0

46

47

t

48

49

50

–0.2 –0.4

d) 388.94; 491.57 e) 161.63; The carbon dioxide concentration will reach 500 parts per million sometime in the year 2120.

d) The graph shows an oscillating behavior for the concentration of carbon dioxide.

450

ANSWERS TO EVEN-NUMBERED EXERCISES

e)

Chapter 2

378 376

1. 0.25 2. a)

374 372 370 45

46

47

t

48

49

50

f) The graph behavior shows that there is a periodic fluctuation in the concentration of carbon dioxide.

h F+ (h) b) h F − (h)

CUBIC MODEL: This model also resembled the original data set’s scatter plot indicates that there is a level after which the concentration of carbon dioxide will not increase. It is the only model that did not allow the concentration level of carbon dioxide to reach 500 parts per million. This model suggests that as time increased the concentration of carbon dioxide will begin to decrease indefinitely.

PERIODIC MODEL: This model, as the other, modeled the data set to a very good degree of accuracy. It was the only model that allowed for oscillating behavior in the future, which is more likely to happen than what the other models suggested.

1 4

1 8

1 16

0.242641

0.24621

0.24808

0.24903

1 2

1 4

1 8

1 16

0.258342

0.25403

0.25098

0.25198

3.a) Forward difference: 301. Backward difference: 339.5. b) Forward difference: 294. Backward difference: 315. It seems that the answers from part a) are more accurate since they use points closer to the point under consideration (March 19) and therefore represent the slope of the tangent line more closely than points away from the point corresponding to March 19.

5. LINEAR MODEL : This model is the easiest mathematically to compute and explain. It does resemble the scatter plot of the original data sets. Under this model, the concentration levels of carbon dioxide will increase with time indefinitely.

QUADRATIC MODEL: This model also resembles the original data set’s scatter plot. At relatively small values of tit allows for a longer time for the increase in the concentration level of carbon dioxide since it is a parabola.

As time increases though the level at which the concentration of carbon dioxide will increase will be quicker than the linear model.

1 2

4. Fc = 1 ⎡ F+ (h)+ F−(h) ⎤⎦ 2⎣ 1 ⎛ f (h) − f (0) f(0)− f (− h)⎞ ⎟ = ⎜ + ⎟ 2⎜ h h ⎠ ⎝ =1 f f (h) − f (0)+ f (0)− (−h)) 2( = 1 f h − f − h) ( ⎞ ⎛ ( ) 2 ⎜⎜ ⎟⎟ f h ⎝ ⎠ f (h) − (− h) = 2h

5. Since F c(h)cover more of the function domain

containing x=0than does F+(h)orF−(h), c(Fh)will be closer to then it makes sense that f′(0)than either of them.

6. h Fc (h)

1 2

0.25049

1 4

0.25012

1 8

0.25003

1 16

0.250005