V I ExerciseAnswers

7/28/2019 V I ExerciseAnswers

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Chapter A

Suggested Solutions to Volume I Exercises

The probability that all of these answers are correct isnegligible, vanishingly small (zero). I would appreciate yourinforming me of incorrect answers. Thank You, JLC

[email protected]

A.1 Chapter 1 Answers.Explore 1.1.3. B5

.= 2.87

Explore 1.1.4. B4 =

53

4B0

Exercise 1.1.1.

a. B3 = 1.53 4 c. B3 = 1.053 0.2 e. B3 = 1.43 100

Exercise 1.1.2

a. Bt = 4 1.5t c. Bt = 0.2 1.05t e. Bt = 100 1.4t

Exercise 1.1.4.

Exercise 1.1.5.

Step 1. Preliminary Mathematical Model: Description of bacterial growth.

Step 2. Notation.

Step 3. Derive a dynamic equation. Density change, Bt+1 Bt and a graph of Bt+1 Bt vs Btare shown in Figure A.3. Because the first four points fall close to a line, we favor them. The slope of the

668


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CHAPTER A. SUGGESTED SOLUTIONS TO VOLUME I EXERCISES 669

t Bt Bt+1 Bt0 1.99 0.671 2.68 0.952 3.63 1.26

3 4.89 1.744 6.63 2.305 8.93 3.176 12.10

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3

3.5

Bacterial density

Densitychange

0 16 32 48 64 80 960

4

8

12

Time

Bacterialdensity

Figure A.1: Data and graphs for Exercise 1.1.4 a. The graph on the left is Bt+1 Bt vs Bt and slope ofthe line is 0.35. The original data (filled circles) and data computed from Bt = 1.99 1.355t are markedwith +s in the right hand graph..

t Bt Bt+1 Bt0 22.1 2.31 23.4 2.72 26.1 1.43 27.5 3.04 30.5 3.95 34.4 2.26 36.6

0 5 10 15 20 25 30 35 400

0.5

1

1.5

2

2.5

3

3.5

4

1 0 1 2 3 4 5 6 720

22

24

26

28

30

32

34

36

38

40

Figure A.2: Data and graphs for Exercise 1.1.4 c. The (ragged!) graph on the left is Bt+1 Bt vs Bt andslope of the line is 0.080. The original data (filled circles) and data computed from Bt = 22.1 1.08t aremarked with +s in the right hand graph.

.

line shown is 0.099/0.141.

= 0.7. Thus approximately

Bt+1 Bt = 0.7Bt

pH 7.85Time Time Population Density(min) Index Density Change

0 0 0.028 0.01916 1 0.047 0.03532 2 0.082 0.05948 3 0.141 0.09964 4 0.240 0.14180 5 0.381

0 0.05 0.1 0.15 0.2 0.250

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

Bacterial density

Densitychang

e

Figure A.3: Data for Exercise 1.1.5. A Time Index and Density Change have been added to the originalequation. The graph shows Density Change vs Density and the line passes through (0,0) and has slope

0.7.

Step 4. Enhance the mathematical model of Step 1.

Enhanced model. Approximately 70 percent of the cells divide in each time interval.


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pH 7.85Time Time Population Computed(min) Index Density Density

0 0 0.028 0.028

16 1 0.047 0.04832 2 0.082 0.08148 3 0.141 0.13864 4 0.240 0.23480 5 0.381 0.398

0 16 32 48 64 800

0.1

0.2

0.3

0.4

Time

Bacteria

ldensity

Figure A.4: Data for Exercise 1.1.5. The equation Bt = 0.028 1.7t was used for Computed Data. Thegraph shows observed density vs time (filled circles) and computed density vs time (+s).

Step 5. Compute a solution to the dynamic equation.

Bt = B0 1.7t = 0.028 1.7t.

Step 6. Compare predictions of the model with the original data. A table of the original

and computed values and a graph comparing the data is shown in Figure A.4. The computed values are

close to the observed values except, perhaps, for the last value.

Exercise 1.2.1.

a. Bt = 1000

1.2t B100 = 1000

1.2100.

= 82, 817, 974, 522

c. Bt = 138 1.5t B100 = 138 1.5100 .= 5.6 1019

e. Bt = 1000 1.2t B100 = 1000 1.2100 .= 82, 817, 974, 522

g. Bt = 1000 0.9t B100 = 1000 0.9100 .= 0.026

Exercise 1.2.2. B4 = (1 + r)4B0

(a.) Bt = 0.2t

50 B40

.= 73, 489

(c.) Bt = 1.05t 50 B40 .= 352

Exercise 1.2.3. Approximately 2,691,588.

Exercise 1.2.4. Approximately 0.00107 g.

Explore 1.3.1. (a.) At depth 20 m the light intensity will be 14I0.

Explore 1.3.2. Light intensity at 6 meters would be 400 0.93 .= 291.6 w/m2.

Exercise 1.3.1.


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t Bt Bt+1 Bt0 3.01 -0.461 2.55 -0.412 2.14 -0.323 1.82 -0.344 1.48 -0.265 1.22 -0.196 1.03

0 0.5 1 1.5 2 2.5 30.5

0.45

0.4

0.35

0.3

0.25

0.2

0.15

0.1

0.05

0

Light Intensity

LightIntensity

change

0 1 2 3 4 5 60

1

2

3

Depth

LightIntensity

Figure A.5: Data and graphs for Exercise 1.3.1 a. The graph on the left is Bt+1 Bt vs Bt and slope ofthe line is -0.16. The original data (filled circles) and data computed from Bt = 3.01 0.84t are markedwith +s in the right hand graph..

t Bt Bt+1 Bt0 521 -2041 317 -1282 189 -703 119 -444 75 -305 45 -176 28

0 50 100 150 200 250 300 350 400 450 500 550220

200

180

160

140

120

100

80

60

40

20

0

Light Intensity

Lig

htIntensity

change

0 1 2 3 4 5 60

100

200

300

400

500

Depth

LightIntensity

Figure A.6: Data and graphs for Exercise 1.3.1 c. The graph on the left is Bt+1 Bt vs Bt and slope ofthe line is -0.40. The original data (filled circles) and data computed from Bt = 521 0.6t are markedwith +s in the right hand graph..


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Exercise 1.3.2 Id.

= 0.842 0.76dw/m2.

Exercise 1.3.4. A formula for the solution is

Id = I3 0.5(d3)/3 = 256 0.5(d3)/3

Exercise 1.4.1 a. 1, c. 10, d. 0.10034

Exercise 1.4.2. From Equation 1.16,

t2 =log2

log B=

(log2 2)/(log2 10)

(log2 B)/(log2 10)=

1

log2 B.

Exercise 1.4.4. a. t1/2 = 1 c. t1/2.= 3.010 e. t2

.= 0.1435 g. t1/2.= 1.553 i. t1/2

.= 1.505

Exercise 1.4.5. a. P = 50 1.10718t

Exercise 1.4.6. a. A10 = 1 (1. + 0.06)10 = 1.79 c. For R=4 percent the Rule of 72 asserts that thedoubling time should double in 72/4 = 18 years. The doubling time is actually 17.67 years, within 2

percent of 18.

Exercise 1.4.7. a. d1/2 = 10 m. c. 10 m

Exercise ??. b. 269 minutes.

Exercise 1.5.3 0 = 0.28, = 1.24.

Exercise 1.5.4 For

A0 = 4, At+1 At = 5

At, A8 = 370 and A9 = 465.

These exceed the observed data, 326 and 420, respectively, and a value of k

2

smaller than 5 should

be chosen.

Exercise 1.6.1.

At = 2 0.95t A20 = 0.717

Exercise 1.6.2.

St+1 St = Amount added to serum Amount removed from serum

= 0.10It 0.15 St

Exercise 1.6.3. a. Pt.

= 71 0.78t


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Mezlocillin1 g injection

Time Mezl.Index conc.

g/ml0 711 562 453 334 25

A

0 10 20 30 40 50 60 70 8018

16

14

12

10

8

6

4

2

0

Mezlocillin concentration mg

Concentrationc

hange

mg

B

0 1 2 3 40

20

40

60

80

Time index (5 min intervals)

Mezlocillinconce

ntrationmg

Figure A.7: Graphs for 1 g injection of mezlocillin. A. Change in serum mezlocillin concentration vsmezlocillin concentration. The slope of the line is -0.22. B. Mezlocillin concentration vs time index.Original data (filled circles) and data computed from Pt = 71 0.78t (+s).

Exercise 1.6.4. Ct = 15(1 k)t where k is a permeability constant of the membrane.

Exercise 1.6.5.

ht+1 = A + (1 B)ht,where A = KP0/(r

2) and B = K 980 /(r2).

Exercise 1.7.1. a. W0, , W4 = 0, 1, 1.2, 1.24, 1.248. E = 1.25,

Wt = 1.25 (0.2)t 1.25 t = 0, 1, W100 = 1.25

T1/2 = 0.43

c. W0, , W4 = 0, 100, 120, 124, 124.8. E = 125,

Wt = 125 (0.2)t 125 t = 0, 1, W100 = 125

T1/2 = 0.43

e. W0, , W4 = 0, 10, 10.5, 10.525, 10.52625. E = 100.95.

= 10.52631,

Wt = 10.52631 (0.05)t 10.52631 t = 0, 1, W100 = 10.52631.T1/2 = 0.23

Exercise 1.7.2. a. W0, , W4 = 0, 1, 0.8, 0.84, 0.832. E .= 0.83333,

Wt = 0.83333 (0.2)t 0.83333 t = 0, 1, W100 .= 0.83333

c. W0, , W4 = 0, 100, 80, 84, 83.2. E .= 83.333,

Wt = 83.333 (0.2)t

83.333 t = 0, 1, W100 = 83.333e. W0, , W4 = 0, 10, 9.5, 9.524, 9.52375. E .= 9.25238,

Wt = 9.25238 (0.05)t 9.25238 t = 0, 1, W100 .= 9.25238


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Exercise 1.7.3. a. W0, , W4 = 0, 1, 0 , 1, 0. The sequence continues to alternate between 0 and1.

c. W0, , W4 = 0, 1, 2 , 3, 4. The sequence is the sequence of non-negative integers.

e. W0, , W4 = 0, 1, 3, 7, 15. The sequence is Wt = 2t

1.Exercise 1.7.4. Lt = 0.5 0.5 0.95t, t = 0, 1, 10.

Exercise 1.7.5. Suppose an amount A of pollutant is released on day t and 10 percent of it is spread

throughout the lake (or at least to the outlet of the lake) by day t + 1, 10 more percent is spread by day

t + 2, and so on. We might assume that

Ct =Wt10 + 1.0 A + 0.9 A + 0.1 A

V=

Wt10 + 4.5A

V.

This would yield

Wt+1 Wt = 100 F Ct Wt+1 Wt = 100 F Wt10 + 4.5AV

Because of the time lag, t 10, we do not have a good way to solve this problem. However, after 10 daysof this, the increase each day in the amount of pollutant throughout the lake between day t and t + 1 is

simply A.

Exercise 1.7.6. Pt = 50

50

(0.8)t.

Exercise 1.7.7.

P0 = 0.8

Pt+1 = Pt + K (2.4 Pt), K = 0.067

Pt = 2.4

1.6

(0.933)t

Pt = 1.6 when t = 10 minutes. The half-life of nitrogen flow to this muscle is 10 minutes.

Exercise 1.8.1

a. Bt = 25 + (100 + 25) 1.2t, B100 .= 1.035 1010c. Bt = 200 + (138 200) 1.05t, B100 .= 9268e. Bt = 25 + (100 25) 1.2t, B100 .= 6.211 109g. Bt = 100 + (100 + 100) 0.9t, B100 .= 99.995

Exercise 1.8.2 Pt = P0 + t b

Explore 1.9.1. The missing data point is (1/302, 0.061). (0,0) is not a data point of the experiment.


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Chapter Exercise 1.11.1 R3 = 1.23.

Chapter Exercise 1.11.2 P3 = 0.844.

Chapter Exercise 1.11.3 R3 = 0.432.

Chapter Exercise 1.11.4. At = A0 (1 K)t.

Chapter Exercise 1.11.5. Tt = 17 + 20 (0.992)t.

Chapter Exercise 1.11.7. Ph = 0.99874h for 0 h 548.6, altitude = 10h meters.

A.2 Chapter 2 Answers

Exercise 2.1.1. We may say that between 22 and 28 C incubation temperatures about 10% of the

embryos will develop to be male and outside that range the embryos will all develop to be female. We

may express this as

Percent Female =

0 if Temp < 22 C10 if 22 C < Temp < 28 C0 if < 28 C < Temp

Exercise 2.2.1. The second and third tables listed in Exercise Table 2.2.1 are functions.

Exercise2.2.2. a. The number of coyotes is the independent variable, the number of rabbits is the

dependent variable, and the number of rabbits decreases as the number of coyotes increases.

Exercise 2.2.3. The two functions are the first and third columns and the second and third columns.

The implied domain of the first function is all times in the interval 0 to 80 minutes. Some people will

consider Time Index to be limited to integers in which case the domain of the second function is {0, 1, 2,3, 4, 5 }.

Exercise 2.2.4. The graph in C is a simple graph. The graph in D is not a simple graph. Two maximal

subgraphs of D are shown in Figure A.8

Exercise 2.2.5. There are eight maximal subgraphs.

Exercise 2.2.6. a. Yes. c. Yes.

Exercise 2.2.8. a. The domain of a bird identification guide book is the collection of all birds, the range

is a set of order-family-genus-species names.

Exercise 2.3.1. e. The range is the set of numbers 1/4.


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D1

2 1.5 1 0.5 0 0.5 1 1.5 22

1

0

1

2

D2

2 1.5 1 0.5 0 0.5 1 1.5 22

1

0

1

2

Figure A.8: Exercise 2.2.4. Graphs of the spiral in Exercise Figure 2.2.4 D are shown as dashed curveswith maximal subgraphs shown as solid overlay. In D2 the point of D1 on the y-axis in the upper spiralhas been omitted and the point on the second spiral arc and the y-axis is included.

Exercise 2.3.2.

F(1 + 2) = F(1) + F(2), F(0 + 4) = F(0) + F(4)

Exercise 2.3.5. For F(x) = x2 + x, a. 9, c. b+a+1.

Exercise 2.3.6. iii. For F(x) = x3,

F(5) F(3)5 3 =

53 335 3 = 58,

F(3 + 2) F(3)2

=(3 + 2)3 33

2= 58,

F(b) F(a)b

a=

b3 a3b

a=

(b a)(b2 + ab + a2)b

a= b2 + ab + a2

F(a + h) F(a)h

=(a + h)3 a3

h=

a3 + 3a2h + 3ah2 + h3 a3h

= 3a2 + 3ah + h2

Exercise 2.3.7. Equations of two such functions are

A. y = x2 for 2 x 2

B. y =

x2 for 2 x 0

x2 for 0 x 2and their graphs appear in Figure A.9

Exercise 2.3.8. The implied domain of

f(x) =1 + x2

1 x2 is all x = 1 and = 1.

Exercise 2.4.1. Graphs of F and P3 are shown in Figure A.10. The relative error in approximating

F(2) with P3(2) is

Relative error = F(2) P3(2)

F(2) =

2

5

8+

15

32 2 5

128 22 + 1

512 23

2

=1.414 1.4221.414

= 0.0057


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A

2 1.5 1 0.5 0 0.5 1 1.5 24

3

2

1

0

1

2

3

4

B

2 1.5 1 0.5 0 0.5 1 1.5 24

3

2

1

0

1

2

3

4

Figure A.9: Graphs of two functions contained in the set A of all number pairs (x, y) for which y2 = x2

and 2 x 2.

0 1 2 3 4 5 6 7 8

0

0.5

1

1.5

2

2.5

3

F(x) = x1/2

P3(x) = 5/8 + (15/32) x (5/128) x

2

+ (1/512) x3

Figure A.10: Graphs of F(x) =

x and P3(x) =58

+ 1532

x 5128

x2 + 1512

x3.

Exercise 2.4.2. (a.) Graphs of F(x) and P2 are shown in Figure A.11. The relative error in

approximating F(2) with P2(2) is

Relative error =F(2) P2(2)F(2)

=

3

2

5

9+

5

9 2 1

9 22

3

2

=

1.26 1.221.26 = 0.032

Exercise 2.5.2. y.

= 9.357142857 + 1.659523810x 0.259523810x2.

Exercise 2.5.3. D(T) = 1.00004105 + 0.00001627T 0.000005850T2 + 0.000000015324T3.

Explore 2.6.3. Yes.

Exercise 2.6.1. E+ S is graph d. and E S is graph a.

Exercise 2.6.3. One sequence is AAAUAUUUAGAAUUU

Exercise 2.6.4. Only c. is invertible.


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0 0.5 1 1.5 2 2.5 3

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

F(x)=x1/3

P2(x) = 5/9 + (5/9)x (1/9)x

2

Figure A.11: Graphs of F(x) = 3

x and P2(x) =59

+ 59

x 19

x2.

Exercise 2.6.5. (0,0), (0.8,2), (and 1.0,5) are three of the ordered pairs of F1

.

Exercise 2.6.6.

2 1 0 1 2 3 4 5

2

1

0

1

2

3

4

5

(1/4,2)

(1,0)

(4,2)

Figure A.12: Exercise 2.6.6 The graph of F1 for F(x) = 2x.

Exercise 2.6.10. The inverse of the function F(x) = x1 is F.

Exercise 2.6.12. a. For F1(x) =1

x+1, F11 (x) = 1 + 1x .

c. For F3(x) = 1 + 2x

, F1

3 (x) = log2(x 1)e. For F5(x) = 10

x2 x 0, F15 (x) = log10 x.

Note: The domain of F15 is (0, 1] and the range is [0,).g. For

F7(x) =2x + 2x

2x > 0, F17 = log2

x +

x2 1

Exercise 2.6.13. 97,000 years.

Exercise 2.7.2.


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Rabbit Coyote--M. cunuculi - - -Coyote

abundance

Exercise 2.7.3. Area = ((1.13 + 2.4 t)/2)2.

Exercise 2.7.4. It appears that there was an abrupt increase in fitness between generations 200 and 300.

Exercise 2.7.5.

a. For F(z) = z3 and G(x) = 1 + x2, F(G(x)) = (1 + x2)3

c. For F(z) = log z and G(x) = 2x2 + 1, F(G(x)) = log(2x2 + 1)

e. For F(z) = 1z1+z

and G(x) = x2, F(G(x)) = 1x2

1+x2

Exercise 2.7.6.

. F(u) =

u G(v) = 1 v H(x) = x, F(G(H(x))) =

1 x F(u) = log u G(v) = 2v + 1 H(x) = x2, F(G(H(x))) = log(2x2 + 1) F(u) = u3 G(v) = 1 v H(x) = 2x, F(G(H(x))) = (1 2x)3

Exercise 2.7.7.

f(z) g(x) f(g(x)) Domain Rangea. 1

1+zx2 1

1+x2 < x < 0 < y 1

c. 5z log x x log 5 0 < x 0 < ye. z

1zx

1+xx x = 1 y = 1

g. 2z x2 2x2 < x < 0 < y 1

Exercise 2.7.8.

a. F(z) =

1 z G(x) = x F G(x) =

1 xF(z) =

z G(x) = 1 x F G(x) =

1 x

c. F(z) = (1 + z)3 G(x) = x2 F

G(x) = (1 + x2)3

F(z) = z3 G(x) = 1 + x2 F G(x) = (1 + x2)3

e. F(z) = 2z G(x) = x2 F G(x) = 2(x2)F(z) = 2(z

2) G(x) = x F G(x) = 2(x2)


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A

2 1 0 1 22

1

0

1

2

C

2 1 0 1 22

1

0

1

2

Figure A.13: A. Graph of f(g(x)) = 1/(1 + x2). C. Graph of f(g(x)) = xlog5.

E

2 1 0 1 22

1

0

1

2

G

2 1 0 1 22

1

0

1

2

Figure A.14: E. Graph of f(g(x)) = x for x = 1. G. Graph off(g(x)) = 2x2

.

Exercise 2.7.9. r = 3

30t4 .

Exercise 2.7.12. Q(P(x)) = 4x6 28x5 + 49x4 + 18x3 63x2 + 20

Exercise 2.7.13. See Figure A.15.

Exercise 2.8.2. F(1004) = F(200 5 + 4) = F(4) = 1.

Exercise 2.8.4. P(x) = 13 13cos x3 .

Exercise 2.8.7. a. P(t) = sin(3

t) = sin( 26

t). The period of P is 22/6

= 6.

c. The period is 2.

e. P(t) = sin(22

t) + sin( 23

t). sin( 22

t) has period 2 and sin( 23

t) has period 3. The sum has period

2 3 = 6.


Exercise 2.8.9. Subtract 2.62 from each x-coordinate and 59.5 from each y-coordinate.


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4 2 0 2 4 6 84

2

0

2

4

6

8

a

c

e

g

Figure A.15: Exercise 2.7.13. Graphs of G (solid curve) and Ga, Gc, Ge and Gg (dashed curves).

5

5

(0,0) 8 16

Figure A.16: Exercise 2.8.8 b. The graph of y = 5cos18

t + /6


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500 400 300 200 100 0800

600

400

200

0

200

400

600

Million Years Ago

ResidualsofDiversity

Figure A.17: Exercise 2.8.12. The graphs of residuals of diversity after subtracting a cubic and of H(t) =162 sin(2

62t 1.09)

Exercise 2.8.11. H(t) = 0.6 + 0.09 cos(2(t 2.3)

Exercise 2.8.12. H(t) = 162 sin(262

t 1.09) fits pretty well as shown in Figure A.17.

Exercise 2.8.13. The relative error in P5(/2) as an approximation to F(/2) is 0.00452.

Exercise 2.8.14. Because cos /2 is zero, computation of relative error of P4(/2) as an approximation

to cos /2 is hazardous. The absolute error is

|cos(/2) P4(/2)| = 0.020

Also see Figure A.18.

1

0.5

0

0.5

1

0 /4

/2

3/4

Figure A.18: Exercise 2.8.14. Graphs of the cosine function and P(x) = 1 x2/2 + x4/24.

A.3 Chapter 3 Answers.

Explore 3.1.1. The human growth rate in 1920 was approximately 7,000,000 people per year.

Explore 3.1.2. Your vote counts.


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Explore 3.1.3. No.

Explore 3.1.4. a. 12, c. 140, e. 1, g. 1, h. 0.69371,

Explore 3.1.5. If b is close to zero, sin bb is close to 1.


Exercise 3.1.4. a. Approximately -0.000365 (Kg/m3)/m.

Exercise 3.1.5. a. Between 570/10 and 1200/10 km/year. c. Approximately 1200/2 km/year.

Exercise 3.1.6. a. 3, c. , e. 39, g. 8, 1.

Exercise 3.1.7. a. (180,000 - 80,000)/1 = 100,000 (RNA copies/ml)/day. b. The patients immune

system was destroying 100,000 (RNA copies/ml)/day. c. The virus reproduced at the rate of 100,000

(RNA copies/ml)/day.

Exercise 3.1.8. a. The line is tangent to itself. b. 0.5 feet per hour.

Exercise 3.1.9. -4/3.

Exercise 3.1.10. See Figure A.19. c. [(-3.7484)+(-3.7480)]/2 (g/ml)/minute = -3.7482(g/ml)/minute.

a.

4.85 4.9 4.95 5 5.05 5.1 5.15179.8

180

180.2

180.4

180.6

180.8

Time t minutes

PlasmaPenicillin

2002

0.0

3t

b.b P(b)P(5)b54.9 -3.75214.95 -3.75014.99 -3.74864.995 -3.7484

5.005 -3.7480

5.01 -3.74785.05 -3.74625.1 -3.7443

Figure A.19: Graph and data for Exercise 3.1.10.

Exercise 3.1.11. To maintain 200 g/ml concentration the patient should be infused at the rate of 20.8

milligrams penicillin per minute.

Exercise 3.1.12. a. y 2 = 4(t 4), c. y 3 = t 2, e. y 1/2 = (1/4) (t 1),

Exercise 3.1.13. a. 12, c. 6, e. 1/4,


17/95


Exercise 3.1.14. Zero.

Explore 3.2.2. 1.7722 x 1.7727

Explore 3.2.3. No. Choose = 1. With x < 1, x2 > 2.14 which is not less than = 1.

Explore 3.2.4. Yes.

Exercise 3.2.1. a. = 0.005. c. = .03. e. = 0.07. g. = 0.0003.

Exercise 3.2.2. a. = 0.01, c. = 0.07,

Exercise 3.2.3.

a. limxa x2

= limxa (x x)

=

limxa

x

limxa

x

Equation 3.15

= a a = a2 Equation 3.11c. For n = 1, 2, 3, , let Sn be the statement, lim

xaxn = an. S1 is Equation 3.11. S2 and S3 are parts a.

and b. of this Exercise, respectively. Suppose n 1 and Sn is true. We prove that Sn+1 is true and theargument is complete.

limxa xn+1

= limxa (xn

x)

=

limxa

xn

limxa

x

Equation 3.15

= an

limxa

x

Sn

= an a = an+1 Equation 3.11

Exercise 3.2.4.

1. Suppose that L1 < L2.

2. Let = (L2 L1)/2.

3. There is a number 1 such that

if x is in the domain of G and |x a| < 1 then |G(x) L1| < .

4. There is a number 2 such that

if x is in the domain of G and |x a| < 2 then |G(x) L1| < .


18/95


5. There is a number x0 in the domain of G such that |x0 a| < 1 and |x0 a| < 2.

6.

L2 L1 = |L2 L1| = |L2 G(x0) + G(x0) L1|

|L2 G(x0)| + |G(x0) L1|

< 2 = L2 L1

The supposition that there are two limits has lead to a contradiction.

Exercise 3.2.5. a. 0, c. , e. -, g. 1, i. 4, k. Not defined.

Exercise 3.2.6. Your vote counts.

Exercise 3.2.7. a. Equation 3.14, Limit of a sum.

c. Equation 3.14, Limit of a sum.

e. Equation 3.19, Limit of xn. in Exercise 3.2.3.

Exercise 3.2.9.

a. 0 = limxa (F1(x) F2(x)) = limxa F1(x) limxa F2(x) .Therefore lim

xaF1(x) = lim

xaF2(x)

c. For all x, let F1(x) = 1 and F2(x) = 1/x. Then

limx0

F1(x)

F2(x)= lim

x0x = 0, but lim

xaF1(x) = 1 = 0.

e. 0 = limxa

(F1(x) F2(x)) =

limxa

F1(x)

limxa

F2(x)

.

Therefore limxa F1

(x) = 0 or limxa F2

(x) = 0.

Exercise 3.2.10. a. -4, c. 24, e. -4, g. 3/4, i. -1, k. -3/16, m. -5/16,

Explore 3.3.1. b. -24.6

Exercise 3.3.1. a. 2x, c. 2x, e. 12x2, g. 2x + 1, i. 3, k. 2x

, m. 0, o. x2, q. 10x3,

Exercise 3.3.2. a. 2x, c. 2x, e. 2x2

.

Exercise 3.3.3. a. Graph 1 is the F. At x = 0.1 the slope of graph 2 is negative but the y-value of

graph 1 is positive. Graph 1 is not the slope of graph 2.


19/95


c. Graph 2 is F. At x = 0.5 the slope of graph 1 is negative and the y-value of graph 2 is positive.Graph 2 is not F.

Exercise 3.3.4. The slopes in graph A are -4, -2, 0-, 2, and 4. The slopes and a graph interpolating the

slopes are shown in Figure A.20A.

A

2 1 0 1 2

4

2

0

2

4

x

F(x)

Figure A.20: Plots of the slopes and interpolating graph for Exercise 3.3.4A.

Exercise 3.3.5. a. One point of the graph of F is (0,2) and the slope of F is consistently 1. The

function F(x) = 2 + x has that property. A graph of F appears in Figure A.21A.

A

0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

x

F(x)

Figure A.21: Exercise 3.3.5A. Plot of a graph that contains (0,2) and has slope equal to 1.

Exercise 3.3.6. 750.

Exercise 3.3.7. F

is a maximum at t = 3.15.

Exercise 3.3.8. The slopes of R(t) =

t are the reciprocals of the slopes of S(t) = t2 (at corresponding

points).


20/95


Exercise 3.3.9. a. 1, c. The angle of reflection, B, is equal to the angle of incidence, A.

Exercise 3.3.10. a. For a transversal crossing parallel lines, alternate interior angles are equal.

c. Slope = 1/

a.

e. tan B = tan(C ) = tan C tan 1 + tan Ctan

=(2

a)/(a 1) 1/a

1 + (2

a/(a 1)) (1/a) =1

a

Exercise 3.4.2. I(x) = KI(x) where I(x) is light intensity at depth x K is a constant.

Exercise 3.4.3. P(t) = KP(t)V where P(t) is the amount of penicillin in the body, V is the volume ofthe vascular pool, and K is a constant.

Exercise 3.4.4.

B

0 100 200 300 400 500 600 700 8002

1

0x 10

4

Time seconds

Rateofconcentrationchange

(M/s)

Figure A.22: Exercise 3.4.4. B. Rate of change of butyl chloride concentration.

Exercise 3.4.5.

B

0 20 40 60 80 100 120 140 1605

4.5

4

3.5

3

2.5

2

1.5

1

0.5

0x 10

5

Time seconds


(M/

s)

Figure A.23: Exercise 3.4.5. B. Rate of change of phenolphthalein concentration.

Exercise 3.4.6.


21/95


BCO

5 0 5 10 15 20 25 30 355

4.5

4

3.5

3

2.5

2

1.5

1

0.5

0x 10

3

Time seconds


(g/l)/s

BN2O5

0 10 20 30 40 50 608

7

6

5

4

3

2

1

0x 10

4

Time seconds

Rateofconcentrationc

hange

(g/l)/s

Figure A.24: Exercise 3.4.6. BCO . Rate of change of carbon dioxide concentration. BN2O5. Rate ofchange of N2O5 concentration.

Explore 3.5.1.d

dtC = 0,

d

dttn = ntn1

d

dt[C F(t)] = C

d

dtF(t)

Explore 3.5.2.

a.

t2 + t3

= limbt

b2 + b3 (t2 + t3)b t

= limbt

(b2 t2) + (b3 t3)b t

= limbt

(b2 t2)b t + limbt

(b3 t3)b t

= limbt

(b + t) + limbt

(b2 + bt + t2) = 2t + 3t2

Explore 3.5.3. a. Equation 3.30, c. Equation 3.31.

Exercise 3.5.1.

a.

1 + t2

= limbt

(1 + b2) (1 + t2)b t = limbt

b2 t2b t = limbt

(b t)(b + t)b t = limbt (b + t) = 2t.

c.

t2 t

= limbt

(b2 b) (t2 t)b t = limbt

(b2 t2) (b t)b t = limbt (b + t 1) = 2t 1.

e. [ 5 3] = limbt

5 3 5 3b t = limbt

0

b t = 0

g.

1 + t + t2

= limbt

1 + b + b2 (1 + t + t2)b t = limbt

(b t) + (b2 t2)b t

= limbt

(b t)b t + limbt

(b2 t2)b t = limbt 1 + limbt (b + t) = 1 + 2t


22/95


i.

5 + 3t 2t2

= limbt

5 + 3b 2b2 (5 + 3t 2t2)b t

= limbt (3b 3t) (2b

2

2t2

)b t = limbt 3(b t)b t + limbt (2)(b

2

t2

)b t

= 3 limbt

b tb t + (2)limbt

b2 t2b t = 3 limbt 1 2limbt (b + t) = 3 4t

Exercise 3.5.4.

a. [ C ] = limba

C Cb a = limba 0 = 0.

b. [ C F(t) ] = limba

CF(b)

CF(a)

b a = ClimbaF(b)

F(a)

b a = C [ F(t) ].

Exercise 3.5.6. a. Definition of A(W). c. Constant factor, Equation 3.31.

Exercise 3.5.8. The two pens combined should be twenty feet by sixty feet. Each pen should have a

side adjacent to the barn.

Exercise 3.5.10. The two pens combined should be sixty feet by eighty feet. The barn is one side of

one pen; the other pen is adjacent to the first.

Exercise 3.5.12.

[a + bt + ct2 + dt3]

= [a] + [bt] + [ct2]

+ [dt3]

Sum Rule, Equation 3.30

= [a] + b [t] + c [t2]

+ d [t3]

Constant Factor, Equation 3.31

= 0 + b [t] + c [t2]

+ d [t3]

Constant Rule, Equation 3.27

= b + c [t2]

+ d [t3]

t Rule,Equation 3.28

= b + 2ct + 3dt2 tn Rule, Equation 3.29

Exercise 3.5.13. The catcher will have 5.5 seconds to catch it.

Exercise 3.5.15. 2/3 R.

Exercise 3.5.16. a. Definition of E. c. Constant Factor Rule, Equation 3.31.

Exercise 3.5.17. a. Definition of P. c. Constant Rule, Equation 3.27. e. t Rule, Equation 3.28.

Exercise 3.5.18.


23/95


a. [15t2 32t6] = [15t2] [32t6] Sum Rule= 15 [t2]

32 [t6] Constant Factor Rule= 15 (2t) 32(6t

5

) t

n

Rule

c.

t44 +

t33

=

t44

+

t33

Sum Rule

=

14t

4

+

13 t

3

Algebra

= 14 [t4]

+ 13 [t3]

Constant Factor Rule

= 14 (4t3) + 13 (t

3) tn Rule

e. [31t52 82t241 + t314] = [31t52] [82t241] + [t314] Sum Rule= 31 [t52]

82 [t241]

+ [t314]


= 31 (52t51) 82(241t240) + (314t313) tn Rule

g.

2 t7427 + 18t35

=

2 t7427

+ [18t35]

Sum Rule

= 0

t7427

+ [18t35]

Constant Rule

=

1427 t

7

+ [18t35]

Algebra

= 1427 [t7]

+ 18 [t35]


= 1427 (7t6) + 18 (35t34] tn Rule

Exercise 3.5.19. a. t = 5, c. t = 0.1. e. t = 2 or t = 0 or t = 2. g. t = 2 or t = 2.

Exercise 3.5.20. d. The highest point between 0 and 100 is (17.1, 25).

Exercise 3.6.1. a. P(t) = 0, P(t) = 0, c. P(t) = 2t, P(t) = 2, e. P(t) = 12

t1/2, P(t) = 14

t3/2, g.

P(t) = 8t7, P(t) = 56t6, i. P(t) = 52

t3/2, P(t) = 154

t1/2.

Exercise 3.6.2. a. 0, c. -9.8.

Exercise 3.6.3.P(t) = a + bt + ct2 + dt3 Cubic.

P(t) = 2c + 6dt Linear.P(4)(t) = 0 Zero.

Exercise 3.6.4.

A. a. Decreasing, b. Negative, c. Increasing, d. Positive, e. Concave up.

Exercise 3.6.5. a. Zero, c. Positive, e. Positive.

Exercise 3.6.6. a. Zero, c. Negative.


24/95


Exercise 3.6.8. s(t) = g2

t2 + v0t + .

s(t) =

g

2t2 + v0t +

=

g2

t2 + [v0t] + []

=g

2

t2

+ v0 [t] + []

=g

2(2t) + v0 1 + []

= gt + v0

Exercise 3.6.9. a. 15, c. 7.

Exercise 3.6.10.

a. P(t) = 5t + 3P(0) = 5 0 + 3 = 3P(t) = [5t + 3] = 5

c. P(t) = t2 + 3t + 7P(0) = 02 + 3 0 + 7 = 7P(t) = [t2 + 3t + 7]

= 2t + 3

e. P(t) = (3t + 4)

2

P(0) = (3 0 + 4)2 = 16P(t) = [(3t + 4)2]

= [9t2 + 24t + 16]

= 18t + 24.


25/95


g. P(t) = (1 2t)1P(0) = (1 2 0)1 = 1P(t) =

1

12t

= limbt

112b

1

12t

b t= lim

bt

2(t b)(1 2t)(1 2b)

1

b t= lim

bt

2

(1 2t)(1 2b) =2

(1 2t)2 = 2

1

(1 2t)

2= 2 ( P(t) )2

i. P(t) = (6t + 9)1/2

P(0) = (6 0 + 9)1/2 = 3.P(t) =

(6t + 9)1/2

= limbt

(6b + 9)1/2

(6t + 9)1/2

b t= lim

bt

(6b + 9) (6t + 9)((6b + 9)1/2 + (6t + 9)1/2) (b t)

= limbt

6(b t)((6b + 9)1/2 + (6t + 9)1/2) (b t) =

6

2(6t + 9)1/2= 3/P(t)

k. P(t) = (4t + 4)3/2

P(0) = (4 0 + 4)3/2 = 8P(t) =

(4t + 4)3/2

= limbt(4b + 4)1/2 (4t + 4)

1/2 (4b + 4) + (4b + 4)1/2(4t + 4)1/2 + (4t + 4)b t

= limbt

((4b + 4) (4t + 4))

(4b + 4) + (4b + 4)1/2(4t + 4)1/2 + (4t + 4)

((4b + 4)1/2 + (4t + 4)1/2) (b t)= lim

bt

4

(4b + 4) + (4b + 4)1/2(4t + 4)1/2 + (4t + 4)

(4b + 4)1/2 + (4t + 4)1/2=

12(4t + 4)

2(4t + 4)1/2

= = 6(4t + 4)1/2 = 6 3

P(t)

Exercise 3.6.13.

Q(t) = a + bt + ct2, Q(t) = b + 2ct, Q(u + v

2) = b + 2c

u + v

2

Q(u) Q(v)u v =

(a + bu + cu2) (a + bv + cv2)u v = b + c(u + v) = b + 2c

u + v

2.

Explore 3.7.1. a. 3, b. Does not exist. Suppose limx3

F(x) = L. Choose = 0.5. For any > 0,

F(3 /2) = 1 and F(3 + /2) = 3. One of

|F(3

/2)

L

|=

|1

L

|< = 0.5 and

|F(3 + /2)

L

|=

|3

L

|< = 0.5

is incorrect. c. , e. 5/3, f. Does not exist. Suppose

H(3) = limb3

H(b) H(3)b 3 = M.


26/95


Then

H+(3) = limb3+

H(b) H(3)b 3 also is M, and

H(3) = limb3

H(b) H(3)b 3

is M.

But from the graph H+(3) is a positive number and H(3) is a negative number. Thus M is both

positive and negative, which is a contradiction.

i. .

Explore 3.7.3. We do not publish a solution to this problem. It is a good problem and you may

eventually solve it. Your education will not be damaged if you do not find a solution

Exercise 3.7.1.

1. a. limt3

F(t) = 2, b. limt3

F(t) = 2, c. limt3+

F(t) = 2, d. F(3) is not defined.

Exercise 3.7.2.

E

2 1 0 1 22

1

0

1

2

E

H

2 1 0 1 22

1

0

1

2

H

Figure A.25: Graphs of the functions, E and H for Exercise 3.7.2.

DN E stands for does not exist.

E. limx0

E(x) = 1, limx0+

E(x) = 1, limx0

E(x) = DNE,

E

(0) = DNE, E +

(0) = DN E, E

(0) = DN E

H. limx0

H(x) = 0, limx0+

H(x) = 0, limx0

H(x) = 0,

H(0) = 1, H+(0) = 0, H(0) = DN E

Exercise 3.7.3. If F is a function and a is a number in the domain of F and

limba

f(b) f(a)b a exists and is equal to m,

then the line with equationy F(a) = m (x a)

is the tangent to the graph of F at a from the left.


27/95


If the tangent to F at a from the left is the same as the tangent to F at a from the right, then that

line is the tangent to F at a.

Chapter Exercise 3.8.1.

a. P(t) = t3

P(t) = limbt

b3 t3b t

= limbt

(b t)(b2 + bt + t2)b t

= limbt

(b2 + bt + t2) = 3t2

c. P(t) = t3/4

P(t) = limbt

b3/4 t3/4b t

= limbt

(1/4)(b t)(b2 + bt + t2)b t

= limbt

(1/4)(b2 + bt + t2) = 3t2/4

e. P(t) = 2

t

P(t) = limbt

2

b 2

t

b t= lim

bt2

b t(

b +

t)(b t)= lim

bt 2 1b + t = 1t

g. P(t) = 7

P(t) = limbt

7 7b t = 0


28/95


i. P(t) = 11 + t

P(t) = limbt

1

1 + b 1

1 + tb t= lim

bt

t b(1 + b)(1 + t)(b t)

= limbt

1(1 + b)(1 + t)

=1

(1 + t)2

k. P(t) = 5t7

P(t) = limbt

5b7 5t7b t

= limbt

5b t)(b6 + b5t + + bt5 + t6)

(b t)= lim

bt5(b6 + b5t + + bt5 + t6) = 35t6

m. P(t) = 1/(3t + 1)2

P(t) = limbt

1/(3b + 1)2 1/(3t + 1)2b t

= limbt

(3t + 1)2 (3b + 1)2(3b + 1)2(3t + 1)2(b t)

= limbt

9(t + b) 6(3b + 1)2(3t + 1)2

= (18t + 6)/(3t + 1)4 = 6/(3t + 1)3

o. P(t) = 1/

t + 1

P(t) = limbt

1/

b + 1 1/t + 1b t

= limbt

t + 1

b + 1

b + 1

t + 1(b t)= lim

bt

(t + 1) (b + 1)(

t + 1 +

b + 1)

b + 1

t + 1(b t)= lim

b

t

1

(

t + 1 +

b + 1)

b + 1

t + 1

=1

2(

t + 1)

3


a. [3t2 2t + 7] = [3t2] [2t] + [7] = 3 [t2] 2 [t] + [7] = 3(2t) 2 + 0

c.

2t

=

2 t1/2

=

2t1/2

=

2(1

2t1/2) = 1/

2t

e.1/

2t

=21/2t1/2

= 21/2

t1/2

= 21/2(1/2t1/21) = 1/

2t3


29/95


g.

5 + tt

= [5t1 + 1]

= 5 [t1]

+ [1] = 5(1t2) + 0 = 5t2

i. [(1 + 2t)2]

= [1 + 4t + 4t2]

= [1] + [4t] + [4t2]

= [1] + 4 [t] + 4 [t2] = 0 + 4(1) + 4(2t) = 4 + 8t

k.

1 + 3

tt

=

t1/2 + t1/6

=

t1/2

+t1/6

= (1/2)t3/2 + (1/6)t7/6

A.4 Chapter 4 Answers

Explore 4.1.1.

b. One of the more interesting places to examine zonation of plant and animal species is theintertidal zone of a rocky beach. Zonation is so distinct and broadly consistent that it is divided into

three zones 1. The barnacles mark the middle zone.

b. The gastropods, Littorina littorea is larger than the gastropod Olivella biplacata and consequently

L. littorea absorbs heat less rapidly than does O. biplacata. L. littorea occurs above O. biplacata in the

intertidal zone.

Exercise 4.1.2. a. 0.5-1/2.1.

Exercise 4.1.3. a. 9, c. 0, , e. 12.

Exercise 4.1.5. b. 6.

Exercise 4.1.8. a. 1 - 0.93 = 0.271.

Exercise 4.1.10. t1.

= 1.37.

Exercise 4.1.11. =

10 3.

Exercise 4.1.14. No.


Exercise 4.2.1. a. C(0) does not exist. b. C is continuous at 0. c. Your vote counts.

Exercise 4.2.3. P(0) = 1.

Explore 4.2.3. We do not give a solution to this problem. It is worth your time to consider, but your

education will not be stunted if you do not find the answer to this problem.

1T. A. Stephenson and A. Stephenson, The universal features of zonation between tide-marks on rocky shores, J. Ecol.37 (1949), 507-522.


30/95


Exercise 4.3.1. a. 8t ( 2 + t2 )3, c. 8t3 ( t4 + 5),

e. 2

18 + 2t

t8 + t

2

, g. 2

1t + t

1t2

+ 1

,

i. (t + 5)2,

Exercise 4.3.2. a. Definition of Derivative, Eq. 3.22, b. Algebra, c. Algebra, d. Eq. 3.15,

e. Eq. 3.22, f. Eq. 3.14, g. Eq. 3.13,

h. Eq. 3.10, i. Theorem 4.2.1, j. Eq: 3.15.

Exercise 4.4.1.

a. 6x2 c. 10(x + 1)3

e. x1 x2

g. 4 (2 x)3 i. x (7 x2)3/2 k. 4.5 (1 + 3x)0.5

m. (9 (x 4)2)1/2(x 2) o. (1 3x)2/3

Exercise 4.4.2. a. -2/3.

Exercise 4.4.3. The tangents intersect at (5,5/3).

Exercise 4.5.1. x.

= 3.82 km, T(3.82).

= 1.75 hours.

Exercise 4.5.2.

a. Linda probably can not travel faster than 60 mph, but she should travel as fast as she can.

b. At 60 mph the 10 miles will require 10 minutes and she will frostbite in 5 minutes. At 50 mph, the

trip will take 12 minutes and she will frostbite in 10 minutes. The best she can hope for is at 20

mph it takes 30 minutes and she will frostbite in 30 minutes; maybe she should go 22 mph and

hope for the best.

c. She should not make the trip if the temperature is -20F.

Exercise ??. 165 lb/acre.

Exercise 4.5.4. c. x

.

= 0.63.

Exercise 4.6.1. a. -2/3 and 2/3.


31/95


Exercise 4.6.2. -1/48, No.

Exercise 4.6.4. Graphs are shown in Figure A.26. Slopes: a. 3/4 and 3/4, c. -2/3, 4, e. 0, , g. -1, 0.

A

4 3 2 1 0 1 2 3 44

3

2

1

0

1

2

3

4

C

30 20 10 0 10 20 3030

20

10

0

10

20

30

E

4 3 2 1 0 1 2 3 44

3

2

1

0

1

2

3

4

G

4 3 2 1 0 1 2 3 44

3

2

1

0

1

2

3

4

Figure A.26: Exercise 4.6.4. Graphs of A. x2 2y2 = 1, C.|x|+

|y| = 5, E. x2 + y2 = (x + y)2, and

G. x3 + y3 = (x + y)3.

Exercise 4.6.5. a. m = (b2x)/(a2y),


a. 6t 2 c. 12

t + 2e. 1

2t + 1

g. 5(t + 5)2 i. 10 ( 1 + 2t )4 k. 12

t(1 +

t)2


Neil de Grass Tyson is referencing the Intermediate Value Property of continuous functions. To do

so, he implies that for every number between the mass of a planet and the mass of a star, there is a

celestial body of mass equal to that number.

Chapter Exercise 4.7.3. Each pen should be 100 by 100.

Chapter Exercise 4.7.4. You should travel 992 meters along the river before crossing.

Chapter Exercise 4.7.5. (8,16/7).


Explore 5.1.2. The pattern is explained in the text.

Explore 5.1.3. For E(t) = 3t,


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Centered diff quot E(t).

= E(0) E(0) E(t)

E(0).

=30+0.0001

300.0001

0.0002 = 1.09861229 1 1.09861229

E(2).

= 32+0.0001 320.0001

0.0002 = 9.88751063 9 9.88751061

Exercise 5.1.1. (b) For P(t) = 5t2 3t + 7, P(a + h) P(a h)2h = 10a 3 = P(a).


A

0 0.5 1 1.5 2 2.50

1

2

3

4

5

6

t

4+2.7

725887(t2)and2

tC

1.8 1.85 1.9 1.95 2 2.05 2.1 2.15 2.2

3.4

3.6

3.8

4

4.2

4.4

t

4+2.7

725887(t2)and2

t

Figure A.27: Exercise 5.1.2. Graphs of y = 2t and y = 4 + 2.7725887(t 2) at increasing magnification.

Exercise 5.1.3. For E(t) = 10t,

i. E(0).

=E(0 + 0.0001) E(0 0.0001)

0.002= 2.302585.

ii. E(1) .= 0.2302585, E(2) .= 23.02585, iii. See Figure A.28.

iii.

1.5 1 0.5 0 0.5 1 1.5 2 2.50

50

100

150

200

250

300

350

400

t

10

t

and

2.

3025851

0t

Figure A.28: Exercise 5.1.3. Graphs of 10t and 2.302585 10t.

Exercise 5.1.4. For E(t) = (1/2)t,

a. E(0).

= 0.693147. c.SeeFigure A.29.


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c.

1.5 1 0.5 0 0.5 1 1.5 2 2.52

1.5

1

0.5

0

0.5

1

1.5

2

2.5

t

(1/2)ta

nd

0.6

93147(

1/2)t

Figure A.29: Exercise 5.1.4. Graphs of (1/2)t and 2.302585 10t.

Exercise 5.1.5.

y = t0 = 1 t0 = 0 t0 = 1

a. 1.5t y 2/3 = 0.2703(t + 1) y 1 = 0.4055t y 1.5 = 0.6082(t 1)c. 3t y 1/3 = 0.3662(t + 1) y 1 = 1.099t y 3 = 3.296(t 1)

Exercise 5.1.6. b. See Figure A.30.

b.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

C(t), V. natrigensconcentration

C(t),V.natrigensgrowthrate

Figure A.30: Exercise 5.1.6. Graph ofC(t) vs C(t).

Exercise 5.1.7. For P(t) = 200 0.96t, b. See Figure A.31.

b.

0 20 40 60 80 100 120 140 160 180 20010

9

8

7

6

5

4

3

2

1

0

P(t), Penicillin concentration

P(t),

Penicillin

clearance

Figure A.31: Exercise 5.1.7. Graph of P(t) vs P(t).

Explore 5.2.1. See Figure A.32.


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1 0.5 0 0.5 10

0.5

1

1.5

2

1.5t

2t

3t

5t

(0, 1)

Figure A.32: Explore 5.2.1. The line through (0,1) with slope 1 and an exponential function with thatline as a tangent.

Explore 5.2.2.

For h = 0.0001, B = (1 + h)1/h.

= 2.71814593

Exercise 5.2.1.

a. Sum rule, constant factor rule, power rule, exponential rule.

c. epq = (ep)q, power chain rule, exponential rule, (ep)q = epq, ep eq = ep+q.

Exercise 5.2.2.

. [5t2 + 32et]

= [5t2] + [32et] S= 5 [t2]

+ 32 [et]

CF= 5t + 32 [et]

tn

= 10t + 32et E

c. 7 8 (et)1

= [7]

8 (et)1

S

= [7] 8

(et)1

CF

= 0

8 (et)1

C

= 8(1) (et)2 [et] PC= 8 (et)

2et E

= 8et Algebra

e. [t25 + 3e]

= [t25] + [3e] S= [t25]

+ 0 C= 25t24 tn

g.

1 + t +

t2

2 et

= [1] + [t] +

t22

[et] S= [1] + [t] + 12 [t

2] [et] CF

= 0 + [t] + 12 [t2] [et] C

= 1 + 12 (2t) [et]

tn

= 1 + t et E


Exercise 5.2.5.

ForB0.01 1

0.01= 1, B = 2.70481383 e = 2.718281828


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1 0.5 0 0.5 1 1.5 20

1

2

3

4

5

6

7

8

9

t

e

t

and

1+t

+

t2/2+t

3/6

+

24/24

Figure A.33: Exercise 5.2.4. Graphs of y = et (solid curve) and y = 1 + t + t2/2 + t3/6 + t4/4 (dashedcurve). The two graphs are indistinguishable on [-1,1].

Exercise 5.2.8.

a. 2x + ex c. 2(1 + ex)ex e. 2e2x g. 3e3x

i. 3(5 + ex)2ex k. 12

(ex)1

2 m. 0.6e0.6x

Exercise 5.2.9.

a. 7

t4 + t16

t4 + t1

= 7

t4 + t16

t4

+t1

Use ( )s

Exercises 5.2.11 5.2.16 are too important to have answers posted. After you have

solved one of these exercises, your instructor may tell you whether your solution is correct.

Exercise 5.3.1. a. P(T) = 200 e0.26T,

Exercise 5.3.2. log2 10.

= 3.22

Exercise 5.3.3. Id = 0.4

e0.2d

Exercise 5.3.4.

a. f(t) = 5 e2.30t c. f(t) = 7 e0.69t e. f(t) = 5 e0.69t

Exercise 5.3.5.

a. f(t) = 52.17 ln t c. f(t) = 7 +7

1.61 ln t e. f(t) = 4.5 + 2.16ln t

Explore 5.4.1. a. Definition of Derivative, b. Algebra, c. lim C F(t) = C lim F(t), andsee Explore 5.4.2.


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Exercise 5.4.2. a. b = eln b, Equation 5.14, c. ekt Rule, Equation 5.16,

Exercise 5.4.3. For b = 1, bt = 1 for all t and ln b = 0. [ bt ]

= [ 1 ] = 0 and bt ln b = 0. Thus,

[ bt ]

= 0 = bt ln b

Exercise 5.4.4.

a. [3e5 t + ]

= [3e5 t]

+ [] Sum rule= 3 [e5 t]

+ [] Constant Factor rule

= 3e5 t 5 + [] ekt rule.= 3e5 t 5 + 0 Constant rule

c. [5t]

= 5t ln 5 Eq. 5.19

Exercise 5.4.5. IF stands for Insufficient Formulas.a. e5x 5 c. IF e. IF

g. ex ex

2 i. 0.03e0.6x 0.3e0.1x k. 12

ex

l. IF n. 0 p. 10x ln 0.1

r. 5 (e5x + e3x)4 (5e5x 3e3x)

Exercise 5.4.6. Let = a2 and = b2 and E(t) = et. Then

limba

e(b2) e(a2)b2 a2 = lim

e e = E

() = e = e(a2).

Exercise 5.4.8.

a.e2t

= lim

ba

e2b e2ab a = limba

e2b e2a2b 2a 2 = limba e

2a 2

c.e2t = lim

bae2

b

e2

a

b a = limba e2

b

e2

a

2

b 2a 2b 2a

b a

= e2

a limba

21

b +

a= e2

a 2 1

2

a= e2

a 1

a

e.

e1

t

= lim

ba

e1

b e 1ab a = limba

e1

b e 1a1

b

1

a

1

b 1

ab a

= limba

e1

b e 1a1

b 1

a

limba

a ba b (b a) = e

1

a limba

a ba b (b a) = e

1

a 1a2


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Exercise 5.4.9.

C(t) = 10e2t + 15e3t ln 1.5 .= 0.4055 C(0.4055) = 0, C(0.4055) = 0.7407

Exercise 5.4.10. Z = 0 or Z = 1 or Z = 1/2. For no value of t is Z = e.1t = 0. For Z = e.1t = 1, t = 0

and the concentration in tissue is 0. The maximum concentration in tissue occurs for Z = e.1t = 1.2,

t = 10 ln 2 = 6.93, and is equal to 0.031 gm/ml.

Exercise 5.5.1.

a. P(t) = 5e2 t tDbl = 0.35 c. P(t) = 2e0.1 t tDbl = 3.47

e. P(t) = 10et tDbl = 0.69 g. P(t) = 0 tDbl = Undefined or 0

Exercise 5.5.2.

a. S(t) = 5 5e2 t t1/2 = 0.35 c. S(t) = 10 tDbl = Undefined or 0

e. S(t) = 20 20et t1/2 = 0.69 g. S(t) = 20 tDbl = Undefined or 0

Exercise 5.5.3.

a. S(t) = 5e0.35 t c. S(t) = 2e0.32 t

e. S(t) = 5e0.458 t g. S(t) = 2.5e0.693 t

Exercise 5.5.5. S(t) = 300 + 100e0.05 t.

Exercise 5.5.6. Let At be the amount of Sotolol in the body at time t with t = 0 hours the time of the

first pill. Then

A0 = H

A+0 = H+ 40

A12 =1

2A+0 =

1

2(H+ 40)

A+12 = A12 + 40 =

1

2H+ 60

A24 =1

2A+12 =

1

4H+ 30

H = A24 =1

4H+ 30 = 40.

The amount of Sotolol in the body fluctuates between 40 and 80 mg, a two to one ratio, substantially

less than the four to one ratio from 26.7 to 106.7 fluctuation when taking 80 mg once per day.

Exercise 5.5.8. The semilog graph of data set a appears to be nonlinear. The semilog graph of data c.

appears to be linear and P(t) = 2e0.173 t fits the data.


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A

1 0 1 2 3 4 5 610

0

101

102

t

P(t)

C

5 0 5 10 15 20 25 3010

0

101

102

t

P(t)

Figure A.34: Exercise 5.5.8. Semilog graphs of data sets A and C.

Exercise 5.5.9. P(t) = 0.022e0.0319 t

Exercise 5.5.12.

a. 105.75 103.75

105.75= 0.99 = 99 percent.

b.

log10 V(t) = 5.9 0.19tV(t) = 105.90.19t = 105.9 100.19t

V(t) = 790000

eln100.19t

= 790000e0.43t

c. t1/2 = (ln 2)/0.43 = 1.6 days.

d. The immune system is eliminating virus at the rate of 43 percent per day.

e. 15.8 6 106 = 95 106 CD4 cells per day.

Exercise 5.5.13. c. P(t) = 2e0.04 t.

Exercise 5.5.14. [(ln(8.02/76.06))/( ln2)] 1.28 109 = 4.15 109 years old.

Exercise 5.5.16. a. [(ln(2.5/2.65))/( ln2)] 50 109 = 4.20 109 years old.

Exercise 5.5.17. a. [(ln(3/10.0))/( ln2)] 5730 = 9953 years old.

Exercise 5.5.18. b. 206 meters.

Exercise 5.5.19. In L1, the intensity of the light that penetrates the surface is higher (800 vs 700) than

for L2 and the water is clearer (smaller decay constant, -0.04 vs -0.05) in L1 than in L2.

Exercise 5.5.20. Increased opacity of the sample over that of the standard is proportional to cell

density. Let ksm and kst be the opacities of the sample and standard, respectively and be the thickness

of each.


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Ist = I0ekst

Ism = I0eksm

ksm kst = 1

ln

IsmIst

Exercise 5.5.21. f. For the patient with the kidney that removes only 15% of the penicillin that passes

through it, P(t) = 10 10e0.02t, saturation level = 10 g.

Exercise 5.5.22. f. The hen should return within 21.8 minutes.

Exercise 5.5.23. Let h(t) be the height of water in the stem of the thistle tube t minutes after the

beginning of the experiment, POz the osmotic pressure across the membrane (assumed to be constant),

the density of the salt water, and g the acceleration of gravity. The pressure, P(t), due to water in the

stem is g h(t). The rate, R(t) at which water crosses the membrane is proportional to POz P(t)and h(t) is proportional to R(t).

Conclude that

h(t) =

POz

g

1 egK t

where K is a proportionality constant.


2 0 2 4 6 8 10 120

0.5

1

1.5

2

2.5

Time days

AmountofRotenone,

kg

Figure A.35: Exercise 5.5.24. Graphs ofpt = 2 0.75t, t = 0, 1, , 10 and P(t) = 2e0.25t, 0 t 10.

Exercise 5.5.25. P(h) = e0.00012634 h.

Exercise 5.5.27. g. She could ascend to d = 8.1 meters.

Exercise 5.5.28. Wilsons statement is a very interesting statement describing a relation between the

number of species to be found on an island and the area of the island and has been confirmed in several


40/95


locations. It does not, however, describe the competition or other interactions between species that

might lead to the relationship and therefore does not meet our definition of a Mathematical Model. The

relation has been derived from very elementary first principles by J. Harte, T. Zillio, E. Conlisk, and A.

B. Smith, Maximum entropy and the state-variable approach to macroecology, Ecology 89 (2008),2700-2711, that applies when the number of individuals per species is small.

Exercise 5.6.1.

a. [3ln t + e3t]

= [3 ln t] + [e3t]

Sum

= 3 [ln t] + [e3t]

Constant Factor

= 3(1/t) + [e3t]

L

= 3(1/t) + e3t

[3t] Exp Chain= 3(1/t) + e3t3 [t] Constant Factor= 3(1/t) + e3t3 1 tn

c. [ln5]

= 0 Constant

e. [ln(t2 + t)]

= 1/(t2 + t) [t2 + t]

Log Chain

= 1/(t2 + t)

[t2]

+ [t]

Sum

= 1/(t2 + t) (2t + 1) tn

g.e1/x

= e1/x [1/x] Exp Chain= e1/x(1/x2) tn

i. [ln((t + 1)2)]

= (1/(t + 1)2) [(t + 1)2]

Log Chain= (1/(t + 1)2)2(t + 1) [(t + 1)] Poly Chain

= (1/(t + 1)2)2(t + 1)

[t] + [1]

Sum

= (1/(t + 1)2)2(t + 1)

1 + [1]

tn

= (1/(t + 1)2)2(t + 1) (1 + 0) Constant

Exercise 5.6.2.

a. e(t2)2t c. 2 (et)2 e. 3 g. 0 i. et+1

Exercise 5.6.3. a. Equation 5.14, c. Exponential chain rule,

Exercise 5.6.4. a. Equation 5.12, c. Derivative of ln t.

Exercise 5.6.5. a. 1, c. 0 or 2.

Exercise 5.6.6.Let v(t) = (u(t))n

ln v(t) = n ln u(t) Equation 5.11


41/95


Exercise 5.6.7.

a. v(t) = ln y(t) = ln t = ln t,

v(t) =1

y(t)

y(t) = 1

ty = y(t)

1

t= t

1

t= t1

c. v(t) = ln y(t) = ln(1 + t2) = ln(1 + t2),

v(t) =1

y(t)y(t) =

1

1 + t22t

y = y(t)1

1 + t22t = (1 + t2)

1

1 + t22t = 2t(1 + t2)1

e. v(t) = ln y(t) = ln esin t = sin t,

v(t) =1

y(t)y(t) = cos t

y = y(t)cos t = esin t cos t

Exercise 5.6.8.

a. y(t) =(t + 1)t 2 +

(t 1)t 2

(t + 1)(t 1)(t 2)2 c. y

(t) = te t22

e. y(t) = 2t3

t2 + 1 2t3

(t2 + 1)2g. y(t) = (ln b) bt

i. y(t) = 1t et

ln tet


a. 5 e5t c. (3/2)t1/2 (et)3/2 e. 1t ln t

g. (1 + et

)2

et

i. (1 + et

) et

k. 3

e

t2

e

t 1

2t


a. 10t ln10, c. 3(t 1)2 (t3 1) + 3t2 t 1)3 e. u v w + u vw + uv w

Chapter Exercise 5.7.3. Only data set a. is exponential.

Chapter Exercise 5.7.4. See Figure A.36.


y = te3t, y = e3t + 3te3t, y = 6e3t + 9te3t.


42/95


0 1 2 3 4 5 6 70

1

2

3

4

5

6

7

8

9

10

11

Time t

y

=

10/(9e

t

+

1),

andy

(2.2,5)

Figure A.36: Chapter Exercise 5.7.4. Graph of logistic population growth, P(t) = 10/(9et + 1) andP. P(t) is maximum for t = 2.2 and P(2.2) = 5, one-half the asymptotic limit of P.

y 6y + 9y = 6e3t + 9te3t 6

e3t + 3te3t

+ 9

te3t

= 6e3t 6e3t + 9te3t 18te3t + 9te3t = 0.


Exercise 6.1.1.

a. e3t (3t + 2) /t3c. (et)/t + et ln t

e. 2t ln 2g. 5e5t

i. 3

Exercise 6.1.2.

a. t2et c. t2 1

(1 + t2)2e. et

1 + t + 1

2

1 + t

g. tet i. 12

tln t 2(ln t)2

k. 1/(t(ln t)2)

m. 10 9e0.2t

(9 + e0.2t)2

Exercise 6.1.3.

a. Definition of derivative, Definition 3.3.1, Equation 3.22.

b. Algebra, subtract and add u(a) v(b) in the numerator.

c. Algebra.

d. Equation 3.14 which states that the limit of the sum of two functions is the sum of the limits of the

two functions, and Equation 3.13, which states that the limit of a constant times a function is the

constant times the limit of the function.


43/95


e. Equation 3.15 which states that the limit of the product of two functions is the product of the

limits of the two functions, and Theorem 4.2.1, The Derivative Requires Continuity.

f. Definition of derivative, Definition 3.3.1, Equation 3.22.

Exercise 6.1.7. Yes, let u(x) = 0 and v(x) = ex. A more interesting example is, let u(x) = x/(x 1)and v(x) = x.

Exercise 6.1.8.

a. L(p) =

1000

41

(41p40 (1 p)959 +p41 959(1 p)958 (1))

b. p = 411000 , L(p) = 0.0635

Exercise 6.1.10. The bird should search each bush 2 minutes.

Exercise 6.10.

a. dPdT

V=constant

= nRv nb

Exercise 6.1.14.

P(t) = u(t)v(t)

ln P(t) = ln u(t) ln v(t)

[ln P(t)] = [ln u(t) ln v(t)]

[ln P(t)] = [ln u(t)] [ln v(t)]

P

P(t)

=u

u(t) v

v(t)

P(t) = P(t)v(t)u(t) v(t)u(t)

u(t)v(t)=

u(t)

v(t)

v(t)u(t) v(t)u(t)u(t)v(t)

=v(t)u(t) v(t)u(t)

v(t)v(t)

Exercise 6.1.15.

P(t) =1

u(t) ln P(t) = ln u(t)

[ln P(t)] = [ln u(t)]


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a

0 5 10 15 20 250

5

10

15

20

25

30

35

40

Time

Po

pulation

1

15

20

30

P0

c

0 10 20 30 40 500

5

10

15

20

25

Time

Population

0.10.20.3r = 0.4

Figure A.37: Exercise 6.1.18. a. Graphs of the logistic equation, P(t) = P0Mert/(M P0 + P0ert),

for M = 20, r = 0.5, and P0 values of 1, 15, 20, and 30. c. Graphs of the logistic equation, P(t) =P0Me

rt/(M P0 + P0ert), for M = 20, P0 = 1 and r = 0.1, 0.2, 0.3, and 0.4.

P

P(t)= u

u(t)

P(t) = P(t)

u

u(t)

=

1

u(t)

u

u(t)

P(t) = u

u2(t)

Exercise 6.1.16.

P(t) = (t2 1) 1t2 + 1

P(t) =4t

(t2 + 1)2

Exercise 6.1.17. b. Constant multiplier, Equation 3.31, d. Derivative of ert, Equation 5.16.


Exercise 6.1.19. The steepest part of the logistic curve occurs where the population size is one-half the

maximum supportable population.

Exercise 6.1.20. a. Approximately, (H,W) = (10.9,44.9)

b.dW

dH=

(0.103 + 0.0389H)2 0.103(0.103 + 0.0389H)2

The last fraction is zero when the numerator is zero, or when H = 10.9. d. The hyperbola,

N =

1 +1

0.139 + 0.0836 H

H 2.


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60 40 20 0 20 40 60

499.8

499.9

500

500.1

500.2

Time, seconds

Observersound

frequency,cycles/sec

Figure A.38: Exercise 6.2.4. Frequency of sound reaching an observer from a passing train.

matches the data reasonably well. f. The number of drops required to break a large whelk from a height

of 6.12 meters is 22.5/6.12 = 3.7.

Exercise 6.2.1.

a. 2te(t2) c. 2t

e. 1/(2t) g. et

i. 4 (t + e2t)3

(1 2e2t) k. 364

x (1 x2/16)1/2

Exercise 6.2.2.

a. 2te(t2) c. 2t e. 4t 12

2t2 t + 1

Exercise 6.2.3.

a. 2 sin(2t) c. ( sin t)ecos t e. ( sin t) cos(cos t)

g. csc t i. et tan et

Exercise 6.2.4.

a. y = (x2 + 1002)1/2

b. y(t)|x(t)=200 = 26.83 m/s

c. x(t) = 30t

e. See Figure A.38

Exercise 6.2.5. 0.9441 cm/sec.


46/95


0 20 40 60 80 100 1200

500

1000

1500

2000

2500

3000

3500

y=3059(1 t/100)2.263

Time minutes

WeightofSpherical

IceBallgrams

Figure A.39: Exercise 6.2.6. Graph of the weight of a melting ice ball and the function, y = 3059(1 t/100)2.263.

Exercise 6.2.6.

a. dVdt = K 4r2

c. dVdt = K 4r2 = 4r2(t) drdt

Divide by 4r2 K = drdt

e. r(t) =

Kt + C

dr

dt

=

K

f. See Figure A.39

Exercise 6.3.1. a. P(t) = 4 t/2, 0 t 4. (2,3) is reflected to (3,2), P(2) = 1/2. P1(t) = 8 2t,2 t 4, [P1] (3) = 2. See Figure A.40a.

c. P(t) =

4 t, 0 t 4, P(t) = 1/(24 t). (2,2) is reflected to (2, 2). P(2) = 1/(22)P1(t) = 4 t2, 1/3 t 1. [P1] (1) = 1/4 See Figure A.40c.

e. P(t) = t2 + 1, 2 t 0, P(t) = 2t. (-1,2) is reflected to (2,-1). P(1) = 2.

P1

(t) = t 1, 1 t 5. [P1

]

(t) = 1/(2t 1) [P1

]

(2) = 1/2 See Figure A.41e.Exercise 6.3.2. h(t) = t1/3, g(x) = x3.

h(t) =1

3(t2/3)

Exercise 6.3.3.

a. The graph of F is the graph labeled c. The graph of F is a, and the graph of F1 is b.


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a

1 0 1 2 3 4 51

0

1

2

3

4

5

P

P1

(2,3)

(3,2)

c

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.5

1

1.5

2

2.5

3

3.5

4

P

P1

(2,21/2

)

(21/2

,2)

Figure A.40: Exercise 6.3.1. a. Graph of P(t) = 4 1/(2t) and of P1. The slope of P at (2,3) is-1/2; the slope of P1 at (3,2) is -2. c. Graph of P(t) =

4

t and of P1. The slope of P at (2,

2) is

1/(22); the slope of P1 at (2, 2)) is 22.

e

2 1 0 1 2 3 4 5

2

1

0

1

2

3

4

5

P

P1

(1,2)

(2,1)

Figure A.41: Exercise 6.3.1. e. Graph of P(t) = t2 + 1, 2 t 0, and of P1. P(t) = 2t. The slopeof P at (1, 2) is 2; the slope of P1 at (2,1)) is 1/2.


a. 4t3 + 2e2t c. 4t3e2t 2t4e2t

e. 8t3e2t4

g. 3(ln t)2/t

i. 1 ln tt2

k. 1t(ln t)2

m. 2t2 + 8t 2(t2 + 1)2


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Exercise 7.1.1.

a. 0.1 sin0.1 = 0.00016658,b.

sin0.001

0.001= 0.9999998.

Exercise 7.1.2.

a. 0.1

tan0.1 = 0.000334672,

b.

tan0.001

0.001= 1.0000003333.

Exercise 7.1.3.

a. Then

sin(A B) = sin(A + (B))

= sin A cos(B) + cos A sin(B)

= sin A cos B cos A sin B.

b. Subtract

sin(A + B) = sin A cos B + cos A sin B

sin(A B) = sin A cos B cos A sin B

sin(A + B) sin(A B) = 2 cos A sin Bc.

A + B = xA B = y

d. By substitution,

sin x sin y = 2 cos x + y2

sinx y

2.


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e.

cos(A B) = cos(A + (B))

f. Subtract

cos(A + B) cos(A B) = 2sin A sin B

Substitute

cos x cos y = 2sin x + y2

sinx y

2

Exercise 7.1.4. c. Choose = .

Exercise 7.1.5. Begin:

|cos(z+ h) cos z| = 2 sin

2z+ h

2

sin

h2

2 1

sin

h2

| sin z| 1Then for any positive number , choose = .

Exercise 7.2.1. See Figure A.42

a

00.2

0

0.2

0.4

0.6

0.8

1

1.2

x

cos

x

and

(sin(x+0.

2)

sin

x)/0.

2

/2 /2

Figure A.42: a. Graph of y = cos x (solid curve) and of y = sin(x+0.2)sinx0.2

(dashed curve).

Exercise 7.2.2.

y(t) = 2 35t6 + 6e3t 1t

+ 2 cos t + 3 sin t


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Exercise 7.2.3.a. 2cos2 t 2sin2 t c. sec t tan t

e. tan t g. csc t cot t

i. ecos t sin t k. et sin t + et cos t

m. e2t

200 o.1

15t

Exercise 7.2.4. See Figures A.43 - A.44

a. y(t) = cos t sin t, y(/4) = 0

c. y(t) =

3 cos t

sin t, y(2/3) = 0

e. y(t) = e3t(cos t sin t ), y(3/4) = 0

a

0

1

0.5

0

0.5

1

1.5

/2

c

0

1

0.5

0

0.5

1

1.5

2

/2

Figure A.43: a. Graph of y = sin t + cos t, max at (/4,

2) and min at (,1). c. Graph of y =3sin t + cos t, max at (/3, 2) and min at (,1).

e

0

1

0.5

0

0.5

1

1.5

2

/2

Figure A.44: e. Graph ofy = et

cos t, max at (0, 1) and min at (3/4,e3/4

2/2).


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Exercise 7.2.5.

a. [cot t ] =

cos tsin t

cos t [sin t] [cos t] sin tsin2 t

= cos2 t + sin2 tsin2 t

= csc2 t

c. [csc t ] = [ (sin t)1]

(1)(sin t)2 [sin t] = (1)(sin t)2(cos t) = csc t cot t

Exercise 7.2.7.

See Figure A.45. The points (-0.75 -0.75) and (0.75 0.75) are on the tangent, the slope of the

tangent is 1 and and [sin t]

t=0 is 1.

1.5 1 0.5 0 0.5 1 1.5

1

0.5

0

0.5

1

x

y

y = sin x

(0, 0)

(0.75, 0.75)

(0.75, 0.75)

Figure A.45: Graphs of y(x) = sin x and a tangent drawn to the graph.

Exercise 7.2.9.

See Figure A.46

a

0

1

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

x

sin

x

and(

cos(x+0.2

)c

os

x)/0.2

/2 /2

Figure A.46: a. Graph of y = sin x (solid curve) and of y = cos(x+0.2)cosx0.2 (dashed curve).


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Exercise 7.2.12.

cos2 x + sin2 x

= [ 1 ]

2 cos x [cos x] + 2 sin x [sin x] = [ 1 ]

2 cos x [cos x] + 2 sin x (cos x) = 0

[cos x] = sin x if cos x = 0.

Exercise 7.3.1.

a. 2sin t c. 2t cos t2 e. sin(t /2)

g. 8t3 sin(t4)cos(t4) i. cos (cos t) ( sin t) k. 4t tan(t2)

m. tan t o. cot t

Exercise 7.3.2.

a. The product rule for derivatives.

b. The exponential chain rule and the cosine chain rule.

c. The sum rule for derivatives.

d. The product rule for derivatives.

e. The exponential chain rule and the cosine chain rule.


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Exercise 7.3.3.a. y + y = 0

y = cos t

y = sin ty = cos ty + y = ( cos t) + (cos t) = 0

c. y + y = 0y = 3 sin t + 2 cos t

y = 3 cos t 2sin ty = 3sin t 2cos t

y + y = (3sin t 2cos t) + (3 sin t + 2 cos t) = 0

e. y + 2y + y = 0y = et

y = ety = et

y + 2y + y = (et) + 2(et) + (et) = 0

Exercise 7.3.4. At the point A.

Exercise 7.3.5.a.

tan (t) = 600(x(t))1

[tan (t)] =600(x(t))1

sec2 (t) (t) = 600(1)(x(t))2x(t)

b.

Time, t Elevation, (t) x(t)Sec Degrees radians meters

0 40 40180

5 45 45180

600

10 51 51180

c. (5).

=(10) (5)

5 = 0.02094

d.

x(5) = x2(5)

600sec2((5))(5)

.= 1200 0.02094 = 25.128 meters/sec


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Exercise 7.3.6.

a. x.

= 23.04 m/s.

b. 1.2 m/s.

c. 0.01745 < (5) < 0.0209. 20.94 < x < 25.13.

d. Let x(5) be the estimate of x using (5) =

(5) + .

x(5) = = x2(5)

600sec2 (5)

x(5) = = x2(5)

600sec2 ((5) + )

|x(5) x(5)| =x

2(5)

600sec2

||

= 1200||

Exercise 7.3.7.

a. 20

2

= 5

2

+ x

2

2 5 x cos .b. = 100 revolutions per minute = 100 2 radians per minute.


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Exercise 7.4.1.

a. y(t) = 2 sin t + cos t y(0) = 2 0 + 1 = 1

y = 2 cos t

sin t y(0) = 2

1

0 = 2

y(t) = 2sin t cos t

y(t) + y(t) = (2sin t cos t) + (2 sin t + cos t) = 0

c. y(t) = cos 3t sin3t y(0) = 1 0 = 1

y = 3sin3t 3cos3t y(0) = 3 0 3 1 = 3

y

(t) =

9cos3t + 9 sin 3t

y(t) + 9y(t) = (9cos3t + 9 sin 3t) + 9(cos 3t sin3t) = 0

e. y(t) = 4 cos(3t + /3) y(0) = 4 1/2 = 2

y = 12 sin(3t + /3) y(0) = 12 3/2 = 63

y(t) = 36 cos(3t + /3)

y

(t) + 9y(t) = (

36 cos(3t + /3)) + 9(4 cos(3t + /3)) = 0

Exercise 7.4.1, Continued

g. y(t) = 2 sin 3t + 3cos 3t y(0) = 2 0 + 3 1 = 3

y = 6 cos 3t 9sin3t y(0) = 6 1 9 0 = 6

y(t) = 18sin3t 27cos3t

y(t) + 9y(t) = (18sin3t 27cos3t) + 9(2 sin 3t + 3 cos3t) = 0

i. y(t) = et sin t y(0) = 1 0 = 0

y = et cos t et sin t y(0) = 1 1 1 0 = 1

y(t) = et sin t 2et cos t + et sin t= 2et cos t

y(t) + 2y(t) + 2y(t) = (2et cos t) + 2(et cos t et sin t) + 2(et sin t) = 0


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k. 4.01 y(t) = e0.1t(0.1sin2t + 2 cos 2t) y(0) = 1 (0 + 2) = 2

+0.2 y = 4.01e0.1t sin2t y(0) = 4.01 1 0 = 0

+1 y(t) = 4.01e0.1t

(2cos2t 0.1sin2t)= e0.1t ((8.02 8.02)cos2t + (0.401 0.802 + 0.401)sin2t) = 0

Exercise 7.4.2. a. = 5. c. = .

Exercise 7.4.3. a. k = 4. c. k = 4.

Exercise 7.4.5.

y(t) = A sin(t) + B cos(t) y(0) = A sin(0) + B cos(0) = B

y(t) = A cos(t) B sin(t) y(0) = A cos(0) B sin(0) = A

y(t) = A2 sin(t) B2 cos(t)

y(t) + 2y(t) =

A2 sin(t) B2 cos(t) + 2(A sin(t) + B cos(t)) = 0

Exercise 7.4.7.

y(t) = ebt cos

c b2 t

y(t) =

c b2 ebt sin

c b2 t bebt cos

c b2 t

y(t) = (c b2)ebt cos

c b2 t + 2b

c b2 ebt sin

c b2 t + b2ebt cos

c b2 t

= (

c + 2b2)ebt cos

c

b2 t + 2b

c

b2 ebt sin

c

b2 t

y(t) + 2by(t) + cy(t) =

+1

(c + 2b2)ebt cos c b2 t + 2bc b2 ebt sin c b2 t

+2bc b2 ebt sin c b2 t bebt cos c b2 t

+c ebt cos c b2 t

= (

c + 2b2

2b2 + c)ebt cos

c

b2 t

+(2bc b2 + 2bc b2 )ebt sin c b2 t

= 0


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0 1 2 3 4 5 6 7 8 9 106

4

2

0

2

4

6

Time, t

Population,

P(t)

Figure A.47: The solid curve is the graph of u(t) = 4sin t + 3 cos t. The dashed curve is the graph ofu(t) = 5 sin(t + ) translated 0.4 units upward; without translation, the two graphs are identical.

Explore 7.5.3.

av(t) = u(t) = v0

a

b

ab cos(

ab t) u0

ab sin(

ab t)

v(t) =

Exercise 7.5.1.

a. See the solid curve in Figure A.47.

b.

u(t) = 4sin t + 3 cos t cos = 4/5, sin = 3/5

= 5

4

5 sin t +

3

5 cos t

= 5 (cos sin t + sin cos t)

= 5 sin(t + ) Use formula sin(A + B) = sin A cos B + cos A sin B.

c. The dashed curve in Figure A.47 is the graph of y = 5sin(t + ) translated vertically 0.4 units.

Without translation, the graphs are identical.

Exercise 7.5.3. For a = b = 1, u0 = 3 and v0 = 4, v(t) = v0 cos(

ab t) + u0

ba sin(

ab t) becomes

v(t) = 4 cos(t) + 3 sin(t). With = arccos(3/5), v(t) = 5(sin cos t + cos sin t) = 5sin(t + ).


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0 20 40 60 80 100 120 1400

20

40

60

80

Hare Pelts Harvested (thousands)

LynxPeltsHarvested

(th

ousands)

75

76

77

78

79

8081

82

83

84

85

86

87

88

Figure A.48: Phase graph of pelts harvested by the Hudson Bay Company, 1875 to 1888. There is clearlya clockwise rotation.

Exercise 7.5.5. 1. At time t1 the predator is at its maximum and the prey is decreasing at its

maximum rate of decrease. 2. At time t2 the prey is at its minimum and the predator is decreasing at its

maximum rate of decrease.

Exercise 7.5.6. a. 1887: Lynx 29, Hare 57; 1888: Lynx 17, Hare 18.

b. See Figure A.48.c. There is clearly a clockwise rotation to the path. This means that low hare population leads to

increased lynx population and high hare population leads to decreased lynx population, for example,

which is contrary to real population forces.


Explore 7.6.2. The light switches off and on rapidly. When farther from the mirror, the light continues

to switch off and on, but at a less rapid pace, because the light intensity striking the photoreceptor on

the light is lower.

Exercise 7.6.1.

a. The voltage is analogous to the predator.

b. Let v(t) be the voltage and i(t) be the light intensity. Then

i(t) = a v(t), and v(t) = b i(t)

Exercise 7.6.2.

a. There are numbers a and b such that

v(t) = a i(t) b


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0 1 2 3 4 5

0

1

2

3

4

5

u(t), Prey

v(t),

Predator

Ve

Ue

Figure A.49: Exercise 7.5.8. Phase graph of u(t) = 2et/10 cos t 0.2et/10 sin t vs v(t) = et/10 sin t.The graph is drawn around an hypothesized equilibrium point (Ue, Ve) = (3, 2).

b.

vk+1 vk

= a ik b, or vk+1 = vk + (a ik b)

c. Matlab code to generate Figure 7.6.2 follows.

close all;clc;clear

a=2; b=1; Von=1.5; Voff=2.5; delta=0.1;

volt(1)=1.25; illum(1)=0;

for k=1:230

volt(k+1) = volt(k) +(a*illum(k)-b)*delta;

if volt(k+1)


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Exercise 7.6.4.

a. y(t) = v.

b. y(t) = L + vt.

c. At time zero, the force on the block is zero.

d. At time t1 = Fstart/(kv) the block will first move.

Exercise 7.44.

x + 2x =k

mvt fslide

m2 =

k

mx(t1) = 0, x

(t1) = 0

x(t) = Fstart + Fslidek

cos( (t t1)) v

sin((t t1)) + vt Fslidek

x(t) = Fstart + Fslidek

sin( (t t1)) v

cos((t t1)) + v

x(t) = 2Fstart + Fslidek

cos( (t t1)) + 2 v

sin((t t1))

x(t1) = Fstart + Fslide

k + vt1 Fslide

k

=Fstart + Fslide

k+ vFstart/(kv) Fslide

k= 0

x(t1) = v

+ v = 0

x + 2x = 2Fstart + Fslidek

cos( (t t1)) + 2 v

sin((t t1))

+2Fstart + Fslidek cos( (t t1))

v

sin((t t1)) + vt F

slidek

= 2vt 2Fslidek

=k

mvt Fslide

m

Exercise 7.6.6. Graphs of

x(t)5 + 4

1cos

1/1 (t 50)

0.1

1/1sin

1/1(t 50)

+ 0.1t 4

1(A.1)

and its translations in time are shown in Figure A.50. At time t = 53.33, x.

= 0 and x

.

= 2.33. Theslippage is complete as the movement was 2.33. The next slippage begins at t

.= 73.33.



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a

48 49 50 51 52 53 54 550.5

0

0.5

1

1.5

2

2.5

3

Time t

Earth

slippage

x(t)

b

0 10 20 30 40 50 60 70 80 90 100 110

0

1

2

3

4

5

6

7

Time t

Earth

slippage

x(t)

Figure A.50: a. Graph of earthquake Equation A.1 and b. repeated instance of slippage.

a

0

1

2

18 0 6 12 18

Circadian Time

ConcentrationoffrqandofFRQ

b

0 1 20

1

2

Concentration of frq

ConcentrationFRQ

1821

0

2

610

Figure A.51: a. Hypothetical graphs of the levels of messenger RNA, frq, (solid curve) and its proteinproduct FRQ (dashed curve). b. Phase plane of frq and FRQ.


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Explore 8.1.2.

b. Suppose

limba

P(b) P(a)b a = L < 0.

Let = L/2 > 0. There is a number > 0 such that

if |b a| < thenP(b) P(a)b a L

< = L/2.Let b = a + /2. Then |b a| < and

L (L/2) 0, ex is an increasing function by Theorem 8.1.2.

Exercise 8.1.3. The maximum concentration occurs at t = 5 ln 2 and is equal to C(5 ln 2) = 2.

Exercise 8.1.5.

f f f f a. [0, 2] [2, 2] c. [2, 0], [1/3, 2]

e. Decreases [2, 2] g. [2, 0] [2, 1/2][

3, 2]

i. (0 < x 2] Decreases k. [, 0] [, /2]

Exercise 8.1.6.

According to Theorem 5.2.2, if a1, a2, , an are positive numbers thena1 + a2 + + an

n na1 a2 an.

Let ak = k. Then1 + 2 + + n

n n1 2 n.

From

1 + 2 + (n 1) + n =n(n + 1)

2

we can write1 + 2 + + n

n=

n + 1

2.


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Thenn + 1

2 n1 2 n, and

n + 1

2

n 1 2 n = n!.

Exercise 8.1.7. Waste concentration will increase in Moose lake for the first 20 days and then decrease.

Exercise 8.2.1. Fold the ends of width 3 up 1 and make a 3 by 2 box of volume 6.

Exercise 8.2.3. Chose x = (7 13)/6 .= 0.5674 and the volume will be approximately 3.032, which is0.032 larger than the box in Example 8.2.1.

Exercise 8.2.5. From Example 8.2.3 of the optimal dimensions of a tuna can, if only one material is

used to make the sides and ends of the can, the can that has a fixed volume and requires the least

material should have height equal to the diameter of the can. The diatom in Exercise Figure 8.2.5a hasheight remarkably close to its diameter. However, the ends are punctuate (have holes in them) and the

silicon per square micron is less than that of the sides.

Suppose the silicon per square micron in the ends is a fraction F of the material in the sides. Then

the silicon S required for the diatom of volume V is

S = 2 r2 F + 2r h, and V = r2h

S = 2

r2

F + 2r

V

r

2

S = 4rF 2Vr2

S = 0 implies that V = 2r3F

From V = r2h and V = 2r3F, we find that h = F 2r or h = F d. The diatom in ExerciseFigure 8.2.5b seems to have h

.= 0.25d.

Exercise 8.2.7. Area =3.

Exercise 8.2.8. Maximize Each pen 15 by 20 meters, total area = 600 square meters.

Exercise 8.2.9. Pens are 4 by 7/4 meters; cost = $560.

Exercise 8.2.10. x = 1 + 3

4, L =

(1 + 3

4)2 + (2 + 2/ 3

4)2.

Exercise 8.2.11. b. L =

(1 + 3

4)2 + (2 + 2/ 3

4)2 + 32.

Exercise 8.2.12. The box should be 1 m by 1 m at the base and 3/2 m high and have volume 3/2 m3.

Exercise 8.2.13. The box should use 36 m3 of material.


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Exercise 8.2.14. x =.

= 4.16

Exercise 8.2.15. The height should.

= 0.9306.

Exercise 8.2.16. r = 2.

Exercise 8.2.17. w = 1/(2

3).

Exercise 8.2.18. h =

3/4.

Exercise 8.2.19. Run 70 60 .= 62.25m, then swim.

Exercise 8.2.20. Minimize

T(x) =

h21 + x2va

+

h22 + (d x)2va

where h1 is the height of the mountain top above the lake, h2 is your height above the lake, and d is the

horizontal distance from the mountain top to you. T(x) is minimized when

xh21 + x

2=

d xh2s + (d x)2

cos(Angle of incidence) = cos(Angle of reflection)

Exercise 8.2.21. Minimize

x = d3

I1

3

I2 +

3

I2

Exercise 8.2.22. P = M/2.

Exercise 8.2.23. P = .

Exercise 8.2.25. r = 3

15A/ where A is the area of a buffalo skin.

Exercise 8.2.26. The optimum tepee dimensions are height = 3.16h and radius = 2.41h where h2 is the

area of a buffalo skin. The angle at the base is arctan(3.16/2.41).

= 53. If the tepee is made with only

15 buffalo skins, the base angle is approximately 60.

Exercise 8.3.1.Vmax =4

27. H(Vmax) =4

27 /2.

Exercise 8.3.2. Examine the graph in Figure A.52. Find rationale in the statement of Exercise 8.3.2 for

the relative locations of the two horizontal lines.

Exercise 8.3.3.


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0 1 2 3 4 5 60

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

Gastropod Size

NutrientConversion/

BodySize

High Nutrient Level

Low Nutrient Level

Figure A.52: The solid line depicts high nutrient level; the dashed line depicts low nutrient level.

a.

P(u) = Au3 + BL2/u

P(u) = 3Au2 BL2/u2

0 = 3Au2 BL2/u2, u = 4

BL2/(3A)

b.

E(u) = Au2

+ BL2

/u2

P(u) = 2Au 2BL2/u3

0 = 2Au2 2BL2/u3, u = 4

BL2/A

c.

Pfolded = (Ab)u3 + B

mg

x

2 1u

d.

P(x) = Abu3 + xAwu3 + B m

2

g

2

xu

P

(x) = Awu3 B m

2g2

x2u

0 = Awu3 B m

2g2

x2u

x =

B

Aw

mg

u2

e.

E(x) = Abu2 + xAwu

2 + Bm2g2

xu2


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E(x) = Awu2 B m

2g2

x2u2

0 = Awu2 B m

2g2

x2u2

x =

B

Aw

mg

u2

Exercise 8.3.4. Kakapo live on average 95 years and even in the unlikely events that times are so good

that all offspring are male, or so poor that all offspring are female, either condition is unlikely to last 95

years.

A(2).

= 2.01 m2/s. Exercise 8.4.1.

Exercise 8.4.2. x.

= 5.4m/s.

Exercise 8.4.3. h(t).

= 5.24 m/minute.

Exercise 8.4.4. km/min,

2 km/min.

Exercise 8.4.5. d =

704km/hr.

Exercise 8.4.6. After 0.0162 hours = 0.97 minutes the plane will be 0.172 km apart.

Exercise 8.4.7. Approximately 36.14 m per minute.

Exercise 8.4.8. p.

= 7.41 newtons/(cm2-min).

Exercise 8.4.9.

a. x =

5

c. x = 2, y1 = 1 +

2, x = 5(1 + y1)4y1 + y

21

.= 0.3232

Exercise 8.4.10. d = 7/

20.

Exercise 8.4.11.

a. x = 1/

2or 1/

2. Speed =

4.5. c. x = 2or 2.

Exercise 8.5.1.

a. Iterates 2.5000, 2.4365, 2.2866, 0.25917110182 (25 iterations).


67/95


b. 2.6000, 2.6927, 2.9544, OVERFLOW.

c. 2.5000, 2.5257, 2.5360, 2.5400, 2.54264135777 (24 iterations).

Exercise 8.5.3.

an f(an) bn f(bn) mn f(mn) an+1 bn+12.5 0.005212 2.6 0.006889 2.55 0.00089 2.5 2.55

2.5 0.005212 2.55 0.00089 2.525 0.00214 2.525 2.55

2.525 0.002147 2.55 0.00089 2.5375 0.00006 2.5375 2.55

2.5375 0.00006 2.55

0.00089 2.54375

0.00001 2.5375 2.54375

Exercise 8.5.4.

a. f(x) = 2 x2. Bisection, a0 = 1.4, b0 = 1.5.an f(an) bn f(bn) mn f(mn) an+1 bn+1

1.4 0.04 1.5 0.25 1.45 0.1025 1.4 1.45

1.4 0.04 1.45 0.1025 1.425 0.03063 1.4 1.425

1.4

0.04 1.425 0.03063 1.4125

0.00484 1.4125 1.425

Newtons method. f(x) = 2 x2, f(x) = 2x, x0 = 1.4, xn+1 = xn (2 (xn)2)/(2xn).

x0 = 1.4, x1 = 1.41428571429, x2 = 1.41421356421, x3 = 1.41421356237

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