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Answers to Odd Exercises

Chapter 1Section 1.2, page 21

1. The variables are the altitude and the wombat density, whichwe can call a and w, respectively. The parameter is the rainfall,which we can call R.

3. f(0) = 5, f(1) = 6, f(4) = 9.

0

8

10

6

4

2

0 1 2 3 4 5x

f(x)

5. h(1) = 1

5, h(2) = 1

10, h(4) = 1

20.

0

8

10

6

4

2

0 1 2 3 4 5z

h(z)

7. (2, 2).

–2

6

8

4

2

01 2 3 4

x

y

9. (4, 12).

0

8

10

6

2

0 1 2 3 4x

y11. f(a) = a + 5, f(a + 1) = a + 6, f(4a) = 4a + 5.

13. h( c

5

)= 1

c, h

(5

c

)= c

25, h(c + 1) = 1

5c + 5.

15. I put them in increasing order to look nice.

10000

14000

15000

13000

12000

11000

OR ID NV UT WA CAState

Hig

hest

alti

tude

17.x f (x ) g(x ) (f + g) (x )

−2 −1 −11 −12

−1 1 −8 −7

0 3 −5 −2

1 5 −2 3

2 7 1 8

–15

15

10

5

–10

–2 –1 2x

f (x)

g(x)

sum

19.x F (x ) G (x) (F + G ) (x )

−2 5 −1 4

−1 2 0 2

0 1 1 2

1 2 2 4

2 5 3 8

–2

10

12

864

–1 1 2x

sum

F(x)

G(x)

803

804 Answers to Odd Exercises

21.x f (x ) g(x ) ( f · g) (x )

−2 −1 −11 11

−1 1 −8 −8

0 3 −5 −15

1 5 −2 −10

2 7 1 7

–20

10

15

5

–10

–1 2x

product

f(x)

g(x)

23.x F (x ) G (x ) (F · G ) (x )

−2 5 −1 −5

−1 2 0 0

0 1 1 1

1 2 2 4

2 5 3 15

–5

10

15

5

–1 1 2x

product

F(x)G(x)

25. If we write y = 2x + 3, we can solve for x with the steps

y − 3 = 2x subtract 3 from both sides

y − 3

2= x divide both sides by 2.

Therefore f −1(y) = y − 3

2. Also, f(1) = 5, and f −1(5) =

5 − 3

2= 1.

27. The function F(y) fails the horizontal line test because, forexample, F(−1) = F(1) = 2. Therefore it has no inverse.

29.

–10

10

15

5

–5–1–2–3–4–5 1 2 3 4 5

x

y

the point (1, 5)

The function

––5

2

–1

1

–3–4

–1–2–3–4–5–6 1 3 4 5 6x

y

the point (5, 1)

The inverse

31. This function doesn’t have an inverse because it fails the hor-izontal line test. From the graph, we couldn’t tell whetherF−1(2) is 1 or −1.

10

15

20

25

30

5

–1–2–3–4–5 1 2 3 4 5x

y

the point (1, 2)

The function

–2.5

21.5

2.5

10.5

0

–1–1.5

2

–0.5 1 2 543 6x

y

the point (2, 1)

What the inverse would look like

the point (2, –1)

33. ( f ◦ g)(x) = f(3x − 5) = 2 · (3x − 5) + 3 = 6x − 7 and(g ◦ f )(x) = g(2x + 3) = 3 · (2x + 3) − 5 = 6x + 4. Thesedon’t match, so the functions do not commute.

35. (F ◦ G)(x) = F(x + 1) = (x + 1)2 + 1 = x2 + 2x + 2 and(G ◦ F)(x) = G(x2 + 1) = x2 + 2. These do not match, so thefunctions do not commute.

37. The cell volume is generally increasing but decreases duringpart of its cycle. The cell might get smaller when it gets readyto divide or during the night.

39. The height increases up until about age 30 and then decreases.

41.

initial

tiny

intermediate

start tiny levels out

Population

Time

Answers to Odd Exercises 805

43.

average temp

minimum

maximum

day night day night

Temperature

Time

45. When f = 0, b = 1; when f = 10, b = 21; when f = 20,b = 41. Perhaps one bee will check out the plant even if thereare no flowers.

0

40

50

30

20

10

0 5 10 15 20 25Number of flowers

Num

ber

of b

ees

47. When T = 10, A = 35; when T = 20, A = 30; when T = 30,A = 25; when T = 40, A = 20. The insect develops mostquickly (in the shortest time) at the highest temperatures.

0

30

40

20

10

10 20 30 40Temperature

Dev

elop

men

t tim

e

49.

0

8

10

4

6

2

0.50 1 1.5 2 2.5 3Age

Length as a function of age

Len

gth

51.

0

1

0.6

0.8

0.4

0.2

20 4 6 8 10Length

Tail length as a function of length

Tai

l len

gth

53. B(M) = B(5W + 2) = 2.5W + 1. Plugging in W = 10 gives26 bites.

55. F(V(I )) = F(5I 2) = 37 + 2I 2. The fever is 39◦C if I = 1.

57. The formula is P(t) = (1 + t2) + (1 − 2t + t2) = 2 − 2t +2t2.

t a (t ) b(t ) P (t )

0.00 1.00 1.00 2.00

0.50 1.25 0.25 1.50

1.00 2.00 0.00 2.00

1.50 3.25 0.25 3.50

2.00 5.00 1.00 6.00

2.50 7.25 2.25 9.50

3.00 10.00 4.00 14.00

0

14

6

12108

24

0 1 2 3Time

Popu

latio

n

P(t)

a(t)

b(t)

Here a is increasing, b decreases to 0 at time 1.0 and then in-creases, and the sum P decreases slightly and then increases.

59. Because the mass is the same at ages 2.5 days and 3.0 days,the function relating a and M has no inverse. Knowing themass does not give us enough information to estimate theage.

0

6

2

5

4

3

1

0 1 2 3Age

Mas

s

61. Glucose production is 8.2 mg at ages 2.0 days and 3.0 days.We cannot figure out the age from this measurement.

0

10

4

8

6

2

0 1 2 3Age

Glu

cose

pro

duct

ion


63. Denote the total mass by T(t). Then T(t) = P(t)W(T ) =(2.0 × 106 + 2.0 × 104t)(80 − 0.5t). Measuring populationin millions gives

t P (t ) W (t ) T (t )

0.0 2.0 80.0 160.0

20.0 2.4 70.0 168.0

40.0 2.8 60.0 168.0

60.0 3.2 50.0 160.0

80.0 3.6 40.0 144.0

100.0 4.0 30.0 120.0

0

4

1

3

2

0 4020 60 80 100Year

Popu

latio

n (m

illio

ns)

Population

0

100

40

20

80

60

0 4020 60 80 100Year

Mass per individual

Mas

s pe

r in

divi

dual

0

200

80

40

160

120

0 4020 60 80 100Year

Total mass

Mas

s (m

illio

ns o

f ki

logr

ams)

The population increases, the mass per individual decreases,and the total mass increases and then decreases.

65. We denote the total mass by T(t). Then T(t) = P(t)W(T ) =(2.0 × 106 + 1000t2)(80 − 0.5t). Measuring population inmillions gives

t P (t ) W (t ) T (t )

0.0 2.0 80.0 160.0

20.0 2.4 70.0 168.0

40.0 3.6 60.0 216.0

60.0 5.6 50.0 280.0

80.0 8.4 40.0 336.0

100.0 12.0 30.0 360.0

The population decreases, the mass per individual increases,and the total mass increases.

0

12

4

2

10

6

0 4020 60 80 100Year

Popu

latio

n (m

illio

ns)

8

Population

30

80

50454035

75

6055

0 4020 60 80 100Year

Mas

s pe

r in

divi

dual

7065

Mass per individual

0

400

250

200

300

0 4020 60 80 100Year

Mas

s (m

illio

ns o

f ki

logr

ams)

350

Total mass

Section 1.3, page 38

1. 3.4 lb ×16oz

lb× 28.35

g

oz≈ 1542.24 g.

3. 60 yr ×365.25days

yr× 24

hr

day= 525960 hr.

5. 2.3g

cm3× 1

28.35

oz

g× 1

16

lb

oz× 2.543 cm3

in.3× 123 in.3

ft3≈ 141.58

lb/ft3.

7. 2.3 cm is 0.023 m, so the final height is 1.34 + 0.023 =1.363 m.

9. The total weight of apples is 6 · 145 = 870 g and the totalweight of oranges is 7 · 123 = 861 g, for a grand total of870 + 861 = 1731 g.


11. The area of the square is 1.72 = 2.89 cm2, but the area of thedisk is πr2 = π · 12 cm2 ≈ 3.1415 cm2. The disk has a largerarea.

13. The volume of the sphere is4

3· π · 1003 ≈ 4.189 × 106 m3.

The lake has area 3.0 × 106 m2 and a depth of 0.5 m, giving avolume of merely 1.5 × 106 m3. The sphere is much larger.

15. Pressure is force per unit area, or

force

area=

M L

T 2

L2= M

LT 2

17. Rate of spread of bacteria on a plate has dimensions of L2/T ,or area per time.

19. This checks, because length = length

time× time.

21. This checks, becausemass × length

time2= mass × length

time2.

23. The vertical axis is scaled by a value greater than 1.

55.5

6

2

11.5

4

2.53

3.5

–1–2–3–4–5 0 4 5x

4.5

321

4g(x)

g(x)

25. The horizontal axis is scaled by a value less than 1.

1.31.4

0.80.7

1.1

0.91

–1–2–3–4–5 0 4 5x

1.2

321

g(x/3)

g(x)

27. The tree with height 23.1 m will have volume of π · 23.1 ·0.52 ≈ 18.14 m3. When the height is 24.1, the volume is

18.93 m3. The ratio is18.93

18.14≈ 1.043.

29. The bottom portion of the tree 23.1 m tall will have half the vol-ume found earlier, or 9.07 m3. The top part is 23.1/2 = 11.55 mhigh and is thus a sphere with radius of 11.55/2 ≈ 5.78 m. Sub-stituting into the formula for the volume of a sphere, we find806.76 m3. The total volume is 815.83 m3. The 24.1-m treehas a total volume of 9.46 + 916.13 ≈ 925.59 m3. The ratio

of the volumes is925.59

815.83≈ 1.134.

31. The volume is 2.0 · 0.2 · 1.5 = 0.6 m3 = 6.0 × 105 cm3. Thisgives a mass of 6.0 × 105 g = 600 kg.

33. 3200 · 0.45g

individual= 1440 g = 1.44 kg.

35. Let p be the number of petals. Then f = p/4, so b = p

2+ 1.

When p = 0, b = 1; when p = 10, b = 6; when p = 20, b = 11.The number of bees goes up more slowly as a function of thenumber of petals.

0

40

50

30

20

10

0 5 10 15 20 25Number of flowers

Original graph

Num

ber

of b

ees

0

40

50

30

20

10

0 5 10 15 20 25Number of petals

With new units

Num

ber

of b

ees

37. Let H be the development time in hours. Then H = 24A. ThenH = 24(40 − T/2) = 960 − 12T . When T = 10, H = 840;when T = 20, A = 720; when T = 30, A = 600; when T = 40,H = 480.

0

30

40

20

10


Original graph

Dev

elop

men

t tim

e (d

ays)

0

700800900

1000

500600

100200300400


Dev

elop

men

t tim

e (h

ours

)

39.186,000

mile

s≈ 200,000

mile

s≈ 200,000

mile

s× 60,000

in.

mile

= 1.2 × 1010 in.

s≈ 1.2 × 1010 in.

s× 2.5

cm

in.

= 3.0 × 1010 cm

s= 30

cm

ns


If a computer is supposed to do an operation in 0.3 ns, it hadbetter not need to move information for more than the distancelight can travel in that time, or about 9 cm.

41. The volume is

4

3πr3 ≈ 3 · 6.53 × 109 km3

≈ 3 · 300 × 109 km3

≈ 1 × 1012 km3

One kilometer is 105 cm, so the mass of a cubic kilometer ofwater is 1015 m = 1012 kg. Multiplying this by the volume andthe density gives a total of 5 × 1024 kg.

43. You’d catch 6 movies per day, or about 2000 per year, for atotal of 120,000 in your life.

45. The volume of a sphere of radius r is 4πr3/3 ≈ 4r3. The radiusof the cell is 10−3 cm, so the volume is about 4 × 10−9 cm3.The mass of a cell is therefore around 4 × 10−9 g. I weighabout 60 kg, which is 6 × 104 g. The number of cells is then

6 × 104 g

4 × 10−9 g/cell= 1.5 × 1013 cells.

47. The brain is about 2% of my weight and should have about2% of my cells, or 3 × 1011 cells. The number of neurons is1 × 1011, but the total number of cells in the brain is between1 × 1012 and 5 × 1012, a bit higher than the previous estimates.

49. The length of the string would be 2πr ≈ 40840.704 km.Adding 1 m would make it 40840.705 km. The radius corre-

sponding to this is r = 40840.705

2π≈ 6500.0002 km. The string

would be 0.0002 km, or 0.2 m, above the earth. It is amazingthat such a relatively tiny change in the length of the stringwould produce such a big effect.

Section 1.4, page 511. The points are (1, 5) and (3, 9). The change in input is 2, the

change in output is 4, and the slope is 2. This is not a pro-portional relation because the ratio of output to input changesfrom 5 at the first point to 3 at the second point. This relationis increasing because larger values of x lead to larger valuesof y (and the slope is positive).

0

8

10

12

6

2

4

1 2 3 4x

y

3. The points are (1, 3) and (3, 13). The change in input is 2, thechange in output is 10, and the slope is 5. This is not a pro-portional relation because the ratio of output to input changesfrom 3 at the first point to 4.333 at the second point. This re-lation is increasing because larger values of w lead to largervalues of z (and the slope is positive).

–5

0

10

15

20

5

1 2 3 4w

z

5. The point lies on the line because f(2) = 2 · 2 + 3 = 7. Thepoint-slope form is f(x) = 2(x − 2) + 7. Multiplying outgives f(x) = 2x − 4 + 7 = 2x + 3, as it should.

7. Multiplying out, we find that f(x) = 2x + 1. The slope is 2and the y-intercept is 1. The original point is (1, 2).

0

8

10

12

6

2

4

10 32

original pointintercept

4 5x

f(x)

9. In point-slope form, this line has equation f(x) = −2(x −1) + 6. Multiplying out, we find that f(x) = −2x + 8. Theslope is −2 and the y-intercept is 8.

0

6

8

10

2

4

10 32

original point

intercept

4x

f(x)

11. The slope between the two points is

slope = change in output

change in input= 3 − 6

4 − 1= −1

In point-slope form, the line has equation f(x) = −1 · (x −1) + 6. In slope-intercept form, it is f(x) = −x + 7. This linehas slope −1 and y-intercept 7.

0

6

8

2

4

0 42

(1, 6)

(4, 3)

intercept

6x

f(x)

13. This is not linear because the input z appears in the denomi-nator.


15. This is linear because the input q is multiplied only by con-stants and has constants added to it.

17. h(1) = 1

5, h(2) = 1

10, h(4) = 1

20. The slope between z = 1 and

z = 2 is −1/10, and that between z = 2 and z = 4 is −1/40.

0

0.2

0 3 41

slope = –1/10

slope = –1/40

2 5z

h(z)

0.4

19. 2x = 7 − 3 = 4, so x = 4/2 = 2. Plugging in, 2 · 2 + 3 = 7.

21. 2x − 3x = 7 − 3 = 4, so −x = 4 or x = −4. Plugging in,2 · (−4) + 3 = −5 = 3 · (−4) + 7.

23. Multiplying out, we get 10x − 4 = 10x + 5. This has nosolution.

25. 2x = 7 − b, so x = 7 − b

2.

27. (2 − m)x = 7 − b, so x = 7 − b

2 − m. There is no solution if m = 2.

However, if m = 2 and b = 7, both sides are identical and anyvalue of x works.

29. 1 in. = 2.54 cm. The slope is 2.54cm

in..

0

12

10

8

6

4

2

0 31 2 4Inches

Cen

timet

ers

31. 1 g = 1

453.6lb ≈ 0.0022 lb. The slope is 0.0022

lb

g.

0

2.5

2

1.5

1

0.5

0 600 800200 400 1000Grams

Poun

ds

33. ( f ◦ g)(x) = (mx + b) + 1 and (g ◦ f )(x) = m(x + 1) +b = mx + m + b. These match only if the intercepts are equal,or b + 1 = m + b. This is true for any b as long as m = 1. Inthis case, both f and g have slope 1, meaning that each justadds a constant to its input. The order cannot matter becauseaddition is commutative.

35. The slope is 1.0 cm, and the equation is V = 1.0A.

37. The slope is 5.0 × 10−9 g, and the equation is M = 5.0 ×10−9b.

39. The line has slope −0.2 and intercept 10,000. The equation isthus a = −0.2d + 10,000.

41. a = −0.2 · 2000 + 10000 = 9600 ft.

43.

4190

4210

4205

4200

4195

1965 1970 1975 1980 1985 1990 1995Year

Surf

ace

elev

atio

n (f

t)

45. The slope is 3 ft every 5 years. It would have been up to 4208by 1990, 5 ft higher than the actual level.

47. Using the first two rows for mass, we find a slope of

slope = change in mass

change in age= 4.0 − 2.5

1.0 − 0.5= 3.0.

Using (1.0, 4.0) as the base point,

M = 3.0(a − 1.0) + 4.0 = 3.0a + 1.0.

The y-intercept is 1.0, meaning that the mass was 1.0 g at age0. This might be the mass of a new seedling. Interpolating ata = 1.75, M = 3.0 · 1.75 + 1.0 = 6.25.

0

10

8

6

4

2

0 1 2 3Age

Mas

s

49. The glucose production changes by 3.4 when the mass changesby 1.5, giving a slope of 2.27. The point-slope form of theline, using the last data point, is G = 2.27(m − 10.0) + 17.0,which simplifies to G = 2.27m − 5.7 in slope-intercept form.The negative intercept must mean that this plant would notstart making glucose until the mass got beyond a certain value(around 2). When m = 20.0, G = 39.7.

0

1618

1412108642

0 2 4 6 8 10Mass

Glu

cose

pro

duct

ion


51. The point (2.0, 175) lies below the line.

180200

1601401201008060

0 0.5 1 1.5 2Weight W

Num

ber

N

53. N = 70(0.72 − 0.5) + 80 = 95.4.

55. The time decreases from 216.8 to 215.9 seconds, giving achange of −0.9 second in 16 years or a slope of −0.9/16 ≈−0.0563 second per year. In point-slope form, the men’s timem as a function of the year y is

m = −0.0563(y − 1972) + 216.8

57. Setting equal to 0 and solving for y,

−0.469(y − 1972) + 241.4 = 0

0.469(y − 1972) = 241.4

(y − 1972) = 241.4

0.469

y = 241.4

0.469+ 1972 = 2486.

It seems likely that this will never happen, so we can assumethat women will not improve this quickly forever.

Section 1.5, page 641. The updating function is f(pt ) = pt − 2, and f(5) = 3,

f(10) = 8, f(15) = 13. This is a linear function.

3. The updating function is f(xt ) = x2t + 2, and f(0) = 2, f(2) =

6, f(4) = 18. This is not a linear function because the input xt

is squared.

5. Denote the updating function by f(v) = 1.5v. Then ( f ◦f )(v) = f(1.5v) = 1.5(1.5v) = 2.25v, so vt+2 = 2.25vt . Ap-plying f to the initial condition twice gives f(1220) = 1830and f(1830) = 2745, which is equal to 2.25 · 1220.

7. Denote the updating function by h(n) = 0.5n; then (h ◦h)(n) = h(0.5n) = 0.5(0.5n) = 0.25n and nt+2 = 0.25nt . Ap-plying the updating function to the initial condition twice givesh(1200) = 600 and h(600) = 300, matching (h ◦ h)(1200) =0.25 · 1200 = 300.

9. Solving for vt gives vt = vt+1

1.5. Then v0 = 1220

1.5= 813.3.

11. Solving for nt gives nt = 2nt+1. Then n0 = 2 · 1200 =2400.

13.( f ◦ f )(x) = f( f(x)) = f

(x

1 + x

)=

x

1 + x

1 + x

1 + x

=x

1 + x1 + 2x

1 + x

= x

1 + 2x.

To find the inverse, set y = f(x) and solve

y = x

1 + x

(1 + x)y = x

y + xy = x

y = x − xy

y

1 − y= x .

Therefore, f −1(y) = y

1 − y.

15. vt = 1.5t · 1220 µm3.

v1 = 1.5 · 1220 = 1830

v2 = 1.5 · 1830 = 2745

v3 = 1.5 · 2745 = 4117.5

v4 = 1.5 · 4117.5 = 6176.25

v5 = 1.5 · 6176.25 = 9264.375.

10000

8000

6000

4000

2000

00 1 2 3 4 5

t

Solution

v t

10000

8000

6000

4000

2000

00 2000 4000 6000 800010000

vt

Updating function

v t +

1

17. nt = 0.5t · 1200.

n1 = 0.5 · 1200 = 600

n2 = 0.5 · 600 = 300

n3 = 0.5 · 300 = 150

n4 = 0.5 · 150 = 75

n5 = 0.5 · 75 = 37.5.


1500

1000

500

00 1 2 3 4 5

t

Solution

n t

1500

1000

500

00 500 1000 1500

nt

Updating function

n t +

1

19. Plugging t = 20 into the solution vt = 1.5t · 1220 µm3, we getv20 = 4.05 × 106 µm3. This might be reasonable.

21. Plugging t = 20 into the solution nt = 0.5t · 1200, we getn20 = 0.0011. This is an unreasonably small population.

23. x1 = 1/2, x2 = 1/3, x3 = 1/4, x4 = 1/5. It looks like xt =1

1 + t.

25. x1 = 3, x2 = 1, x3 = 3, x4 = 1. It seems to be jumping backand forth between 1 and 3. If I start at x0 = 0, the results jumpback and forth between 0 and 4.

27. These do not commute. If you started with 100, doubled (giv-ing 200), and then removed 10, you would end up with 190.If you started with 100, removed 10 (leaving 90), and thendoubled, you’d have only 180. In general, if we call the start-ing population Pt , if we double first and then remove 10, weend up with Pt+1 = 2Pt − 10. If we first remove 10 and thendouble, we end up with Pt+1 = 2(Pt − 10) = 2Pt − 20, whichnever matches the result in the other order.

29. These do commute. Either way, it ends up 1.0 cm taller.

31. h20 = 10.0 + 20 = 30.0 m, a reasonable height for a tree.

33. 1.05 × 106 million bacteria. These will weigh about 10−6 g,which sounds reasonable.

35. The solution is x1 = 50, x2 = 130, x3 = 290. Adding 30, wesee that x0 + 30 = 40, x1 + 30 = 80 = 40 · 2, x2 + 30 = 160 =40 · 22, and x3 + 30 = 320 = 40 · 23. It looks like xt + 30 =40 · 2t , so xt = 40 · 2t − 30.

37. 3000

2500

2000

1500

10001000 1200 1400 1600 1800 2000

vt

v t +

1

The discrete-time dynamical system is vt+1 = 1.5vt and themissing value is 2130.

39. 1500

1000

500

00 500 3000

Old number

New

num

ber

1000 150020002500

The discrete-time dynamical system is nt+1 = 0.5nt and themissing value is 4.0 × 102.

41. The length increases by 1.5 cm each half-day, so lt+1 =lt + 1.5 cm.

10

6

8

2

4

00 2 10

lt

l t +

1

4 6 8

43. The mass doubles each half-day, so mt+1 = 2mt .

50

30

40

10

20

00 10 50

mt

mt +

1

20 30 40

45. The argument is the initial score. The value is the final score.

47. 100

30405060708090

1020

00 30 100

Initial score

Fina

l sco

re

2010 5040 7060 9080

49. Let vt+1 and vt be the total volume before and after the exper-iment. Then

vt = 104bt and vt+1 = 104bt+1.

The original discrete-time dynamical system is

bt+1 = 2.0bt .


Therefore,

vt+1 = 104bt+1

= 104(2.0bt )

= 2.0 · 104bt

= 2.0vt .

51. Vt = πht 0.52, and Vt+1 = πht+10.52. Therefore,

Vt+1 = π(ht + 1)0.52 = πht · 0.52 + π · 0.52 = Vt + π · 0.52.

53. The points for the first patient are (20.0, 16.0), (16.0, 13.0),and (13.0, 10.75).

5

10

15

20

00

Mt

Mt +

1

5 10 15 20

Let the level be Mt at the beginning of the day. The two pointsare (20.0, 16.0) and (16.0, 13.0). The slope is

slope = change in output

change in input= 13.0 − 16.0

16.0 − 20.0= 0.75.

In point-slope form, the discrete-time dynamical system is

Mt+1 = 0.75 (Mt − 20.0) + 16.0 = 0.75Mt + 1.0.

55. The solution for the first is bt = 2.0t · 1.0 × 106, and the so-lution for the second is bt = 2.0t · 3.0 × 105. The differenceis 2.0t · 0.7 × 106, but the ratio is always approximately 3.33.Both populations are growing at the same rate, but the first hasa head start. It is always 3.33 times bigger, which becomes alarger difference as the populations become larger.

57. a. b1 = 2b0 − 1.0 × 106 = 2 · 3.0 × 106 − 1.0 × 106 = 5.0 ×106, b2 = 9.0 × 106, b3 = 17.0 × 106.

b. There were three harvests of 1.0 × 106 bacteria, for a totalof 3.0 × 106 bacteria.

c. bt+1 = 2.0bt − 1.0 × 106.

d. The population would have doubled three times, so therewould be 24.0 × 106 bacteria. You could harvest 7.0 × 106

bacteria and still have a population of 17.0 × 106, whichmeans harvesting 4.0 × 106 more than in part b. The bac-teria removed early never had a chance to reproduce.

59. a. If the old fraction is ft , then ft+1 = ft + 0.1 ft , or ft+1 =1.1 ft .

b. ft = 0.001 · 1.1t .

c. The fraction gets larger and larger and will eventually ex-ceed 1. The discrete-time dynamical system doesn’t makesense when ft+1 > 1 because a fraction can’t be biggerthan 1.

61. We have that bt+1 = 2bt and mt+1 = 3mt . Then the total massMt+1 = mt+1bt+1 = 3mt · 2bt = 6mt bt = 6Mt .

Section 1.6, page 761.

5

10

00

b t +

1

5 10bt

bc

d e

f

a

10

5

00 3

Time

Solution

Popu

latio

n (m

illio

ns)

1 2

3. The solution was vt = 1.5t · 1220 µm3, consistent with a cob-web diagram that predicts a solution that increases faster andfaster.

6000

4000

2000

00 3

t

Solutionv t

1 2

4000

2000

6000

00

v t +

1

2000 4000

Cobwebbing

6000vt

5. The solution was nt = 0.5t · 1200, consistent with a cobwebdiagram that decays toward 0.

1500

1000

500

00 3

t

Solution

n t

1 2


1000

500

1500

00

n t +

1

500 1000

Cobwebbing

1500nt

7.

20

10

30

00

x t +

1

10 20 30xt

9.5

4

3

2

1

6

00

wt +

1

1 2 3 4 5 6wt

11. 1

0.8

0.6

0.4

0.2

00

x t +

1

0.2 0.4 0.6 0.8 1xt

13. The equilibrium seems to be at about 1.3.

10

8

6

4

2

00

Fina

l val

ue

2

equilibrium

4 86 10Initial value

15. The equilibria seem to be at about 0.0 and 7.5.

10

8

6

4

2

00

New

val

ue

2

equilibrium

equilibrium

4 86 10

Old value

17. f(x) = x when x2 = x , or x2 − x = 0, or x(x − 1) = 0, whichhas solutions at x = 0 and x = 1.

2

1.5

1

0.5

00

Fina

l val

ue

0.5 1 1.5 2Initial value

19. The equilibrium is where c∗ = 0.5c∗ + 8.0, or c∗ = 16.0.

30

20

10

00

c t +

1

10 20 30ct

equilibrium

diagonal

updating functionlies below diagonal

updatingfunction

lies abovediagonal

21. The equilibrium is b∗ = 0.

10

5

00

b t +

1

5 10bt

equilibrium

diagonal

updating function alwayslies below diagonal

23. v∗ = 1.5v∗ if v∗ = 0.

25. x∗ = 2x∗ − 1 has solution x∗ = 1.

27. w∗ = −0.5w∗ + 3 has solution w∗ = 2.

29. x∗ = x∗

1 + x∗ has solution x∗ = 0.

31. w∗ = aw∗ + 3

w∗ − aw∗ = 3

w∗ = 3

1 − a.

This solution does not exist if a = 1, and it is negative if a > 1.

33. x∗ = ax∗

1 + x∗

x∗(1 + x∗) = ax∗

x∗(1 + x∗) − ax∗ = 0

x∗(1 + x∗ − a) = 0.

There are two solutions, x∗ = 0 and x∗ = a − 1. The secondis negative if a < 1. The two solutions are equal (leaving onlyone) when a = 1.


35.

30252015105

3540

00

h t +

1

15105 35302520 40ht

37.

8000

6000

4000

2000

10000

00

v t +

1

40002000 80006000 10000vt

39.

800

600

400

200

1000

00

n t +

1

400200 800600 1000nt

41.

15

10

5

20

00

Mt +

1

105 15 20Mt

The equilibrium is

M∗ = 0.75M∗ + 1

M∗ − 0.75M∗ = 1

0.25M∗ = 1

M∗ = 4.

43.

15

10

5

20

00 105 15 20

bt

b t +

1

The equilibrium is

b∗ = 2.0b∗ − 1.0 × 106

b∗ − 2.0b∗ = −1.0 × 106

−b∗ = −1.0 × 106

b∗ = 1.0 × 106.

The population grows, as we found in Section 1.5, Exercise57, and seems to be moving away from the equilibrium.

45. The equilibrium is

b∗ = 2.0b∗ − h

b∗ − 2.0b∗ = −h

−b∗ = −h

b∗ = h.

It is strange that the equilibrium gets larger as the harvest getslarger. However, the cobwebbing diagram indicates that onlypopulations above the equilibrium will grow, and those belowit will shrink. The equilibrium in this case is the minimumpopulation required for the population to survive.

Section 1.7, page 891. Law 6: 43.20 = 1.

3. Law 3: 43.2−1 = 1/43.2 ≈ 0.023.

5. Law 4: 43.27.2/43.26.2 = 43.27.2−6.2 = 43.21 = 43.2.

7. Law 2: (34)0.5 = 34 · 0.5 = 32 = 9.

9. To use law 1, we must first multiply out the exponents, finding223 · 222 = 28 · 24 = 212 = 4096.

11. Law 6: ln(1) = 0.

13. Law 5: log43.2 43.2 = 1.

15. Law 1: log10(5) + log10(20) = log10(5 · 20) = log10(100) =log10(102) = 2.

17. Law 4: log10(500) − log10(50) = log10(500/50) = log10(10)

= 1.

19. Law 2: log43.2(43.27) = 7.

21. Law 3: log7

( 1

43.2

)= − log7 43.2 ≈ −1.935.

23. e3x = 21

7= 3. Taking logs of both sides, 3x = ln(3) ≈ 1.099,

and x ≈ 0.366. Checking, 7e3 · 0.366 ≈ 21.0.

25. Taking logs of both sides gives ln(4) − 2x + 1 = ln(7) + 3x .Moving the x’s to one side gives 5x = ln(4) + 1 − ln(7) =1 + ln

(4

7

)≈ 0.440, and x = 0.088. Checking, 4e−2 · 0.088+1 =

7e3 · 0.088 ≈ 9.11.

27. e2x = 7 at 2x = ln(7.0) ≈ 1.946 or x ≈ 0.973. This function

is increasing, and it doubles after a “time” of x = ln(2)

2≈

0.346.


50

40

30

20

10

60

–2 0–1 1 2x

Therefore, e2x = 14.0 when x ≈ 0.973 + 0.346 = 1.319, ande2x = 3.5 when x ≈ 0.973 − 0.346 = 0.627.

29. 5e0.2x = 7 when e0.2x = 1.4, or 0.2x = ln(1.4) ≈ 0.336 or x ≈1.68. This function is increasing, and it doubles after a “time”

of x = ln(2)

0.2≈ 3.46.

12108642

14

–5 0–1–2–3–4 1 2 3 4 5x

Therefore, 5e0.2x = 14.0 when x ≈ 1.68 + 3.46 = 5.14, and5e0.2x = 3.5 when x ≈ 1.68 − 3.46 = −1.78.

31.

0.5

1

00

Fina

l val

ue

10.5 1.5 2

Initial value

33. The solution is bt = 1.5t · 1.0 × 106. In exponential nota-tion, bt = 1.0 × 106eln(1.5)t ≈ 1.0 × 106e0.405t . The populationreaches 1.0 × 107 when e0.405t = 10, or 0.405t = ln(10) ≈2.302, or t ≈ 2.302

0.405≈ 5.865.

2

4

6

8

10

12

00

b t

21 3 4 5 6

t

35. The solution is vt = 1.5t · 1350. In exponential notation, vt =1350eln(1.5)t ≈ 1350e0.405t . The volume reaches 3250 whene0.405t = 3250

1350= 2.407, or 0.405t = ln(2.407) ≈ 0.878, or t ≈

0.878

0.405≈ 2.17.

600040002000

800010000

1400012000

1600018000

00

v t

21 3 4 5 6

t

37. td ≈ 0.6931/1.0 = 0.6931 days. It will take twice this long toquadruple, or 1.386 days.

39. td ≈ 0.6931/0.1 = 6.931 hr. It will quadruple in 13.86 hr.

41. Converting to base e, we have that S(t) = 2.34eln(10) · 0.5t ≈2.34e1.151t . The doubling time is td ≈ 0.6931

1.151= 0.602 days.

It will have increased by a factor of 10 when αt = 1, whichoccurs when t = 2.

43. Q(50000) = 6.0 × 1010e−0.000122 · 50000 ≈ 1.34 × 108. This is1.34 × 108

6.0 × 1010≈ 0.00223 of the original amount.

45. th ≈ 0.693

0.000122≈ 5680 years.

47. It will have doubled twice and will be 2000.

49. If the doubling time is 24 years, the parameter α is0.693

24≈

0.0289. Therefore, P(t) ≈ 500e0.0289t .

51. It will have halved three times, which takes 129 years.

53. If the half-life is 43 years, the parameter α is −0.693

43≈

−0.0161, and the equation is P(t) ≈ 1600e−0.0161t .

55. We plot the line ln(S(t)) = ln(2.0et ) ≈ 0.693 + t .

4

2

6

8

10

12

00

log(

S(t)

)

42 6 8 10

t

57. ln(P(t)) = ln(500e0.0289t ) ≈ 6.21 + 0.0289t .

76.5

7.58

8.59

9.5

60

log(

P(t

))

4020 60 80 100

t


Section 1.8, page 981. sin(π/2) = 1, cos(π/2) = 0.

0.5

–0.5

–1

1

–p–2p p 2px

sin(x)

sin(x)

5

3

1

–1

1

–p–2p p 2px

cos(x)

cos(x)

1

3

5

On circles

1

3

5

3. sin(π/9) ≈ 0.342, cos(π/9) ≈ 0.940.

5. sin(−2.0) ≈ −0.909, cos(−2.0) ≈ −0.416.

7. π/6 rad.

9. 0.017 rad.

11. 114.6◦.

13. −36◦ = 324◦.

15. 200 grads, after multiplying by400 grads

360◦ .

17. 50.0 grads, after multiplying by400 grads

2π.

19. 135◦, after multiplying by360◦

400 grads.

21. No answer, 0, no answer, 1.

–10

–5

5

–p–2p p 2px

tan(x)10

25

23

2523

–10

–5

5

–p–2p p 2px

cot(x)10

21

3

25 23

–10

–5

5

–p–2p p 2px

sec(x)10

3

25

2321

–10

–5

5

–p–2p p 2px

csc(x)10

23. tan(π/9) ≈ 0.36397, cot(π/9) ≈ 2.74748, sec(π/9) ≈1.06418, csc(π/9) ≈ 2.92380.

25. tan(−2.0) ≈ 2.18504, cot(−2.0) ≈ 0.45766, sec(−2.0) ≈−2.40300, csc(−2.0) ≈ −1.09975.

27. cos(0) = √1, cos

(π

4

)=

√2

2=

√1

2, cos

(π

2

)= 0 =

√0

2.

29. cos(0 − π) = cos(−π) = −1 = − cos(0), cos(π/4 − π) =cos(−3π/4) = −√

2/2 = − cos(π/4), cos(π/2 − π) =cos(−π/2) = 0 = − cos(π/2), cos(π − π) = cos(0) = 1 =− cos(π).

31. cos(2 · 0) = cos2(0) − sin2(0) = 1 − 0 = 1 = cos(0),cos(2 · π/4) = cos2(π/4) − sin2(π/4) = 1/2 − 1/2 = 0 =cos(π/2), cos(2 · π/2) = cos2(π/2) − sin2(π/2) = 0 − 1 =−1 = cos(π), cos(2 · π) = cos2(π) − sin2(π) = 1 − 0 = 1 =cos(2π).

33. Multiplying the factor of 5.0 through gives r(t) = 10.0 +5.0 cos(2π t) with average 10.0, amplitude 5.0, period 1.0,and phase 0.

5

14131211

15

109876

0 1 2 3 4t

r(t)

35. Use the fact that cos(t − π) = − cos(t). Then f(t) = 2.0 +1.0 cos(t − π) with average 2.0, amplitude 1.0, period 2π ,and phase π .

3

4

1

–4 –2 420 6 8 10t

r(t)


37. Average is 6.0, minimum is 4.0, maximum is 8.0, amplitudeis 2.0, period is 5.0, and phase is 2.0.

39. Average is 4.0, minimum is -1.0, maximum is 9.0, amplitudeis 5.0, period is 2.0, and phase is 0.0.

41. The average is 3.0, the amplitude is 4.0, the maximum is 7.0,the minimum is −1.0, the period is 5.0, and the phase is 1.0.

�1

3

7

02 4 6 8 10

x

Amplitude

avg

max

min

phasef(x)

period

43. The average is 1.0, the amplitude is 5.0, the maximum is 6.0,the minimum is −5.0, the period is 4.0, and the phase is 3.0.

Amplitude

�4

1

6

02 4 6 8 10

zavg

max

min

h(z)

periodphase

45. This function increases overall but wiggles around quite a bit.It might describe the size of an organism that grows on aver-age, but grows quickly during the day and shrinks down a bitat night.

321

4567

00 21 3 4 5

t

47. This function wiggles up and down with increasing amplitude.Perhaps it describes the insane temperature oscillations thatwill precede the next ice age.

50

–5–10

10152025

–15

1.5

1

0.5

2 2.5 3t

49. This function wiggles up and down with increasing periodbut with constant amplitude. Perhaps it describes the insanetemperature oscillations that will precede the next ice age.

10.8

1.5

1

0.5

2 2.5 3t

0.60.40.2

0–0.2–0.4–0.6–0.8

–1

51. Let Sc(t) represent the circadian rhythm. Then Sc(t) =cos

(2π t

24

).

�1.5

�1

�0.5

0

0.5

1

1.5

6 12 18 24

S c t

t

53. To plot, compute the value of the combined function Sc(t) +Su(t) every hour, and smoothly connect the dots.

�1.5

�1

�0.5

0

0.5

1

1.5

6

12

18

24

S c t

+ S

ut

t

Section 1.9, page 1091. 2/3 of the water is at 100◦C, and 1/3 is at 30◦C. The fi-

nal temperature is then the weighted average T = 1

3· 30◦C +

2

3· 100◦C ≈ 76.7◦C.

3. A fraction20

52got 50,

18

52got 75, and

14

52got 100, for an average

of20

52· 50 + 18

52· 75 + 14

52· 100 ≈ 72.1.

5. 1/3 of the water is at T1, and 2/3 is at T2. The final tem-

perature is then the weighted average T = 1

3· T1 + 2

3· T2. If

T1 = 30 and T2 = 100, we get T = 1

3· 30 + 2

3· 100 = 76.7 as

before.

7. A fraction V1/(V1 + V2) is at T1, and a fraction V2/(V1 + V2)

is at T2. The final temperature is the weighted average

T = T1V1

V1 + V2+ T2

V2

V1 + V2.


9. The 100◦C water cools to 50◦C, so 2/3 of the water is at50◦C, and 1/3 is at 15◦C. The final temperature is then the

weighted average T = 1

3· 15◦C + 2

3· 50◦C ≈ 38.3◦C. This is

indeed half the value in Exercise 1.

11. After the deduction, a fraction20

52got 40,

18

52got 65, and

14

52got 90, for an average of

20

52· 40 + 18

52· 65 + 14

52· 90 =

62.1. This is indeed 10 less than the average before thededuction.

13. a. amount = volume times concentration, or V · c0 =2.0 · 1.0 = 2.0 mmol.

b. 0.5 L at 1.0 mmol/L = 0.5 mmol.

c. 1.5 L at 1.0 mmol/L = 1.5 mmol.

d. 0.5 L at 5.0 mmol/L = 2.5 mmol.

e. 1.5 mmol + 2.5 mmol = 4.0 mmol.

f. 4.0 mmol/2.0 L = 2.0 mmol/L.

g. q = 0.5/2.0 = 0.25. Then ct+1 = (1 − q)ct + qγ =0.75 · ct + 0.25 · 5.0. When c0 = 1.0, c1 = 0.75 · 1.0 +0.25 · 5.0 = 2.0 mmol/L.

15. Start with 9.0 mmol, breathe out 8.1 mmol, leaving 0.9mmol, breathe in 4.5 mmol, ending with 5.4 mmol and aconcentration of 5.4 mmol/L. This checks with the discrete-time dynamical system. In this case, q = 0.9 and γ = 5.0,so ct+1 = 0.1ct + 0.9 · 5.0. Substituting c0 = 9.0, we findc1 = 5.4.

17. The discrete-time dynamical system has q = 0.5/2.0 =0.25 and γ = 5.0 and thus has formula ct+1 = (1 −0.25)ct + 0.25 · 5.0 = 0.75ct + 1.25. We want to start from1.0 mmol/L.

5

10

00

c t +

1

5 10ct

Cobwebbing

5

10

00

c t

1 2 3t

Solution

19. The discrete-time dynamical system is ct+1 = 0.1ct + 4.5,starting from c0 = 9.0.

5

10

00

c t +

1

5 10ct

Cobwebbing

5

10

00

c t

1 2 3t

Solution

21. The discrete-time dynamical system is ct+1 = 0.75 · ct + 1.25.Solving for the equilibrium, we find c∗ = 0.75 · c∗ + 1.25, or0.25c∗ = 1.25, or c∗ = 5.0. This matches the value of γ .

23. The discrete-time dynamical system is ct+1 = 0.1 · ct + 4.5.Solving for the equilibrium, we find c∗ = 0.1 · c∗ + 4.5, or0.9c∗ = 4.5, or c∗ = 5.0. This matches the value of γ .

25. Using the equation c∗ = qγ

1 − (1 − q)(1 − α)for the equilib-

rium, we find that c∗ = 0.4 · 0.21

1 − 0.6 · 0.9≈ 0.183. The concentra-

tion is higher because more of the air in the lung at any onetime comes from outside.

27. The concentration after absorption is ct − 0.02. Using theweighted-average idea,

ct+1 = (1 − q) (ct − 0.02) + qγ = 0.8 (ct − 0.02) + 0.042

= 0.8ct + 0.026.

The equilibrium solves

c∗ = 0.8c∗ + 0.026

0.2c∗ = 0.026

c∗ = 0.13.

This function does not really make sense if ct < 0.02 becausethere not would be enough there to absorb.

29. The concentration after absorption is ct − 0.2(ct − 0.05) =0.8ct + 0.01. Then

ct+1 = (1 − q)(0.8ct + 0.01) + qγ = 0.8(0.8ct + 0.01)

+ 0.042 = 0.64ct + 0.05.

The equilibrium is then

c∗ = 0.64c∗ + 0.05

0.36c∗ = 0.05

c∗ = 0.0139.


31. The concentration after absorption is ct − A. Then

ct+1 = (1 − q)(ct − A) + qγ = 0.8(ct − A) + 0.042.

The equilibrium solves

c∗ = 0.8c∗ + 0.042 − 0.8A

0.2c∗ = 0.042 − 0.8A

c∗ = 0.21 − 4A.

Then c∗ = 0.15 if 0.21 − 4A = 0.15, or A = 0.015. The lungmust reduce the oxygen concentration by 1.5%. In Exam-ple 1.9.9, we found that the lung absorbs 10% of the equilib-rium concentration of 0.15, which is equivalent to A = 0.15.

33. The concentration before exchanging air is ct + 0.001, so thediscrete-time dynamical system is the weighted average

ct+1 = (1 − q)(ct + 0.001) + qγ

ct+1 = 0.8(ct + 0.001) + 0.2 · 0.0004 = 0.8ct + 0.00088.

The equilibrium is

c∗ = 0.8c∗ + 0.00088

0.2c∗ = 0.00088

c∗ = 0.0044.

This is about 11 times higher than the external concentration.

35. a. Population after reproduction is 0.6 · 3.0 × 106 = 1.8 ×106.

b. Population after supplementation is 1.8 × 106 + 1.0 ×106 = 2.8 × 106.

c. bt+1 = 0.6bt + 1.0 × 106.

37. The discrete-time dynamical system is bt+1 = 0.6bt + 1.0 ×106. The equilibrium satisfies b∗ = 0.6b∗ + 1.0 × 106, or0.4b∗ = 1.0 × 106, or b∗ = 2.5 × 106.

39. The discrete-time dynamical system is bt+1 = 0.5bt + S. Theequilibrium satisfies b∗ = 0.5b∗ + S, or 0.5b∗ = S, or b∗ = 2S.The equilibrium becomes larger when S is large. This makessense because the population will be larger when more bacte-ria are added.

41. Let st be the concentration of salt before inflow and lossthrough outflow. There is then 3.3 × 107st salt in the lake,which receives 3.0 × 103 of salt from inflow, for a total of3.3 × 107st + 3.0 × 103 of salt in 3.6 × 107 cubic meters ofwater. The concentration is

st+1 = 3.3 × 107st + 3.0 × 103

3.6 × 107≈ 0.917st + 8.33 × 10−5.

Because water that flows out is well mixed, this gives thediscrete-time dynamical system.

43. As before, there are 3.3 × 107st + 3.0 × 103 cubic meters ofsalt in 3.6 × 107 cubic meters of water. After evaporation,there are 3.3 × 107 cubic meters of water, in which the con-centration is

st+1 = 3.3 × 107st + 3.0 × 103

3.3 × 107≈ st + 1.0 × 10−4.

45. The discrete-time dynamical system is st+1 = 0.917st +8.33 × 10−5. The equilibrium solves

s∗ = 0.917s∗ + 8.33 × 10−5

0.083s∗ = 8.33 × 10−5

s∗ = 0.001.

The water ends up like the water that flows in.

47. The discrete-time dynamical system is st+1 = st + 1.0 × 10−4.The equilibrium equation is s∗ = s∗ + 1.0 × 10−4, which hasno solution. This lake has no equilibrium and will get saltierand saltier.

49. The discrete-time dynamical system is bt+1 = 1.5bt − 1.0 ×106, and the equilibrium is b∗ = 2.0 × 106.

8

6

4

2

10

00

b t +

1

2 4 6 8 10bt

Section 1.10, page 1181. There are now 400 red birds and 800 blue birds, or 1/3 red

birds and 2/3 blue birds. These fractions add to 1.

3. There are now 200r red birds and 800 blue birds. The frac-tion of red birds is 200r out of 200r + 800, or a fraction

r

r + 4.

There are 800 blue birds out of 200r + 800, or a fraction4

r + 4.

These fractions add to 1 no matter what r is.

5.

0.8

0.6

0.4

0.2

1

00

f(x)

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x

7.pt = mt

mt + bt= 1.2 × 105

1.2 × 105 + 3.5 × 106≈ 0.033

mt+1 = 1.2mt = 1.44 × 105

bt+1 = 2.0bt = 7.0 × 106

pt+1 = mt+1

mt+1 + bt+1= 1.44 × 105

1.44 × 105 + 7.0 × 106≈ 0.020.


9. pt ≈ 0.033, mt+1 = 0.36 × 105, bt+1 = 1.75 × 106, pt+1 ≈0.020.

11. The equation for the equilibria is p∗ = p∗

p∗ + 2.0(1 − p∗). Then

p∗(p∗ + 2.0(1 − p∗)) = p∗ multiply both sidesby the denominator

p∗(p∗ + 2.0(1 − p∗)) − p∗ = 0 subtract p∗ from bothsides

p∗(p∗ + 2.0(1 − p∗) − 1) = 0 factor out p∗

p∗(1.0 − p∗) = 0 simplify

p∗ = 0

or p∗ = 1.0 solve each piece.

13. The equation for equilibria is x∗ = x∗

1 + ax∗ . Solving,

x∗(1 + ax∗) = x∗ multiply both sides by thedenominator

x∗(1 + ax∗) − x∗ = 0 subtract x∗ from both sides

x∗(1 + ax∗ − 1) = 0 factor out x∗

x∗(ax∗) = 0 simplify.

The only equilibrium is at x∗ = 0, as long as a = 0. If a = 0,the system is xt+1 = xt , which has every value of x∗ as anequilibrium.

15. The equilibrium seems to be at about 1.3.

8

6

4

2

10

00

Fina

l val

ue

2 4

stableequilibrium

6 8 10

Initial value

17. The equilibria seem to be at about 0.0 and 7.5.

8

6

4

2

10

00

Fina

l val

ue

2 4

unstable equilibrium

can't tell

6 8 10

Initial value

19. The discrete-time dynamical system is

pt+1 = 1.2pt

1.2pt + 2.0(1 − pt ).

The equilibrium at p∗ = 0 is stable, and the one at p∗ = 1 isunstable. This makes sense because the wild type are repro-ducing faster than the mutants (r > s) and should dominatethe population.

0.8

0.6

0.4

0.2

1

00

p t +

1

0.2 0.4 0.6 0.8 1

pt


pt+1 = 0.3pt

0.3pt + 0.5(1 − pt ).

The equilibrium at p∗ = 0 is stable, and the one at p∗ = 1 isunstable. The picture looks the same as in a because the ratioof r to s is the same. But in this case, both populations aredecreasing.

0.8

0.6

0.4

0.2

1

00

p t +

1

0.2 0.4 0.6 0.8 1

pt

23.

0

200

400

600

800

1000

0 200 400 600 800 1000

Fina

l pop

ulat

ion

Initial population

stable

unstable

25. a. 2.0 × 105 mutate and 1.0 × 104 revert.


b. There are 1.0 × 106 − 2.0 × 105 + 1.0 × 104 = 8.1 × 105

wild type and 1.0 × 105 − 1.0 × 104 + 2.0 × 105 = 2.9 ×105 mutants.

c. The total number before and after is 1.1 × 106. It does notchange because the bacteria are not reproducing or dying,just changing their type.

d. The fraction before is 1.0 × 105/1.1 × 106 = 0.091. Thefraction after is 2.9 × 105/1.1 × 106 ≈ 0.264.

27. a. 0.2bt mutate and 0.1mt revert.

b. bt+1 = bt − 0.2bt + 0.1mt = 0.8bt + 0.1mt . mt+1 = mt −0.1mt + 0.2bt = 0.9mt + 0.2bt .

c. The total number before is bt + mt . The total number afteris

bt+1 + mt+1 = (0.8bt + 0.1mt ) + (0.9mt + 0.2bt )

= 0.8bt + 0.2bt + 0.1mt + 0.9mt

= bt + mt .

d. Divide the discrete-time dynamical system for mt+1 bybt+1 + mt+1,

pt+1 = mt+1

bt+1 + mt+1= 0.9mt + 0.2bt

bt+1 + mt+1= 0.9mt + 0.2bt

bt + mt

= 0.9mt

bt + mt+ 0.2bt

bt + mt= 0.9pt + 0.2(1 − pt )

= 0.2 + 0.7pt .

e. The equilibrium solves p∗ = 0.2 + 0.7p∗, or p∗ ≈ 0.667.

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

p t +

1

pt

f. The equilibrium seems to be stable. The fraction of mutantswill increase until it reaches 66.7%.

29. a. 1.0 × 105 mutate.

b. There are 1.0 × 106 − 1.0 × 105 = 9.0 × 105 wild typeand 1.0 × 105 + 1.0 × 105 = 2.0 × 105 mutants.

c. There are 1.8 × 106 wild type and 3.0 × 105 mutants.

d. The total number after is 2.1 × 106.

e. The fraction before is 1.0 × 105/1.1 × 106 ≈ 0.091. Thefraction after is 3.0 × 105/2.1 × 106 ≈ 0.143.

31. a. 0.1bt mutate.

b. There are 0.9bt wild type and mt + 0.1bt mutants aftermutation.

c. There are bt+1 = 1.8bt wild type and mt+1 = 1.5mt +0.15bt mutants after reproduction.

d. The total number after is 1.95bt + 1.5mt .

e. Divide the discrete-time dynamical system for mt+1 bybt+1 + mt+1,

pt+1 = mt+1

bt+1 + mt+1

= 1.5mt + 0.15bt

1.95bt + 1.5mt

=1.5mt

mt + bt+ 0.15bt

mt + bt

1.95bt

mt + bt+ 1.5mt

mt + bt

= 1.5pt + 0.15(1 − pt )

1.5pt + 1.95(1 − pt ).

f. The equilibrium solves

p∗ = 1.5p∗ + 0.15(1 − p∗)1.5p∗ + 1.95(1 − p∗)

.

Following the algebra gives

p∗(1.5p∗ + 1.95(1 − p∗)) = 1.5p∗ + 0.15(1 − p∗)p∗(1.5p∗ + 1.95(1 − p∗))

−1.5p∗ + 0.15(1 − p∗) = 0

0.15 − 0.6p∗ + 0.45(p∗)2 = 0

0.45(p∗ − 1/3)(p∗ − 1.0) = 0.

Therefore, p∗ = 1/3 or p∗ = 1.0.

0

0.2

0.4

0.6

0.8

1

0 10.80.60.40.2

p t +

1

pt

equilibrium

g. The equilibrium at 1/3 seems to be stable. The fraction ofmutants will end up at about 33.3%.

33. x1 = 100 − 20 + 30 = 110. y1 = 100 − 30 + 20 = 90. x2 =115 and y2 = 85.

35. xt+1 = xt − 0.2xt + 0.3yt = 0.8xt + 0.3yt . yt+1 = yt −0.3yt +0.2xt = 0.7yt + 0.2xt .

37. a. Consider the first island. After migration, there are 80 but-terflies that started on the first island and 30 that startedon the second. The 80 reproduce, making a total of 190.On the second island, there are 20 from the first and 70from the second after migration. The 20 reproduce, mak-ing a total of 110. Following the same reasoning, x2 = 337and y2 = 153.


b. xt+1 = 2(xt − 0.2xt ) + 0.3yt = 1.6xt + 0.3yt . yt+1 = yt −0.3yt + 2 · 0.2xt = 0.7yt + 0.4xt .

c. xt+1 + yt+1 = 2xt + yt . Then

pt+1 = xt+1

xt+1 + yt+1= 1.6xt + 0.3yt

xt+1 + yt+1

= 1.6xt + 0.3yt

2xt + yt

=1.6xt

xt + yt+ 0.3yt

xt + yt

2xt

xt + yt+ yt

xt + yt

= 1.6pt + 0.3(1 − pt )

2pt + 1 − pt.

d. We must solve the equation

p∗ = 1.6p∗ + 0.3(1 − p∗)2p∗ + 1 − p∗ = 0.3 + 1.3p∗

1 + p∗ .

Multiplying out and using the quadratic formula, wefind p∗ ≈ 0.718. This is larger than the result in Exer-cise 1.10.36 because the butterflies from the first islandreproduce.

e.

0

0.2

0.4

0.6

0.8

1.0

0 1

p t +

1

pt

Starting from 0.1


Mt+1 = Mt − 0.5

1.0 + 0.1MtMt + 1.0.

To find the equilibrium,

M∗ = M∗ − 0.5

1.0 + 0.1M∗ M∗ + 1.0

0.5

1.0 + 0.1M∗ M∗ = 1.0

0.5M∗ = 1.0 + 0.1M∗

0.4M∗ = 1.0.

The equilibrium is therefore 2.5. It is larger because the frac-tion absorbed is always less than 0.5.


Mt+1 = Mt − β

1.0 + 0.1MtMt + 1.0.

To find the equilibrium,

M∗ = M∗ − β

1.0 + 0.1M∗ M∗ + 1.0

β

1.0 + 0.1M∗ M∗ = 1.0

βM∗ = 1.0 + 0.1M∗

(β − 0.1)M∗ = 1.0.

The equilibrium is therefore

M∗ = 1.0

1.0β − 0.1.

This becomes smaller as β becomes larger. Large values ofβ indicate that the fraction absorbed is large, which makessense. The fact that the equilibrium is negative when β ≤ 0.1indicates that there is no equilibrium. The body cannot useup the dose of 1.0 each day, and the level just keeps buildingup.

0

10

20

30

40

50

0 10.8

Equ

ilibr

ium

b

0.60.40.2

The second diagram looks a lot like the tree growth model.Once the concentration becomes large, it simply increases by1.0 per day.

0

2

4

6

8

10

0 108

Mt +

1

Mt

b = 0.05

642

0

2

4

6

8

10

0 108

Mt +

1

Mt

b = 0.5

642


43. a.

0

0.5

1

1.5

2

0 10.8

Per

capi

ta p

rodu

ctio

n

Population (millions)

0.60.40.2

b. bt+1 = 2.0bt

(1 − bt

1.0 × 106

).

c. The equilibria are b∗ = 0 and b∗ = 5.0 × 105.

0

0.2

0.6

0.4

0.8

1

0 10.8

b t +

1

bt

0.60.40.2

45. a.

0

0.5

1

1.5

2

0 21.5

Per

capi

ta p

rodu

ctio

n

Population (millions)

10.5

b. bt+1 = 2.0bt e−bt

1.0 × 106 .

c. The equilibria are b∗ = 0 and b∗ = ln(2)1.0 × 106 ≈6.93 × 105.

0

0.5

1

1.5

2

0 11.5

b t +

1

bt

10.5

Section 1.11, page 1271. V̂t = 0.5 · 30.0 = 15.0 < Vc. The heart will beat, and Vt+1 =

25.0 mV.

3. V̂t = 0.7 · 30.0 = 21.0 > Vc. The heart will not beat, andVt+1 = 21.0 mV.

5. V ∗ = u/(1 − c) = 20.0. Because cV ∗ = 10.0 < Vc = 20.0, theinequality in Equation 1.11.2 is satisfied and the equilibriummakes sense. This heart will beat every time.

7. V ∗ = 33.3. Because cV ∗ = 23.1 > Vc = 20.0, the inequality inEquation 1.11.2 is not satisfied. The equilibrium does not makesense. Is this a case of 2 : 1 AV block? We find that V̄ = 19.61from Equation 1.11.5. However, cV̄ < Vc, so that the secondinequality in Equation 1.11.4 is not satisfied. This is not a caseof 2 : 1 AV block. In fact, this turns out to be an example of theWenckebach phenomenon where the heart skips every fourthbeat.

9. c = e−ατ = 0.3678. The heart beats every time.

11. c = e−ατ = 0.3678. The heart beats every time, twice as fastas in part a, but with recovery also twice as fast.

13. The dynamical system is

Vt+1 = cVt + 2(1 − c)

1 + V 2t

.

Substituting Vt = 1 gives Vt+1 = 1, so this is an equilibrium.

0

1

2

3

4

5

0 54

Vt +

1

Vt

321

From the cobweb, it seems to be stable.

15. I got the following strange results. These are jumping all overthe place.

Time Solution 1 Solution 2 Solution 3

0 500 600.0 800.0

1 750.0 900.0 1200.0

2 1125.0 1350.0 800.0

3 687.50 1025.0 1200.0

4 1031.25 537.5 800.0

5 546.88 806.25 1200.0

6 820.31 1209.37 800.0

7 1230.47 814.062 1200.0

8 845.70 1221.10 800.0

9 1268.55 831.64 1200.0

10 902.83 1247.46 800.0

17. I got the following strange results. The first jumps around likethe earlier ones, but the second seems to be shooting off toinfinity.


Time Solution 1 Solution 2 Solution 3

0 500.0 600.0 800.0

1 825.0 990.00 1320.0

2 1361.25 1633.50 1178.0

3 1246.06 1695.27 943.70

4 1056.00 1797.20 1557.10

5 742.40 1965.39 1569.22

6 1224.97 2242.89 1589.22

7 1021.20 2700.76 1622.21

8 684.98 3456.26 1676.65

9 1130.21 4702.83 1766.47

10 864.85 6759.67 1914.67

Supplementary Problems, page 1281. a. 3.0 × 1010 bacteria. b. 2.0 × 1010 ≤ number ≤ 5.0 × 1010.

3. a. f −1(y) = − ln(y)/2, g−1(y) = 3√

y − 1. f(x) = 2 when

x = f −1(2) = − ln(2)

2≈ −0.345. Similarly, we find g(1) =

2.

b. ( f ◦ g)(x) = e−2(x3+1), (g ◦ f )(x) = e−6x + 1, ( f ◦ g)(2)

= e−18 = 1.5 × 10−8, (g ◦ f )(2) = 1 + e−12 = 1.00006.

c. (g ◦ f )−1(y) = − ln(y − 1)/6, domain is y > 1.

5. a.

0

2

4

6

8

10

0 1 2 3 4 5Time

Num

ber

(hun

dred

s of

mill

ions

)

b. Line has slope of 1.5million bacteria

hourand equation b(t) =

1.5 + 1.5t . At t = 3, the point 5.0 lies below the line.

c. Substituting t = 3 into the equation, we get 6.0 million.

d. Substituting t = 7 into the equation, we get 12.0 million.

7. a. π/3 radians. sin(θ) = √3/2, cos(θ) = 1/2.

b. −π/3 radians. sin(θ) = −√3/2, cos(θ) = 1/2.

c. 1.919 radians. sin(θ) ≈ 0.9397, cos(θ) ≈ −0.3420.

d. −3.316 radians. sin(θ) ≈ 0.1736, cos(θ) ≈ −0.9848.

e. 20.25 radians. sin(θ) ≈ 0.9848, cos(θ) ≈ 0.1736.

9. a. Let Bt be the number of butterflies in the late summer.There are then 1.2Bt eggs, leading to 0.6Bt new but-terflies from reproduction plus 1000 from immigration.The discrete-time dynamical system is Bt+1 = 0.6Bt +1000.

b.

0

1000

2000

3000

0 1000 2000 3000

Bt +

1

Bt

c. Equilibrium has 2500 butterflies.

11. a. The size at t = 30 is S(30) = 1.0e0.1 · 30 ≈ 20.08. The sizeafter treatment, which we can denote ST (t), is a line throughthe point (30, 20.08) with slope −0.4, so

ST (t) = −0.4(t − 30) + 20.08.

b.

0

5

10

15

20

0 10 20 30 40 50 60 70 80

S T (t

)t

c. We solve ST(t) = 0, or

−0.4(t − 30) + 20.08 = 00.4(t − 30) = 20.08

t − 30 = 20.08

0.4

t = 30 + 20.08

0.4≈ 80.2.

13. a. 2.66cm2. b. The discrete-time dynamical system is At+1 =1.1At . c. 1.82cm2. d. 0.55. e. When 2.0 × 1.1t = 10 or in16.9 hours.

0

2

4

6

8

10

0 2 4 6 8 10

At +

1

At

15. a.

0

50

100

150

200

250

300

0 50 100 150 200 250 300

Fina

l pop

ulat

ion

Initial population (millions)

Cobwebbing


b.

0

50

100

150

200

250

300

0 1 2 3 4 5 6

Fina

l pop

ulat

ion

Initial population (millions)

Solution

c. The only equilibrium is at 0.

17. a. xt+1 = 0.5 + xt/(1 + xt ).

b.

0

0.5

1.0

1.5

2.0

0 0.5 1.0 1.5 2.0

x t +

1

xt

Cobwebbing

0

0.5

1.0

1.5

2.0

0 1 2 3 4 5

Solution

x t

t

c. The equilibrium is at 1.0.

19. a. She has $1120, and the casino has $10,880.

b. Let g represent the money the gambler has and c the amountthe casino has. Then

gt+1 = gt − 0.1gt + 0.02ct = 0.9gt + 0.02ct

ct+1 = ct + 0.1gt − 0.02ct = 0.1gt + 0.98ct .

c.

pt+1 = gt+1

gt+1 + ct+1= 0.9gt + 0.02ct

gt + ct

= 0.9pt + 0.02(1 − pt )

pt + (1 − pt )= 0.88pt + 0.02.

d. The equilibrium is at p = 1

6.

e. They start out with $12,000, and the gambler ends up with1/6, or $2000.

21. a. D(0) = 10.0, H(0) = 10.0, D(7.5) = 12.5, H(7.5) = 98.2,D(15) = 15.7, H(15) = 193.9.

b.

0

40

80

120

160

200

0 5 10 15

Size

head diameter

height

Age

c.

0

2

4

6

0 5 10 15

log

size

Age

head diameter

height

d. Doubling time is 23.1 for head diameter and 7.7 for height.

23. a. $148,643.

b. In 16.88 yr, or in about 2012.

c. Mt+1 = 1.1Mt .

d. Mt = 1.000001 × 106 · 1.1t where t is measured in yearsbefore or after 1995.

0

0.5

1.0

1.5

2.0

0 0.5 1.0 1.5 2.0

Mt +

1

Mt

Cobwebbing

0

1

2

3

4

5

0 5 10 15

Mt

t

Solution

25. a. It will have $460 million.

b.

340

380

420

460

500

0 2 4 6 8 10

End

owm

ent (

mill

ions

)

Years from present


c. Mt+1 = Mt + 15.

300

400

500

300 400 500

Fina

l end

owm

ent (

mill

ions

)

Initial endowment (millions)

d. I’d hire the second Texan—the money keeps piling up.

27. a. The distance a car moves in 2 seconds is 40 m.

b. There will be 25 vehicles per kilometer.

c. There will be 25 · 72 = 1800 vehicles passing a point inone hour, carrying 2700 people.

d. This oscillation has period 24 hours, amplitude of 900 (halfthe difference between the minimum of 900 and the max-imum of 2700), a mean of 1800, and a phase of 8.0, withformula

p(t) = 1800 + 900 cos

(t − 8.0

24

).

29. a. 40,000 · 1.6 − 10,000 = 54,000.

b. Tt+1 = 1.6Tt − 10,000.

c. Solving Tt+1 = Tt gives T ∗ = 16,667.

d.

0

200000

150000

100000

50000

250000

0 50000 100000 150000 200000 250000

Tt +

1

Tt

e. This one will grow faster because traffic goes up by 60%rather than 50%. Once the numbers get really big, the10,000 car reduction won’t matter much.

answers to odd exercises - colorado state universityshipman/math155/suppprobs/ch1.pdf · answers to...

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