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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
806 Answers to Odd Exercises
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.
Answers to Odd Exercises 807
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
10 20 30 40Temperature
Original graph
Dev
elop
men
t tim
e (d
ays)
0
700800900
1000
500600
100200300400
10 20 30 40Temperature
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
808 Answers to Odd Exercises
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.
Answers to Odd Exercises 809
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
810 Answers to Odd Exercises
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.
Answers to Odd Exercises 811
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 .
812 Answers to Odd Exercises
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
Answers to Odd Exercises 813
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.
814 Answers to Odd Exercises
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.
Answers to Odd Exercises 815
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
816 Answers to Odd Exercises
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)
Answers to Odd Exercises 817
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.
818 Answers to Odd Exercises
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.
Answers to Odd Exercises 819
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.
820 Answers to Odd Exercises
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
21. The discrete-time dynamical system is
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.
Answers to Odd Exercises 821
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.
822 Answers to Odd Exercises
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
39. The discrete-time dynamical system is
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.
41. The discrete-time dynamical system is
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
Answers to Odd Exercises 823
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.
824 Answers to Odd Exercises
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
Answers to Odd Exercises 825
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
826 Answers to Odd Exercises
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.