Solução_Calculo_Stewart_6e

F.
392 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS
TX.10
9.4 Models for Population Growth
1. (a) dP/dt =0.05P − 0.0005P 2 =0.05P (1 − 0.01P )=0.05P (1 − P/100). Comparing to Equation 4,
dP/dt = kP(1 − P/K), we see that the carrying capacity is K = 100 and the value of k is 0.05.
(b) The slopes close to 0 occur where P is near 0 or 100. The largest slopes appear to be on the line P =50. The solutions
are increasing for 0 100.
(c)
All of the solutions approach P =100as t increases. As in
part (b), the solutions differ since for 0 100 they are decreasing. Also, some
have an IP and some don’t. It appears that the solutions which
have P 0 =20and P 0 =40have inflection points at P =50.
(d) The equilibrium solutions are P =0(trivial solution) and P = 100. The increasing solutions move away from P =0and
all nonzero solutions approach P =100as t →∞.
3. (a) dy 1
dt = ky − y
K
K
K − y(0)
⇒ y(t) =
with A = .WithK =8× 10 7 , k =0.71,and
1+Ae−kt y(0)
y(0) = 2 × 10 7 ,wegetthemodely(t) = 8 × 107
8 × 107
,soy(1) =
1+3e−0.71t 1+3e ≈ 3.23 × −0.71 107 kg.
(b) y(t) =4× 10 7
⇒
8 × 10 7
1+3e −0.71t =4× 107 ⇒ 2=1+3e −0.71t ⇒ e −0.71t = 1 3
⇒
−0.71t =ln 1 3
⇒ t = ln 3 ≈ 1.55 years
0.71
5. (a) We will assume that the difference in the birth and death rates is 20 million/year. Let t =0correspond to the year 1990
anduseaunitof1 billion for all calculations. k ≈ 1 dP
P dt = 1
5.3 (0.02) = 1
265 ,so
dP
1
dt = kP − P
= 1
K 265 P 1 − P
, P in billions
100
(b) A = K − P0 100 − 5.3
= = 947
P 0 5.3 53 ≈ 17.8679. P (t) = K
1+Ae = 100
,soP (10) ≈ 5.49 billion.
−kt e−(1/265)t
1+ 947
53
(c) P (110) ≈ 7.81,andP (510) ≈ 27.72. The predictions are 7.81 billion in the year 2100 and 27.72 billion in 2500.
(d) If K =50,thenP (t) =
1+ 447
53
50
.SoP (10) ≈ 5.48, P (110) ≈ 7.61,andP (510) ≈ 22.41. The predictions
e−(1/265)t
become 5.48 billion in the year 2000, 7.61 billion in 2100, and 22.41 billion in the year 2500.
7. (a) Our assumption is that dy = ky(1 − y),wherey is the fraction of the population that has heard the rumor.
dt