Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 9.4 MODELS FOR POPULATION GROWTH ¤ 393
(b) Using the logistic equation (4), dP 1
dt = kP − P
, we substitute y = P dP
, P = Ky,and
K
K dt = K dy
dt ,
to obtain K dy
dt
dy
= k(Ky)(1 − y) ⇔ = ky(1 − y), our equation in part (a).
dt
Now the solution to (4) is P (t) =
We use the same substitution to obtain Ky =
K
K − P0
,whereA = .
1+Ae−kt P 0
K
y 0
1+ K − ⇒ y =
Ky0 e
Ky −kt y 0 +(1− y 0 )e . −kt
0
Alternatively, we could use the same steps as outlined in the solution of Equation 4.
(c) Let t be the number of hours since 8 AM. Theny 0 = y(0) = 80
1000 =0.08 and y(4) = 1 2 ,so
1
2 = y(4) = 0.08
0.08 + 0.92e −4k . Thus, 0.08 + 0.92e−4k =0.16, e −4k = 0.08
0.92 = 2
23 ,ande−k = 2
23
1/4
,
so y =
0.08
0.08 + 0.92(2/23) = 2
. Solving this equation for t,weget
t/4 t/4
2 + 23(2/23)
t/4 2
⇒
= 2 − 2y t/4 2
⇒
= 2 23 23y 23 23 · 1 − y
y
t/4 2
2y +23y =2
23
It follows that t 4
When y =0.9, 1 − y
y
the rumor by 3:36 PM.
9. (a) dP 1
dt = kP − P
K
ln[(1 − y)/y]
− 1=
ln 2 ,sot =4
23
= 1 9 ,sot =4
1 − ln 9
ln 2 23
⇒
d2 P
dt 2
1+
= k P − 1 K
= k kP 1 − P K
ln((1 − y)/y)
ln 2
23
.
⇒
2
23
t/4−1
= 1 − y .
y
≈ 7.6 hor7 h 36 min. Thus, 90% of the population will have heard
dP
+ 1 − P dt K
dP
= k dP
dt dt
1 − P K
1 − 2P
= k 2 P
K
− P K +1− P
K
1 − 2P
K
(b) P grows fastest when P 0 has a maximum, that is, when P 00 =0. From part (a), P 00 =0 ⇔ P =0, P = K,
or P = K/2. Since0