Solução_Calculo_Stewart_6e

F.
dy
dt =2−
3y
dy
. Rewriting this equation as
100 + 2t
I(t) =exp
TX.10
SECTION 9.6 PREDATOR-PREY SYSTEMS ¤ 399
dt + 3
y =2, we see that it is linear.
100 + 2t
=exp 3
ln(100 + 2t) = (100 + 2t) 3/2
2
3 dt
100 + 2t
Multiplying the differential equation by I(t) gives (100 + 2t) 3/2 dy
dt +3(100+2t)1/2 y =2(100+2t) 3/2
[(100 + 2t) 3/2 y] 0 =2(100+2t) 3/2 ⇒ (100 + 2t) 3/2 y = 2 5 (100 + 2t)5/2 + C ⇒
y = 2 (100 + 2t)+C(100 + 5 2t)−3/2 .Now0=y(0) = 2 (100) + C · 5 100−3/2 =40+ 1 C ⇒ C = −40,000,so
1000
2
y = (100 + 2t) − 40,000(100 + 5 2t)−3/2 kg. From this solution (no pun intended), we calculate the salt concentration
at time t to be C(t) =
y(t)
100 + 2t = −40,000
(100 + 2t) + 2 kg
−40,000
. In particular, C(20) = 5/2 5 L 140 + 2 5/2 5 ≈ 0.2275 kg L
and y(20) = 2 5 (140) − 40,000(140)−3/2 ≈ 31.85 kg.
35. (a) dv
dt + c m v = g and I(t) =e (c/m) dt = e (c/m)t , and multiplying the differential equation by
I(t) gives e (c/m)t dv
dt + vce(c/m)t
0
= ge (c/m)t ⇒ e v (c/m)t = ge (c/m)t . Hence,
m
v(t) =e −(c/m)t ge (c/m)t dt + K = mg/c + Ke −(c/m)t . But the object is dropped from rest, so v(0) = 0 and
K = −mg/c. Thus, the velocity at time t is v(t) =(mg/c)[1 − e −(c/m)t ].
(b) lim
t→∞
v(t) =mg/c
(c) s(t) = v(t) dt =(mg/c)[t +(m/c)e −(c/m)t ]+c 1 where c 1 = s(0) − m 2 g/c 2 .
s(0) is the initial position, so s(0) = 0 and s(t) =(mg/c)[t +(m/c)e −(c/m)t ] − m 2 g/c 2 .
⇒
9.6 Predator-Prey Systems
1. (a) dx/dt = −0.05x +0.0001xy. Ify =0,wehavedx/dt = −0.05x, which indicates that in the absence of y, x declines at
a rate proportional to itself. So x represents the predator population and y represents the prey population. The growth of
the prey population, 0.1y (from dy/dt =0.1y − 0.005xy), is restricted only by encounters with predators (the term
−0.005xy). The predator population increases only through the term 0.0001xy; that is, by encounters with the prey and
not through additional food sources.
(b) dy/dt = −0.015y +0.00008xy. Ifx =0,wehavedy/dt = −0.015y, which indicates that in the absence of x, y would
decline at a rate proportional to itself. So y represents the predator population and x represents the prey population. The
growth of the prey population, 0.2x (from dx/dt =0.2x − 0.0002x 2 − 0.006xy =0.2x(1 − 0.001x) − 0.006xy), is
restricted by a carrying capacity of 1000 [from the term 1 − 0.001x =1− x/1000] and by encounters with predators (the
term −0.006xy). The predator population increases only through the term 0.00008xy;thatis,byencounterswiththeprey
and not through additional food sources.