Math 3400
Homework 3
Section 2.3
# 1 A tank contains 200l of dye solution with a concentration of 1g/l. The tank is
rinsed with fresh water flowing at 2l/min. When will the concentration reach 1 percent of
its original value.
Solution Let Q be the amount of dye. The equation is
dQ
= (rate in) − (rate out) = 0 − (2)(Q/200),
dt
or
dQ
= −.01Q.
dt
The form of the solution is
Q(t) = Q0 e−.01t .
The concentration reach 1 percent of its original value when
e−.01t = .01,
or
−.01t = log(.01) = − log(100),
t = 460 minutes.
Notice that the initial concentration is irrelevant.
# 13
a) Let Q0 = −rQ with r > 0. The general solution is
Q(t) = Q0 e−rt .
If the half life is 5730 years, then
Q0 /2 = Q0 exp(−5730r),
so −5730r = log(1/2), or
r = 1.2097 × 10−4 .
b) The amount of Carbon-14 at time t (years) is
Q(t) = Q0 exp(−1.2097 × 10−4 t).
c) If Q(t) = Q0 /5, then
Q0 /5 = Q0 exp(−1.2097 × 10−4 t),
or t = 13, 304 years.
# 16
Newton’s law of cooling states that
dT1
= k(T1 − T0 ),
dt
1
where T0 is the environmental temperature. Fresh coffee is 200F . One minute later it is
190F . The room temperature is 70F . When does the coffee reach 150F ?
Solution We have
dT1
− kT1 = −kT0 = −70k,
dt
d −kt
(e T1 ) = −70ke−kt .
dt
Integration gives
e−kt T1 = C + 70e−kt ,
T1 = Cekt + 70.
We then have
T1 (0) = 200 = C + 70,
C = 130.
T1 (1) = 190 = 130ek + 70,
k = −.08.
Finally,
150 = 130e−.08t + 70,
t = 6 minutes.
Section 2.5
# 20
y
dy
= r(1 − )y − Ey
dt
K
a. Equilibrium points occur when
r(1 −
y
)y − Ey = 0.
K
Obviously y1 = 0 is an equilibrium point, and dividing by y leads to
K(1 − E/r) = y2 ,
which has a positive value if 0 < E < r.
b. The right hand side of the equation may be rewritten as
f (y) =
r
y(K[1 − E/r] − y).
K
The function f (y) satisfies
f (y) < 0,
f (y) > 0,
y > K(1 − E/r),
y < 0,
0 < y < K(1 − E/r).
Thus small positive solutions move away from 0 (unstable), while solutions starting near
y2 = K(1 − E/r) move toward y2 , asymptotically stable.
c. Sustainable yield Y is given by
Y = Ey2 (E) = EK(1 − E/r) = EK − E 2 K/r.
2
d. The value of E maximizing Y is given by
dY
= K − 2EK/r = 0,
dE
or
E1 = r/2.
The maximum sustainable yield is
Ym = E1 y2 (E1 ) = rK/4.
3