James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

604 Chapter 9 Differential Equations
Mixing Problems
A typical mixing problem involves a tank of fixed capacity filled with a thoroughly
mixed solution of some substance, such as salt. A solution of a given concentration enters
the tank at a fixed rate and the mixture, thoroughly stirred, leaves at a fixed rate, which
may differ from the entering rate. If ystd denotes the amount of substance in the tank at
time t, then y9std is the rate at which the substance is being added minus the rate at which
it is being removed. The mathematical description of this situation often leads to a firstorder
sepa rable differential equation. We can use the same type of reasoning to model a
variety of phenomena: chemical reactions, discharge of pollutants into a lake, injection
of a drug into the bloodstream.
Example 6 A tank contains 20 kg of salt dissolved in 5000 L of water. Brine that
contains 0.03 kg of salt per liter of water enters the tank at a rate of 25 Lymin. The
solution is kept thoroughly mixed and drains from the tank at the same rate. How much
salt remains in the tank after half an hour?
SOLUtion Let ystd be the amount of salt (in kilograms) after t minutes. We are given
that ys0d − 20 and we want to find ys30d. We do this by finding a differential equation
satisfied by ystd. Note that dyydt is the rate of change of the amount of salt, so
5
dy
dt
− srate ind 2 srate outd
where (rate in) is the rate at which salt enters the tank and (rate out) is the rate at which
salt leaves the tank. We have
rate in −S0.03 kg L
− 0.75
LDS25
minD kg
min
The tank always contains 5000 L of liquid, so the concentration at time t is ystdy5000
(measured in kilograms per liter). Since the brine flows out at a rate of 25 Lymin, we
have
rate out −S ystd
5000
kg
LDS25
L
−
minD ystd
200
kg
min
Figure 10 shows the graph of the
function ystd of Example 6. Notice
that, as time goes by, the amount of
salt approaches 150 kg.
y
150
100
50
0 200 400
FIGURE 10
t
Thus, from Equation 5, we get
dy
dt
− 0.75 2
ystd
200
Solving this separable differential equation, we obtain
y
dy
150 2 y − y
2ln | 150 2 y | −
Since ys0d − 20, we have 2ln 130 − C, so
2ln | 150 2 y | −
−
150 2 ystd
200
dt
200
t
200 1 C
t 2 ln 130
200
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