Calculus II, Section 9.3, #48
Separable Equations
A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank
at the rate of 5 L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at the rate of
10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 l/min. How much
salt is in the tank (a) after t minutes and (b) after one hour?1
We want a function that gives the amount of salt at time t, but all the information is about how the amount
of salt is changing.
Let s(t) = amount, in kg of salt at time t. Then we have
ds
= (rate of salt into tank) − (rate of salt out of tank)
dt
s kg
= (0.05 kg/L · 5 L/min) + (0.04 kg/L · 10 L/min) −
· 15 L/min
1000 L
15s
= 0.25 kg/min + 0.4 kg/min −
kg/min
1000
So we get the differential equation
15s
ds
= 0.65 −
dt
1000
15s
65
−
=
100 1000
130 − 3s
ds
=
dt
200
We separate s and t to get
1
1
ds =
dt
130 − 3s
200
Integrate
Z
1
ds =
130 − 3s
Z
1
dt
200
1
1
− · ln |130 − 3s| =
t + C1
3
200
3
t + C2
ln |130 − 3s| = −
200
3
|130 − 3s| = e− 200 t+C2
|130 − 3s| = C3 e−3t/200
130 − 3s = C4 e−3t/200
−3s = −130 + C4 e−3t/200
s=
130 − C4 e−3t/200
3
Since we begin with pure water, we have s(0) = 0. Substituting,
130 − C4 e−3·0/200
3
0 = 130 − C4
0=
C4 = 130
1 Stewart,
Calculus, Early Transcendentals, p. 607, #48.
Calculus II
Separable Equations
So our function is
s(t) =
130 − 130e−3t/200
3
After one hour (60 min), we have
130 − 130e−3·60/200
3
s(60) ≈ 25.7153
s(60) =
Thus, after one hour there is about 25.72 kg of salt in the tank.