A tank contains 160 liters of fluid in which 30 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

A(t) = 160 - 130 e^(-t/40)

Explanation:

At the start, the tank contains A(0) = 30 g of salt.

Salt flows in at a rate of

(1 g/L) * (4 L/min) = 5 g/min

and flows out at a rate of

(A(t)/160 g/L) * (4 L/min) = A(t)/40 g/min

so that the amount of salt in the tank at time t changes according to

A'(t) = 4 - A(t)/40

Solve the ODE for A(t):

A'(t) + A(t)/40 = 4

e^(t/40) A'(t) + e^(t/40)/40 A(t) = 4e^(t/40)

(e^(t/40) A(t))' = 4e^(t/40)

e^(t/40) A(t) = 160e^(t/40) + C

A(t) = 160 + Ce^(-t/40)

Given that A(0) = 30, we find

30 = 160 + C

C = -130

so that the amount of salt in the tank at time t is

A(t) = 160 - 130 e^(-t/40)

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