Question

A tank contains 120 liters of fluid in which 50 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 3 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.

Answer 1

If A(t) is the amount of salt in the tank at time t, then A(0) = 50 g and

A'(t) = (1 g/L)*(3 L/min) - (A(t)/120 g/L)*(3 L/min)

$\implies A'+\dfrac A{40}=3$

Multiply both sides by $e^{t/40}$, so that the left side can be condensed as the derivative of a product:

$e^{t/40}A'+\dfrac{e^{t/40}}{40}A=3e^{t/40}$

$\left(e^{t/40}A\right)'=3e^{t/40}$

Integrate both sides and solve for A(t).

$e^{t/40}A=120e^{t/40}+C$

$\implies A(t)=120+Ce^{-t/40}$

Given that A(0) = 50 g, we have

$50=120+C\implies C=-70$

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

$\boxed{A(t)=120-70e^{-t/40}}$