Answer to Question #247099 in Differential Equations for nin

Question

A large tank is filled to capacity with 700 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 7 gal/min. The well-mixed solution is pumped out at a rate of 14 gals/min. Find the number A(t) of pounds of salt in the tank at time t.

Answer 1

$\displaystyle \frac{dA}{dt} = R_{in}-R_{out}\\ R_{in} = (2lb/gal \cdot 7gal/min) =14 lb/min\\ R_{out} = (\frac{A(t)}{700}lb/gal).(14gal/min)=\frac{A(t)}{50}lb/min\\ \text{Hence, } \frac{dA}{dt} = 14- \frac{A(t)}{50}\\ \text{Integrating, we have that}\\ \ln |14- \frac{A}{50}| = -0.02t +\ln c\\ 14-\frac{A}{50} = ce^{-0.02t}\\ A(t) = 700 - 50ce^{-0.02t}\\ \text{Applying the initial conditions, we have}\\ A(0) = 700 - 50ce^0=0\\ \implies c = 50\\ \therefore A(t) = 700 - 700e^{-0.02t}$