Question

A tank with a capacity of 100 gallons initially contains 50 gallons of water with 10 pounds of salt in solution. fresh water enters at a rate of 2 gallons per minute and a well-stirred mixture is pumped out at the rate of 1 gallon per minute. compute the amount of salt (in pounds) in the tank at the first moment when the tank is filled.

Answer 1

Let $A(t)$ be the amount of salt (in pounds) in the tank at time $t$. We're given that$A(0)=10\text{ lb}$.

The rate at which the amount of salt in the tank changes is given by the ODE

$A'(t)=\dfrac{2\text{ gal}}{1\text{ min}}\cdot\dfrac{0\text{ lb}}{1\text{ gal}}-\dfrac{1\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{50+(2-1)t\text{ gal}}$

$A'(t)+\dfrac{A(t)}{50+t}=0$

$(50+t)A'(t)+A(t)=0$

$\bigg((50+t)A(t)\bigg)'=0$

$(50+t)A(t)=C$

$A(t)=\dfrac C{50+t}$

Given that $A(0)=10$, we find that

$10=\dfrac C{50+0}\implies C=500$

so that the amount of salt in the tank is described by

$A(t)=\dfrac{500}{50+t}$

The tank will be filled when $50+t=100$, or after $t=50$ minutes. At this time, the amount of salt in the tank is

$A(50)=\dfrac{500}{50+50}=5\text{ lb}$