Q:

# 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.

Accepted Solution

A:
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

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