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)=\frac{2\text{ gal}}{1\text{ min}}\cdot\frac{0\text{ lb}}{1\text{ gal}}-\frac{1\text{ gal}}{1\text{ min}}\cdot\frac{A(t)\text{ lb}}{50+(2-1)t\text{ gal}}`


`A'(t)+(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)=\frac C{50+t}`

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


`10=\frac C{50+0}\implies C=500`

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


`A(t)=(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)=(500)/(50+50)=5\text{ lb}`