Question

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11 pounds of salt per gallon flows into the tank at the rate of 44 gallons per minute, and the well-stirred mixture flows out of the tank at the rate of 11 gallon per minute. set up a differential equation for the amount of salt a(t)a(t) in the tank at time tt. how much salt is in the tank when it is full? (round your answer to the 2 decimal places).

Answer 1

Let
`A(t)` denote the amount of salt in the tank at time
`t`. We're given that the tank initially holds
`A(0)=100` lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation


`(\mathrm dA)/(\mathrm dt)=\underbrace{\frac{11\text{ lb}}{1\text{ gal}}*\frac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{(A(t))/(200+(44-11)t)*\frac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}`

`\implies A'(t)+(11)/(200+33t)A(t)=484`

Find the integrating factor:


`\mu(t)=\exp\left(\displaystyle\int(11)/(200+33t)\,\mathrm dt\right)=(200+33t)^(1/3)`

Distribute
`\mu(t)` along both sides of the ODE:


`(200+33t)^(1/3)A'(t)+11(200+33t)^(-2/3)A(t)=484(200+33t)^(-1/3)`

`\bigg((200+33t)^(1/3)A(t)\bigg)'=484(200+33t)^(-1/3)`

`A(t)=484\displaystyle\int(200+33t)^(-1/3)\,\mathrm dt`

`A(t)=22(200+33t)^(2/3)+C`

Since
`A(0)=100`, we get


`100=22(200)^(2/3)+C\implies C\approx-652.39`

so that the particular solution for
`A(t)` is


`A(t)=22(200+33t)^(2/3)-652.39`

The tank becomes full when the volume of solution in the tank at time
`t` is the same as the total volume of the tank:


`200+(44-11)t=500\implies 33t=300\implies t\approx9.09`

at which point the amount of salt in the solution would be


`A(9.09)\approx733.47\text{ lb}`