Question

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 3 gal/min. determine a(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by cin(t) = 2 + sin(t/4) lb/gal.

Answer 1

`A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})`

`\implies A'(t)=\frac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\frac t4\right)\frac{\text{lb}}{\text{gal}}-\frac{3\text{ gal}}{1\text{ min}}\cdot\frac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}`

`A'(t)+\frac1{100}A(t)=6+3\sin\frac t4`

We're given that
`A(0)=30`. Multiply both sides by the integrating factor
`e^(t/100)`, then


`e^(t/100)A'(t)+\frac1{100}e^(t/100)A(t)=6e^(t/100)+3e^(t/100)\sin\frac t4`

`\left(e^(t/100)A(t)\right)'=6e^(t/100)+3e^(t/100)\sin\frac t4`

`e^(t/100)A(t)=600e^(t/100)-(150)/(313)e^(t/100)\left(25\cos\frac t4-\sin\frac t4\right)+C`

`A(t)=600-(150)/(313)\left(25\cos\frac t4-\sin\frac t4\right)+Ce^(-t/100)`

Given that
`A(0)=30`, we have


`30=600-(150)/(313)\cdot25+C\implies C=-(174660)/(313)\approx-558.02`

so the amount of salt in the tank at time
`t` is


`A(t)\approx600-(150)/(313)\left(25\cos\frac t4-\sin\frac t4\right)-558.02e^(-t/100)`