Question

(2 points) A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 4 gal/min. The well-mixed solution is pumped out at the same rate. Find the number A(t) of pounds of salt in the tank at time t. What is the concentration c(t) of the salt in the tank at time t? At t=5? What is the concentration of the salt in the tank after a long time t→[infinity]? A(t)= , c(t)= , c(5)= lb/gal limt→[infinity]c(t)=

Answer 1

(a)
`A(t)=1000-1000e^{-(t)/(125)}`

(b)
`C(t)=2-2e^{-(t)/(125)}`

(c) C(5)=0.07842 lb/gal

(d)
`\lim_(t \to \infty) =2$ lb/gal`

Explanation:

Amount of Salt in the Tank

`(dA)/(dt)=R_(in)-R_(out)`

`R_(in)` =(concentration of salt in inflow)(input rate of brine)

`=(2(lbs)/(gal))( 4(gal)/(min))=8(lbs)/(min)`

`R_(out)` =(concentration of salt in outflow)(output rate of brine)

`=((A(t))/(500))( 4(gal)/(min))=(A)/(125)`

Therefore:

`(dA)/(dt)=8-(A)/(125)`

We then solve the resulting differential equation by separation of variables.

`(dA)/(dt)+(A)/(125)=8\\$The integrating factor: e^{\int (1)/(125)}dt =e^{(t)/(125)}\\$Multiplying by the integrating factor all through\\(dA)/(dt)e^{(t)/(125)}+(A)/(125)e^{(t)/(125)}=8e^{(t)/(125)}\\(Ae^{(t)/(125)})'=8e^{(t)/(125)}`

Taking the integral of both sides

`\int(Ae^{(t)/(125)})'=\int 8e^{(t)/(125)} dt\\Ae^{(t)/(125)}=8*125e^{(t)/(125)}+C, $(C a constant of integration)\\Ae^{(t)/(125)}=1000e^{(t)/(125)}+C\\$Divide all through by e^{(t)/(125)}\\A(t)=1000+Ce^{-(t)/(125)}`

Recall that when t=0, A(t)=0 (our initial condition)

`0=1000+Ce^(0)\\C=-1000\\$Therefore:\\`

`A(t)=1000-1000e^{-(t)/(125)}`

(b)

Concentration c(t) of the salt in the tank at time t

Concentration, C(t)=
`(Amount)/(Volume)`

Therefore:

`C(t)=\frac{1000-1000e^{-(t)/(125)}}{500}\\\\C(t)=2-2e^{-(t)/(125)}`

Therefore:

`C(5)=2-2e^{-(5)/(125)}\\=0.07842$ lb/gal`

As t tends to infinity

`\lim_(t \to \infty) C(t)=\lim_(t \to \infty) \left(2-2e^{-(t)/(125)}\right)=2$ lb/gal`