Question

A tank contains 350 liters of fluid in which 50 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Let the A(t) be the number of grams of salt in the tank at time t. Find the number A(t).)

Answer 1

`A(t)=50e^{-(t)/(70)}`

Step-by-step explanation:

The volume of fluid in the tank =350 liters

Initial Amount of Salt, A(0)=50 grams

Amount of Salt in the Tank

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

Pure water is then pumped into the tank at a rate of 5 L/min. Therefore:

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

`=(0(g)/(L))( 5(L)/(min))\\\\R_(in)=0(g)/(min)`

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

`=((A(t))/(350))( 5(L)/(min))\\R_(out)=(A(t))/(70)`

Therefore:

`(dA)/(dt)=0-(A)/(70)`

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

`(dA)/(dt)+(A)/(70)=0\\$The integrating factor: e^{\int (1)/(70)dt} =e^{(t)/(70)}\\$Multiplying by the integrating factor all through\\(dA)/(dt)e^{(t)/(70)}+(A)/(70)e^{(t)/(70)}=0\\(Ae^{(t)/(70)})'=0`

Taking the integral of both sides

`\int(Ae^{(t)/(70)})'=\int 0 dt\\Ae^{(t)/(70)}=C\\$Divide both sides by e^{(t)/(70)}\\A(t)=Ce^{-(t)/(70)}`

Recall that when t=0, A(0)=50 grams (our initial condition)

`A(t)=Ce^{-(t)/(70)}\\50=Ce^{-(0)/(70)}\\C=50`

Therefore, the number of grams of salt in the tank at time t is:

`A(t)=50e^{-(t)/(70)}`