Answer:
[tex]A=10e^{\frac{-t}{350} }[/tex]
Step-by-step explanation:
Given that a tank contains 350 liters of fluid in which 10 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 A(t) be the grams of salt at time t.
Then A(0) = 10 gm
Inflow of fluid = outflow of fluid = 5l/min
So volume of tank at time t = 350 litres
Rate of change of salt = A'(t) = incoming rate - outgoing rate
= 0-[tex]A(t)/350[/tex]
i.e. [tex]\frac{dA}{dt} =\frac{-A}{350} \\ln A = -t/350 +C\\A = Pe^{\frac{-t}{350} }[/tex]
Use the fact that when t =0, A = 10
P = 10
So we get
[tex]A=10e^{\frac{-t}{350} }[/tex]
