Answer:
A(t) = 120 - 100 e^(-t/40)
Step-by-step explanation:
Since The Fluid is been pumped both in and out, there will always be a 120 Liters of fluid in a tank
The amount of salt at time t in tank is= A(t) (in grams)
the amount of salt in a tank at time t is given as = A(t) grams
Level of concentrations of salt is = A/120 g/L
The brine quantity that is pumped in the tank =3L (of salt concentration = 1 g/L)
The salt pumped in: 3L x 1g/L = 3g rams
The level of Amount of brine that is pumped out at time t: 3L (of salt concentration = A/120 g/L)
The quantity of salt pumped out at time t: 3L x A/120 g/L = A/40 g
Therefore, dA/dt = 3 - A/40
40 dA/dt = 120 - A
40/(120 - A) dA = dt
-40 ln|A-120| = t + C₀
ln|A-120| = -t/40 + C₁ , where C₁ = -C₀/40
A - 120 = Ce^(-t/40) .. where C = e^C₁
A = 120 + C e^(-t/40)
Recall that, the initial brine contains 20 grams of salt
so,
A(0) = 20
120 + C e^0 = 20
C = -100 or 100
Then
A(t) = 120 - 100 e^(-t/40)
