AP Calculus BC - Mixing Problem

Question

A 1400 gallon tank initially contains 600 gallons of water with 4 lbs of salt dissolved in it. Water with a salt concentration of 2 lbs/gallon enters the tank at a rate of 10 gal/h, while the solution leaves the tank at a rate of 6 gal/h. Find the amount of salt in the tank when it overflows.

Answer 1

Let A = A(t) = amount (in pounds) of salt in the tank at time t hours. We are given that A(0) = 4.

dA/dt = (Rate of salt in) - (Rate of salt out)

Rate in = (2 lb/gal)(10 gal/hr) = 20 lb / hr

Rate out = (A/600 lb/gal)(6 gal/hr) = A(t)/100 lb/hr

So, dA/dt = 20 - A(t)/100

dA/dt + (1/100)A(t) = 20 (First order linear differential equation)

Multiply by the integrating factor e^∫(1/100)dt = e^(t/100):

e^(t/100)(dA/dt) + (1/100)e^(t/100)A(t) = 20e^(t/100)

(e^(t/100)A(t))' = 20e^(t/100)

e^(t/100) A(t) = 2000e^(t/100) + C

A(t) = 2000 + Ce^-(t/100)

SInce A(0) = 4, we have 2000 + C = 4. So, C = -1996

A(t) = 2000 - 1996e^-(t/100)

Each hour, 10 gal enters the tank while 6 gal leaves the tank, so each hour, the amount of water in the tank increases by 4 gal.

The tank overflows after (1400 - 600) / 4 hr = 200 hr

Amount of salt when the tank overflows = A(200)