Differential Equation Problem?

Differential equation word problem?

• A 500 gallon tank initially contains 100 gallons of alcohol. Starting at t=0, pure water is pumped into the tank at a rate of 2 gallons per minute, and the completely mixed contents of the tank are pumped out at a rate of 1 gallon per minute. Since water is being pumped in faster that the mixture is being pumped out, the tank will get fuller and fuller and will eventually overflow. a. write an expression for A(t), the volume of alcohol in the tank at time t. As usual, this will require solving a differential equation. b. How many gallons of alcohol remain in the tank when it starts to overflow? I am just stuck on this problem.. Please help!! thank you

• Answer:

  A(t) = volume of alcohol in tank at time t -----> A(0) = 100 V(t) = volume of mixture in tank at time t -----> V(0) = 100 Liquid is pumped in at a rate of 2 gal/min, and pumped out at a rate ot 1 gal/min So liquid in tank increases at a net rate of 2 gal/min − 1 gal/min = 1 gal/min dV/dt = 1 V(t) = t + 100 Liquid pumped in is pure water, so it contains no alcohol Alcohol is being pumped in at rate of 0 gal/min Liquid pumped out of tank has alcohol concentration of A(t)/V(t) = A/(t+100) at time t Liquid is being pumped out at rate of 1 gal/min Therefore alcohol is being pumped out at rate of A/(t+100) * 1 gal/min = A/(t+100) gal/min dA/dt = 0 − A/(t+100) dA/dt = −A/(t+100) dA/A = −1/(t+100) dt Integrate both sides: ln|A| = −ln(t+100) + ln(C) ln|A| = ln|C/(t+100)| A = C / (t+100) A(0) = 100 C/100 = 100 C = 10,000 A(t) = 10,000 / (t+100) -------------------- Tank starts to overflow when it reaches maximum capacity (500 gal) So we find time at which this happens: V(t) = 500 t + 100 = 500 t = 400 Now we find amount of alcohol at that time A(t) = 10,000 / (t+100) A(400) = 10,000 / 500 = 20 20 gallons of alcohol remain in tank when it starts to overflow