Differential Equation mixing problem

[jordan123]

Differential Equation mixing problem!

Homework Statement

A vat with 500 gallons of beer contains 4% alcohol (by volume). Beer with 6% alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour?

The Attempt at a Solution

This is what I've been doing.

dA/dt = inflow - out flow
= (.06)(5) - (A/500)(5)
= .3 - (A/100)
= (30 - A)/100

etc but I feel I must be setting it up wrong or something ?? But it never gets the correct answer which is.

4.9%

If anyone can point out what's up that would be much appreciated!

 

[Dick]

Your solution never uses the "4% alcohol (by volume)." initial condition. Doesn't that make you feel uncomfortable? Must it not make a difference? If the initial value were 6% then nothing would ever change.

 

[jordan123]

Your solution never uses the "4% alcohol (by volume)." initial condition. Doesn't that make you feel uncomfortable? Must it not make a difference? If the initial value were 6% then nothing would ever change.

Well yes, id use the initial condition after I have done the differential equation.

So,

-ln |30 - A | = t/100 + C

then put in initial condition. Solve for c = -ln10 (because 4% of 500 is 20)

No? Its not working for me, eeek

 

[Dick]

It looks like it's working fine to me. Now solve for A. Why do you think it's not working? Show the rest. What are you putting in for t? What are you getting for A? What's that in terms of percentages?

 

[viciousp]

make sure when you do e^(-ln|30-A|) you get 1/(30-A), everything else is right