# Homework Help: Differential Equation: Mixing Problem

1. Jun 16, 2010

### qw111

1. The problem statement, all variables and given/known data

The air in a small room 12x8x8 ft (768 ft^3 is the volume) is 3% carbon monoxide. Starting at t=0, fresh air containing no carbon monoxide is blown into the root at rate of 100 ft^3/min. If the air in the room flows out at same rate (100 ft^3/min), when will air in room be 0.01%?

2. Relevant equations

s'(t) = (rate entering)(concentration entering) - (rate exiting)(concentration exiting)

concentration = amount/volume

3. The attempt at a solution

I don't need you to solve the problem for me, but I am having trouble getting started and figuring out what the concentration is.

s'(t) = (rate entering)(concentration entering) - (rate exiting)(concentration exiting)
s'(t) = (100)(?) - (100)(?)

concentration = amount/volume
Since it said at time t=0, fresh air containing *NO* carbon monoxide..., will the concentration entering be 0?
And for the concentration exiting, will it be s(t)/768?

so the s'(t) is: 0 - 100s(t)/768

Is this correct?

I am contemplating about the 3% carbon monoxide, could this be the concentration entering??

2. Jun 16, 2010

### HallsofIvy

Yes. But notice that it does NOT say "at time t= 0"- the problem says "starting at time t= 0". There is fresh air containing NO carbon monoxide entering the room for all positive t.

Well, that would depend upon what "s(t)" means- and you didn't tell us that!

If s(t) is the amount of carbon monoxide in the room at time t, then the concentration will be s(t)/768 per cubic foot. But s'= ds/dt is amount of carbon monoxide per minute. Since air is flowing out at 100 ft^3/min, it carries with it (s(t)/768 carbon monoxide/ft^3)(100 ft^2/min) carbon monoxide per minute.
so the s'(t) is: 0 - 100s(t)/768

No, it is the initial value: If s(t) is the amount of carbon monoxide in the room then s(0)= .03(768).

I cannot comment on what