# DE word problem: fluid in a tank

1. Sep 7, 2013

### oneamp

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

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dye solution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 L / min. The well stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

2. Relevant equations

This is my first differential equations word problem, so I'm trying to learn how to do them. I don't need the answer to the 1% question; just some advice on what I did wrong below.

3. The attempt at a solution

$\frac{dQ}{dt} = -(\frac{2L}{min})(\frac{Q(t)}{200L}) = \frac{-1}{2} Q(t)$

Initial value: Q(0) = 1g/L

$\frac{dQ}{dt} + \frac{1}{2} Q = 0$

$\mu = e^{1/2*t}$

$Q(t) = \frac{C}{e^{1/2*t}}$

At this point I solve for C using the initial value, and get Q(t) = 1 :( Where am I going wrong?

Thank you

2. Sep 7, 2013

### vela

Staff Emeritus
For one thing, 2/200 = 1/100, not 1/2.

3. Sep 7, 2013

### oneamp

I fixed that, thanks. But regardless I end up in the same situation, with c/ e^... = 1. What else can I try?

4. Sep 7, 2013

### vela

Staff Emeritus
Initial means t=0, so initial value means Q(0)=1. Remember C is a constant. It can't depend on t.

5. Sep 7, 2013

### oneamp

When I solve for: 1= C/ e^(1/100 t)
I still get 1 as a solution... Plugging in 0 for t to satisfy the initial condition Q(0)=1g/L

Thanks

6. Sep 7, 2013

### vela

Staff Emeritus
You found $Q(t) = C e^{-t/100}$. That's fine. The problem is that you're solving for C by saying Q(t)=1, but that's not true for all t. The equation you want to solve is Q(0)=1.

7. Sep 7, 2013

### oneamp

That simply means I plug 0 for t, and get 1=1 right? I must be confused...

8. Sep 7, 2013

### oneamp

I guess i understand....

Last edited: Sep 7, 2013