# Diff. Eq. Mixture problem. Did most of the work, Need to solve for C

1. Feb 22, 2013

### Jeff12341234

I need to solve for C. I know it's probably simple but i don't remember how to. This is what I have so far:

2. Feb 22, 2013

### Mangoes

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

Use the initial conditions given.

You only have 100L of solution at one point in time, so at only one point in time will the concentration leaving the tank be A/100. There's a net change in volume as time passes.

3. Feb 22, 2013

### Jeff12341234

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

so what should it be instead of A/100?? A/35?

4. Feb 22, 2013

### Mangoes

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

The rate out will be the the concentration times the rate at which the fluid flows out per minute.

You already know the fluid flows out at 5L/min, but your concentration's wrong.

It's correct that the concentration will depend on the amount A of chemical X, but the concentration also depends on the total volume of the solution. Five grams in 1 L is much more concentrated than five grams in 100L.

You just need to write the concentration in a way that's dependent on time since your volume also depends on time.

Last edited: Feb 22, 2013
5. Feb 22, 2013

### Jeff12341234

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

so it will be A/(100+5t)?

or maybe A/(100-2t) <--- -2t came from Rate in rate out

Last edited: Feb 22, 2013
6. Feb 22, 2013

### Mangoes

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

No, but you're getting closer.

V(t) = 5t + 100 would imply that the volume is increasing with time. It's not.

Look at the net change in volume.

7. Feb 22, 2013

### hogrampage

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

Close, but you need to consider the rate into the tank as well.

8. Feb 22, 2013

### Jeff12341234

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

A/(100-2t)? <--- -2t came from Rate in - rate out

9. Feb 22, 2013

### Mangoes

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

A/(100 - 2t) is correct, but don't forget to multiply it by 5 representing the liters/min to get the total rate out.

10. Feb 22, 2013

### Jeff12341234

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

Right. Thanks

11. Feb 22, 2013

### Jeff12341234

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

I get a non-real result now when I try to solve for c. Where did I go wrong?

12. Feb 23, 2013

### hogrampage

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

That shouldn't be giving you a non-real solution. It looks fine to me.

13. Feb 23, 2013

### Jeff12341234

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

no, it doesn't work out.. :/

14. Feb 23, 2013

### hogrampage

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

Oh, I read it wrong. It should be 50-t, not t-50.

15. Feb 23, 2013

### Jeff12341234

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

are you saying it should be A*(t-50)^(-5/2)=-8(50-t)^(-3/2)+c ?
How do you figure??

16. Feb 23, 2013

### hogrampage

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

The integral of 5/(100-2t) is (-5/2)ln(50-t).

17. Feb 23, 2013

### Jeff12341234

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

Is there another source we can check?

18. Feb 23, 2013

### hogrampage

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

That's strange that it's giving you that answer. I checked by doing it by hand, as well as wolfram alpha and I get (-5/2)ln(50-t).

19. Feb 23, 2013

### Jeff12341234

Re: Diff. Eq. Mixture problem. Did most of the work, Need to solve for

do you have a wolfram alpha link? I want to try it.

20. Feb 23, 2013