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

[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:

 

[Mangoes]

Use the initial conditions given.

Also, your solution's incorrect.

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.

 

[Jeff12341234]

Use the initial conditions given.

Also, your solution's incorrect.

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.

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

 

[Mangoes]

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.

 

[Jeff12341234]

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

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

 

[Mangoes]

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.

 

[hogrampage]

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

 

[Jeff12341234]

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

 

[Mangoes]

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

 

[Jeff12341234]

Right. Thanks

 

[Jeff12341234]

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

 

[hogrampage]

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

 

[Jeff12341234]

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

 

[hogrampage]

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

 

[Jeff12341234]

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

 

[hogrampage]

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

 

[Jeff12341234]

I get a different answer:

Is there another source we can check?

 

[hogrampage]

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).

 

[Jeff12341234]

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

 

[hogrampage]

http://wolframalpha.com/

Here is the result for the integration:

http://wolframalpha.com/input/?i=in...ntegral.rangeend--.**Integral.variable---.*--

 

[Jeff12341234]

wolfram says: (-5*Log[-2*(-50 + x)])/2

 

[Jeff12341234]

That's not the answer I got, nor is it the answer you got...