Diff EQ: Mixing Problem! What am I doing wrong?

  • Thread starter space_yes
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  • #1
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Homework Statement



Alright, I know this has been done in various incarnations but I haven't found an example exactly similar to mine..

A tank contains 60kg of salt and 1000L of water. Pure water enters the tank at the rate of 8L/min. The solution is thoroughly mixed and drains at a rate of 4L/min. Find the amount of salt in the tank after 2 hours.

Homework Equations



y(t) = amount of salt in the tank after t minutes
v(t) = volume of water in the tank after t minutes

v(t) = 1000 + 8t-4t = 1000 + 4t

dy/dt = rate in - rate out

dy/dt = 8L/min * 0kg/L - y kg/(1000 + 4t L) * 4L/min

dy/dt = - 4y/(1000+4t)

The Attempt at a Solution



Separating...

dy/4y = - dt/(1000 + 4t)

Integrating both sides...

(1/4)Ln(y) = -(1/4)Ln(250+t)+c

y^(1/4) = (e^c)/(250+4t)^(1/4)

Let k be the arbitrary constant = e^c

y^(1/4) = k/(250 + 4t)^(1/4)

y = k/(250+4t)

Using the initial conditions...

60 = k/(250 + 4*0)
k = 15000

y = 15000/(250 + 4t) <--- this is not right
y(120) = 20.5479 <--- this is not right

The correct answer = 40.5405405405...
 

Answers and Replies

  • #2
ideasrule
Homework Helper
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(1/4)Ln(y) = -(1/4)Ln(250+t)+c

y^(1/4) = (e^c)/(250+4t)^(1/4)
I'm not sure how you got from the first equation to the second, but if you fix this step, you'd get the right answer.
 
  • #3
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I raised both sides to powers of e to remove the natural logs so I can get an expression for y...

...should I be doing something else there?
 
  • #4
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I figured it out... the step where I let k = e^c is the problem. Leaving e^c and continuing we would have:

y = e^(4c)/(250+4t)

Letting k = e^(4c)...

y = k/(250+4t)

Solving for k = 15000...

Now solving for c to find the numerator in the expression for y...

15000 = e^(4c)
c = ln(15000)/4

Putting c back into e^(4c)/(250+4t) yields the correct expression:

y(t) = (e^(ln(15000))/(250+4t)

y(120) = 40.5405...
 
  • #5
ideasrule
Homework Helper
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No, y(t) = (e^(ln(15000))/(250+4t) simplifies to 15000/(250+4t), because e^x and ln x are like addition and subtraction; if you take the log of a number and then raise e to the power of the result, you'll get the original number back. If you evaluate (e^(ln(15000))/(250+4t) correctly for t=120, you still get 20.5, not 40.5.

At this step:

(1/4)Ln(y) = -(1/4)Ln(250+t)+c

You should notice that you can multiply both sides by 4 to get:

ln y = -ln(250+t) + C
y=k/(250+t), where k is a constant.
You got y=k/(250+4t) instead of y=k/(250+t), which isn't right.
 
  • #6
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Thanks for the reply! I actually ended up getting the right answer b/c I worked out the problem out in mathematica and it correctly simplified. However, I didn't notice (what you and Mathematica did) the discrepancy between my written work (posted) and my code ergo my erroneous function y(120) = 40... when it ='d 20...

:biggrin:
 

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