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

  1. Jun 18, 2009 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant 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)

    3. 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...
     
  2. jcsd
  3. Jun 18, 2009 #2

    ideasrule

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    Homework Helper

    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.
     
  4. Jun 18, 2009 #3
    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?
     
  5. Jun 18, 2009 #4
    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...
     
  6. Jun 18, 2009 #5

    ideasrule

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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.
     
  7. Jun 19, 2009 #6
    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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