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Diffy Qs, mixing problem set up

  1. Mar 24, 2009 #1
    Hey, I have this problem that is for a grade, so I want to make sure I set it up correctly. I don't care about solving for the function or anything(I know I did that part correctly).

    1. The problem statement, all variables and given/known data
    A container has 10 pounds of salt dissolved in 100 liters of water. Pure water is pumped in at 4 L/hour and perfectly mixed solution is pumped out at 3L/hour.

    How much salt remains in the tank after 2 hours?

    3. The attempt at a solution

    change in pounds of salt = pounds of salt divided by the ever growing amount of liters times the amount of liters leaving should equal the pounds of salt leaving.
    [tex]
    \frac{dP}{dt} = \frac{P}{100 + 4t}(-3)t
    [/tex]
    where P is a function for pounds of salt in terms of hours.

    OR should the total volume be the difference of the liters entering and leaving? I thought it makes more sense to say the liters are added at 4L as 3L leave, so we should think of the total volume as increasing by 4L regardless of the 3L leaving.
     
  2. jcsd
  3. Mar 24, 2009 #2
    So can anyone clarify: should the denominator be 100+4t or 100+t?
     
  4. Mar 24, 2009 #3

    Mark44

    Staff: Mentor

    Can you explain in words what each factor on the right side of the equation above represents?
     
  5. Mar 24, 2009 #4
    P = function that gives pounds of salt as a function of time.
    t = times in hours
    -3t = flow of water out
    4t = flow of water in
    100 = initial number of liters

    pounds per liter times liters results in pounds traveling due to the flow of liters.

    so i said pounds (the function p gives pounds) divided by total liters(100 + 4t... or is it t) times liters leaving equals pounds leaving aka change in pounds.
     
  6. Mar 24, 2009 #5

    djeitnstine

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    Gold Member

    If 4L are entering and 3L are leaving then how can the volume increase by 4L? The net volume obviously increases by 1L for every time
     
  7. Mar 24, 2009 #6
    Are you sure? I was thinking the equation represents what is leaving by using what hasn't left yet. Thus, 4L add to the volume BEFORE the 3L leave. It's as if the 4L were added before the 3L leave by an infinitesimally small time.
     
  8. Mar 24, 2009 #7

    djeitnstine

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    You're over analyzing the problem. Lets imagine the reverse situation, entering at 3L and leaving at 4L, would the decrease in Volume as time goes be 4L also?
     
  9. Mar 24, 2009 #8
    noted. Everything else looks correct?
     
  10. Mar 24, 2009 #9

    Mark44

    Staff: Mentor

    I don't know. You have
    What does P/(100 +4t) represent, and why is it being multiplied by -3t?

    Your differential equation should model what's going on in the system. IOW, one should be able to translate directly from English into the symbols in the differential equation, and I'm not seeing it.
     
  11. Mar 24, 2009 #10

    djeitnstine

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    That extra 't' after (-3) has to go.
     
  12. Mar 24, 2009 #11
    [tex]

    \frac{dP}{dt} = \frac{P}{100 + t}(-3)

    [/tex]

    gives the solution P = c * (t+150)^2, which results in the pounds of salt boundlessly increasing. That doesn't make sense, because the salt is leaving the tank (and definitely isn't increasing).

    with the t, there is a bunch of really high powers(like to the 300) and low powerse(like -600), but it looks smooth and heads to zero smoothly and stays there.
     
    Last edited: Mar 24, 2009
  13. Mar 24, 2009 #12

    djeitnstine

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    Show your working because I got something completely different.
     
  14. Mar 24, 2009 #13
    I did my calculations with 2 instead of -3. I'm getting

    [tex]

    P =\frac{C}{(t+100)^3}
    [/tex]

    is that right?
     
  15. Mar 24, 2009 #14

    djeitnstine

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    Partially correct, you have the integration constant in the wrong place.
     
  16. Mar 24, 2009 #15
    i checked my work and i checked with a calculator. The calculator says i'm right. are you sure about your answer?
     
  17. Mar 24, 2009 #16

    djeitnstine

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    Ok lets do this step by step

    [tex]\frac{dP}{dt} =\frac{-3P}{(t+100)}[/tex]

    [tex]\int \frac{dP}{P} = \int \frac{-3dt}{(t+100)}[/tex]

    [tex]ln(P) = ln(t+100)^-3 + c[/tex] <--- integrating constant

    [tex]P= \frac{1}{(t+100)^3} + e^c[/tex]


    [tex]P(t)= \frac{1}{(t+100)^3} + k[/tex] <---e to the power of a constant is just another constant
     
  18. Mar 24, 2009 #17
    e^(a + b) = e^(a) * e^(b)

    just like x^3 * x^10 = x^13 or x^(3 + 10)

    but thanks for your insights on taking that t out and replacing that 4t with a t. You saved me points :O
     
  19. Mar 24, 2009 #18

    djeitnstine

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    I forgot about that for a moment, my bad =]
     
  20. Mar 25, 2009 #19

    Mark44

    Staff: Mentor

    The left side above is correct since [itex]e^{ln(P)} = P[/itex], but the right side is not.
    You are exponentiating each side, so the right side is
    [tex]e^{ln\frac{1}{(t + 100)^3 } + c}[/tex]
    [tex]= e^{ln\frac{1}{(t + 100)^3 }}e^c[/tex]
    [tex]= \frac{1}{(t + 100)^3}K[/tex]
    [tex]= \frac{K}{(t + 100)^3}[/tex]
    where K = e^c

    In the 2nd line above, I am using the fact that [itex]e^{a + b} = e^a * e^b[/itex]

    Edit: I didn't see that you had corrected your work.

     
    Last edited: Mar 25, 2009
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