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Mixture Problem ODE

  1. Jun 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A 500 gallon tank is filled with pure water. A solution containing 4lbs of salt per gallon is added at a rate of 2 gallons per minute. The well-mixed solution is drained at a rate of 3 gallons per minute.

    A) How long does it take for the container to achieve a concentration of 2lbs/gal?

    B) What is the domain of x(t)?

    2. Relevant equations
    ODE


    3. The attempt at a solution
    I found x(t) to be (500-t)^2*(4000-t^2)

    then I tried using x(t)/v(t)=2 lbs/gal and this is where I am stuck
     
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  3. Jun 21, 2009 #2

    Dick

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    If x(t)=(500-t)^2*(4000-t^2), then if t=0 then x(0)=500^2*4000. What does that mean? What is x(t) supposed to be? What ODE did you solve? How did you get that strange answer?
     
  4. Jun 21, 2009 #3
    x'+(3/(500-t))x=8 was the DE I used
     
  5. Jun 21, 2009 #4

    Dick

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    I'm going to guess x(t) is the amount of salt. The ODE looks right. The solution doesn't. Can you show us more of how you solved it?
     
    Last edited: Jun 21, 2009
  6. Jun 22, 2009 #5
    Integrating Factor

    e^[tex]\int(3/(500-t))[/tex] dt = (500-t)^(-3)

    Dt[tex]\int [(500-t)^(-3)*x]dt[/tex]=8[tex]\int(500-t)^(-3)[/tex]


    (500-t)^(-3) X=4/(t-500)^2 + C
     
  7. Jun 22, 2009 #6

    Dick

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    That seems ok again. Now what's C? What is x(0) supposed to be?
     
    Last edited: Jun 22, 2009
  8. Jun 22, 2009 #7

    HallsofIvy

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    Surely this is not the exact statement of the problem! There is NO "x(t)" in the original statement. What does x(t) mean here?

     
  9. Jun 22, 2009 #8

    The initial conditions are x(0)=0 since it is all pure water at time 0

    After solving using the integrating factor my x(t)(pounds of salt)= -4(t-500)+c(t-500)^3
     
  10. Jun 22, 2009 #9

    Dick

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    Ok, but now you have to find C using x(0)=0. Then find where the concentration x(t)/(500-t)=2lbs/gallon, right?
     
  11. Jun 22, 2009 #10
    Yes that is where I am caught up I tried several ways to get c and then plug it into x(t)/(500-t)=2lbs/gallon


    The c I come up with is .000016
     
  12. Jun 22, 2009 #11

    Dick

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    That's gives you x(0)=0 doesn't it? What's wrong with that?
     
  13. Jun 22, 2009 #12
    Yes but when I plug that into my X(t)/v(t)

    (-4(t-500)+.000016(t-500)^3)/((500-t)=2

    which I can reduce to 4-.000016(t-500)^2=2

    but the answer t=853.55 is not correct I just need some help verifying my answer
     
  14. Jun 22, 2009 #13

    Dick

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    It's a quadratic equation. It has two solutions, not just one. Can you find the other one?
     
  15. Jun 22, 2009 #14
    Yea i am sorry t=146.44 t= 853.553
    However doesnt t have to be less then 500 going all the way back to the integrating factor.

    e^(-3*ln(500-t)) t<500???
     
  16. Jun 22, 2009 #15

    Dick

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    Sure, t has to be less than 500. At t=500 the tank is drained. I think that's the sense to the question about what is the domain of x(t). The ODE has a solution outside of that range, but it doesn't make physical sense.
     
    Last edited: Jun 22, 2009
  17. Jun 22, 2009 #16

    Yes thank you so the Domain is [0,500] and when t=146.446609 min our concentration is 2lbs/gal
     
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