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Modeling problem

  1. Aug 27, 2005 #1


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    I was hoping that I could find help solving this problem. I've been working on it for a while and haven't been able to solve it correctly.

    So far I have:
    \frac{dQ}{dt} = r - \frac{rQ}{100}

    Where Q is the amount of salt, r is the rate in and out.
    I think that this may work for the first 10 minutes while the salt is part of what is getting mixed in, but after that I'm lost. Any help will be appriciated.
    Thank you.
    Last edited: Aug 27, 2005
  2. jcsd
  3. Aug 27, 2005 #2
    you first need an expression for dy/dt

    dy / dt = rate_in - rate_out

    rate_in = .5 lb/gal * 2 gal/min = 1lb/min

    rate_out = y(t) / 100 lb/gal * 2 gal / min = y(t) / 50 lb/min

    so dy/dt = 1 lb/min - y(t) / 50 lb/min

    then you find the IF and go from there
  4. Aug 28, 2005 #3


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    No, not quite. There are 2 gallons of water coming in every minute (I don't know why you call that "r") and each gallon carries 1/2 pound of salt so 2*(1/2) = 1 pound of salt comes in every minute ((1/2)r, not r). Each gallon of water going out contains [itex]\frac{Q}{100}[/itex] pounds of salt so the amount going out is [itex]\frac{rQ}{100}[/itex] as you have (with, of course, r= 2). The differential equation is:
    [tex]\frac{dQ}{dt}= 1- \frac{Q}{50}[/tex]
    Also, of course, Q(0)= 0.

    Yes, so solve that differential equation for Q(t) and use it to find Q(10).
    Now, start all over again. Now, there is NO salt coming in so the differential equation is
    [tex]\frac{dQ}{dt}= -\frac{Q}{50}[/tex].
    Q(10)= whatever you found before and then find Q(20) (ten more minutes).

    (Since t does not appear explicitely in that equation, you could just "restart" the clock: Let Q(0)= whatever you found before and then find Q(10) from this new equation.)
  5. Aug 28, 2005 #4


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    Thanks to both of you for replying. I've got it all worked out.
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