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DE word problem: fluid in a tank

  1. Sep 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dye solution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 L / min. The well stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

    2. Relevant equations

    This is my first differential equations word problem, so I'm trying to learn how to do them. I don't need the answer to the 1% question; just some advice on what I did wrong below.

    3. The attempt at a solution

    [itex]
    \frac{dQ}{dt} = -(\frac{2L}{min})(\frac{Q(t)}{200L}) = \frac{-1}{2} Q(t)
    [/itex]

    Initial value: Q(0) = 1g/L

    [itex]
    \frac{dQ}{dt} + \frac{1}{2} Q = 0
    [/itex]

    [itex]
    \mu = e^{1/2*t}
    [/itex]

    [itex]
    Q(t) = \frac{C}{e^{1/2*t}}
    [/itex]

    At this point I solve for C using the initial value, and get Q(t) = 1 :( Where am I going wrong?

    Thank you
     
  2. jcsd
  3. Sep 7, 2013 #2

    vela

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    For one thing, 2/200 = 1/100, not 1/2.
     
  4. Sep 7, 2013 #3
    I fixed that, thanks. But regardless I end up in the same situation, with c/ e^... = 1. What else can I try?
     
  5. Sep 7, 2013 #4

    vela

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    Initial means t=0, so initial value means Q(0)=1. Remember C is a constant. It can't depend on t.
     
  6. Sep 7, 2013 #5
    When I solve for: 1= C/ e^(1/100 t)
    I still get 1 as a solution... Plugging in 0 for t to satisfy the initial condition Q(0)=1g/L

    Thanks
     
  7. Sep 7, 2013 #6

    vela

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    You found ##Q(t) = C e^{-t/100}##. That's fine. The problem is that you're solving for C by saying Q(t)=1, but that's not true for all t. The equation you want to solve is Q(0)=1.
     
  8. Sep 7, 2013 #7
    That simply means I plug 0 for t, and get 1=1 right? I must be confused...
     
  9. Sep 7, 2013 #8
    I guess i understand....
     
    Last edited: Sep 7, 2013
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