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Modeling Growth and Decay Problem

  1. Apr 1, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose water leaks out of a barrel at a rate proportional to the square root of the depth of the water. If the level starts at 36 in. and drops to 35 in. after 1 hour, how long will it take for all the water to leak out of the barrel?

    I have to choose and implement a differential equation solution method to determine a solution.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    dL/dt = -√(L), where L is the water level and t is time in hours
    ∫dL/-√(L) = ∫1dt
    -2√(L) = t + c
    L = ((t+c)/2)^2
    36 = (-c/2)^2
    c = 12

    So now that I've solved for the constant using the initial condition, I'm not sure what to do. I think I might've even set up this problem wrong to begin with because when I substitute 1 in for t, I don't get 35, which is what L should equal at that t.

    Help.
     
  2. jcsd
  3. Apr 1, 2014 #2

    haruspex

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    It says proportional to, not equal to.
     
  4. Apr 1, 2014 #3
    I'm not sure how to translate that to an equation. Proportional is a tricky word in this problem.
     
  5. Apr 1, 2014 #4

    Mark44

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    If, for example, x is proportional to y, then the ratio of these quantities is a constant. IOW, x/y = k.

    Equivalently, x = ky, where k is the constant of proportionality.
     
  6. Apr 1, 2014 #5

    haruspex

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    To expand on that, this gives you another constant that needs to be adjusted to fit the data. This should resolve your problem of getting the wrong answer when you set t = 1.
     
  7. Apr 2, 2014 #6
    Ok so (dp/dt)/20-p = k

    then dp/dt = k(20-p)

    The problem is there are no conditions I can use in this problem that allow me to solve for k. I have conditions for p, but none for dp/dt.
     
  8. Apr 2, 2014 #7

    haruspex

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    p? 20?

    Your ODE was dL/dt = -√(L). Inserting the missing constant of proportionality we have dL/dt = -k√(L). Solve that.
     
  9. Apr 2, 2014 #8
    I was thinking of the other problem I posted. I think this is the same problem with that one too.
     
  10. Apr 2, 2014 #9
    Ok so solving for k.

    The water level dropped 1 inch in an hour at L=35 so

    1 = -k√35
    k=-1/√35

    Does that look right?
     
  11. Apr 2, 2014 #10

    haruspex

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    It might be near enough, but it's not exactly right. The rate is constantly changing. That 1 inch in an hour is the average rate over the first hour. It will be the rate at some L value between 35 and 36, but it won't be the rate at L=35.
    Correct is to solve the integral first then plug in the numbers.
     
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