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Another Modeling Growth and Decay Problem

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

    Consider an item that is initially sold at a market price of $10 per unit. Over time, market forces push the price toward the equilibrium price, $20, at which supply balances demand. After 6 months, the price has increased to $15. the Evans Price Adjustment model says that the rate at which the market price changes is proportional to the difference between the market price and the equilibrium price.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    dp/dt = 20-p, where p=market price in dollars and t=time in months
    ∫dp/20-p = ∫1dt
    ln (20-p) = t+c
    20-p = e^(t+c)
    20-p = e^t*e^c
    20-p = Ae^t, where A=e^c
    p = 20-Ae^t
    10 = 20-Ae^0
    10 = 20-A
    A=10

    Now that I've solved for A with the initial condition, I have hit a roadblock. I'm pretty sure my initial setup of this problem is off. When I substitute 15 in for p, I get a negative t, which is impossible. Please help.
     
  2. jcsd
  3. Apr 1, 2014 #2

    LCKurtz

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    $$\int \frac{dp}{20-p}\ne \ln(20-p)$$
     
  4. Apr 1, 2014 #3
    Whoops that's a negative. Now this problem makes sense.
     
  5. Apr 1, 2014 #4
    Sorry for the double post, but actually it looks like the sign doesn't even matter. Can someone please give me a little more insight? I'm pretty sure at this point it has to do with my initial setup.
     
  6. Apr 2, 2014 #5
    Ok so I've become aware that I need to solve for the constant of proportionality k.

    dp/dt = k(20-p)

    The price rises $5 in 6 months, so dp/dt=5/6 when p=10

    5/6=k(20-10)
    k=1/12
    dp/dt=(20-p)/12

    From here, I get what appears to be a faulty equation:

    p = 20 - 10e^t/12

    The reason I say this is because when I plug in 6 for t I don't get 15.
     
  7. Apr 2, 2014 #6

    pasmith

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    In general [itex]p'(6) \neq \dfrac{p(6) - p(0)}{6}[/itex].

    You need to solve
    [tex]
    \frac{dp}{dt} = k(20 - p)
    [/tex]subject to [itex]p(0) = 10[/itex], and then find [itex]k[/itex] from the condition that [itex]p(6) = 15[/itex].
     
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