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Population growth carrying capacity on a logistic model

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Here is the equation I am given. I'm supposed to find the carrying capacity.

    dp/dt=.05P-6.6666667e-5P^2

    I know the general solution is rp-rp^2/k with k=carrying capacity, but the addition of the middle term has thrown me off.

    3. The attempt at a solution

    I tried ignoring the middle term and did r/p=5. Since r=.05, p would be .01 in this case, but that doesn't show up as the correct answer. What does the middle term change?
     
  2. jcsd
  3. Aug 29, 2011 #2

    lanedance

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    can you explain which is the middle term you're talking about?
     
  4. Aug 29, 2011 #3

    lanedance

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    guessing what you're asking, first consider the simple first order exponential DE
    [tex]
    \frac{dP}{dt}=0.05P
    [/tex]

    for any positive starting value P_0, [itex] \frac{dP}{dt}=0.05P[/itex] will always be positive and [itex] P(t) [/itex] will increase exponentially

    now consider
    [tex]
    \frac{dP}{dt}=0.05P - \frac{2}{3} 10^{-5} P^2
    [/tex]

    for small [itex] P [/itex], this will behave the same as the first equation as [itex] P<1 \implies P^2<<1 [/itex]

    however as [itex] P [/itex] increases, the [itex]P^2[/itex] term will get larger decreasing the growth rate, until at some point [itex] \frac{dP}{dt} = 0 [/itex], the population will assymtotically approach this point -

    so can you find where [itex] \frac{dP}{dt} = 0 [/itex]?
     
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