Another one on logistics equations?

In ecology, the logistic equation is often written in the form:

dN/dt = rN (1 – N/K),

where N = N(t) stands for the size of the population at time t, the constants r and K stand for the intrinsic growth rate and the carrying capacity of the species, respectively.

If a pond on a fish farm has a carrying capacity of 1000 fish and an intrinsic growth rate of 0.3 (when time is measured in months), and if the pond was originally stocked with 100 fish.

(a) Set up a logistic equation for the fish population N (t) in the pond, with t measured in months.

(b) Find the size of the population when t = 100

(c) Can the population reach 1200 at any future time?

1 Answer

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  • ?
    Lv 7
    9 years ago
    Favourite answer

    a) dN/dt = 0.3N (1 - N/1000)

    so, dN/[N(1 - N/1000)] = 0.3 dt

    Now, 1/N(1 - N/1000) = 1000/N(1000 - N)

    Using partial fractions we can express this as 1/N + 1/(1000 - N)

    Then, integrating gives:

    ln N - ln(1000 - N) = 0.3t + C

    Now, N = 100 when t = 0

    so, ln 100 - ln 900 = C

    => ln N - ln(1000 - N) = 0.3t + ln 100 - ln 900

    => ln[ N/(1000 - N) ] + ln 9 = 0.3t

    i.e. ln[ 9N/(1000 - N) ] = 0.3t

    so, 9N/(1000 - N) = e^0.3t

    when t = 100, 9N/(1000 - N) = e^30 = 1.069 x 10^13

    so, 9N = (1.069 x 10^13)(1000 - N) = 1.069 X 10^16 - N(1.069 x 10^13)

    => 9N + N(1.069 x 10^13) = 1.069 x 10^16

    Then, N = [(1.069 x 10^16)/( 9 + 1.069 x 10^13)] = 1000....the carrying capacity!!

    How can this be exceeded by 200?....Help!!

    :)>

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