Consider a rabbit population​ P(t) satisfying the logistic equation StartFraction dP Over dt EndFraction equals aP minus bP squared ​, where

Question

Consider a rabbit population​ P(t) satisfying the logistic equation StartFraction dP Over dt EndFraction equals aP minus bP squared ​, where Upper B equals aP is the time rate at which births occur and Upper D equals bP squared is the rate at which deaths occur. If the initial population is 220 rabbits and there are 9 births per month and 15 deaths per month occurring at time tequals​0, how many months does it take for​ P(t) to reach 110​% of the limiting population​ M?

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Trung Dũng 5 years 2021-09-05T10:11:41+00:00 1 Answers 31 views 0

Answers ( )

  1. ngocdiep
    0
    2021-09-05T10:13:06+00:00
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    Solution:

    Given :

    $\frac{dP}{dt}= aP-bP^2$         ………….(1)

    where, B = aP = birth rate

                D = $bP^2$  =  death rate

    Now initial population at t = 0, we have

    $P_0$ = 220 ,  $B_0$ = 9 ,  $D_0$ = 15

    Now equation (1) can be written as :

    $ \frac{dP}{dt}=P(a-bP)$

    $\frac{dP}{dt}=bP(\frac{a}{b}-P)$    ……………..(2)

    Now this equation is similar to the logistic differential equation which is ,

    $\frac{dP}{dt}=kP(M-P)$

    where M = limiting population / carrying capacity

    This gives us M = a/b

    Now we can find the value of a and b at t=0 and substitute for M

    $a_0=\frac{B_0}{P_0}$    and     $b_0=\frac{D_0}{P_0^2}$

    So, $M=\frac{B_0P_0}{D_0}$

              = $\frac{9 \times 220}{15}$

              = 132

    Now from equation (2), we get the constants

    k = b = $\frac{D_0}{P_0^2} = \frac{15}{220^2}$

            = $\frac{3}{9680}$

    The population P(t) from logistic equation is calculated by :

    $P(t)= \frac{MP_0}{P_0+(M-P_0)e^{-kMt}}$

    $P(t)= \frac{132 \times 220}{220+(132-220)e^{-\frac{3}{9680} \times132t}}$

    $P(t)= \frac{29040}{220-88e^{-\frac{396}{9680} t}}$

    As per question, P(t) = 110% of M

    $\frac{110}{100} \times 132= \frac{29040}{220-88e^{\frac{-396}{9680} t}}$

    $ 220-88e^{\frac{-99}{2420} t}=200$

    $ e^{\frac{-99}{2420} t}=\frac{5}{22}$

    Now taking natural logs on both the sides we get

    t = 36.216

    Number of months = 36.216

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