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?

Answer 1

Given :

`$(dP)/(dt)= aP-bP^2$` .............(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 :

`$ (dP)/(dt)=P(a-bP)$`

`$(dP)/(dt)=bP((a)/(b)-P)$` .................(2)

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

`$(dP)/(dt)=kP(M-P)$`

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

`$a_0=(B_0)/(P_0)$` and
`$b_0=(D_0)/(P_0^2)$`

So,
`$M=(B_0P_0)/(D_0)$`

=
`$(9 * 220)/(15)$`

= 132

Now from equation (2), we get the constants

k = b =
`$(D_0)/(P_0^2) = (15)/(220^2)$`

=
`$(3)/(9680)$`

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

`$P(t)= (MP_0)/(P_0+(M-P_0)e^(-kMt))$`

`$P(t)= \frac{132 * 220}{220+(132-220)e^{-(3)/(9680) *132t}}$`

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

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

`$(110)/(100) * 132= \frac{29040}{220-88e^{(-396)/(9680) t}}$`

`$ 220-88e^{(-99)/(2420) t}=200$`

`$ e^{(-99)/(2420) t}=(5)/(22)$`

Now taking natural logs on both the sides we get

t = 36.216

Number of months = 36.216