Question

8. Scientists use the Logistic Growth P.K P(t) = function P. +(K-P.)e FC to model population growth where P. is the population at some reference point, K is the carrying capacity which is a theoretical upper bound of the population and ro is the base growth rate of the population.

Answer 1

Given Logistic Growth Function of population is:

$P(t)=\frac{P_{0}K}{P_{0}+(K-P_{0})e^{-r_{0}t}}$

where P0 is the population at some reference point, K is the carrying capacity, r0 is the base growth rate of the population and t is the time.

e)

The growth rate function is given by P'(t). Therefore:

$P'(t)=\frac{\mathrm{d} (\frac{P_{0}K}{P_{0}+(K-P_{0})e^{-r_{0}t}})}{\mathrm{d} t}$

$=>P'(t)=P_{0}K\frac{\mathrm{d} (\frac{1}{P_{0}+(K-P_{0})e^{-r_{0}t}})}{\mathrm{d} t}$

$=>P'(t)=-P_{0}K\frac{\frac{\mathrm{d} (P_{0}+(K-P_{0})e^{-r_{0}t})}{\mathrm{d} t}}{(P_{0}+(K-P_{0})e^{-r_{0}t})^2}$

$=>P'(t)=-P_{0}K\frac{\frac{\mathrm{d} (e^{-r_{0}t}))}{\mathrm{d} t}(K-P_{0})}{(P_{0}+(K-P_{0})e^{-r_{0}t})^2}$

$=>P'(t)=-P_{0}K\frac{(-r_{0})e^{-r_{0}t}(K-P_{0})}{(P_{0}+(K-P_{0})e^{-r_{0}t})^2}$

$=>P'(t)=\frac{e^{-r_{0}t}P_{0}Kr_{0}(K-P_{0})}{(P_{0}+(K-P_{0})e^{-r_{0}t})^2}$

On simplification:

$P'(t)=-\frac{KP_{0}(P_{0}-K)r_{0}e^{r_{0}t}}{(P_{0}e^{r_{0}t}-P_{0}+K)^2}$

f)

There seems to be insufficient data for plotting of the graph, the values of P₀ , K and r₀ are needed for graphing.However, the graph looks somewhat like this (values are taken for showing the way the graph looks)

+ P'(T) X -40 + y KP.(P. - K)reot (Poet +K) 2 - Pot 2 -30 X K= 100 -10 100 3 x Po = 5 20 T! -10 10 x = 1.5 -10 112 -10 10 5 0 10 20 30 T

g)

The growth rate is maximum at time t when P''(t)=0 and P'''(t)>0:

On doing the necessary differentiation:

$t_{max.\:growth}=\frac{ln(\frac{K}{P_{0}}-1)}{r_{0}}$