Question

Suppose that a population y grows according to the logistic model given by formula: y=(L)/1+Ae^-kt . a. At what rate is y increasing at time t=0 ? b. In words, describe how the rate of growth of y varies with time. c. At what time is the population growing most rapidly?

Answer 1

$y=\frac{l}{1+Ae^{-kt}}$

a) $\frac{dy}{dt}=\frac{-lAe^{-kt}(-k)}{(1+Ae^{-kt})^2}$

$=\frac{kAle^{-kt}}{(1+Ae^{-kt})^2}$

$y'(t=0)=\frac{kAl}{1+A)^2}$

At time t y is increasing at a rate $=\frac{kAl}{1+A)^2}$

b) the rate of growth of y increases as t increases this is because it is always positive.

c) $y''=\frac{(1+Ae^{-kt})^2(-k^2lAe^{-kt})-kAle^{-kt}\cdot2(1+Ae^{-kt})(-kA)}{(1+Ae^{-kt})^4}=0$

$\implies (1+Ae^{-kt})(-1)+2=0$

$\implies 1+Ae^{-kt}=2$

$e^{-kt}=\frac{1}{A}$

$t=\frac{ln\frac{1}{A}}{-k}$

The population is growing most rapidly at time, $t=\frac{ln\frac{1}{A}}{-k}$