Question

Another model for a growth function for a limited population is given by the Gompertz function, which is a solution of the differential equation dPdt=cln(KP)P where c is a constant and K is the carrying capacity.(a) Solve this differential equation for c=0.1, K=2000, and initial population P0=500. P(t)= .(b) Compute the limiting value of the size of the population. limt→[infinity]P(t)= .(c) At what value of P does P grow fastest? P= .

Answer 1

Answer:

A) $P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}$

B) P(t→∞)=2000

C) $P=\frac{K}{e}=\frac{carrying capacity}{e}$

Step-by-step explanation:

Given differential eq is

                      $\frac{dP}{dt}=c ln (\frac{K}{P})P$ --- (1)

Eq is separable

                     $\frac{1}{ln (\frac{K}{P})P}dP=cdt$ --- (2)

                     $let \\u = ln\frac{K}{P}\\du= \frac{1}{\frac{K}{P}}(\frac{-K}{P^{2}}).dP\\du=\frac{-1}{P}.dP\\dP=-P.du$

substituting in  (2)

$-\frac{du}{u}=dt$

Integrating both sides

$\int {-\frac{1}{u}} \, du=\int{c}\,dt\\-ln|u|=ct +B\\ln|u|=-ct -B$

Back substituting value of u

$ln |ln\frac{K}{P}|=-ct-B\\|ln\frac{K}{P}|=e^{-ct-B}\\ln|\frac{K}{P}|=be^{-ct}$---(3)

at t =0

$ln|\frac{K}{P}|=be^{-ct}\\b=ln|\frac{K}{P}|\\b=ln\frac{2000}{500}\\b=ln|4|$

from (3)

$ln|\frac{K}{P}|=be^{-ct}\\\frac{K}{P}=e^{ln4e^{-ct}}\\P(t)=\frac{K}{e^{ln4e^{-ct}}}$

$P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}$

B) $\lim{t \to \infty}$

$P( {t \to \infty} )=\frac{2000}{e^{ln4e^{-0.1\infty}}}\\e^{-0.1\infty}=0\\\implies P( {t \to \infty} )=\frac{2000}{e^{0}}}\\\\P(\infty)=2000$

which is the carrying capacity.

C) To find the fastest growth rate we have to maximize $\frac{dP}{dt}$

From given differential eq

$\frac{dP}{dt}=cln|\frac{K}{P}|P$

so function to maximize is

$f(P)=cln|\frac{K}{P}|P$

$f'(P)=cln|\frac{K}{P}|+c\frac{1}{\frac{K}{P}}\frac{-K}{P^{2}}.P$

$f'(P)=c[ln|\frac{K}{P}|-1]$

To maximize find f'(P)=0

$c[ln|\frac{K}{P}|-1]=0$

$ln|\frac{K}{P}|=1$

$\frac{K}{P}=e$

$P=\frac{K}{e}=\frac{carrying capacity}{e}$