Chapter 6.5, Problem 34E

$\text{(a)}$

To determine

To find: The formula for the gorilla population in terms of $t$ .

Answer to Problem 34E

The answer: $P\approx \frac{250}{1+7.929e^{-0.1t}}$

Explanation of Solution

Given information:

The rate of growth of the population is:

  $\frac{dP}{dt}=0.0004P(250-P)$

Calculation:

  $\begin{array}{c}\frac{dP}{dt}=0.0004P(250-P)\\ =kP(M-P)\end{array}$

Put,

  $\begin{array}{c}M=250\\ k=0.0004\end{array}$

Into the general form of the logistic function.

  $\begin{array}{c}P=\frac{250}{1+A{e}^{-250(0.0004)t}}\\ =\frac{250}{1+A{e}^{-0.1t}}\end{array}$

Put,

  $P=28$

When

  $t=0$

To find the value of $A$ :

  $\begin{array}{c}28=\frac{250}{1+A{e}^0}\\ \to 28(1+A)=250\\ \to 1+A=\frac{125}{14}\\ \to A=\frac{111}{14}\\ \to A\approx 7.929\end{array}$

Put the values of $A$ into the equation:

  $\begin{array}{c}P=\frac{250}{1+\frac{111}{14}{e}^{-0.1t}}\\ \approx \frac{250}{1+7.929e^{-0.1t}}\end{array}$

Therefore, the required formula for the gorilla population in terms of $t$ is

  $P\approx \frac{250}{1+7.929e^{-0.1t}}$ .

$\text{(b)}$

To determine

To find: When will the gorilla population in the preserve achieve its carrying capacity.

Answer to Problem 34E

The more years pass, the closer the carrying capacity will be.

Explanation of Solution

Given information:

The rate of growth of the population is:

  $\frac{dP}{dt}=0.0004P(250-P)$

Calculation:

  $\begin{array}{c}P(t)=\frac{250}{1+7.93e^{-0.1t}}\\ \to 250=\frac{250}{1+7.93e^{-0.1t}}\\ \to 1=\frac{1}{1+7.93e^{-0.1t}}\\ \to 7.93e^{-0.1t}=0\\ t\to \infty \end{array}$

Therefore, the required the more years pass, the closer the carrying capacity will be.