Chapter 7, Problem 68RE

a. $$

To determine

Tofind $P\left(t\right)$ in terms of $t$ and $k$ if $P\left(0\right)=200$ .

Answer to Problem 68RE

The value of $P\left(t\right)$ is $P=600-400e^{-kt}$ .

Explanation of Solution

Given information:

The given population $P$ of wolves at time $t$ years is increasing at a rate directly proportional to $600-P$ .

Formula:

The logistic differential equation is

  $\frac{dP}{dt}=kP\left(M-P\right)$ where $M$ -carrying capacity, $k$ - constant.

The logistic differential equation is

  $\frac{dP}{dt}=kP\left(M-P\right)$ where $M$ -carrying capacity, $k$ - constant.

From given population,

  $\begin{array}{l}\frac{dP}{dt}=k\left(600-P\right)\\ \frac{dP}{600-P}=kdt\end{array}$

Integrating on both sides,

  $\begin{array}{l}-\ln \left|P-600\right|=kt+c\\ P-600=c{e}^{-kt}\\ 200-600=c\\ c=-400\\ \Rightarrow P=600-400e^{-kt}\end{array}$

Therefore, the value of $P\left(t\right)$ is $P=600-400e^{-kt}$ .

b. $$

To determine

To find thevalue of $k$ if $P\left(2\right)=500$ .

Answer to Problem 68RE

The value of $k$ is $k=0.69$ .

Explanation of Solution

Given information:

The given population $P$ of wolves at time $t$ years is increasing at a rate directly proportional to $600-P$ .

Formula:

The logistic differential equation is

  $\frac{dP}{dt}=kP\left(M-P\right)$ where $M$ -carrying capacity, $k$ - constant.

From part (a) we know that $P=600-400e^{-kt}$ ,

  $\begin{array}{l}P=600-400e^{-kt}\\ 500=600-400e^{-2k}\\ e^{-2k}=\frac{1}{4}\\ k=0.69\end{array}$

Therefore, the value of $k$ is $k=0.69$ .

c. $$

To determine

To find the value when $\underset{t\to \infty }{\lim }P\left(t\right)$ .

Answer to Problem 68RE

Thevalue is found to be $\underset{t\to \infty }{\lim }P\left(t\right)=600$ .

Explanation of Solution

Given information:

The given population $P$ of wolves at time $t$ years is increasing at a rate directly proportional to $600-P$ .

From part (a) and part (b) we know that $P=600-400e^{-kt}$ and $k=0.69$ ,

Applying limit we get,

  $\begin{array}{l}\underset{t\to \infty }{\lim }P\left(t\right)=600-400e^{-0.69\times \infty }\\ \underset{t\to \infty }{\lim }P\left(t\right)=600\end{array}$

Therefore, the value is found to be $\underset{t\to \infty }{\lim }P\left(t\right)=600$ .