Chapter 7.5, Problem 36E

a.

To determine

To show that $P=M-A{e}^{-kt}$ , where $A$ is a constant determined by an appropriate initial condition.

Answer to Problem 36E

It is proved $P=M-A{e}^{-kt}$ , where $A$ is a constant determined by an appropriate initial condition.

Explanation of Solution

Given:

It is given that another differential equation that models limited growth of a population $P$ in an environment with carrying capacity $M$ is $dP/dt=k\left(M-P\right)$ where $k>0$ and $M>0$ .

Calculation:

The differential equation is

  $dP/dt=k\left(M-P\right)$

Using separation of variable method

  $\begin{array}{l}\frac{dP}{dt}=k\left(M-P\right)\\ \frac{1}{M-P}dP=kdt\end{array}$

integrating both side

  $\begin{array}{l}{\displaystyle \int \frac{1}{M-P}dP}={\displaystyle \int kdt}\\ -\ln \left|M-P\right|=kt+C\\ \ln \left|M-P\right|=-kt-C\\ M-P=e^{-kt-C}=e^{-kt}{e}^{-C}\\ let{e}^{-C}=A\\ M-P=A{e}^{-kt}\\ \therefore P=M-A{e}^{-kt}\end{array}$

where $A$ is a constant determined by an appropriate initial condition.

Therefore, it is proved $P=M-A{e}^{-kt}$ , where $A$ is a constant determined by an appropriate initial condition.

b.

To determine

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

Answer to Problem 36E

The value of $\underset{t\to \infty }{\lim }P\left(t\right)$ is $M$ .

Explanation of Solution

Given:

It is given that another differential equation that models limited growth of a population $P$ in an environment with carrying capacity $M$ is $dP/dt=k\left(M-P\right)$ where $k>0$ and $M>0$ .

Calculation:

Since, $P\left(t\right)=M-A{e}^{-kt}$

  $\begin{array}{l}\underset{t\to \infty }{\lim }P\left(t\right)\\ \underset{t\to \infty }{\lim }M-A{e}^{-kt}\\ =M-A{e}^{-k\times \infty }=M-A{e}^{-\infty }=M-A\left(\frac{1}{e^{\infty }}\right)\\ =M-A\left(\frac{1}{\infty }\right)=M-A\left(0\right)=M\\ \therefore \underset{t\to \infty }{\lim }P\left(t\right)=M\end{array}$

c.

To determine

To find for what time $t\ge 0$ is the population growing the fastest.

Answer to Problem 36E

Population growth is fastest at $t=0$ .

Explanation of Solution

Given:

It is given that another differential equation that models limited growth of a population $P$ in an environment with carrying capacity $M$ is $dP/dt=k\left(M-P\right)$ where $k>0$ and $M>0$ .

Calculation:

Since, the solution of differential equation $dP/dt=k\left(M-P\right)$ is $P=M-A{e}^{-kt}$

Now, by second derivative test

  $\begin{array}{l}\frac{dP}{dt}=k\left(M-P\right)=0\\ P=M\\ \frac{d^{2}P}{d{t}^2}=-K\\ \because \frac{d^{2}P}{d{t}^2}<0\\ \therefore whenP=Mpopulationgrowthisfastest\\ P=M-A{e}^{-kt}\\ M=M-A{e}^{-kt}\\ e^{-kt}=0\\ t=0\end{array}$

Therefore , population growth is fastest at $t=0$ .

d.

To determine

To explain how does the growth curve in this model differ from the growth curve in the logistic model.

Answer to Problem 36E

The graph of logistic curve has inflection point whereas the graph of limited growth does not have any inflection point.

Explanation of Solution

Given:

It is given that another differential equation that models limited growth of a population $P$ in an environment with carrying capacity $M$ is $dP/dt=k\left(M-P\right)$ where $k>0$ and $M>0$ .

The graph of logistic curve looks like the graph below

  

The graph of limited growth curve looks like the graph below

  

The graph of logistic curve has inflection point whereas the graph of limited growth does not have any inflection point.