Chapter 6.5, Problem 36E

(a)

To determine

To prove: The statement: $P=M=A{e}^{-kr}$

Explanation of Solution

Given information:

The differential equation is $dP/dt=k(M-P)$

Calculation:

  $\begin{array}{c}\frac{1}{M-P}dP=kdt\\ \to {{\displaystyle \int }}^{\text{}}\frac{1}{M-P}dP={{\displaystyle \int }}^{\text{}}kdt\\ \to -\ln |M-P|=kt+C\\ \to M-P=e^{-kt-C}\\ \to M-P=e^{-kt}\cdot e^{-C}\\ \to -P=-M+e^{-kt}\cdot e^{-C}\\ \to P=M-e^{-kt}\cdot A\\ \to P=M-A{e}^{-kt}\end{array}$

Therefore, the required statement $P=M=A{e}^{-kr}$

  $\text{(}$ proved $)$ .

(b)

To determine

To find: The value of

Answer to Problem 36E

The answer:

Explanation of Solution

Given information:

The differential equation is $dP/dt=k(M-P)$

Calculation:

  

Therefore, the required value of

(c)

To determine

To find: The time that the population growing the fastest.

Answer to Problem 36E

The answer: The population is growing the fastest at $t=0$ .

Explanation of Solution

Given information:

The differential equation is $dP/dt=k(M-P)$

Calculation:

When the second derivative is $0$ , growth is the quickest,

  $\begin{array}{c}\frac{dP}{dt}=k(M-P)\\ \frac{d^{2}P}{d{t}^2}=-k\frac{dP}{dt}\\ =0\\ -k^2(M-P)=0\\ P=M\\ =M-A{e}^{-kt}\\ e^{-kt}=0\end{array}$

Therefore, the required population is growing the fastest at $t=0$ .

$(d)$

To determine

To find: How is the growth curve of this model different from the logistic model's growth curve

Answer to Problem 36E

The answer: The restricted growth curve does not have an inflection point.

Explanation of Solution

Given information:

The differential equation is $dP/dt=k(M-P)$

Explain:

The restricted growth curve does not have an inflection point.