Chapter 7.1, Problem 9E

a.

To determine

To find : The value of $P$ for which the population is increasing.

Answer to Problem 9E

At $P=4200$ population is increasing.

Explanation of Solution

Given information :

  $\frac{dP}{dt}=1.2P\left(1-\frac{P}{4200}\right)$

Formula used :

For increasing or decreasing population $\frac{dP}{dt}=0$

If, ${\frac{d^{2}P}{d{t}^2}|}_{\text{at some P}}<0$ ,then the population is increasing.

Calculation :

  $\frac{dP}{dt}=1.2P\left(1-\frac{P}{4200}\right)$

Differentiating both sides with respect to $t$ , we get

  $\begin{array}{l}\frac{d^{2}P}{d{t}^2}=1.2\left(1-\frac{P}{4200}\right)+1.2P\left(-\frac{1}{4200}\right)\\ \frac{d^{2}P}{d{t}^2}=1.2-\frac{2.4P}{4200}\end{array}$

Now, for population increasing or decreasing

  $\frac{dP}{dt}=0\Rightarrow 1.2P\left(1-\frac{P}{4200}\right)=0$

i.e. either $P=0$ or, $P=4200$

Now, ${\frac{d^{2}P}{d{t}^2}|}_{\text{at P}=0}=1.2>0$

  $\begin{array}{c}{\frac{d^{2}P}{d{t}^2}|}_{\text{at P}=4200}=1.2-\frac{2.4\times \cancel{4200}}{\cancel{4200}}\\ =-1.2<0\end{array}$

At $P=4200$ , $\frac{d^{2}P}{d{t}^2}<0$

So, at $P=4200$ population is increasing.

b.

To determine

To find : The value of $P$ for which the population is decreasing.

Answer to Problem 9E

At $P=0$ population is decreasing.

Explanation of Solution

Given information :

  $\frac{dP}{dt}=1.2P\left(1-\frac{P}{4200}\right)$

Formula used :

For increasing or decreasing population $\frac{dP}{dt}=0$

If, ${\frac{d^{2}P}{d{t}^2}|}_{\text{at some P}}>0$ ,then the population is decreasing.

Calculation :

  $\frac{dP}{dt}=1.2P\left(1-\frac{P}{4200}\right)$

Differentiating both sides with respect to $t$ , we get

  $\begin{array}{l}\frac{d^{2}P}{d{t}^2}=1.2\left(1-\frac{P}{4200}\right)+1.2P\left(-\frac{1}{4200}\right)\\ \frac{d^{2}P}{d{t}^2}=1.2-\frac{2.4P}{4200}\end{array}$

Now, for population increasing or decreasing

  $\frac{dP}{dt}=0\Rightarrow 1.2P\left(1-\frac{P}{4200}\right)=0$

i.e. either $P=0$ or, $P=4200$

Now, ${\frac{d^{2}P}{d{t}^2}|}_{\text{at P}=0}=1.2>0$

  $\begin{array}{c}{\frac{d^{2}P}{d{t}^2}|}_{\text{at P}=4200}=1.2-\frac{2.4\times \cancel{4200}}{\cancel{4200}}\\ =-1.2<0\end{array}$

At $P=0$ , $\frac{d^{2}P}{d{t}^2}>0$

So, at $P=0$ population is decreasing.

c.

To determine

To find : The equilibrium solution.

Answer to Problem 9E

At $P=2100$ population is in equilibrium.

Explanation of Solution

Given information :

  $\frac{dP}{dt}=1.2P\left(1-\frac{P}{4200}\right)$

Formula used :

For increasing or decreasing population $\frac{dP}{dt}=0$

If, ${\frac{d^{2}P}{d{t}^2}|}_{\text{at some P}}=0$ ,then the population is in equilibrium.

Calculation :

  $\frac{dP}{dt}=1.2P\left(1-\frac{P}{4200}\right)$

Differentiating both sides with respect to $t$ , we get

  $\begin{array}{l}\frac{d^{2}P}{d{t}^2}=1.2\left(1-\frac{P}{4200}\right)+1.2P\left(-\frac{1}{4200}\right)\\ \frac{d^{2}P}{d{t}^2}=1.2-\frac{2.4P}{4200}\end{array}$

For equilibrium solution,

  $\begin{array}{l}\frac{d^{2}P}{d{t}^2}=0\\ \Rightarrow 1.2-\frac{2.4P}{4200}=0\\ \Rightarrow P=2100\end{array}$

So, at $P=2100$ population is in equilibrium.