Chapter 7.5, Problem 1E

a.

To determine

To define : the carrying capacity and also find the value of $k$

Answer to Problem 1E

The carrying capacity is $10$ and the value of $k=0.005$

Explanation of Solution

Given information : The question is

A population develops according to the logistic equation $\frac{dP}{dt}=0.05P-0.0005P^2$

Calculation :

The logistic equation is $\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)$

Where $P$ is the population

  $k$ Is the growth constant

  $M$ is carrying capacity

  $\begin{array}{l}\frac{dP}{dt}=0.05P-0.0005P^2\\ \frac{dP}{dt}=0.005P\left(1-0.1P\right)\\ \frac{dP}{dt}=0.005P\left(1-\frac{P}{10}\right)\\ \therefore M=10\\ k=0.005\end{array}$

b.

To determine

that where is the slope is $0$ ,largest, increasing or decreasing

Explanation of Solution

Given information :

A direction field of graph is shown below

  

The direction field to see over which range the slopes increase or decrease. The more flat the lines are the closer they are to $0$

The graph is shown below

  

c.

To determine

that which solutions have inflection point and at what population level they do occur

Explanation of Solution

Given information : The initial populations are $20,40,60,80,120\&140$

Calculation :

When the initial population is $20$

  

When the initial population is $40$

  

When the initial population is $60$

  

When the initial population is $80$

  

When the initial population is $120$

  

When the initial population is $140$

  

d.

To determine

To define : the equilibrium solutions and how they are different from other

Explanation of Solution

Given information : The question is

A population develops according to the logistic equation $\frac{dP}{dt}=0.05P-0.0005P^2$

The equilibrium solutions are $P=0\&P=100$

At these population levels, there is no change with time

The population levels are constant. These are the asymptotic points for other solutions. When enough time passes, the other solution will converge to one