Chapter 7.5, Problem 26E

(a)

To determine

To find the carrying capacity of the population.

Answer to Problem 26E

The carrying capacity is $5000$ bears.

Explanation of Solution

Given:

The logistic differential equation describes the growth of a population $P$ ,where $t$ is measured in years

  $\frac{dP}{dt}={10}^{-5}P\left(5000-P\right)$ .

Calculation:

The logistic differential equation is given by

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

Therefore, on comparing it from given logistic differential equation it can be observed that the carrying capacity is $5000$ bears.

(b)

To determine

To find the size of the population when it is growing fastest.

Answer to Problem 26E

The size of population is $2500\text{ bears}$ when it is growing fastest.

Explanation of Solution

Given:

The logistic differential equation describes the growth of a population $P$ ,where $t$ is measured in years

  $\frac{dP}{dt}={10}^{-5}P\left(5000-P\right)$ .

Calculation:

The logistic differential equation is given by

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

Since, the growth rate is maximum when the population reaches half the carrying capacity.

Therefore , $P=\frac{M}{2}=\frac{5000}{2}=2500\text{ bears}$

Hence, the size of population is $2500\text{ bears}$ when it is growing fastest.

(c)

To determine

To find the rate at which the population is growing fastest.

Answer to Problem 26E

The rate at which the population is growing when it is growing fastest is $\text{62}\text{.5 bears/years}$ .

Explanation of Solution

Given:

The logistic differential equation describes the growth of a population $P$ ,where $t$ is measured in years

  $\frac{dP}{dt}={10}^{-5}P\left(5000-P\right)$ .

Calculation:

The logistic differential equation is given by

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

Since, for fastest growing rate $P=2500\text{ bears}$ .

So, $\frac{dP}{dt}={10}^{-5}P\left(5000-P\right)={10}^{-5}\times 2500\left(5000-2500\right)=\text{ 62}\text{.5 bears/years}$

Therefore, the rate at which the population is growing when it is growing fastest is $\text{62}\text{.5 bears/years}$ .