Chapter 7.5, Problem 34E

a.

To determine

To find a formula for the gorilla population in terms of t.

Answer to Problem 34E

The required formula is ,

  $P=\frac{250}{1+\frac{11}{14}{e}^{-250\left(0.0004\right)t}}\approx \frac{250}{1+7.929e^{-0.1t}}$ .

Explanation of Solution

Given information: A certain wild animals preserve can support no more than 250 lowland gorillas. Twenty eight gorillas were known to be in the preserve in 1970. Assume that the rate of growth of the population is.

  $\frac{dp}{dt}=0.0004P\left(250-P\right)$

Calculation:

As per the given problem,

Compare the given differential equation to $\frac{dp}{dt}=kP\left(M-P\right)$ .

  $\frac{dp}{dt}=0.0004P\left(250-P\right)=kP\left(M-P\right)$

Substitute in M = 250 and k = 0.0004.into the logistic function $p=\frac{M}{1+A^{-Mkt}}$ .

  $P=\frac{250}{1+A{e}^{-250\left(0.0004\right)t}}=\frac{250}{1+A{e}^{-0.1t}}$

It is given that there were 28 gorillas in 1970 so let t = 0 represent 1970 and P = 28 when t = 0 to find A.

  $\begin{array}{l}28=\frac{250}{1+A{e}^{\left(0\right)}}\\ 28=\frac{250}{1+A}\\ 28\left(1+A\right)=250\\ 1+A=\frac{125}{14}\\ A=\frac{11}{14}\\ A\approx 7.929\end{array}$

Substitute in the value for A into the equation $P=\frac{250}{1+A{e}^{-0.1t}}$ .

  $P=\frac{250}{1+\frac{11}{14}{e}^{-250\left(0.0004\right)t}}\approx \frac{250}{1+7.929e^{-0.1t}}$

Hence, the required formula is , $P=\frac{250}{1+\frac{11}{14}{e}^{-250\left(0.0004\right)t}}\approx \frac{250}{1+7.929e^{-0.1t}}$ .

b.

To determine

To find the number of years it take for the gorilla population to reach the carrying capacity of the preserve.

Answer to Problem 34E

It take 82.8 years for the gorilla population to reach the carrying capacity of the preserve.

Explanation of Solution

Given information: A certain wild animals preserve can support no more than 250 lowland gorillas. Twenty eight gorillas were known to be in the preserve in 1970. Assume that the rate of growth of the population is.

  $\frac{dp}{dt}=0.0004P\left(250-P\right)$

Calculation:

As per the given problem,

The equation found in part (a),

  $P=\frac{250}{1+7.929e^{-0.1t}}$

249.5 rounds up to 250, which is the carrying capacity.

  $\begin{array}{l}P=\frac{250}{1+0.12e^{-0.1t}}=249.5\\ \ln \left(\frac{\frac{250}{249.5}-1}{7.93}\right)=\ln \left(e^{-0.1t}\right)\\ -8.28=-0.1t\\ t=82.8\end{array}$

Hence, it take 82.8 years for the gorilla population to reach the carrying capacity of the preserve.