Chapter 3.5, Problem 43E

a.

To determine

Find the population of the conservation organization after given months.

Answer to Problem 43E

The population of the conservation organization after $5$ months are $203$ .

Explanation of Solution

Given:

The growth of the pack is modeled by the logistic curve $p\left(t\right)=\frac{1000}{1+9e^{-0.1656t}}$ . the given number of months are $5$ .

Calculation:

  $p\left(t\right)=\frac{1000}{1+9e^{-0.1656t}}$

Substitute the value of $t=5$ in the given function.

  $\begin{array}{l}\Rightarrow p\left(t\right)=\frac{1000}{1+9e^{-0.1656t}}\\ \Rightarrow p\left(5\right)=\frac{1000}{1+9e^{-0.1656\cdot 5}}\\ =\frac{1000}{1+3.9323}\\ =\frac{1000}{4.9323}\\ =202.745\\ \approx 203\end{array}$

Hence the population of the conservation organization after $5$ months are $203$ .

b.

To determine

Find the year in which the populations of the conservation organizationwill reached at the given number.

Answer to Problem 43E

The populations of conservation organization will reached $500$ after $13$ months.

Explanation of Solution

Given:

The growth of the pack is modeled by the logistic curve $p\left(t\right)=\frac{1000}{1+9e^{-0.1656t}}$ . The number of populations are $500$ .

.Calculation:

The given function is $p\left(t\right)=\frac{1000}{1+9e^{-0.1656t}}$ .

Substitute the value of $p=500$ in the function.

  $500=\frac{1000}{1+9e^{-0.1656t}}$

Multiply both sides by $1+9e^{-0.1656t}$ .

  $\begin{array}{l}\Rightarrow 500\left(1+9e^{-0.1656t}\right)=\frac{1000}{\left(1+9e^{-0.1656t}\right)}\cdot \left(1+9e^{-0.1656t}\right)\\ \Rightarrow 500\left(1+9e^{-0.1656t}\right)=1000\end{array}$

Divide both sides by $500$ .

  $\begin{array}{l}\Rightarrow \frac{500}{500}\left(1+9e^{-0.1656t}\right)=\frac{1000}{500}\\ \Rightarrow 1+9e^{-0.1656t}=2\end{array}$

Subtract $1$ from both sides.

  $\begin{array}{l}\Rightarrow 1+9e^{-0.1656t}-1=2-1\\ \Rightarrow 9e^{-0.1656t}=1\end{array}$

Divide both sides by $9$ .

  $\begin{array}{l}\Rightarrow \frac{9}{9}{e}^{-0.1656t}=\frac{1}{9}\\ \Rightarrow e^{-0.1656t}=\frac{1}{9}\end{array}$

Taking natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln e^{-0.1656t}=\ln \frac{1}{9}\\ \Rightarrow -0.1656t=-2.197224577\\ \Rightarrow 0.1656t=2.197224577\end{array}$

Divided both sides by $0.1656$ .

  $\begin{array}{l}\Rightarrow \frac{0.1656}{0.1656}t=\frac{2.197224577}{0.1656}\\ \Rightarrow t=\frac{2.197224577}{0.1656}\\ \Rightarrow t=13.268\\ \Rightarrow t\approx 13\end{array}$

Hence the populations of conservation organizationwill reached $500$ after $13$ months.

c.

To determine

Draw a graph of the given function using graphing utility.

Answer to Problem 43E

The horizontal asymptotes are $p=0$ and $p=1000$ .

Explanation of Solution

Given:

The growth of the pack is modeled by the logistic curve $p\left(t\right)=\frac{1000}{1+9e^{-0.1656t}}$ .

Calculation:

The given function is $p\left(t\right)=\frac{1000}{1+9e^{-0.1656t}}$ .

By the use of graphing utility, the graph of the function is given below.

  

From the graph, the horizontal asymptotes are

  $\begin{array}{l}p=0\\ p=1000\end{array}$

When the time increases, the population approaches $1000$ in the graph