Chapter 4.6, Problem 12E

Population Growth These exercises use the population growth model.

12. Frog Population Some bullfrogs were introduced into a small pond. The graph shows the bullfrog population for the next few years. Assume that the population grows exponentially.

1. (a) What was the initial bullfrog population?
2. (b) Find a function that models the bullfrog population t years since the bullfrogs were put into the pond.
3. (c) What is the projected bullfrog population after 15 years?
4. (d) Estimate how long it takes the population to reach 75,000.

To determine

To find: The initial bullfrog population.

Answer to Problem 12E

The initial bullfrog population is $100$.

Explanation of Solution

The graph shows the bullfrog population for few years.

Figure (1)

From Figure (1), it noticed that the $x\text{-axis}$ denotes the years after introducing the bullfrogs into the pond and $y\text{-axis}$ denotes the bullfrog population.

At $x$ is $0$ the value of $y$ is $100$ and $0$ denotes the initial year.

Thus, the initial bullfrog population is $100$.

To determine

To find: The function for bullfrog population $t$ years s after the bullfrogs are put into the pond.

Answer to Problem 12E

The function for bullfrog population $t$ years after the bullfrogs are put into the pond is $n\left(t\right)=100\left(e^{0.405t}\right)$.

Explanation of Solution

Given:

The initial population is $100$ from part (a).

The graph shows the bullfrog population for few years.

Figure (1)

Formula used: Exponential growth model (relative growth rate)

$n\left(t\right)=n_0{e}^{rt}$ (I)

Where,

• $n_0$ is the initial size of the population.
• $r$ is relative growth rate.
• $n\left(t\right)$ is size of the population at time $t$.

Calculation:

From figure (1), it is noticed that the deer population after $2\text{years}$ is $225$.

Substitute $100$ for $n_0$, $225$ for $n\left(t\right)$ and $2$ for $t$,

$\begin{array}{c}225=100\left(e^{2r}\right)\\ \frac{225}{100}=e^{2r}\\ \ln \left(2.25\right)=\ln \left(e^{2r}\right)\left(\because \text{Taking}\ln \text{to}\text{each}\text{side}\right)\\ \ln \left(2.25\right)=2r\left(\because \ln \left(e^a\right)=a\right)\end{array}$

Simplify the above expression,

$\begin{array}{c}r=\frac{1}{2}\ln \left(2.25\right)\\ =\frac{0.81093}{2}\\ =0.4054651\\ \approx 0.405\end{array}$

The relative growth rate is $0.405$.

Substitute $100$ for $n_0$ and $0.405$ for $r$ to find the bullfrog population function,

$n\left(t\right)=100\left(e^{0.405t}\right)$.

Thus, the function for bullfrog population $t$ years since the bullfrogs put into the pond is $n\left(t\right)=100\left(e^{0.405t}\right)$.

To determine

To find: The projected population of bullfrogs after $15\text{years}$.

Answer to Problem 12E

The projected population of bullfrogs after $15\text{years}$ is $43,789$.

Explanation of Solution

Given:

The bullfrog population function from part (b) is given below,

$n\left(t\right)=100\left(e^{0.405t}\right)$ (II)

Calculation:

Substitute $15$ for $t$ in equation (II) to find the bullfrog population after $15$ years,(Approximating the value of $r$ will affect the number of bullfrogs, so approx it up to $6$ or more than $6$ decimal places)

$\begin{array}{c}n\left(15\right)=100\left(e^{0.405465\left(15\right)}\right)\left(\because r=0.405465\right)\\ =100\left(e^{6.081975}\right)\\ \approx 100\left(437.89318\right)\left(\because e^{6.081975}\approx 437.89318\right)\\ =43789.318\end{array}$

Thus, the projected population of bullfrogs after $15\text{years}$ is $43,789$.

To determine

To find: The time in which bullfrog population reach to $75000$.

Answer to Problem 12E

The bullfrog population reach to $75000$ in $16.35\text{years}$.

Explanation of Solution

Given:

The bullfrog population function from part (b) is given below,

$n\left(t\right)=100\left(e^{0.405t}\right)$.

Calculation:

Substitute $75000$ for $n\left(t\right)$ in equation (II) to find the time in which bullfrog population reach to $75000$,

$\begin{array}{c}75000=100\left(e^{0.405t}\right)\\ 750=e^{0.405t}\\ \ln \left(750\right)=\ln \left(e^{0.405t}\right)\left(\because \text{Take}\ln \text{to}\text{each}\text{side}\right)\\ \ln \left(750\right)=0.405t\left(\because \ln e^a=a\right)\end{array}$

Simplify the above expression,

$\begin{array}{c}t=\frac{6.6200732}{0.405}\\ =16.34586\\ \approx 16.35\text{years}\end{array}$

Thus, the bullfrog population reach to $75000$ in $16.35\text{years}$.