Chapter 4, Problem 95RE

(a)

To determine

To Find: The function that models the stray cat population $n(t)$ after $t$ years.

Answer to Problem 95RE

  $n(t)=30e^{0.15t}$

Explanation of Solution

Given: The stray cat population in a small town grows exponentially. In $1999$ the town had $30$ stray cats and the relative growth rate was $15\%$ per year.

FORMULA USED:

  $\begin{array}{l}n(t)=n_0{e}^{rt}\\ where\\ n(t)=\text{population at time }t\\ n_0=\text{intial size of population}\\ r=\text{relative rate of growth }\\ t=\text{time}\end{array}$

Given: The population of stray cat is $n_0=30$

The relative growth rate $r=15\%=0.15$

Then

Function is

  $n(t)=30e^{0.15t}$

(b)

To determine

To Find: The projected population after $4$ years.

Answer to Problem 95RE

  $55$ cats

Explanation of Solution

Given: The stray cat population in a small town grows exponentially. In $1999$ the town had $30$ stray cats and the relative growth rate was $15\%$ per year.

FORMULA USED:

  $\begin{array}{l}n(t)=n_0{e}^{rt}\\ where\\ n(t)=\text{population at time }t\\ n_0=\text{intial size of population}\\ r=\text{relative rate of growth }\\ t=\text{time}\end{array}$

Given: The population of stray cat is $n_0=30$

The relative growth rate $r=15\%=0.15$

Then

Function is

  $n(t)=30e^{0.15t}$

Now, number of cats after $4$ years is

Put $t=4$

We get

  $\begin{array}{l}n(4)=30e^{0.15(4)}\\ n(4)=30e^{0.60}\\ n(4)=54.66\approx 55\end{array}$

Hence, population after $4$ years is $55$ cats

(c)

To determine

To Find: The number of years required for the stray cat population to reach $500$.

Answer to Problem 95RE

  $19$ Years

Explanation of Solution

Given: The stray cat population in a small town grows exponentially. In $1999$ the town had $30$ stray cats and the relative growth rate was $15\%$ per year.

FORMULA USED:

  $\begin{array}{l}n(t)=n_0{e}^{rt}\\ where\\ n(t)=\text{population at time }t\\ n_0=\text{intial size of population}\\ r=\text{relative rate of growth }\\ t=\text{time}\end{array}$

Given: The population of stray cat is $n_0=30$

The relative growth rate $r=15\%=0.15$

Then

Function is

  $n(t)=30e^{0.15t}$

Then

  $\begin{array}{l}n(t)=500\\ 30e^{0.15t}=500\\ e^{0.15t}=\frac{500}{30}\\ \ln e^{0.15t}=\ln \frac{500}{30}\\ 0.15t\ln e=\ln 50-\ln 3\\ t=\frac{\ln 50-\ln 3}{0.15}\\ t=18.7560\approx 19\end{array}$

Hence, $19$ years