Chapter 6.4, Problem 63E

a.

To determine

To determine the function that represents the population y after t years.

Answer to Problem 63E

  $y=25000{\left(1+0.055\right)}^t$

Explanation of Solution

Given information :

Initial population is 25000.

Population increases by 5.5% annually for the next decade.

As population increase by 5.5% annually, we get,

  $y=25000{\left(1+0.055\right)}^t$

b.

To determine

To determine monthly percent increase in population.

Answer to Problem 63E

Monthly increase is around 0.45%.

Explanation of Solution

Given information :

Initial population is 25000.

Population increases by 5.5% annually for the next decade.

Let us take x be the monthly percent increase in population.

We get,

  $y=25000{\left(1+x\right)}^{12t}$

Now we shall compare the equation we got in part a.

  $\begin{array}{l}y=25000{\left(1+x\right)}^{12t}=25000{\left(1+0.055\right)}^t\\ {\left(1+x\right)}^{12t}={\left({\left(1+0.055\right)}^{\frac{1}{12}}\right)}^{12t}\\ {\left(1+x\right)}^{12t}={\left({\left(1+0.055\right)}^{\frac{1}{12}}\right)}^{12t}\\ {\left(1+x\right)}^{12t}={\left(1+0.00447\right)}^{12t}\\ x=0.0045\end{array}$

Hence the monthly increase is around 0.45%.

c.

To determine

To graph the function.

Answer to Problem 63E

Population after 4 years shall be around 31000.

Explanation of Solution

Given information :

Initial population is 25000.

Population increases by 5.5% annually for the next decade.

Graph of the function $y=25000{\left(1+0.055\right)}^t$ is,

Population after 4 years shall be around 31000.