Chapter MPS, Problem 13.9MPS

a.

To determine

Determine the period of the function.

Answer to Problem 13.9MPS

  $12\sec onds$

Explanation of Solution

Given information:

In a certain wildlife refuge, the population of field mice can be modelled by $y=3000+1250\sin \frac{\pi }{6}t,$ where $y$ represents the number of mice and $t$ represents the number of months past March $1$ of a given year.

Calculation:

The equation for the population is $y=3000+1250\sin \frac{\pi }{6}t.$

The general equation is $y=k+a\sin b\theta .$

The period is $p=\frac{2\pi }{\left|b\right|}$

The value of $b=\frac{\pi }{6}$ .

Hence, $p=\frac{2\pi \times 6}{\pi }$ .

Hence, the period of the function is $12\sec onds$ . It means that the populations increases in every $12\sec onds$ .

b.

To determine

Find the maximum number of mice and the time.

Answer to Problem 13.9MPS

  $y=4250$ & $t=3$

Explanation of Solution

Given information:

In a certain wildlife refuge, the population of field mice can be modelled by $y=3000+1250\sin \frac{\pi }{6}t,$ where $y$ represents the number of mice and $t$ represents the number of months past March $1$ of a given year.

Calculation:

Compare the equation $y=3000+1250\sin \frac{\pi }{6}t$ with the equation $y=k+a\sin b\theta .$

Then, the value of $k=3000,b=\frac{\pi }{6},anda=1250$ .

The equation of midline is $y=k$ and the maximum value of the function is $y=k+a.$

The midline for the function $y=3000+1250\sin \frac{\pi }{6}t$ is,

  $y=3000$

The maximum value is,

  $\begin{array}{l}y=3000+1250\\ y=4250\end{array}$

Hence, the maximum number of mice are $4250$ .

Substitute $y=4250$ in $y=3000+1250\sin \frac{\pi }{6}t$

Solve as follows:

  $\begin{array}{l}y=3000+1250\sin \frac{\pi }{6}t\\ 4250=3000+1250\sin \frac{\pi }{6}t\\ 1250=1250\sin \frac{\pi }{6}t\end{array}$

  $\sin \frac{\pi }{6}t=1$

Rewrite $1=\sin \frac{\left(2n+1\right)\pi }{2}$ in above equation.

  $\begin{array}{l}\sin \frac{\pi }{6}t=\sin \frac{\left(2n+1\right)\pi }{2}\\ \frac{\pi }{6}t=\frac{\left(2n+1\right)\pi }{2}\\ t=3\left(2n+1\right)\end{array}$

Take $n=0$ . Then, $t=3$

Hence, after $3$ months from March $1$ the population of mice will be maximum $y=4250$ .