Chapter SH, Problem 13.21EP

a.

To determine

To find at what point the toy will be at $4$ inches above its resting position for the first time

Answer to Problem 13.21EP

  $\frac{6}{11}$ Second.

Explanation of Solution

Given:

Consider the following modeled equation of a toy’s height in inches as follows:

  $h=4\sin \frac{11\pi }{12}t$

Here, t represents time in seconds.

Calculation:

To be at $4$ inches above its resting position, the time is calculated as follows:

  $\begin{array}{l}4=4\sin \frac{11\pi }{12}t\\ 1=\sin \frac{11\pi }{12}t\end{array}$

From the angle identity:

  $\sin \frac{\pi }{2}=1$

Thus:

  $\begin{array}{l}\sin \frac{\pi }{2}=\sin \frac{11\pi }{12}t\\ \frac{\pi }{2}=\frac{11\pi }{12}t\\ t=\frac{6}{11}\end{array}$

Hence, at time of $\frac{6}{11}$ second, the toy will be at 4 inches above its resting position for the first time.

b.

To determine

To find at what point the toy will be at $2$ inches above its resting position for the first time

Answer to Problem 13.21EP

  $\frac{2}{11}$ Second.

Explanation of Solution

Given:

Consider the following modeled equation of a toy’s height in inches as follows:

  $h=4\sin \frac{11\pi }{12}t$

Here, t represents time in seconds.

Calculation:

To be at $2$ inches above its resting position, the time is calculated as follows:

  $\begin{array}{l}2=4\sin \frac{11\pi }{12}t\\ \frac{1}{2}=\sin \frac{11\pi }{12}t\end{array}$

From the angle identity:

  $\sin \frac{\pi }{6}=\frac{1}{2}$

Thus:

  $\begin{array}{l}\sin \frac{\pi }{6}=\sin \frac{11\pi }{12}t\\ \frac{\pi }{6}=\frac{11\pi }{12}t\\ t=\frac{2}{11}\end{array}$

Hence, at time of $\frac{2}{11}$ second, the toy will be at $2$ inches above its resting position for the first time.