Chapter 13.4, Problem 58PPSE

Solution

To find: The function among the given option whose period is $4\pi$ and amplitude of $8$ .

The period of the function $\left(G\right)y=-8\sin \left(\frac{1}{2}\theta \right)$ is $4\pi$ and its amplitude is $8$ . So, option $\left(G\right)$ is correct.

Given:

The given options of functions are written as:

  $\begin{array}{l}\left(F\right)y=-8\sin \left(8\theta \right)\left(G\right)y=-8\sin \left(\frac{1}{2}\theta \right)\\ \left(H\right)y=8\sin \left(2\theta \right)\left(I\right)y=4\sin \left(8\theta \right)\end{array}$

Concept used:

The amplitude $\left(a\right)$ of the sine curve of the form $y=a\sin \left(b\theta \right)$ where $a\ne 0\text{ and }b>0$ is $\left|a\right|$ and its period $\left(T\right)=\frac{2\pi }{b}$ .

Calculation:

With the help of concept stated above, the amplitude $\left(a\right)$ and period $\left(T\right)$ of the given sine curves whose period is $4\pi$ and amplitude of $8$ are evaluated as:

For the function $\left(F\right)y=-8\sin \left(8\theta \right)$ , the amplitude and period is evaluated as:

  $\begin{array}{c}\text{Amplitude}\left(a_F\right)=\left|-8\right|\mathrm{....}\left(\because a=-8\right)\\ =8\\ \text{Period}\left(T_F\right)=\frac{2\pi }{b}\\ =\frac{2\pi }{\left(8\right)}\mathrm{....}\left(\because b=8\right)\\ =\frac{\pi }{4}\mathrm{....}\left(\begin{array}{l}\text{Neglected because}\\ \text{its period isn't }4\pi \end{array}\right)\end{array}$

For the function $\left(G\right)y=-8\sin \left(\frac{1}{2}\theta \right)$ , the amplitude and period of sine curve is evaluated as:

  $\begin{array}{c}\text{Amplitude}\left(a_G\right)=\left|-8\right|\mathrm{....}\left(\because a=-8\right)\\ =8\\ \text{Period}\left(T_G\right)=\frac{2\pi }{b}\\ =\frac{2\pi }{\left(\frac{1}{2}\right)}\mathrm{....}\left(\because b=\frac{1}{2}\right)\\ =4\pi \mathrm{....}\left(\begin{array}{l}\text{True, because its period }\\ \text{is }4\pi \text{ and amplitude is }8\end{array}\right)\mathrm{....}\left(1\right)\end{array}$

For the function $\left(H\right)y=8\sin \left(2\theta \right)$ , the amplitude and period of sine curve is evaluated as:

  $\begin{array}{c}\text{Amplitude}\left(a_H\right)=\left|8\right|\mathrm{....}\left(\because a=-8\right)\\ =8\\ \text{Period}\left(T_H\right)=\frac{2\pi }{b}\\ =\frac{2\pi }{\left(2\right)}\mathrm{....}\left(\because b=2\right)\\ =\pi \mathrm{....}\left(\begin{array}{l}\text{Neglected, because }\\ \text{its period isn't }4\pi \end{array}\right)\end{array}$

For the function $\left(I\right)y=4\sin \left(8\theta \right)$ , the amplitude and period of sine curve is evaluated as:

  $\begin{array}{c}\text{Amplitude}\left(a_I\right)=\left|4\right|\mathrm{....}\left(\because a=4\right)\\ =4\\ \text{Period}\left(T_I\right)=\frac{2\pi }{b}\\ =\frac{2\pi }{\left(8\right)}\mathrm{....}\left(\because b=8\right)\\ =\frac{\pi }{4}\mathrm{....}\left(\begin{array}{l}\text{Neglected, because its period }\\ \text{isn't }4\pi \text{ and amplitude isn't }8\end{array}\right)\end{array}$

From $\left(1\right)$ the period of the function $\left(G\right)y=-8\sin \left(\frac{1}{2}\theta \right)$ is $4\pi$ and its amplitude is $8$ . So, option $\left(G\right)$ is correct.

Conclusion:

The period of the function $\left(G\right)y=-8\sin \left(\frac{1}{2}\theta \right)$ is $4\pi$ and its amplitude is $8$ . So, option $\left(G\right)$ is correct.