Chapter 4.5, Problem 72E

To determine

To graph: the given function.

Explanation of Solution

Given information:

Given function is

  $y=\frac{1}{100}\sin 120\pi t$

Concept used:

For the standard equation $y=a\sin \left(bt+c\right)$ .

Amplitude is $\left|a\right|$ and period is $\frac{2\pi }{b}$ .

Horizontal shift is $\frac{c}{b}$

Calculation:

The functionis $y=\frac{1}{100}\sin 120\pi t$ .

Comparing $y=\frac{1}{100}\cos 120\pi t$ with the equation $y=a\sin \left(bt+c\right)$ .

  $b=120\pi$

So,

  $\begin{array}{c}\text{period=}\frac{\text{2π}}{\text{b}}\\ \text{=}\frac{\text{2π}}{120\pi }\\ \text{=}\frac{1}{60}\end{array}$

Then two full period

  $\begin{array}{c}\text{=}\frac{2}{60}\\ =\frac{1}{30}\end{array}$

Press the button $Y=$

Type $Y_1=\frac{1}{100}\sin 120\pi t$

Press the button TRACE or GRAPH to obtain the graph:

It will show following graph: