Chapter 6.4, Problem 32E

To determine

To state: the amplitude and period of the function and then graph that function.

Answer to Problem 32E

Amplitude = $\frac{1}{3}$

Period = $6\pi$

Explanation of Solution

Given information:

  $y=\frac{1}{3}\sin \frac{\theta }{3}$

Calculation:

The equation of sine function is as follows:

  $y=A\sin k\theta$..............(1)

Here $A$ is an amplitude which is an absolute value of the coefficient and $k=\frac{2\pi }{\text{Period}}$ .

Compare the given equation with equation (1) to get,

  $\begin{array}{c}\text{Amplitude}=\left|A\right|\\ =\left|\frac{1}{3}\right|\\ =\frac{1}{3}\end{array}$

Thus the amplitude of the function is $\frac{1}{3}$ .

Since $k=\frac{1}{3}$ , so the period will be evaluated as follows:

  $\begin{array}{c}\text{Period}=\frac{2\pi }{k}\\ =\frac{2\pi }{\left(\frac{1}{3}\right)}\left[\text{Put the value of }k\right]\\ =2\pi \left(3\right)\\ =6\pi \end{array}$

Thus the period is $6\pi$ .

The graph of the function for the given amplitude and period is as follows: