Chapter 6.4, Problem 10CFU

To determine

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

Answer to Problem 10CFU

Amplitude = 0.5

Period = $12\pi$

Explanation of Solution

Given information:

  $y=0.5\sin \frac{\theta }{6}$

Calculation:

The equation of sine function is as follows:

  $y=A\sin k\theta$..........(1)

Here $A$ is amplitude which is an absolute value of the coefficient and $k$ is the absolute value of coefficient $\theta$ .

  $k=\frac{2\pi }{\text{Period}}$ .

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

  $\begin{array}{c}\text{Amplitude}=\left|A\right|\\ =\left|0.5\right|\\ =0.5\end{array}$

Thus the amplitude of the function is 0.5.

Since $k=\frac{1}{6}$ , so the period will be evaluated as follows:

  $\begin{array}{c}\text{Period}=\frac{2\pi }{k}\\ =\frac{2\pi }{\left(\frac{1}{6}\right)}\left[\text{Put the value of }k\right]\\ =2\pi \left(6\right)\\ =12\pi \end{array}$

Thus the period is $12\pi$ .

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