Chapter 6.4, Problem 27E

To determine

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

Answer to Problem 27E

Amplitude = 3

Period = $4$

Explanation of Solution

Given information:

  $y=-3\sin \frac{\pi }{2}\theta$

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,

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

Thus the amplitude of the function is 3.

Since $k=\frac{\pi }{2}$ , so the period will be evaluated as follows:

  $\begin{array}{c}\text{Period}=\frac{2\pi }{k}\\ =\frac{2\pi }{\left(\frac{\pi }{2}\right)}\left[\text{Put the value of }k\right]\\ =\frac{2\pi \left(2\right)}{\pi }\\ =4\end{array}$

Thus the period is 4.

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