Chapter 13.4, Problem 6WE

(a).

To determine

The amplitude of given description.

Answer to Problem 6WE

The amplitude is 1/2.

Explanation of Solution

Given:

  $y=\frac{1}{2}\sin 2x$

Calculations:

Suppose a function.

  $y=a\sin bx$ ,

Here, a is the amplitude of the function.

From the given descriptions:

  $y=\frac{1}{2}\sin 2x$

The amplitude is 1/2.

(b).

To determine

The maximum and minimum value of given description.

Answer to Problem 6WE

The value of maximum is 1/2 and minimum is -1/2.

Explanation of Solution

Given:

  $y=\frac{1}{2}\sin 2x$

Calculations:

Suppose a function.

  $y=c+a\sin bx$ ,

Here, a is the amplitude of the function.

Maximum is given by;

  $M=c+a$

And,

Minimum is given by;

  $m=c-a$

From the given descriptions:

  $y=\frac{1}{2}\sin 2x$

Here,

  $\begin{array}{l}c=0\\ a=1/2\end{array}$

Therefore,

Maximum:

  $\begin{array}{l}M=0+\frac{1}{2}\\ M=\frac{1}{2}\end{array}$

Minimum:

  $\begin{array}{l}m=0-\frac{1}{2}\\ m=-\frac{1}{2}\end{array}$

The value of maximum is 1/2 and minimum is -1/2.

(c).

To determine

The period of given description.

Answer to Problem 6WE

The period is $\pi$ .

Explanation of Solution

Given:

  $y=\frac{1}{2}\sin 2x$

Calculations:

Suppose a function.

  $y=a\sin bx$ ,

Here, a is the amplitude of the function.

And period is $\frac{2\pi }{b}$ .

From the given descriptions:

  $y=\frac{1}{2}\sin 2x$

Here,

  $b=2$

Therefore,

The period is $\frac{2\pi }{2}=\pi$ .