Chapter 12, Problem 57SGR

To determine

To find: Amplitude, Vertical shift, Phase shift and period of each function.

Answer to Problem 57SGR

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

Period $=6\pi$

Phase shift is ${90}^0$ units Right.

Vertical shift is $D=2$ Two units Up

Explanation of Solution

Given information:

  $y=\frac{1}{3}\sin \left[\frac{1}{3}\left(\theta -{90}^0\right)\right]+2$

Concept:

If equation in the form

  $y=A\sin \left(B\left(x-C\right)\right)+D$

Then

Amplitude $=A$

Period $=\frac{2\pi }{B}$

Phase shift $=C$ .

Vertical shift $=D$

Given

  $y=\frac{1}{3}\sin \left[\frac{1}{3}\left(\theta -{90}^0\right)\right]+2$

On comparing with general equation

We get

  $\begin{array}{l}A=\frac{1}{3}\\ B=\frac{1}{3}\\ C=-{90}^0\\ D=2\end{array}$

Graph of given equation is

  

X-axis is represented by $\theta$

Therefore,

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

Period $=\frac{2\pi }{B}$

  $\Rightarrow$ Period $=\frac{2\pi }{\frac{1}{3}}$

  $\Rightarrow$ Period $=2\pi \times 3$

  $\Rightarrow$ Period $=6\pi$

Phase shift

Since $C={90}^0$

Phase shift is ${90}^0$ units Right.

Vertical shift is $D=2$ Two units Up

Therefore,

We got

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

Period $=6\pi$

Phase shift is ${90}^0$ units Right.

Vertical shift is $D=2$ Two units Up