Chapter 12, Problem 55SGR

To determine

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

Answer to Problem 55SGR

Amplitude $=undefined$

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

Phase shift is $\frac{\pi }{2}$ units right.

Vertical shift is $D=2$ two units Up

Explanation of Solution

Given information:

  $y=2\sec \left[2\left(\theta -\frac{\pi }{2}\right)\right]+2$

Concept:

If equation in the form

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

Then

Amplitude $=undefined$

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

Phase shift $=C$ .

Vertical shift $=D$

Given

  $y=2\sec \left[2\left(\theta -\frac{\pi }{2}\right)\right]+2$

On comparing with general equation

We get

  $\begin{array}{l}A=2\\ B=3\\ C=\frac{\pi }{2}\\ D=2\end{array}$

Graph of given equation is

  

X-axis is represented by $\theta$

Therefore,

Amplitude $=A=undefined$

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

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

Phase shift

Since $C=\frac{\pi }{2}$

Phase shift is $\frac{\pi }{2}$ units right.

Vertical shift is $D=2$ two units Up

Therefore,

We got

Amplitude $=undefined$

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

Phase shift is $\frac{\pi }{2}$ units right.

Vertical shift is $D=2$ two units Up