Chapter 12.7, Problem 35PPS

To determine

To find: Amplitude , period and graph of the function.

Answer to Problem 35PPS

Amplitude $=undefined$

Period $={450}^0$

Explanation of Solution

Given information:

  $y=2\sec \frac{4}{5}\theta$

Concept:

If function is in the form

  $y=a\sec b\theta$

Then

Amplitude $=undefined$

Period $=\frac{{360}^0}{\left|b\right|}$

Given

  $y=2\sec \frac{4}{5}\theta$

On comparing with $y=a\sec b\theta$

We get

  $\begin{array}{l}a=2\\ b=\frac{4}{5}\end{array}$

Therefore,

Amplitude $=undefined$

Period $=\frac{{360}^0}{\left|b\right|}$

  $\Rightarrow$ Period $=\frac{{360}^0}{\left|\frac{4}{5}\right|}$

  $\Rightarrow$ Period $={360}^0\times \frac{5}{4}$

  $\Rightarrow$ Period $={450}^0$

Therefore,

Amplitude $=undefined$

Period $={450}^0$

Graph of given function is

  

X-axis represents $\theta$ .