Chapter 13.6, Problem 22PPSE

Solution

To sketch: The graph of the function $y=\tan \left(\theta \right)$ in the interval from $0\text{ to }2\pi$ .

The period of the function is $\frac{3{\pi }^2}{2}$ and two vertical asymptotes of the function are $-\frac{3{\pi }^2}{4},\frac{3{\pi }^2}{4}$ .

Given:

The tangent function is $y=\tan \left(\theta \right)$ .

Concept used:

The tangent curve of the form $y=a\tan \left(b\theta \right)$ where $a\ne 0$ and $b>0$ has period $\frac{\pi }{b}$ , the number of cycles in the interval from $-\frac{\pi }{2b}\text{ to }\frac{\pi }{2b}$ is $1$ and two vertical asymptotes $-\frac{\pi }{2b},\frac{\pi }{2b}$ at each end of the cycle.

Graph:

With the help of concept stated above, evaluate the period $\left(T\right)$ , one cycle in the interval from $-\frac{\pi }{2b}\text{ to }\frac{\pi }{2b}$ and two vertical asymptotes of the function $y=\tan \left(\theta \right)$ as:

  $\begin{array}{c}\text{Period}\left(T\right)=\frac{\pi }{b}\\ =\frac{\pi }{\left(1\right)}\mathrm{....}\left(\because b=1\right)\\ =\pi \end{array}$

  $\begin{array}{c}\text{One cycle of the interval}=-\frac{\pi }{2b}\text{ to }\frac{\pi }{2b}\\ =-\frac{\pi }{2\left(1\right)}\text{ to }\frac{\pi }{2\left(1\right)}\mathrm{....}\left(\because b=1\right)\\ =-\frac{\pi }{2}\text{ to }\frac{\pi }{2}\end{array}$

  $\begin{array}{c}\text{Two vertical asymptotes}\left(\theta \right)=-\frac{\pi }{2b},\frac{\pi }{2b}\\ =-\frac{\pi }{2\cdot \left(1\right)},\frac{\pi }{2\cdot \left(1\right)}\mathrm{....}\left(\because b=1\right)\\ =-\frac{\pi }{2},\frac{\pi }{2}\end{array}$

So, three consecutive vertical asymptotes occur at $-\frac{\pi }{2},\frac{\pi }{2},\frac{3\pi }{2}$ among which $\frac{\pi }{2},\frac{3\pi }{2}$ lie in the interval as shown in graph.

Now divide the period into fourths and locate three points between the period as shown below in graph and draw vertical asymptotes with dotted lines.