Chapter 13.6, Problem 32PPSE

Solution

To find: The period of the function $y=\tan \left(-\frac{3}{2\pi }\theta \right)$ and sketch the graph of the function in the interval from $-2\pi \text{ to }2\pi$ .

The period of the function is $\frac{2{\pi }^2}{3}$ and its graph is sketched as shown below.

  

Given:

The function is $y=\tan \left(-\frac{3}{2\pi }\theta \right)$ which can be re-written as:

  $\begin{array}{c}y=\tan \left(-\frac{3}{2\pi }\theta \right)\\ y=-\tan \left(\frac{3}{2\pi }\theta \right)\mathrm{....}\left(\because \tan \left(-x\right)=-\tan \left(x\right)\right)\end{array}$

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.

Calculation:

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 $y=\tan \left(2.5\theta \right)$ as:

  $\begin{array}{c}\text{Period}\left(T\right)=\frac{\pi }{\left(\frac{3}{2\pi }\right)}\mathrm{....}\left(\because b=\frac{3}{2\pi }\right)\\ =\frac{2{\pi }^2}{3}\end{array}$

  $\begin{array}{c}\text{One cycle of the interval}=-\frac{\pi }{2b}\text{ to }\frac{\pi }{2b}\\ =-\frac{\pi }{2\left(\frac{3}{2\pi }\right)}\text{ to }\frac{\pi }{2\left(\frac{3}{2\pi }\right)}\mathrm{....}\left(\because b=\frac{3}{2\pi }\right)\\ =-\frac{{\pi }^2}{3}\text{ to }\frac{{\pi }^2}{3}\end{array}$

  $\begin{array}{c}\text{Two vertical asymptotes}\left(\theta \right)=-\frac{\pi }{2b},\frac{\pi }{2b}\\ =-\frac{\pi }{2\cdot \left(\frac{3}{2\pi }\right)},\frac{\pi }{2\cdot \left(\frac{3}{2\pi }\right)}\mathrm{....}\left(\because b=\frac{3}{2\pi }\right)\\ =-\frac{{\pi }^2}{3},\frac{{\pi }^2}{3}\end{array}$

So, with the help of period of the function, the consecutive vertical asymptotes occur at $-\frac{{\pi }^2}{3},\frac{{\pi }^2}{3}$ which lie in the interval $\left(-2\pi ,2\pi \right)$ as shown in graph.

Now divide the period into fourths and locate three points between the vertical asymptotes as shown below in graph and draw vertical asymptotes $-\frac{{\pi }^2}{3},\frac{{\pi }^2}{3}$ with dotted lines.

  

Conclusion:

The period of the function is $\frac{2\pi }{5}$ and its graph is sketched as shown above.