Chapter 13.6, Problem 31PPSE

Solution

To find: The period of the function $y=\tan \left(2.5\theta \right)$ and sketch the graph of the function in the interval from $-2\pi \text{ to }2\pi$ .

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

  

Given:

The function is $y=\tan \left(2.5\theta \right)$ .

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.

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:

  $\begin{array}{c}\text{Period}\left(T\right)=\frac{\pi }{\left(2.5\right)}\mathrm{....}\left(\because b=\frac{\pi }{6}\right)\\ =\frac{2\pi }{5}\end{array}$

  $\begin{array}{c}\text{One cycle of the interval}=-\frac{\pi }{2b}\text{ to }\frac{\pi }{2b}\\ =-\frac{\pi }{2\left(2.5\right)}\text{ to }\frac{\pi }{2\left(2.5\right)}\mathrm{....}\left(\because b=2.5\right)\\ =-\frac{\pi }{5}\text{ to }\frac{\pi }{5}\end{array}$

  $\begin{array}{c}\text{Two vertical asymptotes}\left(\theta \right)=-\frac{\pi }{2b},\frac{\pi }{2b}\\ =-\frac{\pi }{2\cdot \left(2.5\right)},\frac{\pi }{2\cdot \left(2.5\right)}\mathrm{....}\left(\because b=2.5\right)\\ =-\frac{\pi }{5},\frac{\pi }{5}\end{array}$

So, with the help of period of the function, the consecutive vertical asymptotes occur at $-\frac{9\pi }{5},-\frac{7\pi }{5},-\pi ,-\frac{3\pi }{5},-\frac{\pi }{5},\frac{\pi }{5},\frac{3\pi }{5},\pi ,\frac{7\pi }{5},\frac{9\pi }{5}$ 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 $-\frac{9\pi }{5},-\frac{7\pi }{5},-\pi ,-\frac{3\pi }{5},-\frac{\pi }{5},\frac{\pi }{5},\frac{3\pi }{5},\pi ,\frac{7\pi }{5},\frac{9\pi }{5}$ with dotted lines.

  

Conclusion:

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