Chapter 13.4, Problem 15PPSE

Solution

To sketch: One cycle of the sine function whose amplitude is $2$ and period $\frac{2\pi }{3}$ assuming $a>0$ .

Given:

The amplitude $\left(a\right)$ of sine curve is $2$ and period $\frac{2\pi }{3}$ .

Concept used:

The sine curve is of the form $y=a\sin \left(b\theta \right)$ where $a\ne 0$ and $b>0$ has an amplitude of $\left(a\right)=\left|a\right|$ , period $\left(T\right)=\frac{2\pi }{b}$ and number of cycles $\left(b\right)$ in the interval from $0\text{ to }2\pi$ . And if $a>0$ , use the five-point pattern of $\left(\theta \right)$ values which follows a pattern for $\left(y\right)$ -values as $0,\max ,0,\min \text{ and }0$ .

Graph:

In order to sketch the graph of sine function, evaluate the missing parameters using the concept stated above.

As $\left|a\right|=2$ and period $\left(T\right)=\frac{2\pi }{3}$ . Therefore,

  $\begin{array}{c}\left(T\right)=\frac{2\pi }{3}\text{ gives,}\\ \frac{2\pi }{b}=\frac{2\pi }{3}\\ b=3\end{array}$

So, the equation of sine curve $y=a\sin \left(b\theta \right)$ is written as $y=2\sin \left(3\theta \right)$ .

Since the number of cycles are $\left(b\right)=3$ , therefore the number of cycles from $0\text{ to }2\pi$ is $3$ . And, we have to draw the sketch for the period $\left(T\right)=\frac{2\pi }{3}$ . Therefore, the number of cycles for the period $\left(T\right)=\frac{2\pi }{3}$ is $\frac{3}{3}=1$ .

Now, in order to sketch the graph of one cycle of the sine curve divide the period into fourths using the five-point pattern of $\left(\theta \right)$ values which follows a pattern for $\left(y\right)$ -values as $0,\max ,0,\min \text{ and }0$ .

Hence, the five-point pattern of $\left(\theta \right)$ values is written as $0,\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{2}\text{ and }\frac{2\pi }{3}$ .

The sketch of the curve $y=2\sin \left(3\theta \right)$ from $0\text{ to }\frac{2\pi }{3}$ is drawn as:

  

Interpretation:

From the graph of sine curve $y=2\sin \left(3\theta \right)$ , it can be interpreted that the five-point pattern of $\left(\theta \right)$ values follows a pattern for $\left(y\right)$ -values as $0,\max ,0,\min \text{ and }0$ where the curve cuts the $\left(\theta \right)$ -values.