Chapter 13, Problem 12CT

Solution

To find: The number of cycles, amplitude, periodand midline of the sine curve shown below in the interval from $0\text{ to }2\pi$ .

  

The number of cycles of the sine curve in the interval from $0\text{ to }2\pi$ is $1$ , amplitude is $2$ , period is $2\pi$ and the equation of midline is $y=0$ .

Given:

The sine curve is shown below.

  

Concept used:

The amplitude $\left(a\right)$ of the sine curve is $\left(a\right)=\frac{1}{2}\left(y_{\max }-y_{\min }\right)$ , period $\left(T\right)=\frac{2\pi }{b}$ , where $\left(b\right)$ is the number of cycles in the interval from $0\text{ to }2\pi$ and $y_{\max }\text{ and }{y}_{\min }$ is the maximum and minimum value of the function. And the equation of midline is written as $y=\frac{1}{2}\left(y_{\max }+y_{\min }\right)$ .

Calculation:

From the graph of sine curve, it can be interpreted that the number of cycles $\left(b\right)$ of the sine curve in the interval from $0\text{ to }2\pi$ is $1$ .

And the maximum and minimum value of the function is $2\text{ and }-2$ respectively. Therefore, period $\left(T\right)$ of the sine curve using the concept stated above is evaluated as:

  $\begin{array}{c}\text{Period}\left(T\right)=\frac{2\pi }{b}\\ =\frac{2\pi }{1}\mathrm{....}\left(\because b=1\right)\\ =2\pi \end{array}$

Now, the amplitude $\left(a\right)$ of the curve is evaluated as:

  $\begin{array}{c}a=\frac{1}{2}\left(y_{\max }-y_{\min }\right)\\ =\frac{1}{2}\left(2-\left(-2\right)\right)\\ =2\end{array}$

The equation of midline $\left(y\right)=\frac{1}{2}\left(y_{\max }+y_{\min }\right)$ is evaluated as:

  $\begin{array}{c}y=\frac{1}{2}\left(y_{\max }+y_{\min }\right)\\ =\frac{1}{2}\left(2+\left(-2\right)\right)\\ =0\end{array}$

Conclusion:

The number of cycles of the sine curve in the interval from $0\text{ to }2\pi$ is $1$ , amplitude is $2$ , period is $2\pi$ and the equation of midline is $y=0$ .