Chapter 13.4, Problem 1LC

Solution

i.

To find: The number of cycles in the given graph between $0$ to $2\pi$ .

  $2$

Given:

A graph as follows:

  

Concept used:

The number of times the curve repeats its structure is known as the cycle.

Calculation:

In the given graph, the curve starts from zero and goes to the positive peak, and from there it falls and goes to the negative peak and further it returns to its actual point which is zero. Here the curve completes one cycle. Again, the curves repeat the same process and completes the second cycle that can be seen in the graph.

Conclusion:

Hence, the number of cycles is $2$ .

ii.

To find: The amplitude and the period.

Amplitude= $3$

Period= $\pi$

Given:

A graph as follows:

  

Concept used:

1. The length of each cycle is known as the period.
2. The amplitude can be computed by using the formula $A=\frac{\left(\max -\min \right)}{2}$ .

Calculation:

The amplitude of the graph is,

  $\begin{array}{l}A=\frac{\left(3-\left(-3\right)\right)}{2}\\ A=3\end{array}$

Each cycle is $\pi$ unit long so the period the graph is $\pi$ .

Conclusion:

The amplitude and period are $3$ and $\pi$ respectively.

iii.

To find: The equation for the function.

  $$

Given:

A graph as follows:

  

Concept used:

Substitute the known values in $y=A\sin \left(B\left(x+C\right)\right)+D$ to get the equation where $A=$ amplitude, $\frac{2\pi }{B}=$ period, $C=$ phase shift and $D=$ vertical shift.

Calculation:

It is understood that amplitude $\left(A\right)=3,C=0$ and $D=0$ . And,

  $\begin{array}{l}B=\frac{2\pi }{\text{period}}\\ B=\frac{2\pi }{\pi }\\ B=2\end{array}$

So, the equation is,

  

Conclusion:

The equation for the function is $$ .