Chapter 6.4, Problem 8CFU

To determine

To state: the amplitude and period of the function and then graph that function.

Answer to Problem 8CFU

Amplitude = 10

Period = $\pi$

Explanation of Solution

Given information:

  $y=10\sin 2\theta$

Calculation:

The equation of sine function is as follows:

  $y=A\sin k\theta$...........(1)

Here $A$ is an amplitude which is an absolute value of the coefficient and $k=\frac{2\pi }{\text{Period}}$ .

Compare the given equation with equation (1) to get,

  $\begin{array}{c}\text{Amplitude}=\left|A\right|\\ =\left|10\right|\\ =10\end{array}$

Thus the amplitude of the function is 10.

Since $k=2$ , so the period will be evaluated as follows:

  $\begin{array}{c}\text{Period}=\frac{2\pi }{k}\\ =\frac{2\pi }{2}\left[\text{Put the value of }k\right]\\ =\pi \end{array}$

Thus the period is $\pi$ .

The graph of the function for the given amplitude and period is as follows: