Q. 27 - Precalculus Enhanced with Graphing Utilities

Question

The current I, in amperes, flowing through an ac (alternating current) circuit at time t, in seconds, is $I\left(t\right)=120\sin (30\mathrm{\pi t}-\frac{\mathrm{\pi }}{3}),t\ge 0$

What is the period? What is the amplitude? What is the phase shift? Graph this function over two periods.

Step-by-Step Solution

Answer

The amplitude of the function $I\left(t\right)=120\sin (30\mathrm{\pi t}-\frac{\mathrm{\pi }}{3}),t\ge 0$ is $120$ and the period is $\frac{1}{15}$ and the phase shift is $\frac{1}{90}$

The graph of the function $I\left(t\right)=120\sin (30\mathrm{\pi t}-\frac{\mathrm{\pi }}{3}),t\ge 0$ is:

Step 1. Given

The function $I\left(t\right)=120\sin (30\mathrm{\pi t}-\frac{\mathrm{\pi }}{3}),t\ge 0$

To find the amplitude, period and phase shift of the function.

To graph the function with at least two cycles.

Step 2. Find the amplitude, period and phase shift.

Compare the given function $y=120\sin \left(30\mathrm{\pi }\right(\mathrm{t}-\frac{1}{90}\left)\right)$ with  $y=A \sin (\omega x-\varphi )+B$

Amplitude, $\left|A\right|=120$

Period, $$\begin{gathered} \frac{2\mathrm{\pi }}{\omega }=\frac{2\mathrm{\pi }}{30\mathrm{\pi }} \\ =\frac{1}{15} \end{gathered}$$

Phase shift, $\frac{\varphi }{\omega }=\frac{1}{90}$

Step 3. Graph the function

Sketch the function using the amplitude and period.