Q. 53 - Precalculus Enhanced with Graphing Utilities

Question

The current I, in amperes, flowing through an ac (alternating current) circuit at time t is

$I\left(t\right)=220\sin (30\mathrm{\pi t}+\frac{\mathrm{\pi }}{6}), \mathrm{t}\ge 0$

(a) What is the period?

(b) What is the amplitude?

(c) What is the phase shift?

(d) Graph this function over two periods.

Step-by-Step Solution

Answer

Part (a) The period is $\frac{1}{15}$.

Part (b) The amplitude is $4$.

Part (c) The phase shift is $-\frac{1}{180}$.

Part (d) The graph of the given function is

Step 1. Given information

Let I be the current flowing in a circuit at time t. 

The function is: $I\left(t\right)=220\sin \left(30\pi t+\frac{\pi }{6}\right) , t\ge 0$

We know that the general form of the sine function is,

$y=A\sin (\omega x-\varphi )+B$

Where, A is amplitude, $T=\frac{2\pi }{\omega }$ is period and $\frac{\varphi }{\omega }$ is phase shift.

Part (a) Step 1. Finding the period of given function.

We know that $\omega =30\pi$.

So the graph is sine curve with period :

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

Part (b) Step 1. Finding the amplitude of the given function.

We get $A=4$.

So the graph is sine curve with amplitude,

$$\begin{gathered} \left|A\right|=\left|4\right| \\ =4 \end{gathered}$$

Part (c) Step 1. Finding the phase shift.

We have $\omega =30\pi$ and $\varphi =-\frac{\mathrm{\pi }}{6}$.

So the graph is sine curve with phase shift :

$$\begin{gathered} \frac{\varphi }{\omega }=\frac{-{\displaystyle \frac{\mathrm{\pi }}{6}}}{30\mathrm{\pi }} \\ =-\frac{1}{180} \end{gathered}$$

Part (d) Step 1. The graph of I ( t ) = 220 sin ( 30 πt + π 6 ) is as follows.