Problem 18 - Precalculus with Calculus Previews

Question

In a certain kind of electrical circuit, the current \(I(t)\) measured in amperes at time \(t\) seconds is given by $$ I(t)=10 \cos \left(120 \pi t+\frac{\pi}{3}\right) $$ (a) Change the form of the function \(I(t)\) to the sine function form given in (7). (b) Give the period of the sine function in part (a) and the phase shift. Use the sine function to sketch two cycles of the graph of \(I(t)\).

Step-by-Step Solution

Answer

The sine form of the function is \( I(t) = 10 \sin(120 \pi t + \frac{5\pi}{6}) \), with a period of \(\frac{1}{60}\) and a phase shift of \(-\frac{1}{144}\).

Step 1: Identify the Given Function

We are given the function in the cosine form:\[ I(t) = 10 \cos\left(120 \pi t + \frac{\pi}{3}\right) \] We need to convert this into the sine form for part (a).

Step 2: Convert Cosine to Sine Function

To convert \(\cos\) to \(\sin\), we use the identity: \[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]Therefore, \[ I(t) = 10 \sin\left(120 \pi t + \frac{\pi}{3} + \frac{\pi}{2}\right) \]Next, simplify the angle:\[ 120 \pi t + \frac{\pi}{3} + \frac{\pi}{2} = 120 \pi t + \frac{\pi}{3} + \frac{\pi}{2} = 120 \pi t + \frac{\pi}{3} + \frac{\pi}{2} = 120 \pi t + \frac{5\pi}{6} \]Thus, the function in sine form is:\[ I(t) = 10 \sin\left(120 \pi t + \frac{5\pi}{6}\right) \]

Step 3: Determine the Period of the Sine Function

The general form of a sine function is \(A \sin(Bt + C)\). The period is given by \[ \text{Period} = \frac{2\pi}{B} \]In our example, \(B = 120\pi\). Therefore, the period is:\[ \frac{2\pi}{120\pi} = \frac{1}{60} \]

Step 4: Calculate the Phase Shift

The phase shift is determined by \(-\frac{C}{B}\), where \(C = \frac{5\pi}{6}\) and \(B = 120\pi\).\[ \text{Phase shift} = -\frac{\frac{5\pi}{6}}{120\pi} = -\frac{1}{144} \]

Step 5: Sketch Two Cycles of the Graph

To sketch the graph of the function, note the period is \(\frac{1}{60}\) and the phase shift is \(-\frac{1}{144}\). This means the graph will start slightly before \(t = 0\) due to the shift. Plot the sine curve using these parameters, ensuring each cycle spans \(\frac{1}{60}\) seconds. Two complete cycles will cover \(\frac{1}{30}\) seconds on the \(t\)-axis.

Key Concepts

Sine and Cosine Conversion

Understanding how to convert between sine and cosine functions is a pivotal aspect of trigonometry. Essentially, sine and cosine are phase-shifted versions of each other. The key identity to remember is:

• \( \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \)

This identity means that the cosine function can be expressed as a sine function with a phase shift of \(\frac{\pi}{2}\). This conversion is useful in various applications, such as electromagnetism and signal processing. By recognizing this relationship, we can easily interchange between these trigonometric forms.
For example, if you have a function like \( I(t) = 10 \cos(120\pi t + \frac{\pi}{3}) \), it can be rewritten as \( I(t) = 10 \sin(120\pi t + \frac{5\pi}{6}) \) after applying the identity. Simplifying and understanding these transformations allows for a deeper grasp of how these functions behave in different scenarios.

Period of Trigonometric Functions

The period of a trigonometric function is the length of one complete cycle before the function starts repeating itself. For both sine and cosine, the period is determined by the coefficient of \(t\) in the function's argument. In general, for a function of the form \( \sin(Bt + C) \) or \( \cos(Bt + C) \), the period \(T\) is calculated as follows:

• \( T = \frac{2\pi}{B} \)

This means that a larger value of \(B\) results in a shorter period, causing the graph to complete more cycles within a given interval.
In the given function, \( I(t) = 10 \sin(120\pi t + \frac{5\pi}{6}) \), \(B\) equals \(120\pi\). By substituting into the period formula, we find:

• \( T = \frac{2\pi}{120\pi} = \frac{1}{60} \)

Thus, the function repeats itself every \(\frac{1}{60}\) seconds. Grasping this concept helps in predicting the behavior of curves modeled by trigonometric functions.

Phase Shift of Trigonometric Functions

Phase shift is another important feature of trigonometric functions, which involves translating the graph horizontally. The phase shift can be thought of as the horizontal displacement of the function from its usual starting point.
For a function of the form \( \sin(Bt + C) \) or \( \cos(Bt + C) \), the phase shift is determined by the expression:

• \( \text{Phase Shift} = -\frac{C}{B} \)

In simple terms, a positive phase shift moves the graph to the right, while a negative phase shift moves it to the left.
For the function \( I(t) = 10 \sin(120\pi t + \frac{5\pi}{6}) \):

• Here, \(C = \frac{5\pi}{6}\) and \(B = 120\pi\).
• The phase shift is thus computed as \( -\frac{\frac{5\pi}{6}}{120\pi} = -\frac{1}{144} \).

This negative value indicates that the function starts slightly to the left of the usual starting point. By understanding phase shifts, you can accurately sketch and interpret the positions of trigonometric function graphs.