Problem 58 - Precalculus

Question

In Exercises \(49-60\), state the amplitude, period, and phase shift (including direction) of the given function. $$y=\frac{1}{2} \sin \left(\frac{1}{3} x+\pi\right)$$

Step-by-Step Solution

Answer

Amplitude: \(\frac{1}{2}\), Period: \(6\pi\), Phase Shift: \(3\pi\) to the left.

Step 1: Identify the standard sine function form

The standard form of a sine function is given by \( y = a \sin(bx + c) \) or \( y = a \sin(b(x + \frac{c}{b})) \). In this problem, we have \( y = \frac{1}{2} \sin\left(\frac{1}{3} x + \pi\right) \), so we can identify \( a = \frac{1}{2} \), \( b = \frac{1}{3} \), and \( c = \pi \).

Step 2: Determine the amplitude

The amplitude of a sine function is the absolute value of \( a \). So, for \( y = \frac{1}{2} \sin\left(\frac{1}{3} x + \pi\right) \), the amplitude is \( |\frac{1}{2}| = \frac{1}{2} \).

Step 3: Calculate the period

The period of a sine function is given by \( \frac{2\pi}{|b|} \). Plugging in \( b = \frac{1}{3} \), we get the period as \( \frac{2\pi}{\frac{1}{3}} = 6\pi \).

Step 4: Find the phase shift and its direction

The phase shift can be found by solving \( bx + c = 0 \) for \( x \), giving \( x = -\frac{c}{b} \). Substituting \( c = \pi \) and \( b = \frac{1}{3} \), the phase shift is \( x = -\frac{\pi}{\frac{1}{3}} = -3\pi \). The negative sign indicates a shift to the left.

Key Concepts

Amplitude

The amplitude of a trigonometric function is reminiscent of how tall a wave appears. It describes the maximum distance a wave can reach from its central or equilibrium position.

In the case of a sine function, it dictates the height of the peaks and the depth of the troughs from the horizontal axis. For the function \( y = \frac{1}{2} \sin\left(\frac{1}{3} x + \pi\right) \), the amplitude is determined by the coefficient \( a \), which is \( \frac{1}{2} \) in this instance.

To calculate the amplitude, we simply take the absolute value of \( a \), which represents how far above or below the center line the graph of the function will go. Therefore, the amplitude here is:

• \( |\frac{1}{2}| = \frac{1}{2} \)

Thus, every peak of the sine wave will rise to 0.5 units above the center line, and every trough will dip to 0.5 units below.

Period

The period of a trigonometric function reveals how often the function's cycle repeats itself. Understanding the period lets us predict where the wave's peaks and troughs will appear along the x-axis.

For sine functions in the form \( y = a \sin(bx + c) \), calculating the period involves using the formula:

\[ \text{Period} = \frac{2\pi}{|b|} \]

In this problem, the coefficient \( b \) is \( \frac{1}{3} \). By substituting it into the formula, we determine that the period is:

• \( \frac{2\pi}{\frac{1}{3}} = 6\pi \)

This result indicates that the complete cycle of the sine wave, from start to finish, takes \( 6\pi \) units to complete. The wave's pattern resets itself every \( 6\pi \) units on the x-axis.

Phase Shift

Phase shift in trigonometry denotes the horizontal translation of the graph. It defines how far and in which direction the entire function moves along the x-axis.

To discover the phase shift of a given sine function, especially one structured as \( y = a \sin(b(x + \frac{c}{b})) \), solve the equation:

\( bx + c = 0 \)

This will determine how far the function has been shifted. In this circumstance, employ \( b = \frac{1}{3} \) and \( c = \pi \). Solving for \( x \), we have:

\[ x = -\frac{c}{b} = -\frac{\pi}{\frac{1}{3}} = -3\pi \]

The negative result indicates a shift to the left. Consequently, the function shifts \( 3\pi \) units towards the left on the x-axis. Phase shifts play an essential role in determining the starting point of the wave's cycle on a graph.