Problem 16 - Algebra 2

Question

Find the amplitude, if it exists, and period of each function. Then graph each function. \(y=\frac{1}{5} \sin \theta\)

Step-by-Step Solution

Answer

Amplitude: \(\frac{1}{5}\); Period: \(2\pi\).

Step 1: Identify the Amplitude

The amplitude for a sinusoidal function of the form \[y = a \sin b\theta\] or \[y = a \cos b\theta\] is \(|a|\). In our function, \(a = \frac{1}{5}\), so the amplitude is:\[ \text{Amplitude} = \left| \frac{1}{5} \right| = \frac{1}{5}. \]

Step 2: Determine the Period

The period of a sine or cosine function is determined by the coefficient \(b\) in front of \(\theta\) using the formula:\[ \text{Period} = \frac{2\pi}{|b|}. \] For \(y = \frac{1}{5} \sin \theta\), \(b = 1\). Thus, the period is calculated as:\[ \text{Period} = \frac{2\pi}{1} = 2\pi. \]

Step 3: Graph the Function

For the graph of \(y = \frac{1}{5} \sin \theta\), start by plotting the key points of the sine function:1. At \(\theta = 0\), \( y = 0 \)2. At \(\theta = \frac{\pi}{2}\), \( y = \frac{1}{5} \)3. At \(\theta = \pi\), \( y = 0 \)4. At \(\theta = \frac{3\pi}{2}\), \( y = -\frac{1}{5} \)5. At \(\theta = 2\pi\), \( y = 0 \)This will cover one full period from \(0\) to \(2\pi\). The graph will oscillate between \(-\frac{1}{5}\) and \(\frac{1}{5}\) and repeat every \(2\pi\).

Key Concepts

Amplitude

The amplitude of a trigonometric function, specifically a sine or cosine function, reflects how "tall" or "short" the waves appear when graphed. In simpler terms, it indicates the maximum height the function reaches from its central value, either in the positive or the negative direction on the vertical axis. To find the amplitude, identify the coefficient \(a\) in front of the sine or cosine term in the equation. Here's how you do it:

• Check the function form: \(y = a \sin b\theta\) or \(y = a \cos b\theta\).
• The amplitude is the absolute value of \(a\), represented by \(|a|\).

For the function \(y = \frac{1}{5} \sin \theta\), the coefficient \(a\) is \(\frac{1}{5}\). Therefore, the amplitude is \(|a| = \frac{1}{5}\). This tells us that the sine wave will oscillate between \(\frac{1}{5}\) and \(-\frac{1}{5}\) from the horizontal axis over each cycle.

Period

The period of a sine or cosine function represents the horizontal length of one full cycle on its graph. This period is crucial because it tells us how frequently the function repeats its shape over time. Calculating the period is straightforward when you know the coefficient \(b\) in the function. Here’s the step-by-step method to find it:

• Identify the function form: \(y = a \sin b\theta\) or \(y = a \cos b\theta\).
• The period is calculated using the formula \(\frac{2\pi}{|b|}\).

For the function \(y = \frac{1}{5} \sin \theta\), note that \(b = 1\). Applying the formula gives us a period of \(\frac{2\pi}{1} = 2\pi\). This means that every \(2\pi\) units along the horizontal axis, the sine curve will begin repeating itself. Such cyclical behavior is characteristic of sine functions and is critical for tasks involving periodic phenomena.

Sine Function

A sine function is one of the most fundamental trigonometric functions, capturing the essence of oscillatory and wave-like movements. When you graph a sine function, it looks like smooth, continuous waves. Let’s break down a simple sine function and its graph characteristics:

• The basic form is \(y = a \sin b\theta\), where \(a\) affects amplitude and \(b\) affects period.
• The curve starts from the point \( (0,0) \) and follows a predictable pattern.
• The sine wave's peak (maximum value) is reached at \( \frac{\pi}{2} \), and its lowest point (minimum value) at \( \frac{3\pi}{2} \).

For the specific function \(y = \frac{1}{5} \sin \theta\), we observe several key points throughout a single period from \(0\) to \(2\pi\):1. At \( \theta = 0 \), \( y = 0 \)2. At \( \theta = \frac{\pi}{2} \), \( y = \frac{1}{5} \)3. At \( \theta = \pi \), \( y = 0 \)4. At \( \theta = \frac{3\pi}{2} \), \( y = -\frac{1}{5} \)5. At \( \theta = 2\pi \), \( y = 0 \)These points sketch out the sine wave, which symmetrically oscillates around the horizontal axis and repeats this pattern every \(2\pi\). Such understanding of the sine function is pivotal in fields such as physics, engineering, and signal processing.