Problem 62 For $$f(x)=A \sin \left(\frac{... [FREE SOLUTION]

Chapter 1: Problem 62

For $$f(x)=A \sin \left(\frac{2 \pi}{B}(x-C)\right)+D$$ identify $A, B, C,$ and $D$ for the sine functions in Exercises $61-64$ and sketch their graphs (see Figure $1.76 ) .$ $$ y=\frac{1}{2} \sin (\pi x-\pi)+\frac{1}{2} $$

Short Answer

Expert verified

A = 1/2, B = 2, C = 1, D = 1/2.

Step by step solution

Identify the amplitude (A)

The given function is in the form: \[ y = A \sin\left(\frac{2\pi}{B}(x - C)\right) + D \] The sine function we have is: \[ y = \frac{1}{2} \sin(\pi x - \pi) + \frac{1}{2} \] In this function, $A$ represents the amplitude of the sine wave, which is the coefficient of $\sin$. Here, $A = \frac{1}{2}$.

Determine the period factor (B)

The term $\frac{2\pi}{B}$ inside the sine function relates to the period of the sine wave. In the function $\sin(\pi x)$, $\pi$ can be equated to $\frac{2\pi}{B}$. Solving for $B$ gives: \[ \pi = \frac{2\pi}{B} \] This implies: \[ B = 2 \]

Identify the phase shift (C)

The phase shift $C$ is the horizontal shift from the origin and comes from the expression $(x - C)$. In our function, the expression inside the sine function is $\pi x - \pi$. By factoring out $\pi$, we have: \[ \pi(x - 1) \] Thus, $C = 1$.

Find the vertical shift (D)

The vertical shift $D$ is indicated by the constant addition outside the sine function. In $y = \frac{1}{2} \sin(\pi x - \pi) + \frac{1}{2}$, the vertical shift $D = \frac{1}{2}$.

Sketch the Graph

To sketch the graph:1. Start at the phase shift $x = 1$ and consider the vertical shift upwards by $\frac{1}{2}$.2. The amplitude is $\frac{1}{2}$ and the period is determined by $B = 2$ (one complete cycle occurs every 2 units).3. Plot key points by calculating sine values at critical points within the given period and shift.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude

When talking about sinusoidal functions, the amplitude is a measure of how high and low the wave goes. Imagine it as the wave's height.

In a sine function like $ y = A \sin(\text{something}) + D $, $ A $ represents the amplitude.
The amplitude is always a positive value.
In our example function, $ y = \frac{1}{2} \sin(\pi x - \pi) + \frac{1}{2} $, the amplitude $ A $ is $ \frac{1}{2} $.
This means the wave will reach a maximum height of $ \frac{1}{2} $ above and below the central axis of the graph, which in this case is raised by the vertical shift.

Amplitude provides insight on how energetic or oscillating the wave is. If $ A $ were larger, the wave would appear taller.

Period

The period of a sine function determines how quickly the wave completes one full cycle. It helps us understand the timing of the wave.

In the function $ y = A \sin\left(\frac{2\pi}{B}(x - C)\right) + D $, the period is related to the value $ B $.
Mathematically, the period $ T $ is calculated using $ T = \frac{2\pi}{B} $.
In our specific function, $ y = \frac{1}{2} \sin(\pi x - \pi) + \frac{1}{2} $, we found that $ B = 2 $.
Therefore, the period $ T $ is $ 2 $. This indicates the sine wave completes one full cycle every 2 units along the x-axis.

Understanding the period allows you to predict where the wave will return to its starting point, providing an overview of its frequency and repetition.

Phase Shift

The phase shift of a trigonometric function refers to the horizontal shifting along the x-axis. It's all about where the wave starts.

For a function $ y = A \sin\left(\frac{2\pi}{B}(x - C)\right) + D $, the term $ C $ determines this shift.
By rewriting the expression inside the sine in $ \sin(\pi x - \pi) $ as $ \pi(x - 1) $
We find that $ C = 1 $.
This tells us the graph of the sine wave starts at $ x = 1 $ instead of the usual origin $ x = 0 $.

Knowing the phase shift helps you pinpoint where the wave's key points will appear on the graph.

Vertical Shift

The vertical shift moves the entire sine wave up or down along the y-axis, altering its central alignment.

In the framework $ y = A \sin\left(...\right) + D $, $ D $ accounts for the vertical shift.
In the function$ y = \frac{1}{2} \sin(\pi x - \pi) + \frac{1}{2} $, $ D = \frac{1}{2} $.
This shift indicates the entire sine wave is lifted upward by $ \frac{1}{2} $ unit.
The midline, which the wave oscillates around, is now at $ y = \frac{1}{2} $ instead of the x-axis ($ y = 0 $).

Recognizing the vertical shift provides insights on how the wave fits relative to the baseline or other functions.

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91影视

For $$f(x)=A \sin \left(\frac{2 \pi}{B}(x-C)\right)+D$$ identify \(A, B, C,\) and \(D\) for the sine functions in Exercises \(61-64\) and sketch their graphs (see Figure \(1.76 ) .\) $$ y=\frac{1}{2} \sin (\pi x-\pi)+\frac{1}{2} $$

Short Answer

Step by step solution

Identify the amplitude (A)

Determine the period factor (B)

Identify the phase shift (C)

Find the vertical shift (D)

Sketch the Graph

Key Concepts

Amplitude

Period

Phase Shift

Vertical Shift

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