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Chapter 6: Problem 51

Graph each function over a two-period interval. $$y=-2+\frac{1}{2} \sin 3 x$$

Short Answer

Expert verified
The interval is from \(0\) to \( \frac{4 \pi}{3} \). Key points repeat every \( \frac{2 \pi}{3} \) due to periodicity of sine function.

Step by step solution

01

Understand the function components

The given function is written as \[ y = -2 + \frac{1}{2} \sin (3x) \] Identify the components: - Amplitude: \( \frac{1}{2} \) - Angular frequency: \(3\) - Vertical shift: \(-2\) These components will help in sketching the graph.
02

Determine the period

The period of a sine function \( \sin (bx) \) is given by \[ \text{Period} = \frac{2 \pi}{b} \] Here, \(b = 3\), so the period is \[ \text{Period} = \frac{2 \pi}{3} \] One period of the function is \( \frac{2 \pi}{3} \). To graph over two periods, the interval will be from \(0\) to \(\frac{4 \pi}{3} \).
03

Define key points within one period

Identify key points within one period (from \(0\) to \( \frac{2 \pi}{3} \)): - \( x = 0 \): \( y = -2 + \frac{1}{2} \sin (0) = -2 \) - \( x = \frac{\pi}{6} \): \( y = -2 + \frac{1}{2} \sin \left( 3 \cdot \frac{\pi}{6} \right) = -2 + \frac{1}{2} \sin \left( \frac{\pi}{2} \right) = \frac{1}{2} - 2 = -1.5 \) - \( x = \frac{\pi}{3} \): \( y = -2 + \frac{1}{2} \sin (\pi) = -2 + 0 = -2 \) - \( x = \frac{\pi}{2} \): \( y = -2 + \frac{1}{2} \sin \left( \frac{3 \pi}{2} \right) = -2 - \frac{1}{2} = -2.5 \) - \( x = \frac{2 \pi}{3} \): \( y = -2 + \frac{1}{2} \sin (2 \pi) = -2 \)
04

Extend to two periods

For the second period, extend the points from \( \frac{2 \pi}{3} \) to \( \frac{4 \pi}{3} \): - Calculate points similarly as those for the first period. - These points repeat because sine function is periodic.
05

Draw the graph

Plot the points on graph paper or a graphing tool. Draw smooth sinusoidal curves that pass through these points for the interval \( 0 \) to \( \frac{4 \pi}{3} \), recognizing the pattern repeats every period.

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Key Concepts

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

sine function
The sine function, denoted as \(\text{sin}(x)\), is one of the fundamental trigonometric functions. It describes a smooth, periodic oscillation. The standard sine function has some characteristic properties:
  • It oscillates between -1 and 1.
  • It is periodic with a period of \[2\pi\].
  • It starts at 0 when \(x = 0\).
In the given problem, the sine function has been modified to \(\frac{1}{2} \text{sin}(3x)\). Despite these modifications, it still possesses the fundamental properties of the sine function but with adjusted frequency and amplitude.
amplitude
Amplitude refers to the maximum value that a periodic function reaches from its central or equilibrium value. For the standard sine function \(\text{sin}(x)\), the amplitude is 1. However, when a function takes the form \(A \text{sin}(Bx)\), the amplitude becomes the absolute value of \(A\). In our problem:
  • The function is \(\frac{1}{2} \text{sin}(3x)\).
  • Therefore, the amplitude is \(\frac{1}{2}\).
This means the graph of the function will oscillate between \(-\frac{1}{2}\) and \(\frac{1}{2}\) from its midline. The given function has a vertical shift of -2, which means its midpoint is lowered by 2 units from the x-axis.
periodic function
A periodic function is one that repeats its values in regular intervals or periods. The most common example is the sine function, which repeats every \[2\pi\] radians.
For a sine function of the form \( \text{sin}(Bx)\), the period is altered by the value of \(B\). It is calculated using the formula:
\[\text{Period} = \frac{2\fft \pi}{B}\]. In this problem, B is 3, making the period:
\[\text{Period} = \frac{2 \pi}{3}.\] Hence, the function \( \frac{1}{2} \text{sin}(3x)\) will repeat every \(\frac{2\pi}{3} \) units. This modified period affects how the graph stretches or compresses along the x-axis.
graphing techniques
Graphing a trigonometric function involves a few basic steps:
  • Identify key components: amplitude, period, phase shift, and vertical shift.
  • Determine key points: start, peak, midline, trough, and end of one period.
  • Extend these points to the required number of periods.
For the function \(-2 + \frac{1}{2} \text{sin}(3x)\), we:

  • Identify amplitude (\(\frac{1}{2}\)), period (\(\frac{2\pi}{3}\)), vertical shift (-2).
  • Calculate key points within one period.
  • Repeat the process for a two-period interval (0 to \(\frac{4\pi}{3}\)).
  • Plot these points and draw smooth, sinusoidal curves through them.
Remember: plotting trigonometric functions accurately ensures better understanding of their behavior and properties.

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Most popular questions from this chapter

A weight attached to a spring is pulled down 3 in. below the equilibrium position. (a) Assuming that the frequency is \(\frac{6}{\pi}\) cycles per sec, determine a model that gives the position of the weight at time \(t\) seconds. (b) What is the period?

Over the interval \([0,2 \pi],\) compare the graphs of $$ y=\cos 3 x \quad \text { and } \quad y=3 \cos x $$ Can we say that, in general, \(\cos b x=b \cos x ?\) Explain.

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The formula \(\omega=\frac{\theta}{t}\) can be rewritten as \(\theta=\omega t\). Substituting \(\omega t\) for \(\theta\) converts \(s=r \theta\) to \(s=r \omega t .\) Use the formula \(s=r \omega t\) to find the value of the missing variable. $$s=\frac{12 \pi}{5} \mathrm{m}, r=\frac{3}{2} \mathrm{m}, \omega=\frac{2 \pi}{5} \text { radians per sec }$$
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