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Chapter 5: Problem 6

Write the equation of the sine function in the form \(y=a \sin b(x-c)+d\) given its characteristics. a) amplitude \(4,\) period \(\pi,\) phase shift \(\frac{\pi}{2}\) to the right, vertical displacement 6 units down b) amplitude 0.5, period \(4 \pi,\) phase shift \(\frac{\pi}{6}\) to the left, vertical displacement 1 unit up c) amplitude \(\frac{3}{4},\) period \(720^{\circ},\) no phase shift, vertical displacement 5 units down

Short Answer

Expert verified
Part (a): y = 4 \sin 2(x - \frac{\pi}{2}) - 6; Part (b): y = 0.5 \sin \frac{1}{2}(x + \frac{\pi}{6}) + 1; Part (c): y = \frac{3}{4} \sin \frac{1}{2}x - 5.

Step by step solution

01

Understand the General Form of the Sine Function

The general form of a sine function is given by \[ y = a \ \sin b(x-c) + d \] where:- \(a\) is the amplitude- \(\frac{2\pi}{b}\) is the period- \(c\) is the phase shift- \(d\) is the vertical displacement.
02

Plug in the Values for Part (a)

For part (a):- Amplitude \(a = 4\)- Period \(\pi\) implies \( b = \frac{2\pi}{\pi} = 2\)- Phase shift of \(\frac{\pi}{2}\) to the right implies \(c = \frac{\pi}{2}\)- Vertical displacement 6 units down implies \(d = -6\)Therefore, the equation is \[ y = 4 \ \sin 2(x - \frac{\pi}{2}) - 6 \]
03

Plug in the Values for Part (b)

For part (b):- Amplitude \(a = 0.5\)- Period \(4\pi\) implies \( b = \frac{2\pi}{4\pi} = \frac{1}{2}\)- Phase shift of \(\frac{\pi}{6}\) to the left implies \(c = -\frac{\pi}{6}\)- Vertical displacement 1 unit up implies \(d = 1\)Therefore, the equation is \[ y = 0.5 \ \sin \frac{1}{2}(x + \frac{\pi}{6}) + 1 \]
04

Plug in the Values for Part (c)

For part (c):- Amplitude \(a = \frac{3}{4}\)- Period \(720^{\circ}\) implies \( b = \frac{360^{\circ}}{720^{\circ}} = \frac{1}{2}\) (converted to degrees)- No phase shift implies \(c = 0\)- Vertical displacement 5 units down implies \(d = -5\)Therefore, the equation is \[ y = \frac{3}{4} \ \sin \frac{1}{2}x - 5 \]

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Key Concepts

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

Amplitude and Its Role in Sine Functions
The amplitude of a sine function determines the height of its peaks and the depth of its troughs. It is represented by the variable \(a\) in the function \(y = a \sin b(x-c) + d\).
Amplitude is a measure of how much the function's value deviates from the centerline. For example:
  • If the amplitude is 4, the sine wave will reach values of 4 and -4 at its highest and lowest points, respectively.
  • If the amplitude is 0.5, the sine wave will only reach values of 0.5 and -0.5 at its extremes.
This means that larger amplitudes result in taller waves, while smaller amplitudes result in more compressed waves.
Amplitudes are never negative; however, a negative amplitude would reflect the sine wave across the horizontal axis. Always remember:
  • A larger amplitude means a more pronounced wave.
  • A smaller amplitude means a less pronounced wave.
Understanding the Period of Sine Functions
The period of a sine function defines how long it takes for the sine wave to complete one full cycle. In the general sine equation \(y = a \sin b(x-c) + d\), the period is calculated as \(\frac{2\pi}{b}\).
The period influences the width of the cycles. For instance:
  • A period of \(\pi\) implies that the sine wave completes a full cycle within a horizontal interval of \(\pi\).
  • A period of \(4\pi\) means the wave stretches out to complete one cycle over \(4\pi\). This results in a more spread-out wave.
This can be visualized as:
  • A smaller period results in a more frequent or tightly packed wave.
  • A larger period results in a less frequent or more stretched-out wave.
The conversion between angles and radians often comes into play, as seen in part (c) where the period in degrees is converted for consistency.
Never forget:
  • The 'b' value in the sine equation stretches or compresses the wave by adjusting the period.
Phase Shift of Sine Functions
The phase shift of a sine function determines where the sine wave begins along the x-axis. This is given by the variable \(c\) in the general form \(y = a \sin b(x-c) + d\).
Phase shift essentially 'moves' the wave left or right:
  • A positive phase shift (\(c\) positive) moves the wave to the right.
  • A negative phase shift (\(c\) negative) moves the wave to the left.
For example:
  • In part (a), a phase shift of \(\frac{\pi}{2}\) to the right means the wave starts at \(x = \frac{\pi}{2}\).
  • In part (b), a phase shift of \(\frac{\pi}{6}\) to the left means the wave starts at \(x = -\frac{\pi}{6}\).
Understanding phase shifts is crucial for graphing sine functions accurately and ensuring the wave starts from the intended point along the x-axis.
Key points to remember:
  • Phase shift moves the wave along the x-axis without changing its amplitude or period.
Vertical Displacement in Sine Functions
The vertical displacement of a sine function is represented by the variable \(d\) in the equation \(y = a \sin b(x-c) + d\). This shifts the entire sine wave up or down along the y-axis.
Vertical displacement is responsible for where the midline of the wave is positioned:
  • A negative vertical displacement moves the wave downward.
  • A positive vertical displacement moves the wave upward.
In practice:
  • A vertical displacement of -6, as in part (a), shifts the wave 6 units downward.
  • A vertical displacement of 1, as in part (b), lifts the wave 1 unit upward.
This affects the overall positioning of all the peaks and troughs, but not their relative distances.
Always bear in mind:
  • Vertical displacement adjusts the baseline or midline of the sinusoidal wave, not the pattern of the wave itself.

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Most popular questions from this chapter

Noise-cancelling headphones are designed to give you maximum listening pleasure by cancelling ambient noise and actively creating their own sound waves. These waves mimic the incoming noise in every way, except that they are out of sync with the intruding noise by \(180^{\circ}\). Suppose that the amplitude and period for the sine waves created by the outside noise are 4 and \(\frac{\pi}{2},\) respectively. Determine the equation of the sound waves the headphones produce to effectively cancel the ambient noise.

A security camera scans a long straight fence that encloses a section of a military base. The camera is mounted on a post that is located \(5 \mathrm{m}\) from the midpoint of the fence. The camera makes one complete rotation in 60 s. a) Determine the tangent function that represents the distance, \(d\), in metres, along the fence from its midpoint as a function of time, \(t,\) in seconds, if the camera is aimed at the midpoint of the fence at \(t=0\) b) Graph the function in the interval \(-15 \leq t \leq 15\) c) What is the distance from the midpoint of the fence at \(t=10 \mathrm{s},\) to the nearest tenth of a metre? d) Describe what happens when \(t=15 \mathrm{s}\)

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If \(\sin \theta=0.3,\) determine the value of \(\sin \theta+\sin (\theta+2 \pi)+\sin (\theta+4 \pi)\)

The frequency of a wave is the number of cycles that occur in 1 s. Adding two sinusoidal functions with similar, but unequal, frequencies results in a function that pulsates, or exhibits beats. Piano tuners often use this phenomenon to help them tune a piano. a) Graph the function \(y=\cos x+\cos 0.9 x\). b) Determine the amplitude and the period of the resulting wave.
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