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Chapter 10: Problem 8

Find the amplitude and period of each function and then sketch its graph. $$y=-\frac{1}{5} \sin 5 x$$

Short Answer

Expert verified
Amplitude: \( \frac{1}{5} \); Period: \( \frac{2\pi}{5} \). Reflects the sine wave vertically.

Step by step solution

01

Identify the General Form

The general form of the sine function is given by \( y = a \sin(bx + c) + d \). For the function \( y = -\frac{1}{5} \sin 5x \), we identify: - \( a = -\frac{1}{5} \) which affects the amplitude.- \( b = 5 \) which affects the period. - \( c = 0 \) and \( d = 0 \) meaning there are no horizontal or vertical shifts.
02

Calculate the Amplitude

The amplitude of a sine function \( y = a \sin(bx) \) is the absolute value of \( a \). Here, \( a = -\frac{1}{5} \), so the amplitude is \( |-\frac{1}{5}| = \frac{1}{5} \).
03

Calculate the Period

The period of a sine function is given by \( \frac{2\pi}{b} \). For \( b = 5 \), the period is \( \frac{2\pi}{5} \). This is the length of one complete cycle of the sine wave along the x-axis.
04

Sketch the Graph

To sketch the graph of \( y = -\frac{1}{5} \sin 5x \), note the following:- Amplitude is \( \frac{1}{5} \), so the graph will oscillate between \( \frac{1}{5} \) and \( -\frac{1}{5} \).- Period is \( \frac{2\pi}{5} \), so one full wave cycle will occur within this length on the x-axis.- The negative \( a \) causes a reflection in the x-axis. Start at the origin and draw a sine curve considering these values.

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Key Concepts

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

Amplitude
The amplitude of a trigonometric function describes its vertical stretch or compression, specifically the maximum distance from the function's midline to its peak. In the function we are working with, \[ y = -\frac{1}{5} \sin 5x \]we identify the amplitude using the coefficient of the sine function. Here, the coefficient is \(-\frac{1}{5}\). Although it is negative, when finding amplitude, we take the absolute value.
  • The amplitude is \(\left|-\frac{1}{5}\right| = \frac{1}{5}\).
  • This means the sine wave will oscillate, reaching up to \(\frac{1}{5}\) above, and \(-\frac{1}{5}\) below its midline.
The negative sign indicates a reflection over the x-axis, but it does not affect the amplitude's magnitude. This simply flips the wave vertically, but does not change how far up or down from the midline the wave goes.
Period
The period of a sine function such as \[ y = -\frac{1}{5} \sin 5x \]is the length of one complete wave cycle along the x-axis. It is determined by the coefficient \(b\) of the \(x\) term in the sine function. The formula to compute the period is:\[ \text{Period} = \frac{2\pi}{b} \]In this exercise, with \(b = 5\), we calculate:
  • Period = \(\frac{2\pi}{5}\).
  • This tells us that the sine function will complete one full cycle over an interval of \(\frac{2\pi}{5}\) units on the x-axis, giving it a relatively fast oscillation rate compared to a standard sine wave.
This smaller period means there will be more waves, packed closer together along the x-axis, compared to the standard \(2\pi\) period of a regular sine wave.
Sine Wave Graph
Sketching the graph of the function \[ y = -\frac{1}{5} \sin 5x \]requires understanding both the amplitude and period, alongside general properties of the sine function. Key details for sketching include:
  • Amplitude: Since it is \(\frac{1}{5}\), the wave oscillates between \(\frac{1}{5}\) and \(-\frac{1}{5}\).
  • Period: The function completes a full cycle over the length \(\frac{2\pi}{5}\).
  • Reflection: The negative sign in front of the amplitude reflects the wave, flipping it upside down across the x-axis.
When starting the sketch at the origin,
- the wave begins at 0, moves downward first due to the negative sign, - hits \(-\frac{1}{5}\) at a quarter of the period, - returns to 0 at half the period value, - reaches \(\frac{1}{5}\), and - finally returns to 0 completing the cycle. Keep in mind these trigonometric waves continue indefinitely in both directions unless otherwise specified.

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Most popular questions from this chapter

View at least two cycles of the graphs of the given functions on a calculator. $$y=18 \csc \left(3 x-\frac{\pi}{3}\right)$$

Solve the given problems. If the period is \(\frac{1}{5} \mathrm{s},\) find \(\omega\)

Sketch the appropriate curves. A calculator may be used. An object oscillating on a spring has a displacement (in ft) given by \(y=0.4 \sin 4 t+0.3 \cos 4 t,\) where \(t\) is the time (in s). Sketch the graph.

Sketch the appropriate curves. A calculator may be used. A normal person with a pulse rate of 60 beats/min has a blood pressure of "120 over 80." This means the pressure is oscillating between a high (systolic) of \(120 \mathrm{mm}\) of mercury (shown as \(\mathrm{mmHg}\) ) and a low (diastolic) of 80 mmHg. Assuming a sinusoidal type of function, find the pressure \(p\) as a function of the time \(t\) if the initial pressure is \(120 \mathrm{mmHg} .\) Sketch the graph for the first \(5 \mathrm{s}\).

A graphing calculator may be used in the following exercises. Sketch two cycles of the curve given by \(d=R\) sin \(\omega t\) for the given values. $$R=18.5 \mathrm{ft}, f=0.250 \mathrm{Hz}$$
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