/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Sketch the graphs and label the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Sketch the graphs and label the axes for $$\text { (a) } \quad y=0.2 \cos \left(\frac{2 \pi}{0.8} t\right) \quad \text { and } \quad \text { (b) } \quad y=5 \cos \left(\frac{1}{8} t+\pi / 6\right)$$

Short Answer

Expert verified
Function (a) has amplitude 0.2, period 0.8; function (b) has amplitude 5, period 16, phase shift \(-\frac{4\pi}{3}\). Graph accordingly.

Step by step solution

01

Identify the amplitude, period, and phase shift for (a)

For the function \(y=0.2 \cos \left(\frac{2\pi}{0.8} t\right)\), the amplitude is \(0.2\), since it is the coefficient in front of the cosine. The period is calculated using \(\frac{2\pi}{B}\), where \(B=\frac{2\pi}{0.8}\), resulting in a period of \(0.8\). There is no phase shift since there is no horizontal shift term.
02

Sketch graph of (a)

To sketch the graph of \(y=0.2 \cos \left(\frac{2\pi}{0.8} t\right)\), start with the amplitude marking 0.2 and -0.2 on the y-axis. The period is 0.8, so label the x-axis for one complete period as 0 to 0.8. Divide graph into four equal segments for key points: maximum, zero crossing, minimum, zero crossing. Plot these points and draw a smooth cosine curve through them.
03

Identify the amplitude, period, and phase shift for (b)

For the function \(y=5 \cos \left(\frac{1}{8} t + \frac{\pi}{6}\right)\), the amplitude is \(5\), the period is \(\frac{2\pi}{\frac{1}{8}} = 16\), and the phase shift is calculated by setting \(\frac{1}{8} t + \frac{\pi}{6} = 0\); this gives \(t = -\frac{\pi}{6}\times 8 = -\frac{4\pi}{3}\).
04

Sketch graph of (b)

Begin sketching \(y=5 \cos \left(\frac{1}{8} t + \frac{\pi}{6}\right)\) with the amplitude markers at 5 and -5 on the y-axis. Set the period on the x-axis from 0 to 16 and apply the phase shift of \(-\frac{4\pi}{3}\), approximately -4.19. Start the graph \(\frac{4\pi}{3}\) units to the right of where it normally begins and plot key cosine points within the period. Connect the points with a smooth curve.
05

Label axes on the graphs

For both sketches, ensure that the x-axes are labeled with time (t) while the y-axes are labeled with displacement (y). Indicate the key periods for each function on the x-axis: 0 to 0.8 for (a) and 0 to 16 for (b). Both amplitudes should be annotated on the y-axis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Amplitude
The concept of amplitude in trigonometric functions is crucial because it tells us how tall the waves are. Amplitude refers to the maximum distance the wave peaks reach from the middle or baseline of the wave. For a cosine or sine wave, which oscillates above and below a central line, the amplitude is simply the coefficient of the sine or cosine function.

In simple terms:
  • Amplitude is the 'height' of the wave.
  • It affects the vertical stretch of the graph.
  • To find the amplitude, look for the number before \cos\ or \sin\ in the equation.
For example, in the function \( y = 0.2 \cos\left(\frac{2\pi}{0.8} t\right) \), the amplitude is \( 0.2 \). Similarly, in \( y = 5 \cos\left( \frac{1}{8}t + \frac{\pi}{6} \right) \), the amplitude is \( 5 \). These values show the maximum and minimum points the wave can reach on the graph, providing a simple way to visualize how far the wave stretches along the y-axis.
Period
The period of a trigonometric function is the distance over which the function repeats its shape. To understand the period, consider how long it takes for a pattern in the graph to start over. The formula to calculate the period for sine and cosine functions is \( \frac{2\pi}{B} \), where \( B \) is the coefficient of \( t \) inside the function.

Key points about period:
  • The period dictates how "wide" a cycle of the wave is on the x-axis.
  • It's the length for one complete cycle of the wave.
  • For a function \( f(t) = \cos(Bt) \), compute the period using: \( \text{Period} = \frac{2\pi}{B} \).
In our exercises, for the first function \( y = 0.2 \cos\left( \frac{2 \pi}{0.8} t \right) \), the period is \( 0.8 \). That means every 0.8 units, the wave begins its cycle again. For the second function \( y = 5 \cos\left( \frac{1}{8}t + \frac{\pi}{6} \right) \), the calculation gives a period of \( 16 \). This means on the graph, the wave pattern stretches over 16 units before repeating. Understanding the period helps in accurately plotting and predicting future points of the wave's cycle.
Phase Shift
Phase shift in trigonometric functions refers to the horizontal movement of the wave along the x-axis. This shift can slide the graph left or right depending on the direction indicated by the function's equation. Simply put, the phase shift indicates where the wave starts.

Important facts about phase shift:
  • Phase shift shows how far to move the graph horizontally from its standard position.
  • Calculated by resolving \( Bt + C = 0 \) for \( t \) in \( f(t) = \cos(Bt + C) \).
  • A positive phase shift moves the graph to the left, while a negative shift moves it to the right.
For instance, in our function \( y = 0.2 \cos\left(\frac{2\pi}{0.8} t \right) \), there is no additional term added to \( t \), so no phase shift occurs. The graph starts from its standard position. However, in \( y = 5 \cos\left( \frac{1}{8}t + \frac{\pi}{6} \right) \), the phase shift is calculated by setting the inside term to zero: \( \frac{1}{8}t + \frac{\pi}{6} = 0 \). Solving this results in a phase shift of \(-\frac{4\pi}{3}\); when converted roughly, this is a rightward shift of about 4.19 units. Phase shifts are essential for marking the wave's correct starting point on a graph.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Shown in Figure Ex. 2.6 .6 is the graph of \(F(x)=2^{x} .(-2,1 / 4),(0,1),\) and (2,4) are ordered pairs of \(F\). What are the corresponding ordered pairs of \(F^{-1}\) ? Plot those points and draw the graph of \(F^{-1}\).

What are the implied domains of the functions $$ \begin{array}{ll} F(x)=\sqrt{x-1} & F(x)=\frac{1+x^{2}}{1-x^{2}} \\ F(x)=\sqrt{4-x^{2}} & F(x)=\log _{10}\left(x^{2}\right) \end{array} $$

A bit of a difficult exercise. For any location, \(\lambda\) on Earth, let Annual Daytime at \(\lambda, A D(\lambda),\) be the sum of the lengths of time between sunrise and sunset at \(\lambda\) for all of the days of the year. Find a reasonable formula for \(A D(\lambda)\). You may guess or find data to suggest a reasonable formula, but we found proof of the validity of our formula a bit arduous. As often happens in mathematics, instead of solving the actual problem posed, we found it best to solve a 'nearby' problem that was more tractable. The \(365.24 \ldots\) days in a year is a distraction, the elliptical orbit of Earth is a downright hinderance, and the wobble of Earth on its axis can be overlooked. Specifically, we find it helpful to assume that there are precisely 366 days in the year (after all this was true about 7 or 8 million years ago), the Earth's orbit about the sun is a circle, the Earth's axis makes a constant angle with the plane of the orbit, and that the rays from the sun to Earth are parallel. We hope you enjoy the question.

Which of the following functions are invertible? a. The distance a DNA molecule will migrate during agarose gel electrophoresis as a function of the molecular weight of the molecule, for domain: \(1 \mathrm{~kb} \leq\) Number of bases \(\leq 20 \mathrm{~kb}\). b. The density of water as a function of temperature. c. Day length as a function of elevation of the sun above the horizon (at, say, 40 degrees North latitude). d. Day length as a function of day of the year.

Find a function, L, defined for all numbers (domain is all numbers) such that for all numbers a and \(\mathrm{b}, \mathrm{L}(\mathrm{a}+\mathrm{b})=\mathrm{L}(\mathrm{a})+\mathrm{L}(\mathrm{b})\). Is there another such function?

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.