/*! 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 9 Determine the amplitude and peri... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the amplitude and period of each function. Then graph one period of the function. $$y=3 \sin \frac{1}{2} x$$

Short Answer

Expert verified
The amplitude of the function \(y = 3 \sin \frac{1}{2} x\) is 3 and its period is \(4\pi\).

Step by step solution

01

Identifying the amplitude

The amplitude of a sine function is given by the absolute value of its coefficient. In this case, the coefficient is 3. Therefore, the amplitude of given function \(y = 3 \sin \frac{1}{2} x\) is 3.
02

Identifying the period

The period of the sine function can be calculated using the formula \(2\pi \div |B|\), where \(B\) is the coefficient of \(x\) inside the sine function. In this case, the coefficient of \(x\) is \frac{1}{2}. So, the period of the function is \(2\pi \div \frac{1}{2} = 4\pi\).
03

Graphing one period of function

To graph one period of the function, first note that a typical sine function starts from the origin (0,0), reaches its highest point at \(\frac{1}{4}\) of its period, returns to the origin at half of its period, reaches its lowest point at \(\frac{3}{4}\) of its period, and returns to the origin at the end of its period. Given the amplitude and period of this function, the highest point becomes (0,3), the 'half period' point becomes \(2\pi , 0\), the lowest point becomes \(3\pi , -3\) and the end of the period is \(4\pi , 0\). Then, graph these points and join them smoothly in the shape of a sine curve, considering these key points.

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!

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

Describe the restriction on the sine function so that it has an inverse function.

For years, mathematicians were challenged by the following problem: What is the area of a region under a curve between two values of \(x ?\) The problem was solved in the seventeenth century with the development of integral calculus. Using calculus, the area of the region under \(y=\frac{1}{x^{2}+1},\) above the \(x\) -axis, and between \(x=a\) and \(x=b\) is \(\tan ^{-1} b-\tan ^{-1} a\). Use this result, shown in the figure, to find the area of the region under \(y=\frac{1}{x^{2}+1}\) above the \(x\) -axis, and between the values of a and b given in Exercises \(97-98\). (GRAPH CANNOT COPY) \(a=0\) and \(b=2\)

Explain what is meant by one radian.

Graph \(y=\tan ^{-1} x\) and its two horizontal asymptotes in a \([-3,3,1]\) by \(\left[-\pi, \pi, \frac{\pi}{2}\right]\) viewing rectangle. Then change the viewing rectangle to \([-50,50,5]\) by \(\left[-\pi, \pi, \frac{\pi}{2}\right] .\) What do you observe?

Your neighborhood movie theater has a 25 -foot-high screen located 8 feet above your eye level. If you sit too close to the screen, your viewing angle is too small, resulting in a distorted picture. By contrast, if you sit too far back, the image is quite small, diminishing the movie's visual impact. If you sit \(x\) feet back from the screen, your viewing angle, \(\theta,\) is given by $$\theta=\tan ^{-1} \frac{33}{x}-\tan ^{-1} \frac{8}{x}$$ (GRAPH CANNOT COPY) Find the viewing angle, in radians, at distances of 5 feet, 10 feet, 15 feet, 20 feet, and 25 feet.

See all solutions

Recommended explanations on Math 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.