/*! 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 13 Test for symmetry and then graph... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 13

Test for symmetry and then graph each polar equation. $$r=2 \cos \theta$$

Short Answer

Expert verified
The function \(r=2 \cos \theta\) is symmetric with respect to the x-axis. The graph is a circle with a radius of 2 centered at the origin.

Step by step solution

01

Checking for Symmetry

First, verify if the function is symmetric with respect to the x and y axis or the origin. This can be done by plugging \(-\theta\) into \(\theta\) and checking if the function remains unchanged. For the function \(r = 2 \cos \theta\), if \(-\theta\) is replaced in place of \(\theta\), we get \(r = 2 \cos (-\theta)\). This simplifies to \(r = 2 \cos \theta\) again due to the cosine function being even, meaning the function is symmetric about the x-axis.
02

Graphing the Function

After confirming the function's symmetry, you should move to the graphing of the function. As this is a cosine function, the general shape will resemble that of a circle with radius 2 since \(r=2\cos \theta\) is a circle centered at the origin with a radius of 2, but only the top half due to the symmetry concluded earlier.
03

Finishing the graph

The function looks like a semi-circle residing on the positive x-axis due to the earlier determined line of symmetry (x-axis). The radius of this semi-circle is 2. For \(\theta\) ranging from \(0\) to \(\pi\), the graph forms a round semi-circle. Beyond this, for \(\theta\) from \(\pi\) to \(2\pi\), the graph repeats itself in the lower half due to the previously established x-axis symmetry. Thus the final graph is a full circle of radius 2 at the origin.

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 91影视!

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

The components of \(\mathbf{v}=180 \mathbf{i}+450 \mathbf{j}\) represent the respective number of one-day and three-day videos rented from a video store. The components of \(\mathbf{w}=3 \mathbf{i}+2 \mathbf{j}\) represent the prices to rent the one-day and three-day videos, respectively. Find \(\mathbf{v} \cdot \mathbf{w}\) and describe what the answer means in practical terms.

Determine whether v and w are parallel, orthogonal, or neither. $$\mathbf{v}=-2 \mathbf{i}+3 \mathbf{j}, \quad \mathbf{w}=-6 \mathbf{i}-4 \mathbf{j}$$

Explain how to find the dot product of two vectors.

Use the dot product to determine whether v and w are orthogonal. $$\mathbf{v}=2 \mathbf{i}+8 \mathbf{j}, \quad \mathbf{w}=4 \mathbf{i}-\mathbf{j}$$

Find \(\text {pro}_{\mathbf{w}} \mathbf{V}\) Then decompose v into two vectors, \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2},\) where \(\mathbf{v}_{1}\) is parallel to \(\mathbf{w}\) and \(\mathbf{v}_{2}\) is orthogonal to \(\mathbf{w}.\) $$\mathbf{v}=2 \mathbf{i}+4 \mathbf{j}, \quad \mathbf{w}=-3 \mathbf{i}+6 \mathbf{j}$$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation
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.