Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Sketch a graph of the polar equation. $$ r^{2}=4 \cos 2 \theta $$

Short Answer

Expert verified
The polar equation \(r^{2}=4 \cos 2 \theta\) represents a limaçon oriented on the x-axis. The graph is a circle centred at (2, 0) with radius 2.

Step by step solution

01

Understand the Standard Form of Polar Equation

A polar equation of the form \(r^{2}=a^{2} \cos(2 \theta)\) represents a limaçon. The length \(a^{2}\) determines the size of the limaçon, and the form \(\cos\) or \(\sin\) determines the orientation. In this case, the polar equation \(r^{2}=4 \cos 2 \theta\) is a limaçon oriented on the x-axis.
02

Transform into Cartesian Coordinates

To create a graph more easily, convert the polar equation into Cartesian coordinates. In Cartesian coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\). Given the polar equation \(r^{2}=4 \cos 2 \theta\), it can be rewritten as \(x^{2}+ y^{2} = 4 x\).
03

Sketch the Graph

The graph of the equation \(x^{2}+ y^{2} = 4 x\) is a circle centred at (2, 0) with radius 2. This circle is the limaçon described by the polar equation \(r^{2}=4 \cos 2 \theta\). Thus, the graph is a circle centred at (2, 0) with radius 2.

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 Vaia!

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

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.

Sign-up for free