Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.1 Polar Cordinates - 8.1 Assess Your Understanding - Page 592: 69


$r^2 \ \cos^2(\theta)-4r \ \sin(\theta) =0$

Work Step by Step

The conversion of polar co-ordinates $(r, \theta)$ to rectangular coordinates $(x,y)$ can be expressed as: $x=r \ \cos(\theta)$, $y=r \ \sin(\theta) $ where, $r=\sqrt{x^2+y^2}$ We have: $x^2=4y$ After making the substitutions above, we get: $r^2 \ \cos^2(\theta)=4 r \ \sin(\theta)$ This implies that $r^2 \ \cos^2(\theta)-4r \ \sin(\theta) =0$ Therefore, $r^2 \ \cos^2(\theta)-4r \ \sin(\theta) =0$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.