Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 730: 34


$(0, \frac{\pi}{2}),(\sqrt 3, \frac{\pi}{6}), (-\sqrt 3, \frac{5\pi}{6})$

Work Step by Step

$r=cot \theta$, and $r= 2cos \theta$ $cot \theta= 2cos \theta$ $cot \theta- 2cos \theta=0$ $\frac{cos \theta}{sin \theta}- 2cos \theta=0$ $sin \theta=\frac{1}{2}$ and $ \theta=\frac{\pi}{6}, \frac{5 \pi}{6}$ Now we will find the points. $(r, \theta): (2 cos \frac{\pi}{2}, \frac{\pi}{2})=(0, \frac{\pi}{2})$ $ (2 cos \frac{\pi}{6}, \frac{\pi}{6})=(\sqrt 3, \frac{\pi}{6})$ $ (2 cos \frac{5\pi}{6}, \frac{5\pi}{6})=(-\sqrt 3, \frac{5\pi}{6})$ Hence, $(0, \frac{\pi}{2}),(\sqrt 3, \frac{\pi}{6}), (-\sqrt 3, \frac{5\pi}{6})$
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