Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 42

Answer

$(\frac{1}{\sqrt[4]{2}}, \pi/8), (\frac{1}{\sqrt[4]{2}}, 9\pi/8)$, and the pole

Work Step by Step

$sin2\theta=cos2\theta$ $\theta=\frac{\pi}{8},\frac{9\pi}{8}$ Thus the points of intersection are: $(\frac{1}{\sqrt[4]{2}}, \pi/8), (\frac{1}{\sqrt[4]{2}}, 9\pi/8)$, and the pole
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