Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 10 - Section 10.4 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 700: 38

Answer

$(1-\frac{\sqrt{2}}{2},\frac{3\pi}{4})$ and $(1+\frac{\sqrt{2}}{2},\frac{7\pi}{4})$

Work Step by Step

Eliminate $r$ or $\theta$: $r=1+\cos\theta$ and $r=1-\sin\theta$ $r-1=\cos\theta$ and $r-1=-\sin\theta$ $(r-1)^2=\cos^2\theta$ and $(r-1)^2=\sin^2\theta$ (Substract both equations) $0=\cos^2\theta-\sin^2\theta$ $0=\cos2\theta$ Find $\theta\in[0,2\pi]$ satisfying the last equation: $\cos2\theta=0$ $\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},$ and $\frac{7\pi}{4}$ Back substitute $\theta$ above into the given polar equations: $\theta=\frac{\pi}{4}$, $r_1=1+\frac{\sqrt{2}}{2}$ and $r_2=1-\frac{\sqrt{2}}{2}$ $\Rightarrow r_1\neq r_2$ (not satisfied) $\theta=\frac{3\pi}{4}$, $r_1=1-\frac{\sqrt{2}}{2}$ and $r_2=1-\frac{\sqrt{2}}{2}$ $\Rightarrow r_1= r_2$ (satisfied) $\theta=\frac{5\pi}{4}$, $r_1=1-\frac{\sqrt{2}}{2}$ and $r_2=1+\frac{\sqrt{2}}{2}$ $\Rightarrow r_1\neq r_2$ (not satisfied) $\theta=\frac{7\pi}{4}$, $r_1=1+\frac{\sqrt{2}}{2}$ and $r_2=1+\frac{\sqrt{2}}{2}$ $\Rightarrow r_1= r_2$ (satisfied) Thus, the intersection points are $(1-\frac{\sqrt{2}}{2},\frac{3\pi}{4})$ and $(1+\frac{\sqrt{2}}{2},\frac{7\pi}{4})$.
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