University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.6 - Conics in Polar Coordinates - Exercises - Page 592: 64

Answer

$ r=\displaystyle \frac{4}{3}\sin\theta$

Work Step by Step

The circle passes through the origin, as (0,0) satisfies the Cartesian equation. Completing the square, $x^{2}+(y^{2}-2(\displaystyle \frac{2}{3})y+(\frac{2}{3})^{2})+y^{2}=0+(\frac{2}{3})^{2}$ $x^{2}+(y-\displaystyle \frac{2}{3})^{2}=\frac{4}{9}$ The radius is $\displaystyle \frac{2}{3}$, and the center is at $(0,\displaystyle \frac{2}{3})$. In polar coordinates, the center lies at $r_{0}=\displaystyle \frac{2}{3}, \theta_{0}=\pi/2,\qquad P(\frac{2}{3},\pi/2)$ A circle passing through the origin, of radius $a$, centered at $P_{0}(r_{0}, \theta_{0}),$ has the polar equation $r=2a\cos(\theta-\theta_{0})$ So this circle has the equation $r=2(\displaystyle \frac{2}{3})\cos(\theta-\frac{\pi}{2})$ or $r=\displaystyle \frac{4}{3}\cos(\theta-\frac{\pi}{2})$ using the identity $\displaystyle \cos(\theta\pm\frac{\pi}{2})=\mp\sin\theta$, $r=\displaystyle \frac{4}{3}\sin\theta$
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