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: 47

Answer

$x-\sqrt{3}y=-6$ .

Work Step by Step

Apply identity for $\cos(\alpha-\beta)$ $\displaystyle \cos(\theta-\frac{2\pi}{3})=\cos\theta\cos\frac{2\pi}{3}+\sin\theta\sin\frac{2\pi}{3}$ $\displaystyle \cos(\theta-\frac{2\pi}{3})=-\frac{1}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta\qquad/\times r$ $ r\displaystyle \cos(\theta-\frac{2\pi}{3})=-\frac{1}{2}r\cos\theta+\frac{\sqrt{3}}{2}r\sin\theta$ Apply the conversion formula$\quad (x,y)=(r\cos\theta,r\sin\theta)$ $r\displaystyle \cos(\theta-\frac{\pi}{4})=-\frac{1}{2}x+\frac{\sqrt{3}}{2}y$ In Cartesian coordinates, the line equation is $-\displaystyle \frac{1}{2}x+\frac{\sqrt{3}}{2}y=3\qquad/\times(-2)$ $x-\sqrt{3}y=-6$
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