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: 50

Answer

$r \cos (\theta +\dfrac{\pi}{6})=\dfrac{1}{2}$

Work Step by Step

Conversion of polar coordinates and Cartesian coordinates are as follows: a) $r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$ b) $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$ c) $x=r \cos \theta$ d) $y=r \sin \theta$ We are given that $\sqrt 3 x-\sqrt y=1$ This implies that $\cos (\dfrac{\pi}{6}) (r \cos \theta)+\sin (\dfrac{\pi}{6}) (r \sin \theta)=\dfrac{1}{2}$ or, $r[\cos (\dfrac{\pi}{6}) (\cos \theta)+\sin (\dfrac{\pi}{6}) (\sin \theta)]=\dfrac{1}{2}$ Thus, $r \cos (\theta +\dfrac{\pi}{6})=\dfrac{1}{2}$
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