University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 10 - Practice Exercises - Page 594: 35

Answer

$r=3\cos \theta$

Work Step by Step

Here, we have $x^2+y^2-3x=0$ This implies that $(x-\dfrac{3}{2})^2+y^2=\dfrac{9}{4}$ Thus, we have an equation of a circle with center $(\dfrac{3}{2},0)$ and radius $\dfrac{3}{2}$ So, $r^2-3r \cos \theta=0$ Hence, $r=3\cos \theta$
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