Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 33


$r=-5 \sin \theta$

Work Step by Step

Consider $x^2+y^2+5y=0$ or, $x^2+(y+\dfrac{5}{2})^2=(\dfrac{25}{4})$ Here, we get the equation of a circle with center $(0,-\dfrac{5}{2})$ and radius $\dfrac{5}{2}$ Thus, $r^2+5r \sin \theta=0 \implies r=-5 \sin \theta$
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