University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 10 - Section 10.3 - Polar Coordinates - Exercises - Page 578: 49

Answer

The graph is a circle whose center is at $(1,1)$ with radius $\sqrt 2$.

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) $y=r \sin \theta$ Here, we have $r^2=2r \cos \theta +2r \sin \theta$ Therefore, our Cartesian equation is $x^2+y^2=2x+2y \implies (x-1)^2+(y-1)^2=2$ This shows that the graph is a circle whose center is at $(1,1)$ with radius $\sqrt 2$.
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