Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.3 Polar Coordinates - Exercises - Page 618: 21



Work Step by Step

Given $$ e^{\sqrt{x^2+y^2}}=1$$ Since $ y=r\sin\theta,\ \ x= r\cos\theta,\ \ r^2=x^2+y^2 $, then \begin{align*} e^{\sqrt{x^2+y^2}}&= 1\\ e^{r}&=1\\ r&=\ln 1\\ r&=0 \end{align*}
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