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 - Section 11.3 - Polar Coordinates - Exercises 11.3 - Page 663: 41

Answer

An exponential function with base $e$.

Work Step by Step

The conversion of polar coordinates and Cartesian coordinates are described as follows: 1. $r^2=x^2+y^2$ and $r=\sqrt {x^2+y^2}$ 2. $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$ 3. $x=r \cos \theta$ and 4. $y=r \sin \theta$ Multiply the given equation by $\sin \theta$ Now, we have $r \sin \theta=e^{r \cos \theta}$ Thus, the Cartesian equation is $y=e^x$ Hence, this shows an exponential function with base $e$.
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