Calculus 8th Edition

Published by Cengage

Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 730: 23

Answer

$-1$

Work Step by Step

Given: $r=e^{-\theta}$ and $\theta =\pi$ In Cartesian Coordinates, $x=rcos\theta$ Then $x=e^{-\theta}cos(\theta)$ $\frac{dx}{d\theta}=\frac{d(e^{-\theta}cos(\theta)}{d\theta}$ Use Product rule of differentiation. $\frac{dx}{d\theta}=-e^{-\theta}cos(\theta)-e^{-\theta}sin(\theta)$ At $\theta=\pi$ $=-e^{-\pi}(-1)-e^{-\pi}(0)$ Thus, $\frac{dx}{d\theta}=e^{-\pi}$ In Cartesian Coordinates, $y=rsin\theta$ Then $y=e^{-\theta}sin(\theta)$ $\frac{dy}{d\theta}=\frac{d(e^{-\theta}sin(\theta)}{d\theta}$ Use Product rule of differentiation. $\frac{dy}{d\theta}=-e^{-\theta}sin(\theta)+e^{-\theta}cos(\theta)$ At $\theta=\pi$ $=-e^{-\theta}sin(\pi)+e^{-\theta}cos(\pi)$ $=-e^{-\pi}(0)+e^{-\pi}(-1)$ Thus, $\frac{dy}{d\theta}=-e^{-\pi}$ Let $m$ be the slope of the tangent line to the given curve. $m=\frac{dy}{dx}|_{\theta=\pi}$ $=\frac{-e^{-\pi}}{e^{-\pi}}$ $=-1$ Hence, $m=-1$

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