Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises: 24

Answer

$-1$

Work Step by Step

Given: $r=(3+cos3\theta)$ and $\theta =\pi/2$ In Cartesian Coordinates, $x=rcos\theta$ Then $x=(3+cos3\theta)cos(\theta)$ $\frac{dx}{d\theta}=\frac{d[(3+cos3\theta)cos(\theta)]}{d\theta}$ Use Product rule of differentiation. $\frac{dx}{d\theta}=(-3sin3\theta)cos(\theta)-(3+cos3\theta)sin\theta$ At $\theta=\pi/2$ $=(-3sin3\pi/2)cos(\pi/2)-(3+cos3\pi/2)sin(\pi/2)$ Thus, $\frac{dx}{d\theta}=((-3)(-1))(0)-(3+0)(1)=-3$ In Cartesian Coordinates, $y=rcos\theta$ Then $y=(3+cos3\theta)cos(\theta)$ $\frac{dy}{d\theta}=\frac{d[(3+cos3\theta)cos(\theta)]}{d\theta}$ Use Product rule of differentiation. $\frac{dy}{d\theta}=(-3sin3\theta)sin(\theta)+(3+cos3\theta)cos\theta$ At $\theta=\pi/2$ $=(-3sin3\pi/2)sin(\pi/2)+(3+cos3\pi/2)cos(\pi/2)$ Thus, $\frac{dy}{d\theta}=((-3)(-1))(1)+(3+0)(0)=3$ Let $m$ be the slope of the tangent line to the given curve. $m=\frac{dy}{dx}|_{\theta=\pi/2}$ $=\frac{3}{-3}$ $=-1$ Hence, $m=-1$
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