Calculus 8th Edition

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

Chapter 10 - Parametric Equations and Polar Coordinates - 10.3 Polar Coordinates - 10.3 - Page 707: 61

Answer

Vertical Tangent Lines at: $(0,\frac{\pi}{2})$, $(3,0)$. Horizontal Tangent Lines at: $(\frac{-3\pi}{2},\frac{3\pi}{4})$, $(\frac{3\pi}{2},\frac{\pi}{4})$.

Work Step by Step

Given: $r=3cos(\theta)$ Use the equation for tangent line slope for polar coordinates: $TLS=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta}{\frac{dr}{d\theta}cos\theta-rsin\theta}$ Find $\frac{dr}{d\theta}$: $\frac{dr}{d\theta}=(3cos(\theta))^{\prime}={-3sin(\theta)}$ Plug in for $r$ and $\frac{dr}{d\theta}$: $TLS\vert_{\theta}=\frac{(-3sin(\theta))sin(\theta)+(3cos(\theta))cos(\theta)}{({-3sin(\theta)})cos(\theta)-(3cos(\theta))sin(\theta)}=\frac{(-3sin(\theta))sin(\theta)+(3cos(\theta))cos(\theta)}{({-3sin(\theta)})cos(\theta)-(3cos(\theta))sin(\theta)}$ $TLS\vert_{\theta}=\frac{-sin^2\theta+cos^2\theta}{{-2sin{\theta}}cos\theta}=\frac{-sin^2\theta+cos^2\theta}{{-2sin(2\theta)}}$ Find the horizontal tangent lines (when the denominator = 0): $-sin^2\theta+cos^2\theta=0$ -->$sin^2\theta=cos^2\theta$ -->$sin\theta=cos\theta$ This is true when $\theta=\frac{\pi}{4}$ and $\frac{3\pi}{4}$ Now find the r-values for these points and form the ordered pairs in $(r,\theta)$: $r=3cos(\frac{\pi}{4})=\frac{3\pi}{2}$ $r=3cos(\frac{3\pi}{4})=\frac{-3\pi}{2}$ Horizontal Tangent Lines at: $(\frac{-3\pi}{2},\frac{3\pi}{4})$, $(\frac{3\pi}{2},\frac{\pi}{4})$. Now find the vertical tangent lines (when the numerator = 0): ${{-2sin(2\theta)}}=0$ -->$sin(2\theta)=0$ This is true when $\theta=0$ and $\frac{\pi}{2}$ Now find the r-values for these points and form the ordered pairs in $(r,\theta)$: $r=3cos(0)=3$ $r=3cos(\frac{\pi}{2})=0$ Vertical Tangent Lines at: $(0,\frac{\pi}{2})$, $(3,0)$.
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