Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 655: 19

Answer

Horizontal asymptotes at $ (-\frac{1}{2}, 1)$ and $(\frac{1}{2}, -1) $, and no vertical asymptotes

Work Step by Step

$x= cos(θ)$, $y=cos(3θ)$ Boundaries of the parametric are $-\pi \leq θ \leq \pi$ To find horizontal asymptotes, we find the derivative of $y$ and set it to zero. $\frac{dy}{dθ}= -3sin(3θ) $ $-3sin(3θ)=0$ $sin(3θ)=0$ $3θ= \frac{arcsin(0)}{3}=0$ $θ=0$ $sin(0) = 0$ This then also becomes: $θ= \frac{k\pi}{3}$, where $k$ is an integer. $sin(k\pi)=0$, so $sin(\frac{3k\pi}{3})=0$ Now to find the points where the asymptotes exist by plugging in the $θ$ values for x and y $θ=0$ $X: cos(0)=1$ $Y: cos(0)=1$ There is no asymptote here because the function is discontinuous at this point. $θ=\frac{\pi}{3}$ $X: cos(\frac{\pi}{3})= \frac{1}{2}$ $Y: cos(\frac{3\pi}{3})=-1$ Horizontal asymptote at ($\frac{1}{2}, -1)$ $θ=\frac{2\pi}{3}$ $X: cos(\frac{2\pi}{3})= -\frac{1}{2}$ $Y: cos(\frac{6\pi}{3})=1$ Horizontal asymptote at ($-\frac{1}{2}, 1)$ $θ=\pi$ $X: cos(0)=-1$ $Y: cos(0)=-1$ There is no asymptote here because the function is discontinuous at this point. To find vertical asymptotes, we find the derivative of $x$ and set it to zero. $\frac{dx}{dθ}= -sin(θ)$ $sin(θ)=0$ $θ = k\pi$, where $k$ is an integer Now to find the points where the asymptotes exist by plugging in the $θ$ values for x and y $θ=0$ $X: cos(0)=1$ $Y: cos(0)=1$ There is no asymptote here because the function is discontinuous at this point. $θ=\pi$ $X: cos(0)=-1$ $Y: cos(0)=-1$ There is no asymptote here because the function is discontinuous at this point. Horizontal asymptotes at $ (-\frac{1}{2}, 1)$ and $(\frac{1}{2}, -1) $, and no vertical asymptotes
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