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 656: 54

Answer

$T o t a l\;l e n g t h=4*\frac{3a}{2}=6a$

Work Step by Step

$$ \begin{gathered} \frac{d x}{d \theta}=-3 a \cos ^2(\theta) \sin (\theta) \\ \frac{d y}{d \theta}=3 a \sin ^2(\theta) \cos (\theta) \end{gathered} $$ $$ \begin{gathered} \left(\frac{d x}{d \theta}\right)^2+\left(\frac{d y}{d \theta}\right)^2=9 a^2\left[\cos ^4(\theta) \sin ^2(\theta)+\sin ^4(\theta) \cos ^2(\theta)\right] \\ =9 a^2 \sin ^2(\theta) \cos ^2(\theta)\left[\cos ^2(\theta)+\sin ^2(\theta)\right] \\ =9 a^2 \sin ^2(\theta) \cos ^2(\theta) \end{gathered} $$ $$ \begin{gathered} \sqrt{\left(\frac{d x}{d \theta}\right)^2+\left(\frac{d y}{d \theta}\right)^2}=\sqrt{9 a^2 \sin ^2(\theta) \cos ^2(\theta)}=3 a \sin (\theta) \cos (\theta)=\frac{3}{2} \mathrm{a} \sin (2 \theta) \\ s=\int_0^{\pi / 2} \frac{3 a}{2} \sin (2 \theta) d \theta=-\frac{3}{4}[\cos (\pi)-\cos (0)]=\frac{3 a}{2} \end{gathered} $$ Total length $=4 * \frac{3 a}{2}=6 a$
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