Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 10 - Parametric Equations and Polar Coordinates - 10.3 Exercises - Page 688: 74

Answer

Graph estimate: $0.77,$ Exact value: $\displaystyle \frac{4\sqrt{3}}{9}$

Work Step by Step

Estimating from the graph, the highest points have $y \approx 0.77$. For the exact value, we solve $\displaystyle \frac{dy}{d\theta}=0$. $y =r\sin\theta=\sin\theta\sin 2\theta \qquad(*)$ $\displaystyle \frac{dy}{d\theta} =2\sin\theta\cos 2\theta+\cos\theta\sin 2\theta$ ... use the double angle identities .. $=2\sin\theta(2\cos^{2}\theta-1)+\cos\theta(2\sin\theta\cos\theta)$ $=2\sin\theta(2\cos^{2}\theta-1)+2\sin\theta\cos^{2}\theta$ $=2\sin\theta(3\cos^{2}\theta-1)$ In the first quadrant, $3\cos^{2}\theta-1$= $0$ when $\displaystyle \cos\theta=\frac{1}{\sqrt{3}} \qquad $ $\sin\theta=\sqrt{1-\cos^{2}\theta}=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}} \Leftrightarrow$ We go back to (*) $ y=\sin\theta\sin 2\theta$ $=\sin\theta$($2\sin\theta\cos\theta)$ $=2\sin^{2}\theta\cos\theta$ $=2\displaystyle \cdot\frac{2}{3}\cdot\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{4\sqrt{3}}{9}$ $(\approx 0.77)$.
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