University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 81



Work Step by Step

$$r\cos2s+\sin^2s=\pi$$ To find $dr/ds$, we would use the methods of implicit differentiation. Differentiate both sides of the equation with respect to $s$: $$\Big(\frac{dr}{ds}\cos2s+r\frac{d(\cos2s)}{ds}\Big)+\frac{d(\sin^2s)}{ds}=\frac{d\pi}{ds}$$ $$\cos2s\frac{dr}{ds}+r(-\sin2s)\frac{d(2s)}{ds}+2\sin s\frac{d(\sin s)}{ds}=0$$ $$\cos2s\frac{dr}{ds}-2r\sin2s+2\sin s\cos s=0$$ $$\cos2s\frac{dr}{ds}-2r\sin2s+\sin2s=0$$ $$\cos2s\frac{dr}{ds}+\sin2s(1-2r)=0$$ Next, separate the elements with $dr/ds$ and those without it into 2 sides of the equation: $$\cos2s\frac{dr}{ds}=-\sin2s(1-2r)$$ $$\cos2s\frac{dr}{ds}=\sin2s(2r-1)$$ Then calculate for $dr/ds$: $$\frac{dr}{ds}=\frac{\sin2s(2r-1)}{\cos2s}$$ $$\frac{dr}{ds}=\tan2s(2r-1)$$
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