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: 82



Work Step by Step

$$2rs-r-s+s^2=-3$$ To find $dr/ds$, we would use the methods of implicit differentiation. Differentiate both sides of the equation with respect to $s$: $$2\Big(s\frac{dr}{ds}+r\frac{ds}{ds}\Big)-\frac{dr}{ds}-\frac{ds}{ds}+\frac{d(s^2)}{ds}=\frac{d(-3)}{ds}$$ $$2\Big(s\frac{dr}{ds}+r\Big)-\frac{dr}{ds}-1+2s=0$$ $$2s\frac{dr}{ds}+2r-\frac{dr}{ds}-1+2s=0$$ Next, separate the elements with $dr/ds$ and those without it into 2 sides of the equation: $$2s\frac{dr}{ds}-\frac{dr}{ds}=1-2s-2r$$ $$\frac{dr}{ds}(2s-1)=1-2s-2r$$ Then calculate for $dr/ds$: $$\frac{dr}{ds}=\frac{1-2s-2r}{2s-1}$$
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