University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 160: 101


See below for detailed proof.

Work Step by Step

As $v$ is inversely proportional to $\sqrt s$, a formula for $v$ would have this form: $$v=\frac{n}{\sqrt s}=ns^{-1/2}$$ with $n$ being a constant. Acceleration $a$ is the derivative of $v$ with respect to $t$: $$a=\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}$$ (The Chain Rule) The derivative of $s$ with respect to $t$ equals $v$, so $$\frac{ds}{dt}=v=ns^{-1/2}$$ The derivative of $v$ with respect to $s$ equals $v'$, so $$\frac{dv}{ds}=(ns^{-1/2})'=-\frac{1}{2}ns^{-3/2}$$ Therefore, $$a=-\frac{1}{2}ns^{-3/2}ns^{-1/2}=-\frac{n^2s^{-2}}{2}=-\frac{n^2}{2s^2}$$ As a result, $a$ is inversely proportional to $s^2$.
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