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: 100


The proof is detailed below.

Work Step by Step

$$v(s)=k\sqrt s=ks^{1/2}(m/sec)$$ ($k$ is a constant) Acceleration $a$ is the derivative of velocity $v$ with respect to time $t$. In other words, $$a(t)=\frac{dv}{dt}$$ Applying the Chain Rule: $$a(t)=\frac{dv}{ds}\frac{ds}{dt}$$ Now $ds/dt$, or the derivative of position $s$ with respect to time $t$, is the velocity $v$. $$a(t)=\frac{dv}{ds}\times v=\frac{dv}{ds}\times ks^{1/2}$$ On the other hand, $dv/ds$ is the derivative of $v(s)$ here: $$\frac{dv}{ds}=(ks^{1/2})'=\frac{1}{2}ks^{-1/2}$$ Therefore, $$a(t)=\frac{1}{2}ks^{-1/2}\times ks^{1/2}=\frac{k^2}{2}(m/sec^2)$$ As $k$ is a constant, $a(t)$ is a constant as a result.
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