Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Practice Exercises - Page 776: 3


highest speed $=1$

Work Step by Step

$v(t)=\dfrac{dr}{dt}=\dfrac{-it+j}{(t^2+1)^{3/2}}$ and $|v(t)|=|\dfrac{-it+j}{(t^2+1)^{3/2}}|=\dfrac{1}{t^2+1}$ and $\dfrac{d|v(t)|}{dt}=\dfrac{-2t}{(t^2+1)^2}$ Set $\dfrac{d|v(t)|}{dt}=\dfrac{-2t}{(t^2+1)^2}=0$ For $t \gt 0$, we can see that $ \dfrac{-2t}{(t^2+1)^2} \lt 0$. This implies that the maximum occurs at $t=0$ with highest speed $=1$.
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