Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 720: 37


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Work Step by Step

Since $r(t)=\lt t,1,1 \gt$, then $r'(t)=\lt 1,0,0 \gt$. $$\|r(t)\|=\sqrt{2+t^2}, \quad \|r'(t)\|=1$$ and moreover $$\|r(t)\|'=\frac{d}{dt} \|r(t)\|=\frac{t}{\sqrt{2+t^2}}.$$ That is $$\|r(t)\|'\neq \|r'(t)\|.$$
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