Answer
$v(1)=\sqrt{14}$
Work Step by Step
We have $r'(t) =\lt 1, 2 t, 3t^2\gt$. Hence the speed $v(t)$ is given by
$$v(t)=\|r'(t)\|=\sqrt{1+4t^2+9t^4} . $$ At $t=1$, $v(1)=\sqrt{14}$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 14 - Calculus of Vector-Valued Functions - 14.3 Arc Length and Speed - Exercises - Page 725: 12
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