## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 14 - Calculus of Vector-Valued Functions - 14.3 Arc Length and Speed - Exercises - Page 725: 13

#### Answer

$v(1)=\sqrt{2}$.

#### Work Step by Step

Using the fact that $(\ln t)'=\frac{1}{t}$ and using the chain rule, we have $r'(t) =\lt 1, \frac{1}{t}, \frac{2\ln t}{t}\gt$. Hence the speed $v(t)$ is given by $$v(t)=\|r'(t)\|=\sqrt{1+ \frac{1}{t^2}+\frac{4(\ln t)^2}{t^2}} .$$ At $t=1$, since $\ln 1=0$, we get that $v(1)=\sqrt{2}$.

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