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