## Calculus (3rd Edition)

$v(3) =\sqrt{\frac{10}{9}}$.
Since $r(t) =\lt e^{t-3},12, 3t^{-1}\gt$, then by taking the derivative for each component, we have $r'(t) =\lt e^{t-3},0, -\frac{3}{t^2}\gt$. Hence the speed $v(t)$ is gievn by $$v(t)=\|r'(t)\|=\sqrt{e^{2t-6}+ \frac{9}{t^4} } .$$ At $t=3$, $v(3)=\sqrt{1+\frac{1}{9} }=\sqrt{\frac{10}{9}}$.