Answer
$3\sqrt{61}$
Work Step by Step
$r(t) = \langle 3t, 4t-3, 6t+1 \rangle$, so
$r'(t) = \langle 3, 4, 6 \rangle$.
$length=\int_0^3 \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}dt$
$=\int_0^3 \sqrt{3^2 + 4^2 + 6^2}dt$
$=\int_0^3 \sqrt{61}dt$
$=t\sqrt{61}|_0^3$
$=3\sqrt{61}$ - $0\sqrt{61}$
$=3\sqrt{61}$ - $0\sqrt{61}$
$=3\sqrt{61}$
