## Calculus (3rd Edition)

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