Answer
The length of path:
$s = 2\sqrt {13} $
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {\sin 2t,\cos 2t,3t - 1} \right)$, for $1 \le t \le 3$. So,
${\bf{r}}'\left( t \right) = \left( {2\cos 2t, - 2\sin 2t,3} \right)$
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {2\cos 2t, - 2\sin 2t,3} \right)\cdot\left( {2\cos 2t, - 2\sin 2t,3} \right)} $
$ = \sqrt {4{{\cos }^2}2t + 4{{\sin }^2}2t + 9} $
$ = \sqrt {13} $
By Theorem 1 of Section 14.3, the length of path is given by
$s = \mathop \smallint \limits_1^3 ||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$s = \sqrt {13} \mathop \smallint \limits_1^3 {\rm{d}}t = 2\sqrt {13} $
