Answer
$$L = 26\pi $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {13\sin 2t,12\cos 2t,5\cos 2t} \right\rangle {\text{, for }}0 \leqslant t \leqslant \pi \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {13\sin 2t,12\cos 2t,5\cos 2t} \right\rangle \cr
& {\text{differentiating}} \cr
& {\bf{r}}'\left( t \right) = \left\langle {26\cos 2t, - 24\sin 2t, - 10\sin 2t} \right\rangle \cr
& {\text{The speed is given by}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \left| {\left\langle {26\cos 2t, - 24\sin 2t, - 10\sin 2t} \right\rangle } \right| \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{\left( {26\cos 2t} \right)}^2} + {{\left( { - 24\sin 2t} \right)}^2} + {{\left( { - 10\sin 2t} \right)}^2}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{26}^2}{{\cos }^2}2t + {{24}^2}{{\sin }^2}2t + 100{{\sin }^2}2t} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {676{{\cos }^2}2t + 676{{\sin }^2}t} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {676} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = 26 \cr
& {\text{Find the length of the trajectory}} \cr
& {\text{Use the Definition of Arc Length for Vector Functions }} \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& L = \int_0^\pi {26} dt \cr
& {\text{simplifying}} \cr
& L = 26\left( {\pi - 0} \right) \cr
& L = 26\pi \cr} $$
