Answer
$$L = 4\sqrt {194} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {5\cos {t^2},5\sin {t^2},12{t^2}} \right\rangle {\text{, for }}0 \leqslant t \leqslant 2 \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {5\cos {t^2},5\sin {t^2},12{t^2}} \right\rangle \cr
& {\text{differentiating}} \cr
& {\bf{r}}'\left( t \right) = \left\langle { - 10t\sin {t^2},10t\cos {t^2},24t} \right\rangle \cr
& {\text{The speed is given by}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \left| {\left\langle { - 10t\sin {t^2},10t\cos {t^2},24t} \right\rangle } \right| \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{\left( { - 10t\sin {t^2}} \right)}^2} + {{\left( {10t\cos {t^2}} \right)}^2} + {{\left( {24t} \right)}^2}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {100{t^2}{{\sin }^2}{t^2} + 100{t^2}{{\cos }^2}{t^2} + 576{t^2}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {200{t^2} + 576{t^2}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {776{t^2}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = 2\sqrt {194} t \cr
& {\text{Find the length of the trajectory}} \cr
& {\text{Use the Definition of Arc Length for Vector Functions }} \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& L = \int_0^2 {2\sqrt {194} t} dt \cr
& {\text{simplifying}} \cr
& L = \sqrt {194} \left[ {{t^2}} \right]_0^2 \cr
& L = \sqrt {194} \left( {4 - 0} \right) \cr
& L = 4\sqrt {194} \cr} $$
