Answer
$$L = 64\sqrt {30} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {2{t^3}, - {t^3},5{t^3}} \right\rangle {\text{, for }}0 \leqslant t \leqslant 4 \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {2{t^3}, - {t^3},5{t^3}} \right\rangle \cr
& {\text{differentiating}} \cr
& {\bf{r}}'\left( t \right) = \left\langle {6{t^2}, - 3{t^2},15{t^2}} \right\rangle \cr
& {\text{The speed is given by}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \left| {\left\langle {6{t^2}, - 3{t^2},15{t^2}} \right\rangle } \right| \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{\left( {6{t^2}} \right)}^2} + {{\left( { - 3{t^2}} \right)}^2} + {{\left( {15{t^2}} \right)}^2}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {36{t^4} + 9{t^4} + 225{t^4}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {270{t^4}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = 3\sqrt {30} {t^2} \cr
& {\text{Find the length of the trajectory}} \cr
& {\text{Use the Definition of Arc Length for Vector Functions }} \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& L = \int_0^4 {3\sqrt {30} {t^2}} dt \cr
& {\text{simplifying}} \cr
& L = \sqrt {30} \left[ {{t^3}} \right]_0^4 \cr
& L = \sqrt {30} \left( {{4^3} - 0} \right) \cr
& L = 64\sqrt {30} \cr} $$
