Answer
$$L \approx 32.50$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {t,4{t^2},10} \right\rangle {\text{, for }} - 2 \leqslant t \leqslant 2 \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {t,4{t^2},10} \right\rangle \cr
& {\bf{r}}'\left( t \right) = \left\langle {1,8t,0} \right\rangle \cr
& {\text{The speed is given by}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \left| {\left\langle {1,8t,0} \right\rangle } \right| \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( {8t} \right)}^2} + {{\left( 0 \right)}^2}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {1 + 64{t^2}} \cr
& {\text{Find the length of the trajectory}} \cr
& {\text{Use the Definition of Arc Length for Vector Functions }} \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& L = \int_{ - 2}^2 {\sqrt {1 + 64{t^2}} } dt \cr
& {\text{Calculate using a computer or calculator}} \cr
& L \approx 32.50 \cr} $$
