Answer
$$L = 5$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {3{t^2} - 1,4{t^2} + 5} \right\rangle ,\,\,\,\,\,{\text{for }}0 \leqslant t \leqslant 1 \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {3{t^2} - 1,4{t^2} + 5} \right\rangle \cr
& {\text{differentiating}} \cr
& {\bf{r}}'\left( t \right) = \left\langle {6t,8t} \right\rangle \cr
& {\text{use the Definition of Arc Length for Vector Functions }}\left( {{\text{see page 832}}} \right) \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& \,{\text{for }}0 \leqslant t \leqslant 1 \to a = 0{\text{ and }}b = 2.{\text{ then}} \cr
& L = \int_0^1 {\left| {\left\langle {6t,8t} \right\rangle } \right|} dt \cr
& L = \int_0^1 {\sqrt {{{\left( {6t} \right)}^2} + {{\left( {8t} \right)}^2}} } dt \cr
& {\text{simplifying}} \cr
& L = \int_0^1 {\sqrt {36{t^2} + 64{t^2}} } dt \cr
& L = \int_0^1 {\sqrt {100{t^2}} } dt \cr
& L = \int_0^1 {10t} dt \cr
& {\text{integrating}} \cr
& L = \left( {\frac{{10{t^2}}}{2}} \right)_0^1 \cr
& L = 5\left( {{t^2}} \right)_0^1 \cr
& {\text{evaluating}} \cr
& L = 5\left( {{1^2} - 0} \right) \cr
& L = 5\left( 1 \right) \cr
& L = 5 \cr} $$
