Answer
$$L = 28$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = {t^3}{\bf{i}} + t{\bf{j}} + \frac{1}{2}\sqrt 6 {t^2}{\bf{k}};\,\,\,\,\,\,\,\,\,\,\,\,1 \leqslant t \leqslant 3 \cr
& \cr
& {\text{Calculate }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{t^3}{\bf{i}} + t{\bf{j}} + \frac{1}{2}\sqrt 6 {t^2}{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{t^3}} \right]{\bf{i}} + \left[ t \right]{\bf{j}} + \left[ {\frac{1}{2}\sqrt 6 {t^2}} \right]{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = 3{t^2}{\bf{i}} + {\bf{j}} + \sqrt 6 t{\bf{k}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \left\| {3{t^2}{\bf{i}} + {\bf{j}} + \sqrt 6 t{\bf{k}}} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( {3{t^2}} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( {\sqrt 6 t} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {9{t^4} + 1 + 6{t^2}} \cr
& \cr
& {\text{Find the arc length using }}L = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& {\text{for the interval }}1 \leqslant t \leqslant 3,\,\,{\text{ }}a = 1{\text{ and }}b = 3.{\text{ Then}}{\text{,}} \cr
& L = \int_1^3 {\sqrt {9{t^4} + 1 + 6{t^2}} } dt \cr
& {\text{factoring}} \cr
& L = \int_1^3 {\sqrt {{{\left( {3{t^2} + 1} \right)}^2}} } dt \cr
& L = \int_1^3 {\left( {3{t^2} + 1} \right)} dt \cr
& \cr
& {\text{Integrate and evaluate}} \cr
& L = \left[ {{t^3} + t} \right]_1^3 \cr
& L = \left[ {{{\left( 3 \right)}^3} + 3} \right] - \left[ {{{\left( 1 \right)}^3} + 1} \right] \cr
& L = 30 - 2 \cr
& L = 28 \cr} $$
