Answer
$$s = 3\sqrt {10} + \ln \left( {3 + \sqrt {10} } \right)$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = {t^2}{\bf{i}} + 2t{\bf{k}},{\text{ }}\left[ {0,3} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{t^2}{\bf{i}} + 2t{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = 2t{\bf{i}} + 2{\bf{k}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( {2t} \right)}^2} + {{\left( 2 \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {4{t^2} + 4} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = 2\sqrt {{t^2} + 1} \cr
& {\text{Find the arc length }}s{\text{, using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^3 {2\sqrt {{t^2} + 1} } dt \cr
& s = 2\int_0^3 {\sqrt {{t^2} + 1} } dt \cr
& {\text{Integrate by tables formula 26}}{\text{.}} \cr
& \int {\sqrt {{u^2} + {a^2}} du} = \frac{1}{2}\left( {u\sqrt {{u^2} + {a^2}} + {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right) + C \cr
& s = \left[ {t\sqrt {{t^2} + 1} + \ln \left| {t + \sqrt {{t^2} + 1} } \right|} \right]_0^3 \cr
& s = \left[ {3\sqrt {{3^2} + 1} + \ln \left| {3 + \sqrt {{3^2} + 1} } \right|} \right] - \left[ {0 + \ln \left| 1 \right|} \right] \cr
& s = 3\sqrt {10} + \ln \left( {3 + \sqrt {10} } \right) \cr
& {\text{Graph}} \cr} $$
