Answer
$${\bf{T}}\left( 1 \right) = \frac{1}{{\sqrt {10} }}{\bf{i}} + \frac{3}{{\sqrt {10} }}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 3t{\bf{i}} + 3{t^3}{\bf{j}},{\text{ }}t = 1 \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {3t{\bf{i}} + 3{t^3}{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = 3{\bf{i}} + 9{t^2}{\bf{j}} \cr
& {\text{Evaluate at }}t = 1 \cr
& {\bf{r}}'\left( 1 \right) = 3{\bf{i}} + 9{\left( 1 \right)^2}{\bf{j}} \cr
& {\bf{r}}'\left( 1 \right) = 3{\bf{i}} + 9{\bf{j}} \cr
& {\text{The unit tangent vector }}{\bf{T}}\left( t \right){\text{ at }}t{\text{ is defined as}} \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{\left\| {{\bf{r}}{\text{'}}\left( t \right)} \right\|}} \cr
& {\text{at }}t = 1 \cr
& {\bf{T}}\left( 1 \right) = \frac{{{\bf{r}}'\left( 1 \right)}}{{\left\| {{\bf{r}}{\text{'}}\left( 1 \right)} \right\|}} \cr
& {\bf{T}}\left( 1 \right) = \frac{{3{\bf{i}} + 9{\bf{j}}}}{{\left\| {3{\bf{i}} + 9{\bf{j}}} \right\|}} = \frac{{3{\bf{i}} + 9{\bf{j}}}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 9 \right)}^2}} }} = \frac{{3{\bf{i}} + 9{\bf{j}}}}{{\sqrt {90} }} \cr
& {\bf{T}}\left( 1 \right) = \frac{{3{\bf{i}}}}{{\sqrt {90} }} + \frac{{9{\bf{j}}}}{{\sqrt {90} }} \cr
& {\bf{T}}\left( 1 \right) = \frac{1}{{\sqrt {10} }}{\bf{i}} + \frac{3}{{\sqrt {10} }}{\bf{j}} \cr} $$
