Answer
$$ \mathbf{T}=\frac{1}{\sqrt{2} } \mathbf{i} +\frac{1}{\sqrt{2} } \mathbf{j} $$
Work Step by Step
Given $$\mathbf{r}= t^2\mathbf{i} +2t\mathbf{j} $$
Since
$$ \mathbf{r'}= 2t\mathbf{i} +2\mathbf{j}$$
and $$ \| \mathbf{r'} \|=\sqrt{ 4t^2+4} =2\sqrt{t^2+1}$$
Then unit tangent vector given by
\begin{align*}
\mathbf{T}(t)&=\frac{\mathbf{r'}(t)}{\|\mathbf{r'}(t)\|}\\
&=\frac{2t\mathbf{i} +2\mathbf{j}}{2\sqrt{t^2+1}}\\
&=\frac{ t\mathbf{i} + \mathbf{j}}{ \sqrt{t^2+1} }
\end{align*}
At $t=1$
$$ \mathbf{T}=\frac{1}{\sqrt{2} } \mathbf{i} +\frac{1}{\sqrt{2} } \mathbf{j} $$
