Answer
$$ \mathbf{T}=\frac{3}{5 } \mathbf{i} +\frac{4}{5} \mathbf{j} $$
Work Step by Step
Given $$\mathbf{r}= t^3\mathbf{i} +2t^2\mathbf{j} $$
Since
$$ \mathbf{r'}= 3t^2\mathbf{i} +4t\mathbf{j}$$
and $$ \| \mathbf{r'} \|=\sqrt{ 9t^4+16t^2} =t\sqrt{9t^2+16}$$
Then unit tangent vector given by
\begin{align*}
\mathbf{T}(t)&=\frac{\mathbf{r'}(t)}{\|\mathbf{r'}(t)\|}\\
&=\frac{3t^2\mathbf{i} +4t\mathbf{j}}{t\sqrt{9t^2+16}}\\
&=\frac{3t\mathbf{i} +4\mathbf{j}}{\sqrt{9t^2+16}}
\end{align*}
At $t=1$
$$ \mathbf{T}=\frac{3}{5 } \mathbf{i} +\frac{4}{5} \mathbf{j} $$
