Answer
$K=0$
Work Step by Step
\[\begin{align}
& \mathbf{r}\left( t \right)={{t}^{2}}\mathbf{i}+\mathbf{j},\text{ at }t=2 \\
& \text{We use the Theorem which states}\\
& \text{If }C\text{ is a smooth curve given by }\mathbf{r}\left( t \right),\text{ then the curvature }K\text{ of} \\
& C\text{ at }t\text{ is }K=\frac{\left\| \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right) \right\|}{{{\left\| \mathbf{r}'\left( t \right) \right\|}^{3}}} \\
& \mathbf{r}\left( t \right)={{t}^{2}}\mathbf{i}+\mathbf{j} \\
& \text{Differentiating} \\
& \mathbf{r}'\left( t \right)=2t\mathbf{i} \\
& \mathbf{r}''\left( t \right)=2\mathbf{i} \\
& \text{Find the cross product and then use the curvature formula} \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left| \begin{matrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2t & 0 & 0 \\
2 & 0 & 0 \\
\end{matrix} \right| \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left| \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right|\mathbf{i}-\left| \begin{matrix}
2t & 0 \\
2 & 0 \\
\end{matrix} \right|\mathbf{j}+\left| \begin{matrix}
2t & 0 \\
2 & 0 \\
\end{matrix} \right|\mathbf{k} \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=0 \\
& \text{Therefore}\text{,} \\
& K=\frac{\left\| \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right) \right\|}{{{\left\| \mathbf{r}'\left( t \right) \right\|}^{3}}}=\frac{0}{{{\left\| 2t\mathbf{i} \right\|}^{3}}} \\
& K=0 \\
\end{align}\]
