Answer
\[\frac{3}{2{{t}^{3}}{{\left( \sqrt{9t+1} \right)}^{3}}}\]
Work Step by Step
\[\begin{align}
& \mathbf{r}\left( t \right)=2\sqrt{t}\mathbf{i}+3t\mathbf{j} \\
& \text{By Theorem 12}\text{.8 } \\
& \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)=2\sqrt{t}\mathbf{i}+3t\mathbf{j} \\
& \mathbf{r}'\left( t \right)=\frac{1}{\sqrt{t}}\mathbf{i}+3\mathbf{j} \\
& \mathbf{r}''\left( t \right)=-\frac{1}{2}{{t}^{-3/2}}\mathbf{i}+0\mathbf{j} \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left| \begin{matrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
{{t}^{-1/2}} & 3 & 0 \\
-\frac{1}{2}{{t}^{-3/2}} & 0 & 0 \\
\end{matrix} \right| \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left| \begin{matrix}
3 & 0 \\
3 & 0 \\
\end{matrix} \right|\mathbf{i}-\left| \begin{matrix}
{{t}^{-1/2}} & 0 \\
-\frac{1}{2}{{t}^{-3/2}} & 0 \\
\end{matrix} \right|\mathbf{j}+\left| \begin{matrix}
{{t}^{-1/2}} & 3 \\
-\frac{1}{2}{{t}^{-3/2}} & 0 \\
\end{matrix} \right|\mathbf{k} \\
& \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\frac{3}{2}{{t}^{-3/2}}\mathbf{k} \\
& \left\| \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right) \right\|=\frac{3}{2}{{t}^{-3/2}} \\
& and \\
& \left\| \mathbf{r}'\left( t \right) \right\|=\left\| \frac{1}{\sqrt{t}}\mathbf{i}+3\mathbf{j} \right\| \\
& \left\| \mathbf{r}'\left( t \right) \right\|=\sqrt{{{\left( \frac{1}{\sqrt{t}} \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
& \left\| \mathbf{r}'\left( t \right) \right\|=\sqrt{\frac{1}{t}+9} \\
& \text{Therefore,} \\
& K=\frac{\left\| \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right) \right\|}{{{\left\| \mathbf{r}'\left( t \right) \right\|}^{3}}}=\frac{\frac{3}{2}{{t}^{-3/2}}}{{{\left( \sqrt{\frac{1}{t}+9} \right)}^{3}}} \\
& \text{Simplifying} \\
& K=\frac{\frac{3}{2}{{t}^{-3/2}}}{{{\left( \frac{\sqrt{9t+1}}{{{t}^{1/2}}} \right)}^{3}}}=\frac{3}{2{{t}^{3}}{{\left( \sqrt{9t+1} \right)}^{3}}} \\
\end{align}\]
