Answer
$$\rho = \frac{{\sqrt 2 }}{2}$$
Work Step by Step
$$\eqalign{
& x = \sin t,\,\,\,\,y = \cos t,\,\,\,\,z = \frac{1}{2}{t^2};\,\,\,\,\,\,\,t = 0 \cr
& {\text{Let }}{\bf{r}}\left( t \right) = x\left( t \right){\bf{i}} + y\left( t \right){\bf{j}} + z\left( t \right){\bf{k}} \cr
& {\text{Then}} \cr
& {\bf{r}}\left( t \right) = \sin t{\bf{i}} + \cos t{\bf{j}} + \frac{1}{2}{t^2}{\bf{k}} \cr
& {\text{Calculate the derivatives }}{\bf{r}}'\left( t \right){\text{ and }}{\bf{r}}{\text{''}}\left( t \right){\text{ and evaluate at }}t = 0 \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\sin t} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {\cos t} \right]{\bf{j}} + \frac{d}{{dt}}\left[ {\frac{1}{2}{t^2}} \right]{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = \cos t{\bf{i}} - \sin t{\bf{j}} + t{\bf{k}} \cr
& {\bf{r}}'\left( 0 \right) = \cos \left( 0 \right){\bf{i}} - \sin \left( 0 \right){\bf{j}} + \left( 0 \right){\bf{k}} \cr
& {\bf{r}}'\left( 0 \right) = {\bf{i}} \cr
& and \cr
& {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ {\cos t} \right]{\bf{i}} + \frac{d}{{dt}}\left[ { - \sin t} \right]{\bf{j}} + \frac{d}{{dt}}\left[ t \right]{\bf{k}} \cr
& {\bf{r}}''\left( t \right) = - \sin t{\bf{i}} - \cos t{\bf{j}} + {\bf{k}} \cr
& {\bf{r}}''\left( 0 \right) = - \sin \left( 0 \right){\bf{i}} - \cos \left( 0 \right){\bf{j}} + {\bf{k}} \cr
& {\bf{r}}''\left( 0 \right) = - {\bf{j}} + {\bf{k}} \cr
& \cr} $$
\[\begin{gathered}
{\text{Calculate the cross product }}{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) \hfill \\
{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
1&0&0 \\
1&{ - 1}&1
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = \left| {\begin{array}{*{20}{c}}
0&0 \\
{ - 1}&1
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
1&0 \\
1&1
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
1&0 \\
1&{ - 1}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = \left( 0 \right){\mathbf{i}} - \left( {1 - 0} \right){\mathbf{j}} + \left( { - 1 - 0} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = - {\mathbf{j}} - {\mathbf{k}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\text{Use the formula }}\left( 3 \right)\,\,\,\kappa \left( t \right) = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}}.{\text{ Thus}}{\text{, at }}t = 0 \cr
& \kappa \left( 0 \right) = \frac{{\left\| { - {\bf{j}} - {\bf{k}}} \right\|}}{{{{\left\| i \right\|}^3}}} \cr
& \kappa \left( 0 \right) = \frac{{\sqrt {1 + 1} }}{{{{\left( {\sqrt 1 } \right)}^3}}} \cr
& \kappa \left( 0 \right) = \sqrt 2 \cr
& \cr
& {\text{Calculate the radius of curvature }}\rho {\text{ at }}t = 0 \cr
& \rho = \frac{1}{{\kappa \left( 0 \right)}} = \frac{1}{{\sqrt 2 }} \cr
& \rho = \frac{1}{{\sqrt 2 }}\left( {\frac{{\sqrt 2 }}{{\sqrt 2 }}} \right) \cr
& \rho = \frac{{\sqrt 2 }}{2} \cr} $$
