Answer
$$\rho = \frac{5}{2}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 3\cos t{\bf{i}} + 4\sin t{\bf{j}} + t{\bf{k}};\,\,\,\,\,\,\,t = \frac{\pi }{2} \cr
& {\text{Calculate the derivatives }}{\bf{r}}'\left( t \right){\text{ and }}{\bf{r}}{\text{''}}\left( t \right){\text{ and evaluate at }}t = \frac{\pi }{2} \cr
& {\bf{r}}'\left( t \right) = - 3\sin t{\bf{i}} + 4\cos t{\bf{j}} + {\bf{k}} \cr
& {\bf{r}}'\left( {\frac{\pi }{2}} \right) = - 3\sin \left( {\frac{\pi }{2}} \right){\bf{i}} + 4\cos \left( {\frac{\pi }{2}} \right){\bf{j}} + {\bf{k}} \cr
& {\bf{r}}'\left( {\frac{\pi }{2}} \right) = - 3{\bf{i}} + {\bf{k}} \cr
& and \cr
& {\bf{r}}''\left( t \right) = - 3\cos t{\bf{i}} - 4\sin t{\bf{j}} \cr
& {\bf{r}}''\left( {\frac{\pi }{2}} \right) = - 3\cos \left( {\frac{\pi }{2}} \right){\bf{i}} - 4\sin \left( {\frac{\pi }{2}} \right){\bf{j}} \cr
& {\bf{r}}''\left( {\frac{\pi }{2}} \right) = - 4{\bf{j}} \cr} $$
\[\begin{gathered}
{\text{Calculate the cross product }}{\mathbf{r}}'\left( {\frac{\pi }{2}} \right) \times {\mathbf{r}}''\left( {\frac{\pi }{2}} \right) \hfill \\
{\mathbf{r}}'\left( {\frac{\pi }{2}} \right) \times {\mathbf{r}}''\left( {\frac{\pi }{2}} \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{ - 3}&0&1 \\
0&{ - 4}&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( {\frac{\pi }{2}} \right) \times {\mathbf{r}}''\left( {\frac{\pi }{2}} \right) = \left| {\begin{array}{*{20}{c}}
0&1 \\
{ - 4}&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 3}&1 \\
0&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{ - 3}&0 \\
0&{ - 4}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( {\frac{\pi }{2}} \right) \times {\mathbf{r}}''\left( {\frac{\pi }{2}} \right) = 4{\mathbf{i}} + 0{\mathbf{j}} + 12{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( {\frac{\pi }{2}} \right) \times {\mathbf{r}}''\left( {\frac{\pi }{2}} \right) = 4{\mathbf{i}} + 12{\mathbf{k}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\text{Use the formula }}\left( 3 \right)\,\,\,\kappa \left( t \right) = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}}.{\text{ Thus}}{\text{, at }}t = \frac{\pi }{2} \cr
& \kappa \left( {\frac{\pi }{2}} \right) = \frac{{\left\| {4{\bf{i}} + 12{\bf{k}}} \right\|}}{{{{\left\| { - 3{\bf{i}} + {\bf{k}}} \right\|}^3}}} \cr
& \kappa \left( {\frac{\pi }{2}} \right) = \frac{{\sqrt {16 + 144} }}{{{{\left( {\sqrt {9 + 1} } \right)}^3}}} \cr
& \kappa \left( {\frac{\pi }{2}} \right) = \frac{{\sqrt {160} }}{{{{\left( {\sqrt {10} } \right)}^3}}} \cr
& \kappa \left( {\frac{\pi }{2}} \right) = \frac{{4\sqrt {10} }}{{{{\left( {\sqrt {10} } \right)}^{3/2}}}} \cr
& \kappa \left( {\frac{\pi }{2}} \right) = \frac{4}{{10}} = \frac{2}{5} \cr
& \cr
& {\text{Calculate the radius of curvature }}\rho {\text{ at }}t = \frac{\pi }{2} \cr
& \rho = \frac{1}{{\kappa \left( {\pi /2} \right)}} = \frac{1}{{2/5}} \cr
& \rho = \frac{5}{2} \cr} $$
