Answer
\[\kappa = \frac{2}{{{{\left( {4{t^2} + 1} \right)}^{3/2}}}}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = \left\langle {4 + {t^2},t,0} \right\rangle \hfill \\
{\text{Calculate }}{\mathbf{v}}\left( t \right){\text{ and }}\left| {{\mathbf{v}}\left( t \right)} \right| \hfill \\
{\mathbf{v}}\left( t \right) = {\mathbf{r}}'\left( t \right) \hfill \\
{\mathbf{v}}\left( t \right) = \frac{d}{{dt}}\left\langle {4 + {t^2},t,0} \right\rangle \hfill \\
{\mathbf{v}}\left( t \right) = \left\langle {2t,1,0} \right\rangle \hfill \\
\left| {{\mathbf{v}}\left( t \right)} \right| = \sqrt {{{\left( {2t} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 0 \right)}^2}} \hfill \\
\left| {{\mathbf{v}}\left( t \right)} \right| = \sqrt {4{t^2} + 1} \hfill \\
\hfill \\
{\text{Calculate }}{\mathbf{a}}\left( t \right) \hfill \\
{\mathbf{a}}\left( t \right) = {\mathbf{v}}'\left( t \right) \hfill \\
{\mathbf{a}}\left( t \right) = \frac{d}{{dt}}\left\langle {2t,1,0} \right\rangle \hfill \\
{\mathbf{a}}\left( t \right) = \left\langle {2,0,0} \right\rangle \hfill \\
\hfill \\
{\text{Calculate }}\left| {{\mathbf{v}}\left( t \right) \times {\mathbf{a}}\left( t \right)} \right| \hfill \\
{\mathbf{v}}\left( t \right) \times {\mathbf{a}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{2t}&1&0 \\
2&0&0
\end{array}} \right| \hfill \\
{\mathbf{v}}\left( t \right) \times {\mathbf{a}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
1&0 \\
0&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{2t}&0 \\
2&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{2t}&1 \\
2&0
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{v}}\left( t \right) \times {\mathbf{a}}\left( t \right) = 0{\mathbf{i}} + 0{\mathbf{j}} - 2{\mathbf{k}} \hfill \\
\left| {{\mathbf{v}}\left( t \right) \times {\mathbf{a}}\left( t \right)} \right| = 2 \hfill \\
\hfill \\
{\text{Use the alternative curvature formula}} \hfill \\
\kappa = \frac{{\left| {{\mathbf{v}}\left( t \right) \times {\mathbf{a}}\left( t \right)} \right|}}{{{{\left| {{\mathbf{v}}\left( t \right)} \right|}^3}}} \hfill \\
\kappa = \frac{2}{{{{\left( {\sqrt {4{t^2} + 1} } \right)}^3}}} \hfill \\
\kappa = \frac{2}{{{{\left( {4{t^2} + 1} \right)}^{3/2}}}} \hfill \\
\end{gathered} \]
