Answer
$$K = \frac{3}{{25}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 4t{\bf{i}} + 3\cos t{\bf{j}} + 3\sin t{\bf{k}} \cr
& {\text{By Theorem 12}}{\text{.8 }} \cr
& {\text{If }}C{\text{ is a smooth curve given by }}{\bf{r}}\left( t \right),{\text{ then the curvature }}K{\text{ of}} \cr
& C{\text{ at }}t{\text{ is }}K = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}} \cr
& {\bf{r}}\left( t \right) = 4t{\bf{i}} + 3\cos t{\bf{j}} + 3\sin t{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = 4{\bf{i}} - 3\sin t{\bf{j}} + 3\cos t{\bf{k}} \cr
& {\bf{r}}''\left( t \right) = 0{\bf{i}} - 3\cos t{\bf{j}} - 3\sin t{\bf{k}} \cr} $$
\[\begin{gathered}
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
4&{ - 3\sin t}&{3\cos t} \\
0&{ - 3\cos t}&{ - 3\sin t}
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{ - 3\sin t}&{3\cos t} \\
{ - 3\cos t}&{ - 3\sin t}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
4&{3\cos t} \\
0&{ - 3\sin t}
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
4&{ - 3\sin t} \\
0&{ - 3\cos t}
\end{array}} \right|{\mathbf{k}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) = \left( {9{{\sin }^2}t + 9{{\cos }^2}t} \right){\bf{i}} + 12\sin t{\bf{j}} - 12\cos t{\bf{k}} \cr
& {\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) = 9{\bf{i}} + 12\sin t{\bf{j}} - 12\cos t{\bf{k}} \cr
& \left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\| = \sqrt {{9^2} + {{12}^2}{{\sin }^2}t + {{12}^2}{{\cos }^2}t} \cr
& \left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\| = \sqrt {225} = 15 \cr
& and \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \left\| {4{\bf{i}} - 3\sin t{\bf{j}} + 3\cos t{\bf{k}}} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{4^2} + 9{{\sin }^2}t + 9{{\cos }^2}t} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {16 + 9} = 5 \cr
& {\text{Therefore,}} \cr
& K = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}} = \frac{{15}}{{{{\left( 5 \right)}^3}}} \cr
& K = \frac{3}{{25}} \cr} $$
