Answer
$\kappa \left( {{t_0}} \right) = \frac{{13}}{{16}}$
Work Step by Step
We have at ${t_0}$, the speed and acceleration vector given by
$v\left( {{t_0}} \right) = 4$ ${\ \ }$ and ${\ \ }$ ${\bf{a}}\left( {{t_0}} \right) = {\bf{r}}{\rm{''}}\left( {{t_0}} \right) = \left( {5,4,12} \right)$
Since at ${t_0}$, the path of a moving particle is tangent to the $y$-axis in the positive direction, so
${\bf{v}}\left( {{t_0}} \right) = {\bf{r}}'\left( {{t_0}} \right) = 4\left( {0,1,0} \right) = \left( {0,4,0} \right)$
By Eq. (3) of Theorem 1 (Section 14.4), the curvature is given by
$\kappa \left( {{t_0}} \right) = \frac{{||{\bf{r}}'\left( {{t_0}} \right) \times {\bf{r}}{\rm{''}}\left( {{t_0}} \right)||}}{{||{\bf{r}}'\left( {{t_0}} \right)|{|^3}}}$
$\kappa \left( {{t_0}} \right) = \frac{{||\left( {0,4,0} \right) \times \left( {5,4,12} \right)||}}{{||\left( {0,4,0} \right)|{|^3}}}$
Evaluate
$\left( {0,4,0} \right) \times \left( {5,4,12} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
0&4&0\\
5&4&{12}
\end{array}} \right| = 48{\bf{i}} - 20{\bf{k}}$
$||\left( {0,4,0} \right) \times \left( {5,4,12} \right)|| = \sqrt {{{48}^2} + {{20}^2}} = 52$
$||\left( {0,4,0} \right)|| = \sqrt {\left( {0,4,0} \right)\cdot\left( {0,4,0} \right)} = 4$
So,
$\kappa \left( {{t_0}} \right) = \frac{{||\left( {0,4,0} \right) \times \left( {5,4,12} \right)||}}{{||\left( {0,4,0} \right)|{|^3}}} = \frac{{52}}{{{4^3}}} = \frac{{13}}{{16}}$
