Answer
Using Eq. (3) we prove that
$\kappa \left( t \right) = \frac{{\left| {x'\left( t \right)y{\rm{''}}\left( t \right) - x{\rm{''}}\left( t \right)y'\left( t \right)} \right|}}{{{{\left( {x'{{\left( t \right)}^2} + y'{{\left( t \right)}^2}} \right)}^{3/2}}}}$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$.
By Eq. (3) of Theorem 1, the curvature is given by
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
We treat ${\bf{r}}\left( t \right)$ as a curve in ${\mathbb{R}^3}$ by setting the $z$-component equal to zero. Thus, ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right),0} \right)$. The derivatives are
${\bf{r}}'\left( t \right) = \left( {x'\left( t \right),y'\left( t \right),0} \right)$
${\bf{r}}{\rm{''}}\left( t \right) = \left( {x{\rm{''}}\left( t \right),y{\rm{''}}\left( t \right),0} \right)$
Omitting the parameter $t$ for convenience, we evaluate
${\bf{r}}' \times {\bf{r}}{\rm{''}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{x'}&{y'}&0\\
{x{\rm{''}}}&{y{\rm{''}}}&0
\end{array}} \right|$
${\bf{r}}' \times {\bf{r}}{\rm{''}} = \left( {x'y{\rm{''}} - x{\rm{''}}y'} \right){\bf{k}}$
So,
$||{\bf{r}}' \times {\bf{r}}{\rm{''}}|| = \sqrt {\left( {x'y{\rm{''}} - x{\rm{''}}y'} \right)\cdot\left( {x'y{\rm{''}} - x{\rm{''}}y'} \right)} $
$||{\bf{r}}' \times {\bf{r}}{\rm{''}}|| = \left| {x'y{\rm{''}} - x{\rm{''}}y'} \right|$
Next, we evaluate $||{\bf{r}}'||$
$||{\bf{r}}'|| = \sqrt {\left( {x',y',0} \right)\cdot\left( {x',y',0} \right)} $
$||{\bf{r}}'|| = \sqrt {{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}} $
Substituting $||{\bf{r}}' \times {\bf{r}}{\rm{''}}||$ and $||{\bf{r}}'||$ in $\kappa$ gives
$\kappa = \frac{{\left| {x'y{\rm{''}} - x{\rm{''}}y'} \right|}}{{{{\left( {{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}} \right)}^{3/2}}}}$
Hence,
(11) ${\ \ \ }$ $\kappa \left( t \right) = \frac{{\left| {x'\left( t \right)y{\rm{''}}\left( t \right) - x{\rm{''}}\left( t \right)y'\left( t \right)} \right|}}{{{{\left( {x'{{\left( t \right)}^2} + y'{{\left( t \right)}^2}} \right)}^{3/2}}}}$
