Answer
Using Eq. (6), we show that the normal vector is given by
${\bf{N}}\left( t \right) = \frac{{v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)}}{{||v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)||}}$
Work Step by Step
Let ${\bf{r}}\left( t \right)$ be the parametrization for the curve. The tangent vector is ${\bf{r}}'\left( t \right)$. So, the unit tangent vector is given by ${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}}$. Since ${\rm{v}}\left( t \right) = ||{\bf{r}}'\left( t \right)||$, so, ${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{v\left( t \right)}}$.
The derivative of ${\bf{T}}\left( t \right)$ is
${\bf{T}}'\left( t \right) = \frac{{v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)}}{{v{{\left( t \right)}^2}}}$
We can write
$v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right) = v{\left( t \right)^2}{\bf{T}}'\left( t \right)$
Since $v{\left( t \right)^2}$ is always positive, thus $v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)$ is a positive multiple of ${\bf{T}}'\left( t \right)$.
By Eq. (6), the normal vector ${\bf{N}}\left( t \right)$ is given by
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = \left( {\frac{{v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)}}{{v{{\left( t \right)}^2}}}} \right)\left( {\frac{{v{{\left( t \right)}^2}}}{{||v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)||}}} \right)$
Hence, ${\bf{N}}\left( t \right) = \frac{{v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)}}{{||v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)||}}$
