Answer
At $t=0$, the osculating circle can be parametrized by
${\bf{o}}\left( \theta \right) = \left( {1 - \cos \theta ,0,\sin \theta } \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {\frac{{{t^2}}}{2},\frac{{{t^3}}}{3},t} \right)$. The derivatives are ${\bf{r}}'\left( t \right) = \left( {t,{t^2},1} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( {1,2t,0} \right)$.
Step 1. Find the radius of the osculating circle
We compute the curvature by using Eq. (3) of Theorem 1:
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
t&{{t^2}}&1\\
1&{2t}&0
\end{array}} \right|$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = - 2t{\bf{i}} + {\bf{j}} + {t^2}{\bf{k}}$
So, $||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)|| = \sqrt {4{t^2} + 1 + {t^4}} $.
$\kappa \left( t \right) = \frac{{\sqrt {{t^4} + 4{t^2} + 1} }}{{{{\left( {\sqrt {\left( {t,{t^2},1} \right)\cdot\left( {t,{t^2},1} \right)} } \right)}^3}}}$
$\kappa \left( t \right) = \frac{{\sqrt {{t^4} + 4{t^2} + 1} }}{{{{\left( {\sqrt {{t^2} + {t^4} + 1} } \right)}^3}}} = \frac{{\sqrt {{t^4} + 4{t^2} + 1} }}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}}$
At $t=0$, $\kappa \left( 0 \right) = 1$. Since $R = \frac{1}{\kappa }$, at $t=0$ the osculating circle has radius $1$.
Step 2. Find the normal vector
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}}$
${\bf{T}}\left( t \right) = \frac{{\left( {t,{t^2},1} \right)}}{{\sqrt {\left( {t,{t^2},1} \right)\cdot\left( {t,{t^2},1} \right)} }} = \frac{{\left( {t,{t^2},1} \right)}}{{\sqrt {{t^2} + {t^4} + 1} }}$
${\bf{T}}\left( t \right) = \left( {\frac{t}{{\sqrt {{t^4} + {t^2} + 1} }},\frac{{{t^2}}}{{\sqrt {{t^4} + {t^2} + 1} }},\frac{1}{{\sqrt {{t^4} + {t^2} + 1} }}} \right)$
To find the derivative of ${\bf{T}}\left( t \right)$:
Evaluate $\frac{d}{{dt}}\left( {\frac{t}{{\sqrt {{t^4} + {t^2} + 1} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{t}{{\sqrt {{t^4} + {t^2} + 1} }}} \right) = \frac{{\sqrt {{t^4} + {t^2} + 1} - \frac{{t\left( {4{t^3} + 2t} \right)}}{{2\sqrt {{t^4} + {t^2} + 1} }}}}{{{t^4} + {t^2} + 1}}$
$\frac{d}{{dt}}\left( {\frac{t}{{\sqrt {{t^4} + {t^2} + 1} }}} \right) = \frac{{{t^4} + {t^2} + 1 - 2{t^4} - {t^2}}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}}$
$\frac{d}{{dt}}\left( {\frac{t}{{\sqrt {{t^4} + {t^2} + 1} }}} \right) = \frac{{1 - {t^4}}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}}$
Evaluate $\frac{d}{{dt}}\left( {\frac{{{t^2}}}{{\sqrt {{t^4} + {t^2} + 1} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{{{t^2}}}{{\sqrt {{t^4} + {t^2} + 1} }}} \right) = \frac{{2t\sqrt {{t^4} + {t^2} + 1} - \frac{{{t^2}\left( {4{t^3} + 2t} \right)}}{{2\sqrt {{t^4} + {t^2} + 1} }}}}{{{t^4} + {t^2} + 1}}$
$\frac{d}{{dt}}\left( {\frac{{{t^2}}}{{\sqrt {{t^4} + {t^2} + 1} }}} \right) = \frac{{2t\left( {{t^4} + {t^2} + 1} \right) - 2{t^5} - {t^3}}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}}$
$\frac{d}{{dt}}\left( {\frac{{{t^2}}}{{\sqrt {{t^4} + {t^2} + 1} }}} \right) = \frac{{{t^3} + 2t}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}}$
Evaluate $\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {{t^4} + {t^2} + 1} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {{t^4} + {t^2} + 1} }}} \right) = \frac{d}{{dt}}\left( {{{\left( {{t^4} + {t^2} + 1} \right)}^{ - 1/2}}} \right)$
$\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {{t^4} + {t^2} + 1} }}} \right) = - \frac{{4{t^3} + 2t}}{{2{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}} = - \frac{{2{t^3} + t}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}}$
Thus, the derivative of ${\bf{T}}\left( t \right)$ is
${\bf{T}}'\left( t \right) = \left( {\frac{{1 - {t^4}}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}},\frac{{{t^3} + 2t}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}}, - \frac{{2{t^3} + t}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}}} \right)$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{{{\left( {1 - {t^4}} \right)}^2} + {{\left( {{t^3} + 2t} \right)}^2} + {{\left( {2{t^3} + t} \right)}^2}}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^3}}}$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{1 - 2{t^4} + {t^8} + {t^6} + 4{t^4} + 4{t^2} + 4{t^6} + 4{t^4} + {t^2}}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^3}}}$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{{t^8} + 5{t^6} + 6{t^4} + 5{t^2} + 1}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^3}}} = \frac{{\left( {{t^4} + 4{t^2} + 1} \right)\left( {{t^4} + {t^2} + 1} \right)}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^3}}}$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{\left( {{t^4} + 4{t^2} + 1} \right)}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^2}}}$
$||{\bf{T}}'\left( t \right)|| = \frac{{\sqrt {{t^4} + 4{t^2} + 1} }}{{\left( {{t^4} + {t^2} + 1} \right)}}$
The normal vector at $t$ is
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = \frac{{\left( {{t^4} + {t^2} + 1} \right)}}{{\sqrt {{t^4} + 4{t^2} + 1} }}\left( {\frac{{1 - {t^4}}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}},\frac{{{t^3} + 2t}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}}, - \frac{{2{t^3} + t}}{{{{\left( {{t^4} + {t^2} + 1} \right)}^{3/2}}}}} \right)$
${\bf{N}}\left( t \right) = \frac{1}{{\sqrt {{t^4} + 4{t^2} + 1} }}\left( {\frac{{1 - {t^4}}}{{\sqrt {{t^4} + {t^2} + 1} }},\frac{{{t^3} + 2t}}{{\sqrt {{t^4} + {t^2} + 1} }}, - \frac{{2{t^3} + t}}{{\sqrt {{t^4} + {t^2} + 1} }}} \right)$
At $t=0$, we get ${\bf{N}}\left( 0 \right) = \left( {1,0,0} \right)$.
Step 3. Find the the center of the osculating circle
By Eq. (9), the center of the osculating circle is given by
$\overrightarrow {OQ} = {\bf{r}}\left( {{t_0}} \right) + \frac{1}{{{\kappa _P}}}{\bf{N}}$
At $t=0$, the center of curvature is
$\overrightarrow {OQ} = {\bf{r}}\left( 0 \right) + \frac{1}{{\kappa \left( 0 \right)}}{\bf{N}}\left( 0 \right)$
$\overrightarrow {OQ} = \left( {0,0,0} \right) + 1\left( {1,0,0} \right) = \left( {1,0,0} \right)$
Step 4. Parametrize the osculating circle
Since the osculating plane is spanned by ${\bf{T}}\left( t \right)$ and ${\bf{N}}\left( t \right)$, the osculating circle of radius $R$ can be parametrized by
${\bf{o}}\left( \theta \right) = \overrightarrow {OQ} + R\cos \theta \left( { - {\bf{N}}\left( t \right)} \right) + R\sin \theta \left( {{\bf{T}}\left( t \right)} \right)$,
for $0 \le \theta \le 2\pi $, where $\overrightarrow {OQ} $ is the center of the osculating circle and ${\bf{N}}$ is the normal vector that points toward the center of the circle.
At $t=0$, the osculating circle has radius $R=1$ and center $\left( {1,0,0} \right)$. So, it can be parametrized by
${\bf{o}}\left( \theta \right) = \left( {1,0,0} \right) - {\bf{N}}\left( 0 \right)\cos \theta + {\bf{T}}\left( 0 \right)\sin \theta $
${\bf{o}}\left( \theta \right) = \left( {1,0,0} \right) - \left( {1,0,0} \right)\cos \theta + \left( {0,0,1} \right)\sin \theta $
${\bf{o}}\left( \theta \right) = \left( {1 - \cos \theta ,0,\sin \theta } \right)$
