Answer
At $t=0$, the osculating circle can be parametrized by
${\bf{o}}\left( t \right) = \left( {2\cos \theta - 1,\sqrt 2 \sin \theta ,\sqrt 2 \sin \theta } \right)$
Work Step by Step
Step 1. Find the radius of the osculating circle
We have ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,t} \right)$. The first and the second derivatives are ${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t,1} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( { - \cos t, - \sin t,0} \right)$, respectively.
We compute the curvature by using Eq. (3) of Theorem 1:
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
Evaluate ${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)$:
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{ - \sin t}&{\cos t}&1\\
{ - \cos t}&{ - \sin t}&0
\end{array}} \right|$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \sin t{\bf{i}} - \cos t{\bf{j}} + \left( {{{\sin }^2}t + {{\cos }^2}t} \right){\bf{k}}$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \sin t{\bf{i}} - \cos t{\bf{j}} + {\bf{k}}$
So,
$||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)|| = \sqrt {{{\sin }^2}t + {{\left( { - \cos t} \right)}^2} + 1} = \sqrt 2 $
$\kappa \left( t \right) = \frac{{\sqrt 2 }}{{{{\left( {\sqrt {\left( { - \sin t,\cos t,1} \right)\cdot\left( { - \sin t,\cos t,1} \right)} } \right)}^3}}} = \frac{{\sqrt 2 }}{{{{\left( {\sqrt 2 } \right)}^3}}} = \frac{1}{2}$
Since $R = \frac{1}{\kappa }$, so the osculating circle has radius $2$.
Step 2. Find the normal vector
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}}$
${\bf{T}}\left( t \right) = \frac{{\left( { - \sin t,\cos t,1} \right)}}{{\sqrt {\left( { - \sin t,\cos t,1} \right)\cdot\left( { - \sin t,\cos t,1} \right)} }} = \frac{1}{{\sqrt 2 }}\left( { - \sin t,\cos t,1} \right)$
The derivative of ${\bf{T}}\left( t \right)$ is
${\bf{T}}'\left( t \right) = \frac{1}{{\sqrt 2 }}\left( { - \cos t, - \sin t,0} \right)$
So, $||{\bf{T}}'\left( t \right)|| = \frac{1}{{\sqrt 2 }}$.
The normal vector at $t$ is
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = \frac{{\left( { - \cos t, - \sin t,0} \right)}}{{\sqrt {\left( { - \cos t, - \sin t,0} \right)\cdot\left( { - \cos t, - \sin t,0} \right)} }} = \left( { - \cos t, - \sin t,0} \right)$
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
$\overrightarrow {OQ} = {\bf{r}}\left( 0 \right) + \frac{1}{{\kappa \left( 0 \right)}}{\bf{N}}\left( 0 \right)$
$\overrightarrow {OQ} = \left( {1,0,0} \right) + 2\left( { - 1,0,0} \right) = \left( { - 1,0,0} \right)$
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 $R=2$ and center $\left( { - 1,0,0} \right)$. So, it can be parametrized by
${\bf{o}}\left( t \right) = \left( { - 1,0,0} \right) - 2\cos \theta {\bf{N}}\left( 0 \right) + 2\sin \theta {\bf{T}}\left( 0 \right)$
${\bf{o}}\left( t \right) = \left( { - 1,0,0} \right) - 2\cos \theta \left( { - 1,0,0} \right) + \frac{2}{{\sqrt 2 }}\sin \theta \left( {0,1,1} \right)$
${\bf{o}}\left( t \right) = \left( {2\cos \theta - 1,\sqrt 2 \sin \theta ,\sqrt 2 \sin \theta } \right)$
