Answer
At $t = \frac{\pi }{4}$, the osculating circle has radius $1$ and centered at $\left( {0,0} \right)$. Thus, the osculating circle can be parametrized by $\left( {\cos t,\sin t} \right)$. Notice that this parametrization coincides with ${\bf{r}}\left( t \right)$.
Work Step by Step
1. Find the radius of the osculating circle
We have ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t} \right)$. The first and the second derivatives are ${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( { - \cos t, - \sin t} \right)$, respectively.
We compute the curvature by using Eq. (3) of Theorem 1:
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
To apply Theorem 1, we treat ${\bf{r}}'\left( t \right)$ and ${\bf{r}}{\rm{''}}\left( t \right)$ as vectors in ${\mathbb{R}^3}$ by setting the $z$-components equal to zero. Thus,
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{ - \sin t}&{\cos t}&0\\
{ - \cos t}&{ - \sin t}&0
\end{array}} \right|$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left( {{{\sin }^2}t + {{\cos }^2}t} \right){\bf{k}} = {\bf{k}}$
So, $||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)|| = 1$.
$\kappa \left( t \right) = \frac{1}{{{{\left( {\sqrt {\left( { - \sin t,\cos t} \right)\cdot\left( { - \sin t,\cos t} \right)} } \right)}^3}}} = 1$
Since $R = \frac{1}{\kappa }$, so the osculating circle has radius $1$.
2. Find the normal vector
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}}$
${\bf{T}}\left( t \right) = \frac{{\left( { - \sin t,\cos t} \right)}}{{\sqrt {\left( { - \sin t,\cos t} \right)\cdot\left( { - \sin t,\cos t} \right)} }} = \left( { - \sin t,\cos t} \right)$
The derivative of ${\bf{T}}\left( t \right)$ is
${\bf{T}}'\left( t \right) = \left( { - \cos t, - \sin t} \right)$
The normal vector at $t$ is
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = \frac{{\left( { - \cos t, - \sin t} \right)}}{{\sqrt {\left( { - \cos t, - \sin t} \right)\cdot\left( { - \cos t, - \sin t} \right)} }} = \left( { - \cos t, - \sin t} \right)$
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 $t = \frac{\pi }{4}$, the center of curvature is
$\overrightarrow {OQ} = {\bf{r}}\left( {\frac{\pi }{4}} \right) + \frac{1}{{\kappa \left( {\frac{\pi }{4}} \right)}}{\bf{N}}\left( {\frac{\pi }{4}} \right)$
$\overrightarrow {OQ} = \left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right) + \left( { - \frac{1}{2}\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right) = \left( {0,0} \right)$
4. Parametrize the osculating circle
The osculating circle has radius $1$ and centered at $\left( {0,0} \right)$. Thus, the osculating circle can be parametrized by $\left( {\cos t,\sin t} \right)$. Notice that this parametrization coincides with ${\bf{r}}\left( t \right)$.
