Answer
At $t=0$, the osculating circle can be parametrized by
${\bf{o}}\left( t \right) = \left( {3 - 2\cos \theta ,\sqrt 2 \sin \theta ,\sqrt 2 \sin \theta } \right)$
Work Step by Step
Step 1. Find the radius of the osculating circle
We have ${\bf{r}}\left( t \right) = \left( {\cosh t,\sinh t,t} \right)$. The derivatives are ${\bf{r}}'\left( t \right) = \left( {\sinh t,\cosh t,1} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( {\cosh t,\sinh t,0} \right)$.
Evaluate $||{\bf{r}}'\left( 0 \right) \times {\bf{r}}{\rm{''}}\left( 0 \right)||$
${\bf{r}}'\left( 0 \right) \times {\bf{r}}{\rm{''}}\left( 0 \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
0&1&1\\
1&0&0
\end{array}} \right|$
${\bf{r}}'\left( 0 \right) \times {\bf{r}}{\rm{''}}\left( 0 \right) = {\bf{j}} - {\bf{k}}$
$||{\bf{r}}'\left( 0 \right) \times {\bf{r}}{\rm{''}}\left( 0 \right)|| = \sqrt 2 $
We compute the curvature by using Eq. (3) of Theorem 1:
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
At $t=0$, we get
$\kappa \left( 0 \right) = \frac{{||{\bf{r}}'\left( 0 \right) \times {\bf{r}}{\rm{''}}\left( 0 \right)||}}{{||{\bf{r}}'\left( 0 \right)|{|^3}}}$
$\kappa \left( 0 \right) = \frac{{\sqrt 2 }}{{{{\left( {\sqrt 2 } \right)}^3}}} = \frac{1}{2}$
Since $R = \frac{1}{\kappa }$, so the osculating circle has radius $2$.
Step 2. Find the normal vector
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}}$
${\bf{T}}\left( t \right) = \frac{{\left( {\sinh t,\cosh t,1} \right)}}{{\sqrt {\left( {\sinh t,\cosh t,1} \right)\cdot\left( {\sinh t,\cosh t,1} \right)} }}$
${\bf{T}}\left( t \right) = \frac{1}{{\sqrt {{{\sinh }^2}t + {{\cosh }^2}t + 1} }}\left( {\sinh t,\cosh t,1} \right)$
Since ${\cosh ^2}t - {\sinh ^2}t = 1$, so
${\bf{T}}\left( t \right) = \frac{1}{{\sqrt {2{{\cosh }^2}t} }}\left( {\sinh t,\cosh t,1} \right) = \frac{1}{{\sqrt 2 }}\left( {\tanh t,1,secht} \right)$
The derivative of ${\bf{T}}\left( t \right)$ is
${\bf{T}}'\left( t \right) = \frac{1}{{\sqrt 2 }}\left( {sec{h^2}t,0, - secht\tanh t} \right)$
So,
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{1}{2}\left( {sec{h^4}t + sec{h^2}t{{\tanh }^2}t} \right)$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{1}{2}sec{h^2}t\left( {sec{h^2}t + {{\tanh }^2}t} \right)$
Since $sec{h^2}t + {\tanh ^2}t = 1$, so
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{1}{2}sec{h^2}t$
$||{\bf{T}}'\left( t \right)|| = \frac{1}{{\sqrt 2 }}secht$
The normal vector at $t$ is
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
At $t=0$, we get
${\bf{N}}\left( 0 \right) = \frac{{\frac{1}{{\sqrt 2 }}\left( {1,0,0} \right)}}{{\sqrt {\frac{1}{2}\left( {1,0,0} \right)\cdot\left( {1,0,0} \right)} }} = \left( {1,0,0} \right)$
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
$\overrightarrow {OQ} = {\bf{r}}\left( 0 \right) + \frac{1}{{\kappa \left( 0 \right)}}{\bf{N}}\left( 0 \right)$
$\overrightarrow {OQ} = \left( {1,0,0} \right) + 2\left( {1,0,0} \right) = \left( {3,0,0} \right)$
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 $\left( {3,0,0} \right)$. So, it can be parametrized by
${\bf{o}}\left( t \right) = \left( {3,0,0} \right) - 2\cos \theta {\bf{N}}\left( 0 \right) + 2\sin \theta {\bf{T}}\left( 0 \right)$
${\bf{o}}\left( t \right) = \left( {3,0,0} \right) - 2\cos \theta \left( {1,0,0} \right) + \frac{2}{{\sqrt 2 }}\sin \theta \left( {0,1,1} \right)$
${\bf{o}}\left( t \right) = \left( {3 - 2\cos \theta ,\sqrt 2 \sin \theta ,\sqrt 2 \sin \theta } \right)$
