Answer
The center of curvature at $t=1$:
$\overrightarrow {OQ} = \left( { - \frac{{11}}{2},\frac{{16}}{3}} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{t^2},{t^3}} \right)$. The first and the second derivatives are ${\bf{r}}'\left( t \right) = \left( {2t,3{t^2}} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( {2,6t} \right)$, respectively.
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}}$
${\bf{T}}\left( t \right) = \frac{{\left( {2t,3{t^2}} \right)}}{{\sqrt {\left( {2t,3{t^2}} \right)\cdot\left( {2t,3{t^2}} \right)} }} = \frac{1}{{\sqrt {4{t^2} + 9{t^4}} }}\left( {2t,3{t^2}} \right)$
${\bf{T}}\left( t \right) = \left( {\frac{2}{{\sqrt {4 + 9{t^2}} }},\frac{{3t}}{{\sqrt {4 + 9{t^2}} }}} \right)$
To find the derivative of ${\bf{T}}\left( t \right)$:
1. Evaluate $\frac{d}{{dt}}\left( {\frac{2}{{\sqrt {4 + 9{t^2}} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{2}{{\sqrt {4 + 9{t^2}} }}} \right) = \frac{d}{{dt}}\left( {2{{\left( {4 + 9{t^2}} \right)}^{ - 1/2}}} \right)$
$\frac{d}{{dt}}\left( {\frac{2}{{\sqrt {4 + 9{t^2}} }}} \right) = - \frac{{18t}}{{{{\left( {4 + 9{t^2}} \right)}^{3/2}}}}$
2. Evaluate $\frac{d}{{dt}}\left( {\frac{{3t}}{{\sqrt {4 + 9{t^2}} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{{3t}}{{\sqrt {4 + 9{t^2}} }}} \right) = \frac{{3\sqrt {4 + 9{t^2}} - \left( {3t} \right)\left( {\frac{{18t}}{{2\sqrt {4 + 9{t^2}} }}} \right)}}{{4 + 9{t^2}}}$
$\frac{d}{{dt}}\left( {\frac{{3t}}{{\sqrt {4 + 9{t^2}} }}} \right) = \frac{{3\left( {4 + 9{t^2}} \right) - 27{t^2}}}{{{{\left( {4 + 9{t^2}} \right)}^{3/2}}}} = \frac{{12}}{{{{\left( {4 + 9{t^2}} \right)}^{3/2}}}}$
Thus,
${\bf{T}}'\left( t \right) = \left( { - \frac{{18t}}{{{{\left( {4 + 9{t^2}} \right)}^{3/2}}}},\frac{{12}}{{{{\left( {4 + 9{t^2}} \right)}^{3/2}}}}} \right)$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{{{18}^2}{t^2} + {{12}^2}}}{{{{\left( {4 + 9{t^2}} \right)}^3}}} = \frac{{36\left( {4 + 9{t^2}} \right)}}{{{{\left( {4 + 9{t^2}} \right)}^3}}} = \frac{{36}}{{{{\left( {4 + 9{t^2}} \right)}^2}}}$
$||{\bf{T}}'\left( t \right)|| = \frac{6}{{4 + 9{t^2}}}$
The normal vector at $t$ is
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = \left( {\frac{{4 + 9{t^2}}}{6}} \right)\left( { - \frac{{18t}}{{{{\left( {4 + 9{t^2}} \right)}^{3/2}}}},\frac{{12}}{{{{\left( {4 + 9{t^2}} \right)}^{3/2}}}}} \right)$
${\bf{N}}\left( t \right) = \left( { - \frac{{3t}}{{\sqrt {4 + 9{t^2}} }},\frac{2}{{\sqrt {4 + 9{t^2}} }}} \right)$
Next, 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,
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{2t}&{3{t^2}}&0\\
2&{6t}&0
\end{array}} \right|$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left( {12{t^2} - 6{t^2}} \right){\bf{k}} = 6{t^2}{\bf{k}}$
So, $||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)|| = 6{t^2}$.
$\kappa \left( t \right) = \frac{{6{t^2}}}{{{{\left( {\sqrt {\left( {2t,3{t^2}} \right)\cdot\left( {2t,3{t^2}} \right)} } \right)}^3}}} = \frac{{6{t^2}}}{{{{\left( {4{t^2} + 9{t^4}} \right)}^{3/2}}}}$
By Eq. (9), the center of the osculating circle is given by
$\overrightarrow {OQ} = {\bf{r}}\left( t \right) + \frac{1}{{{\kappa _P}}}{\bf{N}}$
$\overrightarrow {OQ} $ is also called the center of curvature of ${\bf{r}}\left( t \right)$.
At $t=1$, the center of curvature is
$\overrightarrow {OQ} = {\bf{r}}\left( 1 \right) + \frac{1}{{\kappa \left( 1 \right)}}{\bf{N}}\left( 1 \right)$
$\overrightarrow {OQ} = \left( {1,1} \right) + \frac{{{{13}^{3/2}}}}{6}\left( { - \frac{3}{{\sqrt {13} }},\frac{2}{{\sqrt {13} }}} \right)$
$\overrightarrow {OQ} = \left( {1,1} \right) + \left( { - \frac{{13}}{2},\frac{{13}}{3}} \right) = \left( { - \frac{{11}}{2},\frac{{16}}{3}} \right)$
