Answer
Using Eq. (9), the center of the osculating circle at $\left( {{x_0},{x_0}^2} \right)$ is given by
$\overrightarrow {OQ} = \left( { - 4{x_0}^3,\frac{1}{2} + 3{x_0}^2} \right)$.
Work Step by Step
We have $f\left( x \right) = {x^2}$. The plane curve can be parametrized by setting $x=t$. So, ${\bf{r}}\left( t \right) = \left( {t,{t^2}} \right)$.
1. Evaluate the curvature
By Eq. (5), the curvature at the point $\left( {t,f\left( t \right)} \right)$ is
$\kappa \left( t \right) = \frac{{\left| {f{\rm{''}}\left( t \right)} \right|}}{{{{\left( {1 + f'{{\left( t \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( t \right) = \frac{{\left| 2 \right|}}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}} = \frac{2}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}}$
2. Evaluate the derivative of ${\bf{T}}\left( t \right)$
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}}$
${\bf{T}}\left( t \right) = \frac{{\left( {1,2t} \right)}}{{\sqrt {\left( {1,2t} \right)\cdot\left( {1,2t} \right)} }} = \frac{1}{{\sqrt {1 + 4{t^2}} }}\left( {1,2t} \right)$
${\bf{T}}\left( t \right) = \left( {\frac{1}{{\sqrt {1 + 4{t^2}} }},\frac{{2t}}{{\sqrt {1 + 4{t^2}} }}} \right)$
To find the derivative of ${\bf{T}}\left( t \right)$:
Evaluate $\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {1 + 4{t^2}} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {1 + 4{t^2}} }}} \right) = \frac{d}{{dt}}\left( {{{\left( {1 + 4{t^2}} \right)}^{ - 1/2}}} \right)$
$\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {1 + 4{t^2}} }}} \right) = - \frac{{8t}}{{2{{\left( {1 + 4{t^2}} \right)}^{3/2}}}} = - \frac{{4t}}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}}$
Evaluate $\frac{d}{{dt}}\left( {\frac{{2t}}{{\sqrt {1 + 4{t^2}} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{{2t}}{{\sqrt {1 + 4{t^2}} }}} \right) = \frac{{2\sqrt {1 + 4{t^2}} - \left( {2t} \right)\left( {\frac{{8t}}{{2\sqrt {1 + 4{t^2}} }}} \right)}}{{1 + 4{t^2}}}$
$\frac{d}{{dt}}\left( {\frac{{2t}}{{\sqrt {1 + 4{t^2}} }}} \right) = \frac{{2\left( {1 + 4{t^2}} \right) - 8{t^2}}}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}} = \frac{2}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}}$
Thus,
${\bf{T}}'\left( t \right) = \left( { - \frac{{4t}}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}},\frac{2}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}}} \right)$
$||{\bf{T}}'\left( t \right)|{|^2} = \left( { - \frac{{4t}}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}},\frac{2}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}}} \right)\cdot\left( { - \frac{{4t}}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}},\frac{2}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}}} \right)$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{16{t^2}}}{{{{\left( {1 + 4{t^2}} \right)}^3}}} + \frac{4}{{{{\left( {1 + 4{t^2}} \right)}^3}}} = \frac{{4 + 16{t^2}}}{{{{\left( {1 + 4{t^2}} \right)}^3}}}$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{4\left( {1 + 4{t^2}} \right)}}{{{{\left( {1 + 4{t^2}} \right)}^3}}} = \frac{4}{{{{\left( {1 + 4{t^2}} \right)}^2}}}$
$||{\bf{T}}'\left( t \right)|| = \frac{2}{{1 + 4{t^2}}}$
3. Evaluate the normal vector
The normal vector at $t$ is
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = \left( {\frac{{1 + 4{t^2}}}{2}} \right)\left( { - \frac{{4t}}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}},\frac{2}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}}} \right)$
${\bf{N}}\left( t \right) = \left( { - \frac{{2t}}{{\sqrt {1 + 4{t^2}} }},\frac{1}{{\sqrt {1 + 4{t^2}} }}} \right)$
By Eq. (9), the center of the osculating circle is given by
$\overrightarrow {OQ} = {\bf{r}}\left( {{t_0}} \right) + \frac{1}{{{\kappa _P}}}{\bf{N}}$
In this exercise, we have ${t_0} = {x_0}$ and ${\bf{r}}\left( {{x_0}} \right) = \left( {{x_0},{x_0}^2} \right)$. So,
$\overrightarrow {OQ} = \left( {{x_0},{x_0}^2} \right) + \frac{{{{\left( {1 + 4{x_0}^2} \right)}^{3/2}}}}{2}\left( { - \frac{{2{x_0}}}{{\sqrt {1 + 4{x_0}^2} }},\frac{1}{{\sqrt {1 + 4{x_0}^2} }}} \right)$
$\overrightarrow {OQ} = \left( {{x_0},{x_0}^2} \right) + \left( { - {x_0}\left( {1 + 4{x_0}^2} \right),\frac{1}{2}\left( {1 + 4{x_0}^2} \right)} \right)$
$\overrightarrow {OQ} = \left( {{x_0},{x_0}^2} \right) + \left( { - {x_0} - 4{x_0}^3,\frac{1}{2} + 2{x_0}^2} \right)$
Hence, $\overrightarrow {OQ} = \left( { - 4{x_0}^3,\frac{1}{2} + 3{x_0}^2} \right)$.
