Answer
The curvature at $t=2$ is $\kappa \left( 2 \right) \simeq 0.012$
Work Step by Step
We have $\left( {x\left( t \right),y\left( t \right)} \right) = \left( {{t^2},{t^3}} \right)$.
The derivatives are
$\left( {x'\left( t \right),y'\left( t \right)} \right) = \left( {2t,3{t^2}} \right)$
$\left( {x{\rm{''}}\left( t \right),y{\rm{''}}\left( t \right)} \right) = \left( {2,6t} \right)$
Using Eq. (11) we compute the curvature
$\kappa \left( t \right) = \frac{{\left| {x'\left( t \right)y{\rm{''}}\left( t \right) - x{\rm{''}}\left( t \right)y'\left( t \right)} \right|}}{{{{\left( {x'{{\left( t \right)}^2} + y'{{\left( t \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( t \right) = \frac{{\left| {12{t^2} - 6{t^2}} \right|}}{{{{\left( {4{t^2} + 9{t^4}} \right)}^{3/2}}}} = \frac{{6{t^2}}}{{{{\left( {4{t^2} + 9{t^4}} \right)}^{3/2}}}}$
The curvature at $t=2$ is
$\kappa \left( 2 \right) = \frac{{6\cdot4}}{{{{\left( {4\cdot4 + 9\cdot16} \right)}^{3/2}}}} = \frac{3}{{80\sqrt {10} }} \simeq 0.012$
