Answer
The curvature at $t=1$:
$\kappa \left( 1 \right) = \frac{{n\left( {n - 1} \right)}}{{{{\left( {1 + {n^2}} \right)}^{3/2}}}}$
Work Step by Step
We have $y = {t^n}$.
Write $y = f\left( t \right) = {t^n}$. So, $f'\left( t \right) = n{t^{n - 1}}$ and $f{\rm{''}}\left( t \right) = n\left( {n - 1} \right){t^{n - 2}}$.
By Eq. (5) of Theorem 2, the curvature is given by
$\kappa \left( t \right) = \frac{{\left| {f{\rm{''}}\left( t \right)} \right|}}{{{{\left( {1 + f'{{\left( t \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( t \right) = \frac{{\left| {n\left( {n - 1} \right){t^{n - 2}}} \right|}}{{{{\left( {1 + {n^2}{t^{2n - 2}}} \right)}^{3/2}}}}$
The curvature at $t=1$:
$\kappa \left( 1 \right) = \frac{{n\left( {n - 1} \right)}}{{{{\left( {1 + {n^2}} \right)}^{3/2}}}}$
