Answer
The curvature at $t = \pi $ is $\kappa \left( \pi \right) = \frac{\pi }{{2\sqrt 2 }}$
Work Step by Step
We have $\left( {x\left( t \right),y\left( t \right)} \right) = \left( {t\cos t,\sin t} \right)$.
The derivatives are
$\left( {x'\left( t \right),y'\left( t \right)} \right) = \left( {\cos t - t\sin t,\cos t} \right)$
$\left( {x{\rm{''}}\left( t \right),y{\rm{''}}\left( t \right)} \right) = \left( { - \sin t - \sin t - t\cos t, - \sin t} \right)$
$\left( {x{\rm{''}}\left( t \right),y{\rm{''}}\left( t \right)} \right) = \left( { - 2\sin t - t\cos t, - \sin t} \right)$
Using Eq. (11) we compute the curvature
$\kappa \left( t \right) = \frac{{\left| {x'\left( t \right)y{\rm{''}}\left( t \right) - x{\rm{''}}\left( t \right)y'\left( t \right)} \right|}}{{{{\left( {x'{{\left( t \right)}^2} + y'{{\left( t \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( t \right) = \frac{{\left| {\left( {\cos t - t\sin t} \right)\left( { - \sin t} \right) - \left( { - 2\sin t - t\cos t} \right)\cos t} \right|}}{{{{\left( {{{\left( {\cos t - t\sin t} \right)}^2} + {{\cos }^2}t} \right)}^{3/2}}}}$
$\kappa \left( t \right) = \frac{{\left| { - \cos t\sin t + t{{\sin }^2}t + 2\sin t\cos t + t{{\cos }^2}t} \right|}}{{{{\left( {{{\cos }^2}t - 2t\cos t\sin t + {t^2}{{\sin }^2}t + {{\cos }^2}t} \right)}^{3/2}}}}$
$\kappa \left( t \right) = \frac{{\left| {\cos t\sin t + t} \right|}}{{{{\left( {2{{\cos }^2}t - 2t\cos t\sin t + {t^2}{{\sin }^2}t} \right)}^{3/2}}}}$
The curvature at $t = \pi $ is
$\kappa \left( \pi \right) = \frac{\pi }{{{{\left( 2 \right)}^{3/2}}}} = \frac{\pi }{{2\sqrt 2 }}$
