Answer
$\dfrac{\sqrt {\pi^2+5}}{(1+\pi^2)^{3/2}}$
Work Step by Step
The curvature $\kappa$ for a plane curve system is:
$\kappa (t)= \dfrac{||r'(t) \times r''(t)||}{||r'(t)||^3}$
We have:
$r'(t) =\lt -\sin t, \cos t, 2t \gt$ and $r''(t) =\lt -\cos t ,-\sin t, 2 \gt$
Thus,
$\kappa(t) = \dfrac{||2(\cos t+t\sin t)i-2(t \cos t -\sin t )j+k|}{||\lt -\sin t, \cos t, 2t \gt||^3} \\=\dfrac{\sqrt {4(\cos t+t\sin t)^2+4(t \cos t -\sin t )^2+1}}{[\sin^2t+\cos^2 t+4t^2]^{3/2}}$
Now, we will compute $\kappa (t)$ at $t=\dfrac{\pi}{2}$
Therefore,
$\kappa(t)= \dfrac{\sqrt {4 (\dfrac{\pi^2}{4})+4(1)+1}}{(1+\pi^2)^{3/2}}=\dfrac{\sqrt {\pi^2+5}}{(1+\pi^2)^{3/2}}$
