Answer
$\kappa(t)= \dfrac{2t^{-3}}{(t^{-4}+1)^{3/2}}$
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:
$r'(t) =\lt -t^{-2}, 0, 1 \gt$ and $r''(t) =\lt 2t^{-3}, 0, 0 \gt$
Thus,
$\kappa(t) = \dfrac{||2t^{-3} j ||}{||\lt -t^{-2}, 0, 1 \gt||^3} \\=\dfrac{2t^{-3}}{[(-t^{-2})^2+0^2+1^2]^{3/2}}$
Therefore,
$\kappa(t)= \dfrac{2t^{-3}}{(t^{-4}+1)^{3/2}}$
