= Solution
This <plane curve> is an <Archimedean spiral>. Put $\mathbf e_r=(\cos u,\sin u,0)$ and $\mathbf e_\theta=(-\sin u,\cos u,0)$, so $\mathbf e_r'=\mathbf e_\theta$ and $\mathbf e_\theta'=-\mathbf e_r$. Then
$$
\mathbf x'=\mathbf e_r+u\mathbf e_\theta,\qquad
\mathbf x''=2\mathbf e_\theta-u\mathbf e_r,\qquad
|\mathbf x'|^2=1+u^2,\qquad
\mathbf x'\times\mathbf x''=(u^2+2)\mathbf e_z.
$$
Therefore the <curvature of a plane curve> is
$$
\boxed{\kappa(u)=\frac{u^2+2}{(1+u^2)^{3/2}},\qquad u\ge0.}
$$
The parametrization is regular even at $u=0$, where the <curvature of a plane curve> is $2$.
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