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An Archimedean spiral has radius linear in polar angle, $r=a+b\theta$. For the special parametrization $(u\cos u,u\sin u)$, the curvature of a plane curve is $(u^2+2)/(1+u^2{)}^{3/2}$. Its parameter speed is $\sqrt{1+u^2}$, so it remains a regular curve at the origin.

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 ${\mathbf{e}}_r^{\prime }={\mathbf{e}}_{\theta }$ and ${\mathbf{e}}_{\theta }^{\prime }=-{\mathbf{e}}_r$. ThenTherefore the curvature of a plane curve isThe parametrization is regular even at $u=0$, where the curvature of a plane curve is $2$.

For the parametrized curve $\mathbf{x}(u)=(u,f(u))$, the curvature of a plane curve isFor the upper semicircle $f(u)=\sqrt{a^2-u^2}$,and therefore $\kappa =1/a$, as required for a circle of radius $a$.