Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 3 / 4A (Vector Calculus)

For a regular curve parametrized by $u$, put $v(u)=|{\mathbf{x}}^{\prime }(u)|>0$ and orient arc length in the direction of increasing $u$. Since $ds=|d\mathbf{x}|=vdu$, the unit tangent vector isRegularity is the condition that permits division by $v$.

Differentiate the unit tangent vector with $d/ds=v^{-1}d/du$ and $v^{\prime }=({\mathbf{x}}^{\prime }\cdot {\mathbf{x}}^{\prime \prime })/v$. This givesThe curvature of a space curve is $\kappa =|d\mathbf{t}/ds|$. Squaring the displayed expression and simplifying givesThe last equality is the cross product norm identity. Equivalently, the hinted vector triple product rewrites $d\mathbf{t}/ds=-{\mathbf{x}}^{\prime }\times ({\mathbf{x}}^{\prime }\times {\mathbf{x}}^{\prime \prime })/v^4$.

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 $2$.