Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 3 / 3C / ii / Solution

Use the arc length coordinate oriented in the direction of increasing $t$: $s(t)={\int }_0^{t}2e^{\tau }d\tau =2(e^t-1)$, with $s>-2$. Also ${\mathbf{r}}^{\prime \prime }=e^t(\sqrt{2},-2\cos t,-2\sin t)$, giving $|{\mathbf{r}}^{\prime \prime }{|}^2=6e^{2t}$ and ${\mathbf{r}}^{\prime }\cdot {\mathbf{r}}^{\prime \prime }=4e^{2t}$. The Lagrange identity for the cross product yieldsThe curvature of a space curve is thereforeFor the forward portion $t\ge 0$, this has ordinary nonnegative arc length $s\ge 0$. If instead nonnegative distance is measured backwards from $t=0$, put $s_b=2(1-e^t)$ for $t\le 0$; that branch has $\kappa =1/[\sqrt{2}(2-s_b)]$, $0\le s_b<2$. Choosing orientation avoids confusing the two points at the same unoriented distance.