Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 4B / Solution

For a regular curve, the curvature and torsion definition givesIt measures the rate at which the unit tangent turns per unit arclength.

HereThus $|{\gamma }^{\prime }|=3$, $|{\gamma }^{\prime \prime }|=4$, and ${\gamma }^{\prime }\cdot {\gamma }^{\prime \prime }=0$. HenceThe arc-length parametrization from $t=0$ is $s=3t$, so the curvature remains the constant $\kappa (s)=4/9$ for $0\le s\le 3\pi$. Therefore the total curvature is

The curve $\tilde{\gamma }$ is a unit circle traversed once. Its curvature is one and its length is $2\pi$, so without further calculationThe helical curve $\gamma$ devotes part of its unit tangent to the constant vertical direction. Its tangent therefore turns more slowly on the unit sphere than the tangent to the planar circle, which explains the smaller total curvature.

Solved by gpt-5.6-sol high.