Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 / 3

Under the Euclidean isometry $\tilde{\mathbf{r}}=\mathbf{a}+R\mathbf{r}$, where $R^{T}R=I$,ThereforeSince the integration parameter is unchanged,Both the induced metric and arc length are invariant.

For an oriented planar curve, the Frenet-Serret formulas areThe frame is orthonormal, so differentiating each scalar product givesIn matrix form this says that the connection matrix plus its transpose is zero; it must be antisymmetric.

The signed curvature isAn orientation-preserving isometry sends $\mathbf{t},\mathbf{n}$ to $R\mathbf{t},R\mathbf{n}$ and leaves $s$ unchanged, soAn orientation-reversing isometry reverses the chosen normal and hence the sign convention for signed curvature, while the geometric curvature $|k|$ remains invariant.

Because ${\partial }_u=\sqrt{g}{\partial }_s$ and ${\mathbf{r}}_t=U\mathbf{n}+W\mathbf{t}$, commuting ${\partial }_t$ and ${\partial }_u$ givesPreservation of orthonormality then givesThe tangential derivative of the velocity issoFinally commute the $s$ and $t$ derivatives in ${\partial }_s\mathbf{t}=k\mathbf{n}$, accounting for the evolving metric throughThe normal component gives

Since ${\partial }_t\sqrt{g}=\sqrt{g}({\partial }_{s}W-kU)$,Thus

Local force balance and moment balance on a slender rod areIn a planar rod, $\mathbf{M}$ and $\mathbf{m}$ point perpendicular to the plane.

Substituteinto force balance. With $\mathbf{\mu }={\mathbf{\gamma }}^{-1}$,Projection onto the local frame gives

ResolveThe Frenet-Serret formulas giveUsingtherefore yields