Because $\partial_u=\sqrt g\,\partial_s$ and $\mathbf r_t=U\mathbf n+W\mathbf t$, commuting $\partial_t$ and $\partial_u$ gives
$$
\partial_t\mathbf t=(\partial_sU+kW)\mathbf n.
$$
Preservation of orthonormality then gives
$$
\boxed{
\partial_t
\begin{pmatrix}\mathbf t\\\mathbf n\end{pmatrix}
=
\begin{pmatrix}
0&\partial_sU+kW\\
-\partial_sU-kW&0
\end{pmatrix}
\begin{pmatrix}\mathbf t\\\mathbf n\end{pmatrix}
}.
$$
The tangential derivative of the velocity is
$$
\partial_s\mathbf r_t
=(\partial_sW-kU)\mathbf t
+(\partial_sU+kW)\mathbf n,
$$
so
$$
\boxed{\partial_tg=2g(\partial_sW-kU)}.
$$
Finally commute the $s$ and $t$ derivatives in $\partial_s\mathbf t=k\mathbf n$, accounting for the evolving metric through
$$
[\partial_t,\partial_s]
=-(\partial_sW-kU)\partial_s.
$$
The normal component gives
$$
\boxed{
\partial_tk=(\partial_s^2+k^2)U+W\partial_sk
}.
$$


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-344/3/b/i/solution