At $x\in\Sigma$, keep the <unit normal> $N(x)$ fixed and define the $j$th <normal derivative> by
$$
\partial_N^ju(x)
=\left.\frac{d^j}{ds^j}u\bigl(x+sN(x)\bigr)\right|_{s=0}.
$$
For $j=1$, the <chain rule> and $N=\nabla\phi/|\nabla\phi|$ give
$$
\partial_Nu
=N\mathbin\cdot\nabla u
=\frac{\phi_tu_t+\phi_ru_r+\phi_zu_z}
{\sqrt{\phi_t^2+\phi_r^2+\phi_z^2}}
\qquad\text{on }\Sigma.
$$

Solved by gpt-5.6-sol high.


 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-105/1/b/solution