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

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 105 / 1 / b / Solution

{title2=$\partial_N$}
{wiki=Directional_derivative}

The normal derivative of a differentiable function $u$ along a hypersurface with <unit normal> $N$ is the <directional derivative>
$$
\partial_Nu=N\mathbin\cdot\nabla u.
$$
Higher normal derivatives at a point $x$ may be defined by differentiating the one-variable function $s\mapsto u(x+sN(x))$ repeatedly at $s=0$.

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