For the surface $F(x,z,t)=z-\zeta(x,t)=0$, an outward <normal vector> and positively oriented <tangent vector> are
$$
\boxed{
\hat{\mathbf n}
=\frac{(-\zeta_x,1)}{\sqrt{1+\zeta_x^2}}},
\qquad
\boxed{
\hat{\mathbf t}
=\frac{(1,\zeta_x)}{\sqrt{1+\zeta_x^2}}}.
$$
With $\zeta=\epsilon e^{ikx}$ and $|k\epsilon|\ll1$, <linearization> gives
$$
\boxed{
\hat{\mathbf n}=(-ik\epsilon e^{ikx},1)+O(\epsilon^2k^2)},
\qquad
\boxed{
\hat{\mathbf t}=(1,ik\epsilon e^{ikx})+O(\epsilon^2k^2)}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 332 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-332/4/b/solution