= Solution
{c}
Use Earth's rotation rate $\Omega=7.292\times10^{-5}\,\mathrm{s}^{-1}$ and the <Coriolis parameter> $f=2\Omega\sin\phi$. At $|\phi|=10^\circ$,
$$
|f|=2.53\times10^{-5}\,\mathrm{s}^{-1},\qquad
\boxed{c=|f|L_d\simeq2.53\,\mathrm{m\,s}^{-1}.}
$$
The <Kelvin waves> propagate polewards along the eastern boundary: northwards at $10^\circ$ N and southwards at $10^\circ$ S. The positive <Rossby deformation radius> uses $|f|$; the source's expression $c/f$ must be interpreted as a magnitude in the southern hemisphere.
The east-west e-folding distance is approximately $100$ km, or
$$
\Delta\lambda\simeq\frac{100\,\mathrm{km}}{R_E\cos10^\circ}
=0.0159\,\mathrm{rad}\simeq0.91^\circ
$$
of longitude. With the coast at $x=0$ and <ocean> at $x<0$, a northern-hemisphere snapshot can be written
$$
\eta=\eta_0e^{x/L_d}\cos[\ell_K(y-ct)],\qquad
u=0,\qquad v=\frac c{h_0}\eta.
$$
At a positive thickness crest, the <thermocline> is depressed and the alongshore <velocity> points polewards; a negative thickness trough reverses that <velocity>. In the southern hemisphere replace $c$ by $-c$ in the propagation and in $v/\eta$. The envelopes of both thickness and alongshore <velocity> decay offshore, and the cross-shore <velocity> vanishes.
\Image[/past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2022-coastal-kelvin-wave.png]
{title=Eastern-boundary Kelvin-wave thickness and horizontal velocity in both hemispheres}
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