= Solution
\b[Balances and density gradients.] The first two relations describe horizontal <geostrophic balance> between <pressure gradient> and <Coriolis force> accelerations. The vertical relation is <hydrostatic balance> between the vertical pressure force and gravity. They describe a slowly varying large-scale interior circulation, excluding the frictional surface/bottom layers and large ageostrophic accelerations.
Differentiate the horizontal balances vertically and use $p_z=-\rho g$, treating $f,\rho_0$ as vertically constant:
$$
\boxed{u_z=\frac{g}{f\rho_0}\rho_y,\qquad v_z=-\frac{g}{f\rho_0}\rho_x}.
$$
These are the oceanic <thermal wind> relations. With no motion at $z=-H$, the exact surface relations involve the depth-integrated density gradients. If the measured density is representative of the vertical mean, or if those gradients are approximately uniform with depth,
$$
u_g\simeq\frac{gH}{f\rho_0}\rho_y,\qquad
v_g\simeq-\frac{gH}{f\rho_0}\rho_x.
$$
Equivalently these hold exactly for the horizontal gradients of $\bar\rho=H^{-1}\int_{-H}^0\rho\,dz$ in this fixed-depth reference model. A single arbitrary surface-density gradient does not otherwise determine a vertical integral. In the Northern Hemisphere the inferred surface current runs along horizontal <isopycnals> with denser water on its left when the no-motion reference and this vertical-gradient approximation apply.
For a steady density pattern and a ship whose actual track velocity is $v_s(\cos\theta,\sin\theta)$,
$$
\frac{d\rho}{dt}=v_s(\rho_x\cos\theta+\rho_y\sin\theta)
=\boxed{\frac{f\rho_0v_s}{gH}(u_g\sin\theta-v_g\cos\theta)}.
$$
If the field changes during the survey an Eulerian time derivative must also be included. Here $H=\Delta z>0$ is the upward separation from reference depth to surface, so its sign must not be reversed.
\b[Inversion and velocity estimate.] Let $C=f\rho_0v_s/(gH)$ and denote the two ship measurements by $y_\theta,y_0$, with the east-going ship assigned the subscript zero. The observation system and its inverse are
$$
\mathbf y=C\begin{pmatrix}\sin\theta&-\cos\theta\\0&-1\end{pmatrix}\begin{pmatrix}u_g\\v_g\end{pmatrix},\qquad
\boxed{v_g=-\frac{y_0}C,\qquad u_g=\frac{y_\theta-y_0\cos\theta}{C\sin\theta}}.
$$
It is invertible for $\sin\theta\ne0$. For two measurements and two unknowns this inverse is also the stated least-squares estimate; the nearly parallel-track geometry explains its poor conditioning at small angle.
The supplied parameters give $C=5.654\times10^{-5}$ in units converting velocity to the measured density tendency. With $y_\theta=1.9\times10^{-5}$ and $y_0=1.8\times10^{-5}\ \mathrm{kg\,m^{-3}\,s^{-1}}$,
$$
\boxed{u_g=+0.217\ \mathrm{m\,s^{-1}},\qquad v_g=-0.318\ \mathrm{m\,s^{-1}}}.
$$
The current therefore has eastward and southward components. Signs follow from the positive-upward, east–north coordinate convention and the specified no-motion reference.
\b[Uncertainty and track design.] The <two-track thermal-wind covariance> exposes the geometrical limitation. The stated error already applies to the difference of successive density observations. Dividing by the sampling interval, rather than adding an extra factor of $\sqrt2$, gives $\sigma_y=7.2\times10^{-4}/3600=2.0\times10^{-7}\ \mathrm{kg\,m^{-3}\,s^{-1}}$. Assume independent errors of that size for the two ships, as in the supplied <covariance> model. Then
$$
\boxed{\mathsf S_{uv}=\frac{\sigma_y^2}{C^2}
\begin{pmatrix}(1+\cos^2\theta)/\sin^2\theta&\cot\theta\\\cot\theta&1\end{pmatrix}}.
$$
At $5^\circ$,
$$
\boxed{\sigma_u=0.0573\ \mathrm{m\,s^{-1}},\quad\sigma_v=0.00354\ \mathrm{m\,s^{-1}},\quad
\operatorname{Cov}(u,v)=1.43\times10^{-4}\ \mathrm{m^2\,s^{-2}}}.
$$
The <correlation> is positive,
$$
\boxed{\operatorname{Corr}(u,v)=\frac{\cos\theta}{\sqrt{1+\cos^2\theta}}=0.706\quad\text{at }5^\circ}.
$$
Both estimates depend on the east-going ship; its positive measurement error lowers $v_g$ and also lowers $u_g$, explaining the sign of the <covariance>. If the other ship goes north instead, $\theta=90^\circ$ gives orthogonal observations: $u_g=y_{90}/C$ and $v_g=-y_0/C$. Both <standard errors> become $0.00354\ \mathrm{m\,s^{-1}}$ and their <covariance> vanishes. The north-going track improves the poorly constrained zonal component by about a factor of sixteen, while leaving the meridional precision unchanged under the same noise assumptions.
\Image[/past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2016-thermal-wind-track-geometry.png]
{title=Density gradients, geostrophic current direction and the uncertainty caused by nearly parallel ship tracks}
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