= Solution
Write $q^1=u$, $q^2=v$, $\sigma_i=\partial_i\sigma$, and let $g_{ij}=\sigma_i\mathbin{\cdot}\sigma_j$ be the <first fundamental form>. Since
$$
\dot\gamma=\sigma_i\dot q^i,
\qquad
\ddot\gamma=\sigma_i\ddot q^i+\sigma_{ij}\dot q^i\dot q^j,
$$
differentiating $\sigma_u\mathbin{\cdot}\dot\gamma$ and $\sigma_v\mathbin{\cdot}\dot\gamma$, or equivalently taking the two <inner products> of the second formula with $\sigma_k$, gives
$$
\sigma_k\mathbin{\cdot}\ddot\gamma
=g_{k\ell}
\left(\ddot q^\ell+\Gamma^\ell_{ij}\dot q^i\dot q^j\right).
$$
The <matrix> $(g_{k\ell})$ is invertible because $\sigma$ is a regular <embedded surface parametrization>. Consequently the two <geodesic equations> hold exactly when $\ddot\gamma$ is <orthogonal> to both <tangent vectors> $\sigma_u$ and $\sigma_v$, which says precisely that $\ddot\gamma$ is a <normal vector>. This is the <ambient acceleration criterion for a surface geodesic>.
Because $\dot\gamma$ is tangent and $\ddot\gamma$ is normal,
$$
\frac d{dt}|\dot\gamma|^2
=2\dot\gamma\mathbin{\cdot}\ddot\gamma=0.
$$
Thus an affinely parametrized <geodesic> has <constant speed>, as recorded by <constant speed of an affinely parametrized geodesic>.
For a <surface of revolution>, use profile <arc length> $s$ and azimuth $\phi$. Its <Riemannian metric> is
$$
ds^2+\rho(s)^2d\phi^2.
$$
The azimuth is an <ignorable coordinate>, so the corresponding <geodesic equation> has the <first integral>
$$
\rho^2\dot\phi=C.
$$
If $v=|\dot\gamma|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel is
$$
v\cos\theta=\rho\dot\phi.
$$
Since $v$ is constant,
$$
\boxed{\rho\cos\theta=\frac Cv=\text{constant}.}
$$
This is the <Clairaut first integral for a surface of revolution>.
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