= Solution
For the <surface of revolution>, the arc-length condition on its profile gives $E=1$, $F=0$, $G=f(u)^2$. Thus the <geodesic equations> reduce to
$$
u''=ff'(v')^2,\qquad\frac d{ds}(f^2v')=0.
$$
Consequently
$$
\boxed{f(u)^2v'=\text{constant},}
$$
the <Clairaut first integral for a surface of revolution>.
On a meridian, take $v=b$ and $u=\pm s+u_0$. Both equations hold, so \b[every meridian is a <geodesic>]. On a cross-section $u=c$, <arc length> requires $v'=\pm1/f(c)$. The first equation then holds exactly when $f'(c)=0$, while the second holds automatically. Hence
$$
\boxed{u=c\text{ is a geodesic}\iff f'(c)=0.}
$$
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