Surface of revolution, arc length approximates distance?

Question

Consider an even function $z=f(x)$ and the surface of revolution $S$ by the graph around $z$-axis. If $f$ is a $C^1$ ($C^2$) function with $f'(0)=0$, then $S$ is a $C^1$ ($C^2$) manifold with the chart $\phi :R^2\to S$: $$ \phi(v,u)=(v\cos u, v\sin u, f(v)). $$ If $S$ is equipped with the induced metric from $R^3$, then it's a Riemannian manifold, the metric $g$ is continuous ($C^1$). The geodesic equation is $$ u''+\frac{2u'v'}{v}=0, $$ $$ v''-\frac{v}{1+(f')^2}(u')^2+\frac{f'f''}{1+(f')^2}(v')^2=0. $$ Consider two points $$ s_1=(v_0,0,f(v_0)), s_2=(v_0\cos u, v_0\sin u, f(v)) \in S. $$ The arc of the parallel $u=u(s), v \equiv v_0$ from $s_1$ to $s_2$ is not a geodesic in general. Denote the length of the arc by $L(s_1,s_2)$. So How to compute the distance from $s_1$ to $s_2$? Do we have $$ \lim_{u\to 0}\frac{d(s_1,s_2)}{L(s_1,s_2)}=1? $$