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Take fixed-endpoint variations $y_{\epsilon }=y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. The first variation isThe boundary term vanishes by the endpoint conditions. The fundamental lemma of the calculus of variations gives the Euler-Lagrange equation $f_y-d(f_{y^{\prime }})/dx=0$. Along its solutions,When $f_x=0$, this proves the Beltrami identity $f-y^{\prime }{f}_{y^{\prime }}=\text{constant}$.

For the surface $z=F(y)$, the line element is $d{s}^2=d{x}^2+[1+F^{\prime }(y{)}^2]d{y}^2$. The arc-length integrand for a graph $y(x)$ is $\mathcal{L}=\sqrt{1+[1+F^{\prime }(y{)}^2]y^{\prime 2}}$. It has no explicit $x$ dependence and satisfies $\mathcal{L}-y^{\prime }{\mathcal{L}}_{y^{\prime }}=1/\mathcal{L}$. Thus $[1+F^{\prime }(y{)}^2]y^{\prime 2}$ is constant. Choosing the fixed direction along a smooth nonconstant segment and integrating yieldsThe same result follows by defining the transverse arc coordinate $u(y)={\int }^y\sqrt{1+F^{\prime }(s{)}^2}ds$, which changes the metric to $d{s}^2=d{x}^2+d{u}^2$. Its geodesics are straight lines in $(x,u)$ coordinates. This proves the formula for geodesics on a generalized cylinder and shows it gives locally shortest paths. The additional straight lines $x=\text{constant}$ in the unrolled plane are also geodesics, but cannot be represented as graphs $y(x)$. Classical differential-geometric statements assume the surface has sufficient smoothness; the arc-length coordinate itself only requires the displayed integral to be well defined.