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In spherical coordinates on a sphere of radius $R$, the position is $X=R(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. Its coordinate tangent vectors satisfy $X_{\theta }\cdot X_{\varphi }=0$, $|X_{\theta }{|}^2=R^2$, and $|X_{\varphi }{|}^2=R^2{\sin }^2\theta$. Expanding $|dX{|}^2$ proves the displayed line element, the first fundamental form of the sphere. Along a parametrized path it gives the arc length by integrating $R\sqrt{{\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2}$. The vanishing azimuthal coefficient at a pole is a coordinate singularity, not a singularity of the sphere.

Choose normal coordinates ${\xi }^{\alpha }$ centered at an event $p$. At $p$, the metric tensor in these coordinates is the constant Minkowski metric ${\eta }_{\alpha \beta }=\mathrm{diag}(-1,1,1,1)$, with ${\xi }^0$ measured in length units, and its first partial derivatives vanish. In arbitrary coordinates $x^{\mu }$, the chain rule for the line element givessoThis construction is pointwise. At nearby events, the metric tensor in the chosen normal coordinates is $g_{\alpha \beta }^{(\xi )}(\xi )={\eta }_{\alpha \beta }+O(|\xi {|}^2)$, with the quadratic corrections determined by the Riemann curvature tensor. Replacing $g^{(\xi )}$ by one constant Minkowski metric throughout an open region would make that region flat, so it cannot describe general curved spacetime exactly.

On the circular cylinder, $\rho =R$ is constant, so the given line element reduces to $d{s}^2=R^{2}d{\varphi }^2+d{z}^2$. Unroll the circular cylinder with $u=R\varphi$. The line element is then $d{s}^2=d{u}^2+d{z}^2$, so the usual Euclidean geometry formula measures arc length in the $(u,z)$ plane.

The endpoint has infinitely many lifts $(R({\varphi }_1+2\pi k),z_1)$, with $k\in \mathbb{Z}$. For each lift, the straight line segment is shortest; its length is $\sqrt{R^2({\varphi }_1+2\pi k{)}^2+z_1^2}$. Comparing the lifts gives the unrestricted shortest path on a circular cylinder:If $\delta$ is a minimizing angular increment, a shortest geodesic is parametrized by $(\rho ,\varphi ,z)=(R,\delta t,z_{1}t)$, $0\le t\le 1$. This is a helix when both increments are nonzero, and a circular arc or vertical generator in the limiting cases. A nonminimal lift also gives a geodesic, but it is only shortest within its fixed winding class.

For $0\le {\varphi }_1\le 2\pi$, put $m=\min ({\varphi }_1,2\pi -{\varphi }_1)$. Then $\boxed{L_{\min }=\sqrt{R^2{m}^2+z_1^2}}$. At ${\varphi }_1=\pi$, both winding directions give distinct shortest paths of equal length.

Along the circle with increasing $\varphi$, the line element is $d\mathbf{r}=a\sin \alpha {\mathbf{e}}_{\varphi }d\varphi$. Thus the two circulations areThe half-angle identities simplify them toHere $0<\alpha <\pi$ so both potentials are defined on the whole curve.

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.

For an oriented piecewise smooth surface $S$ with boundary curve $C=\partial S$, and a continuously differentiable vector field $\mathbf{F}$ near $S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is positive surface area, and $d\mathbf{r}$ is the directed tangent line element. The positive traversal of $C$ is fixed by the right-hand rule relative to $\mathbf{n}$. Thus the flux of the curl through the surface equals the circulation along its oriented boundary.

Differentiate the position vector, using the angular factor from the scale factors of orthogonal coordinates:The basis is an orthonormal basis, so all cross terms in its squared length vanish. The Euclidean line element is thereforeIn particular, an angular change $d\varphi$ at fixed radius traverses a physical distance $\rho |d\varphi |$.

For a finite separation use the Cartesian distance, rather than substituting finite coordinate differences into the local line element. Expanding the two horizontal components givesThe vertical contribution is $(z-z^{\prime }{)}^2$, independently. ThusandEquivalently, the horizontal contribution is $(\rho -{\rho }^{\prime }{)}^2+4\rho {\rho }^{\prime }{\sin }^2[(\varphi -{\varphi }^{\prime })/2]$, which also makes its nonnegativity explicit.

The downward drop from the starting rim to angle $\theta$ is $R\cos \theta$. With no dissipative forces, conservation of energy gives $m{v}^2/2=mgR\cos \theta$, so $v=\sqrt{2gR\cos \theta }$. Dividing the line element by this speed yieldsThe endpoint singularities are integrable for the minimizing path. The integrand is independent of $z$; its Euler-Lagrange equation gives a constant $K$, after absorbing the overall prefactor:Solving for $z^{\prime }$ and taking the positive branch required by $z_2>z_1$ givesThe constant is fixed implicitly byThis integral is strictly increasing from zero to infinity as $K$ increases from zero to one, so every finite positive separation selects a unique $K$. The time integrand has second derivative $R^2/[\sqrt{\cos \theta }(R^2+z^{\prime 2}{)}^{3/2}]>0$ with respect to $z^{\prime }$. Its supporting-line inequality, integrated between fixed endpoints with constant first derivative with respect to $z^{\prime }$, proves this stationary path is the unique global minimum within the stated single-valued class.

As $\theta \downarrow -\pi /2$, $z^{\prime }\to 0$. Therefore the skater should initially head directly down the curved ramp, perpendicular to the cylinder axis, and gain axial motion later. The path differs from the constant-pitch shortest route because extra distance is cheaper in time near the faster, lower part of the ramp. This is the brachistochrone on a horizontal half-cylinder.

On the cylindrical surface $r=R$, the line element is $d{s}^2=R^{2}d{\theta }^2+d{z}^2$. For a single-valued path parametrized by increasing $\theta$,The integrand has no explicit $z$, so its Euler-Lagrange equation gives the constant conjugate quantity $z^{\prime }/\sqrt{R^2+z^{\prime 2}}$. Its derivative with respect to $z^{\prime }$ is strictly positive, hence $z^{\prime }$ itself is constant. The two endpoints determineThis is a constant-pitch helix on the cylinder. To verify global minimality, unroll the cylinder with planar coordinates $(R\theta ,z)$. The path becomes a straight line between points separated by $\pi R$ and $z_2-z_1$, so the triangle inequality for arc length givesThis is the standard description of geodesics on a circular cylinder.

Unroll a circular cylinder of radius $R>0$ using the coordinate $u=R\varphi$. Its line element becomes $d{s}^2=d{u}^2+d{z}^2$, so a geodesic lifts to a straight line in the plane. An endpoint has infinitely many lifts differing by $2\pi Rk$. Minimize the ordinary Euclidean distance over those lifts to obtain the unrestricted shortest path. A fixed lift instead specifies a winding class and need not give the global shortest path.

For angular difference $0\le \delta \le 2\pi$, put $m=\min (\delta ,2\pi -\delta )$. Then $d=\sqrt{R^2{m}^2+(z_1-z_0{)}^2}$. At $\delta =\pi$, two different geodesics have the same shortest length. This is the source of the corner in the distance-versus-angle graph at the opposite generator of the circular cylinder.