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Let $g>0$ be the gravitational acceleration, and let $y$ denote positive downward depth. For a frictionless particle released from rest, conservation of energy gives $v^2=2gy$. The arc length element of $(x,-y(x))$ is $ds=\sqrt{1+y^{\prime 2}}dx$, so the travel time and length areFor a finite-time single arch, require $y>0$ in $0<x<1$, $y(0)=y(1)=0$, and interpret the endpoint integrals improperly. At a regular constrained minimizer with $S[y]=L$, the Lagrange multiplier rule for an isoperimetric constraint makes $\sqrt{2g}T+\lambda (S-L)$ stationary. Dropping the constant $-\lambda L$ givesFor $L>1$ the constraint is regular under interior variations: if its first variation vanished for every compactly supported variation, the Euler-Lagrange equation for $S$ would force $y^{\prime }$ constant, hence $y\equiv 0$ and $S=1$, a contradiction. Thus the multiplier is justified for a nondegenerate arch.

The printed case $L=1$ is degenerate. Since $\sqrt{1+y^{\prime 2}}\ge 1$, equality of the length with the horizontal separation forces $y^{\prime }=0$ and $y=0$. A particle initially at rest then has no tangential gravitational acceleration and cannot complete the trip. Its travel time is infinite. The finite-time calculation therefore concerns $L>1$, with endpoint regularity qualified below.

With $H>0$, setThen the first integral gives $1/\sqrt{1+y^{\prime 2}}=s$. On the descending and ascending halves of the physical track choose respectively $0<\varphi <\pi$ and $\pi <\varphi <2\pi$, so that $y^{\prime }=\cot (\varphi /2)$. Solving for $y$ givesThe unsquared equation requires $1-bs>0$. At $s=1$ this says $b<1$; it is not permissible to retain other values merely because the displayed formula for $y$ has a squared denominator.

For completeness, the Euler-Lagrange equation also gives $y^{\prime \prime }=(A^{\prime }(y)/A(y))(1+y^{\prime 2})<0$. Thus $y^{\prime }$ decreases through zero exactly once, so the chosen parameter runs through the whole single arch without an extra branch. Differentiating the formula for $y$ and dividing by $y^{\prime }=\cot (\varphi /2)$ gives, continuously also at $\varphi =\pi$,Using $x(0)=0$ yieldsSince $\sqrt{1+y^{\prime 2}}=1/\sin (\varphi /2)$, the arc length is obtained by cancelling one factor of $\sin (\varphi /2)$ in $dx/d\varphi$. The conditions $x(2\pi )=1$ and $S=L$ becomeThese determine a unique parameter pair for each $L>1$. To see uniqueness, regard $L$ as the expected value of $1/s$ under a probability density function proportional to $s^2/(1-bs{)}^3$ on $0<u<2\pi$. Differentiation under the integral sign gives the covariance identityThe first argument decreases with $s$ and the second increases. The strict sign follows by applying $(f(s)-f(t))(g(s)-g(t))<0$ to two independent unequal values and averaging; the integrable endpoint singularity of $1/s$ causes no problem. As $b\uparrow 1$, the weight concentrates at $s=1$, so $L\downarrow 1$. As $b=-k\to -\infty$, endpoint estimates give numerator $\sim 2/k^2$ and denominator $\sim 4\log k/k^3$, so $L\sim k/(2\log k)\to \infty$. Continuity then covers every $L>1$. At $b=0$ one recovers the cycloid and $L=4/\pi$.

Smoothness qualification. Near $\varphi =0$,and the other endpoint has $y^{\prime }\to -\infty$. These are smooth graphs on $0<x<1$, not $C^1$ graphs on the closed interval. Indeed a finite endpoint slope would make the conserved $H$ diverge as $y\to 0$. The travel time remains finite becauseThus the printed phrase “smooth solution” must be understood on the interior, with improper endpoint limits. The variational calculation supplies necessary stationary arches; it does not by itself prove global minimality in every larger class of tracks.

On the upper semicircular contour, $|z|=R$ and the reverse triangle inequality gives $|z^4+1|\ge R^4-1>0$. Its arc length is $\pi R$, so the estimation lemma givesThe arc is open, so an ordinary integral sign is appropriate even though the source uses a closed-integral symbol there. Subtracting the vanishing arc contribution from the closed contour integral gives the real-line improper integral $\pi /\sqrt{2}$. Its integrand is an even function, and its tail is $O(x^{-2})$, ensuring convergence. Thus

Replacing $L$ by the angular momentum operator gives the Hamiltonian operatorThe time-dependent Schrödinger equation is thereforeFor a stationary state, separate variables as $\Psi (\varphi ,t)=\psi (\varphi )T(t)$. Dividing by the product separates a function of $t$ from one of $\varphi$:Thus $T=e^{-iEt/\hslash }$ up to a constant, and the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2I)=E\psi$.

Diagonalize the commuting angular momentum operator at the same time: $-i\hslash {\psi }^{\prime }=\ell \psi$ has solutions $A{e}^{i\ell \varphi /\hslash }$. A single-valued wavefunction requires $e^{2\pi i\ell /\hslash }=1$, hence $\ell =n\hslash$ with $n\in \mathbb{Z}$. Normalizing with the angular probability measure $d\varphi$ givesThese form an orthonormal basis under ${\int }_0^{2\pi }{\psi }_m^{\ast }{\psi }_{n}d\varphi ={\delta }_{mn}$. The periodic boundary conditions apply to both the wavefunction and its first derivative, ensuring the physical self-adjoint Hamiltonian operator. For probability per arc length $ds=rd\varphi$, the normalization instead reads ${\psi }_n=e^{in\varphi }/\sqrt{2\pi r}$; the two conventions describe the same probabilities. General states are superpositions of these energy eigenstates.

Ideal and self-avoiding walks. An ideal chain has independent isotropic coarse-grained steps and permits self-intersections in its mathematical model. For $N$ steps of length $b$, the cross terms average to zero, so its root-mean-square end-to-end distance of a polymer chain obeysA self-avoiding walk excludes repeated occupancy or intersection at the segment scale. This represents positive excluded volume in a dilute good solvent. The nonlocal constraint correlates otherwise distant portions of the chain and swells it:The Flory approximation replaces this exponent by $3/5$, by balancing stretching free energy $R^2/(N{b}^2)$ with an interaction contribution $v{N}^2/R^3$. The result is a more extended and less internally dense coil than the ideal walk; self-avoidance changes the statistics, not merely the local step length.

Two ideal-chain regimes. At the theta temperature for a dilute polymer–solvent pair, leading effective segment attractions and repulsions cancel. Large-scale chain statistics are approximately ideal, even though microscopic atoms still have finite size. In a homogeneous polymer melt, surrounding chains screen long-range excluded volume interactions of a chosen chain, again making its equilibrium large-scale size approximately ideal. Local bond-angle correlations must be included in an effective statistical length such as the Kuhn length. Polymer entanglements can slow the dynamics without invalidating this leading equilibrium size law.

Semiflexible DNA. The nucleic acid double helix gives double-stranded DNA a bending stiffness and substantial local orientational persistence. Independent step orientations are therefore a poor description on short scales, while self-avoidance alone does not specify the energy cost of bending. The worm-like chain describes the molecule as a continuously bendable, locally inextensible curve $\mathbf{r}(s)$ parameterized by arc length $s$, with unit tangent $\mathbf{t}(s)=d\mathbf{r}/ds$. Its bending energy iswhere $A$ is bending stiffness. In the free three-dimensional model, the persistence length is ${\ell }_p=A/(k_{B}T)$ andThus ${\ell }_p$ is the contour separation over which tangent alignment falls by the factor $e^{-1}$; it quantifies stiffness relative to thermal bending. The model bridges locally rod-like and globally coil-like conformations. It omits long-range self-avoidance in its simplest form, so sufficiently long dilute DNA can require excluded-volume corrections.

Flexible and rigid limits. Integrating the tangent correlation over the two contour positions gives the end-to-end distance of a worm-like chain:For a very flexible chain on the scale of its total length, $L/{\ell }_p\gg 1$, the exponential is negligible andThis is the ideal chain result with Kuhn length $b_K=2{\ell }_p$ and effective step count $N_K=L/(2{\ell }_p)$. Many persistent sections have decorrelated orientations. The flexible limit means a large ratio $L/{\ell }_p$ within the continuum model's range, not an extrapolation of the persistence length below all molecular length scales.

For a nearly rigid chain, $L/{\ell }_p\ll 1$, the constant and linear terms in the exponential cancel those outside it. The Taylor series givessoThis is the expected rod limit: all tangents are nearly aligned, and the end-to-end distance approaches the arc length.

Helix–coil curves. A helix-coil transition model allows ordered helical runs and disordered coil regions. Use $s$ and $\sigma$ with exactly the free-energy conventions defined in the paper. A negative $\Delta F_{\mathrm{hc}}$ gives $s>1$ and favors the helical state; a positive junction cost gives $0<\sigma <1$ and penalizes starting new domains. The supplied helical fraction is the long-chain form associated with the Zimm-Bragg model.

Its main features follow directly without assuming a sharp transition. Put $Q=(s-1{)}^2+4s\sigma$. For every $\sigma >0$,and differentiating givesThus all curves rise monotonically and cross at $(1,1/2)$. Increasing the boundary free-energy cost reduces $\sigma$, suppresses frequent coil–helix boundaries and makes longer runs favorable. The transition region becomes narrower and steeper: balancing $(s-1{)}^2$ against $4s\sigma$ gives a width of order $\sqrt{\sigma }$ near $s=1$.

For $\sigma =1$, the fraction simplifies to $\boxed{f_h=s/(1+s)}$, the noncooperative two-state result. For $0<\sigma \ll 1$, the steep curve describes cooperative helix formation. In the singular limit $\sigma \downarrow 0$, it approaches zero for $s<1$ and one for $s>1$; its midpoint remains $1/2$ when approached through positive $\sigma$. Setting $\sigma =0$ directly makes the formula indeterminate at $s=1$. For any fixed positive $\sigma$, the given long-chain curve is smooth: it is a cooperative crossover, not a true discontinuity. The plot varies $s$ at fixed $\sigma$; in an actual temperature scan both parameters may change, and a finite chain can further round the crossover.

The disk exit time is finite almost surely; for example, either coordinate of the planar Brownian motion eventually leaves $(-1,1)$. By path continuity, the exit lies on the unit circle. Starting at the origin, rotational invariance of planar Brownian motion makes the exit law invariant under every circle rotation. The unique rotation-invariant probability measure is normalized arc length. Consequently $Z_{T_D}$ is uniform on the unit circle, or equivalently its angle has density $1/(2\pi )$ on $[0,2\pi )$.

By Brownian scaling, $r^{-1}{Z}_{r^{2}t}$ is planar Brownian motion started at $w_0/r$. Its unit-disk exit angle is exactly the original radius-$r$ exit angle. Substitute $z_0=w_0/r$ into the supplied Poisson kernel for the unit disk:This density is with respect to angle $d\theta$. Its density with respect to arc length would be smaller by a factor $r$.

A topological space $X$ is path-connected if, for every $a,b\in X$, there is a continuous mapThis continuous path must stay within $X$ for its entire duration. Its parameter need not represent physical time, and continuity does not require a derivative or finite arc length. ^[ref]

In the closed interval $[0,1]$, any $a,b$ are joined by $\gamma (t)=(1-t)a+tb$. For $a=1/4$ and $b=3/4$, this becomes $\gamma (t)=1/4+t/2$, so $\gamma (0)=1/4$, $\gamma (1/2)=1/2$, and $\gamma (1)=3/4$. All its values stay in $[0,1]$. In the union $D=[-2,-1]\cup [1,2]$, however, a continuous path from $-1$ to $1$ would attain $0$ by the intermediate value theorem, although $0\notin D$. Thus $D$ is not path-connected.

Every path-connected space is a connected space. Proof outline: if $X=U\cup V$ admitted a separation of a topological space, choose $a\in U$ and $b\in V$. A continuous path between them would have nonempty open inverse images ${\gamma }^{-1}(U),{\gamma }^{-1}(V)$ with empty intersection and union $[0,1]$. This contradicts the connectedness of that closed interval.

The converse fails: the topologist's sine curve is connected but has two path components. Path connectedness is useful when we need an actual continuous path of allowed positions, rather than only the absence of a separation of a topological space. For instance, when modelling allowed positions as a topological space, membership in the same path component answers whether one can move continuously between two positions while remaining in that topological space.

For a piecewise differentiable curve given in polar coordinates by $r=r(\theta )$, differentiate $x=r\cos \theta$ and $y=r\sin \theta$. The cross terms cancel in $d{x}^2+d{y}^2$, leaving $d{s}^2=d{r}^2+r^{2}d{\theta }^2$. Therefore the arc length on an increasing parameter interval isThis formula remains valid through an isolated zero of the speed when the parametrization is piecewise smooth. A square root of a square must retain its absolute value.

Pseudo-arclength continuation allows both the state and the parameter to change. Write $z=(u,\lambda )$ and choose a unit vector $t_k$ in the kernel of the full Jacobian matrix $DF(z_k)$. A continuation predictor-corrector step seeks the next solution on a hyperplane perpendicular to $t_k$, a distance $h>0$ along the old tangent vector. This adds one equation to $F(z)=0$ and makes the correction a square system. The constraint approximates arc length; $h$ is the displacement projected onto the old tangent vector, rather than the exact curved distance.^[ref]

Let $\Gamma$ be a piecewise smooth contour forming a simple closed curve, positively oriented: travel counterclockwise, with the interior on the left. Suppose $f$ is holomorphic on an open set containing $\Gamma$ and its interior, except at finitely many isolated singularities $a_1,\dots ,a_m$ strictly inside. ThenNo isolated singularity may lie on $\Gamma$: the ordinary contour integral must be defined there. Reversing the direction of the contour reverses the sign of the answer. The theorem applies to essential singularities as well as poles. ^[ref]

Return to $f(z)=1/(z^2+1)$. A counterclockwise circle $|z|=2$ encloses both simple poles, soYet $f$ is nonzero everywhere it is defined. A zero contour integral does not imply a zero function, or even absence of poles: the enclosed residues can cancel. By contrast, the counterclockwise circle $|z-i|=1/2$ encloses only $i$, givingThe picture compares these two selections. Changing the shape of a contour while keeping the same enclosed poles leaves the contour integral unchanged, provided the contour deformation stays where $f$ is holomorphic.

This is the practical appeal of the residue theorem: local Laurent coefficients determine a global contour integral. For a rational function, a few simple pole computations can replace a long parametrized calculation. It also makes real improper integrals accessible by treating the real line as part of a closed contour in the complex plane.

To compute $I={\int }_{-\infty }^{\infty }dx/(1+x^2)$, take $R>1$ and travel along the real line from $-R$ to $R$, then return along the upper semicircle $A_R$, parametrized by $z=R{e}^{it}$ for $0\le t\le \pi$. This contour is positively oriented and encloses the simple pole $i$ only; $-i$ is outside. HenceThe added arc must be controlled before taking a limit. On $A_R$, the reverse triangle inequality gives $|z^2+1|\ge R^2-1$, and the arc length is $\pi R$. The estimation lemma therefore yieldsThe denominator grows quadratically while the arc length grows only linearly. This quantitative comparison, rather than a sketch alone, explains why closing the contour does not alter the limiting real improper integral. It even supplies an error bound: the difference between $\pi$ and the finite real integral is at most $\pi R/(R^2-1)$.

For a numerical check, $R=10$ gives the finite real integral $2\arctan (10)\approx 2.94226$, leaving about $0.19934$ to reach $\pi$. The arc bound is $10\pi /99\approx 0.31733$, comfortably larger than that error. The graph shows both quantities approaching zero. The exact tail here can be checked with the antiderivative $\arctan x$; the residue theorem method remains useful for improper integrals where a convenient real antiderivative is unavailable.

Finally, both one-sided improper integrals converge absolutely: for $|x|\ge 1$, $0<1/(1+x^2)\le 1/x^2$, whose tails are integrable. Thus the symmetric limit computes the ordinary improper integral, not merely a Cauchy principal value, andClosing below instead would enclose the simple pole $-i$. With the same left-to-right traversal of the real line, that contour is clockwise, so its contour integral is $-2\pi i\mathrm{Res}(f,-i)=\pi$ again. This provides a useful sign check: the opposite residue is accompanied by the opposite direction around the enclosed region.

The same strategy extends to many other rational functions: choose a contour, identify its enclosed isolated singularities, compute their residues, and bound the added pieces. The choice of contour and the bound are as much part of the calculation as the residues. ^[ref]

The distance on a connected Riemannian manifold is the infimum of the arc lengths of piecewise smooth curves joining the endpoints:It answers “how far must I travel while staying in this geometry?” The Riemannian metric measures an instantaneous tangent vector, whereas Riemannian distance is the global result of comparing complete routes.

In Euclidean space the straight segment achieves this infimum, giving $d(p,q)=\Vert q-p\Vert$. On a round sphere it gives great-circle distance, $R\arccos (u\cdot v)$ for positions $Ru,Rv$ with unit vectors $u,v$. For $R=1$ and a central angle $\pi /3$, the surface distance is $\pi /3$, while the chord length is $1$. On a circular cylinder, routes with different winding lead to the minimum in Riemannian distance on a circular cylinder.

The word infimum matters: a shortest route need not exist. In the Euclidean plane with the origin removed, the points $(-1,0)$ and $(1,0)$ still have Riemannian distance $2$, approached by paths detouring around an arbitrarily small circle. Every allowed route has length strictly greater than $2$, since the only Euclidean straight segment passes through the missing point. The Hopf-Rinow theorem guarantees a minimizing geodesic between any two points when a connected Riemannian manifold is complete.

On the lateral surface of an infinite circular cylinder of radius $R$, the induced metric is $R^{2}d{\theta }^2+d{z}^2$. Lifting a curve to the unrolled plane $(x,z)=(R\theta ,z)$ preserves its arc length. If the endpoints differ by angular coordinate $\Delta \theta$ and height $\Delta z$, the possible lifted endpoint displacements are $(R(\Delta \theta +2\pi k),\Delta z)$ for integers $k$. Each straight segment gives a geodesic, and the shortest of their lengths isChoose the representative of $\Delta \theta$ with smallest absolute value modulo $2\pi$; if its absolute value is $\pi$, the two angular directions tie. This minimum is attained because the candidate lengths tend to infinity as $|k|\to \infty$. The formula concerns travel on the lateral surface; it does not allow a shortcut through the cylinder's interior or across end caps.

For $R=1$, take $p=(1,0,0)$ and $q=(0,1,3)$ in surrounding Euclidean space. The angular difference is $\pi /2$ and height difference is $3$. The $k=0$ route has length $\sqrt{9+{\pi }^2/4}\approx 3.386$. The $k=-1$ route goes the other way around and has length $\sqrt{9+9{\pi }^2/4}\approx 5.586$. Both are among the geodesics on a circular cylinder, but only the first is a minimizing geodesic between these endpoints.

This is a global consequence of intrinsic flatness of a circular cylinder: locally everything is Euclidean, yet the periodic angular coordinate leaves several competing routes. A shortest-route problem on the curved surface becomes a family of ordinary planar distance calculations.

Riemannian geometry studies smooth manifolds equipped with a Riemannian metric: an inner product at each tangent space, varying smoothly with position. This local measuring rule determines angles, arc length, Riemannian distance and geodesics. An inhabitant of a surface can discover its geometry by measuring within it, without knowing how it sits in a larger Euclidean space.

A circular cylinder is a particularly useful example. Its surface bends in space, yet unrolling it preserves the induced metric: a small patch has the geometry of a flat sheet. By contrast, a round sphere has positive Gaussian curvature. The distinction between intrinsic flatness of a circular cylinder and extrinsic curvature separates the geometry an inhabitant measures from the way an outside observer sees the surface bend.

Geodesics supply the analogue of straight lines. On a round sphere they follow great circles; on a circular cylinder they become straight lines when lifted to an unrolled sheet. Their arc length lets us compare candidate routes, as in great-circle distance and Riemannian distance on a circular cylinder. The same measuring rule also defines differential operators such as the Laplace-Beltrami operator, connecting geometry with diffusion and spectral geometry.

Bernhard Riemann's 1854 inaugural lecture proposed studying spaces of arbitrary dimension through their infinitesimal measuring rules, including a positive quadratic expression for the squared line element. This moved the subject beyond a particular surrounding space: the measuring rule itself became part of the geometry. ^[ref].

A Riemannian metric is a smoothly varying positive-definite inner product $g_p$ on each tangent space $T_{p}M$ of a smooth manifold. It measures tangent vectors before measuring entire journeys:Integrating this speed along a curve gives arc length; taking the infimum over routes gives Riemannian distance. A smooth manifold together with this measuring rule is a Riemannian manifold.

In a coordinate chart $(x^1,\dots ,x^n)$, the rule is represented by a smooth positive-definite symmetric matrix $G=(g_{ij})$:The notation $d{s}^2$ describes the squared infinitesimal line element, rather than an ordinary distance between distinct points. Changing coordinates changes the matrix entries; it preserves the measured norm of the same tangent vector.

For the plane with $d{s}^2=4d{x}^2+d{y}^2$, the vector $(1,0)$ has norm $2$ and $(0,1)$ has norm $1$. The coordinate change $X=2x,Y=y$ turns this into $d{s}^2=d{X}^2+d{Y}^2$: coefficients that look different can describe the same geometry. On a circular cylinder the induced metric $R^{2}d{\theta }^2+d{z}^2$ similarly becomes Euclidean when unrolled. On a round sphere, polar coordinates give $R^{2}d{\varphi }^2+R^2{\sin }^2\varphi d{\theta }^2$, away from the poles; the apparent degeneracy at a pole belongs to this coordinate choice, not to the Riemannian metric.

The point of the construction is that measurements make sense intrinsically, even without a chosen surrounding Euclidean space. It gives a common starting point for geodesics, curvature encoded by the Riemann curvature tensor, and the Laplace-Beltrami operator. ^[ref].

Let an Archimedean spiral have polar radius $r=at$ and angle $t$ in polar coordinates, with $0\le t\le R/a$, where $R>0$ is fixed. Parameterizing by $r$ gives the arc lengthThe bounds $r\le \sqrt{a^2+r^2}\le r+a$ yieldThe squeeze theorem proves the displayed limit. Thus the arc length grows like $R^2/(2a)$ as the turns become closely spaced, without requiring an explicit antiderivative.

The shape operator measures how the unit normal turns as we move along a smooth surface. A plane has a constant unit normal, whereas the outward unit normal of a sphere turns in every direction in the tangent plane. Keeping the direction of this change, rather than just its magnitude, lets one recover the principal curvatures and their principal directions.

Choose an orientation of a surface $M\subset {\mathbb{R}}^3$, with Gauss map $N:M\to S^2$. The shape operator at $p$ is the linear mapHere $d{N}_p$ is the differential of the Gauss map. Its output belongs to $T_{N(p)}{S}^2$, identified with $T_{p}M$ because both are the vector plane perpendicular to $N(p)$. The minus sign is our convention throughout this branch. ^[ref]

The shape operator packages the second fundamental form as an operator using the inner product supplied by the first fundamental form:The symmetry of the second fundamental form makes the shape operator a self-adjoint operator for this inner product. Thus an orthonormal basis of principal directions diagonalizes it. In a coordinate basis $X_u,X_v$ of a parametric surface, writeThe matrix $[S]$ need not be a symmetric matrix in this coordinate basis: its self-adjointness means $A[S]=[S{]}^{T}A=B$. Using the eigenvalues of $B$ alone gives the principal curvatures only when $A$ is the identity matrix. ^[ref]

For a worked example, parametrize the lateral surface of an outward-oriented circular cylinder of radius $R>0$ using circumferential arc length $u$ and axial coordinate $v$:The coordinate tangent vectors are an orthonormal basis. Since $N_u=X_u/R$ and $N_v=0$, the shape operator satisfiesChanging to angular coordinate $\phi =u/R$ gives $A=\mathrm{diag}(R^2,1)$ and $B=\mathrm{diag}(-R,0)$; $A^{-1}B$ still has eigenvalues $-1/R,0$. This is why the first fundamental form must participate in the calculation.

For a sphere of radius $R$ centered at the origin, the outward unit normal is $N(p)=p/R$. Hence $S_{p}v=-v/R$, so both principal curvatures are $-1/R$. For a plane, $dN=0$ and the shape operator vanishes. Reversing the unit normal negates the shape operator and both principal curvatures; the Gaussian curvature remains unchanged while the mean curvature changes sign, as in orientation reversal of surface curvature.

Fix the standard orientation of a vector space on the plane. For an oriented regular curve in that plane, let $T=(T_x,T_y)$ be its unit tangent vector and let $N_L=(-T_y,T_x)$ be its left unit normal, obtained by an anticlockwise rotation through $\pi /2$. Its signed curvature $k$, also commonly written ${\kappa }_s$, is defined byHere $s$ increases with the chosen traversal and $\det$ is the determinant. Positive signed curvature means turning left; negative signed curvature means turning right. The left unit normal exists even when $k=0$; when $k<0$ it points opposite to the principal normal vector. See Ghomi's university lecture notes, section 1.10.

For an arbitrary regular parametric curve $r(t)=(x(t),y(t))$ traversed with increasing $t$, the signed curvature isFor the anticlockwise circle $r(t)=(R\cos t,R\sin t)$, the numerator is $R^2$ and the denominator is $R^3$, so $k=1/R$. For the clockwise traversal $q(t)=(R\cos t,-R\sin t)$, the numerator is $-R^2$, so $k=-1/R$, while unsigned curvature is $1/R$ in both cases. On this circle $N_L$ points inward for anticlockwise traversal and outward for clockwise traversal. For any simple closed regular curve traversed anticlockwise, $N_L$ is the inward unit normal, but signed curvature can still be negative along nonconvex portions.

To see why the formula uses the cube of the speed, write $r^{\prime }=vT$ and $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$. The term parallel to $T$ contributes zero to the determinant, giving $\det (r^{\prime },r^{\prime \prime })=v^3\det (T,dT/ds)=v^{3}k$. Under a reparametrization $q(u)=r(\varphi (u))$ with ${\varphi }^{\prime }\ne 0$, the numerator gains a factor $({\varphi }^{\prime }{)}^3$ and the denominator gains $|{\varphi }^{\prime }{|}^3$. Thus orientation-preserving reparametrization preserves signed curvature, while reversing traversal changes its sign at corresponding points. Unsigned curvature is preserved in either case.

Locally write the unit tangent vector as $T=(\cos \theta ,\sin \theta )$. Differentiating with respect to arc length gives $k=d\theta /ds$: signed curvature records the direction as well as the amount of turning. Its integral therefore measures net turning, as in the turning tangent theorem; the integral of $|k|$ is total curvature and counts both left and right turns positively.

Speed is the nonnegative magnitude of velocity, a scalar rather than a vector. In orthonormal Cartesian coordinates,Its SI unit is $m/s$. In the reversal example under velocity, the endpoint velocities $-2m/s$ and $+2m/s$ both have speed $2m/s$, but the speed falls to zero between them. A speed reading alone therefore cannot tell which way the body is moving.

For a smooth path, the total distance traveled is its arc length, $L={\int }_{t_1}^{t_2}v(t)dt$. Unlike the integral of a signed velocity component, this integral cannot cancel outward and return motion. The reversal example travels $2\mathrm{m}$ in $2\mathrm{s}$, giving an average speed of $1m/s$, while its average velocity is zero. This makes speed useful for estimating travel duration or fuel use, while velocity also tracks progress toward a destination. ^[ref]

For a sphere with metric $a^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$, physical arc length $s$ gives $h={\sin }^2\theta d\varphi /ds$. The cyclic coordinate $\varphi$ proves conservation of $a^{2}h$, while unit-speed normalization givesThis azimuthal form requires $h\ne 0$. The solutions are great circles; meridians have $h=0$ and require another parameter.