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The arc length of a piecewise smooth curve $\gamma :[a,b]\to M$ on a Riemannian manifold is the integral of its speed:An odometer measures this accumulated travel rather than displacement between the endpoints. In Euclidean space, the Riemannian metric is the ordinary inner product, so the formula becomes ${\int }_a^b\sqrt{{\sum }_i{\dot{\gamma }}_i(t{)}^2}dt$.

For a circle of radius $R$ traversed through an angle $\alpha$, use $\gamma (t)=(R\cos (\alpha t),R\sin (\alpha t))$, $0\le t\le 1$. Its speed is $R|\alpha |$, hence its arc length is $R|\alpha |$. Traversing a full circle twice gives $4\pi R$, even though the starting and ending points coincide. The same formula measures the two great circle arcs pictured in geodesic.

The curve parameter controls how fast the journey happens, not its arc length. A regular increasing or decreasing reparametrization preserves the integral by change of variables; reversing direction leaves the speed norm unchanged. Retracing part of a route does add length, because it changes the actual journey.

For the cycloid $x=a(t-\sin t)$, $y=a(1-\cos t)$, $a>0$, $0\le t\le 2\pi$, the area between the arch and its baseline isIts arc length isThe positive sign of $\sin (t/2)$ inside this parameter interval removes the absolute value in the speed. Omitting the two endpoint cusps does not change either integral.

The part of an elastic rod's elastic energy that penalizes departure of its bending strain of a rod $u$ from the intrinsic value $\hat{u}$. Here $B$ is the bending stiffness and $S$ is reference arc length. For an inextensible rod, the same formula can be written using spatial curvature and current arc length.

For planar director kinematics of a rod measured in reference arc length $S$, define the signed bending strain by $d{\mathbf{d}}_3/dS=u{\mathbf{d}}_1$. If $s$ is current arc length and $\alpha =ds/dS>0$ is the stretch ratio, the signed spatial curvature is $u/\alpha$: the chain rule gives $d{\mathbf{d}}_3/ds=(u/\alpha ){\mathbf{d}}_1$. Thus bending strain and spatial curvature coincide for an inextensible rod, but differ in general. An intrinsic value $\hat{u}$ is the preferred reference-length bending strain in an energy density $B(u-\hat{u}{)}^2/2$ per unit reference length.

A Bézier curve is a parametric curve specified by control points $P_0,\dots ,P_n$ in a Euclidean space and the Bernstein basis:Here $\left(\genfrac{}{}{0ex}{}{n}{i}\right)$ is a binomial coefficient. Each coordinate is a polynomial of polynomial degree at most $n$; special choices of control points can lower the polynomial degree. The ordered control points form the control polygon. A cubic Bézier curve uses four control points and has the explicit formThe curve passes through $P_0$ and $P_3$; the interior control points steer its shape and need not lie on it.^[ref]

For a worked example, chooseSubstituting these control points givesHere $y(t)=1.5-6(t-1/2{)}^2\le 1.5$, so neither interior control point, both at height $2$, lies on the curve. De Casteljau's algorithm computes the same point using only successive linear interpolations.

The endpoint tangent vectors follow from the Bézier derivative control polygon:Thus the curve leaves $P_0$ toward $P_1$ and arrives at $P_3$ from $P_2$. These nonzero derivatives determine the endpoint tangent lines; if an endpoint derivative vanishes, that formula alone does not determine its tangent direction. Endpoint handles are useful when arranging a parametric first-derivative join of Bézier curves.

The parameter $t$ is usually not arc length. For this example,Equal steps in $t$ therefore do not generally travel equal distances. Distance along the curve is instead the arc length, given by the integral $s(t)={\int }_0^t\Vert C^{\prime }(u)\Vert du$; its derivative is the norm of the tangent vector and measures the rate of travel.

Two geometric facts make Bézier curves especially convenient for design. First, the Bernstein basis weights are nonnegative and form a partition of unity, since the binomial theorem gives ${\sum }_i{b}_{i,n}(t)=((1-t)+t{)}^n=1$. Hence $C(t)$ is a convex combination of the control points and lies in their convex hull.^[ref] The convex hull is the smallest convex set containing the control points, not their axis-aligned bounding box. Here it is a trapezoid, strictly smaller than $[0,4]\times [0,2]$: the box corner $(0,2)$ is outside the convex hull. The curve need not fill or touch every edge of either enclosure. The convex hull guarantee concerns $0\le t\le 1$; outside that interval weights can be negative.

Second, applying an affine map to every control point gives exactly the same result as applying it to the entire curve. For $T(x)=Ax+b$,This is affine equivariance of a geometric basis: a translation, rotation or shear mapping can be performed on a handful of control points. For example, $T(x,y)=(2x+y,y+1)$ transforms the four control points to $(0,1),(4,3),(8,3),(8,1)$ and transforms the midpoint to $(5.5,2.5)$, also obtained from the same weights $(1,3,3,1)/8$.

These editable handles explain the usefulness of Bézier curves in computer aided geometric design, vector graphics paths and outline fonts: a small collection of points controls an entire smooth stroke, and resizing the design preserves its mathematical shape. Cambridge's computer-graphics notes describe their use in graphic design, typography and PostScript outlines. The industrial context includes Pierre Bézier's work at Renault and Paul de Casteljau's work at Citroën.^[ref]

// Target: analysis.bigb

Measure polar angle $\theta$ from the downward vertical on the lower hemisphere of radius $R$. A particle released from rest at the rim and sliding without energy loss has speed $\sqrt{2gR\cos \theta }$. The spherical arc length along $\varphi =\varphi (\theta )$ is $R\sqrt{1+{\sin }^2\theta {\varphi }^{\prime 2}}|d\theta |$. Thus its travel time has integrand proportional to $\sqrt{(1+{\sin }^2\theta {\varphi }^{\prime 2})/\cos \theta }$. Since the azimuth is a cyclic coordinate, the Euler-Lagrange equation givesSolving for ${\varphi }^{\prime 2}$ gives the displayed first integral on a segment where $\theta$ is a valid path parameter. For finite $C$, approaching the rim $\theta \uparrow \pi /2$ gives ${\varphi }^{\prime }\to 0$. Thus a regular stationary path begins along the meridian toward the bottom. This is an initial direction condition; a path joining other rim points can turn before reaching the bottom. Monotone segments use positive $|d\theta |$ in the time integral, and a turning point requires another path parameter or a split into segments.

A catenary is the static shape of an ideal flexible, inextensible chain hanging under its own gravitational weight in a uniform gravitational field, with uniform linear mass density along its arc length. With the lowest point above $x=0$, one convenient vertical origin givesHere $a$ has units of length. The supports and the chain's arc length determine the scale and position; horizontal and vertical shifts give the general catenary. ^[ref]

For $a=1$, the three points at $x=-1,0,1$ have heights $\cosh 1\approx 1.543081$, $1$, and $\cosh 1\approx 1.543081$. If heights should instead be measured from the lowest point, subtract $a$:The same three heights are then approximately $0.543081$, $0$, and $0.543081$. The shift changes the vertical origin and preserves the catenary shape.

The scale also measures tension: if $H>0$ is its constant horizontal component, $\mu$ is the linear mass density, and $g$ is the magnitude of gravitational acceleration, then $a=H/(\mu g)$. A short force balance explains this. For the portion from the lowest point to $x\ge 0$, with arc length $s(x)$, vertical tension balances its gravitational weight and horizontal tension givesApplying differentiation yields $a{y}^{\prime \prime }=\sqrt{1+y^{\prime 2}}$. The proposed hyperbolic cosine solves this differential equation, because $y^{\prime }=\sinh (x/a)$ and $a{y}^{\prime \prime }=\cosh (x/a)=\sqrt{1+{\sinh }^2(x/a)}$. This is the mechanism behind the shape, under the stated ideal chain assumptions.

The same identity simplifies the arc length between symmetric supports at $x=\pm b$:For $a=b=1$, $L\approx 2.350402$, exceeding the horizontal span of two. A larger $a$ at fixed $b$ gives a shallower sag: the Taylor series gives $a(\cosh (x/a)-1)=x^2/(2a)+x^4/(24a^3)+\cdots$. Thus a parabola is a local approximation, while the exact ideal chain shape is a catenary. The useful connection is that the same hyperbolic Pythagorean identity controls its coordinates, force balance, and arc length.

Among curves of fixed arc length $L>2a$ joining $(-a,0)$ to $(a,0)$, the hanging catenary globally minimizes $\int yds$. Parametrize a competitor by arc length $s\in [-L/2,L/2]$, so $x^{\prime 2}+y^{\prime 2}=1$. Integration by parts and the fixed horizontal endpoint separation give, for every $b>0$,The inequality is the Cauchy-Schwarz inequality. Equality requires $x^{\prime }=b/\sqrt{b^2+s^2}$ and $y^{\prime }=s/\sqrt{b^2+s^2}$. Integration gives the catenaryThe last equation has a unique positive $b$, since $\sinh q/q$ increases strictly for $q>0$. The equality curve is a single-valued graph below the chord, so the same global bound proves the result under those additional restrictions.

A circular arc is a connected portion of a circle. For radius $R>0$ and subtended central angle $0<\theta <2\pi$ measured in radians, its arc length is $R\theta$. Two distinct endpoints determine two circular arcs: when they are not opposite, the minor arc has $\theta <\pi$ and the major arc has $\theta >\pi$. Opposite endpoints determine two semicircles. The straight line segment joining the endpoints is a chord, rather than the curved circular arc itself.

The circumference of a circle is its boundary arc length. For radius $r$, it equals $2\pi r$. This growing boundary length accounts for the radial factor in the Rayleigh distribution.

For $0\le k\le 1$, define $E(k)={\int }_0^{\pi /2}\sqrt{1-k^2{\sin }^2\theta }d\theta$. An ellipse with semi-major and semi-minor axes $a\ge b>0$ and eccentricity of a conic $e=\sqrt{1-b^2/a^2}$ has perimeter $4aE(e)$. Differentiating its parametrization gives the speed $a\sqrt{1-e^2{\sin }^2\theta }$, and fourfold symmetry reduces the full arc length to a quadrant. The binomial series gives $E(k)=(\pi /2)(1-k^2/4-3k^4/64+O(k^6))$ near zero.

The lateral surface obtained by rotating a straight segment in a plane containing the rotation axis has endpoint radii $a,c$ and slant length $\ell$. Its surface area is $\pi (a+c)\ell$: the radius is linear in the segment's arc length, so the surface area of a surface of revolution is $2\pi {\int }_0^{\ell }\rho (s)ds=2\pi \ell (a+c)/2$. This formula excludes the end disks and assumes that the segment's radial direction is fixed. A straight segment whose endpoint horizontal directions differ generally sweeps a different surface of revolution.

For a parametric curve represented by fixed scalar functions $b_j$ and control points $P_j$ in Euclidean space, writeThe influence function of this handle is $b_i$: the resulting displacement vector is $\Delta C(t)=\tilde{C}(t)-C(t)=b_i(t)h$, and its norm is $|b_i(t)|\Vert h\Vert$. These are exact finite changes, not only a first-order estimate. Where $b_i$ is negative the displacement points opposite to $h$. Where $b_i(t)=0$, the curve is unchanged at that parameter.

For the cubic Bézier curve above, the Bernstein basis influence function is $b_{1,3}(t)=3t(1-t{)}^2$. Its derivative isIt increases up to $t=1/3$ and decreases thereafter on $[0,1]$. Consequently the largest displacement at corresponding parameters isFor $h=(0,1)$ it is $4/9$. This is a comparison at the same parameter. A geometric closest-distance comparison, as in a closest-point search on a subdivision curve, instead minimizes over another parameter:The closest parameter $s$ can differ from $t$, so $4/9$ is not automatically a maximum closest-distance separation of the two curves. Neither parameter is assumed to be arc length.

The support of a function $b_i$ is the closure of the parameter values where $b_i\ne 0$. It bounds where a nonzero handle move can act, but may include boundary points where the actual displacement is zero. Here the support of a function $b_{1,3}$ is all of $[0,1]$ relative to the parameter domain, while the endpoints are unaffected. In contrast, a B-spline of order $k$ has support within $[t_i,t_{i+k}]$ in its spline knot sequence, so its handle affects only that parameter range; the support and Gram bandwidth of B-splines describes the same locality in another application. Changing the spline knots or the basis along with the handle requires a different comparison.

When the basis functions form a partition of unity, ${\sum }_j{b}_j(t)=1$, each evaluation is an affine combination of the control points. Applying an affine map $T(x)=Ax+a$ to all handles givesThis is affine equivariance of a geometric basis, also called affine covariance: affine maps preserve affine combinations. In particular, translating every handle by $h$ translates the whole curve by $h$, whereas moving just one handle produces $b_i(t)h$. The Bernstein basis has this partition of unity property.^[ref] Nonnegative weights further make evaluations convex combinations, giving the convex hull bound illustrated by the control polygon; nonnegativity is not needed for affine covariance.

For a proof outline of the influence formula, subtract the two basis representations: every unchanged coefficient cancels, leaving $b_i(t)(P_i+h)-b_i(t)P_i=b_i(t)h$. The same subtraction with several moved handles gives $\Delta C(t)={\sum }_i{b}_i(t)h_i$. No claim about interpolation, closest-point search on a subdivision curve or endpoint motion follows until the actual basis values are examined.

For a twice continuously differentiable regular curve $r(t)$ in Euclidean space, its unsigned curvature measures how quickly its unit tangent vector changes per unit arc length. WriteThe curvature iswhere the norm is the Euclidean norm. Differentiating with respect to arc length removes the effect of traversal speed; differentiating $T$ with respect to $t$ alone generally does not. See Ghomi's university lecture notes, section 1.7.

For a circle of radius $R>0$, use $r(t)=(R\cos t,R\sin t)$. Then $v=R$, $T=(-\sin t,\cos t)$ and $\Vert dT/dt\Vert =1$, giving $\kappa =1/R$. A straight line $r(t)=p+ta$ with $a\ne 0$ has constant unit tangent vector, so $\kappa =0$.

The same circle traversed as $q(u)=(R\cos (3u),R\sin (3u))$ has speed $3R$ and $\Vert dT/du\Vert =3$: its curvature remains $1/R$. By contrast, enlarging the circle to radius $2R$ changes its geometry and gives curvature $1/(2R)$. More generally, a dilation $r\mapsto ar$ with $a>0$ divides curvature by $a$, because it multiplies arc length by $a$ while leaving unit tangent vectors unchanged. Thus curvature has units of inverse length, and a reparametrization changes neither its value at a corresponding point nor the shape.

Differentiating $T\cdot T=1$, using the dot product, shows $T\cdot (dT/ds)=0$: the change is perpendicular to the unit tangent vector. Wherever $\kappa >0$, the principal normal vector $N_p=(dT/ds)/\kappa$ points toward the bend. For a plane curve, the osculating circle at such a point has radius of curvature $\rho =1/\kappa$ and center $r+\rho N_p$. It matches the point, tangent line, and curvature, providing a local circular approximation. At $\kappa =0$ there is no finite osculating circle defined by this formula. See Ghomi's university lecture notes, section 1.13.

This makes curvature useful for comparing bends without confusing them with speed. If $t$ denotes time, differentiating $r^{\prime }=vT$ gives acceleration $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$, whose component perpendicular to the unit tangent vector has magnitude $v^2\kappa$. The curvature of a plane curve example below computes the bend and osculating circle of a parabola explicitly.

For smooth surfaces, bending has more than one direction. The principal curvatures $k_1,k_2$ instead define Gaussian curvature $K=k_1{k}_2$ and mean curvature $H=(k_1+k_2)/2$. These are surface quantities, distinct from the curvature of an individual curve. In particular, the circle value $1/R$ must not be confused with the Gaussian curvature $1/R^2$ of a sphere of radius $R$. See Ritter's Oxford lecture notes, section 12.2.

For a continuously differentiable nonzero vector field $F$, the curvature of a space curve following its direction isTake the unit tangent vector $t=\pm F/|F|$ and differentiate with respect to arc length. The derivative of $|F{|}^{-1}$ is parallel to $F$ and disappears from $t\times t^{\prime }$. Since $t\cdot t^{\prime }=0$, we have $|t\times t^{\prime }|=|t^{\prime }|=\kappa$, proving the formula. It remains valid with zero curvature, even where the usual Frenet frame is undefined.

A curve parameter labels the points visited by a parametric curve $\gamma (t)$. Choosing time as the parameter determines its velocity; an arc-length parametrization instead measures accumulated arc length. A regular monotone reparametrization changes the labels and speed while preserving the route and its arc length.

Differential geometry uses calculus to study smooth curves, smooth surfaces, and smooth manifolds. A tangent space describes possible instantaneous motions; a Riemannian metric turns those motions into measurable lengths and angles; curvature records how the resulting geometry differs from Euclidean geometry. Local calculations with derivatives can therefore answer geometric questions about whole shapes.

An especially revealing experiment is rolling a flat sheet into a circular cylinder without stretching it. The sheet visibly bends, yet a small creature measuring only arc lengths, angles, and surface areas detects the same geometry. This separates geometry measured within a surface from its bending in the surrounding Euclidean space. The first fundamental form encodes the measurements; the second fundamental form encodes the bending. The Theorema Egregium explains why a sphere behaves differently, while geodesics on a circular cylinder turn a route-planning problem into drawing a straight line.

The differential geometry of smooth surfaces studies their induced Riemannian metrics and intrinsic or extrinsic curvature. The first fundamental form measures lengths, while the second fundamental form and shape operator describe bending in an ambient Euclidean space. A regular parametric surface $X(u,v)$ must have linearly independent tangent vectors $X_u,X_v$: otherwise its parameters cannot provide a manifold chart with two independent directions.

For a sphere of radius $R>0$, use colatitude $0<\theta <\pi$ and longitude $\varphi$:Thus a change $d\varphi$ has arc length $R\sin \theta |d\varphi |$: equally spaced longitudes draw closer together toward a pole. The coordinate degeneracy at the poles belongs to this parametrization, not to the smooth surface.

Both principal curvatures of the sphere have magnitude $1/R$ and the same sign, giving Gaussian curvature $K=1/R^2$. A plane has $K=0$, and a circular cylinder also has $K=0$ because one principal curvature vanishes. These three familiar shapes show why visible bending alone cannot measure intrinsic Gaussian curvature.

For a differentiable material frame of a rod, there is a unique vector $\mathbf{u}$ such that ${\mathbf{d}}_i^{\prime }=\mathbf{u}\times {\mathbf{d}}_i$. To see this, form the orthogonal matrix $R=({\mathbf{d}}_1,{\mathbf{d}}_2,{\mathbf{d}}_3)$. Differentiation of $R{R}^T=I$ shows that $R^{\prime }{R}^T$ is a skew-symmetric matrix; every three-dimensional skew-symmetric matrix has the unique form $\mathbf{v}\mapsto \mathbf{u}\times \mathbf{v}$. Primes may denote differentiation with respect to reference or current arc length, but that choice must be stated. In a planar bend about ${\mathbf{d}}_2$, reference differentiation gives ${\mathbf{d}}_3^{\prime }=u{\mathbf{d}}_1$ and ${\mathbf{d}}_1^{\prime }=-u{\mathbf{d}}_3$.

For real modulus $0\le k\le 1$, the displayed integral is the incomplete elliptic integral of the second kind, with amplitude $\varphi$. The modulus convention matters: some sources instead use the parameter $m=k^2$. The arc length of an ellipse reduces to this integral, because a trigonometric parametrization gives a square root of $1-k^2{\sin }^2\theta$. At $k=0$ it reduces to $\varphi$; at $k=1$ the real integrand is $|\cos \theta |$.