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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 an integral projective curve on a smooth projective surface, blowing up a point of multiplicity $m$ gives $C^{\prime }={\pi }^{\ast }C-mE$. The intersection formula for blowing up a surface and canonical divisor formula for a surface blowup yield $C^{\prime 2}=C^2-m^2$ and $K_{X^{\prime }}\cdot C^{\prime }=K_X\cdot C+m$. Substitute both in the arithmetic adjunction formula on a smooth surface. This does not require the transformed curve to be smooth.

An asymptote of an unbounded curve is a straight line whose distance from the curve tends to zero along a specified end. For a graph, a sloping asymptote $y=mx+b$ satisfies $y(x)-(mx+b)\to 0$ as $x$ tends to the relevant infinity. Factorization of the leading quadratic part locates the two asymptotes of a nondegenerate hyperbola.

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.

Chaikin subdivision generates a rounded subdivision curve by repeatedly cutting the corners off a control polygon. Each edge contributes two new control points, one a quarter of the way along the edge and the other three quarters of the way along it. Connecting these new control points in edge order replaces the old corners by shorter line segments. This makes Chaikin subdivision a useful way to draw a smooth outline from a small, editable set of control points, without evaluating B-spline basis functions at each drawing step. Joy presents both the corner-cutting rule and its closed version.^[ref]

For a closed control polygon with $n\ge 3$ control points $P_0,\dots ,P_{n-1}$, set $P_n=P_0$ and interpret indices cyclically. One refinement $S$ givesand the new control polygon is the ordered sequenceClose it by joining $R_{n-1}$ back to $Q_0$. Use the old control points for every average in the same refinement; updating them in place would define a different algorithm. One refinement produces $2n$ stored control points, and $k$ refinements produce $2^{k}n$, counting entries even if some coordinates coincide. Open control polygons need an explicit endpoint convention: applying only the interior edge rule discards the original endpoints, whereas retaining endpoints changes the boundary rule. The closed analysis below uses cyclic indices throughout.

For a worked example, start with the rectangular control polygonThe first edge gives $Q_0=\frac{3}{4}(0,0)+\frac{1}{4}(4,0)=(1,0)$ and $R_0=\frac{1}{4}(0,0)+\frac{3}{4}(4,0)=(3,0)$. Applying Chaikin subdivision to all four edges, including the closing edge, gives exactly eight control points in this order:Thus the rectangle becomes an octagonal control polygon. Two refinements have $16$ control points, and five have $128$. The illustration shows finite control polygons at stages $0,1,2,5$; the last is already visually rounded but still consists of line segments.

The convex hull property follows directly from the weights. Each new control point is a convex combination of the endpoints of one old edge, because the weights are nonnegative and sum to one. Consequently, if $H_k$ is the convex hull of the stage-$k$ control points, then $H_{k+1}\subseteq H_k$. Every later line segment, and every limit point, stays in the original convex hull. This gives a useful geometric enclosure. It also explains why Chaikin subdivision is approximating rather than an interpolation rule: the rectangle's four original corners are outside the first octagon's convex hull, so the limiting curve cannot pass through them. A refinement control polygon generally lies off the limiting curve as well.

For the cyclic uniform rule, the closed limit is the periodic quadratic uniform B-spline curve determined by the original control points. Uniform refers to equal spacing in the spline knot sequence, not equal physical lengths of edges of the control polygon. An infinite uniform control sequence has the corresponding local description; arbitrary open endpoint conventions need separate treatment. Joy derives the quarter-point refinement by splitting quadratic uniform B-spline spans.^[ref]

Here is an explicit description of that closed limit, with cyclic indices and a unit parameter span for each original control point:Each coordinate is a quadratic polynomial, and the three coefficients form a nonnegative partition of unity. The span runs from $(P_{i-1}+P_i)/2$ to $(P_i+P_{i+1})/2$. Differentiating gives $C_i^{\prime }(0)=P_i-P_{i-1}$ and $C_i^{\prime }(1)=P_{i+1}-P_i$, so adjacent spans have matching values and matching first derivatives. Their second derivatives generally differ. Thus the parameterized curve is $C^1$, generally not $C^2$; degenerate control points can still make its tangent vector vanish. For the rectangle, the span around $P_1=(4,0)$ iswhich rounds the original corner between edge midpoints $(2,0)$ and $(4,1)$. For equally spaced abscissae, the centered Chaikin basis function gives the scalar graph version of this quadratic B-spline representation and has support of width three original grid intervals.

Another useful property is affine equivariance of a geometric basis: refining and applying an affine map commute. For $T(x)=Ax+b$,and the same calculation holds for $R_i$. The sum-to-one weights account for the translation. Therefore $T(S(P))=S(T(P))$ at every refinement, and also for the limit. A designer can apply a shear mapping, translation, or other affine map to the original control polygon and obtain exactly the transformed outline.

For example, compose the shear mapping $(x,y)\mapsto (x+y/2,y)$ with the translation $(1,-1)$ to obtain $T(x,y)=(x+y/2+1,y-1)$. The rectangle's control points become $(1,-1),(5,-1),(6,1),(2,1)$. On its right edge, cutting first gives $Q_1=(4,1/2)$ and then $T(Q_1)=(21/4,-1/2)$. Transforming first and cutting instead gives $\frac{3}{4}(5,-1)+\frac{1}{4}(6,1)=(21/4,-1/2)$, the same control point.

Historically, George M. Chaikin published “An algorithm for high-speed curve generation” in Computer Graphics and Image Processing 3(4), December 1974, pages 346–349. Its emphasis was practical curve generation through a recursive algorithm using simple arithmetic, with applications including machine control and three-dimensional curves. The lasting appeal of Chaikin subdivision is how an elementary local average creates an entire smooth B-spline curve.^[ref]

For a single-valued curve below the chord joining $(-a,0)$ to $(a,0)$, with length $2a<L\le \pi a$, the maximum enclosed area is attained by the lower minor circular arc. Equivalently it minimizes ${\int }_{-a}^{a}ydx$ at fixed length. Its equation and length areTo prove global minimality for $R>a$, augment the objective by $R\int \sqrt{1+y^{\prime 2}}dx$. The augmented integrand is linear in $y$ and strictly convex in $y^{\prime }$. The displayed arc satisfies its Euler-Lagrange equation, so the supporting-line inequality for the convex slope dependence, followed by integration by parts, proves it minimizes the augmented functional among curves with the prescribed endpoints. On the fixed-length class the multiplier term is constant. The semicircle endpoint follows by a limit; its endpoint tangents are vertical but its length is finite.

A curve $\gamma :[0,1]\to X$ whose endpoints agree, $\gamma (0)=\gamma (1)$. This is the same unbased notion as a loop; self-intersections are allowed.

A complex integration contour is an oriented piecewise continuously differentiable curve in the complex plane. For a parametrization $\gamma (s)$, its contour integral is ${\int }_{\Gamma }f(z)dz=\int f(\gamma (s)){\gamma }^{\prime }(s)ds$. Reversing orientation changes the sign. Unbounded contours require a convergent improper or explicitly regularized limit.

Computer aided geometric design studies computational representations, evaluation and manipulation of curves and regular surfaces, especially using splines and subdivision surfaces.

// Target: geometry-and-topology.bigb

For a regular curve with positive curvature of a space curve, the Frenet-Serret formulas give $dT/ds=\kappa N$. If the unit tangent vector has constant component along a fixed vector $a$, differentiation of $T\cdot a$ gives $\kappa N\cdot a=0$. Hence every principal normal vector is perpendicular to that direction. Nonzero curvature is needed to define the principal normal.

A continuation method follows a solution branch of a parameter-dependent equation from a known solution, using nearby solutions to guide the next solve. This is useful when one wants to understand how solutions change, rather than obtain just one isolated answer. A continuation predictor-corrector step follows a tangent vector and then uses a Newton method correction on a transverse section. The same geometry supports both pseudo-arclength continuation through a continuation fold and transversal surface intersection tracing.

Consider the original scalar polynomial exampleIts solution branch bends back twice in $\lambda$. Increasing $\lambda$ alone eventually stops following the same nearby solutions; following the curve in the $(x,\lambda )$ plane can continue around both bends. The continuation folds areFor $|\lambda |<2/(3\sqrt{3})$ there are three distinct real zeros of a function at that fixed $\lambda$, all lying on this one connected curve. A continuation method keeps their geometric relationship visible.

Every continuation method needs a seed, supplied by an independent solve, an explicitly known solution, or another search. Following one solution branch does not guarantee finding every connected component of the solution set. A large step can miss detail or make the correction converge to a distant solution. At branch intersections or a loss of rank of the full Jacobian matrix, the regular-curve argument needs separate analysis. The seed and component limitations also matter in transversal surface intersection tracing.