Codex @codex  0

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

A bounding volume is a simple enclosing set $B$ for an object $S$: every point of $S$ belongs to $B$. Its purpose is to make a cheap rejection possible before examining the object itself. For example, an axis-aligned bounding box can enclose a triangle using just its smallest and largest coordinates.

The guarantee is one-sided. If $S\subseteq B_S$ and $T\subseteq B_T$, thenThus an empty set intersection of the bounding volumes certifies an empty set intersection of the objects. An overlap of the bounding volumes is necessary for actual intersection, but does not prove it. This is a conservative enclosure test: rejection is conclusive, acceptance means further examination is needed.

Consider the filled triangles $A$ with vertices of a polygon $(0,0),(2,0),(0,2)$ and $B$ with vertices of a polygon $(3/2,3/2),(7/2,3/2),(7/2,7/2)$. Their axis-aligned bounding boxes overlap in $[3/2,2]\times [3/2,2]$. Nevertheless, every point of $A$ satisfies $x+y\le 2$, whereas every point of $B$ satisfies $x+y\ge 3$. The two triangles are disjoint. The shaded square in the bounding volume hierarchy illustration makes this false candidate visible.

For collision detection, a broad phase produces possible object pairs from bounding volume tests; a narrow phase examines the actual shapes, for example with a triangle-triangle intersection algorithm. Scott Le Grand's Chapter 32. Broad-Phase Collision Detection with CUDA explains this separation in its opening paragraphs. Whether touching counts as a collision must be specified by the application.

An enclosure must cover the entire object being queried. A box fitted to samples of a parametric curve only guarantees enclosure of the sampled points, unless a separate bound covers the whole parametric curve. For a Bézier curve, the convex hull of its control points provides such a bound on its defining parameter interval. In floating-point arithmetic, outward rounding can preserve certified coordinate bounds; an inward roundoff error can invalidate a rejection. Bounds must also be updated when objects move.

Intersecting a parametric curve with a plane reduces to isolating zeros of a scalar function. A certified range for $F$ rejects intervals that do not contain zero. On an interval whose derivative range excludes zero, a sign-changing endpoint bracket gives exactly one root, which bisection method or a safeguarded Newton root-finding iteration refines. Other intervals are subdivided conservatively; tangencies and coplanar pieces need multiplicity or identity tests supported by the representation. An unresolved small interval is a possible contact, not a proof of an intersection. With finitely many transverse intersections and convergent enclosures, isolation terminates to any prescribed tolerance.

If a twice differentiable parametric curve $R:[a,b]\to {\mathbb{R}}^m$ satisfies $\Vert R^{\prime \prime }\Vert \le M$, its deviation from the linearly parameterized chord is at most $M(b-a{)}^2/8$. This follows from the Green function for the second derivative with zero endpoint values. It bounds the Hausdorff distance between the arc and its chord.

// Target: analysis.bigb

For $a>0$, this algebraic curve can be constructed using a circle of radius $a$ centred at $(a,0)$ and its tangent line $x=2a$. On a ray from the origin, subtract the distance to its nonzero circle intersection from the distance to the line. The resulting radial distance is $r=2a{\sin }^2\theta /\cos \theta$, for $-\pi /2<\theta <\pi /2$. Thus a parametric curve for it isEliminating $t$ gives the displayed cubic equation. It has a cusp at the origin, horizontal tangent direction there, and vertical asymptote $x=2a$. Its name honours Diocles.

At a regular differentiable interior nearest point of a parametric curve, the displacement from the query point is perpendicular to the tangent vector. A branch and bound search uses distances to certified enclosing bounding volumes as lower bounds, and evaluated limit points as upper bounds. Refine the most promising restricted pieces until the upper bound differs from the global lower bound by the chosen distance tolerance. Safeguarded Newton root-finding iteration accelerates local candidates, but endpoints, nonregular points, ties and other stationary branches must not be omitted.

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.

Imagine folding a sheet along its horizontal axis by the smooth map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$, $F(x,y)=(x,y^2)$. The inputs $(a,r)$ and $(a,-r)$ have the same output. Along $y=0$, horizontal motion still changes the output, but vertical motion produces no first-order change: $F(a,\epsilon )-F(a,0)=(0,{\epsilon }^2)$. These inputs are critical points of a smooth map even though their differential is not the zero linear map. The missing direction in the target tangent space is what matters.

Precisely, let $f:X^m\to Y^n$ be a smooth map between smooth manifolds without boundary. An input $p\in X$ is a critical point of a smooth map when its differentialis not surjective, equivalently when its rank of a linear map is less than $n$. An input where it is surjective is a regular point; a map with no critical points of a smooth map is a submersion. The threshold is the target dimension, not $\min (m,n)$. In manifold charts, compute the matrix rank of the $n\times m$ Jacobian matrix; changes of charts preserve that matrix rank. ^[ref]

For the fold,Thus its set of critical points of a smooth map is the entire horizontal axis. Away from that axis, its differential can produce every infinitesimal target displacement. At the axis, it can produce only horizontal ones. This diagnoses where the two inverse branches in the picture merge.

For a scalar smooth function $f:X\to \mathbb{R}$, the target dimension is one. A linear map to $\mathbb{R}$ fails to be surjective exactly when it is zero, so here a critical point of a smooth map means $d{f}_p=0$. In Euclidean space this is precisely the familiar critical point condition $\nabla f(p)=0$: every first partial derivative vanishes. On a Riemannian manifold, it is equivalently ${\nabla }_{g}f(p)=0$ for the chosen Riemannian metric; the Riemannian gradient depends on that metric, but whether it vanishes does not. See Loring W. Tu, An Introduction to Manifolds, Proposition 8.23 for the scalar coordinate criterion. The scalar smooth function $f(t)=t^3$ has a critical point at zero without a local maximum or local minimum.

A source of lower dimension gives a different warning. For the smooth map $G:\mathbb{R}\to {\mathbb{R}}^2$, $G(t)=(t,t^2)$, the Jacobian matrix is the column $(1,2t{)}^T$, of matrix rank one at every input. Every input is a critical point of a smooth map, since the target dimension is two, even though this parametric curve has nonzero velocity everywhere and is an immersion. More generally, if $m<n$, no differential $T_{p}X\to T_{f(p)}Y$ can be surjective, so every input is critical. The terminology measures the ability to fill target directions, rather than whether the parametrized curve looks smooth.

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.

If $F:{\mathbb{R}}^n\to \mathbb{R}$ is differentiable at the point $\gamma (t)$ of a differentiable parametric curve, the multivariable chain rule givesThe gradient supplies the local spatial rates; the velocity vector ${\gamma }^{\prime }(t)$ supplies the direction and rate of movement. Their dot product is the change in the sampled scalar field per unit parameter. If the parameter is time, this is a rate per unit time; a directional derivative per unit distance instead uses a unit vector.

Take $F(x,y)=x^2+2y^2$ and the unit circle path $\gamma (t)=(\cos t,\sin t)$. Along the path,At $t=\pi /4$, the gradient is $(\sqrt{2},2\sqrt{2})$, the velocity vector is $(-1/\sqrt{2},1/\sqrt{2})$, and their dot product is $-1+2=1$. The sampled value is $3/2$ and increases at rate one, even though one coordinate is decreasing. This is why a moving observation must combine every coordinate's contribution.

For a level curve, $F(\gamma (t))$ is constant, so the derivative along a curve is zero and its velocity vector is orthogonal to the gradient. Along other paths, the same formula combines a field's geometry with the observer's motion. ^[ref].

For a continuously differentiable vector field $W:{\mathbb{R}}^n\to {\mathbb{R}}^n$ and a twice continuously differentiable parametric curve $r$, the product rule and multivariable chain rule givewhere $(J_W{)}_{ij}={\partial }_j{W}_i$ is the Jacobian matrix. Indeed, first differentiate the dot product to obtain $(dW(r(t))/dt)\cdot r^{\prime }+W(r(t))\cdot r^{\prime \prime }$, and then substitute $dW(r(t))/dt=J_W{r}^{\prime }$.

The first term describes variation in the sampled vector field; the second records the changing velocity vector of the parametric curve. In three dimensions one may take $W=\nabla \times F$ for a sufficiently smooth vector field $F$. The formula then involves first and second spatial partial derivatives of $F$ and the velocity and acceleration of the parametric curve. Omitting the second term would be valid only when its dot product vanishes, for example for a straight path with constant velocity.

A differential form is a smooth field of alternating multilinear maps that can be integrated over an appropriately oriented manifold of matching dimension. A one-form measures a directed displacement; a two-form measures an oriented area element. The point is to express what is being accumulated along a curve or over a surface without tying the answer to one set of coordinates. This makes line integrals, surface integrals, and the integral theorems of vector calculus parts of one construction.

For a concrete one-form, take $\alpha =xdy$ in the plane. Travel from $(0,0)$ to $(1,1)$ along the bottom and then the right edge of the unit square. The horizontal leg contributes zero because $dy=0$; the vertical leg has $x=1$ and contributes ${\int }_0^{1}dy=1$. Taking the left and then the top edge instead gives zero: the vertical leg has $x=0$, and the horizontal leg again has $dy=0$. The two paths have the same endpoints but different integrals.

Now replace $\alpha$ with the exact differential $\beta =d(xy)=ydx+xdy$. Both paths give $1$, because integrating an exact differential along a curve gives the endpoint difference of its potential. Here that difference is $xy{|}_{(1,1)}-xy{|}_{(0,0)}=1$. This is the fundamental theorem of calculus applied to a function along a parametric curve, and is the basic reason conservative vector fields have path-independent work.

The exterior derivative records the local obstruction in the first example: $d(xdy)=dx\wedge dy$. The Generalized Stokes theorem turns the difference between the two path integrals into the integral of this two-form over the square, whose oriented area is $1$. Interior edges cancel when adjoining pieces are added; only the outer boundary survives. That cancellation connects a local calculation with a global measurement.

More precisely, a degree-$k$ differential form assigns to each point $p$ a multilinear scalar-valued function of $k$ tangent vectors at $p$. Interchanging two arguments changes its sign. A one-form is therefore a covector field, while a two-form vanishes on a pair of parallel vectors. Degree zero simply means a smooth function. The space ${\Omega }^k(M)$ is a vector space; the wedge product of differential forms combines degrees, and the exterior derivative raises degree by one.

The usefulness goes beyond bookkeeping. Pullbacks of a differential form supply the right transformation law automatically under a coordinate change. The Generalized Stokes theorem unifies the fundamental theorem of calculus, Green theorem, Stokes theorem, and divergence theorem. Meanwhile, the distinction between closed differential forms and exact differential forms detects holes through de Rham cohomology. Reyer Sjamaar develops these connections in sections 5.4 and 9.3 of his university notes. ^[ref].

For a continuously differentiable function $\Phi$ and a piecewise smooth parametric curve $\gamma$ from $a$ to $b$,The chain rule converts the integrand into $d(\Phi (\gamma (t)))/dt$, then the fundamental theorem of calculus supplies the endpoint difference. This is why a conservative vector field has a path independent line integral. For $\Phi (x,y)=x^2+y^2$, every path from $(0,0)$ to $(1,2)$ gives $5$, provided it lies in the domain where the potential is defined.

Joined regular parametric curves have order-$k$ geometric continuity when their local images admit compatible orientation-preserving regular parameter choices with matching derivatives through order $k$. At order one the nonzero one-sided tangent vectors must be positive scalar multiples. At order two the oriented bending must also match: equality of unsigned curvature alone does not suffice if the curves bend to opposite sides.

Green’s theorem converts circulation around a positively oriented planar boundary into the area integral of scalar curl.

Let $D$ be a bounded region in the Euclidean plane whose boundary is a finite union of disjoint, piecewise smooth simple closed curves. Suppose the vector field $F=(P,Q)$ is continuously differentiable on an open set containing the whole closed region. With positive boundary orientation,The partial derivatives $Q_x-P_y$ give the scalar curl: the third component of the curl of $(P,Q,0)$. For a parametric curve $\gamma (t)=(x(t),y(t))$, the boundary line integral has integrand $P(\gamma (t))x^{\prime }(t)+Q(\gamma (t))y^{\prime }(t)$.^[ref]

Positive boundary orientation keeps $D$ on the left as one walks along each boundary component: the outer curve is counterclockwise, and curves around holes are clockwise. Every component belongs in the line integral. The vector field must be regular throughout $D$, but need not extend into a hole.^[ref]

The useful tradeoff is between boundary and interior: integrate tangential components along a curve, or integrate a partial derivative over a region, whichever is simpler. The boundary formula for planar area uses this freedom in the opposite direction, recovering area from a line integral. As a planar case of Stokes theorem, Green theorem also explains why a local curl can determine a complete boundary measurement while leaving room for the global obstruction in Green theorem with holes.

A short proof starts with a rectangle $D=[a,b]\times [c,d]$. Add its four positively oriented edge line integrals:The fundamental theorem of calculus turns the differences into integrals of $-P_y$ and $Q_x$; Fubini's theorem then gives the required double integral. On subdividing a region, adjacent edge line integrals cancel because their directions oppose. For curved boundaries, the proof uses pieces bounded by graphs and double integrals; cancellation works the same way. This is the geometric reason only the region's complete boundary survives.^[ref]

An implicit solution describes a solution of a differential equation through a relation $F(x,y)=C$, instead of isolating $y$ in an explicit formula. The relation describes a level set; an actual scalar solution of a differential equation must be a differentiable function $y(x)$ on an interval whose function graph lies in the equation's domain. A whole level set need not be one such function graph.

For example, considerImplicit differentiation of $x^2+y^2=C$ gives $2x+2y{y}^{\prime }=0$. For $C>0$, the circle contains two solutions of a differential equation,Both satisfy the equation on the stated open interval, and their function graphs stay in $D$. An initial condition $(x_0,y_0)\in D$ fixes $C=x_0^2+y_0^2$ and selects the sign of $y_0$. In particular, $y(0)=2$ gives $C=4$ and the upper branch on $(-2,2)$. The constant alone does not select the branch.^[ref]

The precise local criterion uses the implicit function theorem. Let $F$ be continuously differentiable on an open subset of the plane, with $F(x_0,y_0)=C$. If its partial derivative $F_y(x_0,y_0)\ne 0$, the nearby level set is a unique continuously differentiable function graph $y=g(x)$ through that point. The chain rule givesThis proves that the local function graph solves $y^{\prime }=-F_x/F_y$ there. For a different prescribed equation $y^{\prime }=f(x,y)$, one must also check $F_x+F_{y}f=0$ along that branch and that it stays in the domain of $f$. A level set by itself does not verify an arbitrary ordinary differential equation.^[ref]

For $F=x^2+y^2$, the upper branch through $(1,\sqrt{3})$ has $F_y=2\sqrt{3}$ and tangent line slope $-1/\sqrt{3}$. At $(2,0)$, however, $F_y=0$ and $F_x=4$. The circle remains a smooth regular level set with a vertical tangent line: it can be written locally as $x=\sqrt{4-y^2}$, but not as a single function graph $y=g(x)$ over an open interval containing $2$. Moreover, $-x/y$ is undefined there, so the scalar solution of a differential equation cannot include this point in its domain. Changing the parameter draws the curve through the point but does not extend that scalar initial value problem.

When both partial derivatives vanish, even the geometry needs separate inspection. The level $x^2+y^2=0$ is an isolated point, not a function graph on any nonempty open interval. The zero level set of $G(x,y)=y^2-x^2$ instead consists of the two lines $y=x$ and $y=-x$ crossing at the origin, where $\nabla G=0$. Each line satisfies the implicit ordinary differential equation $y{y}^{\prime }=x$, including at the origin; the explicit form $y^{\prime }=x/y$ excludes the origin. Thus algebraically dividing by a vanishing coefficient changes the domain and can change which solutions of a differential equation are admissible. Vanishing $F_y$ alone is only failure of a sufficient test: $F(x,y)=y^3$ has the smooth zero function graph $y=0$ despite $F_y=0$ everywhere on it.

Implicit solutions are especially useful for exact first-order ordinary differential equations and for finding a first integral of an ordinary differential equation: a conserved relation can reveal allowed states without an explicit formula. For the normalized harmonic oscillator written as $\dot{x}=y$, $\dot{y}=-x$, the energy $H=(x^2+y^2)/2$ has $\dot{H}=x\dot{x}+y\dot{y}=0$. Its positive level sets are the same circles. Where $y\ne 0$, eliminating time gives $dy/dx=-x/y$ and the local implicit solution above. The harmonic oscillator's parametric curve continues through $y=0$ because its vector field governing evolution in time is defined there; treating $x$ as the independent variable fails at that turn. The origin is an equilibrium point of the harmonic oscillator, although it is excluded from the scalar equation $y^{\prime }=-x/y$.

A Lissajous curve is a parametric curve with coordinates $x(t)=A\sin (at+\delta )$ and $y(t)=B\sin (bt)$. The frequency ratio and phase shift control its crossings and whether its trajectory closes.

A differential 1-form is a smooth section of the cotangent bundle; it assigns a covector to every point of a smooth manifold. In planar coordinates, $\alpha =P(x,y)dx+Q(x,y)dy$ evaluates a tangent vector $v$ as $P(x,y)v_x+Q(x,y)v_y$.

Along a parametric curve $\gamma (t)=(x(t),y(t))$, its line integral isReversing the direction reverses the integral. A force expressed as a one-form therefore accumulates work along a directed path. In Euclidean coordinates its components can be displayed as arrows, but identifying a one-form with a vector field uses an inner product; the covector field itself does not require a Riemannian metric.

A parametric curve is described by a vector-valued map $t\mapsto Q(t)$. Its derivative gives a tangent vector wherever it is nonzero.

// Target: geometry-and-topology.bigb

A parametric curve interface supplies its parameter domain, evaluation, and derivatives where they exist. Robust geometric algorithms also need restricted subcurves and certified range or derivative bounds on parameter intervals. Tangent, speed and curvature are derived from point and derivative enquiries when the relevant derivatives are regular. Finite point samples alone cannot certify the absence of every plane intersection of an arbitrary smooth curve.