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In the isolated model, conservation of energy and angular momentum conservation reduce the exterior test particle's motion toHere $\epsilon$ is specific orbital energy and $h$ is specific angular momentum. The effective potential is ${\Phi }_{\mathrm{eff}}=h^2/(2r^2)-\mu /r$. A zero radial velocity is a radial turning point, and need not mean zero total speed: at the periapsis or apoapsis of a nonradial Kepler orbit, tangential velocity remains. Thus the vertical gap between $\epsilon$ and $-\mu /r$ represents total kinetic energy per unit mass, whereas the gap above ${\Phi }_{\mathrm{eff}}$ represents its radial part.

For $h>0$, the orbital eccentricity obeys $e^2=1+2\epsilon h^2/{\mu }^2$. Negative $\epsilon$ gives an elliptic orbit, including a circular orbit at its minimum allowed energy; zero $\epsilon$ gives a parabolic trajectory; positive $\epsilon$ gives a hyperbolic Kepler orbit. The effective potential tends to zero from below at infinity, so its outer region creates no extra positive escape barrier. This explains why escape velocity depends on energy even though angular momentum changes the path. The $h=0$ case is radial motion, rather than a nondegenerate parabola or hyperbola; inward radial motion can end in collision.

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.

If the phenomenological single-density shift is continued into the condensed phase and the thermal population remains saturated at $n_T(T)$, the zero minimum requires $\nu [n_0+n_T]=\eta (H-H_g)$. At fixed temperature, $n_0=(\eta /\nu )[H-H_c(T)]$ above onset and zero below it. This prefactor belongs to that closure; a microscopic interaction model can distinguish normal and condensed self-energies. Continuing the normal parabola into negative energies without changing the state is not a consistent Bose distribution.

A conic section is obtained by intersecting a plane with a double circular cone. Cuts away from the vertex produce an ellipse, a parabola or a hyperbola; a perpendicular cut is the circle case of an ellipse. A cut through the vertex can instead produce a degenerate conic. One simple surface therefore connects several familiar plane curves.^[ref]

Consider the original familySubstituting the plane equation into the double circular cone equation givesThus $m=0$ gives a circle, $m=3/5$ an ellipse, $m=1$ a parabola, and $m=7/5$ a hyperbola. The last cut meets both sheets of the double circular cone, producing its two branches. These parameters specify a mathematical family, rather than measurements of a physical object.

The Cartesian coordinate system descriptions of these curves are second-degree equationsA rotation and a translation expose their principal directions and offsets. The conic type discriminant identifies the ordinary nondegenerate real type, while a separate degenerate conic check prevents confusing two crossing straight lines with a hyperbola.

This is useful because a difficult-looking equation can be understood through a small set of geometric features: foci of a conic, directrices of a conic, eccentricity of a conic and the semi-major and semi-minor axes. The same ellipse can be traced by a constant-length string or parametrized by $x=a\cos t$, $y=b\sin t$. These are different descriptions of one curve, with different advantages for construction and calculation.

For a second-degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$, the number ${\Delta }_2=B^2-4AC$ is the discriminant of a binary quadratic form associated with its quadratic part. For a nonempty, nondegenerate real conic section, a negative value gives an ellipse, zero a parabola, and a positive value a hyperbola. It is invariant under a rotation of orthonormal coordinates.^[ref]

A short explanation uses the symmetric matrixAn orthogonal matrix $R$ changes this matrix to $R^{\mathsf{T}}HR$, whose determinant equals $\det H$. The spectral theorem for real symmetric matrices then removes the mixed term. Equal-sign, opposite-sign and one-zero eigenvalues give the three types after the remaining linear terms and offsets are considered.

The qualification matters. The equations $x^2+y^2=0$, $y^2=0$ and $xy=0$ describe a point, a repeated straight line, and two crossing straight lines, respectively. Their quadratic discriminants have the signs expected of ellipse, parabola and hyperbola types, but all three are degenerate conics. Also, $x^2+y^2=-1$ has an empty set of real solutions despite its negative quadratic discriminant.

The constant rank theorem separates the directions a smooth map preserves from the directions it forgets. If a smooth map from an open subset of ${\mathbb{R}}^m$ to ${\mathbb{R}}^n$ has derivative of constant rank $r$ on a neighbourhood of a point, there are local coordinate transformations in source and target in which it becomesIt keeps $r$ coordinates, discards the other $m-r$, and inserts $n-r$ zeros. This is stronger than simplifying the Jacobian matrix at one point: it straightens the actual nonlinear map on a neighbourhood. ^[ref]

For a worked example, takeThe two columns agree, and the first row is nonzero, so the rank is exactly one everywhere. The image of a function is the parabola $b=a^2$. Every line $x+y=a$ is a fiber mapping to the same point $(a,a^2)$.

Choose input coordinates $u=x+y$, $v=x-y$, and output coordinates $\alpha =a$, $\beta =b-a^2$. Both coordinate transformations are diffeomorphisms: their inverses are $x=(u+v)/2$, $y=(u-v)/2$ and $a=\alpha$, $b=\beta +{\alpha }^2$. In these coordinates,The curved image has become a straight line, and the unused coordinate $v$ is explicit.

Constant rank on a neighbourhood is essential. The map $(x,y)\mapsto (x,y^2)$ has rank one when $y=0$ and two when $y\ne 0$, so the constant rank theorem does not supply one rank-one form across that line. By contrast, the full-rank square case of the theorem forgets no coordinates and recovers the inverse function theorem. The usefulness is that curved fibers and images can often be studied through this simpler local form.

A critical point of a smooth map is an input; a critical value is an output reached from at least one such input. For the fold $F(x,y)=(x,y^2)$, all inputs $(a,0)$ are critical points of a smooth map, and all outputs $(a,0)$ are critical values. The identical coordinate pairs live in different copies of ${\mathbb{R}}^2$. To classify an output $(a,b)$, inspect its entire fiber of a function:For $b>0$ both inputs have matrix rank two, so the output is a regular value. For $b=0$ its input has matrix rank one, so the output is a critical value. For $b<0$ the fiber of a function is the empty set, and the output is a regular value despite being outside the image of a function.

In general, for a smooth map $f:X\to Y$, write $C_f$ for its set of critical points of a smooth map. Its set of critical values is $f(C_f)\subseteq Y$. Equivalently, $q\in Y$ is a critical value if some $p\in f^{-1}(q)$ has a differential that is not surjective. A regular value requires a surjective differential at every input in that fiber of a function. The universal condition holds vacuously for an empty fiber of a function: every output outside $f(X)$ is a regular value. These quantifiers are explicit in Loring W. Tu, An Introduction to Manifolds, Definition 8.22 and its two following remarks.

The distinction is useful when solving equations. For a smooth map $f:X^m\to Y^n$ between smooth manifolds without boundary and a regular value $q$ with nonempty fiber of a function, the regular level set theorem makes $f^{-1}(q)$ an embedded submanifold of dimension $m-n$, with tangent spaceThe implicit function theorem explains the freedom count: locally, $n$ independent equations fix $n$ coordinates and leave $m-n$ free. The rank-nullity theorem gives the same count for the displayed kernel of a linear map. See Eckhard Meinrenken, Differentiable Manifolds, Theorems 1.11 and 1.15 for the local projection description and its consequence for level sets.

For example, $H:{\mathbb{R}}^2\to \mathbb{R}$, $H(x,y)=x^2+y^2$, has gradient $(2x,2y)$ and just one critical point, the origin, with critical value zero. For $c>0$, its fiber of a function is the circle of radius $\sqrt{c}$, a one-dimensional embedded submanifold; negative $c$ are regular values with empty fibers. At $c=0$, the fiber of a function is a single point instead of a circle. A critical value does not force its fiber of a function to be nonsmooth: that singleton is itself a zero-dimensional embedded submanifold, but the regular dimension conclusion $2-1=1$ does not apply.

Nor does a single regular input certify a regular value. The scalar smooth function $h(t)=t^2(t-1)$ has $h^{-1}(0)=\{0,1\}$ and derivative $h^{\prime }(t)=3t^2-2t$. The input zero is a critical point, while the input one is regular because $h^{\prime }(1)=1$. Their shared output zero is nevertheless a critical value. Conversely, different critical points of a smooth map can share one critical value: $f(t)=(t^2-1{)}^2$ sends both critical inputs $-1$ and $1$ to zero. Counting inputs and counting critical outputs are different tasks.

Finally, for $G(t)=(t,t^2):\mathbb{R}\to {\mathbb{R}}^2$, every output on the parabola is a critical value, and every output off it is a regular value with empty fiber of a function. There is no nonempty regular fiber of a function here, so the negative expression $1-2$ never becomes a claimed manifold dimension. Both diagrams show exact mathematical examples, not measured data.

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 twice continuously differentiable regular curve $r(t)=(x(t),y(t))$ in the plane, unsigned curvature is the absolute value of its signed curvature:The nonvanishing velocity vector makes the denominator positive. For a graph of a function $(t,f(t))$, this reduces to $|f^{\prime \prime }(t)|/(1+f^{\prime }(t{)}^2{)}^{3/2}$; omitting the absolute value gives the signed curvature of a plane graph for increasing $t$.

For the parabola $r(t)=(t,t^2)$, the derivatives are $r^{\prime }=(1,2t)$ and $r^{\prime \prime }=(0,2)$. ThereforeThe curvature is greatest at the vertex $t=0$, where $\kappa =2$ and the radius of curvature is $1/2$. At $t=\pm 1/2$ the curvature is $1/\sqrt{2}$, so the parabola bends less sharply despite its larger speed $\sqrt{2}$.

Its unit tangent vector and left unit normal areSince $k>0$, $N_L$ is also its principal normal vector. At $t=0$ the osculating circle consequently has center $(0,0)+(1/2)(0,1)=(0,1/2)$ and equationThe lower branch $y=1/2-\sqrt{1/4-x^2}$ has value $0$, derivative $0$, and second derivative $2$ at $x=0$, matching the parabola $y=x^2$ through second order. Away from the vertex the two curves separate: an osculating circle describes a local bend, not the whole curve.

Vanishing curvature at one point does not imply a straight segment. The graph of a function $y=x^3$ has signed curvature $6x/(1+9x^4{)}^{3/2}$, which vanishes at its inflection point $x=0$ and changes sign across it. By contrast, if curvature vanishes throughout a connected regular interval, $dT/ds=0$ throughout that interval: the constant unit tangent vector forces the image to lie on a straight line.

This MacTutor biography gives approximate dates and discusses the surviving evidence for Diocles' work on the cissoid of Diocles and the parabola. The footer identifies John Joseph O'Connor and Edmund Frederick Robertson as authors and April 1999 as its revision date.

Online text: mathshistory.st-andrews.ac.uk/Biographies/Diocles/

Diocles was a Greek mathematician associated with the cissoid of Diocles and the geometry of the parabola. The dates of his life are approximate.^[ref]

Biography: mathshistory.st-andrews.ac.uk/Biographies/Diocles/

A directrix of a conic is the fixed line in its focus-directrix description: distance to the focus of a conic is a constant multiple of distance to this line. For a parabola the multiple is one. The directrix of a conic belongs to the geometry of the conic section, rather than being a tangent or an asymptote.

The parameter $e\ge 0$ in a focus-centred polar coordinates equation $r=\ell /(1-e\cos \theta )$ for a conic section. It distinguishes an ellipse ($e<1$), a parabola ($e=1$), and a hyperbola ($e>1$).

For a noncircular conic section, the eccentricity of a conic is the constant ratiowhere $F$ is a focus of a conic, $L$ the corresponding directrix of a conic, and $X$ a point on the curve. For an ellipse with semi-major axis $a$ and distance $c$ from the focal midpoint to either focus of a conic, $e=c/a=\sqrt{1-b^2/a^2}$. Its circle case has $e=0$; a finite directrix does not give that circle through the displayed ratio. The circle is the limit as $c\to 0$ and its directrices of a conic move arbitrarily far away.

The eccentricity of a conic is intrinsic to the cutting plane, and is not generally its slope in a chosen three-dimensional drawing. For the family $z=1+mx$ in conic section, with $0<m<1$, completing the square givesA displacement $dx$ along this plane has Euclidean norm $\sqrt{1+m^2}|dx|$. Thus the intrinsic semiaxis ratio satisfies $b^2/a^2=(1-m^2)/(1+m^2)$, givingFor $m=3/5$, this is $3/\sqrt{17}$, rather than $3/5$. This normalization separates the curve's geometry from a distorted projection of it onto the $xy$ plane.

Consider the original pair of polynomial equationsTreat them as polynomials in $y$ with coefficients in $\mathbb{R}[x]$. Their Sylvester matrix and resultant areEvery common solution must therefore have $x=-1$ or $x=1$. Back-substitution into $F=0$ gives $y=x^2=1$, and both pairs also satisfy $G=0$. The complete real solution set isHere elimination and direct substitution agree: subtracting $F$ from $G$ gives $2x^2-2$, exactly the same condition.

In general, first solve the one-variable resultant equation, then recover and check the other coordinate in both original polynomial equations. A real coordinate candidate can correspond only to nonreal values of the eliminated variable. Also, vanishing leading coefficients require the separate check in degree drop in a specialized resultant. The general elimination-and-back-substitution method is described in the lecture's §3.1.^[ref]

University of Florence mathematical-museum exhibition, Analytical geometry and the problem of tangents. It gives a translation and reconstruction of Pierre de Fermat's parabola example, using the letters B, C, D, E, I, O of the exam image. The text distinguishes circulation in the 1630s from the 1679 publication. Its final displayed label contains a typographical anomaly; the derivation and geometry give $CE=2CD$.

Online source: archimede.dimai.unifi.it/archimede/archimede_NEW_inglese/mostra_calcolo/guida/node7.html

Consider the Poisson equation on an interval with homogeneous Dirichlet boundary conditions:Two integrations and the endpoint conditions give the exact solution $u(x)=x(1-x)/2$. To approximate it, multiply by a test function $v$ that vanishes at both endpoints. Integration by parts givesThe boundary term is zero. Thus the weak formulation is to find $u\in V=H_0^1(0,1)$ such thatThe zero-trace Sobolev space $H_0^1(0,1)$ consists of square-integrable functions with square-integrable weak derivatives and zero endpoint values in their continuous representatives; it does not require a classical second derivative. This is why continuous piecewise linear functions are admissible despite corners. The Poincaré inequality makes $a$ a coercive bilinear form on $V$. Strang derives this one-dimensional weak formulation, the piecewise-linear hat functions, and their load and derivative-product integrals in the cited lecture; its endpoint example uses a free left endpoint, while both endpoints here are fixed.^[ref]

Take the uniform finite element mesh $x_j=jh$, $j=0,\dots ,n$, with $h=1/n$. The continuous $P_1$ trial space isHere $P_1$ means polynomials of degree at most one on each interval. This is a conforming finite element space in $V$, with $n-1$ interior piecewise-linear hat functions ${\varphi }_i$ as a basis. Their endpoint values enforce the Dirichlet boundary conditions automatically. Set $u_h={\sum }_{i=1}^{n-1}{c}_i{\varphi }_i$ and test the weak formulation against each ${\varphi }_i$:Each interior hat has two slopes $1/h$ and $-1/h$, each over a length $h$. Its squared derivative therefore integrates to $2/h$. Adjacent hats have opposite slopes on their one shared interval, giving $-1/h$. All other derivative products vanish. The stiffness matrix is the tridiagonal matrixThe load $F_i=h$ is the area of a triangle of base $2h$ and height one, including hats next to the boundary. Boundary hats are omitted because the boundary values are prescribed. The piecewise-linear hat function article explains how the same entries arise by adding local two-node contributions.

With four mesh intervals, $h=1/4$, there are three unknowns:For example, the first row gives $8(3/32)-4(1/8)=1/4$ and the middle gives $-4(3/32)+8(1/8)-4(3/32)=1/4$. The coefficients are the nodal values $u_h(x_i)=c_i$. At $x=3/8$, linear interpolation gives $u_h(3/8)=7/64=0.109375$, whereas $u(3/8)=15/128=0.1171875$: exactness at the nodes still leaves a visible error between them.

In this model the Galerkin method solution equals the nodal approximation $I_{h}u$ obtained by linear interpolation on any finite element mesh. Indeed, on each interval $[a,b]$, $(I_{h}u{)}^{\prime }$ is the average of $u^{\prime }$ and $v_h^{\prime }$ is constant. Therefore ${\int }_a^b(u^{\prime }-(I_{h}u{)}^{\prime })v_h^{\prime }dx=0$. Summing proves Galerkin orthogonality for $I_{h}u$, and uniqueness gives $u_h=I_{h}u$. This argument relies on the constant diffusion coefficient, conformity and exact integration; nodal exactness is not a general promise for finite element methods with other operators or integration rules.

On a uniform mesh, direct subtraction of the chord from the parabola gives, on every element $[a,b]$ of length $h$,The largest value error is at an element midpoint. Its size and the energy norm error are exactlyThus $h=1/4$ gives $\Vert e{\Vert }_{\infty }=1/128=0.0078125$ and $\Vert e{\Vert }_a=1/(4\sqrt{12})\approx 0.0721688$. With eight elements, $h=1/8$, these become $1/512=0.001953125$ and $1/(8\sqrt{12})\approx 0.0360844$. Halving the mesh spacing quarters the maximum value error but only halves the energy norm error: values and slopes converge at different rates. The figure shows the small value error at its own scale, where the four-to-one reduction becomes clear.

The hyperbolic cosine is the even functionIt has derivative $\sinh x$ and second derivative $\cosh x>0$, so it is a strictly convex function. Its unique global minimum is $\cosh 0=1$, and its image is $[1,\infty )$. For example, $\cosh (-1)=\cosh 1\approx 1.543081$. On the unit hyperbola, the hyperbolic cosine is the horizontal coordinate; as a graph against $x$, it is a symmetric bowl.

Its Taylor series beginsThe positive fourth-order term explains why the bowl resembles a parabola near its minimum yet differs from one away from it. Restricting to $x\ge 0$ gives the inverse hyperbolic cosine.

Laminar flow is a regime of fluid flow in which the motion remains organized, with neighboring fluid parcels following smooth, coherent paths rather than the irregular interacting fluctuations characteristic of turbulence. In a straight pipe, the simplest example consists of coaxial layers sliding past one another. This describes the character of the motion, not a requirement that every parcel have the same velocity. Nor does it require steady flow: a smooth profile can evolve in time as viscosity transmits motion from a newly moving wall into a resting fluid. The unsteady wall-driven solution provides an explicit example. ^[ref]

The Reynolds pipe-and-dye experiment makes this distinction visible. A thin injected dye band can remain coherent in laminar flow, whereas irregular motion can fold it and carry dye across the pipe. However, molecular diffusion can spread dye even in laminar flow, and an inlet disturbance can move a coherent band without producing turbulence. A dye band is a streakline, not a measurement of the entire velocity field; in steady flow an ideal passive tracer's streakline coincides with a streamline. Reynolds explicitly distinguished a wandering band from one broken into eddies. ^[ref]

For a circular pipe of internal diameter $D$, define the mean axial velocity $U$ using the volumetric flow rate $Q$ and cross-sectional area $A$:Here $\rho$ is mass density, $\mu$ is dynamic viscosity and $\nu =\mu /\rho$ is kinematic viscosity. This particular Reynolds number uses the diameter and mean velocity, not the radius or centreline velocity. Small Reynolds number generally favors laminar flow, but a single universal critical Reynolds number does not distinguish it from turbulence in all geometries and experiments. Even in a given pipe, inlet conditions and the magnitude of disturbances affect the observed onset. Reynolds demonstrated this dependence by deliberately inserting a wire coil. ^[ref]

The quantitative example above assumes steady flow, incompressible flow, a Newtonian fluid with constant dynamic viscosity, a straight horizontal circular pipe of radius $R=D/2$, stationary walls satisfying the no-slip boundary condition, and fully developed flow. The last assumption means that the axial velocity profile no longer changes downstream. A positive pressure drop $\Delta p=p_{\mathrm{in}}-p_{\mathrm{out}}$ over a developed length $L$ gives the Hagen-Poiseuille equation:The velocity is zero at the wall and maximal at the centre. Averaging over the circular area, whose annuli have area $2\pi rdr$, gives $U=u(0)/2$; averaging along one diameter would give a different answer. These assumptions produce a parabola, rather than every laminar flow having a parabolic profile. ^[ref]

As a stipulated calculation, take $D=0.010\mathrm{m}$, $U=0.10\mathrm{m}{\mathrm{s}}^{-1}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, $\mu =0.0010\mathrm{Pa}\mathrm{s}$ and $L=1.0\mathrm{m}$, and assume the developed laminar flow above. Then $\nu =10^{-6}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, ${\mathrm{Re}}_D=1000$, $u(0)=0.20\mathrm{m}{\mathrm{s}}^{-1}$, $Q=7.85\times 10^{-6}{\mathrm{m}}^3{\mathrm{s}}^{-1}$ (about $0.471\mathrm{L}\mathrm{mi}{\mathrm{n}}^{-1}$), and $\Delta p=32\mu LU/D^2=32\mathrm{Pa}$. These are predictions from chosen parameters, not measurements or a claim that ${\mathrm{Re}}_D=1000$ guarantees laminar flow in every apparatus.

The beauty of this example is that a spatially varying velocity field still gives an exact relation between two accessible measurements, pressure drop and volumetric flow rate. It predicts the hydraulic resistance $\Delta p/Q=8\mu L/(\pi R^4)$. Doubling the radius therefore multiplies $Q$ by sixteen at fixed pressure drop, length and dynamic viscosity, provided the model remains applicable. This makes the laminar flow model useful for sizing narrow tubes or determining dynamic viscosity. It also explains why a persistent narrow dye band can be informative without implying that the fluid is moving as a uniform plug.

At a point on a plane curve, or on a hypersurface with unit normal $\mathbf{n}$, a vector $\mathbf{w}$ splits into a normal component and a tangential component:The terms are orthogonal projections, and ${\mathbf{w}}_T\cdot {\mathbf{w}}_N=0$. Changing $\mathbf{n}$ to $-\mathbf{n}$ reverses the scalar coefficient $\mathbf{w}\cdot \mathbf{n}$, but leaves the normal component vector unchanged.

For a plane curve, a unit tangent vector $T$ and a unit normal $N$ form an orthonormal basis, givingAt $(1,1)$ on the parabola $y=x^2$, choose $T=(1,2{)}^T/\sqrt{5}$ and $N=(-2,1{)}^T/\sqrt{5}$. For $\mathbf{w}=(3,1{)}^T$,Their sum is $(3,1{)}^T$ and their inner product is zero. A negative normal coefficient means that the normal component points opposite the chosen $N$, rather than giving it a negative Euclidean norm.

These are local directions: being tangent at a point does not mean that a straight displacement stays on the curve. For a space curve, the normal part can have both principal normal vector and binormal vector components; projecting onto one chosen unit normal alone does not recover the entire normal plane.

At a point of a twice continuously differentiable regular curve with positive curvature $\kappa$, its osculating circle has radius $1/\kappa$ and center $c=r+N_p/\kappa$, where $N_p$ is the principal normal vector. It matches the point, tangent line and curvature. For a space curve, it lies in the plane spanned by the unit tangent vector and principal normal vector. The parabola example under curvature of a plane curve gives center $(0,1/2)$ and radius $1/2$ at the origin.