Sectrix of Maclaurin

Sectrix of Maclaurin

In geometry, a sectrix of Maclaurin is defined as the curve swept out by the point of intersection of two lines which are each revolving at constant rates about different points called poles. Equivalently, a sectrix of Maclaurin can be defined as a curve whose equation in biangular coordinates is linear. The name is derived from the trisectrix of Maclaurin (named for Colin Maclaurin), which is a prominent member of the family, and their sectrix property, which means they can be used to divide an angle into a given number of equal parts. There are special cases are also known as arachnida or araneidans because of their spider-like shape, and Plateau curves after Joseph Plateau who studied them.

Equations in polar coordinates

We are given two lines rotating about two poles P and P_1. By translation and rotation we may assume P = (0,0) and P_1 = (a, 0). At time t, the line rotating about P has angle heta = kappa t + alpha and the line rotating about P_1 has angle heta_1 = kappa_1 t + alpha_1, where kappa, alpha, kappa_1 and alpha_1 are constants. Eliminate t to get heta_1 = q heta + heta_0 where q = kappa_1 / kappa and heta_0 = alpha_1 - q alpha. We assume q is rational, otherwise the curve is not algebraic and is dense in the plane. The Q be the point of intersection of the two lines and let psi be the angle at Q, so psi = heta_1 - heta. If r is the distance from P to Q then, by the law of sines, :{r over sin heta_1} = {a over sin psi} so:r = a frac {sin heta_1}{sin psi} = a frac {sin [q heta + heta_0] }{sin [(q-1) heta + heta_0] } is the equation in polar coordinates.

The case heta_0 = 0 and q = n where n is an integer greater than 2 gives arachnida or araneidan curves:r = a frac {sin n heta}{sin (n-1) heta}

The case heta_0 = 0 and q = -n where n is an integer greater than 1 gives alternate forms of arachnida or araneidan curves:r = a frac {sin n heta}{sin (n+1) heta}

A similar derivation to that above gives:r_1 = (-a) frac {sin [(1/q) heta_1 - heta_0 /q] }{sin [(1/q-1) heta_1 - heta_0 /q] } as the polar equation (in r_1 and heta_1) if the origin is shifted to the right by a. Note that this is the earlier equation with a change of parameters; this to be expected from the fact that two poles are interchangeable in the construction of the curve.

Equations in the complex plane, rectangular coordinates and orthogonal trajectories

Let q = m/n where m and n are integers and the fraction is in lowest terms. In the notation of the previous section, we have heta_1 = q heta + heta_0 or n heta_1 = m heta + n heta_0.If z = x + iy then heta = arg(z), heta_1 = arg(z-a), so the equation becomesn arg(z-a) = m arg(z) + n heta_0 or m arg(z) - n arg(z-a) = arg(z^m (z-a)^{-n} ) = const. This can also be written:frac{operatorname{Re}(z^m (z-a)^{-n})}{operatorname{Im}(z^m (z-a)^{-n})} = const. from which it is relatively simple to derive the Cartesian equation given m and n. The function w = z^m (z-a)^{-n} is analytic so the orthoganal trajectories of the family arg(w) = const. are the curves |w| = const, orfrac = const. These form the Apollonian circles with poles (0, 0) and (a, 0).


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These curves have polar equation :r = a frac {sin (- heta + heta_0)}{sin (-2 heta + heta_0)},complex equation arg(z(z-a)) = const. In rectangular coordinates this becomesx^2 - y^2 - x = c(2xy - y) which is a conic. From the polar equation it is evident that the curves has asymptotes at heta = heta_0 /2 and heta_0 /2 + pi/2 which are at right angles. So the conics are, in fact, rectangular hyperbolas. The center of the hyperbola is always (a/2, 0). The orthogonal trajectories of this family are given by |z||z-a| = c which is the family of Cassini ovals with foci (0, 0) and (a, 0).

References

* [http://www.mathcurve.com/courbes2d/sectrice/sectricedemaclaurin.shtml "Sectrix of Maclaurin" at Encyclopédie des Formes Mathématiques Remarquables] (In French)
*MathWorld |title=Arachnida |urlname=Arachnida
*MathWorld |title=Plateau Curves |urlname=PlateauCurves


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