Plane curve

A plane curve is a curve contained in a plane. In algebraic geometry, one studies plane curves defined by polynomial equations. General plane curves, such as an Archimedean spiral, need not be algebraic curves.

A tractrix is a plane curve whose tangent segment from the point of contact to a fixed axis has constant length. For segment length one and a positive acute tangent angle $\theta$, its height is $r=\sin \theta$. The slope relation $dr/dh=\tan \theta$ gives $dh/d\theta ={\cos }^2\theta /\sin \theta$. Integrating yields the displayed axial coordinate up to an additive constant. Conversely these coordinates have the required slope, and their tangent intercept is at distance $r/\sin \theta =1$. Rotating this profile about its axis gives a tractoid.

The trajectory of a point at fixed distance $a$ from the centre of a circle of radius $r$ rolling along a line. In coordinates aligned with that line it has a linear centre displacement plus sinusoidal components. The case $a=r$ is a cycloid; $a<r$ gives a shortened trajectory and $a>r$ gives loops. An inclined rolling line rotates this geometry relative to horizontal and vertical coordinates.

A point at distance $a$ from the axis of a radius-$r$ cylinder rolling downhill at constant axis speed $v$ has $\dot{x}=v[\cos \alpha +(a/r)\cos \theta ]$, $\dot{y}=-v[\sin \alpha +(a/r)\sin \theta ]$. For $0\le \alpha <\pi /2$, the existence of $\dot{x}=0$, $\dot{y}>0$ is equivalent to $a>r$. At $a=r$ the relevant zero-horizontal-velocity point is stationary, so a point within or on the rolling cylinder never has strictly vertical upward motion.

A hyperbolic spiral is a plane curve whose radial distance in polar coordinates is inversely proportional to the angular displacement from a fixed angle: $r=a/(\theta -{\theta }_0)$, on a branch where $r>0$. For $a>0$ and $\theta >{\theta }_0$, it winds infinitely often towards the origin as $\theta \to \infty$. Its geometric winding alone does not determine whether a moving particle traverses it in finite time: that also requires a law for the angular velocity.

An involute is the curve traced by the end of a taut string as it is unwound from a fixed curve. For a circle of radius $R$, one parametrization is $\mathbf{x}(\theta )=R(\cos \theta +\theta \sin \theta ,\sin \theta -\theta \cos \theta )$. The straight free length is $R\theta$, and differentiation shows that the traced velocity is perpendicular to the taut string. Reversing the motion winds the string back around the circle.

A point mass attached to a light inextensible string winds around a fixed smooth circular pole of radius $R$. Let the initial straight free length be $L$, and assume the string remains taut without slipping over the part already laid on the pole. Since its tension is perpendicular to the point mass's velocity, the speed $v$ stays constant. If the contact point turns through $d\theta$, the free length $s$ obeys $ds=-Rd\theta$, while the involute geometry gives $v=s\dot{\theta }$. Thus $s\dot{s}=-Rv$, or $s^2=L^2-2Rvt$. First contact of the point mass with the pole occurs at $s=0$, after $L^2/(2Rv)$. In the ideal model, the required tension $m{v}^2/s$ diverges at contact; finite string strength or finite particle size limits that idealization.

A figure-eight plane curve with two congruent lobes and a crossing at the origin. With scale $a>0$, its polar equation is $r^2=a^2\cos 2\theta$. One quarter has coordinates $x=at\sqrt{(1+t^2)/2}$, $y=at\sqrt{(1-t^2)/2}$, $0\le t\le 1$. Differentiation gives arc element $adt/\sqrt{1-t^4}$. The total arc length is therefore $4a{\int }_0^{1}dt/\sqrt{1-t^4}$.

An astroid is the four-cusped plane curve parametrized by $x=a{\cos }^{3}t$, $y=a{\sin }^{3}t$ for $a>0$. It appears as the relative trajectory in the astroid motion of a harmonic oscillator in a rotating frame. The parametrization has zero first derivative at the cusps while remaining smooth as a function of its parameter.

This irreducible plane branch has parametrization $x=t^5$, $y=t^2$. Its only singular point is the origin. The resolution of the (2,5) cusp by two blowups makes its strict transform smooth in every characteristic.

Blow up the origin and use the chart $x=uy$, giving the strict transform $u^2-y^3=0$. Blow up its remaining singular point and use $u=vy$, giving $v^2-y=0$, which is smooth because its derivative in $y$ is $-1$. The complementary chart at each stage has no point of the strict transform on its exceptional divisor. These two blowups of a smooth algebraic surface resolve the branch, although the smooth transform is still tangent to an exceptional divisor.

An Archimedean spiral has radius linear in polar angle, $r=a+b\theta$. For the special parametrization $(u\cos u,u\sin u)$, the curvature of a plane curve is $(u^2+2)/(1+u^2{)}^{3/2}$. Its parameter speed is $\sqrt{1+u^2}$, so it remains a regular curve at the origin.

Let an Archimedean spiral have polar radius $r=at$ and angle $t$ in polar coordinates, with $0\le t\le R/a$, where $R>0$ is fixed. Parameterizing by $r$ gives the arc lengthThe bounds $r\le \sqrt{a^2+r^2}\le r+a$ yieldThe squeeze theorem proves the displayed limit. Thus the arc length grows like $R^2/(2a)$ as the turns become closely spaced, without requiring an explicit antiderivative.

A superelliptic curve has an affine equation $y^m=f(x)$. When $f$ is square-free and the characteristic does not divide $m$, projection to the $x$-coordinate is a finite morphism whose finite ramification occurs over the roots of $f$; its behavior at infinity depends on $m$ and $\mathrm{deg}f$.

For a smooth plane curve of degree $d\ge 4$, adjunction givesIts canonical map is induced by the $(d-3)$rd Veronese map and is therefore an embedding.

A semicubical parabola is a cusped cubic curve affinely equivalent to $y^2=x^3$.

The homomorphismhas kernel $(y^2-x^3)$ and image $k[t^2,t^3]$. The quotient is therefore an integral domain, so the curve is irreducible. In characteristic zero its gradient $(-3x^2,2y)$ vanishes at the origin, making the origin its unique singular point of an algebraic variety.