What is the equation of the Semicubical parabola?

Cartesian equation: y^2 = ax^3; Parametric equations: x = t^2, y = √ a t^3. a sets how steeply the branches rise from the cusp at the origin.

What is the arc length of the Semicubical parabola?

Arc length (from the cusp to x): L = (8/27a)[(1 + 9ax/4)^(3/2) - 1]. a sets how steeply the branches rise from the cusp at the origin.

Classic Cartesian curves

Semicubical parabola

y² = ax³, the cusp curve: the first algebraic curve other than a straight line whose arc length was found exactly (Neile, 1657).

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a (steepness)1

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Formulas and key values

Cartesian equation: $y^2 = ax^3$
Parametric equations: $x = t^2$ $y = \sqrt a\,t^3$
Arc length · from the cusp to x: $L = \frac{8}{27a}\left[\left(1 + \frac{9ax}{4}\right)^{3/2} - 1\right]$

a sets how steeply the branches rise from the cusp at the origin.

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01 · Definition

What is the Semicubical parabola?

In 1657 William Neile found the exact length of an arc of y² = ax³, and the result surprised mathematicians who believed that curved lengths could not be expressed algebraically. The answer needs nothing beyond powers and roots.

The curve has a cusp at the origin, where its two branches meet tangent to the x-axis. It turns up again and again: it is the evolute of a parabola, and it is the isochrone of Leibniz’s 1687 challenge, a path along which a falling body descends at a constant vertical speed.

02 · Properties

Key properties

Cusp at the origin, with both branches tangent to the x-axis.
Arc length from the cusp to x: (8/27a)[(1 + 9ax/4)^{^(3/2)} − 1].
Parametrisation x = t², y = √a·t³.
The evolute of a parabola is a semicubical parabola.
Solution of Leibniz’s isochrone problem (1687).

03 · Questions