The Cissoid of Diocles $y = \sqrt{\dfrac{x^3}{1-x}}$

$y = \sqrt{\dfrac{x^3}{1-x}}$ is the upper half of the cissoid of Diocles¹. Squaring both sides gives the equation of the curve, $y^2(1-x) = x^3$. The Greek mathematician Diocles devised it around the second century BC in order to solve the problem of duplicating the cube, and the name comes from a Greek word meaning "ivy-shaped".

Domain and range

The domain is therefore $0 \leq x < 1$ and the range is $y \geq 0$. Together with the lower half the full curve is symmetric about the $x$-axis.

The cusp at the origin

Near the origin $1 - x \approx 1$, so the curve behaves like $y \approx x^{3/2}$ and $y' = \dfrac{3}{2}x^{1/2} \to 0$. The tangent at the origin is therefore the $x$-axis, but because the upper and lower branches both arrive from the same direction they do not join smoothly: the origin is a cusp.

Monotonicity and asymptote

Differentiating $y^2 = \dfrac{x^3}{1-x}$ gives $2yy' = \dfrac{x^2(3-2x)}{(1-x)^2}$, whose right-hand side is positive on $0 < x < 1$, so the curve increases throughout. As $x \to 1^{-}$ the denominator tends to $0$ and $y \to +\infty$, making the line $x = 1$ a vertical asymptote. The curve leaves the origin and climbs along that asymptote.

Geometric construction

Take the circle $x^2 + y^2 = x$ of diameter $1$ and the line $x = 1$ tangent to it at $(1, 0)$. Let a ray from the origin meet the circle at $Q$ and the line at $R$, and mark the point $P$ on that ray whose distance from the origin equals the length $QR$. The locus of $P$ is the cissoid.

Indeed $OQ = \cos\theta$ and $OR = \sec\theta$, so $OP = \sec\theta - \cos\theta$, giving the polar equation below.

Duplicating the cube

Doubling the volume of a given cube means constructing $\sqrt[3]{2}$, which is known to be impossible with straightedge and compass alone². Diocles solved it with this curve. At any point of the cissoid the following identity holds.

Intersecting the curve with the line $y = 2(1-x)$ joining $(1, 0)$ and $(0, 2)$ makes the right-hand side equal to $2$, so the ratio $\dfrac{y}{x}$ at that intersection is exactly $\sqrt[3]{2}$. Replacing $(0, 2)$ by $(0, k)$ yields $\sqrt[3]{k}$ by the same construction.

Relation to the parabola

Inverting the curve in the unit circle about the origin, that is, sending each point $P$ to the point $P'$ on the same ray with $OP \cdot OP' = 1$, turns it into the parabola $y^2 = x$. Conversely the cissoid is the inverse of a parabola with respect to its vertex. As a cubic curve with one asymptote and one cusp, it also appears in Newton's classification of cubics.

1. Cissoid of Diocles, Wikipedia
2. Doubling the cube, Wikipedia