Folium of Descartes

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1 Definition
2 Characteristics of the curve
3 Algebraic components of the folium of Descartes

[edit] Definition

The Folium of Descartes is an algebraic curve first proposed by Descartes in 1638 with an implicit equation:

x 3 + y 3 - 3 a x y = 0

It can also be described explicitly in polar coordinates as:

$r(\theta) = \frac{3 a \sin \theta \cos \theta}{\sin^3 \theta + \cos^3 \theta }$

[edit] Characteristics of the curve

[edit] Equation of the tangent

Using the method of implicit differentiation, we can solve the above equation for y':

$\frac{dy}{dx} = \frac{a y - x^2}{y^2 - a x}$

Using Point-slope form of the equation of a line, we can find an equation to the tangent of the curve at $(x 1, y 1)$ :

$y - y_1 = \frac{a y_1 - x_1^2}{y_1^2 - a x_1}(x - x_1)$

[edit] Horizontal and vertical tangents

The tangent line of Folium of Descartes is horizontal when $a y - x 2 = 0$ . Therefore, the tangent line is horizontal when:

x 6 = 2 a 3 x 3

The tangent line of Folium of Descartes is vertical when $y 2 - a x = 0$ . Therefore, the tangent line is vertical when:

y 6 = 2 a 3 y 3

This is explainable through a fact about the symmetry of the curve. By looking at the graph, we can see that the curve has two horizontal tangents and two vertical tangents. The curve of Folium of Descartes is symmetrical about $y = x$ , so if a horizontal tangent has a coordinate of $(x 1, y 1)$ , there is a corresponding vertical tangent, $(y 1, x 1)$ .

[edit] Asymptote

The curve has an asymptote:

x + y + a = 0

The asymptote has a gradient of -1 and x-intercept and y-intercept of -a.

[edit] Algebraic components of the folium of Descartes

If we solve $x 3 + y 3 = 3 a x y$ for y in terms of x, we obtain the following three functions:

$y = f(x) = \sqrt[3]{-\frac{1}{2} x^3 + \sqrt{\frac{1}{4} x^6 - 8 x^3}} + \sqrt[3]{-\frac{1}{2} x^3 - \sqrt{\frac{1}{4} x^6 - 8 x^3}}$

and

$y = \frac{1}{2} \left[ - f(x) \pm \sqrt{-3} \left( \sqrt[3]{-\frac{1}{2} x^3 + \sqrt{\frac{1}{4} x^6 - 8 x^3}} - \sqrt[3]{- \frac{1}{2} x^3 - \sqrt{\frac{1}{4} x^6 - 8 x^3}} \right) \right ]$

The reader may see that implicit differentiation is a much easier way of obtaining an equation to the tangent of the curve, rather than attempting to differentiate the above equations, which is much more complicated than $x 3 + y 3 = 3 a x y$ . As an additional note, you may be able to see intuitively that it is impossible to find a general formula for the roots of an nth-degree equation, if n is any integer larger than 4.

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Categories: Curves | Algebraic curves