Folium of Descartes
It is symmetrical about .
The curve was featured, along with a portrait of Descartes, on an Albanian stamp in 1966.
The curve was first proposed by Descartes in 1638. Its claim to fame lies in an incident in the development of calculus. Descartes challenged Fermat to find the tangent line to the curve at an arbitrary point since Fermat had recently discovered a method for finding tangent lines. Fermat solved the problem easily, something Descartes was unable to do. Since the invention of calculus, the slope of the tangent line can be found easily using implicit differentiation.
Graphing the curve
Since the equation is degree 3 in both x and y, and does not factor, it is difficult to solve for one of the variables.
However, the equation in polar coordinates is:
which can be plotted easily. By using this formula, the area of the interior of the loop is found to be .
We can see that the parameter is related to the position on the curve as follows:
- p < -1 corresponds to x>0, y<0: the right, lower, "wing".
- -1 < p < 0 corresponds to x<0, y>0: the left, upper "wing".
- p > 0 corresponds to x>0, y>0: the loop of the curve.
Another way of plotting the function can be derived from symmetry over y = x. The symmetry can be seen directly from its equation (x and y can be interchanged). By applying rotation of 45° CW for example, one can plot the function symmetric over rotated x axis.
This operation is equivalent to a substitution:
Plotting in the cartesian system of (u,v) gives the folium rotated by 45° and therefore symmetric by u axis.
Since the folium is symmetrical about , it passes through the point .
Relationship to the trisectrix of MacLaurin
and change variables to find the equation in a coordinate system rotated 45 degrees. This amounts to setting . In the plane the equation is
If we stretch the curve in the direction by a factor of this becomes
which is the equation of the trisectrix of Maclaurin.