The Tip of the Cartesian Leaf

Photo by Sarah Dorweiler on Unsplash

Today’s problem is about the Cartesian leaf, named after the great mathematician and philosopher RenΓ© Descartes. In 1638, Descartes challenged Pierre de Fermat (yes, the one with the famous theorem) to find the tangent line to an arbitrary point of the curve. Fermat solved the problem easily, but Descartes was unable to do so. We are going to do something slightly different. We will find the position of the tip of the leaf by using a technique invented by some other French guy of the same era: Joseph-Louis Lagrange.

The Problem

The Cartesian leaf is the curve defined by the equation

Equation 1: The Cartesian Leaf Equation

If you plot it in the plane, it looks like this (for π‘Ž=1):

The problem for today is: find the position of the tip of the leaf!

Solution with Python

Using Python’s computer algebra package sympy , we start a Jupyter notebook and define the symbols and equation for the Cartesian leaf:

So, where is the tip? Well, it is at a point where the distance from the origin has a maximum. The distance from the origin is simply

But if π‘Ÿ has a maximum, also π‘ŸΒ² has a maximum, too. So we can save ourselves the burden of calculating with square roots. So, we require that

is a maximum. But not for all (π‘₯,𝑦)! Only for those points which satisfy the equation of the Cartesian leaf! So we have to find a maximum with constraints. That should ring a bell, called β€œLagrange multipliers”.

We can put all of the Cartesian leaf equation on one side

and call that 𝑓(π‘₯,𝑦). So this function 𝑓 is the zero function. That means, if you differentiate it with respect to π‘₯ or 𝑦, it is still zero. The total differential therefore reads

In Python, we write

Remember that we wanted to calculate the maximum of π‘ŸΒ². So we demand that

So we have that 𝑑(π‘ŸΒ²)=0 and 𝑑𝑓=0. Therefore, for some constant πœ† (called Lagrange multiplier), we also have

Since both π‘ŸΒ² and 𝑓 are functions of π‘₯ and 𝑦, this equals

Both 𝑑π‘₯ and 𝑑𝑦 are arbitrary, so this can only be true if the terms in braces are zero. Or in Python:

Solve the first equation for πœ† and insert it in the second to eliminate πœ†:

Not only this equation has to be fulfilled, but also the original equation for the curve. So we have to solve

If you look closely at the output, the solution set only contains one real solution, which is

and that’s the tip of the leaf.