Tangent Before Limits

René Descartes (1596-1650)

Pierre de Fermat (1601-1665)

One of the key topics for a freshmen math student taking single variable Calculus is Differentiation. It all starts with words like “velocity”, “rate of change” and “the tangent line”. At first, it all seems like a very simple idea that should not have required a long time to develop. But, this little problem caused one of the most interesting feuds in the history of Mathematics. That too between two giants like Rene Descartes (1596-1650) and Piere De Fermat (1601-1665). Let’s find out what exactly happened.

Before I go into the rivalry, let me explain what the tangent problem is. Given a curve y=f(x) (say) and a point P(a,b) on the curve, find the equation of the tangent line to the curve at P. The problem is surprisingly simple for a modern-day student (thanks to Newton and Differentiation and most importantly Limits), but it wasn’t so for Descartes and Fermat. To solve this problem in his time, Descartes came up with a beautiful geometrical argument. Afterall, Descartes laid foundation for analytical geometry (hence the “Cartes”ian coodinates). Let’s see how Descartes solved the problem.

Descartes' Genius:

Consider a line normal (orthogonal to the tangent line) to the curve at P. Of course, we do not know this line yet, otherwise we would know the slope of tangent line as well. But assuming such a line exists, Descartes looked at the x-intercept of this line Q(q,0) and then went on to construct a circle with center at Q, passing through P. The radius of this circle is the distance between P and Q, so square of it is (a-q)^2+b^2. This gives the equation of the circle to be (x-q)^2+y^2=(a-q)^2+b^2. In general, the circle is going to intersect the curve at two distinct points (secant) but Descartes genius idea was to find q when circle kisses the curve at one point only and that is where the intersection becomes a double root. To put this algebraically, he said that for the chosen line to be normal, (x-a)^2 will completely divide (x-a)^2+(f(x))^2-(a-q)^2-b^2. Equivalently, (x-a)^2+(f(x))^2-(a-q)^2-b^2=(x-a)^2 * g(x), where g(x) is another polynomial. Now to understand this properly, consider y=x^2 at P(1,1). Descartes’ Q could be found using (x-q)^2+x^4-(1-q)^2-1=(x-1)^2 * g(x). Notice how the left hand side is a polynomial of degree, which forces g(x) to also be a polynomial of degree 2. Assuming g(x)=x^2+Bx+C, multiplying and equating coefficients [DIY], one sees that (B-2)=0 (for x^3), C-2B+1=1(for x^2), B-2C=-2q (for x) and C=2q-2 (for constant term). Which gives us q=3. Amazing no? This gives us the two points on the normal line, (1,1) and (3,0), giving us the slop of -1/2. This solves the mystery completely as the slope of tangent line would now be 2 (So is the derivative of x^2 at 1 :D) and its equation be y=2x-1.

The Start of a rivalry and Fermat's solution:

Fermat (a LAWYER and a Mathematician) on the other hand found this method to be quite complicated and overly convoluted. He exchanged letters with Descartes through a mediator voicing his issues with the method. Descartes, however, at times was too proud of his work and would not take criticism lightly. You would often find replies like “Do not send me any more letters from Beaugrand... because we already have enough toilet paper here.” Well, this can give you an idea about how ugly things are going to get because Fermat came up with his own way of finding tangents .

Once again, for a modern day math student, this will look surprisingly similar. Instead of considering the curve in the form y=f(x), he instead considered F(x,y)=0 (for y=f(x), F(x,y)=y-f(x)=0). His method involved moving a small distance (but positive Ha!) from x, so we are now sitting at x+a with true y-value being f(x+a). Let the y-value on the tangent line at x+a be r. We are getting two similar triangles and using similarity, we arrive at the expression r/y=(x+a-q)/(x-q). Fermat noticed that as “a” gets smaller, the “r” is getting closer and closer to the actual value “y_1” of the function. Which would make our expression (y_1)(x-q)=y(x+a-q), and we solve it to find q and hence the slope of the tangent line. Let’s apply this now to the same problem y=x^2 at P(1,1).

In this case, y_1=f(1+a)=(1+a)^2, so we have (1+a)^2(1-q)=(1+a-q), simplifying yields to (2a+a^2-2aq-a^2q)=a, which upon division by a (non-zero, remember?), gives 2+a-2q-aq=1. Fermat was already playing around with limits long before they were formally introduced!. Anyways, since we wanted “a” to be as small as possible, his final argument was (which didn’t sit well with Descartes and to this day, it still does not sit well with freshmen Math students) to let a=0. Doing so, we arrive at q=1/2. Now using P(1,1) and Q(1/2,0), the slope of tangent line would be 2 [Magic?].

To anyone taking or has already taken Single Variable Calculus, Fermat’s solution would seem like a genuine invitation to modern day Calculus tools. Descartes, obviously wasn’t much pleased as the argument was not very rigorous but Fermat believed that his method was far more superior.

To come to a conclusion, as I worry a lot about the attention span of kids these days, Descartes himself failed to find the tangent line equation using his own method on a curve attributed to Descartes himself. The curve is called Folium of Descartes and upon failure Descartes sent this as a challenge to Fermat to find tangent line using his method. Turns out, it wasn’t a good day for Descartes as Fermat was able to solve it using his method, proving the superiority of his method which later on (with some tweaking) became the best way of finding derivatives.

Nevertheless, what Descartes was able to achieve was purely remarkable despite its limitations and Fermat’s work inspired many to come after him. Which method did you like the most? Would you have thought similarly or differently?

For questions, comments or feedback, feel free to write an email to waqas.azhar@lums.edu.pk

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