Differential calculus answers a question: given a curve, what is its tangent at each point? The mathematicians of the seventeenth century soon realized that the inverse question was harder and more useful: given a property the tangents must have, what is the curve? This problem was called the inverse method of tangents, and it is the subject of the fourth and last book of the Instituzioni analitiche. Today we call it solving a differential equation.

A problem from 1638

One of the first problems of this kind was put to René Descartes in 1638 by Florimond de Beaune, a French lawyer and amateur mathematician. Descartes could not solve it completely. The solution only came in 1684, when Leibniz published the first paper on the calculus and showed that the curve sought involved logarithms. It was proof that the new method solved problems that the old geometry could not reach.

Constant subtangent

A classic example, of the kind found in the textbooks of the period, uses the subtangent. At a point $(x, y)$ of the curve, draw the tangent until it cuts the $x$-axis. The subtangent is the horizontal distance between that crossing and the foot of the ordinate. Since the tangent rises by $y$ over that distance, the slope is $y' = \frac{y}{s}$, that is,

$$s = \frac{y}{y'}.$$

Which curve has a constant subtangent, equal to $k$ at every point? The condition is $y' = \frac{y}{k}$. Separating the variables, that is, putting everything that depends on $y$ on one side and what depends on $x$ on the other:

$$\frac{dy}{y} = \frac{dx}{k} \quad\Longrightarrow\quad \ln y = \frac{x}{k} + C \quad\Longrightarrow\quad y = A\,e^{x/k}.$$

The curve with constant subtangent is the exponential, which at the time was called the “logarithmic curve”.

The versiera is a solution too

The curve of Agnesi itself solves a differential equation. With $a = 1$, $y = \frac{1}{1 + x^2}$, and the derivative is

$$y' = -\frac{2x}{(1 + x^2)^2} = -2x \cdot \left(\frac{1}{1 + x^2}\right)^2 = -2x\,y^2.$$

In other words, the versiera is the curve in which the slope of the tangent at each point is $-2x$ times the square of the height. In the exercise below, we go the other way: we start from this property and recover the curve.

Exercise: recovering the versiera

Solve the equation $y' = -2x\,y^2$ with the condition $y(0) = 1$.

Separating the variables: $\frac{dy}{y^2} = -2x\,dx$. Integrating both sides: $-\frac{1}{y} = -x^2 - C$, or $\frac{1}{y} = x^2 + C$. At $x = 0$, $y = 1$, so $C = 1$. Hence $y = \frac{1}{1 + x^2}$, the curve of Agnesi.

Exercise: constant subnormal

The subnormal at a point of a curve is the horizontal distance between the foot of the ordinate and the point where the normal line (perpendicular to the tangent) cuts the $x$-axis. It equals $y\,y'$. Which curve has a constant subnormal equal to $k$?

The condition is $y\,y' = k$. Separating: $y\,dy = k\,dx$, and integrating: $\frac{y^2}{2} = kx + C$. The curve is $y^2 = 2kx + 2C$, a parabola with a horizontal axis. With $k = 2$, for example, we get $y^2 = 4x + 2C$.

Sources: Maria Gaetana Agnesi, Instituzioni analitiche ad uso della gioventù italiana, vol. II (1749), book IV · Gottfried Wilhelm Leibniz, “Nova methodus pro maximis et minimis”, Acta Eruditorum (1684) · Morris Kline, Mathematical Thought from Ancient to Modern Times (1972), ch. 21.

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