The second book of the Instituzioni analitiche is devoted to differential calculus, and one of its main applications is the one every student learns: finding maxima and minima. Agnesi teaches the method with dozens of examples, of curves and of geometric problems, always in the same sequence: write the quantity, differentiate, set the differential equal to zero and check.

The highest point

At a maximum or a minimum, the curve is momentarily horizontal: the tangent line has slope zero. Since the slope of the tangent is the derivative, the candidates for maximum and minimum are the points where $y' = 0$.

On the curve of Agnesi with $a = 1$, $y = \frac{1}{1 + x^2} = (1 + x^2)^{-1}$. By the chain rule,

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

The derivative vanishes only at $x = 0$. For $x < 0$ it is positive (the curve rises) and for $x > 0$ it is negative (the curve falls). So $x = 0$ is the maximum, with $y = 1$.

Where the curve changes its bend

Near the top, the versiera bends downward, like a lid. Far from the top, where it approaches the axis, it bends upward, like a bowl. The point where the bending changes direction is called a point of inflection. Agnesi, like the mathematicians of her time, found them with the second differential: the bending changes direction where $y''$ changes sign.

Differentiating $y'$ by the quotient rule:

$$\begin{aligned} y'' &= \frac{-2(1 + x^2)^2 + 2x \cdot 2(1 + x^2) \cdot 2x}{(1 + x^2)^4} \\ &= \frac{-2(1 + x^2) + 8x^2}{(1 + x^2)^3} = \frac{6x^2 - 2}{(1 + x^2)^3}. \end{aligned}$$

The denominator is always positive, so the sign is that of $6x^2 - 2$, which vanishes at

$$x = \pm\frac{1}{\sqrt{3}} \approx \pm 0.577, \qquad y = \frac{1}{1 + \frac{1}{3}} = \frac{3}{4}.$$

Between the two points, $y'' < 0$ and the curve is concave down; outside them, $y'' > 0$ and it is concave up. With any diameter $a$, the points of inflection lie at $x = \pm\frac{a}{\sqrt{3}}$, always at three quarters of the maximum height.

Move the point and watch the tangent:

Notice that the points of inflection are exactly where the tangent is steepest. This is no coincidence: the slope $y'$ has its maximum (or minimum) there, because the derivative of $y'$, which is $y''$, vanishes.

Exercise: the rectangle of largest area

A rectangle has its two bottom vertices on the $x$-axis and its two top vertices on the curve $y = \frac{1}{1 + x^2}$, symmetric about the $y$-axis. What is the largest possible area?

If the top vertices are $(\pm x, y)$, the area is $A(x) = 2x \cdot \frac{1}{1 + x^2} = \frac{2x}{1 + x^2}$. Differentiating: $A'(x) = \frac{2(1 + x^2) - 2x \cdot 2x}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}$, which vanishes at $x = 1$. The largest area is $A(1) = \frac{2}{2} = 1$: the rectangle with base 2 and height $\frac{1}{2}$.

Exercise: the steepest tangent

What is the slope of the curve $y = \frac{1}{1 + x^2}$ at the point of inflection $x = \frac{1}{\sqrt{3}}$?

$y' = -\frac{2x}{(1 + x^2)^2} = -\frac{2/\sqrt{3}}{(4/3)^2} = -\frac{2}{\sqrt{3}} \cdot \frac{9}{16} = -\frac{9}{8\sqrt{3}} = -\frac{3\sqrt{3}}{8} \approx -0.650$. No other tangent to the curve is as steep.

Sources: Maria Gaetana Agnesi, Instituzioni analitiche ad uso della gioventù italiana, vol. II (1749), book II · John Colson (trans.), Analytical Institutions, ed. John Hellins (1801) · Judith V. Grabiner, The Origins of Cauchy’s Rigorous Calculus (1981).

All topics on Maria Gaetana Agnesi · ← The Instituzioni analitiche · The area under the curve and the number π →

Image credits.