464 16. THE CALCULUSFIGURE 1. Fermat's method of finding the subtangent.take up the story of calculus at the point where algebra enters the picture, beginning
with some elementary problems of finding tangents and areas.1.1. Tangent and maximum problems. The main problem in finding a tangent
to a curve at a given point is to find some second condition, in addition to passing
through the point, that this line must satisfy so as to determine it uniquely. It
suffices to know either a second point that it must pass through or the angle that it
must make with a given line. Format had attacked the problem of finding maxima
and minima of variables even before the publication of Descartes' Geomctrie. As
his works were not published during his lifetime but only circulated among those
who were in a rather select group of correspondents, his work in this area was
not recognized for some time. His method is very close to what is still taught in
calculus books. The difference is that whereas we now use the derivative to find
the slope of the tangent line, that is, the tangent of the angle it makes with a
reference axis, Fermat looked for the point where the tangent intercepted that axis.
If the two lines did not intersect, obviously the tangent was easily determined as
the unique parallel through the given point to the given axis. In all other cases
Fermat needed to determine the length of the projection of the tangent on the axis
from the point of intersection to the point below the point of tangency, a length
known as the subtangent. In a letter sent to Mersenne and forwarded to Descartes
in 1638 Fermat explained his method of finding the subtangent.
In Fig. 1 the curve DB is a parabola with axis CE, and the tangent at Â
meets the axis at E. Since the parabola is convex, a point Ï between  and
Å on the tangent lies outside the parabola. That location provided Fermat with
two inequalities, one of which was CD : DI > BC : 01. (Equality would
hold here if 01 were replaced by the portion of it cut off by the parabola.) Since
BC : Ol = CE : ̧7, it followsjhat CD : 7JI > CE^2 : El^2. Then abbreviating by
setting CD = g, CE = x, and CI - y, we have g : g - y > x^2 : x^2 -f y^2 - 2xy, and
cross-multiplying,
gx^2 + gy^2 - 2gxy > gx^2 — x^2 y.Canceling the term gx^2 and dividing by y, we obtain gy — 2gx > —x^2. Since this
inequality must hold for all y (no matter how small), it follows that x^2 > 2gx, that
is, ÷ > 2g if ÷ > 0. Choosing a point Ï beyond  on the tangent and reasoning in
the same way would give ÷ < 2g, so that ÷ = 2g. Since ÷ was the quantity to be
determined, the problem is solved.