CHAPTER XIX.

THE METHOD OF INFINITESIMALS.

390. REFERRING the reader to other works where it is shown how the differential calculus enables us readily to draw tangents to curves, and to determine the magnitude of their areas and arcs, we wish here to give him some idea of the manner in which these problems were investigated by geometers before the invention of that method. The geometric methods are not merely interesting in a historical point of view; they afford solutions of some questions more concise and simple than those furnished by analysis, and they have even recently led to a beautiful theorem (Art. 400) which had not been anticipated by those who have applied the integral calculus to the rectification of conic sections.

If a polygon be inscribed in any curve, it is evident that the more the number of the sides of the polygon is increased, the more nearly will the area and perimeter of the polygon approach to equality with the area and perimeter of the curve, and the more nearly will any side of the polygon approach to coincidence with the tangent at the point where it meets the curve. Now, if the sides of the polygon be multiplied ad infinitum, the polygon will coincide with the curve, and the tangent at any point will coincide with the line joining two indefinitely near points on the curve. In like manner, we see that the more the number of the sides of a circumscribing polygon is increased, the more nearly will its area and perimeter approach to equality with the area and perimeter of the curve, and the more nearly will the intersection of two of its adjacent sides approach to the point of contact of either. Hence, in investigating the area or perimeter of any curve, we may substitute for the curve an inscribed or circumscribing polygon of an indefinite number of sides; we may consider any tangent of the curve as the line joining two indefinitely near points on the curve, and any point on the curve as the intersection of two indefinitely near tangents.

391. Ex. 1. To find the direction of the tangent at any point of a circle.

B

D

In any isosceles triangle AOB, either base angle OBA is less than a right angle by half the vertical angle; but as the points A and B approach to coincidence, the vertical angle may be supposed less than any assignable angle, therefore the angle OBA which the tangent makes with the radius is ultimately A equal to a right angle. We shall frequently have occasion to use the principle here proved, viz., that two indefinitely near lines of equal length

are at right angles to the line joining their extremities.

Ex. 2. The circumferences of two circles are to each other as their radii.

If polygons of the same number of sides be inscribed in the circles, it is evident, by similar triangles, that the bases ab, AB, are to each other as the radii of the circles, and, therefore, that the whole perimeters of the polygons are to each other in the same ratio; and since this will be true, no matter how the number of sides of the polygon be increased, the circumferences are to each other in the same ratio.

Ex. 3. The area of a circle is equal to the radius multiplied by the semi-circumference.

For the area of any triangle OAB is equal to half its base multiplied by the perpendicular on it from the centre; hence the area of any inscribed regular polygon is equal to half the sum of its sides multiplied by the perpendicular on any side from the centre; but the more the number of sides is increased, the more nearly will the perimeter of the polygon approach to equality with that of the circle, and the more nearly will the perpendicular on any side approach to equality with the radius, and the difference between them can be made less than any assignable quantity; hence ultimately the area of the circle is equal to the radius multiplied by the semi-circumference; or = r2.

392. Ex. 1. To determine the direction of the tangent at any point on an ellipse.

Let P and P' be two indefinitely near points on the curve,

then FP+ PF" = FP' + P'F"; or,

T

PP

T'

F

N

taking FR=FP, FR' = F'P', we have P'R=PR'; but in the triangles PRP', PR'P', we have also the base PP' common, and (by Ex. 1, Art. 391) the angles PRP' PR'P' right; hence the angle PP'R=P'PR'. Now TPF is ultimately equal to PP'F, since their difference PFP' may be supposed less than any given angle; hence TPF= T'PF", or the focal radii make equal angles with the tangent.

Ex. 2. To determine the direction of the tangent at any point on a hyperbola.

We have

or, as before,

P'R= P'R'.

Hence the angle

R'

P

PP'R=PP'R',

or, the tangent is the internal bisector of the angle FPF".

Ex. 3. To determine the direction of the tangent at any point of a parabola.

We have _FP=PN, and FP' = P'N'; hence P'R=P'S, or the angle N'P'P=FP'P. The tangent, therefore, bisects the angle FPN.

393. Ex. 1. To find the area of the parabolic sector FVP.

=

Since PS PR, and PN= FP, we have the triangle FPR half the parallelogram PSNN'. Now if we take a number of points P'P", &c. between V and P, it is evident that the closer we take them, the more nearly will the sum of all the parallelograms PSNN', &c., approach

N'

N

S P

R

to equality with the area DVPN, and the sum of all the triangles PFR, &c., to the sector VFP; hence ultimately the sector PFV is half the area DVPN, and therefore one-third of the quadrilateral DFPN.

Ex. 2. To find the area of the segment of a parabola cut off by any right line.

T

R

R

N

M M'

Draw the diameter bisecting it, then the parallelogram PR' is equal to PM', since they are the complements of parallelograms about the diagonal; but since TM is bisected at V', the parallelogram PN' is half PR'; if, therefore, we take a number of points P, P', P", &c., it follows that the sum of all the parallelograms PM' is double the sum of all the parallelograms PN', and therefore ultimately that the space V'PM is double V'PN; hence the area of the parabolic segment V'PM is to that of the parallelogram V'NPM in the ratio 2: 3.

394. Ex. 1. The area of an ellipse is equal to the area of a circle whose radius is a geometric mean between the semi-axes of the ellipse.

Α

D d

B

"

A

Cm m' 222"

For if the ellipse and the circle on the transverse axis be divided by any number of lines parallel to the axis minor, then since mb: md:: m'b': m'd' :: b:a, the quadrilateral mbb'm' is to mdd'm' in the same ratio, and the sum of all the one set of quadrilaterals, that is, the polygon Bbb'b" A inscribed in the ellipse is to the corresponding polygon Ddd'd" A inscribed in the circle, in the same ratio. Now this will be true whatever be the number of the sides of the polygon: if we suppose them, therefore, increased indefinitely, we learn that the area of the ellipse is to the area of the circle as b to the area of the circle being = "a", the area of the ellipse = πab. COR. It can be proved, in like manner, that if any two figures be such that the ordinate of one is in a constant ratio to the corresponding ordinate of the other, the areas of the figures are in the same ratio.

D'

Ꮓ Ꮓ

a;

but

Ex. 2. Every diameter of a conic bisects the area enclosed by the curve.

For if we suppose a number of ordinates drawn to this diameter, since the diameter bisects them all, it also bisects the trapezium formed by joining the extremities of any two adjacent ordinates, and by supposing the number of these trapezia increased without limit, we see that the diameter bisects the area.

395. Ex. 1. The area of the sector of a hyperbola made by joining any two points of it to the centre, is equal to the area of the segment made by drawing parallels from them to the asymptotes. For since the triangle PRC=QLC, the area PQC=PQKL. Ex. 2. Any two segments PQLK, RSNM, are equal, if PK: QL:: RM: SN.

For

PK: QL:: CL: CK,

but (Art. 197)

CL=MT", CK=NT;

we have, therefore,

RM: SN:: MT": NT,

CK L

R

M TNT

and therefore QR is parallel to PT. We can now easily prove that the sectors PCQ, RCS are equal, since the diameter bisecting PS, QR will bisect both the hyperbolic area PQRS, and also the triangles PCS, QCR.

If we suppose the points Q, R to coincide, we see that we can bisect any area PKNS by drawing an ordinate QL, a geometric mean between the ordinates at its extremities.

Again, if a number of ordinates be taken, forming a continued geometric progression, the area between any two is constant.

396. The tangent to the interior of two similar, similarly placed, and concentric conics cuts off a constant area from the

exterior conic.

For we proved (p. 213) that this tangent is always bisected at the point of contact; now if we draw any two tangents, the angle AQA' will be equal to BQB' and the nearer we suppose the point Q to P, the more nearly will the sides. AQ, A'Q approach to equality with the sides BQ, B'Q; if, therefore, the two

B

PP

B