Scholium.

If the preceding lemmas had been treated on the supposition that the density was variable along the line ks, fig. 1, (which is the same as making the density f (t), or more generally f (a, T,t) a function of a, T, and t) their application to the problem we have been considering, would give an indefinite number of different equations, for the curve pr, fig. 15, according to the nature of the assumed function f (a, T, t): every one of which equations will, however, have this peculiarity, that if we make ro, it will become a*x'= (x*+y2)3. For when r = 0, t = o, and f (a, T, t) becomes a function of a and Tonly, and the case enters into Prop. 33 just now considered.

It may be worth while to see an example of this; we should have had, in general, for the action of the polygonal prismatic element of the solid, by Prop. 1,

A = 2nd s

a f (a, T', t) Ti (a2+T2+12)2

and the mass of the same element would have been

;

2nd fff (a, T, t) Ti.

These must be integrated, with respect to t, from t = 0 to t=rT: which cannot be done till we assign a form for the function f (a, T, t). Let this be a2+ T2 + ť, that is to say, let the density at any point q, in the triangle vrm, be as the square of its distance pq from the attracted point p. This will give

A =

aTi

2nd

1å SS (@2 + T2 + 1) = 2naåƒ{L(rT+√æ2+(1+ro) To) —L√a+T}İ; and for the mass

2nåf (a2rT+rT3+”TM3) 1.

2nåff (a2 + T2+ ť2) İi If therefore we solve Prop. 32, on this supposition of density, we have for the equation of the curve pr, fig. 15, when the solid has the greatest attraction,

xL (ry √x2 + (1 + ¿a ) y2 ) − xL √x° + y2 + C (x2ry + ry2 + "'313)

= 0.

Now, when r is infinitely small, we shall have, by neglecting all the higher powers thereof,

=

3

2

L (ry + √x2 + (1 + r2) y⋅ ) = L√x® + y2+

by substituting which our equation becomes

x

Va2+y2 + C (x2+ y2), or a*x*— (x2+y')'= 0, as we shewed

a priori must necessarily happen.

I shall just remark here, that, as the results of Prop. 32, are not altered by conceiving the density any function of a and T, such is also the case with respect to Problem 31, if T there represent the distance of any particle from a plane passing through the attracted point and the axis of the cylinder. This the reader may easily convince himself of.

The proposition just mentioned (31) is only a particular case of the following very general one.

Prop. 34.

Let uvv'u', fig. 15, be a rectangle, whose plane is perpendicular to the line pm, and its centre in that line. Let this rectangle move parallel to itself, in the direction pm, and vary in such a manner, that the middle points r and r' of its sides may continually touch two different curves.

The quantity of matter in the solid so generated being given, and the nature of one of the curves as pr', to find what must be the other curve pr, so that the action of the solid, on a point at its vertex p, may be the greatest possible.

Put r for the absciss pm, y' and y for the ordinates mr′, mr; then by Prop. 4, the action of the solid will be

; and its mass is 4/yy'x.

yy' 4fx arc (tang. x √x2 + y2+y22 But y' is a given function of x, suppose f(x). The quantity Arc (tang.: x √x2+ y2+ f (x)2 is therefore to have its fluxion, with respect to y, made 0: and this gives, for the equation of the curve pr,

yf (x)

=

) + Cyf (x)

+C=0, or x2— C2 (x2+ y2)2 (x2+yˆ+ f

(.x2+y2) √ x2+y2+1(x)2 (x)') = 0.

Ex. 1. Let f (x) = ax, or pr' be a straight line, the equation of pr must be xa — C′(x2+y°)' { ( 1+a*) x*+y3}

= 0.

Ex. 2. Let pr' be a circle, or f (x)2 = 2kx — x2, k being the radius, then xa — C2 (x2 + y2)' (2kx + ya) = o, is the equation of the other curve.

Ex. 3. If pr' is a parabola, or f(x) = ax, the equation of pr is xa — C2 (x* + y2)' (ax + x' + yo) = 0.

Scholium.

In Prop. 27, after having found the action of the solid there treated of, we derived, as corollaries, the action of parabolic and circular cylinders of infinite length, by separately making infinite the diameter of the circle and the parameter of the parabola. Perhaps it might therefore be supposed, that if we made k infinite in the second of the preceding examples, or a

infinite in the third, the result would be the equation of the base of the infinitely long cylinder of greatest attraction; which however is by no means the case; for that was found to be a circle, whereas the equation we get here is x C' (x2+ yo) = 0,

and if we make a infinitely great in the first example, the equation becomes C′ = x2+ y2, or the line pr is a circle with its centre at the attracted point.

We might resolve this problem, on a variety of hypotheses respecting the density; or we might add other conditions of a different kind; for instance, not only the mass of the solid, but the area of the section, passing through the required curve pr and axis pm, might be supposed constant. But I pass on. to other suppositions respecting the force of attraction; which will be treated with as much brevity as possible,

Lemma 3.

To find the attraction of the triangle vrm, fig. 1, on the point p, in the direction pm, supposing the force to be inversely as the mth power of the distance.

172

; which being re

Keeping the same notation as in Prop. 1, we have, for the Ti attraction of an element at q, (a2+T2+t2) * solved, gives, for the force of the whole triangle, in the direction pm, A =SS ; the fluent is to be taken from

aTi (a3+T2+t2) 2 to to trT, and we have

arTT

A =√

(4-m)

14T4
3-5 (a2+7')

(a2+T2) (a2+(1+r2) T2) 3

&c.}.

2T2

- (2 —m) 3 (a2 +T2) + (2—m)

{ 1 —

The further integration, with respect to T, is not necessary for our purpose.

Cor. 1. If we multiply this by 2n, it will be the attraction of a regular polygon of n sides; and making n infinitely great and r infinitely small, the attraction of a circle to the radius T

is found to be

={

A = 2π

-2nra

(m− 1) (a2 + T2)TM —

&c. }

+

a

(m-1) a

(m—1) (a2 + T2)TM —1

the same as is found differently by other writers.

Cor. 2. When r becomes infinite, the triangle vrm is changed into a parallelogram, infinitely extended in the direction rv; and we have

(m—1) a'

ar

r2T2

A=ƒ (2+1) (75={1— (2—m) 3 (a2+213) † (2—m) (4—m)

+

(a2+T2) (rT)

++T+ 3·5 (a2+T2)2

which, when m is an

2, is reduced to A =

even positive whole number greater than
ai
2.4.6... (m2) S.
35·5·7..... (m—1). (a2+1'2)—=—=—=*
T12)TMTM

Cor. 3. If instead of the action of the triangle vrm, that of a rectangle, whose sides are rm (y) and rv (y′), had been required, we must have proceeded exactly in the same manner, but the fluent, with respect to t, must have been taken from t = 0, to t=y'; so that we have only to substitute yʻ for rT, in the value found by the lemma, and there results A =√ (4-m) 5.5 (+7) - &c.}

ayt

1

(a2 + T2) (a2+ T*+ y2)′′—1 { 1 — ( 2—m) 5 (+15) + (2—m)

T2+y12)

3

3'4

3·5 MDCCCXII.

Q q