A History of Mathematics From Mesopotamia to Modernity

TheCalculus 187

therefore also to the triangleSAB. By the like argument, if the centripetal force acts successively in

C,D,E, &c., and makes the body, in each single particle of time, to describe the right linesCD,DE,

EF, &c., they will all lie in the same plane; and the triangleSCDwill be equal to the triangleSBC,

andSDEtoSCD, andSEFtoSDE. And therefore, in equal times, equal areas are described in one

immovable plane: and, by composition any sumsSADS,SAFS, of those areas, are one to the other

as the times in which they are described. Now let the number of those triangles be augmented, and

their breadth diminishedin infinitum; and (by Cor. 4, Lem. III.) their ultimate perimeterADFwill

be a curve line: and therefore the centripetal force, by which the body is perpetually drawn back

from the tangent of this curve, will act continually; and any described areasSADS,SAFS, which

are always proportional to the times of description, will, in this case also, be proportional to those

times. Q.E.D.

Solutions to exercises

1. If the ‘Sine’ is equivalent toRsinθ(whereθis the angle, expressed in radians), then the ‘Cosine’
   isRcosθ, and the arc isRθ.Now,xin the series (1) is the tangent, tanθ=sinθ/cosθ; and
   this clearly is the same as the quotient of the Sine by the Cosine, since theRs cancel. Now the
   first term is the product of the Sine and the radius divided by the Cosine (i.e.Rx). To get from
   each term to the next, we multiply by the square of the Sine and divide by the square of the
   Cosine; that is, we multiply by the square of the tangent, orx^2. We then divide successively by
   1, 3, 5,...and alternate the signs; the result is:

Rθ=Rx−

(Rx)x^2

3

+

(Rx)x^4

5

−

(Rx)x^6

7

+···

which is the series (1).

2. The gradient of the tangent is (increase iny)/(increase inx)=AT/TA′=qo/po=q/p.


3. The strength is the (relative) simplicity of calculation. Once you have grasped the sequence
   (subtract the expression for A from the expression for A′; divide byo; cross out any terms left
   which still haveoin them), it becomes automatic. The weakness is the fuzzy logic. Why is it all
   right to seto=0 at the end of the argument but not at the beginning?


4. We have(x+po)^3 −ab(x+po)+a^3 −d(y+qo)^2 for the equation of A′. Subtracting the
   equation of A, we get

3 x^2 (po)+ 3 x(p^2 o^2 )+p^3 o^3 −abpo−d( 2 yqo+q^2 o^2 )= 0

Divide byoand discard the terms which still have anooro^2 in them; you get 3x^2 p−abp− 2 dyq=

0 which gives the ratio ofqtop, or, as we would say,dy/dx:

q/p=( 3 x^2 −ab)/ 2 dy.

5. By the rule:d(x^2 )=xdx+dx x= 2 xdx. Generally,d(xn)=d(xn−^1 ·x)=xn−^1 dx+d(xn−^1 )x.
   From this we deduce by induction (which was not very commonly used or well formalized in
   the 1680s, but still...) the usual formula:

d(xn)=nxn−^1 dx