CASE 1. If the figure DES is a circle, or a rectangular hyperbola, bisect its transverse diameter AS in O, and SO will be half the latus rectum. And because TC is to TD as Cc to Dd, and TD to TS as CD to SY; ex æquo TC will be to TS as CD × Cc to SY × Dd. But (by Cor. 1, Prop. XXXIII) TC is to TS as AC to AO; to wit, if in the coalescence of the points D, d, the ultimate ratios of the lines are taken. Wherefore AC is to AO or SK as CD × Cc to SY × Dd. Farther, the velocity of the descending body in C is to the velocity of a body describing a circle about the centre S, at the interval SC, in the subduplicate ratio of AC to AO or SK (by Prop. XXXIII); and this velocity is to the velocity of a body describing the circle OKk in the subduplicate ratio of SK to SC (by Cor. 6, Prop IV); and, ex æquo, the first velocity to the last, that is, the little line Cc to the arc Kk, in the subduplicate ratio of AC to SC, that is, in the ratio of AC to CD. Wherefore CD × Cc is equal to AC × Kk, and consequently AC to SK as AC × Kk to SY × Dd, and thence SK × Kk equal to SY × Dd, and equal to , that is, the area KSk equal to the area SDd. Therefore in every moment of time two equal particles, KSk and SDd, of areas are generated, which, if their magnitude is diminished, and their number increased in infinitum, obtain the ratio of equality, and consequently (by Cor. Lem. IV), the whole areas together generated are always equal. Q.E.D. [Pg 164] CASE 2. But if the figure DES is a parabola, we shall find, as above, CD × Cc to SY × Dd as TC to TS, that is, as 2 to 1; and that therefore , is equal to . But the velocity of the falling body in C is equal to the velocity with which a circle may be uniformly described at the interval (by Prop. XXXIV). And this velocity to the velocity with which a circle may be described with the radius SK, that is, the little line Cc to the arc Kk, is (by Cor. 6, Prop. IV) in the subduplicate ratio of SK to ; that is, in the ratio of SK to . Wherefore is equal to , and therefore equal to ; that is, the area KSk is equal to the area SDd, as above. Q.E.D. PROPOSITION XXXVI. PROBLEM XXV. To determine the times of the descent of a body falling from a given place A. Upon the diameter AS, the distance of the body from the centre at the beginning, describe the semi-circle ADS, as likewise the semi-circle OKH equal thereto, about the centre S. From any place C of the body erect the ordinate CD. Join SD, and make the sector OSK equal to the area ASD. It is evident (by Prop. XXXV) that the body in falling will describe the space AC in the same time in which another body, uniformly revolving about the centre S, may describe the arc OK. Q.E.F. PROPOSITION XXXVII. PROBLEM XXVI. To define the times of the ascent or descent of a body projected upwards or downwards from a given place.