Page:Newton's Principia (1846).djvu/291

 DTV, by which the moment of the time, always equal to itself, is expressed, there be put any determinate rectangle, as BD $$\scriptstyle \times$$ m, the area DPQ, that is, ½BD $$\scriptstyle \times$$ PQ, will be to BD $$\scriptstyle \times$$ m as CK $$\scriptstyle \times$$ Z to BD². And thence PQ $$\scriptstyle \times$$ BD³ becomes equal to 2BD $$\scriptstyle \times$$ m $$\scriptstyle \times$$ CK $$\scriptstyle \times$$ Z, and the moment KLON of the area AbNK, found before, becomes $$\scriptstyle \frac{BP\times BD\times m}{AB}$$. From the area DET subduct its moment DTV or BD $$\scriptstyle \times$$ m, and there will remain $$\scriptstyle \frac{AP\times BD\times m}{AB}$$. Therefore the difference of the moments, that is, the moment of the difference of the areas, is equal to $$\scriptstyle \frac{AP\times BD\times m}{AB}$$; and therefore (because of the given quantity $$\scriptstyle \frac{BD\times m}{AB}$$) as the velocity AP; that is, as the moment of the space which the body describes in its ascent or descent. And therefore the difference of the areas, and that space, increasing or decreasing by proportional moments, and beginning together or vanishing together, are proportional. Q.E.D.

. If the length, which arises by applying the area DET to the line BD, be called M; and another length V be taken in that ratio to the length M, which the line DA has to the line DE; the space which a body, in a resisting medium, describes in its whole ascent or descent, will be to the space which a body, in a non-resisting medium, falling from rest, can describe in the same time, as the difference of the aforesaid areas to $$\scriptstyle \frac{BD\times V^{2}}{AB}$$; and therefore is given from the time given. For the space in a non-resisting medium is in a duplicate ratio of the time, or as V²; and, because BD and AB are given, as $$\scriptstyle \frac{BD\times V^{2}}{AB}$$. This area is equal to the area $$\scriptstyle \frac{DA^{2}\times BD\times M^{2}}{DE^{2}\times AB}$$ and the moment of M is m; and therefore the moment ot this area is $$\scriptstyle \frac{DA^{2}\times BD\times2M\times m}{DE^{2}\times AB}$$. But this moment is to the moment of the difference of the aforesaid areas DET and AbNK, viz., to $$\scriptstyle \frac{AB\times BD\times m}{AB}$$, as $$\scriptstyle \frac{DA^{2}\times BD\times M}{DE^{2}}$$ to ½BD $$\scriptstyle \times$$ AP, or as $$\scriptstyle \frac{DA^{2}}{DE^{2}}$$ into DET to DAP; and, therefore, when the areas DET and DAP are least, in the ratio of equality. Therefore the area $$\scriptstyle \frac{BD\times V^{2}}{AB}$$ and the difference of the areas DET and AbNK, when all these areas are least, have equal moments; and are therefore equal. Therefore since the velocities, and therefore also the spaces in both mediums described together, in the beginning of the descent, or the end of the ascent, approach to equality, and therefore are then