Page:Newton's Principia (1846).djvu/276

 $$\scriptstyle \frac{3S\sqrt{1+QQ}}{4RR}$$ directly and $$\scriptstyle \frac{1+QQ}{R}$$ inversely; that is, as $$\scriptstyle \frac{S}{R\sqrt{1+QQ}}$$. Q.E.I.

. 1. If the tangent HN be produced both ways, so as to meet any ordinate AF in T$$\scriptstyle \frac{HT}{AC}$$ will be equal to $$\scriptstyle \sqrt{1+QQ}$$; and therefore in what has gone before may be put for $$\scriptstyle \sqrt{1+QQ}$$. By this means the resistance will be to the gravity as 3S $$\scriptstyle \times$$ HT to 4RR $$\scriptstyle \times$$ AC; the velocity will be as $$\scriptstyle \frac{HT}{AC\sqrt{R}}$$, and the density of the medium will be as $$\scriptstyle \frac{S\times AC}{R\times HT}$$.

. 2. And hence, if the curve line PFHQ be defined by the relation between the base or abscissa AC and the ordinate CH, as is usual, and the value of the ordinate be resolved into a converging series, the Problem will be expeditiously solved by the first terms of the series; as in the following examples.

1. Let the line PFHQ be a semi-circle described upon the diameter PQ, to find the density of the medium that shall make a projectile move in that line.

Bisect the diameter PQ in A; and call AQ, n; AC, a; CH, e; and CD, o; then DI² or AQ² - AD² = nn - aa - 2ao - oo, or ee - 2ao - oo; and the root being extracted by our method, will give $$\scriptstyle DI=e-\frac{ao}{e}-\frac{oo}{2e}-\frac{aaoo}{2e^{3}}-\frac{ao^{3}}{2e^{3}}-\frac{a^{3}o^{3}}{2e^{5}}-$$, &c. Here put nn for ee + aa, and DI will become $$\scriptstyle =e-\frac{ao}{e}-\frac{nnoo}{2e^{3}}-\frac{anno^{3}}{2e^{5}}-$$, &c

Such series I distinguish into successive terms after this manner: I call that the first term in which the infinitely small quantity o is not found; the second, in which that quantity is of one dimension only; the third, in which it arises to two dimensions; the fourth, in which it is of three; and so ad infinitum. And the first term, which here is e, will always denote the length of the ordinate CH, standing at the beginning of the indefinite quantity o. The second term, which here is $$\scriptstyle \frac{ao}{e}$$, will denote the difference between CH and DN; that is, the lineola MN which is cut off by completing the parallelogram HCDM; and therefore always determines the position of the tangent HN; as, in this case, by taking MN to HM as $$\scriptstyle \frac{ao}{e}$$ to o, or a to e. The third term, which here is $$\scriptstyle \frac{nnoo}{2e^3}$$, will represent the lineola IN, which lies between the tangent and the curve; and therefore determines the angle of contact IHN, or the curvature which the curve line