Page:Newton's Principia (1846).djvu/280

 term $$\scriptstyle \frac{bb}{\overline{A-O|}^{n}}$$ be resolved into an infinite series $$\scriptstyle \frac{bb}{A^{n}}+\frac{nbb}{A^{n+1}}\times O+\frac{nn+n}{2A^{n+2}}\times bb\ O^{2}+\frac{n^{3}+3nn+2n}{6A^{n+3}}\times bb\ O^{3}$$,&c.,and GD will be equal to $$\scriptstyle C-\frac{d}{e}A+\frac{bb}{A^{n}}+\frac{d}{e}\ O-\frac{nbb}{A^{n+1}}\ O-\frac{+nn+n}{2A^{n+2}}bb\ O^{2}-\frac{+n^{3}+3nn+2n}{6A^{n+3}}bbO^{3}$$, &c. The second term $$\scriptstyle \frac{d}{e}\ O-\frac{nbb}{A^{n+1}}\ O$$ of this series is to be used for Qo, the third $$\scriptstyle \frac{nn+n}{2A^{n+2}}bb\ O^{2}$$ for Roo, the fourth $$\scriptstyle \frac{n^{3}+3nn+2n}{6A^{n+3}}bbO^{3}$$ for So³. And thence the density of the medium $$\scriptstyle \frac{S}{R\sqrt{1+QQ}}$$, in any place G, will be $$\scriptstyle \frac{n+2}{3\sqrt{A^{2}+\frac{dd}{ee}A^{2}-\frac{2dnbb}{eA^{n}}A+\frac{nnb^{4}}{A^{2n}}}}$$, and therefore if in VZ you take VY equal to n $$\scriptstyle \times$$ VG, that density is reciprocally as XY. For A² and $$\scriptstyle \frac{dd}{ee}A^{2}-\frac{2dnbb}{eA^{n}}A+\frac{nnb^{4}}{A^{2n}}$$ are the squares of XZ and ZY. But the resistance in the same place G is to the force of gravity as 3S $$\scriptstyle \times\frac{XY}{A}$$ to 4RR, that is, as XY to $$\scriptstyle \frac{2nn+2n}{n+2}$$ VG. And the velocity there is the same wherewith the projected body would move in a parabola, whose vertex is G, diameter GD, and latus rectum $$\scriptstyle \frac{1+QQ}{R}$$ or $$\scriptstyle \frac{2XY^{2}}{nn+n\times VG}$$. Q.E.I.

In the same manner that the density of the medium comes out to be as $$\scriptstyle \frac{S\times AC}{R\times HT}$$, in Cor. 1, if the resistance is put as any power Vn of the velocity V, the density of the medium will come out to be as $$\scriptstyle \frac{S}{R^{\frac{4-n}{2}}}\times\left(\frac{AC}{HT}\right)^{n-1}$$

And therefore if a curve can be found, such that the ratio of $$\scriptstyle \frac{S}{R^{\frac{4-n}{2}}}$$ to