Methodus inveniendi lineas curvas

282 DE CURVIS Coordinatæ vero $x$ & $y$ ita quantitatibus constantibus augeri diminuive possunt, ut æquatio pro curva hujusmodi faciem ac- quirat $xx+yy = A + \frac{B}{R}$. Hæc autem æquatio si multipli- cetur per $xdx+ydy$, fiet integrabilis; est enim $\int \frac{xdx+ydy}{R}$ $= -\int \frac{x+y\omega}{(1+\omega\omega)^{\frac{3}{2}}} d\omega$, [posito $dy=\omega dx$] $= \int \frac{y-\omega x}{\sqrt{(1+\omega\omega)}}$ $= \frac{ydx-xdy}{ds}$. Hanc ob rem, post integrationem constantibus mutatis, prodibit $(xx+yy)^2 = A(xx+yy) + \frac{B(ydx-xdy)}{ds}$ $+C$. Sit $\sqrt{(xx+yy)}=z$ & $y=uz$; erit $x=z\sqrt{(1-uu)}$; unde $ydx-xdy = \frac{zzdu}{\sqrt{(1-uu)}}$, & $ds = \sqrt{(dz^2 + \frac{zzdu^2}{1-uu})}$. Ergo posito $\frac{du}{\sqrt{(1-uu)}} = dr [TRUNCATED]