Methodus inveniendi lineas curvas

280 DE CURVIS 59. Sint laminæ, loco potentiarum verticalium & horizontalium $p$ & $q$, in singulis punctis $M$ applicatæ duæ potentiæ, altera normalis $MN=dv$ & altera tangentialis $MT=dt$. Hinc erit $dp = \frac{dx dv}{ds} + \frac{dy dt}{ds}$ & $dq = \frac{dx dt}{ds} - \frac{dy dv}{ds}$, & ob $dy=wdx$ & $ds=dx\sqrt{(1+ww)}$, habebitur $dp = \frac{dv}{\sqrt{(1+ww)}} + \frac{wdt}{\sqrt{(1+ww)}}$ & $dq = \frac{dt}{\sqrt{(1+ww)}} - \frac{wdv}{\sqrt{(1+ww)}}$: quibus in præced. §. æquatione ultima substitutis, proveniet sequens æquatio, $-Ekk d.\frac{\frac{dR}{RRdx}}{dv} = \frac{-dt}{\sqrt{(1+ww)}} + \frac{2wdv}{\sqrt{(1+ww)}} + \sqrt{(1+ww)}d.\frac{dv}{dw}$; quæ multiplicata per $\sqrt{(1+ww)}$ fit integrabilis: posito enim, brevitatis gratia, $z = \frac{dR}{RRdx}$, reperietur integrale, $A-t+\frac{dv\sqrt [TRUNCATED]