Methodus inveniendi lineas curvas

MAX. ET MIN. RELATIVA. 239 mulæ $\int dx\sqrt{(1+pp)}$ valor differentialis est $= -m.d. \frac{p}{\sqrt{(1+pp)}}$: & tertio formulæ $\int yydx$ valor differentialis est $= 2m.ydx$. Ex quibus tribus valoribus differentialibus conficietur pro curva quæsita ista æquatio, $0 = adx - bd. \frac{p}{\sqrt{(1+pp)}} + 2\gamma ydx$, seu $cd. \frac{p}{\sqrt{(1+pp)}} = bdx + 2ydx = \frac{ccdp}{(1+pp)^{\frac{3}{2}}}$. Multiplicetur hæc æquatio per $p$ & integretur; habebitur $ff+by+yy = \frac{cc}{\sqrt{(1+pp)}}$, ubi tam $cc$ quam $ff$ pro arbitrio sive affirmative ve five negative accipere licet. Hinc porro fiet $(ff+by+yy)^2 (1+pp) = c^4$; & $p = \frac{\sqrt{c^4 - (ff+by+yy)^2}}{ff+by+yy} = \frac{dy}{dx}$; ideoque $dx = \frac{(ff+by+yy)dy}{\sqrt{c^4 - (ff+by+yy)^2}}$; quæ est æquatio pro curva Elastica. Ingredietur autem, per integrationem unam reliquam, nova quarta constans arbitraria; atque hisce qu [TRUNCATED]