Methodus inveniendi lineas curvas

MAX. ET MIN. RELATIVA. 197 Inter omnes igitur curvas proprietate communi $\int dx \sqrt{(1+pp)}$ gaudentes, ea quæritur in qua fit $\int yy dx$ maximum. Quoniam ergo formulæ $\int dx \sqrt{(1+pp)}$ valor differentialis est $= -ny.d.\frac{p}{\sqrt{(1+pp)}}$; formulæ vero $\int yy dx$ valor differentialis est $= 2nv.ydx$; habebitur pro curva quæsita ista æquatio $2ydx = \pm bbd.\frac{p}{\sqrt{(1+pp)}}$, quæ, multiplicata per $p$ & integrata, dabit $yy+bc = \pm \frac{bb}{\sqrt{(1+pp)}}$; seu $\sqrt{(1+pp)} = \frac{\pm bb}{yy+bc}$; hincque $p = \frac{\sqrt{(b^4-(yy+bc)^2)}}{yy+bc} = \frac{dy}{dx}$; ex qua fit $x = \int \frac{(yy+bc)dy}{\sqrt{(b^4-(yy+bc)^2)}}$. Hæc curva hanc habet proprietatem ut ejus radius osculi; qui generaliter est $= dx : d.\frac{p}{\sqrt{(1+pp)}}$, fiat $= \frac{bb}{2y}$; hoc est, proportionalis est applicatæ $y$ inversæ; unde pat [TRUNCATED]