Methodus inveniendi lineas curvas

'MAX. ET MIN. RELATIVA. 221 $+b$, seu $x-c = \frac{b\sqrt{(1+pp)}}{p}$, & $dx = \frac{-bdp}{pp\sqrt{(1+pp)}}$. Hinc ergo fiet $y=\int p dx = -b\int \frac{dp}{p\sqrt{(1+pp)}}$; ex quibus æquationibus cum construetur, eritque curvæ longitudo $\int dx\sqrt{(1+pp)}$ $= s = -\frac{b}{p} + \text{Const.}$ Hinc alia Constructio, definiendis $x$ & $y$ per $s$ formari poterit: erit nempe $p = \frac{b}{s-f}$ &, si initium capiatur in A, ubi fit $p=\infty$, ponendum est $f=0$, ita ut fit $p=\frac{b}{s}$; unde fit $\sqrt{(1+pp)} = \frac{\sqrt{(bb+ss)}}{s}$ & $dx\sqrt{(1+pp)} = ds = \frac{dx\sqrt{(bb+ss)}}{s}$; hincque $dx = \frac{sds}{\sqrt{(bb+ss)}}$ & $x=\sqrt{(bb+ss)}-b$. Porro erit $dy=pdx = \frac{bds}{\sqrt{(bb+ss)}}$ atque $y=b l \frac{s+\sqrt{(bb+ss)}}{b}$. Æquatio autem inter coordinatas orthogonales $x$ & $y$ deducetur ex æquatione $x-c = \frac{b\sqrt{(1+pp)}}{p}$; quæ si desideretur super axe AP, qui est diameter, & pro initio abscissarum in A sumpto, ubi est $p=\infty$, poni oportet $c=-b$; eritque $(x+b)p = b\sqrt{(1+pp)}$, hincque $(x+b)^2 pp = bb+bbpp$, & $p=\frac{b}{\sqrt{(xx+2bx)}}$; ideoque $dy = \frac{bdx}{\sqrt{(xx+2bx)}}$, quæ est æquatio pro Catenaria nota. EXEMPLUM III. 74. Inter omnes curvas ejusdem longitudinis, determinare eam in qua fit $\int S dx \sqrt{(1+pp)}$ minimum; denotante $S$ functionem quamcunque arcus curvæ $s=\int dx \sqrt{(1+pp)}$. In hoc Exemplo continetur inventio curvæ Catenariæ, si catena non fuerit ubique uniformiter crassa, sed cujus crassities ar- cui Ee 3