Methodus inveniendi lineas curvas

158 DE USU METHODI reperitur valor differentialis $dA = -uv.d.\frac{xp}{\sqrt{(1+pp)}}$, & for- mulæ $\int dx\sqrt{(1+pp)}$ valor differentialis $dB = -uv.d.\frac{p}{\sqrt{(1+pp)}}$; quibus in æquatione $BdA = AdB$ substitutis, prodibit $Bd.\frac{xp}{\sqrt{(1+pp)}} = Ad.\frac{p}{\sqrt{(1+pp)}}$, & posito $\frac{A}{B}=c$, erit $d.\frac{xp}{\sqrt{(1+pp)}}$ $= cd.\frac{p}{\sqrt{(1+pp)}}$; unde integrando oritur $\frac{cp}{\sqrt{(1+pp)}}$ $= \frac{xp}{\sqrt{(1+pp)}} - b$, seu $b\sqrt{(1+pp)} = (c-x)p$; hincque eli- citur $p = \frac{b}{\sqrt{((c-x)^2-bb)}} = \frac{dy}{dx}$. Erit ergo $y =$ $\int \frac{bdx}{\sqrt{((c-x)^2-bb)}}; quæ æquatio indicat curvam quæsitam esse Catenariam, initio abscissarum pro $x$ in loco axis AC quocun- que accepto: quin etiam pro axe sumi potest recta quæcunque dia- metro Catenariæ AC parallela, in eaque punctum quodcunque pro axis initio. Quomodocunque autem axis, ejusque initium constituatur, quæstioni satisfiet ea tantum curvæ portio, ubi fit $\int x dx \sqrt{(1+pp)} = c \int dx \sqrt{(1+pp)}$. Ponamus pro axe ip- sam diametrum AC, & verticem A pro initio abscissarum acci- pi. Quia in A, ubi est $x=0$, fit $\frac{dy}{dx}=p=\infty$, necesse est ut fit $cc-bb=0$, ideoque $b=c$. Verum hoc casu fit $y = \int \frac{cdx}{\sqrt{(xx-2cx)}}$, quæ curva sursum directa fit imaginaria, donec fiat $x>2c$. Sit ergo $x=2c+t$, erit $t=$ abscissæ AP, & $y = PM = \int \frac{cdt}{\sqrt{(2ct+tt)}}; curvaque DAD erit catenaria or- dinaria. Quo autem appareat quanta ejus portio quæstioni sa- tisfaciat, notandum est, ob $dx=ds$, esse $p = \frac{c}{\sqrt{(2ct+tt)}}$, & $\sqrt{(1+pp)} = \frac{c+t}{\sqrt{(2ct+tt)}}; hincque \int dx \sqrt{(1+pp)} =$ $\int \frac{(c+t)dt}{\sqrt{(2ct+tt)}} = \sqrt{(2ct+tt)}$. At ipsa expressio $\frac{\int xdx\sqrt{(1+pp)}}{\int dx\sqrt{(1+pp)}}$ fit