Methodus inveniendi lineas curvas

IN RESOLVENDIS QUAESTIONIBUS. 137 arcus BS, radio CB = $t$ descriptus, = $t$; erit, ut ante vidi- mus, Mn = $yds$, & area MCm = $\frac{1}{2}yydt=dx$, unde fit $dt = \frac{2dx}{yy}$. Quia porro est Mm = $\sqrt{(dy^2 + y^2dt^2)} = $ $\sqrt{(dy^2 + \frac{4dx^2}{yy})}$; fit $dy=pdx$, minimumque esse debet $\int \frac{dx}{y} \sqrt{(4+ppyy)}$. Cum igitur fit $Z=\frac{\sqrt{(4+y^2p^2)}}{y}$, erit $M=0$, $N=-\frac{4}{yy\sqrt{(4+y^2p^2)}}$, & $P=\frac{yyp}{\sqrt{(4+y^2p^2)}}$, $Q=0$, &c. Hinc resultat ista aequatio $Z=C+Pp$; propterea quod fit $M=0$: ideoque $\frac{\sqrt{(4+y^2p^2)}}{y}=C+\frac{yyp^2}{\sqrt{(4+y^2p^2)}}$, seu $4=Cy\sqrt{(4+yypp)}$ vel $2b=y\sqrt{(4+yypp)}$; hincque $p=\frac{2\sqrt{(bb-yy)}}{yy}=\frac{dy}{dx}=\frac{2dy}{yydt}$; ac $dt=\frac{dy}{\sqrt{(bb-yy)}}$: itemque integrando $t = \text{A fin. } \frac{y}{b} + \text [TRUNCATED]