Methodus inveniendi lineas curvas

216 DE METHODO fitque $e^{-∫Udx\sqrt{1+pp}} (K+∫e^{∫Udx\sqrt{1+pp}} \frac{d\sqrt{1+pp}}{2v\sqrt{v}}) = T:$ atque erit formulæ $∫\frac{dx\sqrt{1+pp}}{v\sqrt{v}}$ valor differentialis = $-nv.d(\frac{p}{\sqrt{v(1+pp)}} + \frac{WTp}{\sqrt{1+pp}})$. Ex his duobus valoribus differentialibus inventis nascitur pro curva quæsita sequens æquatio, $0 = ad. \frac{W V p}{\sqrt{1+pp}} + d(\frac{p}{v\sqrt{1+pp}} + \frac{WTp}{\sqrt{1+pp}}) + \frac{Wp(aV+T)}{\sqrt{1+pp}}$. & integrando, $B=\frac{p}{v\sqrt{1+pp}} + \frac{WTp}{\sqrt{1+pp}}$. At est $aV+T=e^{-∫Udx\sqrt{1+pp}} (aH+K+∫e^{∫Udx\sqrt{1+pp}} \frac{dx\sqrt{1+pp}}{2v\sqrt{v}})$. Quod si ergo ponatur $aH+K=C$, erit $C$ constans [TRUNCATED]