Methodus inveniendi lineas curvas

IN RESOLVENDIS QUÆSTIONIBUS. 139 $dZ = \frac{+TEdx+TFdy+pdp+VEpdx+VFpdy+TVdp+TFpdx+TGpdy+V'pdp+VFp^2dx+VGp^2dy}{\sqrt{(1+p^2+T^2+2TVp+V^2p^2)}}$ Quæ formula cum ad casum primum pertineat, proveniet ista æquatio inter $x$ & $y$; $\frac{TFdx+VFpdx+TGpdx+VGp^2dx}{\sqrt{(1+p^2+T^2+2TVp+V^2p^2)}} = d.\frac{p+TV+V^2p}{\sqrt{(1+p^2+T^2+2TVp+V^2p^2)}}$. Eft vero $Fdx+Gdy=dV$, unde erit $\frac{TdV+VpdV}{\sqrt{(1+p^2+(T+Vp)^2)}} = d.\frac{p+TV+V^2p}{\sqrt{(1+p^2+(T+Vp)^2)}} = \frac{+dp(1+T^2+V^2)+dT(V-Tp)+dV(T+T^3+3T^2Vp+3TV^2p^2+V^3 [TRUNCATED]