Methodus inveniendi lineas curvas

140 DE USU METHODI $ddz = -Vddy$, quo valore substituto prodibit ista aequatio $d^2x ddy + T^2 dx ddy + V^2 dx ddy = T dy ddz - V dx ddz - T V dy ddy + V^2 dx ddy$, seu $dx ddy + T dz ddy = T dy ddz - V dx ddz$; hincque $ddy:ddz = Tdy-Vdx:dx+Tdz$. Multiplicetur aequatio inventa per $dz$, ac in primo termino scribatur $Tdx+Vdy$ loco $dz$, erit $Tdx^2 ddy + Vdx dy ddy + Tdz^2 ddy = Tdy dxd dz - Vdxdz ddz$. Addatur utrimque $Tdy^2 ddy - Vdx dy ddy$, erit $Tddy(dx^2+dy^2+dz^2) = (dzddz+dyddy)(Tdy-Vdx)$ feu $\frac{dy ddy+dz ddz}{dx^2+dy^2+dz^2} = \frac{Tddy}{Tdy-Vdx} = \frac{Tddz}{dx+Tdz}$. Vel multiplicetur aequatio per $dx$, ac loco $Tdx$ scribatur $dz-Vdy$; obtinebitur $dx^2 ddy + dz^2 d [TRUNCATED]