Methodus inveniendi lineas curvas

IN RESOLVENDIS QUÆSTIONIBUS. 141 occurrit, haec $ \frac{dyddy+dzddz}{dx^2+dy^2+dz^2} = \frac{ddy+Vddz}{dy+Vdz} $, quæ, ob $ V = -\frac{z}{y} $, transit in hanc $ \frac{dyddy+dzddz}{dx^2+dy^2+dz^2} = \frac{zddy-yddz}{zdy-ydz} $, cujus integrale est $ l\sqrt{(dx^2+dy^2+dz^2)} = l\frac{zdy-ydz}{b} $, seu $ zdy-ydz = b\sqrt{(dx^2+dy^2+dz^2)} $. Quoniam nunc est $ z = \sqrt{(X^2-y^2)} $, ponatur $ dX = vdx $, erit $ dz = \frac{Xvdx-ydy}{\sqrt{(X^2-y^2)}} $, & $ zdy-ydz = \frac{X^2dy-Xyvdx}{\sqrt{(X^2-y^2)}} $, & $ \sqrt{(dx^2+dy^2+dz^2)} = \frac{\sqrt{(X^2dx^2-y^2dx^2+X^2dy^2+X^2v^2 [TRUNCATED]