Methodus inveniendi lineas curvas

MAX. ET MIN. RELATIVA. 199 ut $\int \frac{p^3 dx}{1+pp}$, cujus valor differentialis $= \frac{mv \cdot d \cdot \overline{3pp+p^4}}{(1+pp)^2}$. Ex his emergit pro curva ista aequatio $dx = \frac{bd \cdot \overline{3pp+p^4}}{(1+pp)^2}$; quae integrata dat $x = c + \frac{bpp(3+pp)}{(1+pp)^2}$. Aequatio autem diffe- rentialis per $p$ multiplicata, abit in hanc $dy = bpd \cdot \frac{3pp+p^4}{(1+pp)^2}$, quae in hanc formam $dy = bdp \frac{3pp+p^4}{(1+pp)^2} - \frac{bdp \cdot \overline{3pp+p^4}}{(1+pp)^2}$ transmutata, habet integrale $y = f + \frac{bp^3}{(1+pp)} - \frac{bp^3}{(1+pp)^2}$ feu $y = f + \frac{2bp^3}{(1+pp)^2}$; cum igitur fit $x = c + \frac{bpp(3+pp)}{(1+pp)^2}$, curva erit algebraica. Efficiendum est autem, ut, quo casu fit $x= [TRUNCATED]