Methodus inveniendi lineas curvas

248 DE CURVIS $\frac{1}{dx} d. \frac{p}{\sqrt{(1+pp)}} = \frac{q}{(1+pp)^{\frac{3}{2}}}$. Deinde cum fit radius ofculi $= \frac{dx(1+pp)^{\frac{3}{2}}}{dp} = R$, formula $\int \frac{ds}{RR}$, quæ minimum esse debet, abit in $\int \frac{qqdx}{(1+pp)^{\frac{5}{2}}}$. Comparetur hæc cum forma generali $\int Zdx$; erit $Z = \frac{qq}{(1+pp)^{\frac{5}{2}}}$, & posito $dZ = Mdx + Ndy+Pdp+Qdq$, crit $M=0$, $N=0$, $P = \frac{-5pqq}{(1+pp)^{\frac{7}{2}}}$ & $Q = \frac{2q}{(1+pp)^{\frac{5}{2}}}$. Valor ergo differentialis ex hac formula $\int \frac{qqdx}{(1+pp)^{\frac{5}{2}}}$ oriundus, erit $-\frac{dP}{dx} + \frac{ddQ}{dx^2}$. Quamobrem pro curva quæsita hæc habebitur æquatio, $\frac{1}{dx} d. \frac{p}{\sqrt{(1+pp)}} = \frac{dP}{dx} - \frac{ddQ}{dx^2}$; quæ [TRUNCATED]