Methodus inveniendi lineas curvas

102 DE METHODO MAX. ET MIN, EXEMPLUM III. 28. Invenire curvam, in qua pro data abscissa sit $\int \frac{dx}{p} \int ydx$ maximum vel minimum. Hic erit $\Pi = \int ydx$, & $[Z]=y$ & $[N]=1$; deinde cum fit $Z = \frac{\Pi}{p}$, erit $L = \frac{1}{p}$ & $P = -\frac{\Pi}{p^2}$; reliquæ litteræ omnes evanescunt. Hinc ergo prodit ista æquatio. $0 = H - \int \frac{dx}{p} + \frac{1}{dx}d.\frac{\Pi}{p^2}$; seu $0 = H - \int \frac{dx}{p} + \frac{\Pi}{p^2} - \frac{2\Pi dp}{p^3 dx}$. Posito ergo $x=a$, quo casu fit $\int \frac{dx}{p} = H$; erit $ydx = \frac{2\Pi dp}{p}$. Differentietur ea æquatio, eritque $0 = -\frac{dx}{p} + \frac{dx}{p} - \frac{2ydp}{p^2} + \frac{6\Pi dp^2}{p^4 dx} - \frac{2\Pi ddp}{p^3 dx}$. Seu $ [TRUNCATED]