Methodus inveniendi lineas curvas

MAX. ET MIM. RELATIVA. 201 APM = $\int ydx$, quæ est proprietas communis, ejuſque valor differentialis = $nv \cdot dx$. Deinde, cum sit radius ofculi MO $= \frac{(1+pp)^{\frac{3}{2}}}{q}$, fiet MN $= -\frac{(1+pp)}{q}$ & PN $= -\frac{(1+pp)}{q} - y$; ex quo area APN erit $= -\int ydx - \int \frac{1+pp}{q} dx$; quæ debet esse minima, cujus valor differentialis est $= nv \cdot (-dx + d \cdot \frac{2p}{q} + \frac{1}{q} dd \cdot (\frac{1+pp}{q}))$; unde iſta naſcitur æquatio $ndx^2 = dxd \cdot \frac{2p}{q} + dd \cdot (\frac{1+pp}{qq})$; quæ integrata dat $nxdx = \frac{2pdx}{q} + d \cdot \frac{1+pp}{qq} + bdx$. Illa vero eadem æquatio, per $p$ multiplicata, dat $ndxdy = dyd \cdot \frac{2p}{q} + pdd \cdot \frac{1+pp}{qq}$; cujus integrale est $nydx = cdx - \frac{2dx}{q} + pd \cdot \ [TRUNCATED]