Methodus inveniendi lineas curvas

156 DE USU METHODI est, $\int yy dx = A$, & $\int dx\sqrt{(yy+pp)} = B$; erit, si $n=1$, area $ACM = \frac{1}{2} A$, & arcus $AM = B$. Jam formulæ $\int yydx = A$ valor differentialis $dA$ est $yy dx$, & formulæ $\int dx\sqrt{(yy+pp)}$ valor differentialis $dB$ est $= dx\sqrt{(yy+pp)}$. Quare cum generaliter invenerimus pro curva hanc aequationem $BdA = AdB$, erit divisione per $n$, instituta, $2 Bydx = \frac{Aydy}{\sqrt{(yy+pp)}}$ $- \frac{Ad.p}{\sqrt{(yy+pp)}}$. Multiplicetur ea per $p$, ob $pdx=dy$, erit $2 Bydy = A ( \frac{ydy}{\sqrt{(yy+pp)}} - pd \frac{p}{\sqrt{(yy+pp)}} )$. At est $d.\sqrt{(yy+pp)} = \frac{ydy}{\sqrt{(yy+pp)}} + \frac{pdp}{\sqrt{(yy+pp)}}$ & $\frac{ydy}{\sqrt{(yy+pp)}} = d.\sqrt{(yy+pp)} - \frac{pdp}{\sqrt{(yy+pp)}}$; unde fiet $2 Bydy = A(d.\sqrt{(yy+pp)} - [TRUNCATED]