Methodus inveniendi lineas curvas

208 DE METHODO Prius vero membrum si evolvatur, transit in $+ \frac{(C-\int Ldx)pdp}{(1+pp)^{3/2}} - \frac{Lppdx}{\sqrt{(1+pp)}} = \int(\frac{Ldx}{\sqrt{(1+pp)}} + \frac{(C-\int Ldx)pdp}{(1+pp)^{3/2}}),$ cujus integrale est $-\frac{C-\int Ldx}{\sqrt{(1+pp)}}$. Quare, casu quo $M=0$, habebitur ista aequatio $\frac{C-\int Ldx}{\sqrt{(1+pp)}} = A - Z + Pp + Qq - \frac{pdQ}{dx}$. Sin autem tertio fuerit tam $M=0$ quam $N=0$, habebitur primum, ob $N=0$, haec aequatio: $A + \frac{(C-\int Ldx)p}{\sqrt{(1+pp)}} = -P + \frac{dQ}{dx}$; quae, multiplicata per $dp=qdx$, abit in hanc $Adp + \frac{(C-\int Ldx)pdp}{\sqrt{(1+pp)}} = -Pdp + Qdq$. Cum autem tem fit $dZ = Ldx\sqrt{(1+pp)} + Pdp + Qdq$, habebitur $dZ + Adp - Ldx\sqrt{(1+pp)} + \frac{(C-\int Ldx)pdp}{\sqrt{(1+pp)}} = qdQ + Qdq$; quae integrata dabit, $Z+B+Ap+\frac{(C-\int Ldx)}{\sqrt{(1+pp)}} = Qg$, seu $C-\int Ldx = \frac{Qg-B-Ap-Z}{\sqrt{(1+pp)}}$. At ex priore aequatione est $C-\int Ldx = -\frac{A\sqrt{(1+pp)}}{p} - \frac{P\sqrt{(1+pp)}}{p} + \frac{dQ\sqrt{(1+pp)}}{pdx}$; ex quibus conjungendis dis elicitur: $Adx - Bdy = Zdy - Pdx - Ppdy + dQ + ppdQ - Qpdp$, in qua non amplius inest formula integralis $\int Ldx$. Usum igitur horum casuum in Exemplis monstrabimus. EXEMPLUM I. 61. Inter omnes curvas isoperimetras; definire eam, in qua sit $\int s^n dx$ maximum vel minimum, denotante $s$ arcum curvae abscissae $x$ respondentem. Quo-