Methodus inveniendi lineas curvas

224 DE METHODO quippe qui constanter est $= -nv.d.\frac{p}{\sqrt{(1+pp)}}$; at in maximi minimive expressione $\frac{\int Sxdx\sqrt{(1+pp)}}{\int Sdx\sqrt{(1+pp)}}$ ponamus, casu quo $x=a$, fore $\int Sxdx\sqrt{(1+pp)} = A$ & $\int Sdx\sqrt{(1+pp)} = B$: illius vero numeratoris $\int Sxdx\sqrt{(1+pp)}$ valorem differentialem esse $=dA$, denominatoris vero $\int Sdx\sqrt{(1+pp)}$ valorem differentialem esse $=dB$. Hinc igitur maximi minimive expressionis, quae, casu $x=a$, fit $=\frac{A}{B}$, valor differentialis erit $\frac{BdA-AdB}{BB}$, qui multiplo cuicunque formulae communis valoris differentialis $-nv.d.\frac{p}{\sqrt{(1+pp)}}$ aequalis positus dabit aequationem pro curva quaesita. Jam ad valores differentiales $dA$ & $dB$ inveniendos, consideremus primum formulam $\int Sdx\sqrt{(1+pp)}$, quae secundum enumerationem §. 7 Cap. praeced. factam, pertinet ad Casum secundum: quo erit $Z=S\sqrt{(1+pp)}$, & posito $dS=Tds$, erit $dZ=Tds\sqrt{(1+pp)} + \frac{Spdp}{\sqrt{(1+pp)}}$. Comparatione ergo facta, erit $n=s$; $L=T\sqrt{(1+pp)}$, $M=0$, $N=0$, & $P=\frac{Sp}{\sqrt{(1+pp)}}$: rum vero ob $n=s=\int dx\sqrt{(1+pp)}$, erit $[Z]=\sqrt{(1+pp)}$ & $[M]=0$, $[N]=0$, & $[P]=\frac{p}{\sqrt{(1+pp)}}$. Jam sumatur integrale $\int Ldx = \int Tdx\sqrt{(1+pp)} = \int Tds = S$, cujus valor, casu $x=a$, fiat $=G$, eritque $V=G-S$. Quamobrem habebitur formulae $\int Sdx\sqrt{(1+pp)}$ valor differentialis $dB = -nv.d(\frac{Sp}{\sqrt{(1+pp)}} + \frac{p(G-S)}{\sqrt{(1+pp)}}) = -nv.d.\frac{Gp}{\sqrt{(1+pp)}}$. Altera porro formula $\int Sxdx\sqrt{(1+pp)}$ pariter in eodem Casu secundo comprehenditur, eritque $Z=Sx\sqrt{(1+pp)}$ & $dZ=Txds\sqrt{(1+pp)} + Sdx\sqrt{(1+pp)} + \frac{Sxpdp}{\sqrt{(1+pp)}}$ unde fit $n=s$; $L=Tx\sqrt{(1+pp)}$; $M=S\sqrt{(1+pp)}$; $N=0$.