Methodus inveniendi lineas curvas

96 DE METHODO MAX. ET MIN. $dt - zdz - zzdt = \frac{2dt(ddz+2dtdz)}{ds^2+(dz+zdt)^2}$. Sit porro $dt = sdz$, erit $ddt = sddz+dsdz$, hincque $ddz = sdz^2 - \frac{dsdz}{s}$. Habebitur ergo hæc æquatio, $sdz - zdz - szzdz = \frac{z^2dz - zds}{s^2+(1+sz)}$; quæ quidem eft differentialis primi gradus inter duas variabiles $s$ & $z$ tantum; verumtamen ultra integrationem non admittit. Multo minus igitur quicquam effici poterit, fi in genere quæftionem confideremus. SCHOLION III. 18. Hujus exempli cafus, quo curvam inveftigavimus, in qua maximum minimumve fit $\int dx \int ydx \sqrt{1+pp}$, etfi ineft duplex fignum integrale, tamen etiam per methodum præcedentis Capitis poteft refolvi; id quod ideo operæ pretium eft oftendere, ut confenfus utriufque methodi declaretur. Præcipue autem hoc opere nova via patefiet refolvendi plurima alia Problemata circa maxima & minima, quæ adhuc, quantum conftat, non eft tacta. Quæftio fcilicet eft, ut pro data abfciffa $AZ = a$, fiat maximum minimumve hæc expreffio $\int dx \int ydx \sqrt{1+pp}$, quæ tranfmutatur in hanc $x\int ydx \sqrt{1+pp} - \int xydx \sqrt{1+pp}$. Ut hæc forma reddatur maximum minimumve, oportet ut ejus valor, pro abfciffa $AZ = a$, idem fit pro ipfa curva quæfita $az$ & pro eadem puncto $n$ in $r$ tranflato. Ponamus ergo fieri $\int ydx \sqrt{1+pp} = A$, fi ponatur $x=a$, atque eodem cafu $\int xydx \sqrt{1+pp} = B$. Jam, elementis $mno$ in $mro$ tranfmutatis, valor $A$ augebitur fuo valore differentiali, qui, per Caput præcedens, eft $= nv \cdot dx(\sqrt{1+pp} - \frac{1}{dx} d.\frac{p}{\sqrt{1+pp}})$; per eadem præcepta autem quantitatis $B$ valor differentialis prodit $= nv \cdot dx(x\sqrt{1+pp} - \frac{1}{dx} d.\frac{xyp}{\sqrt{1+pp}})$. Quamobrem formulæ propofitæ $\int dx \int ydx \sqrt{1+pp}$, tranflato puncto $n$ in $r$, pro abfciffa $AZ = a$,