Methodus inveniendi lineas curvas

270 DE CURVIS metiri licet, omnium materiarum elasticitates inter se comparari ac determinari poterunt. De curvatura laminae elastice inaequabilis. Fig 2. 41. Quemadmodum igitur hactenus laminæ, cujus curvaturam determinavi, elasticitatem absolutam $Ekk$ per totam longitudinem constantem posui; ita solutio eadem methodo poterit absolvi, si quantitas $Ekk$ utcunque ponatur variabilis. Scilicet si elasticitas absoluta fuerit ut functio quæcunque portionis laminæ AM, quæ functio fit $=S$, posito arcu $AM=s$; atque existente radio ofculi in $M=R$; curva AM, quam lamina induit, ita erit comparata, ut in ea, inter omnes alias ejusdem longitudinis, fit $\int \frac{Sds}{RR}$ minimum. Solvetur ergo iste casus per formulam secundam generalem. Sit $dy=pdx$; $dp=qdx$; at $dS=Tds$, atque, inter omnes curvas in quibus est $\int dx\sqrt{(1+pp)}$ ejusdem magnitudinis, ea determinari debebit, in qua fit $\int \frac{Sqqdx}{(1+pp)^{\frac{5}{2}}}$ minimum. Prior formula $\int dx\sqrt{(1+pp)}$ dat pro formula differentiali $\frac{1}{dx} d. \frac{p}{\sqrt{(1+pp)}}$. Altera vero $\int \frac{Sqqdx}{(1+pp)^{\frac{5}{2}}}$ cum $\int Zdx$ comparata dabit $Z=\frac{Sqq}{(1+pp)^{\frac{5}{2}}}$. Cum igitur positum fit $dZ=Ld\Pi+Mdx+Ndy+Pdp+Qdq$, $\Pi=\int [Z]dx$, & $d[Z]=[M]dx+[N]dy+[P]dp$; erit $Ld\Pi = \frac{qqTds}{(1+pp)^{\frac{5}{2}}}$; unde $L=\frac{qqT}{(1+pp)^{\frac{5}{2}}}$. $d\Pi=ds=dx\sqrt{(1+pp)}$; ideoque $[Z]=\sqrt{(1+pp)}$, $[M]=0$, $[N]=0$, & $[P]=\frac{p}{\sqrt{(1+pp)}}$. Deinde vero est $M=0$, $N=0$, $P=-\frac{5Sqqp}{(1+pp)^{\frac{7}{2}}}$, & $Q=\frac{2Sq}{(1+pp)^{\frac{5}{2}}}$; ita ut fit $dZ = \frac{qqdS}{(1+pp)^{\frac{5}{2}}} + Pdp + Qdq$. 42. Jam sumatur integrale $\int Ldx = \int \frac{qqTdx}{(1+pp)^{\frac{5}{2}}} = \int$