Methodus inveniendi lineas curvas

ELASTICIS. 397 Quinto, quia lamina in extremitate B est libera, ibi curvaturam nullam habere poterit; eritque ideo, posito $x=a$, quoque $\frac{ddy}{dx^2}=0$; unde erit $0 = A e^{\frac{a}{c}} + B e^{-\frac{a}{c}} - C \sin \frac{a}{c} - D \cos \frac{a}{c}$. His igitur quinque conditionibus in computum ductis, non solum quatuor constantes $A$, $B$, $C$ & $D$ determinabuntur; sed etiam fractionis $\frac{a}{c}$ valor reperietur; ex quo proinde longitudo penduli simplicis isochroni $f$ innotescet. 82. Ex harum aequationum secunda & tertia, obtinetur $D=A+B$, & $C=A-B$, qui in sequentibus substituti praebebunt $0 = A e^{\frac{a}{c}} - B e^{-\frac{a}{c}} - (A-B) \cos \frac{a}{c} + (A+B) \sin \frac{a}{c}$ $0 = A e^{\frac{a}{c}} + B e^{-\frac{a}{c}} - (A-B) \sin \frac{a}{c} - (A+B) \cos \frac{a}{c}$ ; ex quibus reperitur: $\frac{A}{B} = \frac{e^{-\frac{a}{c}} - \cos \frac{a}{c} - \sin \frac{a}{c}}{e^{\frac{a}{c}} - \cos \frac{a}{c} + \sin \frac{a}{c}} = \frac{-e^{-\frac{a}{c}} - \sin \frac{a}{c} + \cos \frac{a}{c}}{e^{\frac{a}{c}} - \sin \frac{a}{c} - \cos \frac{a}{c}}$ ex qua aequalitate elicitur ista aequatio $0 = 2 - e^{\frac{a}{c}} \cos \frac{a}{c} - e^{-\frac{a}{c}} \cos \frac{a}{c}$ ; seu $e^{\frac{a}{c}} = \frac{1+\sin \frac{a}{c}}{\cos \frac{a}{c}}$ unde sequentes formabuntur aequationes Euleri De Max. & Min. P p L