Leonhard Euler · 1744 · Internet Archive
Methodus inveniendi lineas curvas
All 336 leaves · leaf 293
Transcription
286 DE CURVIS
tur. Primum posito $x=0$, fieri debet $y=b$; hinc ergo oritur ista aequatio, $b=A+B+D$, quae est prima.
Secundo, cum fit $\frac{c^4 ddy}{dx^2} = \int dx \int ydx$; facto $x=0$, fieri debet $\frac{ddy}{dx^2}=0$; at est $\frac{ddy}{dx^2} = \frac{A}{cc} e^{\frac{x}{c}} + \frac{B}{cc} e^{-\frac{x}{c}} - \frac{C}{cc} \sin \frac{x}{c} - \frac{D}{cc} \cos \frac{x}{c}$: unde oritur haec aequatio secunda, $0=A+B-D$.
Tertio, cum fit $\frac{c^4 d^3 y}{dx^3} = \int ydx$; posito $x=0$, simul $\frac{d^3 y}{dx^3}$ evanescere debet: quia ergo erit $\frac{c^3 d^3 y}{dx^3} = A e^{\frac{x}{c}} - B e^{-\frac{x}{c}} - C \cos \frac{x}{c} + D \sin \frac{x}{c}$: prodit aequatio tertia, $0=A-B-C$.
Quarto autem, si ponatur $x=a$, applicata $y$ evanescit, unde obtinebitur aequatio quarta, $0=Ae^{\frac{a}{c}} + Be^{-\frac{a}{c}} + C \sin \frac{a}{c} + D \cos \frac{a}{c}$.
Quinto, quia AB est tangens curvae in puncto B; facto $x=a$, fieri debet $\frac{dy}{dx}=0$: unde prodit aequatio quinta, $0=Ae^{\frac{a}{c}} - Be^{-\frac{a}{c}} + C \cos \frac{a}{c} - D \sin \frac{a}{c}$.
✝ Ex his ergo quinque aequationibus, primum quatuor constantes $A, B, C, D$ definientur; tum vero, in quo cardo rei versatur, determinabitur valor ipsius $c = \sqrt[4]{\frac{Bkk\,af}{M}}$; ex quo longitudo penduli simplicis isochroni $f$ elicietur; quo ipso, durationes oscillationum cognoscentur.
70. Ex aequationibus secunda & tertia, constantes $C$ & $D$ ex $A$ & $B$ ita definientur, ut fit $C=A-B$, & $D=A+B$.
qui
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