Steps / Historical AI

Leonhard Euler · 1744 · Internet Archive

Methodus inveniendi lineas curvas

All 336 leaves · leaf 303

Transcription

296 DE CURVIS Deinde fit abfciffa $AP = x$, applicata $PM = y$, arcus $aM = s$, qui cum abfciffa $x$ confundetur; ita ut ftatui queat $ds = dx$; ex quo radius ofculi in M orietur $= \frac{ds^3}{dx ddy} = R$. Sit autem porro applicata prima $Aa = b$. His pofitis, ratiocinium ut ante inftituendo, ad eandem pervenietur aequationem $\frac{M}{af}\int fdx \int ydx = \frac{Ekkddy}{dx^2}$. 81. Si igitur ponamus $\frac{Ekk af}{M} = c^4$, ubi $f$ ut ante exprimit longitudinem penduli fimplicis isochroni; habebitur, integrando, pro curva haec aequatio $y = Ae^{\frac{x}{c}} + Be^{\frac{-x}{c}} + C \sin.\frac{x}{c} + D \cos.\frac{x}{c}$; quae ad praefentem cafum ita accommodabitur. Primo, fi ponatur $x = 0$; fieri debet $y = b$; unde fit $b = A + B + D$. Secundo, cum fit $\frac{c^4 ddy}{dx^2} = \int dx \int ydx$; pofito $x = 0$; fieri debet $\frac{ddy}{dx^2} = 0$, unde prodit $0 = A + B - D$. Tertio, cum fit $\frac{c^4 d^3y}{dx^3} = \int ydx$, pofito $x=0$; fieri quoque debet $\frac{d^3y}{dx^3} = 0$, unde nafcitur: $0 = A - B - D$. Quarto, fi ponatur $x = a$, evanefcere debet $\int ydx$, feu $\frac{d^3y}{dx^3}$; propterea quod $\int ydx$ exprimit fummam omnium virium laminam in directione ad axem AB normali trahentium, quae fumma fi non effet $= 0$, ipfa lamina motu locali promoveretur, contra inftitutum; erit ergo, ob hanc rationem, $0 = Ae^{\frac{a}{c}} - Be^{\frac{-a}{c}} - C \cos.\frac{a}{c} + D \sin.\frac{a}{c}$. Quinto
confidence 0.33 28 expressions source scan at the Internet Archive

This text was produced by machine and has not been corrected by hand. Where it matters, read the scan. Confidence is the model's own, not an external check.