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
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