Steps / Historical AI

Leonhard Euler · 1744 · Internet Archive

Methodus inveniendi lineas curvas

All 336 leaves · leaf 224

Transcription

MAX. ET MIN. RELATIVA. 217 $Wdx\sqrt{(1+pp)}$, ac $dW$ loco $Udv$; quo facto, habebitur ista aequatio $\frac{2gBdp}{W\sqrt{v}(1+pp)} + 2Bdp - \frac{gp^2dv}{Wv\sqrt{v}} - \frac{2gppdW}{W^2\sqrt{v}} - \frac{2gBpdW\sqrt{(1+pp)}}{W^2} = 0$; quae divisa per $p^2$ fit in- tegralis; eritque aequatio integrata haec, $2C - \frac{2B}{p} = \frac{2gB\sqrt{(1+pp)}}{Wp} - \frac{2g}{W\sqrt{v}}$, sive $W = \frac{gB\sqrt{v(1+pp)}}{Cpv-B\sqrt{v}} - \frac{gp}{Cp\sqrt{v}-B\sqrt{v}} = \frac{gB\sqrt{v(1+pp)}-gp}{Cpv-B\sqrt{v}}$. Unde nascitur aequatio a resistentia $W$ libera haec, $(Cp-B)dv = gCpdx + gBppdx - \frac{gpdx\sqrt{(1+pp)}}{\sqrt{v}}$. Cum autem $W$ fit functio ipsius $v$ data, ope aequationis $W\sqrt{v} = \frac{gB\sqrt{(1+pp)}}{Cp-B} - \frac{gp}{Cp-B}$, dabitur $p$ per $v$; qui valor si in praecedente aequatione substituatur, dabitur $dx$ per $v$ & $dv$; hincque curva quaesita poterit construi. PROPOSITIO VI. PROBLEMA. 67. Inter omnes curvas proprietate communi A praeditas, de- terminare eam, in qua fit functio quaecunque, cum ipsius illius expres- sionis A, tum alius cujuscunque B, maximum vel minimum. S O L U T I O. Sit $dA$ valor differentialis expressionis A, atque $dB$ valor dif- ferentialis expressionis B; habebit functionis illius ipsarum A & B, quam maximum minimumve esse oportet, valor differentialis hujusmodi formam $\alpha dA + \epsilon dB$; in qua constantes $\alpha$ & $\epsilon$ a ratione compositionis qua expressiones A & B in illa functione inter se permiscentur pendent; ita ut valores obtineant deter- minatos ab abscissa quantitate, cui solutionem accomodatam esse oportet pendentes. Quoniam vero expressionis A, quae proprietatem communem complectitur, valor differentialis est $\delta A$. Euleri de Max. & Min. E e
confidence 0.30 23 expressions source scan at the Internet Archive

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