Newton to Euler

the words moment and fluxion appear to be used as synonymous.

After showing by examples how to solve the first problem, Newton proceeds to the demonstration of his solution:–-

"The moments of flowing quantities (that is, their indefinitely

small parts, by the accession of which, in infinitely small portions of time, they are continually increased) are as the velocities of their flowing or increasing.

"Wherefore, if the moment of any one (as $x$) be represented by the product of its celerity $\dot{x}$ into an infinitely small quantity $0$ (i.e. by $\dot{x}0$), the moments of the others, $v$, $y$, $z$, will be represented by $\dot{v}0$, $\dot{y}0$, $\dot{z}0$; because $\dot{v}0$, $\dot{x}0$, $\dot{y}0$, and $\dot{z}0$ are to each other as $\dot{v}$, $\dot{x}$, $\dot{y}$, and $\dot{z}$.

"Now since the moments, as $\dot{x}0$ and $\dot{y}0$, are the indefinitely little accessions of the flowing quantities $x$ and $y$, by which those quantities are increased through the several indefinitely little intervals of time, it follows that those quantities, $x$ and $y$, after any indefinitely small interval of time, become $x+\dot{x}0$ and $y+\dot{y}0$, and therefore the equation, which at all times indifferently expresses the relation of the flowing quantities, will as well express the relation between $x+\dot{x}0$ and $y+\dot{y}0$, as between $x$ and $y$; so that $x+\dot{x}0$ and $y+\dot{y}0$ may be substituted in the same equation for those quantities, instead of $x$ and $y$. Thus let any equation $x^3-a{x}^2+axy-y^3=0$ be given, and substitute $x+\dot{x}0$ for $x$, and $y+\dot{y}0$ for $y$, and there will arise $\left. {4} &\phantom{a}x^3 &&+ 3x^2\dot{x}0 &&+ 3x\dot{x}0\dot{x}0 &&+ \dot{x}^3 0^3 \ -{}&ax^2 &&- 2ax\dot{x}0 &&- a\dot{x}0\dot{x}0 \ +{}&axy &&+ ay\dot{x}0 &&+ a\dot{x}0\dot{y}0 \ & &&+ ax\dot{y}0 \ -{}&y^3 &&- 3y^2\dot{y}0 &&- 3y\dot{y}0\dot{y}0 &&- \dot{y}^3 0^3

\right} = 0.$

"Now, by supposition, $x^3-a{x}^2+axy-y^3=0$, which therefore, being expunged and the remaining terms being divided by 0, there will remain

3x^2\dot{x} &- 2ax\dot{x} + ay\dot{x} + ax\dot{y} - 3y^2\dot{y} + 3x\dot{x}\dot{x}0 - a\dot{x}\dot{x}0 + a\dot{x}\dot{y}0 \ &- 3y\dot{y}\dot{y}0 + \dot{x}^3 00 - \dot{y}^300 = 0.

But whereas zero is supposed to be infinitely little, that it may represent the moments of quantities, the terms that are multiplied by it will be nothing in respect of the rest (termini in eam ducti pro nihilo possunt haberi cum aliis collati); therefore I reject them, and there remains $3x^2\dot{x}-2ax\dot{x}+ay\dot{x}+ax\dot{y}-3y^2\dot{y}=0,$ as above in Example I." Newton here uses infinitesimals.

Much greater than in the first problem were the difficulties encountered in the solution of the second problem, involving, as it does, inverse operations which have been taxing the skill of the best analysts since his time. Newton gives first a special solution to the second problem in which he resorts to a rule for which he has given no proof.

In the general solution of his second problem, Newton assumed homogeneity with respect to the fluxions and then considered three cases: (1) when the equation contains two fluxions of quantities and but one of the fluents; (2) when the equation involves both the fluents as well as both the fluxions; (3) when the