Page:Newton's Principia (1846).djvu/268

proportion, as nascent. It will be the same thing, if, instead of moments, we use either the velocities of the increments and decrements (which may also be called the motions, mutations, and fluxions of quantities), or any finite quantities proportional to those velocities. The co-efficient of any generating side is the quantity which arises by applying the genitum to that side.

Wherefore the sense of the Lemma is, that if the moments of any quantities A, B, C, &c., increasing or decreasing by a perpetual flux, or the velocities of the mutations which are proportional to them, be called a, b, c, &c., the moment or mutation of the generated rectangle AB will be aB + bA; the moment of the generated content ABC will be aBC + bAC + cAB; and the moments of the generated powers A², A³, A⁴, A^½, A^3/₂, A^⅓, A^⅔, A^-1, A^-2, A^-½ will be 2aA, 3aA², 4aA³, ½aA^-½, ³/₂aA^½, ⅓aA^-⅔, ⅔aA^-⅓, -aA^-2, -2aA^-3, - ½aA^-3/₂ respectively; and in general, that the moment of any power A$\scriptstyle {\frac {n}{m}}$, will be $\scriptstyle {\frac {n}{m}}$aA$\scriptstyle {\frac {n-m}{m}}$. Also, that the moment of the generated quantity A²B will be 2aAB + bA²; the moment of the generated quantity A³ B⁴ C² will be 3aA² B⁴ C² + 4bA³B³C² + 2cA³B⁴C; and the moment of the generated quantity $\scriptstyle {\frac {A^{3}}{B^{2}}}$ or A³B^-2 will be 3aA²B^-2-2bA³B^-3; and so on. The Lemma is thus demonstrated.

Case 1. Any rectangle, as AB, augmented by a perpetual flux, when, as yet, there wanted of the sides A and B half their moments ½a and ½b, was A-½a into B-½b, or AB - ½a B - ½b A + ¼ab; but as soon as the sides A and B are augmented by the other half moments, the rectangle becomes A + ½a into B + ½b, or AB + ½a B + ½b A + ¼ab. From this rectangle subduct the former rectangle, and there will remain the excess aB + bA. Therefore with the whole increments a and b of the sides, the increment aB + bA of the rectangle is generated.   Q.E.D.

Case 2. Suppose AB always equal to G, and then the moment of the content ABC or GC (by Case 1) will be gC + cG, that is (putting AB and aB + bA for G and g), aBC + bAC + cAB. And the reasoning is the same for contents under ever so many sides.   Q.E.D.

Case 3. Suppose the sides A, B, and C, to be always equal among themselves; and the moment aB + bA, of A², that is, of the rectangle AB, will be 2aA; and the moment aBC + bAC + cAB of A³, that is, of the content ABC, will be 3aA². And by the same reasoning the moment of any power Aⁿ is naA^n-1.   Q.E.D

Case 4. Therefore since $\scriptstyle {\frac {1}{A}}$ into A is 1, the moment of $\scriptstyle {\frac {1}{A}}$ drawn into