Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/373

Rh with a continued motion perpendicularly into the ame length, will decribe the area of an hyperbola, which ubducted from the area SL x AB will leave ANB the area ought. Whence aries this contruction of the problem. At the points L, A, B(Fig. 2.) erect the perpendiculars Ll, Aa, Bb; making Aa equal to Ll, and Bb equal to LA. Making Ll, and LB aymptotes, decribe through the points LA, the hyperbolic curve ab. And the chord ba being drawn will incloe the area aba equal to the area ought ANB.

If the centripetal force tending to the everal particles of the phere be reciprocally as the cube of the ditance, or (which is the ame thing) as that cube applied to any given plane; write $$\scriptstyle \frac {PE^2I}{2AS^2}$$ for V, and 2PS x LD for $$\scriptstyle PE^2$$; and DN will become as $$\scriptstyle \frac {SL \times AS^2}{PS \times LD} - \frac {AS^2}{2PS} - \frac {ALB \times AS^2}{2PS \times LD^2}$$ that is (becaue PS, AS, SI are continually proportional) as $$\scriptstyle \frac {LSI}{LD} - \frac 12SI - \frac {ALB \times SI}{2LD^2}$$. If we draw then thee three parts into the length AB, the firt $$\scriptstyle \frac {LSI}{LD}$$ will generate the area of an hyperbole; the econd $$\scriptstyle \frac 12SI$$, the area $$\scriptstyle \frac 12AB \times SI$$; the third $$\scriptstyle \frac {ALB \times SI}{2LD^2}$$, the area $$\scriptstyle \frac {ALB \times SI}{2LA} - \frac {ALB \times SI}{2LB}$$ that is $$\scriptstyle \frac 12AB \times SI$$. Form the firt ubduct the um of the econd and third, and there will remain ANB the area ought. Whence arifes this contruction of the problem. At the points