Page:Newton's Principia (1846).djvu/236

the hyperbolic curve ab. And the chord ba being drawn, will inclose the area aba equal to the area sought ANB.

Example 2. If the centripetal force tending to the several particles of the sphere be reciprocally as the cube of the distance, or (which is the same thing) as that cube applied to any given plane; write $\scriptstyle {\frac {PE^{3}}{2AS^{2}}}$ for V, and 2PS $\scriptstyle \times$ LD for PE²; and DN will become as $\scriptstyle {\frac {SL\times AS^{2}\quad AS^{2}\quad ALB\times AS^{2}}{PS\times LD\quad 2PS\quad 2PS\times LD^{2}}}$ that is (because PS, AS, SI are continually proportional), as $\scriptstyle {\frac {LSI}{LD}}-{\frac {1}{2}}SI-{\frac {ALB\times SI}{2LD^{2}}}$. If we draw then these three parts into the length AB, the first $\scriptstyle {\frac {LSI}{LD}}$ will generate the area of an hyperbola; the second ½SI the area ½AB $\scriptstyle \times$ SI; the third $\scriptstyle {\frac {ALB\times SI}{2LD^{2}}}$ the area $\scriptstyle {\frac {ALB\times SI}{2LA}}$ $\scriptstyle {\frac {ALB\times SI}{2LB}}$, that is, ½AB $\scriptstyle \times$ SI. From the first subduct the sum of the second and third, and there will remain ANB, the area sought. Whence arises this construction of the problem. At the points L, A, S, B, erect the perpendiculars Ll Aa Ss, Bb, of which suppose Ss equal to SI; and through the point s, to the asymptotes Ll, LB, describe the hyperbola asb meeting the perpendiculars Aa, Bb, in a and b; and the rectangle 2ASI, subducted from the hyberbolic area AasbB, will leave ANB the area sought.

Example 3. If the centripetal force tending to the several particles of the spheres decrease in a quadruplicate ratio of the distance from the particles; write $\scriptstyle {\frac {PE^{4}}{2AS^{3}}}$for V, then $\scriptstyle {\sqrt {2PS+LD}}$ for PE, and DN will become as $\scriptstyle {\frac {SI^{2}\times SL}{\sqrt {2SI}}}\times {\frac {1}{\sqrt {LD^{3}}}}-{\frac {SI^{2}}{2{\sqrt {2SI}}}}\times {\frac {1}{\sqrt {LD}}}-{\frac {SI^{2}\times ALB}{2{\sqrt {2SI}}}}\times {\frac {1}{\sqrt {LD^{5}}}}$. These three parts drawn into the length AB, produce so many areas, viz. $\scriptstyle {\frac {2SI^{2}\times SL}{\sqrt {2SI}}}$ into $\scriptstyle {\overline {{\frac {1}{\sqrt {LA}}}-{\frac {1}{\sqrt {LB}}}}}$; $\scriptstyle {\frac {SI^{2}}{\sqrt {2SI}}}$ into $\scriptstyle {\sqrt {LB-{\sqrt {LA}}}}$; and $\scriptstyle {\frac {SI^{2}\times ALB}{3{\sqrt {2SI}}}}$ into $\scriptstyle {\overline {{\frac {1}{\sqrt {LA^{3}}}}-{\frac {1}{\sqrt {LB^{3}}}}}}$. And these after due reduction come forth $\scriptstyle {\frac {2SI^{2}\times SL}{LI}}$, SI², and SI² +