Page:Newton's Principia (1846).djvu/278

From Wikisource
Jump to navigation Jump to search
This page has been validated.
272
the mathematical principles
[Book II.

to the horizon PQ, to find the density of the medium, which will make a projectile move in that line.

From the nature of the parabola, the rectangle PDQ is equal to the rectangle under the ordinate DI and some given right line; that is, if that right line be called b; PC, a; PQ, c; CH, e; and CD, o; the rectangle a + o into c - a - o or ac - aa - 2ao + co - oo, is equal to the rectangle b into DI, and therefore DI is equal to . Now the second term of this series is to be put for Qo, and the third term for Roo. But since there are no more terms, the co-efficient S of the fourth term will vanish; and therefore the quantity , to which the density of the medium is proportional, will be nothing. Therefore, where the medium is of no density, the projectile will move in a parabola; as Galileo hath heretofore demonstrated.   Q.E.I.

Example 3. Let the line AGK be an hyperbola, having its asymptote NX perpendicular to the horizontal plane AK, to find the density of the medium that will make a projectile move in that line.

Let MX be the other asymptote, meeting the ordinate DG produced in V; and from the nature of the hyperbola, the rectangle of XV into VG will be given. There is also given the ratio of DN to VX, and therefore the rectangle of DN into VG is given. Let that be bb: and, completing the parallelogram DNXZ, let BN be called a; BD, o; NX, c; and let the given ratio of VZ to ZX or DN be . Then DN will be equal to a - o, VG equal to , VZ equal to , and GD or NX - VZ - VG equal to . Let the term be resolved into the converging series , &c., and GD will become equal to ,