Motion of bodies I

Two bodies move at the same time, from two given places A and B, and proceed uniformly from thence in given directions, AP and BQ, with celerities in a given ratio; it is proposed to find their position, and how far each has gone, when they are nearest possible to each other (Example XIII, page 28).

Let a = AC, b = BC, c = DC in the adjacent figure. Assume m is the velocity ("celerity") of the body that moves in the direction AP while n is the velocity of the body that moves in the direction BN.

At time t the first body will be at M while the second body will be at N (Simpson calls these two points "cotemporary"). Next draw perpendiculars NE and BD to AP, and let x = CN. Since DECN is similar to DDCB, we can conclude that b/x = c/CE. Because M and N are cotemporary points it follows that AM/m = BN/n, therefore AM = ^m/_n(x - b). So CM = AC - AM = a - ^m/_n(x - b) = d - ^m/_nx, where d = a + ^m/_nb. By the law of cosines we have MN² = CM² + CN² - 2(CM)(CN)cosC. But cosC = EC/CN. Therefore

MN2 = CM2 + CN2 - 2(CM)(CE) = æ è d - m n x ö ø 2   + x2 - 2 æ è d - m n x ö ø cx b = d2 - æ è 2dm n + 2cd b ö ø x + æ è m2 n2 + 2cm nb +1 ö ø x2.

Taking the derivative of MN², which is obviously a function of x, and making it equal to zero we get

x = mnbd + cdn2 bm2 + 2cmn + bn2 .

This is the value where MN², and consequently MN, attains its minimum.

Remark: Maybe Simpson should have mentioned that MN² is a second degree polynomial in the variable x, thus on the Cartesian plane it represents a vertical parabola that opens upwards with minimum at -r/2s where r and s are the coefficients of x and x² respectively. We would get exactly the same value obtained before, namely

x = mnbd + cdn2 bm2 + 2cmn + bn2 .

So, in this problem there is an alternative to the use of derivatives.