# Thomas Simpson and Maxima and Minima - Motion of bodies I

Author(s):
Michel Helfgott

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 $\overline{NE}$ and $\overline{BD}$ to $AP$, and let $x = CN$. Since $\triangle ECN$ is similar to $\triangle DCB$, we can conclude that $\frac{b}{x} = \frac{c}{CE}$. Because $M$ and $N$ are cotemporary points it follows that $\frac{AM}{m} = \frac{BN}{n}$, therefore $AM = \frac{m}{n}(x - b)$. So $CM = AC - AM = a - \frac{m}{n}(x - b) = d - \frac{m}{n}x$, where $d = a + \frac{m}{n}b$. By the law of cosines we have $MN^2 = CM^2 + CN^2 - 2(CM)(CN)cosC$. But $cosC = \frac{EC}{CN}$. Therefore

$MN^2 = CM^2 + CN^2 - 2(CM)(CE) = (d - \frac{m}{n} x)^2 + x^2 - 2(d - \frac{m}{n} x) \frac{cx}{b}$

$= d^2 - (\frac{2dm}{n} - \frac{2cd}{b}) x + (\frac{m^2}{n^2} + \frac{2cm}{nb}) x^2$.

Taking the derivative of $MN^2$, which is obviously a function of $x$, and making it equal to zero we get

$x = \frac{mnbd + cdn^2}{bm^2 + 2cmn}$

This is the value where $MN^2$, and consequently $MN$, attains its minimum.

Remark: Maybe Simpson should have mentioned that $MN^2$ 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 $-\frac{r}{2s}$ where $r$ and $s$ are the coefficients of $x$ and $x^2$ respectively. We would get exactly the same value obtained before, namely

$x = \frac{mnbd + cdn^2}{bm^2 + 2cmn}$

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