# 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.

Editor’s Note: This article was published in 2005.