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}/_{n}x, where d = a + ^{m}/_{n}b. By the law of cosines we have MN^{2} = CM^{2} + CN^{2}  2(CM)(CN)cosC. But cosC = EC/CN. Therefore


= CM^{2} + CN^{2}  2(CM)(CE) = 
æ
è 
d  
m
n

x 
ö
ø 
2

+ x^{2}  2 
æ
è 
d  
m
n

x 
ö
ø 

cx
b






= d^{2}  
æ
è 
2dm
n

+ 
2cd
b

ö
ø 
x + 
æ
è 
m^{2}
n^{2}

+ 
2cm
nb

+1 
ö
ø 
x^{2}. 


Taking the derivative of MN^{2}, which is obviously a function of x, and making it equal to zero we get
x = 
mnbd + cdn^{2}
bm^{2} + 2cmn + bn^{2}

. 

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 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 = 
mnbd + cdn^{2}
bm^{2} + 2cmn + bn^{2}

. 

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