534 Mr DE MORGAN, ON SOME POINTS 



This last is the only primordinal which coexists with y = <£, % = ty : the joint primitives of 

 / = and any other can have no more than two arbitrary constants. Any other primitive of 

 /= must be deducible from y = <t>, % = ^, by making a, b, c, compensatory, which gives, 

 6 and c being functions of a, <fr a + & b b'+ O c c'= 0, ^„+ Sec = 0. Take any surface 



ft> (*, y, *) - 0, 



and eliminate x between $ a|M = 0, "^ a , 6jC = and w (x, $, 't) = 0. We have then two results 

 of the form (a, b, c, b', c) = 0, from which we get b = b (a, G, H), c = c(a, G, H), G and 

 H being arbitrary constants. 



Substitute in y = <!>, s? = -p and we have a new form of the primitive, involving a, G, H: 

 and thus every form of w gives its own variation of y = <!>, « = ^r". But we are not to infer 

 that the curves thus obtained form a shading of the surface u> = from which they are obtained. 

 This will happen only under two relations between a, G, H. 



If we simply eliminate x from 3> a|6)( ,= 0, ¥ o|6)( .= 0, we have (a, b, c, b', c) = 0, of which 

 we might assume a primitive as general as that of f = 0, say b = B (a, L, M, N), 

 c = C(a, L, M, N). From these, by help of 4> ol6(( .= 0, we get x m (a, L, M, N), whence 

 a <= (x, L, M, N), and similarly for b and c. Hence y = $, i*tj give forms of the same 

 character as we began with. Any one of these new primitives, selected by assigning L, M, N, 

 connects, by contact of the first order, all the original primitives which are selected by b = B, 

 c = C : and all ordinary primitives are found, in an infinite number of ways, among the con- 

 necting curves of others. There is a singular solution, a curve of contact to all primitives, 

 when it happens that the six equations <t> a = 0, <J> 6 = 0, 4> c =0, ^ a = 0, < i r b = 0, fym 0, can be 

 satisfied by one assignment of a, b, and c, each in terms of x : the result is obtained by 

 substitution in y = 0, % = *r". 



Let the pole (a, b, c) be restricted to the surface c = c (a, b). If we take <J> a|c da + <S> 6|t .d6 = 0, 

 and ^ aic da + < ^ bic db = 0, which give T = 0, T being O a| ^ 6| - 3> 6 | ^r a[ , we may eliminate a 

 and b between y = <1>, % = ^t", T = 0, and obtain « = % (x, y). When the pole, then, is on 

 c = c(a, b), the polar curve touches the surface ss = ss(x, y) in a point determined by obtaining 

 x, y, #, in terms of a and b from y =$,«== *ir, T = 0. Substitute these values in either of 

 the compensating equations, and integrate, giving b = b(a, const.). This is a mode of shading 

 the surface c = c(a, b), so that if the pole move along one of the lines of shading, the polar 

 curve touches a corresponding line of shading on x = % (x, y). And all these properties are 

 reciprocal. The original equations being O (x, y, z, a, b, c) = 0, if (x, &c.) = 0, if we make 

 (x, y, %) the pole, and restrict it to the surface % = * (x, y) which resulted above, and then 

 eliminate x and y between <£ '= 0, ¥ = 0, 4> x , s ^ , 2 - ^^^^ = we shall find c-c(a,b) 

 as at first assumed : and so on. The main point of the proof is as follows. If in = 0, ^ = 0, 

 we suppose c = c (a, b), % = ss (.v, y), and a, b, therefore, each an implied function of x and y, 

 we have $,, + <J> a| a x + & b[ b x = 0, and *,-, + ^,6^ + ¥„,&, = 0, to determine a x and b x . But if 

 *ai^6i - ^61^1 = ° ancl a x and b x be not infinite, we must have 4> | , 1 > I| - ^ xl ^ M = 0. Similar 

 reasoning on y gives ^ai^i - ^yi^i = °j whence $>,*„ - ♦yt*,, = °- It is not * n our 

 power to take any surface we please, and find another with the properties of a polar recipro- 

 cal : the reason being, that there is a 'primordinal equation, f(x, y, x, y ', %') = 0, necessarily 



