(719) 
3n 
TE, $ 
Doma, =0, van eee Nu mike emg ved: CLA) 
1 
the meaning of which is that the curvature is perpendicular to the path. 
Let P, and Ps be two paths, having in common a position A and 
the direction in this position, so that the direction-vector D, in the 
position A, is likewise the same for the two paths, or 
Dia) = Deva). 
Let us consider elements of the two paths, beginning in A, and 
of egual lengths ds. If Dj and Dg are the direction-vectors at the 
ends of these elements, the vector 
= D1 — Ds ; (15) 
ds 
may appropriately be called the relative curvature of the path P, with 
respect to the path Py. Now, we may replace the numerator in (15) 
by [Di — Dial — De — Daa]; the relative curvature is therefore 
related as follows to the curvatures ¢ and ¢,-of the two paths: 
Cp sha: oe aoe Me ay ar ae (16) 
Like ¢, and co, the relative curvature is perpendicular to both paths. 
Cr 
§ 10. What has been said thus far holds for every imaginable 
path. We shall now consider possible paths, i. e. such as are com- 
posed of possible infinitely small displacements. The direction- 
constants of such a path satisfy the ¢ conditions 
Sn 
Pra a'y = 0). ass Whee) er coe See OA 
1 
as may be deduced from (9). 
Let there be given a position A and a direction in this position, 
so that the values of ay and 2’, are known, and let us seek the 
values of #”,, which make the curvature ¢ a minimum. 
In solving this problem, we have to take into account equation 
(14) and the conditions 
3n 3n _3n aye 
why 
Do an ey + EPE an Bat at Ope eee eee ae ADE 
1 
which are got by differentiating (17). We may therefore write for 
the values of a", that make (13) a minimum 
1 
me, = Dn P, + m, 2, Q, 
1 
P, and Q being i+ 1 quantities whose values can be determined 
by means of (14) and (18). The first of these equations, combined 
with (17) and (11), gives Q=0. The solution becomes therefore 
