166 LIPKA. 



contact with) the corresponding geodesic circles? To answer this we 

 need simply apply the condition 



(41) 1 = 



to the form (39) ; we get 



(42) (1 + v'^)G' - 6'{(X„+ xy) (1 + /2) + SvV'} = 0. 



Substituting in this the value of 6" from (11), and solving for v", 

 we find 



(43) v" = 



{(^.-X„V^+2X,0) + (^.-0»-X.</)+X.V^)/-((/).-X..j)+2X„^/^)/^}{l+/^} 



3(0 +M 



where we have discarded the factor G whose vanishing leads to the 

 geodesies; as has already been noted, these curves form part of every 

 system of trajectories; for the geodesies p, |, ri are infinite. In (43) 

 we note that for a given value of u, v, v' , there corresponds only one 

 value of t" , so that we have 



Theorem 6. Through every 'point and in every direction through that 

 point on a surface, there passes one trajectory which hyperoscidates its 

 corresponding geodesic circle of curvature. 



If, now, we keep the point fixed and vary the initial direction v', 

 the center of geodesic curvature of the hyperosculating trajectories 

 of Theorem 6, will describe a certain locus. We get the equation of 

 this locus by eliminating p, v', v" from (39), (40), and (43). From 

 (39) and (40) we find 



,,,. , % „ (X.g + X.r? + c^) (|^+ rf) 



(44) v'= — ~\v = , 



r? -rf 



and substituting these in (43), we get 



(45) i\<^,- X.0- \u^) + i-n{i>,- (^„+ 2X„0- 2\4,) - r]%^Pu- X^iA 



- X„<^) + Se\H - ^k) = 0, 



a conic passing through the point 0. The equation of the tangent to 

 this conic at is 



(46) (l>7i-i^ = 0, 



whose direction is that of the force vector \l//(f). Hence, we have 

 Theorem 7. The locus of the centers of geodesic curvature of the ooi 



