CHARACTERISTIC FUNCTION TO SPECIAL CASES OF CONSTRAINT. 157 



are the rectangular components of the component of the impressed force per- 

 pendicular to the path. 



But, if R be the force of constraint, X, /jl, v, its direction-cosines, we have by 

 ordinary kinetics 



d 2 x 



- X - RX, &c. 



Hence KX=2 (x - * (x| + t| + Z *)) , to, to, 



and therefore the whole pressure is double that due to the impressed forces. 



From the above follows also the well-known theorem, that the osculating plane 

 of the brachistochrone contains, at each point, the resultant of the impressed forces. 

 For it has been shown that this resultant coincides in direction with the centri- 

 fugal force, and the latter of course lies in the osculating plane. 



14. Another, and perhaps simpler proof of the theorem above is furnished 

 directly by (10). Thus, squaring and adding the three equations of that form, 

 after substituting in them from (11), we have 



(&x\ /AY , fdh\ 2 _ i u<m\ 2 fd®y (<mv 

 \w) + \dt 2 ) + \dty ~ ±w \\dxj + \d y ) + \&J 



1 d& (dr dM fad® ,dr_d&\ 

 ®* dt \ dx dx dy dy dz dz ) 



+ ®* \dt) \\dx) + \dy) + \dz) S 

 - 1 $( M W ( M W ( M \*\ l m (® d ®\4- 1 ( M \ 2 (fh 



= 4p tw + \dy) + v&y i " w dt (® -dirw \di) {W) 



[ by (12) and (7) ] 



I ( /d®\ 2 /d®\ 2 /d®\ 2 ) , , v 



= ip { Km) + Kdy) + (& ) } = X2 + Y2 + Z2 > b y (8). 



Hence the whole acceleration is equal to the resultant of the impressed forces ; and 

 therefore the component of the acceleration, normal to the curve, must be equal 

 to that of the resultant of the impressed forces ; from which the theorem follows 

 at once if we can show independently that the resultant of the impressed forces 

 lies in the osculating plane. This is easily done as follows. We have 



*•=?$*•*■ b y( 9 >- 



Hence "*-£«(£)- $«* * 



