( 192 ) 



Hence 



or 



or 



1 + 2z dT 



d*a 

 2 dx 1 * 



or 



3T 



1 + 2z dT 

 3T dx 



dT 



i 

 d\i 



1 da 1 db 

 dx 3 da a dx b dx 



dx 



'\ da 

 a dx 



1 db' 

 bdx 



dra | 

 a — 

 2 dx 



1 



3 



(£)'J 



1 da 1 db 

 So the value — is equal to 0, first if -— = -—, and secondly 



dx 



a dx b dx 

 If, when drawing T as function of x, we begin 



/da\ 

 if (_ )— _a — 



\dxj 3 dx° 



with small values of x, and if we should admit also negative values 



of x into our consideration, then both factors in the expression for 



dT . „ da ' dT . . . ic . 



- are negative e.g. tor — = U, and so — is positive. It x increases 

 dx dx dx 



a value of x is reached for which one of these factors becomes equal 



