40 Mr. A. W. Riicker on the Critical Mean Curvature 



(11) find the values of fa corresponding to values between 

 37r/2 and Zir ; whence, since (7) and (9) depend on the 

 squares of the sine and cosine of 0, the values of (j> and fa 

 between vr/2 and 37r/2 are also known. 



Before making any numerical calculations it is convenient 

 to discuss (7) more fully. 



Differentiating, we get 



. . 0/) / f *i sin 2 (/> , , 1 + cos 2 f *i sin 2 <j> 



_ sin 2 fa cot 2<fc | g6> ^ | 2A 1 4- cos 2 

 Aj J L Ai 



sin 2 cos 2<f>! . , „.. K. A 



— il 1 _ 4 a x cosec 2 2<^x V ofa = ; 



which, if we use equations (5) and simplify, becomes 



tan 0{ E -F cos 2 — A 1 tan fa}$0 + 4 A x cosec 2 2fa$fa = 0. (12) 



Hence Sfa = if = mr, and if fa*=nir/2. Also, considering 

 the case in which 



E-Fcos 2 ^-A 1 tan^> 1 = 0, .... (13) 



we notice that, if we subtract (13) from (7), we get 



E-F + AiCot^O; (14) 



and these equations are satisfied if = 7r/2 and ^ = 56° 28'. 

 For, if = tt/2, (13) is true identically, and (14) reduces to 



log e tan(j + 2 ) = cosec< £i> 



which holds good when fa = 56° 28'. 



Hence, when = 7r/2, S fa/86 is of the form oo x 0, which 

 is readily shown to be equal to zero. 



To find the corresponding value of8fa/S0, we have from (11), 



sec 2 fa tan fa -^ + sec 2 fa tan fa -S-= — sin sec 2 0; 



and by substituting from (11) for tan<£/ and sec 2 fa, this 

 becomes 



-.nn^^UA^*'- sin (9 sin 2^ 



C0S ^W + Al Iff- 2 



Putting = tt/2, 



^= - tan fa=- tan 56° 28'. 



It must be remembered that corresponds to fa and that if 

 0' corresponds to fa, 0' = 2ir-0, so that dfa/d0 f = -dfa/d0. 



