286 PROFESSOR TAIT ON THE ROTATION OF A 



of which a particular solution is evidently 



2d = on . 



But this must be completed by the addition (to the second member) of a solution 



of the equation 



yr + r£ = 0, 



since any such term in the value of $ would disappear from the differential 



equation. 



Such a solution is easily found, by putting — £ for £ in (17), and attending 



to § 32, in the form 



r = 7 A - a£ (28), 



with (as in § 32) the condition 



S( 7 + Qa = (29). 



Hence, finally, 



2d = h n + 7 a - a£ (30), 



which, by taking the scalar, gives 



S(7-0a = -SAi (31). 



34. By differentiation of (2G) we have 



S( 7 - $3 = sat = s . lip . 



Substituting the value of $ from (30) we have 



S . ( 7 - Qdr, + 2S . 7 £a = 2S . bin , 

 or 



2S. 7 ^a=-S.( 7 + C)^ .... (32). 



From (29), (31), and (32), we find A by the usual quaternion process in the form 



2aS . ( 7 - 0(7 + Wrl = - Vfr - 0(7 + OS • (7 + 0*» - 2V . (7 + ov 7 gs*i , 

 or 



2aV 2 7 £ = - V 7 £S . ( 7 + fyir, + ( 7 - Q ( 7 2 + S 7 0S^ . . (33), 



where, in transforming the last term, we must recollect the equation T£ = Ty. 

 From this we deduce at once 



2( 7 A - a£)V 2 7 £= - ( 7 V 7 £ - V 7 £ . os . (7 + 0ft» + [7(7 - - (7 - 03 (7 2 + S 7 0S*i , 

 or 



2( 7 A-A0V 2 7 C=( 7 -0(7 2 +S70S.(7 + ?% + 2( 7 2 -70(7 2 + S70S^. 



or, finally, remembering that 



V 2 7 £ = S 2 7 ?-7 2 r = S 2 7£-7 4 , 



2( 7 a - a£) (S 7 £ - 7 2 ) = ( 7 - S . ( 7 + a, + 2( 7 2 - 7 l) S8n . 



35. Substituting this in (30), we get, after a slight transformation, consisting 

 in omitting the scalar parts of the right hand side, whose sum is zero, 



2a(S 7 £ - 7 2 ) = (S 7 £ - 7 2 ) Yin + i (7 ~ 5) S • (7 + 0* - V 7 £Sa„ . 



