228 
Proceedings of Royal Society of Edinburgh. [sess. 
must find the particular values of p and v which minimise e'. As 
e may change sign between 0 and 7 r, it is necessary to write the 
above integral 
cos \-m+n) -cos \-m-ri) it 
' = ~f dt . + f 
l 
(18) 
edt 4- J edt 
cos cos ~\-m-n) 
where m and n have the following values. Ascertain the points at 
which e changes sign by putting e equal to zero in equation (15), 
which leads to 
p 2 cos 2 t 4 - 2 pv cos t + v 2 = a - 2/5 cos t. 
pv + [3 + J(ap 2 4- 2 (3vp + /3 2 ) . 
cos t = 
or, say, 
where, 
p* p l 
cos t = - m + n , 
pv + (3 
, and n= 
J(ap 2 4- 2 (3vp 4- (3 2 ) 
(19) 
P“ p“ 
From (18) by partial differentiation, remembering that 
cos _1 ( — m ± n) wdien substituted for t in (15) makes e equal 
to zero, we have — 
de 
dp 
df 2 
dp 
y II V I IV J \j\JO y — u V — I V J It 
= - | f cos t dt - J cos t dt 4 - j cos t dt | 
^ 0 cos _1 (~™'+w) J cos -1 (-m-n) 
~ { s/[l - ( - m + n) 2 ] - ^[1 - (m + ») 2 ] | 
At a minimum value of e', d e /dp must be zero. But there are 
only two ways in which this can occur, either m = 0, or n = 0. In 
the latter case it can be shown that de /dv = 1 ; hence the only way 
to minimise e' is to make m = 0. 
With this value of m, by partial differentiation of (18) we have 
5- 1 ' 
1 dt 
0 J cos -1 n J cos -i (-?i) 
0e 
dv 7 r 
df_2 
dV 7T 
cos 1 n ,.cos k - n) it 
’ = l{ ( ldt - f ldt + fldt i 
’ IT * H " nr\a 1 nn * r»r\c 1/ m \ ' 
| cos 1 n - cos 1 ( - n) 4- ^ | 
Now 0 and therefore when both de/dp = 0 and de/dv = 0, 
we have 
m = 0 .... ( 20 ) 
-1 7T 
cos = 
4 
( 21 ) 
