of the Hyperbola. 
115 
assume V = a \/ | ~~uu ) ’ th 00 * by taking the fluxions on 
both sides, we shall h&veY T ILlIf 
-f- Here u s/f the first term on the 
(1— mm ) 1 ' uu 1 
right hand side of the equation, is evidently the fluxion of an 
arch of an ellipsis of which the transverse semi-axis is 1, the 
eccentricity is e, and the abscissa (counted from the center) 
-■ \ uu) -, by multiplying both 
is u. 
rp 1 . 1 • -| . U Ull t/ ( I “ E £ Ull) 
The third term, 
(1 — liu)‘ 
numerator and denominator by (1 — et uu), becomes 
, which (by division) is very easily resolved 
U UU EE U M 4 
(1— uu ) 2 X V ( 1 — E£ mm) 
EE U UU 
into 
V (1— uu) x VC 1 — f£ uu ) 
-j ; so that the sum 
(1 — uuY X \/( I EE 
of the second and third terms on that side of the equation 
becomes barely 
viously = 
( I EE )ll UU 
(I —uu) 2 X v' ( I— ES uu) 
(i ££ )u 
•, which expression is ob- 
( I £e)m 
; and the 
yqi— uu) x ^(i— ££ uu) 8 (1— mm)|x(i— se mm) 
last of these ( e being by the notation = — ) differs from H, 
the fluxion of the hyperbolic arch, found in the preceding Ar- 
ticle, only in that it is not multiplied by e ; or, in other words, 
it is = — . Let us therefore substitute the values now found 
e 
for their equals in the above fluxional equation, and we shall 
have V — u s/ ( 
I — uu 
( I — es)« 
+ (*) 
p (1 — mm) X vhi — Es mm) 
7. Now it is easy to perceive, from what is done in Art. ig 
and 14 of Landen’s second Memoir, (above referred to,) that 
the fluent of the second term on the right-hand side of this 
equation may be found by means of the fluent of the first term, 
O 2 
