386 
PROFESSOR J. CASEY ON A NEW 
Hence J F'(<p) sin <pdq> = c sec <p, 
v=c |log tan —cosec <p j+c (86) 
(2) Find the tangential equation of the involute of the catenary; that is, the 
tangential equation of the locus of the point Q (see fig. art. 33). The intrinsic equation 
of the involute of the catenary is 
s=cj tan <p d<p. 
Hence we have 
F(<p)=<? j tan <pd<p, 
jF'(<p) sin q>d<p=c{log (sec <p + tan <p) — sin <p}. 
Hence 
J log (sec <p + tan <p) C dip _ 
sin a <p J sin ip ’ 
and integrating the first integral by parts, we find it equal 
- cot <p. log {sec<p+tanpf + J^. 
Hence 
v—C—c cot <p . log (sec ip -f- tan <p), 
where C is the constant of integration, which is evidently equal to c ; therefore the 
required tangential equation is 
{g7) 
(3) Let the intrinsic equation be s=asmncp. Then we find 
2 na . 0 «<b 
sm 2 (88) 
This formula fails when «=1 ; but in that case we have F (p)=a sin p, and we find 
a<p 
V =J' ( 89 ) 
(4) Find the equation of a curve, being given 
v=ns. 
Here we have, if 
v =f($), s=nf(<p ); 
1$ (/W s ^ n2 ( P) =n f'( < P ) s i n <?• See equation (64). 
.\/(p) sin 2 <p=C (tan |p) B , 
