424 
PROFESSOR J. CASEY ON A NEW 
therefore the denominator of the fraction on the right of equation (237) will vanish ; 
and therefore 
ac cos %|/=(g— «)( — A\|/) 
(241) 
Hence A(\}/) must be negative. But A(\}/)=P'K-r-&, .*. P'K is negative ; or the point 
P' must lie between B and K, and therefore IP' is less than IB ; that is, the curve must 
be an intern epicycloid ; and, from what we have proved, we see that the angle IP'B 
must be obtuse. 
BK 
Again, we have <?cos\|/= -y and A4/=-y- ; and therefore the following conditions for 
determining a point of inflection: — 
1st. The angle IP'B must be obtuse. 
2nd. BK : KP' : : g — a : a from equation (241) ; that is, BK : KP' : : Cl : IB. Hence 
the triangle IP B can be constructed ; and it follows that there are two points of inflec- 
tion in each revolution of the generating circle. 
120. Since 
and also 
we have 
but 
<r'=2aQ (see art. 114), 
o^fe-2 a)<p, 
<p = ^ sin _, )sin\f/(ccoS\|/-f- A\}/)( ; 
<p = ^4-<p, 
. fo_ gc eos4> + ( g— g)Afr /oio\ 
• • d\ (g — 2a)A(4f) m jj ; 
Again 
~r = radius of curvature; 
a<p 
ds b(q — ff){ccos\J/ + A (vj/) 
dq> eccos\{/+(g — a) A\(/ ’ 
ds b'q — a) {c cos 4/ + A (4/) \- 
dA>~ (q — 2a)A(4/) 
(243) 
Now, if s' denote the corresponding arc of the extern cycloid, we have, from art. 85, 
ds' 6(cos vj/+ A'\f)) 2 
.'. ds= 
(244) 
lliis formula, which connects the extern epicycloid with the extern cycloid, may be 
also obtained as follows, for we have evidently 
