where 
FORM OF TANGENTIAL EQUATION. 
s=2(a + £)E(c',<p), . . 
411 
. ( 200 ) 
( 201 ) 
Hence the arc of an extern cycloid is equal in length to an elliptic arc. 
If we make d — we find from equation (201) c = ^ ; but from the triangle IBP' 
we have 
a : b :: sin (2<p — 0) : sin 0. 
Hence 
sin (2 <p — $)=c sin 6. 
Hence we can apply Landen’s formula of transformation to the function on the right 
of equation (200), and we get (see Cayley’s ‘ Elliptic Functions,’ p. 329) 
2(a + i)E(c', <p)=2iE(c, 6)+^-F(c, S) + 2« sin 0 ; 
.'. s=25E(<?, F(c, ^)-j-2o!sin^ 
( 202 ) 
Now since B is the centre of instantaneous rotation, the locus of the point P' will 
be at right angles to P'B ; that is, the tangent at P' will be perpendicular to P'B, 
and the tangent at the highest point will be perpendicular to IB; hence the angle 
between these tangents will be equal to 6, and therefore the transcendental equation 
(202) is the intrinsic equation of the extern cycloid. 
84. If the point P' be inside the circle, that is, if the curve be an intern cycloid, the 
formula (202) will still hold, and be the intrinsic equation of the curve ; but the modulus 
c of the functions E and F will be greater than unity. A simple transformation of that 
formula will give one in which c is changed into its reciprocal. If we interchange the 
quantities a and b in equation (200), the value of s remains unaltered; hence when the 
point P' is inside, and b less than a, if we wish that the modulus of the functions E 
and F should be less than unity, instead of formula (202) we shall have the following * — 
s=2«E(c, 0)+^-~ F(c, 0) +2b sin 6. 
In this formula c= f \ and the angle Q is the angle IP'B. 
(203) 
85. If we differentiate the equation (202) we get, after some slight reduction, 
ds b(c cosfl + A(0)) 2 
d0~ A i0) 
Hence if § denote the radius of curvature of an extern cycloid, we have 
b(c cos 0 + A(0)) 2 
(204) 
A(0) 
