FOKM OF TANGENTIAL EQUATION. 
397 
(7) Let s ~ c tan 2 <p. then /<p=c tan 2 <p ; 
jf(<p) sin <p d<p=c(cb8<p+sec <p), 
v=c I* (cos <p-f-sec <p) cosec 2 <p d<p 
~c\ log (sec <p + tan <p) — 2 cosec <p[ (141) 
(8) Find the tangential equation of the involute of a circle. 
The intrinsic equation of the circle is 
s=a<p ; 
.-. f/(<p) sin <p (7<p = a(sin cp-<p cos <p), 
.-. j cosec 2 <p{f(<p) sin <pd<p } (l<p = £^‘> 
the tangential equation of the involute of the circle is 
a< P , 
sin<p 
We can verify the foregoing result geometrically as 
follows: — For in the annexed diagram, which represents 
a circle and its involute, we have PQ = «<p, and OB = r; 
and since QB is parallel to PO, we have at once 
OBsin<p=PQ; that is, v sin <p=a$, 
which proves the proposition. 
(142) 
Fig. 7. 
CHAPTER IV. 
Section I. — Positive Pedals. 
52. If we make the perpendicular to our director line the initial line, it is evident 
that the polar equation of the first positive pedal of the curve 
v =f{<P) 
is ?=/(<P)sin<p (143) 
Hence the tangential equation of any curve is at once transformed into the polar 
equation of its first positive pedal by changing v into g, and multiplying the function 
on the right-hand side by sin <p. 
Thus the tangential equation of the parabola is 
v=a tan <J5 (see art. 25) ; 
