EOE.M OF TANGENTIAL EQUATION. 
385 
we get 
Hence, from equation, 
e°— e °=2 tan <p. 
s=l(e°-e~ l o) 
(82) 
36. If from the point P we let fall the perpendicular P Q on the tangent at C, we 
have evidently 
PQ=ycos<p; but y = c sec <p, 
Again we have 
.\PQ =c. 
CQ 
PQ 
= tan 
<P, 
(83) 
CQ=s. 
(84) 
Hence the locus of the point Q is the involute of the catenary. 
37. From the diagram we have 
and 
T 
— = cosec <p 
CP y 
— = cosec <(> = ,; 
and since s is the length of a portion of the string whose weight is W, y is the length 
of a portion whose weight is T. 
38. The Intrinsic Equation of a curve being given, to find its Tangential Equation. 
This problem is the converse of the one solved in art. 30, Section II. 
Let s=F(p) be the given intrinsic equation, 
Hence from equation (64) we have 
(/'(<P) sin 2 <p) = F \<p) sin <p ; 
/(<?)=! cosec 2 <p{ jF (<p) sin <pdp\d<p. 
Hence the required tangential equation is 
p—Jcosec^-j JF'^) sin <f>d<p\d$ (85) 
Examples. 
(1) Find the tangential equation of the catenary. 
Here F(<p)=c tan <p. See equation (80). 
3 i 2 
