EOKM OF TANGENTIAL EQUATION. 
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or, since C is an arbitrary constant, 
(tan$p)»+* ) 
n + 1 y 
(90) 
(5) It is required to find the equation of two curves A and B, which are so related 
that the Tangential Equation of A is the same as the Intrinsic of B, and the Tangential 
of B the same as the Intrinsic of A. 
Let v=F (<p) be the curve A, 
'=/« B. 
Then by the first condition we have 
F(<p) sin <p + J F(<p) cos <pd<p 
=/'(<£>) sin 2 <pT/"(<p) sin 2 <p+2 J./'(<p) cos *<pdq> sin <p cos <pdq>, 
and by the second condition 
/(<p)=F(<p) sin F 7 (<p) cos <pd<p ; 
s ^ n 2<p+/"(<p) sin 2 <p+2j/'(<p) cos 2 <P+J/"(<P) sin <f> cos <p<7<p. 
And by differentiating and some easy reduction we get 
5 f" (<p) sin 2 <p + 6 f'(<p) cos 2 <p —f” (<p) cos 2<p (<p) = 0, 
or 
^ { 3/(<p) sin 2<p -/" (<p) cos 2<p +/' (<p) } = 0. 
3/(<p) sin 2<p +/"(<?){ 1 — cos 2<p } =4C„ 
Hence 
the multiple 4 being put to the arbitrary constant in order to avoid fractions ; 
(*) 
This may be written 
0 ) 
+C 2 log tani<p + C 3 . 
This is the tangential equation of the curve B. 
To find the equation of A we have, from equation (j3), 
