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Mr. F. W. niU on 



Let the area of the curve be divided into infinitesimal 

 triangles having their vertices at a point inside the curve, 



Fig. 1. 



and for their bases elements of the perimeter, the problem is 

 best attacked by making the tracer move round each of these 

 triangles in turn, travelling along every radius-vector twice, 

 in opposite directions. Taking any triangle OPQ (fig. 2), 



let 6, 6 + d6 be the inclinations of OP, OQ to a fixed line 



OX; <j>. </>', fi + dfi, <f> + d<j>, the inclinations of the rod to the 



fixed line in the positions ^0, S 2 P, S 3 Q, S 4 respectively; r, 



r+dr the lengths of OP, OQ. 



Then by (1) 



, 6-Q -r ; 0-ch 



tan — y-2- =e *tan — 5-^ .... (2) 



and tan e+de-y- d# _-^ im 0+dd-^>-d<f, 



2 2 



for the motions from to P, and from Q to ; hence 



dO-d<f>' __dr 

 sin (0-0') ~ 7 + 



d0-dcf> 

 sin (0-0) ' 



. (3) 



