326 Mr DAVIES on the Equations of Loci 



sin 2 r cos 2 R 



= 



(1 + sin 4 r) 1 _|_ 2 (1 + sin* r) sin r cos r\/ 1'" 

 which, being imaginary, shows that no such relation can exist. 



2. Take R r : then, after reducing in the usual manner, we find the 

 relation 



sin 2 R sin* r = (9.) 



which shews that the said condition is fulfilled by arcs which are either equal 



or supplementary. But if we put this in(5), we shall have dd=.k \/ 1, which 

 shows that d 6 does not now depend upon < or its differentials, and hence, 

 that the length of the epicycloid is not, in this case, a function of <j> ; as in- 

 deed we know it should not, $ being constant, while 6 is variable. In- 

 deed, if we put this value in (6), we find, that whilst it fulfils the conditions 

 imposed upon it, viz. the condition of equation (7), it also renders both the 

 numerator and denominator of the radical in (6) imaginary, indicating that 

 contradictory conditions have been introduced amongst the data : and we 

 readily discover that the contradiction is, the supposed dependence of 6 

 upon 0, whilst sin 2 R sin* r = 0. This result was necessary, in order to 

 fully indicate all the circumstances of the inquiry, as, had the second side of 

 (5) merely vanished, one should have inferred (legitimately too) that was 

 constant ; that is, it would have indicated an error. 



All attempts to accomplish this object by means of a relation between 

 R and r, therefore, failing, we are led to consider the only possible way in 

 which it might be effected, that is, by considering R and r as independent 

 quantities, or, in other words, taking some fixed value of one of them, whilst 

 the other is perfectly arbitrary, which would fulfil the conditions we have 

 been seeking for. To perform this, we have only to expand (7) in terms of 

 some one function of R, and again in terms of some one function of r. We 

 shall thus get, considering R the arbitrary, 



_ ( sin 4 r cos 4 r 2 sin r cos r (1 + sin 2 rY sin 2 R + i 



for R + r. 1 (IQ\ 



I +(1 + sin 2 r) 2 { (1 + sin 2 r) 2 + 4sinVcos 2 r} sin'R = 0. ) 



for R, r, cos 4 r sin 2 R cos 4 r sin 2 r = (11.) 



