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ROYAL SOCIETY OF CANADA 



Next let us consider how the eqiintion 



of the spherical parabola is deduced 



according to this method. Let C D = 



C = a. Take as origin and the line 



cos 6 tan r = tan iS a as directrix. Take 



any point P such that P = P Af = r. 



Let 6 he as represented. Then sin r = 



cos P /i? sin {'-3 a — li) from triangle 



MPY. 



Therefore 



sin r = (sin 2 a cos OR— cos 2 a sin R) cos P R 



sin 2 a cos r — cos 2 a sin R cos P^. 



But 



/n u r> r. Sm P R „ 



sm C ^ cos P R = — ^--^ cos ^ = sm r cos 

 sin ^ 



Therefore 



sin 2 a 

 tan r 



(1 4- cos ^ cos 2 a), 



which ia the equation of the spherical parabola. It, of course, breaks 



2 a 



down into the form = (1 + cos 6) when R becomes infinite. 



r 



The work on Cartesian co-ordinates in connection with the 

 ellipse will lead us to expect that the equation cannot be found in 

 any convenient form. That is, the inverse trigonometrical functions 

 appear. 



XXXIV. Let us now find a formula for giving the inclination of 

 the tangent to a radius rector, in connection with any continuous 

 curve, /(r, d) = o, on the surface of the sphere. Let the co-ordin- 

 ates of P be (r, d), of Q (r 4- A r, 

 ^ + A ^), and of i? (r + Ar,e). PQ is 

 a finite portion of a curve, /(r, 6) = o. 

 The angle ?/' is defined to be the limit 

 of the angle R P XQ âs Q moves up to P, 

 the point X lying on the great circle 

 joining P and Q. Q R is also an arc of a 

 great circle. We shall first prove that 

 the angle R becomes ultimately 90°, 

 cos R = — cos R cos A ^ + sin i^ sin A ^ cos (r -f- A r). 



Therefore 



cos R {I + cos A 0) = sin R sin A 6 cos (r + A r) 



sin A . cos (r 4- A '') 



cot R = 



1 + cos A 



