262 ROYAL SOCIETY OF CANADA 



XV. We shall first take the special case where a = — . The 



differential equation then reduces to the form 



d ^ _ sin Odd 



sin cos cos d 



The solution is here arrived at readilv. 



That is 



f d 2 <f) ^ r d OS d 



J sin 2 à) J cos d 



4>' 0' ^ Q' Q' 



tan , cos d' 



log rrr^' = log 



tan 0' ^ cos d ' 



That is tan <^ cos ^ =tan 0' cos d' , where (<9', </>') were the co-ordinates 

 of the initial position. This we see is the equation of what corres- 

 ponds to a parallel of latitude, since tan cos d = tan (latitude). 



XVI. The general case yet contains the angle /? which is to be 

 eliminated. 



sin PTtan /? = cot 6, 



tan TT = cos 6 tan <j>. 



Therefore 



Q ^ cot e \^l -\- cos'' d tSLiï^cj) ^ s/ 1 -}- cos' e tan' 

 cos 6 tan <^ sin ^ tan 



dé „;„ /!„;„ j> cos I ,^, „•„ J ^- fl cos d!) \/ 1 + cos' ^ tan"^. 



-x _ sm ^ sin 9 r -[" ^'O' ^ sm 9 sm a \ . — ^ -r ; — j- 



d 6 cos u cos a sm p tan 9 



cos ^ , ^ . . 



= sin 6 sin ^7^ + ^'ot oc cos^0 s/ 1 _|- tan' 6 + tan <f>' 



= tan (9 tan ^ -j- cot a v^ J + tan" ^ -|- tan ' <f) 

 1 + tan' <j) 

 Therefore 



d tan ^ _ tan ^ tan (f> -f cot a v^ 1 -f tan' <^ 4- tan' ^ 

 c? tan (9 ~ 1 + tan' 6 



The differential equation 



du u V -\- cot a ^ 1 -\- u* -\- v' 



dv 1 + v' 



will be quite difficult to solve, where we have made the substitution 



u — tan 

 V = tan Q 



