228 Prof. Morton and Mr. Tobin on Times of Descent 



BAO, and the position A for which the time is a minimum, 

 or what may be called the two-line brachistochrone. 



(3) Range of validity of the statement in the Scholium 

 that the arc is quicker than the chord : involving comparison 

 of arc and chord-times in the general case where there is 

 initial velocity. 



(4) It seems obvious that Gralileo's circular arc, having a 

 horizontal tangent at the lowest point, will not in general 

 give a shorter time than any other circular arc which can be 

 drawn. So we may examine the question of the circular 

 brachistochrone. 



(5) Comparison of the actual times taken by the different 

 routes : the books give no idea as to how much quicker the 

 cycloid is than, say, the direct line. 



(1) It is required to compare the times taken by a particle 

 to describe the paths BO and BAO, when its initial velocity , 

 at B is that due to a fall from the level of 0. 



Let the inclinations to the horizontal of the chords OA, 

 OB, OC be a/87, then t ne inclination of BA is x + ft. If a 

 particle starts from rest at the level of its velocity at the 

 levels of BAO will be proportional to (sin 2 7 — sin 2 /3)% 

 (sin 2 7 — sin 2 a)2 5 and sin 7. To get the time along any chord 

 divide the gain of velocity by the acceleration. The con- 

 dition that the path BAO should be quicker than BO is 



cosec /3{ sin 7 — (sin 2 7 -sin 2 /3ji} — cosec a{ sin 7— (sin 2 y — sinV^J- 



— cosec (a + ft) { (sin 2 7 — sin 2 a) 2 — (sin 2 7 - sin 2 ft) I } > 0. 



Multiply across by the positive quantity sin a sin ft sin 

 («4-yQ) and get rid of the roots by writing- 

 sin a = sin 7 sin 6, sin /3=sin7sin <f>. 

 So that cos « =A0, cosft = A<f> (mod. sin 7). 



Dropping the factor sin 3 7 we arrive at 

 (sin 0A(j> + sin <£A0) [sin (</> — 6) — (sin <£- sin 6) } 



— sin 6 sin </> (cos 6 — cos <f) > 0. 



From this the factor sin \0 sin J<£ sin J(0— 9) can be 

 removed and the inequality reduces to 



sin0{2A0-(l + cos0)} + sin0{2A<£-(l + cos£)}>O, 

 or finally 



cosec e { 2A0 - (1 + cos 0) \ -f cosec \2A<j> - (1 + cos <j>) } > 0. 



