1 84 On the Problem of Shortest Twilight. 



Resolving now the quadratic equation (6), we have 

 cos o = cos A l+cosftCQB^gin^rinp, 



COSft+COSft, 



1 + COS p, 4. p„ 



= cos X — n ^ ™ 



cos p, + cos p u 



= cosX co **^P<> (7) 



C0S iP/+ Py/Cosi^c^^ 



But the question now naturally arose in D'Alembert's mind, 

 which of these roots belongs to the minimum! or do they both 

 designate minima ? 



It is difficult to account for that distinguished geometer over- 

 looking the simple and common test furnished by the dif- 

 ferential calculus itself for deciding this question: it is even 

 difficult to conceive how the consideration of the problem it- 

 self, as presented by the diagram on the globe, should have 

 failed to point out to him the proper interpretation of the two- 

 fold result in our last equation : yet such is the case. He 

 devotes considerable space to proving that the lower sign of 

 our equation does not answer to a minimum twilight. What 

 it does answer to he does not attempt to explain ; but he de- 

 nies that it answers to the maximum. 



The general fact, that functions of one independent va- 

 riable which do not admit of indefinite increase and dimi- 

 nution, can never have two maxima without the interme- 

 diate occurrence of a minimum, nor two minima without an 

 intermediate maximum, might have guided him to discover 

 from the figure itself, that the two roots could not both indicate 

 minima; but rather that one of them necessarily indicated the 

 time of maximum twilight. However, the common appeal to 

 the second differential coefficient determines it at once. For 

 differentiating (6), and inserting the values of cos p given in 

 (7) in the equation, we have for the test the value of 



cos a *+«»?/«» ft. + cos x s&timti 



COS p { + cos p n COS p { -f- cos p t i 



-cosx 1+cos ft C( ^", 



cos p t -f cos p n 



the first and third terms of which cancel, and leave, for the 



test, the sign of 



__ sin o sin p.. , . 



-f cos A M r// - (8) 



cos p t + cos p u v ' 



We thus see that the problem has never yet been fully solved, 



