On Maxima and Minima of Functions^ ^c. 93 



In this case let the radius and half the arc be the simple 

 'variable quantities a; and y: make the radius constant and 

 = a; then the chord - 2a sin ?/, or varies as sin. y, and the 

 sector varies as y. Since n=2, the fraction to be made a 

 maximum is -A-. By taking the differential, sin ydy=2y. 

 d{smy). But d(siny)=-^^^; hence by substitution and 

 reduction, ^^=2y. Or tang y, to rad. a=2?/. Hence 

 ^=66° 46' S4''i. 



Prob. XV. 



Having given the area of a circular sector, which is sus-^ 

 pended by its vertex, and vibrates in its own plane, it is re- 

 quired to determine when the time of vibration is a mini- 

 mum. 



The notation remaining as before, the distance of the 



centre of gravity from the vertex ot the sector =—2—-j and 



that of the centre of oscillation is found without difficulty to 



be = -^^. The time of vibration varies as the square 



root of the lineT-^» or as y/-^- If the function u de- 



4 sin J/' "^ sin^ 



note the time, and v the area of the sector, when - is sup- 



posed constant, u o: vi, or u=l ; hence ^-^, or ^r^ = 



min. which leads to the same result as in the last problem. 



It will be unnecessary to add more examples in illustra- 

 tion of this method. If it furnishes no new instrument to 

 the adept in Analysis, it may still perhaps be regarded with 

 some interest by those who are desirous of giving the great- 

 est possible extent to the ordinary method of obtaining 

 maxima and minima, in consequence of not enjoying the 

 opportunity of becoming familiar with all the refinements of 

 the modern calculus. 



Yah College, August, 181 8v 



