On Maxima and Minima of Functions, ^-c. 93 



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

 variable quantities x 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 -^^* By taking the differential, sin yAy=2.y. 



d(siny). But d(sin7/)=:-^^^; hence by substitution and 



reduction, -^—=2}/, Or tang y, to rad. a=2y. Hence 



V=:66^ 46' 54"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 ot gravity from the vertex ol the sector=— g— -;and 

 that of the centre of oscillation is found without difficulty to 



2ay 



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

 root of the line ^ ^'^ , or as J ^ . If the function u de- 

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



posed constant, u ex y?, 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, I SI 8. 



