DR WALLACE ON A FUNCTIONAL EQUATION. 



633 



here a is the value of f{x) when x=0; now, when x=0, then cx=(i, and 



R X 



cos (c<2^) = l, and R=2P, and -p- =f{x)=% therefore a=2, and R=2P cos — . 



There is yet an indeterminate quantity c ; to find the value of which, we 

 must consider that, P being supposed given, while x increases from to |^ tt, the 

 resultant R must decrease continually from 2 P to ; now, this can only happen 

 when c=l, for if c were less than 1, the resultant would vanish before x became 

 a right angle ; and if c were greater than 1, it would not vanish when x was a 

 right angle ; therefore, c=l ; and in the case when the forces are equal, 



R=2Pcosi2;: .... (A) 

 thus the first case of the problem is resolved. 



17. Case 11. Let us next suppose that P and Q are any two forces whose 

 directions are AB and AD, and R their 

 resultant, whose direction is AC, the 

 angle BAD being any whatever ; we have 

 to determine the force R, and the direc- 

 tion of the line AC relatively to AB and 

 AD. 



Put the angles BAC=0, CAD = ^, 

 then BAD = «=^ + ^. At the point A in 

 the line AC, make the angles CAE, CAE', 

 each equal to BAD ; then the angle BAE 

 will be equal to CAD = 0, and DAE'= 

 CAB=0. 



Make the lines AB, AD, AC, propor- 

 tional to the forces P, Q, R, or such, that 

 P : R=AB : AC, R : Q = AC : AD, and, 

 therefore, P : Q = AB : AD. In the line AC, 



take AH 



AD^ 

 AC 



and AK = 



AB^ 

 AC 



and 



make AE = AE' = — j-^ — , and draw the 



lines CB, CD, DE^ DH, BK, BE. 



The triangles BAC, BAK have a common angle, and by constructioia 

 CA : AB=AB : AK, so that the sides about that angle are proportionals; there- 

 fore the triangles are similar, and have their remaining angles equal, viz. ACB= 

 ABK, and ABC=AKB. 



The triangles BAE, CAD have their angles BAE, CAD equal, and by con- 

 struction the sides about these angles proportionals, for AC : AD= AB : AE ; there- 

 fore the triangles are similar, and hence the angles ABE=ACD, and AEB=ADC. 



And since the angle ABK = ACB, and ABE = ACD, therefore EBK = BCD ; 



VOL. XIV. PART II. 6 B 



