308 Application of the Fluxional Ratio, fyc. 



i 



the fluxion of the arc z. CE : CO: \nd : Od: :(a 2 -y 2 ) 2 : a: : 



ay _i /l V 2 3w 4 5y 6 



( fl 2_y2)2 \ a 2a 3 3 a 5 16 a T 



35y 8 63y 10 231y 12 429y 14 \ _ y 2 y 



128a 1 + 256a" + 1 024a ' 3 + 2048a • 5 + / oy * = y ' + 2a" 2 + 

 3y 4 y 5y e y 35y*y* 63y' °y 231?/ 12 y 429# 14 y 

 ~8a 1_+ 16a 8+ 128a 8+ 256a , » + 1024a 12 + 2048a 14 + &; c« 



„ . y y 2 r . y y* 3y*y* . y 



JNovv y* X ratio - =y : ;r-- X ratio 5- —7~~i J "o - T X ratio z— 



y y 2a 2 3y 6a 2 " 8a 4 5y 



3y J 5y 6 y . y 5y 7 35y 8 y . y 35y 9 

 40a 4 ' 16a 6Xratl0 7y-U'2a« J 128a 8 X ratl ° 9y ~ 1 1 52a 8 » &C ' 

 Let the several terms of the fluent AB be represented by A, B, 

 C, D, &c. 



thenBa : A: :0a 7 : y 



Ba:B::Od 



y 



♦ • • 



6a 



Ba : C::Od 



% 



40a 



Ba:D::Od: 



by 



112a 6 



BatA+B+C+fcc.-Od: • • • -y+|f 3 +|^+x^+ 



35y» 63y" 231y 13 429j/ 15 



rT5^ + 2^l6^« + T3Tl2^ 2 + 3^^0^ 4+kc - =arcDa When 

 a=l, and DO = arc of 30°, then the series becomes DO=(.5) 1 + 

 -e(- 5 ) 3 +T 3 o(-5) s + T^(.5) 7 + T ffT(-5) 9 +&;c.=.5235985 which is 

 equal to the arc of 30° when the diameter is 2 ; therefore the diam- 

 eter of a circle is to the periphery as 1 t 3. 141 59+. 



The general equation for the fluxion of any power is D"x n X 



=nDz n ~ , :r, which will be applied in a few examples. 

 I. To find the fluxion of (xz) 8 , which is termed the direct method 



fix 9 



of fluxions. In this example D n z n =(xz)% and in the ratio — , » 

 represents the exponent 8, r represents 2zr, the fluxion of the root, 



fix* 



and x represents zz, the root of the given power ; hence — answers 

 8x2xx' , V 7 '- nx ' , x 8X2zx* 



to ; therefore D n x n X — =(xx) 8 X = 16x' s x\ 



XX X * ' XX 



X 



