308 Application of the Fluxional Ratio, Sfc. 



the fluxion of the arc z. CK : CO: [nd : Od::{a^ —y^)^ : a'.'. 



«2/' , o ..-i /I , y" ,^y"^y" , 



((^2 ~y^Y \a 2a^ Sa^ 16a' 



352/8 GSj/'" 231?/^2 429y^^ \ _ y^?/- 



1280^ "'"2560^' "^ 1024a » ^'^2048a> ^ "'" ^'^"j ^2/' = 2/* + 20^' + 

 3y^y' 5y^y S5y^y 6oy">y' 2Sly^^y 429y^'^y 

 "Sa^'^ 16a^~^ 128a«'^256ai» "'"l024a'2 +2048a'* "^ ^' 



y y^y y y^ '^y'^y y 



Now 2/-Xratio -=y', ^ X ratio ^=g^, ; -g^ X ratio g^ = 

 3j^5 5j^6j^. ^ y 5^7 35j/''?/" . y S5y^ 

 40^' 16^^^^^'°7^- = ri27«' 128^^'^*'°9^-"'ll52a^' ^^• 



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

 C, D, &£c. 

 then Ba : Ay.Od : y 



B„:B::0<Z:..5 

 BatCr.Od: -^ 



Ba : D::Odf : 



• • • • 40a* 



5y' 



112a« 



«3 3«^ 5u'^ 

 Ba 1 A+B+C+&c.::Od : . . . . 3'+^.+i57.+II5^.+ 



36,. , 6%^ , ^^ , J?9|:i^^^,^^^^po_ ^,^„ 



1152a«^28l6a>»^13312a»2 ' 30720a' * 

 a=l, and DO=arc of 30°, then the series becomes D0=(.5)' + 

 K-5)=' + A(-5)' + t^3(.5)^ + tH3(-5)H-&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 : 3. 141 59 + . 



The general equation for the fluxion of any power is D"a;" x 



— =wDx"~'a;'5 which will be applied in a few examples. 



X 



I. To find the fluxion of (xx) % which is termed the direct method 



nx' 

 of fluxions. In this example D''x" = (cc2;)% and in the ratio — ,n 



represents the exponent 8, x' represents 2xz', the fluxion of the root, 



nx' 

 and X represents xx, the root of the given power ; hence — answers 



8X2a;x- , , ^ nx' , ^ 8x2xx' ^^ ,_^ 



to ; therefore D"x" X — ^M^ x — — — = 16x'*x-. 



