问题标题:
【数学题求半立方抛物线y^2=2/3(x-1)^3被抛物线y^2=x/3截得的一段弧长】
问题描述:

数学题求半立方抛物线y^2=2/3(x-1)^3被抛物线y^2=x/3截得的一段弧长

常慧岭回答:
  联立y^2=2/3(x-1)^3和y^2=x/3得   2(x-1)^3=x   令x-1=t,得2t^3=t+1   2t^3-t-1=0   2t^3-2t+t-1=0   2t(t^2-1)+(t-1)=0   2t(t+1)(t-1)+(t-1)=0   (t-1)[2t(t+1)+1]=0   (t-1)(2t^2+2t+1)=0   得t=1(2t^2+2t+1=0无解)   故x=t+1=1+1=2,y=±√(2/3)=±√6/3   半立方抛物线y^2=2/3(x-1)^3,两边对x求导得   2y*dy/dx=2/3*3(x-1)^2=2(x-1)^2   y'=dy/dx=(x-1)^2/y   于是√(1+y'^2)=√[1+(x-1)^4/y^2]=√{1+(x-1)^4/[2/3*(x-1)^3}=√[(3x-1)/2]   则该段圆弧长度L=2∫(x:0,2)√(1+y'^2)dx   =2∫(x:1,2)√[(3x-1)/2]dx   =2*√(3/2)*∫(x:1,2)√(x-1/3)dx   =√6*2/3*(x-1/3)^(3/2)|(x;1,2)   =2(5√10-4)/9
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