求抛物线y=ax^2在x=-b到x=b之间的弧长

Question

求抛物线y=ax^2在x=-b到x=b之间的弧长

Answer 1

由对称性,得
L = 2∫ √[1+(y')^2]dx = 2∫ √[1+(2ax)^2]dx
令 2ax = tant, 则 x = [1/(2a)]tant
L = (1/a)∫(sect)^3dt
= (1/a)∫[cost/(cost)^4]dt
= (1/a)∫[1/(1-sin^2t)^2]dsint
= (1/a)∫ 1/(1-u^2)^2]du
= [1/(4a)]∫[1/(1-u)+1/(1+u)+1/(1-u)^2+1/(1+u)^2]du
= [1/(4a)][ln[(1+u)/(1-u)]+1/(1-u)-1/(1+u)]
= [1/(2a)]{ln{√[1+(2ab)^2]-2ab} + (1/2)b√[1+(2ab)^2]/(1-8a^2b^2).