Question

求y=ax²在某一区间内的弧长

Answer 1

y = ax^2, y' = 2ax
ds = √[1+(y')^2]dx = √(1+4a^2x^2)dx
在区间 [p, q] 上的弧长
s = ∫<p, q> √(1+4a^2x^2)dx
= [x√(1+4a^2x^2)]<p, q> - ∫<p, q> [4a^2x^2/√(1+4a^2x^2)]dx
= q√(1+4a^2q^2) - p√(1+4a^2p^2) - ∫<p, q> [(1+4a^2x^2-1)/√(1+4a^2x^2)]dx
= q√(1+4a^2q^2) - p√(1+4a^2p^2) - s + ∫<p, q> [1/√(1+4a^2x^2)]dx
2s = q√(1+4a^2q^2) - p√(1+4a^2p^2) + [1/(2a)]∫<p, q> [1/√(1+4a^2x^2)]d(2ax)
= q√(1+4a^2q^2) - p√(1+4a^2p^2) + [1/(2a)][arctan(2ax)]<p, q>
= q√(1+4a^2q^2) - p√(1+4a^2p^2) + [1/(2a)][arctan(2aq)-arctan(2ap)]
s = (q/2)√(1+4a^2q^2) - (p/2)√(1+4a^2p^2) + [1/(4a)][arctan(2aq)-arctan(2ap)]