Question
Find the arc length of the curve y^3=8x^2 fr0m x=1 to x=8.
Answers
Steve
since y^3 = 8x^2,
y = 2x^(2/3)
y' = 4/3 x^(-1/3)
y'^2 = 16/(9x^(2/3))
arc length is thus
∫[1,8] √(1+16/(9x^(2/3))) dx
u=x^-2/3
then let v = u^1/2
then let 4w = 3tanv
and you finally wind up with
1/27 √(9+16/x^2/3) (9x + 16x^1/3) [1,8]
= 1/27 ((√(9+4)(72+32))-(√(16+9)(9+16)))
= 1/27 (104√13 - 125)
y = 2x^(2/3)
y' = 4/3 x^(-1/3)
y'^2 = 16/(9x^(2/3))
arc length is thus
∫[1,8] √(1+16/(9x^(2/3))) dx
u=x^-2/3
then let v = u^1/2
then let 4w = 3tanv
and you finally wind up with
1/27 √(9+16/x^2/3) (9x + 16x^1/3) [1,8]
= 1/27 ((√(9+4)(72+32))-(√(16+9)(9+16)))
= 1/27 (104√13 - 125)