determine the length of the perimeter of the hypocycloid x2/3 + y^2/3 = a^2/3

Answer 1

y = (a^(2/3) - x^(2/3))^(3/2)
y' = -√(a^(2/3) - x^(2/3)) /∛x
ds^2 = 1 + y'^2 dx = 1 + (a^(2/3) - x^(2/3)) / x^(2/3) = (a/x)^(2/3) dx
using the symmetry of the figure, the perimeter is
s = 4∫[0,a] ∛(a/x) dx = 6a

Or, using parametric equations,
x = a cos^3θ
y = a sin^3θ
ds^2 = x'^2 + y'^2 = (-3a cos^2θ sinθ)^2 + (3a sin^2θ cosθ) = 3a/2 sin2θ
s = 4∫[0,π/2] 3a/2 sin2θ dθ = -3a cos2θ [0,π/2] = 6a