Find the perimeter of the hypocycloid of four cusps,

Question

x^2/3+y^2/3=1/9

Answer 1

find the arc length of one cusp:

2/3 x^(-1/3) + 2/3 y^(-1/3) y' = 0
y' = -(y/x)^(1/3)
Y' = -√(1/9 - x^(2/3))/(3x^(1/3))

s = ∫[0,1/27] √(1+y'^2) dx
= ∫[0,1/27] √(1+(1/9 - x^(2/3))/9x^(2/3)) dx
= 1/(648x) √(72x+∛x)^(3/2) [0,1/27]
= 1/18

So, the full perimeter is 2/9

It might be easier using parametric equations:

x = 1/27 cos^3(t)
y = 1/27 sin^3(t)

dx/dt = -1/9 cos^2(t)sin(t)
dy/dt = 1/9 sin^2(t)cos(t)

s = ∫[0,π/2] (1/9)sin(2t) dt
= -1/18 cos(2t) [0,π/2]
= 1/18

The full perimeter is thus 2/9