math
posted by Jingchi .
Find the perimeter of the hypocycloid of four cusps,
x^2/3+y^2/3=1/9

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