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Find the perimeter of the hypocycloid of four cusps,
x^2/3+y^2/3=1/9

  • math -

    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

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