Arc length is given by,

L = int sqrt((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2) dt [t=-1 to t=1]

dx/dt = 1/2

dy/dt = (1-t)^2 / 2

dz/dt = (1-t)^2 / 2

(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 = 1/4 + 2(1 - 2t + t^2)/4

(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 = 1/4 + 1/2 - t + t^2 / 2 = 3/4 - t + t^2 / 2

int sqrt((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2) dt [t=-1 to t=1] = int (3/4 - t + t^2 / 2) [-1 to 1]

L = [3t/4 - t^2 / 2 + t^3 / 6][-1 to 1]

L = {(3/4 - 1/2 + 1/6) - (-3/4 - 1/2 - 1/6)} = 3/2 + 1/3 = 9/6 + 2/6 = 11/6

L = 11/6

Please ignore my last answer. I'm afraid I was in a hurry to get to work. There are some changes I should make to the above solution.

x=t/2, y=(1/3)(1-t)^(3/2), z=(1/3)(1-t)^(3/2)

Arc length is given by,

L = ∫ √((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2) dt [t=-1 to t=1]

dx/dt = 1/2

dy/dt = (1/2)(1-t)^(1/2) *(-1) = (-1/2)(1-t)^(1/2)

dz/dt = (1/2)(1-t)^(1/2) *(-1) = (-1/2)(1-t)^(1/2)

(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 = 1/4 + (1/4)(1 - t) + (1/4)(1 - t) = (1/4)(1 + 1 - t + 1 - t) = (1/4)(3 - 2t)

∫ √((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2) dt [t=-1 to t=1] = ∫ √{(1/4)(3 - 2t)} [-1 to 1] = ∫ (1/2)(3 - 2t)^(1/2) [-1 to 1]

L = [(1/2)(2/3)(-1/2)(3 - 2t)^(3/2)][-1 to 1] = (-1/6)[(3 - 2t)^(3/2)][-1 to 1]

L = (-1/6){(1)^(3/2) - (5)^(3/2) = (1/6)(√(125) - 1)

Answer: L = (√(125) - 1)/6