Length of Curve problem?

Question

y = (1-x)sqrt(x/3), show that 1 + y' ^2 = (3x + 1)^2/12x

hence find the length of the curve from the origin at the point A (1,0)

Answer 1

y = (1 - x)SQRT(x/3)

dy/dx = (-1)[SQRT(x/3)] + (1 - x)(1/3)(1/2)[1/SQRT(x/3)]

dy/dx = [-(x/3) + (1 - x)(1/6)] / SQRT(x/3)

dy/dx = (1/6)[-3x + 1] / SQRT(x/3)

(dy/dx)^2 = (1/36)[-3x + 1]^2 / (x/3)

(dy/dx)^2 = [-3x + 1]^2 / 12x

1 + (dy/dx)^2 = 1 + [-3x + 1]^2 / 12x

1 + (dy/dx)^2 = [9x^2 - 6x + 12x + 1] / 12x

1 + (dy/dx)^2 = [9x^2 + 6x + 1] / 12x

1 + (dy/dx)^2 = [3x + 1]^2 / 12x

ds = SQRT[1 + (dy/dx)^2] dx

ds = SQRT{ [3x + 1]^2 / 12x}

ds = [3x + 1] / [2SQRT(3x)]

ds = [SQRT(3)/2][x/SQRT(x)] + [SQRT(3)/6][1/SQRT(x)]

ds = [SQRT(3)/2]SQRT(x) + [SQRT(3)/6][1/SQRT(x)]

s =  [SQRT(3)/3][(x)^(3/2)] + [SQRT(3)/3]SQRT(x)

s = [2SQRT(3)/3] - [0]

s = 1.154