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    A curve has parametric equations x = acos^3 t, y = asin^3 t where 0 <=t <= pi/2 and a = positive constant.

    Prove that the tangent to the curve at the point P where t = pi/3 has equation 2x(SqRt3) + 2y = a(SqRt3)


    Find the coordinates of the turning points on the curve with equation y^3 + 3xy^2 - x^3 = 3

    meh, doing this myself, but can't seem to! Help me!!

    dx/dt = a(-sint)(cos²t)
    dy/dt = a(cost)(sin²t)
    dy/dx = dy/dt/dx/dt = a(cost)(sin²t)/a(-sint)(cos²t) = -tant
    When t=pi/3, dy/dx=-sqrt[3], y=(a/8).3sqrt[3] and x=a/8
    So, equation of the tangent:
    y - (a/8).3sqrt[3] = -sqrt[3](x-a/8)= -xsqrt[3] + asqrt[3]/8
    y + xsqrt[3] = (3a/8)sqrt[3] + (a/8)sqrt[3] = (a/2)sqrt[3]
    => 2y + 2x.sqrt[3] = a.sqrt[3]
    as required.

    y^3 + 3xy^2 - x^3 = 3
    3y^2(dy/dx) + 3[y^2+2xy(dy/dx)] - 3x^2 = 0
    Turning points occur when dy/dx=0:
    3y^2 - 3x^2 = 0
    x^2 = y^2
    x=+y or x=-y
    Substitute these back in the original equation:
    y^3 + 3y^3 - y^3 = 3 => y=1, x=1
    y^3 - 3y^3 + y^3 = 3 => y=-cbrt[3], x=cbrt[3]
    (1, 1) and (cbrt[3], -cbrt[3]) are the turning points.
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Updated: March 3, 2005

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