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    Let f(x,y,z) = x + y + x^2 - z^3 be a polynomial.
    Consider the surface S = (f(x,y,z) = 0) \subset R^3

    Write down the equation of the tangent plane at the origin 0 = (0, 0, 0)




    The answer is x + y = 0.
    Is this because x and y are the only variables in the polynomial that aren't to the power of 2, 3... etc? Or have I got this wrong?

    Thank you
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    Don't you work out the value of the partial derivatives f_x, f_y, f_z at (0,0,0) then use

    f_x( x-0) + f_y(y-0) + f_z(z-0) = 0 to get the tangent plane at the origin?

    This does give you x+y=0
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    (Original post by vc94)
    Don't you work out the value of the partial derivatives f_x, f_y, f_z at (0,0,0) then use

    f_x( x-0) + f_y(y-0) + f_z(z-0) = 0 to get the tangent plane at the origin?

    This does give you x+y=0
    hi, thank you!

    i've tried to do as (i think) you're explaining but it doesn't make sense.

    i derive x + y + x^2 - z^3 = 0 and get 1 + 1 + 2x - 3z^2 = 0.

    substituting (0, 0, 0), i am left with 1 + 1 = 0 which obviously isn't right.

    i realise that this 1 + 1 is from x + y so is that what you've done? i'm probably doing it all wrong!

    thank you again!
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    f_x = 1+2x, f_y = 1, f_z = -3z^2.
    Evaluate at (0,0,0) gives f_x = 1, f_y = 1, f_z = 0.

    Use the equation f_x( x-0) + f_y(y-0) + f_z(z-0) = 0 to get the result.
 
 
 
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Updated: March 20, 2011
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