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Math Help - Tangent plane to a surface

  1. #1
    Newbie ohms's Avatar
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    Tangent plane to a surface

    Is it correct to assume that taking the dot product of the gradient and the derivative of a surface function will give me the equation of the tangent plane at any point on the surface? My notes are not entirely clear about this. Thanks in advance.

    Also, would a question of this nature be better suited in the calculus section?
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  2. #2
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    To answer this question properly, I have be sure that you know how to write the equation of a plane given a point on the plane and its normal. Do you?

    If so then the gradient to a surface at a point is the normal of the tangent plane at that point.
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  3. #3
    Newbie ohms's Avatar
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    maybe i'm confusing the derivative of a vector function curve. I know that it represents the tangent at any point on a curve, and if I know the normal vector as well at that point (the gradient?), then the dot product of the two equal to zero should give me a plane tangent to the curve at that point

    I understand that when looking at surfaces, the tangent plane at a point (x1, y1, z1) of the function F(x,y,z) is going to be:

    ∇F(x,y,z) . <(x - x1), (y - y1), (z - z1)> = 0

    and my visualization of this is the following:
    Tangent plane to a surface-untitled.png
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  4. #4
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    It should be \nabla F(x_1 ,y_1 ,z_1 ) \cdot \left\langle {x - x_1 ,y - y_1 ,z - z_1 } \right\rangle  = 0 for the tangent plane to the field F(x,y,z) at (x_1,y_1,z_1).

    But remember that the gradient involves partial derivatives.
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  5. #5
    Newbie ohms's Avatar
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    super, pretty much what the notes and readings say. Thanks for the help!
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