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Math Help - plane tangent to 3d surface mmn14 3a

  1. #1
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    plane tangent to 3d surface mmn14 3a

    i have a suface \sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{a}

    find the formula of a plane which is tangent to it in (x0,y0,z0)

    i need to find the vector whic is vertical to this surface
    the formula of it is <f_x,f_y,-1>
    but i dont know what is the formula for f(x,y) ,
    i have a formula in other form
    how to write it in the f(x,y) form

    ?
    Last edited by transgalactic; June 7th 2011 at 08:51 PM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    If S\equiv f(x,y,z)=0 then, the tangent plane to S at P_0(x_0,y_0,z_0) is

    \dfrac{\partial f}{\partial x}(P_0)(x-x_0)+\dfrac{\partial f}{\partial y}(P_0)(y-y_0)+\dfrac{\partial f}{\partial z}(P_0)(z-z_0)=0
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  3. #3
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    i need to find the vector whic is vertical to this surface
    the formula of it is <f_x,f_y,-1>
    but i dont know what is the formula for f(x,y) ,
    i have a formula in other form
    how to write it in the f(x,y) form

    ?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by transgalactic View Post
    i need to find the vector whic is vertical to this surface
    the formula of it is <f_x,f_y,-1>
    but i dont know what is the formula for f(x,y) ,
    i have a formula in other form
    how to write it in the f(x,y) form

    Sincerely, it is rather difficult to understand your question. Perhaps you mean that if the surface is written in explicit form z=f(x,y) then, the equation of the tangent plane at P_0(x_0.y_0,z_0) is \pi\equiv <(f_x(P_0),f_y(P_0),-1),(x-x_0,y-y_0,z-z_0)>=0 . For the general case, see my previous post.
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  5. #5
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    Quote Originally Posted by FernandoRevilla View Post
    If S\equiv f(x,y,z)=0 then, the tangent plane to S at P_0(x_0,y_0,z_0) is

    \dfrac{\partial f}{\partial x}(P_0)(x-x_0)+\dfrac{\partial f}{\partial y}(P_0)(y-y_0)+\dfrac{\partial f}{\partial z}(P_0)(z-z_0)=0
    i was told that the norm vector <df/dx,df/dy,-1>
    Last edited by transgalactic; June 8th 2011 at 01:37 AM.
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