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Cal 3; surfaces

  1. Nov 4, 2011 #1
    Given that the surface x^8y^7+y^3z^6+z^8x63+6xyz=9 has the equation z=f(x,y) in a neighbourhod of the point (1,1,1) with f(x,y) differentiable, find the derivatives

    fx(1,1)=

    fy(1,1)=

    Attempt: I tried plugging (1,1,1) in fx but it wasnt rigth. We havent seen this in class and I dont know what to do.

    Thanks
     
  2. jcsd
  3. Nov 4, 2011 #2

    LCKurtz

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    If a funtion z = f(x,y) is defined implicitly by F(x,y,z)= 0 as in your problem, you can use the formulas:

    [tex]f_x = -\frac {F_x}{F_z}\hbox{ and }f_y=-\frac{F_y}{F_z}[/tex]
     
  4. Nov 4, 2011 #3

    I like Serena

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    Hi lisa92! :smile:

    I'd like to show you how to derive the formulas of LCKurtz.

    With F(x,y,z)=0 you have:

    [itex]d(F(x,y,z))=F_x(x,y,z)dx + F_y(x,y,z)dy + F_z(x,y,z)dz=0[/itex]


    From this you can find:

    [itex]dz={-F_x(x,y,z)dx - F_y(x,y,z)dy \over F_z(x,y,z)}[/itex]


    Taking partial derivatives (for which the other coordinate is considered constant):

    [itex]{\partial z \over \partial x}={-F_x(x,y,z) \over F_z(x,y,z)}[/itex]

    [itex]{\partial z \over \partial y}={-F_y(x,y,z) \over F_z(x,y,z)}[/itex]


    This can also be written as the formulas LCKurtz gave:

    [itex]f_x={-F_x \over F_z}[/itex]

    [itex]f_y={-F_y \over F_z}[/itex]
     
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