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Cal 3; surfaces

  • Thread starter lisa92
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  • #1
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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
 

Answers and Replies

  • #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]
 
  • #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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