- #1

- 3

- 0

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

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- Thread starter lisa92
- Start date

- #1

- 3

- 0

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

- 9,557

- 767

[tex]f_x = -\frac {F_x}{F_z}\hbox{ and }f_y=-\frac{F_y}{F_z}[/tex]

- #3

I like Serena

Homework Helper

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- 176

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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