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Prove that the acute angle [tex]\gamma[/tex] between the z axis and the normal to the surface F(x,y,z)=0 at any point is given by [tex]sec \gamma = \frac{\sqrt{F_x^2 +F_y^2+F_z^2}}{|F_z|}}[/tex]

Where i am having trouble is that i do not know what this surface is, can someone help clarify what the surface is. thanks
 
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The point is that it doesn't matter what the surface is, the equation should hold for any F(x,y,z) with continuous partial derivatives.
 
can i use something like Ax + By + Cz = D as the surface and then show this relationship?
 
No, Ax + By + Cz = D is the equation of a plane, a very specific type of surface. This question asks you to prove it for any surface of the form F(x,y,z) = 0. You first need to find the equation of a normal vector to the surface, then work on what [tex] \gamma [/tex] might be.
 
or would it be [tex]AF_x + BF_y + CF_z = D [/tex] where the normal vector would be [A, B, C]?
 
i am not sure how to begin this one, is there any hint for what i should go about doing
 
thenewbosco said:
i am not sure how to begin this one, is there any hint for what i should go about doing
slearch said:
You first need to find the equation of a normal vector to the surface.
And to do this, you should think about what the properties the normal vector to a surface might have.
 
oh does the F sub x mean the x part of the gradient?
 
[tex] F_x [/tex] is the partial of F with respect to x, yes.
 
thanks, it was no problem actually i just had some brain cramp and didnt think of [tex]F_x[/tex] being the x component of the gradient vector
 
glad to help :)
 

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