Acute Angle Between Z-Axis and Surface Normal in Surface and Angles Proof

[thenewbosco]

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

Where i am having trouble is that i do not know what this surface is, can someone help clarify what the surface is. thanks

 

[slearch]

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.

 

[thenewbosco]

can i use something like Ax + By + Cz = D as the surface and then show this relationship?

 

[slearch]

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 $$\gamma$$ might be.

 

[thenewbosco]

or would it be $$AF_x + BF_y + CF_z = D$$ where the normal vector would be [A, B, C]?

 

[thenewbosco]

i am not sure how to begin this one, is there any hint for what i should go about doing

 

[slearch]

i am not sure how to begin this one, is there any hint for what i should go about doing

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.

 

[thenewbosco]

oh does the F sub x mean the x part of the gradient?

 

[slearch]

$$F_x$$ is the partial of F with respect to x, yes.

 

[thenewbosco]

thanks, it was no problem actually i just had some brain cramp and didnt think of $$F_x$$ being the x component of the gradient vector

 

[slearch]

glad to help :)