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Surface and angles

  1. Nov 12, 2006 #1
    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
     
    Last edited: Nov 12, 2006
  2. jcsd
  3. Nov 12, 2006 #2
    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.
     
  4. Nov 12, 2006 #3
    can i use something like Ax + By + Cz = D as the surface and then show this relationship?
     
  5. Nov 12, 2006 #4
    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.
     
  6. Nov 12, 2006 #5
    or would it be [tex]AF_x + BF_y + CF_z = D [/tex] where the normal vector would be [A, B, C]?
     
  7. Nov 12, 2006 #6
    i am not sure how to begin this one, is there any hint for what i should go about doing
     
  8. Nov 12, 2006 #7
    And to do this, you should think about what the properties the normal vector to a surface might have.
     
  9. Nov 12, 2006 #8
    oh does the F sub x mean the x part of the gradient?
     
  10. Nov 12, 2006 #9
    [tex] F_x [/tex] is the partial of F with respect to x, yes.
     
  11. Nov 12, 2006 #10
    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
     
  12. Nov 12, 2006 #11
    glad to help :)
     
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