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I Finding a unit normal to a surface

  1. Sep 13, 2017 #1

    dyn

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    Hi.
    I'm trying to self-study vector calculus and I have to admit I struggle with it. As regards finding normals to a surface I know 2 ways - one involves writing the surface equation as f(x.y.z) = 0 and taking the gradient. The other method involves writing the surface in terms of 2 parameters and taking the vector product of the 2 partial derivatives. Finally, my question. If I have a circle such as x2 + y2 = a2 I know the unit normal is k , but how do I show this ? My 2 methods don't work !
    Thanks
     
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  3. Sep 13, 2017 #2

    Charles Link

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    The circle equation is the boundary of the surface. (## x^2+y^2=a^2 ## is the equation for a cylinder.) The actual equation of the plane in which the circle ## x^2+y^2=a^2 ## resides is simply ## z=0 ##. (I'm presuming you are wanting your surface to be a flat circle with center at the origin in the x-y plane). What you have for a surface is the endface of the cylinder. Take the gradient of ## z=0 ## and you do get ## \hat{n}=\hat{k} ##.
     
    Last edited: Sep 13, 2017
  4. Sep 13, 2017 #3

    dyn

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    Yes i am taking the surface as a flat circle in the x-y plane. To take the gradient I need the form f(x,y,z)=0. What do I take f as ?
     
  5. Sep 13, 2017 #4

    Charles Link

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    ## f(x,y,z)=z ##. Comes from ## z=0 ## is your equation of ## f(x,y,z)=0 ##. ## \nabla z=\hat{k} ##. ## \\ ## If the plane was instead ##z=5 ##, ## f(x,y,z)=z-5 ##, and again ## \nabla (z-5)=\hat{k} ##.
     
    Last edited: Sep 13, 2017
  6. Sep 13, 2017 #5

    dyn

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    Thanks for your help. On a similar topic ; is it possible to find the unit normal to the curved surface of a cylinder orientated along the z-axis in this way. I can't find a way to write this surface as f(x,y,z) = 0 but I can get an answer using my other method using the vector product
     
  7. Sep 13, 2017 #6

    Charles Link

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    The equation of the cylinder is simply ## x^2+y^2=a^2 ##. Thereby ## f(x,y,z)=0 ## is ## x^2+y^2-a^2=0 ##.
     
  8. Sep 13, 2017 #7

    dyn

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    Thanks again. Much appreciated.
     
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