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Vector calculus finding the surface normal

  1. Feb 4, 2009 #1
    okay i was reading the book "div, grad cur and all that"
    i've just started for my exam. but i got stuck in the beginning. i'm attaching the page cuz i really dont get it .
    basically pages 14 and 15. i dont get how they calculate the components of u and v.
    if someone explains u i'll get v obviously.
    thanks
     

    Attached Files:

  2. jcsd
  3. Feb 5, 2009 #2

    chiro

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    U is calculated as a tangent vector. Consider how the rate of change impacts the tangent vector. Think about how a steep surface or how a flat surface would impact the definition of u. Its probably best if visualize a normal one dimensional function and the normal tangent vector to that function.

    If you relate u vector to the function tangent in the x direction and v vector to that in the y direction you have two vectors that are orthogonal to each other and represent the best information to calculate a normal since the normal is u x v.

    On a steep surface we would have a normal that would have a smaller z component than if the surface were flat on the x,y plane.

    In terms of calculation they are converting the x and z components which is df/dx * ux or whatever it says in the book by writing it in terms of its components which are in terms of the orthonormal vectors i j and k which correspond to the unit vectors of the x y and z axis respectively. Just reflect on what units everything is in by looking at the diagram and see how the units for x and z correspond to units or i and k respectively.
     
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