Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    Attached Files:

  2. jcsd
  3. Feb 5, 2009 #2


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook