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Normal vector of curved surface

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data
    2-3.JPG

    2. Relevant equations


    3. The attempt at a solution
    I can understand it intuitively, but can't prove mathematically...Can you help me??
     
  2. jcsd
  3. Mar 15, 2015 #2

    Ray Vickson

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    PF rules forbid us from offering help until you have demonstrated an effort and shown your work.
     
  4. Mar 15, 2015 #3
    I'm so sorry. It was my first time to write in this forum. I know that cross product makes perpendicular vectors. But in this problem, I don't understand how we explain three dimension by using two parameter, u and v. I searched in internet and thought it is related to gradient. Is it right?
     
  5. Mar 15, 2015 #4
    If you use 3 parameters to parametrize a region of space in ℝ3, what you get is a region covering some volume.
    However, if you only use 2 parameters to parametrize a region of space in ℝ3, you get a "2D" curved surface located in that 3D space.
     
  6. Mar 15, 2015 #5
    Ahh, I got what 2 parameters mean. Thanks. then could you help me more to solve that problem??
     
  7. Mar 15, 2015 #6

    Dick

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    You know that the partial derivatives are tangent vectors, right? And you know that the cross product is orthogonal to the two vectors, I hope. And dot products are related to cosines of the included angle and cross products are related to sine? Put all of those ingredients together.
     
  8. Mar 15, 2015 #7
    I can explain that in word and can image in mind, but i can't in numerical expression... That's my problem...
     
  9. Mar 15, 2015 #8

    Dick

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    Look at the denominator in terms of the included angle ##\theta##. Use ##1-\cos^2(\theta)=\sin^2(\theta)##. Now does that help?
     
  10. Mar 15, 2015 #9
    I was so silly. It helps me a lot. Thanks for your help!!
     
  11. Mar 15, 2015 #10

    Dick

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    A nudge is a good as a wink. And you are welcome!
     
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