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[Differential Geometry of Curves] Prove the set f(p) = 0 is a circle

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider a function f that can be put in the form f(p) = g(|p|) where g : [0,+∞) -> ℝ is C1 with g(0) < 0 and g'(t) > 0 for all t ≥ 0

    Assume that |∇f(p)| = 1 for all p ≠ 0 and prove that the set f(p) = 0 is a circle.

    2. Relevant equations

    Given above

    3. The attempt at a solution

    I know i can use |∇f(p)| = 1 because some circle parametrization will be (cos(p), sin(p)) but I can't figure out really where to start.
     
  2. jcsd
  3. Nov 14, 2012 #2

    Dick

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    I don't see anything too subtle here. If you can show there is exactly one value of c>0 such that g(c)=0, then the solution to f(p)=0 is just |p|=c. Which is a circle, yes?
     
  4. Nov 14, 2012 #3
    I'm sorry, I am terrible at this... how is f(p) = 0 a circle for |p| = c?
     
  5. Nov 14, 2012 #4

    Dick

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    Are you asking why |p|=c is a circle?
     
  6. Nov 14, 2012 #5
    Essentially, yes. What coordinate system are we working in?
     
  7. Nov 14, 2012 #6

    micromass

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    The normal coordinate system on [itex]\mathbb{R}^2[/itex], I suppose.
    Then [itex]|(x,y)|=\sqrt{x^2+y^2}[/itex] is the distance to the origin. So |p|=c says that p has distance c to the origin. So all points with |p|=c are all the points with distance c to the origin, which is a circle, right?
     
  8. Nov 14, 2012 #7

    Dick

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    Any coordinate system. What's the definition of a circle around the origin? |p| is the distance from p to origin, isn't it?
     
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