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Can a curve with singular point be a regular curve?

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Given a parameterized curve [tex]\alpha:(a,b)\rightarrow \mathbb{R}^2[/tex], show that this curve is regular except at t = a.

    2. Relevant equations

    I know that according to the defintion that a parameterized curve [tex]\alpha: I \rightarrow \mathbb{R}^3[/tex] is said to be regular if [tex]\alpha'(t) \neq 0[/tex] [tex]\forall t \in I.[/tex]


    3. The attempt at a solution

    I have read that any curve which has a point where the tangent vector is zero cannot be a regular curve, so how is it even possible to just forget about that singular point in such a proof?

    Best regards
    Cauchy
     
  2. jcsd
  3. Feb 5, 2009 #2

    Office_Shredder

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    Do you have more information about alpha? Like, what the formula is?

    Also, a isn't in your interval, so you're fine anyway?
     
  4. Feb 5, 2009 #3
    Hi

    First of all its suppose be t = p and [tex]p \in I [/tex] and the curve is defined as

    [tex]\alpha(t) = (x(t),y(t))[/tex] a parameter curve.

    having a singular point on a regular curve isn't that a contradiction?

    Sincerrely

    Cauchy
     
  5. Feb 5, 2009 #4

    Office_Shredder

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    Where a is not in (a,b) so it doesn't even fail the regularity condition.

    Anywho, if someone says "Show a curve is regular everywhere except point p" it's like if someone said "show f(x)=|x| is differentiable except at 0" By definition, a differentiable function is differentiable everywhere, but you understand what they mean anyway. Same principle applies here.
     
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