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Help with a smooth curve

  1. Aug 18, 2007 #1
    1. The problem statement, all variables and given/known data
    In calculus II, vector valued function in space. The vector function r(t)=f(t)i+g(t)j+h(t)k. The curve traced by r is smooth if dr/dt is continuous and NEVER 0.


    2. Relevant equations

    I don't understand why there is "NEVER 0" in the above statement, in order for the curve traced to be smooth.

    3. The attempt at a solution
    The original statement appear in texts such as Thomas/Finney Calculus, Davis (introduction to vector analysis). But their explanations are very different. In learning calculus I, a smooth curve does not seem to require dy/dx=0 (e.g. y=x^3 is a smooth everywhere). Why in dealthing with vector calculus, there is such restriction.

    Thanks
     
  2. jcsd
  3. Aug 18, 2007 #2

    Dick

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    Consider the curve defined by (t^2,0,0) for t<0 and (0,t^2,0) for t>=0. It has a continuous derivative, but it doesn't look like something you would want to consider a smooth curve.
     
  4. Aug 18, 2007 #3

    quasar987

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    To play devil's advocate, take a plane parabola: (t, t^2, 0). According to your definition, it is smooth, and indeed, a parabola is a curve we would intuitively like to call smooth: unlike Dick's curve, it doesn't take a sudden rough turn.

    However, consider the same parabola, but parametrize slightly differently: (t^3, t^6, 0). This is not smooth at t=0!


    The way I learned diff. geometry of curves, the curve r(t) is smooth if f,g, and h are and it is called regular if r'(t) is never 0.
     
  5. Aug 18, 2007 #4
    Dick,

    Thanks for the reply. Your example seems good. But I am a little worried about it.Should your stuff be considered as TWO curves or ONE curve?And how to dinstinguish these?

    Check my example below.

    A curve defined by (t,t^2,0). This curve has a derivative at (0,0,0). Since the dr/dt=0, therefore the curve is not smooth at (0,0,0) ??

    I think I may have missed some points somewhere and please point them out.
     
  6. Aug 18, 2007 #5
    quasar987,

    You has grasped my worries.But I still don't understand the definitions of the SMOOTH.

    Also, for (t^3, t^6, 0), Why is this curve not smooth at (0,0,0)?

    Thanks
     
  7. Aug 18, 2007 #6

    Dick

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    If you don't like two domains of definition, try (t^2,t^3,0). This a curve with a cusp at t=0. As a function it's differentiable, but as a curve it has a sharp corner at (0,0,0) and doesn't look 'smooth'. That's what r'(t) not equal zero is trying to avoid. Your example is smooth, dr/dt=(1,0,0) at t=0. As quasar987 points out, a smooth curve in one parametrization can still have zero derivatives in other parametrizations. The definition should probably be a little more careful to say that a curve is smooth at a point p if there exists ANY parametrization of the curve r(t) such that r(t0)=p and r'(t0) is not zero.
     
  8. Aug 19, 2007 #7

    quasar987

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    The curve you used:
    is indeed smooth (according to your definition). if I differentiate it, I get dr/dt=(1,2t,0), which is never (0,0,0) because of the 1 in the x position. My curve on the other hand, has dr/dt=(3t²,6t^5,0) for a derivative. It is therefor 0 at t=0.
     
  9. Aug 19, 2007 #8
    Thank Dick and quasar987 again.

    Regarding to the curve determined by (t^3, t^6, 0)

    Understanding1: According to Dick, we can make it equivalent(?) to (u,u^2,0). For such a choice of parameter u, the curve is then smooth. Then Dick will say "dr/dt is NEVER 0" is not true. Therefore, such a curve described by (t^3, t^6, 0) is smooth. This means that "NEVER" (in my original definition) is with respect to the parameter itself in the argument parameter's domain for any choice of parameters.

    Understanding2: quasar987 followed my original definition. "NEVER" is with respect to the parameter itself in the argument parameter's domain for one given choice of the parameter. This seems more straightforward. But (t^3, t^6, 0) is smooth when plotting in x-y-z space?

    A further question, given t, u is any real,
    (t^3, t^6, 0) and (u,u^2,0) are the same curve or NOT (I cannot find the reason for saying NO)? But (f'(0),g'(0),h'(0)) have two different values for them. Then how can they be the same curve? Alternatively, can it be said that the curve's (if they are ONE curve)derivative at (0,0,0) doesnot exist?

    Could sb point out where my trouble comes from?
     
  10. Aug 19, 2007 #9

    quasar987

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    They are different curves, with the same trace. So the moral of the story is that, awkwardly enough, whether a curve is smooth or not is not an intrinsic property of its trace. Rather it is a property of the particular way it is parametrized. So there might be traces that have both smooth parameterizations and non-smooth ones, as it is the case with the parabola.
     
  11. Aug 20, 2007 #10

    So you still followed my original definition of "smooth" (disagree with DICK's). And therefore, for the same curve's trace in space, we can maybe find smooth parameterizations and non-smooth ones. But "smooth" here is defined to be
    the stuff such that r'(t) is not zero anywhere.

    Your statement is different from the following:

    Thomas/Finney Calculus book says: We require dr/dt to be non-zero for a smooth curve to make sure the curve has a continuously turning tangent
    at each point. On a smooth curve there are no sharp corners or cusps.

    "Smooth" mentioned by Thomas/Finney is still the usual meaning.We can think of the reverse process: if there are corners or cusps (not smooth in the usual sense) then there must exist some point where dr/dt=0 (for a continuous curve). This seems what Thomas/Finney want to mean.

    Although Thomas/Finney used the curve traced by r is smooth if dr/dt is continuous and NEVER 0.. "if" seems only to give a sufficient condition, BUT "NEVER" is used there. And I can only accept that the condition IS the DEFINITION of the SMOOTH,which is also indicated by
    We require dr/dt to ...

    I cannot find any mistake in your statements, but I find it hard for me to accept this "fact?".
     
    Last edited: Aug 20, 2007
  12. Aug 20, 2007 #11

    quasar987

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    You seem to be overanalysing our postst and putting words in our mouth.

    All Dick said is he explained to you why it is reasonable to ask for dr/dt to be non zero in the definition of smoothness. And all I said is I explained why it is not so reasonable afterall.

    But who cares if it's "reasonable" or not? It's the definition you're given, and you have to work with it like it or not.

    (You seem to be puzzled by the "if" in the definition. In definitions, "if" always mean "if and only if".)
     
  13. Aug 20, 2007 #12
    In my opinion, the problem is not 100% satisfactorily sorted out, is it? It is why I should let you know what my textbook says, and it is my job to check whether your statement is agreement with my texts and others. Why do you say "put words in your mouth"(my english is not good, I don't understand the english meaning of this phrase actually)??

    :grumpy: :mad:I don't like such an attitude to be honest. What's more, my problem belongs to only basic calculus. Maybe, for advanced maths, I would pick your such unresponsible attitude for the time being.


    Thank you.
     
    Last edited: Aug 20, 2007
  14. Aug 20, 2007 #13

    Dick

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    The point is that as your book says, the intention of the definition is to force the curve to have a continuously varying tangent. Insisting r'(t) be continuous and nonzero suffices for that. quasar987's point is simply that it is a bit overly strict and you can't really say a CURVE is NOT smooth (in the sense of a continuously turning tangent) just because a particular parametrization has a zero derivative. Another equally valid one may not. That's about all there is to it. These are 'facts'. At this point I'm not sure what question is remaining...
     
  15. Aug 21, 2007 #14
    Thanks a lot for both of your explanations.
     
  16. Aug 21, 2007 #15

    HallsofIvy

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    I am puzzled as to why you consider Dick agreeing with you that the definition in your book is the one you have to work with is "irresponsible".
     
  17. Nov 9, 2008 #16
    My book on complex analysis (Bak & Newman) specifies that a piecewise differentiable curve is smooth if the partials don't simultaneously vanish except at a finite number of points, which I think neatly solves the issue of parametrization affecting the "smoothness" of curves.
     
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