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Thread: 'smooth' curves with cusps in 3d

  1. #1 'smooth' curves with cusps in 3d 
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    While reviewing basic calculus, I noticed that the curve (1+t^2,t^2,t^3), which clearly has a cusp at (1,0,0), has a derivative curve (2t,2t,3t^2) which is clearly smooth. This struck me as odd since differentiation usually seems to turn cusps into discontinuities, whereas integration smoothes out a curve, especially a curve described by polynomials. In fact, in general I have always taken a curve to be smooth iff it has a continuous derivative, which this curve has, and yet a cusp cannot be smooth in any sensible sense. I suspect the explanation is relatively simple - just something I'm missing.

    Thx in advance.


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  3. #2 Re: 'smooth' curves with cusps in 3d 
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    Quote Originally Posted by inkliing
    While reviewing basic calculus, I noticed that the curve (1+t^2,t^2,t^3), which clearly has a cusp at (1,0,0), has a derivative curve (2t,2t,3t^2) which is clearly smooth. This struck me as odd since differentiation usually seems to turn cusps into discontinuities, whereas integration smoothes out a curve, especially a curve described by polynomials. In fact, in general I have always taken a curve to be smooth iff it has a continuous derivative, which this curve has, and yet a cusp cannot be smooth in any sensible sense. I suspect the explanation is relatively simple - just something I'm missing.

    Thx in advance.
    Let's simplify this and look at a similar example in 2Dt^3,t^2). This curve has exactly the same behavior as your example.

    If you think of this parameterized curve as the trajectory of aq particle, then the time derivative is the velocity vector, which exists everywhere, and is 0 at t=0. The particle stops at the origin and changes directi0n.

    So as a function of "t" the curve is differentiable.

    However the image of the curve is also, as a set, the graph of the function y=x^(2/3), which is not differentiable at t=0 -- a manifestation of the cusp at (0,0).


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