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Smooth curves

  1. Sep 21, 2009 #1
    Hi

    I have a general question about the definition. I know that a curve parametrized by t is smooth if the derivative r'(t) is not 0. I assume this is the 0 vector?

    So then does that mean that if we have r(t) = f(t)i + g(t)j + r(t)k then any of the two can be 0 simultaneously while the third isn't 0 and the curve is still smooth? For example, can f'(t) = 0 and g'(t) = 0 at some point t0 and with r'(t) not 0 then the curve is smooth? Or must f'(t), g'(t) and r'(t) all not be 0?

    Thank you.
     
  2. jcsd
  3. Sep 21, 2009 #2

    Dick

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    Science Advisor
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    Yes to both. r'(t)=0 means the zero vector and f'(t), g'(t) and h'(t) not all zero is the correct condition.
     
  4. Sep 21, 2009 #3
    Thanks!
     
  5. Sep 21, 2009 #4

    LCKurtz

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    Gold Member

    It means the vector is not the zero vector. So as long as one of the components isn't zero, you are OK. Think of t as time and r(t) representing a moving point. You don't want the point to smoothly come to a stop and then go in a different direction. For example consider:

    [tex]f(t) =

    \left \{
    \begin{array}{}
    t^2, t \ge 0\\
    0, t \leq 0
    \end{array}
    \right.
    [/tex]

    and

    [tex]g(t) =

    \left \{
    \begin{array}{}
    0, t \ge 0\\
    t^2, t \leq 0
    \end{array}
    \right.
    [/tex]

    and look at the curve

    [tex]\vec{r}(t) = < f(t), g(t) > [/tex]

    This [tex]\vec{r}(t)[/tex] has a continuous derivative even at t = 0. The problem is that the point stops at the origin and makes a right angle turn. You don't want to call that "smooth", so you don't want the velocity, which is to say the derivative, to be zero.
     
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