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Math Help - vector-valued function

  1. #1
    Junior Member cinder's Avatar
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    vector-valued function

    Find the open interval(s) on which the curve given by the vector-valued function is smooth.

    r(t)=t^2i+t^3j

    If I follow the book correctly, I get the derivative which is r'(t)=2ti+3t^2j and then after that I have no idea.

    In the example problem it shows r(t)=0i+0j and list some intervals, but I'm not sure what they're doing or what's going on.
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  2. #2
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    Quote Originally Posted by cinder View Post
    Find the open interval(s) on which the curve given by the vector-valued function is smooth.

    r(t)=t^2i+t^3j
    A vector function is smooth iff it is differenciable and \bold{r}'(t)\not = 0.

    Thus,
    \bold{r}'(t) = 2t\bold{i}+3t^2\bold{j}
    It can only be be zero if t=0. Thus any open interval NOT containing zero makes this smooth.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    A vector function is smooth iff it is differenciable and \bold{r}'(t)\not = 0.

    Thus,
    \bold{r}'(t) = 2t\bold{i}+3t^2\bold{j}
    It can only be be zero if t=0. Thus any open interval NOT containing zero makes this smooth.
    Why would smoothness require that \bold{r}'(t)\not = 0?

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    Why would smoothness require that \bold{r}'(t)\not = 0?

    -Dan
    Because the intuitive meaning is that we can draw a tangent vector at every point. Now what type of tangent is a zero vector?


    I believe the mathematical meaning is that we end up dividing by \bold{r}' within the computation. And to keep it valid we require vectors being smooth for certain theorerms.
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