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Math Help - Derivative of an unit vector

  1. #1
    Newbie Gustavis's Avatar
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    Derivative of an unit vector

    Hello

    If I have an unit vector the derivative is an orthogonal vector (orthogonality is enough easy to demonstrate)
    How I can demonstrate that the derivative of unit vector is not an unit vectors ?


    Thanks !
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Gustavis View Post
    Hello

    If I have an unit vector the derivative is an orthogonal vector (orthogonality is enough easy to demonstrate)
    How I can demonstrate that the derivative of unit vector is not an unit vectors ?


    Thanks !
    It can be a unit vector... Look for instance at \vec{u}(\theta)={\cos\theta\choose\sin\theta}. But it does not have to be a unit vector, look at \vec{u}(\theta)={1\choose 0}.
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  3. #3
    Newbie Gustavis's Avatar
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    perfect !

    Can I demonstrate in this way?

    the versor
    \hat{v}(t) = \cos(\theta(t))\hat{i} + \sin(\theta(t))\hat{j}

    and its derivative
    \hat{v}'(t) = \theta'(t)(-\sin(\theta(t))\hat{i} + \cos(\theta(t))\hat{j})

    where :
    (-\sin(\theta(t))\hat{i} + \cos(\theta(t))\hat{j})
    is the orthogonal versor

    and theta(t)
    \theta'(t)
    in general different to 1 ?
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  4. #4
    MHF Contributor

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    Quote Originally Posted by Gustavis View Post
    perfect !

    Can I demonstrate in this way?

    the versor
    \hat{v}(t) = \cos(\theta(t))\hat{i} + \sin(\theta(t))\hat{j}

    and its derivative
    \hat{v}'(t) = \theta'(t)(-\sin(\theta(t))\hat{i} + \cos(\theta(t))\hat{j})

    where :
    (-\sin(\theta(t))\hat{i} + \cos(\theta(t))\hat{j})
    is a unit orthogonal versor

    and
    \theta'(t)
    in general different to 1 ?
    That's quite fine.
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  5. #5
    MHF Contributor

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    1968
    In general \theta(t)' is different from 1 but not always. Your original statement "the derivative of unit vector is not a unit vector" is not true.
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