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Thread: find T(t), N(t), at the given time t for the plane curve

  1. #1
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    find T(t), N(t), at the given time t for the plane curve

    find T(t), N(t), $\displaystyle a_t , a_n $ at the given time t for the plane curve r(t)

    $\displaystyle r(t)= ti + \frac{1}{t} j, t = 1$

    The book gives this formula for T(t)

    $\displaystyle T(t) = \frac {r'(t)}{ || r'(t) ||}$

    So, i find $\displaystyle r' = i - \frac {1}{t^2} j$

    then the magnitude
    $\displaystyle ||r'(t)|| = \sqrt{ (\frac{1^2 + 1^2}{t^2})^2} = \sqrt{ \frac{2}{t^4}} $

    then, combining for T(t) the solutions manual has this which i cant seem to understand.

    $\displaystyle T(t) = \frac {t^2} {\sqrt{ t^4 + 1}} ( i - \frac{1} {\sqrt{ {t^2}}}j) = \frac{1} {\sqrt{ t^4 + 1}}(t^2i - j)$

    I am completely lost!!!
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  2. #2
    Senior Member BAdhi's Avatar
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    Re: find T(t), N(t), at the given time t for the plane curve

    Quote Originally Posted by icelated View Post
    find T(t), N(t), $\displaystyle a_t , a_n $ at the given time t for the plane curve r(t)

    $\displaystyle ||r'(t)|| = \sqrt{ (\frac{1^2 + 1^2}{t^2})^2} = \sqrt{ \frac{2}{t^4}} $
    this is wrong,

    $\displaystyle ||r'(t)||=\sqrt{1^2+\left(\frac{1}{t^2}\right)^2}= \frac{\sqrt{t^4+1}}{t^2}$ as in $\displaystyle r=ai+bj$ and $\displaystyle |r|=\sqrt{a^2+b^2}$
    Last edited by BAdhi; Jul 10th 2012 at 09:10 AM.
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