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Thread: tangent lines to path

  1. #1
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    tangent lines to path

    need some clarification on this because my book isn't always clear with the notation

    let $\displaystyle \vec{c}(t) = <e^t, e^{-t}, cos (t) > $

    the particle will follow this path and flies off at $\displaystyle t = t_{0} = 1$ find where it is a $\displaystyle t_{1} = 2$

    $\displaystyle \vec{c}'(t) = <e^{t} , -e^{-t}, -sin (t) >$

    so $\displaystyle \vec{l}(2) = <e, \frac{1}{e}, cos (1) > + (2-1)<e, - \frac{1}{e}, - sin(1) >$

    $\displaystyle = <e, \frac{1}{e}, cos (1) > + <e, - \frac{1}{e}, - sin(1) > $

    correct?
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  2. #2
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    Re: tangent lines to path

    Yes.
    Thanks from Jonroberts74
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