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Math Help - Another Vector Problem

  1. #1
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    Another Vector Problem

    The parametric form of the tangent line to the image of f(t)=2t^24/tt−5 at t=−1 is ?


    I got the normal vector to the tangent, <4t,4/t^2, 1>.

    Should I plug in the t=-1 to the original equation to find a point, then plug in t=-1, into the tangent line equation as well?

    Thanks!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    There is a mistake:

    f'(t)=(4t,-4/t^2,1) and so, f'(-1)=(-4,-4,1),\;f(-1)=(2,-4,-6)

    The tangent line is:

    r \equiv\begin{Bmatrix}x=2-4t\\y=-4-4t\\z=-6+t\end{matrix}\quad (t\in{\mathbb{R}})

    ---
    Fernando Revilla
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  3. #3
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    Oh duh!


    So I get,

    <-4t,-4-4t,t-6>
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