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Math Help - Help with vector functions

  1. #1
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    Help with vector functions

    Consider the vector function r(t) =< 4t + 1, t2 − 3t, cos(t) >.
    (a) Find the unit tangent vector T at (1, 0, 1)

    I know that T = r'(t)/|r-(t)|
    and r'(t) = < 4, 2t-3, -sin(t)>

    I'm not sure why the book plugs in 0 for t.
    Thanks in advance
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by khuezy View Post
    Consider the vector function r(t) =< 4t + 1, t2 − 3t, cos(t) >.
    (a) Find the unit tangent vector T at (1, 0, 1)

    I know that T = r'(t)/|r-(t)|
    and r'(t) = < 4, 2t-3, -sin(t)>

    I'm not sure why the book plugs in 0 for t.
    Thanks in advance
    Ok \bold{r}(t)=\left\langle4t+1,t^2-3t,\cos(t)\right\rangle


    So as you noted \bold{T}(t)=\frac{\bold{T}'(t)}{|\bold{T}'(t)|}

    So we can compute this to be

    \bold{T}(t)=\frac{\langle4,2t-3,-\sin(t)\rangle}{\sqrt{16+\left(2t-3\right)^3+\sin^2(t)}}

    So \bold{T}(1,0,1)=\frac{\langle4,-3,-\sin(1)\rangle}{\sqrt{16+9+\sin^2(1)}}
    Last edited by Mathstud28; November 14th 2008 at 05:35 PM. Reason: typo
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  3. #3
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    In the answer key, they did this
    T(0) to get <4, -3, 0> as r'(0)
    Why did they use 0?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by khuezy View Post
    In the answer key, they did this
    T(0) to get <4, -3, 0> as r'(0)
    Why did they use 0?
    \bold r(t)=\left<4t+1,t^2-3t,\cos t\right> at the point (1,0,1) implies that there is a t value such that 4t+1=1,~t^2-3t=0, and \cos(t)=1

    It should be a little clearer why they picked t=0...

    --Chris
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