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Math Help - Find velocity vector from position vector

  1. #1
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    Find velocity vector from position vector

    I'm reviewing for my calculus exam on Tuesday and am a bit stuck on this problem. I must be differentiating something wrong.

    The position vector of a particle is r(t). Find the requested vector.
    Find the velocity vector at time (t)=0,

    r(t)=ln(t^{3}-2t^{2}+3)i - (\sqrt{t^{2}+9})j-8cos(t)k

    Derivative of i component:
    (\frac{3t^{2}-4t}{t^{3}-2t^{2}+3})i

    Derivative of j component:
    (\frac{t}{\sqrt{t^{2}+9}})j

    Derivative of j component:
    8sin(t)k

    v(t)=(\frac{3t^{2}-4t}{t^{3}-2t^{2}+3})i- (\frac{t}{\sqrt{t^{2}+9}})j+ 8sin(t)k

    Shouldn't v(0)=0 ?

    What am I doing wrong?
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  2. #2
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    It is zero.. All your differentiations are correct.
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  3. #3
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    Thanks. I guess the professor's practice review has an error. None of his answer choices are 0. Thanks again.
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