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Thread: Normal vector

  1. February 4th 2011, 03:48 PM #1
    Killer
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    Normal vector

    Given the curve C on \mathbb R^3 parametrized by r(t). Find the normal vector on r(1)=(3,1,3) if the osculator plane on that point is 3y-x=0 and r(1)\cdot N(1)>0 with r'(1)=(6,2,2). (I don't know if "osculator" makes sense here.)

    I hope you guys can help, I have an exam on march and I'm scared.

    I don't remember how to solve this, since I lost my calculus notebook.

    Thanks!
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  2. February 4th 2011, 10:52 PM #2
    FernandoRevilla
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    Hint :


    \det [r(t)-r(1),r'(1),r''(1)]=k(3y-x)


    Fernando Revilla
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  3. February 5th 2011, 10:30 AM #3
    Killer
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    Quote Originally Posted by FernandoRevilla View Post
    \det [r(t)-r(1),r'(1),r''(1)]=k(3y-x)
    Okay, this may be related to the osculator plane, but I don't get the determinant and their components.

    Could you elaborate?

    Thanks!
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  4. February 7th 2011, 12:12 PM #4
    Killer
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    Hi Fernando, is it possible that you may help me a bit more?

    I'll appreciate that!
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  5. February 7th 2011, 12:45 PM #5
    zzzoak
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    <br /> r(t)=(x(t),y(t),z(t))<br />

    <br /> r(1)=(x(1),y(1),z(1))<br />

    <br /> r'(t)=(x'(t),y'(t),z'(t))<br />

    <br /> r'(1)=(x'(1),y'(1),z'(1)).<br />
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