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Math Help - Vectors

  1. #1
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    Post Vectors

    Hi again!

    I was doing a vector question until I got stuck! I have done part a) of the question assuming its right I need help on part b) & c)

    In b) what is it meant by x-y plane?

    c) I remember doing this type of question before where you have to get rp = rq not sure if this is the case for this question.

    Hope you guys can help.

    Regards

    dadon
    Attached Thumbnails Attached Thumbnails Vectors-vectors.gif  
    Last edited by dadon; May 2nd 2006 at 03:00 PM.
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  2. #2
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    Quote Originally Posted by dadon
    Hi again

    In b) what is it meant by x-y plane?
    That means the graph in terms of x and y.
    Plane means flat and x-y means you made a graph as a function of x and y and you end up with a curve which is flat in this plane
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  3. #3
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    You have,
    \bold{R}_p(t)=a\cos \omega t \bold{i}+a\sin \omega t \bold{j}
    This is the displacement vector. The velocity vector is its derivative thus,
    \bold{R}'_p(t)=-a\omega \sin \omega t \bold{i}+a\omega \cos \omega t\bold{j}
    To find the acceleration take derivative again,
    \bold{R}''_p(t)=-a\omega^2\cos \omega t \bold{i} -a\omega^2 \sin \omega t\bold{j}
    -------
    For the other vector function ya got,
    \bold{R}_q(t)=bt\bold{i}+ct\bold{j}
    \bold{R}'_q(t)=b\bold{i}+c\bold{j}
    \bold{R}''_q(t)=\bold{0}
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  4. #4
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    The vector function,
    <br />
\bold{R}_p(t)=a\cos \omega t \bold{i}+a\sin \omega t \bold{j}<br />
    Means, parametrically
    x=a\cos \omega t
    y=a\sin \omega t

    Umm, I cannot sketch this you did not specify a,\omega

    The same in vector function,
    \bold{R}_q(t)=bt\bold{i}+ct\bold{j}<br />
    Means, parametrically
    x=bt
    y=at
    This is a line, but again cannot graph it I need you to specify the values.
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  5. #5
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    They collide when,
    <br />
a\cos \omega t \bold{i}+a\sin \omega t \bold{j}=bt\bold{i}+ct\bold{j}<br />
    That happens when,
    a\cos \omega t=bt
    a\sin \omega t=ct

    Divide equation 2 by equation 1 to get,
    \tan \omega t=\frac{c}{b}
    So the problem reduces to solving this equation.
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  6. #6
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    re:

    Hi thanks for that theperfecthacker!

    How can I use: <br />
a\cos \omega t=bt<br />
and <br />
a\sin \omega t=ct<br />

    To get it in terms of t.
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  7. #7
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    Post re:

    OR is it not possible to get it in terms of t?
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  8. #8
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    Quote Originally Posted by dadon
    OR is it not possible to get it in terms of t?
    Look at my previous post,
    it says these equations reduce to,
    \tan \omega t= \frac{c}{b}
    Therefore use the arctangent function to get,
    \omega t=\tan^{-1} (c/b)+\pi k
    Thus,
    t=\frac{1}{\omega}\tan^{-1}(c/b)+\frac{\pi k}{\omega}
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  9. #9
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    Post

    hi thanks for that!

    is it possible to be in terms of a, b and c?
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  10. #10
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    Quote Originally Posted by dadon
    hi thanks for that!

    is it possible to be in terms of a, b and c?
    Of corse you can write,
    <br />
t=\frac{a}{a\omega}\tan^{-1}(c/b)+\frac{a\pi k}{a\omega}<br />
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  11. #11
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    Thank you for spending time to answer my problem.

    Kind Regards,

    dadon
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