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Math Help - Curviture

  1. #1
    Newbie usvn's Avatar
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    Curviture

    x = 2cos(t) and y = 3sin(t), where the parameter t satisfies 0 < t < 2pi.




    This is only part of the problem but i'm having trouble with it, i need to find the curvature for this parametric equation at points (2,0) and (0,3). I was told to use CAS. Can anyone explain what CAS is and how to do this?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    the formula for curvature is |dT/dt|/|dr/dt| where T is the unit tangent

    To see an explanation see Unit Tangent Vectors/Unit Normal Vectors
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by usvn View Post
    x = 2cos(t) and y = 3sin(t), where the parameter t satisfies 0 < t < 2pi.




    This is only part of the problem but i'm having trouble with it, i need to find the curvature for this parametric equation at points (2,0) and (0,3). I was told to use CAS. Can anyone explain what CAS is and how to do this?
    here, we have \bold{r}(t) = \left< 2 \cos t, 3 \sin t \right>

    Now, the unit tangent vector is given by \bold{T}(t) = \frac {\bold{r}'(t)}{|\bold{r}'(t)|}

    the curvature \kappa is given by  \kappa (t) = \frac {|\bold{T}'(t)|}{|\bold{r}'(t)|}

    for the point (2,0), we need t = 0

    for the point (0,3), we need t = \frac {\pi}2
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  4. #4
    Newbie usvn's Avatar
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    Thanks. I knew those formulas just didn't know they where the CAS or whatever. Let me try it
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  5. #5
    Newbie usvn's Avatar
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    Curviture...

    comment removed
    Last edited by usvn; May 13th 2009 at 09:02 AM. Reason: sorry
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  6. #6
    Newbie usvn's Avatar
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    So so far i have

    r'(t)=(-2sint,3cost)

    r''(t)=(-2cost,-3sint)

    |r'(t)| = sqrt(13)

    T(t)=(-2sint/sqrt(13),3cost/sqrt(13))

    T'(t)=(-2cost/sqrt(13),-3sint/sqrt(13))

    |T'(t)| = sqrt(4/13 + 9/13) = 1

    K=1/sqrt(13)

    But then i have nowhere to plug in the 2 given coordinantes, where did i go wrong?
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  7. #7
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    Quote Originally Posted by usvn View Post
    So so far i have

    r'(t)=(-2sint,3cost)

    r''(t)=(-2cost,-3sint)

    |r'(t)| = sqrt(13)

    T(t)=(-2sint/sqrt(13),3cost/sqrt(13))

    T'(t)=(-2cost/sqrt(13),-3sint/sqrt(13))

    |T'(t)| = sqrt(4/13 + 9/13) = 1

    K=1/sqrt(13)

    But then i have nowhere to plug in the 2 given coordinantes, where did i go wrong?
    How did you get the term in red?
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  8. #8
    Newbie usvn's Avatar
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    sqrt(4sin^2(t) + 9cos^2(t)

    =sqrt (4+9)

    =sqrt(13)
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  9. #9
    MHF Contributor Calculus26's Avatar
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    No not quite

    |r '(t)| = sqrt(4+5cos^2(t))

    at the point (2,0) t= 0

    |r '(0)| = 3



    Then T(t) = r'(t)/sqrt(4+5cos^2(t))=(-2sint,3cost)/sqrt(4+5cos^2(t))

    Now compute T ' (t)

    I get then T ' (0) = -2/3 but again check my results

    you then get k = 2/9 at (2,0)
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  10. #10
    MHF Contributor
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    Quote Originally Posted by usvn View Post
    sqrt(4sin^2(t) + 9cos^2(t)

    =sqrt (4+9)

    =sqrt(13)
    Unfortunately, that's not right. Substitute t = 0 which would give

     <br />
\sqrt{4 \sin^2 0 + 9 \cos^2 0} = \sqrt{9} = 3 \ne \sqrt{13}<br />
.
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  11. #11
    Newbie usvn's Avatar
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    how does |r '(t)| = sqrt(4+5cos^2(t)) ??
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  12. #12
    MHF Contributor Calculus26's Avatar
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    4sin^2(t) + 9cos^2(t)

    = 4sin^2(t) + 4 cos^2(t) + 5cos^2(t)

    = 4 + 5*cos^2(t)
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  13. #13
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    To answer the question originally posed, "CAS" means "computer algebra system" such as MAPLE or MATHEMATICA.
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  14. #14
    MHF Contributor Calculus26's Avatar
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    In fact I used Mathcad for T ' (t) even though the derivative wouldn't have been impossible by hand
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  15. #15
    Newbie usvn's Avatar
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    ...hmm i'm not very proficient with maple. Trying to get |T'(t)| i get

    T'(t) = (-2cost/sqrt(4+5cos^2(t)),-3sint/sqrt(4+5cos^2(t)))

    |T'(0)| = (sqrt(4sin^2(t)/sqrt(4+5cos^2(t)) + 9cos^2(t)/sqrt(4+5cos^2(t)) = sqrt(4/3+5/3) = sqrt(3)??
    am i doing the algebra wrong?

    kappa = |T'(t)|/|r'(t)|

    so kappa = sqrt(3)/3 (wrong)
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