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Math Help - Curvature of a parametric curve

  1. #1
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    Curvature of a parametric curve

    Show that the curvature of the plane parametric curve x=f(t), y = g(t) is

    k(t) = (x'y'' - y'x'')/[(x')^2+(y')^2]^3/2

    Again I'm still trying to figure out the notation on this website so if anything is confusing the denominator is to the power 3/2
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  2. #2
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    Quote Originally Posted by Todeezy View Post
    Show that the curvature of the plane parametric curve x=f(t), y = g(t) is

    k(t) = (x'y'' - y'x'')/[(x')^2+(y')^2]^3/2

    Again I'm still trying to figure out the notation on this website so if anything is confusing the denominator is to the power 3/2
    What definition of "curvature" are you using?
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  3. #3
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    I'm not sure, that's the question on my problem set. My professor didn't do a very good job explaining anything this semester
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  4. #4
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    I think you need absolute values in there. I'll do some of it. You want:

    |\frac{d\phi}{ds}|

    which is the rate of change - right out of Liethold - of the measure of the angle giving the direction of the unit tangent vector \bold{T}(t) at a point on a curve with respect to the measure of the arc length along the curve.

    Then if the curve is given parametrically as x=f(t), y=g(t):

    \frac{d\phi}{ds}=\frac{\frac{d\phi}{dt}}{\frac{ds}  {dt}}=\frac{\frac{d\phi}{dt}}{\sqrt{f'(t)^2+g'(t)^  2}}

    and note:

    \tan(\phi)=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\fra  c{dx}{dt}}

    Now try and find "The Calculus" by Leithold in the Library and finish it ok.
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  5. #5
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    Thanks but I'm a little unsure where to go next. Any further hints would be appreciated.
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  6. #6
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    you're in my class TB Moodie.

    hey i went and asked him about this question and he said to do this:


    k=|r' x r''|/|r'|^3

    r= x + y j
    r'=x' i + y j
    r''=x" i + y" j

    r' x r'' = | i j k|
    |x' y' 0|
    |x'' y'' 0|

    = |x'y'' - y'x''| which is the top part.

    to get |r'|^3 -----> |r'|=(x'^2 + y'^2)^(1/2)

    that should give you enough info.
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  7. #7
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    Thanks to both of you. I actually figured it out using the first method, I don't remember us doing any cross product in class but that way looks alot cleaner
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  8. #8
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    ya i don't remember it either, but it's in the textbook as one of the formulas for curvature
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