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Math Help - plane curvature

  1. #1
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    plane curvature

    Let (x(s), y(s)) be a plane curve parametrized by arclength. Show that the curvature is given by the formula:

    C = |x'y'' - x''y'|

    The standard formula for curvature is:

    ||dT/ds||,

    where T is the unit tangent vector and equal to:

    x'(t) / ||x'(t)||

    I don't know where to start here, or even if these are the formulas needed to get to the first equation. Any help would be appreciated!
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    Quote Originally Posted by gralla55 View Post
    Let (x(s), y(s)) be a plane curve parametrized by arclength. Show that the curvature is given by the formula:

    C = |x'y'' - x''y'|

    The standard formula for curvature is:

    ||dT/ds||,

    where T is the unit tangent vector and equal to:

    x'(t) / ||x'(t)||

    I don't know where to start here, or even if these are the formulas needed to get to the first equation. Any help would be appreciated!


    We can express the curvature of a plane curve as \kappa (t)=\frac{\|r'(t)\times r''(t)\|}{\|r'(t)\|^3} , so evaluate this taking into consideration that \|r'(t)\|=1 if we use the arclength parametrization.

    Tonio
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    Could you elaborate? Thanks
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    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by gralla55 View Post
    Let (x(s), y(s)) be a plane curve parametrized by arclength. Show that the curvature is given by the formula:

    C = |x'y'' - x''y'|

    The standard formula for curvature is:

    ||dT/ds||,

    where T is the unit tangent vector and equal to:

    x'(t) / ||x'(t)||

    I don't know where to start here, or even if these are the formulas needed to get to the first equation. Any help would be appreciated!
    r=xi + yj

    r`=x`i + y`j

    r``=x``i + y``j

    Curvature = C = ||\frac{d (T(t))}{ds} || = ||r''(s)||

    C(t) = \frac{ ||r'(t) X r''(t)||}{ ||r'(t)||3 }

    Where,

    ||r'(t)||^3 =1^3

    C= ||r'(t) X  r''(t)||

    Which is just the cross product from what we wrote above, which is

    C=|x`y`` -x``y`|
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