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What is curvature? (multivariable calculus)

  1. Jul 16, 2013 #1
    if space curve C=<f(t),g(t),h(t)>, and

    v=[itex]\frac{dC}{dt}[/itex]=<[itex]\frac{df(t)}{dt}[/itex],[itex]\frac{dg(t)}{dt}[/itex],[itex]\frac{dh(t)}{dt}[/itex]>

    Why is curvature defined this way? κ[itex]\equiv[/itex][itex]\frac{d\widehat{T}}{dS}[/itex]

    [itex]\hat{T}[/itex]=unit tangent vector
    S=arc length

    to elaborate, for a space curve, i understand what [itex]\frac{dT}{dt}[/itex] is, but what is [itex]\frac{d\widehat{T}}{dS}[/itex]? please explain this to me in an intuitive way, as in what it graphically represents.

    wiki says that the "cauchy defined the curvature C as the intersection point of two infinitely close normals to the curve"

    okay so how is this any different from [itex]\frac{dT}{dt}[/itex]?
     
    Last edited: Jul 16, 2013
  2. jcsd
  3. Jul 16, 2013 #2

    lurflurf

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    Is this what you mean?
    $$\kappa = \left\|\frac{d\mathbf{T}}{ds}\right\|$$

    So at a point we fit the curve (locally) to a circle. The curvature is the reciprocal of the radius of that circle. We think of a small circle as very curved so it has high curvature. A large circle is slightly curved so it has low curvature.
     
  4. Jul 16, 2013 #3
    alright, i have no problem with the curvature being defined that way, but could you show me how $$\kappa = \left\|\frac{d\mathbf{T}}{ds}\right\|$$ means that? ie.. could you show me how $$\kappa = \left\|\frac{d\mathbf{T}}{ds}\right\|$$ directly leads 1/r?
     
  5. Jul 16, 2013 #4

    WannabeNewton

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