What is curvature? (multivariable calculus)

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  • #1
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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]?
 
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  • #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.
 
  • #3
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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.

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?
 
  • #4
WannabeNewton
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