if space curve C=<f(t),g(t),h(t)>, and(adsbygoogle = window.adsbygoogle || []).push({});

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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# What is curvature? (multivariable calculus)

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