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And not with respect to time $t$? (or whatever parameter one is using)

$\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{dt}}|$ seems more intuitive to me.

I can also see that $\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{ds}}| = |\frac{d\mathbf{r}'(t)}{dt}|$ (because $\displaystyle |\mathbf{r}'(t)| = \frac{ds}{dt}$, which does make sense, but I don't quite understand the implications of $\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{dt}}|$ vs. $\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{ds}}|$ and why the one was chosen over the other.

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    You can define curvature without reference to any parametrization if you want. The idea being that the tangent line is the line that intersects the curve and is a good approximation in a 1st order sense. You can then talk about circles that intersect the curve and agree with it in a 2nd order sense. I believe this is called the osculating circle. The radius is $1/\kappa$ where $\kappa$ is the curvature. This is a perfectly good definition.2011-10-20
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    What does it mean to divide $T(t)$ by $ds$? Intuitively you are dividing a normal-sized vector by an infinitesimal scalar, which doesn't make sense to me...2011-10-20
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    typo, meant dT. Curvature as a function of the parameter t.2011-10-20
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    You want the notion of curvature to be independent of the parametrization. That is, it should only depend on the image of the curve and not the function defining it. Your definition of curvature depends on the parametrization.2011-10-20
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    @Eric hm.. let me see if understand: if it's independent, then, say (in physical terms) for a path travelled by an object in 2 minutes, and then by an object that travels the same path in 4 minutes, we'd get the same curvature. — but with my equation, the "curvature" would appear different because it would depend on the time t that it took to travel?2011-10-20

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