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I'm learning how to calculate curvature and one of the steps is to calculate $$\frac{dT}{ds}$$

$T$ is the tangent vector to the curve and $ds$ is arc length. But here in this video, he uses says we can calculate this by finding:

$$\frac{||\frac{dT}{dt}||}{||\frac{d\vec{s}}{dt}||}$$ So why can we use $ds$ and $d\vec{s}$ both the same in this context? One is describing arc length, and one is describing the differential vector. I'm thinking they're supposed to be the same but I'd like geometric intuition, because different speeds at which this function is parameterized changes the derivative of the vector valued function. For example, if the function was modified to trace the curve twice as fast, $\frac{d\vec{s}}{dt}$ would be larger, so I don't quite see how it could trace out arc length.

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In this context $ds=\sqrt{(dx)^2+(dy)^2}=\sqrt{x'^2+y'^2}dt=\Vert s'(t)\Vert dt$. He talks in the video about "tiny changes", he is using the idea of differential ($ds$, $dt$) to make calculations.

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  • $\begingroup$ Thank you! Can you explain what you mean by $(ds,dt)$? I'm not exactly sure what that represents. $\endgroup$ – rb612 May 25 '17 at 5:12
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    $\begingroup$ Sorry. I meant only that these differentials are $dt$ and $ds$. It's not and weird operation with them, only a simple parenthesis to indicate what are the diferentials. The formulas I've wrote explain what amounts as a change in arclength $s$, represented this change by $ds$, when a change in $t$ occurs, represented by $dt$. Sincerely, I don't like the way $dt$ is simplified in the video, but it is the old debate about whether or not the symbol $\dfrac{d}{dx}$ can be treated as a quotient. Correct or not, to provide some insight is very useful. $\endgroup$ – Rafa Budría May 25 '17 at 9:58

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