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Suppose $\alpha$ is a time dependent curve so that $\alpha:[0,T]\times I \to \mathbb{R} ^n$. I am a bit confused as to what the meaning of the expression $\partial_t(ds)$ is, where $ds = |\partial_x \alpha|dx$ is the arclength element, I am given.

How do I interpret it? Also how do I interpret $\partial_t(dx)$? Is this identically 0?


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up vote 1 down vote accepted

Yes, the time derivative of $dx$ is zero.

Depending on your sign convention, the time derivative of $|\partial_x\alpha|$ is $\langle F, \vec{k} \rangle$, where $\partial_t\alpha = F$, by direct differentiation. There are a lot of papers which explain this. The classic by Gage and Hamilton on curve shortening flow is a nice read, and has this computation explicitly. It does require a subscription to read however.

At the risk of being overly self-promoting, you can also find this computation in my paper on curve diffusion flow in the plane here. In higher codimension the computation is completely analogous, and can be found in Dziuk, Kuwert, Sch\"atzle's SIAM paper.

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Thanks, from looking at your paper I see that the time derivative of $ds$ is required when we differentiate under the integral sign. In your case, your domain of integration depends on $t$ so you use the transport theorem I think. – soup May 31 '12 at 20:24
@soup This is totally standard when we take as a measure something induced by the map we are studying. You can see this same computation coming up in the first variation of the area element. I'm not sure what you mean by transport theorem. – Glen Wheeler Jun 4 '12 at 9:04

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