# differential and arc length notation question

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?

Thanks

-

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.
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