# Time derivative of unit vector in spherical coordinates

Is it possible to take a time derivative of a vector given in some curvelinear coordinate system (i.e. spherical)? Mathematica would need to take into account the time dependence of the basis vectors.

$$\frac{d}{dt}\vec{r}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}+r\sin[\theta]\dot{\phi}\hat{\phi}$$

I bet that there is something built into mathematica already to get the above result, but I can't figure it out.

Edit: Thanks for your answers, they already helped me a lot. In the end I was hoping for a quick way to get:

$$\frac{d}{dt}\left( \begin{array}{c} f(r,\theta,\phi) \\ 0 \\ 0 \end{array} \right)=\left(\frac{d}{dt}f(r,\theta,\phi)\right) \left( \begin{array}{c} \dot{r} \\ r\dot{\theta} \\ r\sin[\theta]\dot{\phi} \end{array} \right)$$

just by typing

$$Dt[\{f(r,\theta,\phi),0,0\},t]$$

in Mathematica. I could get this behavior by multiplying the time derivative of $f$ with the result from TransformedField, but this could quickly become tedious with higher time derivatives and more vector components. Is there a more direct way to do it?

Thanks a lot

-
Well, if you concluded that there's something built-in why didn't you take a Documentation Centre tour ? As a starter take a look here – Sektor Mar 22 '14 at 10:55
The documentation center tour didn't really show me any examples on what I want to achieve. But my question was not clear in that regard, for which I apologize. I have updated it (see my edit). – ftiaronsem Mar 22 '14 at 12:01

TransformedField["Cartesian" -> "Spherical", Dt[{x, y, z}, t], {x, y, z} -> {r, θ, ϕ}]