# Time derivatives of the unit vectors in cylindrical and spherical

In cylindrical and spherical coordinates, the position vectors are given by $$\mathbf{r}=\rho \widehat{\boldsymbol{\rho}}+z \hat{\mathbf{k}}$$ and $$\mathbf{r}=r \hat{\mathbf{r}}$$, next to next, and their derivatives with respect to time are $$\dot{\mathbf{r}}=\dot{\rho} \hat{\boldsymbol{\rho}}+\rho \dot{\varphi} \widehat{\boldsymbol{\varphi}}+\dot{z} \hat{\mathbf{k}}$$ $$\dot{\mathbf{r}}=\dot{r} \hat{\mathbf{r}}+r \dot{\theta} \widehat{\boldsymbol{\theta}}+r \sin \theta \dot{\varphi} \widehat{\boldsymbol{\varphi}}$$ I wonder what would be the time derivatives of the unit vectors of the basis themselves. Is there any straightforward way to deduce their expressions?

To make this more concrete, think about $$\hat{r}$$ as a vector field: $$\hat{r}(r,\theta,\phi)$$. If we then have a particle that has a trajectory given by $$\mathbf{r}(t) = r(t) \; \hat{r}(r(t),\theta(t),\phi(t))$$, then we can derive the velocity vector as follows: \begin{align*} \frac{d\mathbf{r}}{dt} &= \frac{dr}{dt} \hat{r} + r \frac{ d \hat{r}}{dt} \\ &= \frac{dr}{dt} \hat{r} + r \left( \frac{\partial \hat{r}}{\partial r} \frac{dr}{dt} + \frac{\partial \hat{r}}{\partial \theta} \frac{d\theta}{dt} + \frac{\partial \hat{r}}{\partial \phi} \frac{d\phi}{dt} \right) \end{align*} where we have used the product rule and the multi-variable chain rule.
If you carefully calculate $$\frac{\partial \hat{r}}{\partial \theta}$$ and $$\frac{\partial \hat{r}}{\partial \phi}$$, you can show that they're equal to $$\hat{\theta}$$ and $$\sin \theta \hat{\phi}$$ respectively, and so the expression you have for spherical coordinates is obtained. The key point to note here, though, is that the $$\hat{r}$$ vector field does not depend inherently on $$t$$; it's only a function of your position in space.