Material derivative of a unique vector in spherical coordinates If I have a vector $\vec{a}_{cart} = a\hat{z}$ in cartesian coordinates in spherical coordinates that vector would be $\vec{a}_{sph} = a\hat{r}$ with $\theta = 0$ and $\phi =0$ (I must specify the angles for it to be unique).
If I then take the material derivative, $\vec{a}\cdot\nabla\vec{b}$ where $\vec{b} = (b_{r},b_{\theta},b_{\phi})$, does my result (taken from a Wikipedia identity) have to be evaluated at $\theta = 0$ and $\phi =0$ too?
 A: If $\theta =0$, then $\hat z=\hat r$. But in general $\hat z=\hat r \cos(\theta)-\hat \theta \sin(\theta)$.  Hence, if $\vec a=a\hat z $, then we have $$\vec a=a\left(\hat r \cos(\theta)-\hat \theta \sin(\theta)\right) \tag 1$$  
Applying $(1)$, we see that 
$$\vec (a\cdot \nabla) \vec b=a\cos(\theta)\frac{\partial \vec b}{\partial r}-\frac ar \sin(\theta)\frac{\partial \vec b}{\partial \theta}$$
Writing $\vec b=\hat r b_r+\hat \theta b_\theta+\hat \phi b_\phi$ and using $\frac{\partial \hat r}{\partial \theta}=\hat \theta$ and $\frac{\partial \hat \theta}{\partial \theta}=-\hat r$, we find that
$$\begin{align}
\vec (a\cdot \nabla) \vec b&=a\cos(\theta)\frac{\partial \vec b}{\partial r}-\frac ar \sin(\theta)\frac{\partial \vec b}{\partial \theta}\\\\
&=a\cos(\theta)\left(\hat r \frac{\partial b_r}{\partial r}+\hat \theta \frac{\partial b_\theta}{\partial r}+\hat \phi \frac{\partial b_\phi}{\partial r}\right)\\\\
&-\frac ar\sin(\theta)\left(\hat r \left(-b_\theta+\frac{\partial b_r}{\partial \theta}\right)+\hat \theta \left(b_r+\frac{\partial b_\theta}{\partial \theta}\right)+\hat \phi \frac{\partial b_\phi}{\partial \theta}\right)\\\\
&=a\hat r \left(\cos(\theta)\frac{\partial b_r}{\partial r}-\frac{\sin(\theta)}{r}\left(-b_\theta+\frac{\partial b_r}{\partial \theta}\right)\right)\\\\
&+a\hat \theta \left(\cos(\theta)\frac{\partial b_\theta}{\partial r}-\frac{\sin(\theta)}{r}\left(b_r+\frac{\partial b_\theta}{\partial \theta}\right)\right)\\\\
&+a\hat \phi \left(\cos(\theta)\frac{\partial b_\phi}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial b_\phi}{\partial \theta}\right)\tag 2
\end{align}$$
If we evaluate $(2)$ at $\theta=0$, we find that
$$(\vec a\cdot \nabla) \vec b=\frac{\partial \vec b}{\partial r}=\frac{\partial \vec b}{\partial z}$$
as expected!
