Deriving a vector identity It was stated in a lecture that in $(r,\theta,z)$ coordinates we have that
$$(\mathbf{a}\cdot\nabla)\mathbf{b} = \left(\mathbf{a}\cdot\nabla b_r-\frac{a_\theta b_\theta}{r}\right)\hat{\mathbf{r}} + \left(\mathbf{a}\cdot\nabla b_\theta + \frac{a_\theta b_r}{r}\right)\hat{\boldsymbol {\theta}} + \left(\mathbf{a}\cdot\nabla b_z\right)\hat{\boldsymbol z}$$
How is this identity found? Surely it should be
$$(\mathbf{a}\cdot\nabla)\mathbf{b} = (\mathbf{a}\cdot\nabla)b_r\hat{\boldsymbol r} + (\mathbf{a}\cdot\nabla)b_\theta\hat{\boldsymbol \theta} + (\mathbf{a}\cdot\nabla)b_z\hat{\boldsymbol z}$$
 A: Note that $\vec b=\hat r b_r+\hat \theta b_{\theta}+\hat z b_z$.  Furthermore, note that $\hat r$ and $\hat \theta$ are functions of $\theta$ with
$$\begin{align}
\frac{d\hat r}{d\theta}&=\hat \theta\\\\
\frac{d\hat \theta}{d\theta}&=-\hat r
\end{align}$$
Then, 
$$\begin{align}
(\vec a\cdot \nabla)\vec b&=(\vec a\cdot \nabla)(\hat rb_r)+(\vec a\cdot \nabla)(\hat \theta b_{\theta})+(\vec a\cdot \nabla)(\hat zb_z)\\\\
&=\hat r(\vec a\cdot \nabla)(b_r)+b_r(\vec a\cdot \nabla)(\hat r)+\hat \theta(\vec a\cdot \nabla)(b_{\theta})+b_\theta (\vec a\cdot \nabla)(\hat \theta)+(\vec a\cdot \nabla)(\hat zb_z)\\\\
&=\hat r(\vec a\cdot \nabla)(b_r)+b_r(a_\theta \frac1r)(\hat \theta)+\hat \theta(\vec a\cdot \nabla)(b_{\theta})+b_\theta (a_\theta \frac1r)(-\hat r)+(\vec a\cdot \nabla)(\hat zb_z)\\\\
&=\hat r\left(\vec a\cdot \nabla (b_r)-\frac{a_\theta b_\theta}{r}\right)+\hat \theta \left(\vec a\cdot \nabla (b_\theta)+\frac{a_\theta b_r}{r}\right)+\hat z (\vec a \cdot \nabla (b_z))
\end{align}$$
as was to be shown!
