I was looking for an intuitive explanation for the total derivative in polar coordinates. Let me be somewhat more specific: Take a standard line of reasoning that the gradient w.r.t. polar coordinates takes the form $(\partial_r, \ \frac 1r\partial_\theta)$ because $dr\frac {\partial}{\partial r} + d\theta\frac \partial{\partial \theta}=d\boldsymbol r \cdot\nabla$ where $(dr,d\theta)$ is the change in the parameter space while $d\boldsymbol r$ is the ($L_2$) change in the point of evaluation, i.e. $(dr,rd\theta)$. That is, the gradient is the change of the function w.r.t. unit axes.
I don't follow the reasoning when applying this to the chain rule. In particular I'm stuck on the material derivative of a fluid flow, but I'll give a scalar version of the problem (since I don't want to mix in differentiating the coordinate axes, which I do actually understand). Given a 2D fluid flow $\frac d{dt}(r,\theta)=\boldsymbol u(r,\theta,t)=(u_1,u_2)$ and a scalar function $f(r(t),\theta(t))$, the chain rule dictates that $\frac d{dt}f = \frac {dr}{dt}\frac {\partial f}{\partial r}+\frac {d\theta}{dt}\frac{\partial f}{\partial\theta}=u_1f_r + u_2 f_\theta=\boldsymbol u\cdot(f_r, f_\theta)\neq \boldsymbol u\cdot\nabla f$. I don't see why $\frac d{dt}f=\boldsymbol u\cdot \nabla f$ since $\boldsymbol u$ measures the change in the parameter space, and $\nabla$ measures the change of the function w.r.t. unit vectors, not w.r.t. parameters so the equality shouldn't hold. If we want the rate of change of $f$ as a product of the gradient we should have $\frac d{dt}f =(u_1,ru_2)\cdot\nabla f$ which makes sense precisely as above, since $ru_2$ measures the change in the point of evaluation.
So whatever I'm misunderstanding means I don't understand derivations like this, where he claims $\frac d{dt}\theta=\frac 1r u_2$, which I don't see, since to my understanding the parameters are defined like so, $\frac d{dt}\theta= u_2$.
