Understanding total derivative Given the information that $\vec A= \vec A(\vec r(t),t)$, why is 
$$ \frac{dA}{dt} = \frac{\partial A}{\partial t} + (\vec r' \nabla)\vec A$$
and not
$$ \frac{dA}{dt} = \frac{\partial A}{\partial t} + \vec r' (\nabla \vec A ) $$ ?
 A: Considering a single component, by the chain rule we have$$\frac{dA_x}{dt}=\frac{\partial A_x}{\partial t}+
\frac{\partial A_x}{\partial x}\frac{dx}{dt}+
\frac{\partial A_x}{\partial y}\frac{dy}{dt}+
\frac{\partial A_x}{\partial z}\frac{dz}{dt}=\frac{\partial A_x}{\partial t}+(\nabla A_x)\cdot\vec r'.$$
Your interpretation would yield
$$\frac{dA_x}{dt}=\frac{\partial A_x}{\partial t}+\left(\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)\frac{dx}{dt}=\frac{\partial A_x}{\partial t}+(\nabla\cdot\vec A)\,\vec r'.$$
This is not possible as the derivative of $A_x$ is independent of $A_y$ and $A_z$.
A: I'm assuming $r=(x(t),y(t),z(t))$ or something similar. Work it out component wise. For example, $\frac{\partial A}{\partial x}\frac{\partial x}{\partial t}$, is the $x$ component derivative and $r'(t)=(x'(t),y'(t),z'(t))$. Now add them up and you'll see it's a dot product between $r'(t)$ and the gradient of $A$. 
Also, notice that $\nabla A$ doesn't make much sense when $A$ is a vector, you'd need something like $\nabla\cdot A$ which is the divergence.
A: You are dealing with a composition, but it is not clear exactly what the functions are. 
I will assume $\textbf r:\mathbb R\rightarrow \mathbb R^{2}$ is given and then that 
$\textbf f:\mathbb R\rightarrow \mathbb R^{3}$ is defined by $\textbf f(t)=(r_{1}(t),r_{2}(t),t)$. 
Next, I will suppose that
$A:\mathbb R^{3}\rightarrow \mathbb R$.
You want to find $(A\circ \textbf f)'(t_{0})$ so it's just a chain rule problem:
$(A\circ \textbf f)'(t_{0})=\left ( \frac{\partial A}{\partial x}(\textbf f(t_{0})),\frac{\partial A}{\partial y}(\textbf f(t_{0})),\frac{\partial A}{\partial z}(\textbf f(t_{0})) \right )\cdot \left ( \frac{dr_{1}}{dt}(t_{0}),\frac{dr_{2}}{dt}(t_{0}),1 \right )$. You can then write this as 
$\nabla A(\textbf f(t_{0}))\cdot \textbf r'(t_{0})+\frac{\partial A}{\partial z}(\textbf f(t_{0}))$ where it is understood that $\nabla $ in this case is $\left ( \frac{\partial }{\partial x},\frac{\partial }{\partial y} \right )$
