Question

What is d/dt(A(x,y,z) ?

Answer 1

Is it the partial of A with respect to time plus the velocity times the gradient of A?

Answer 2

we have to use the chain-rule in order to write your derivative

Answer 3

What if A were A(x,y,z,t) and I wanted the derivative, not the partial derivative? Sorry. A in my question is unclear.

Answer 4

we have to recall the partial derivatives of A, here is your solution:
\[\begin{gathered}
\frac{d}{{dt}}A\left( {x,y,z} \right) = \frac{{\partial A}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial A}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial A}}{{\partial z}}\frac{{dz}}{{dt}} = \hfill \\
= \nabla A \cdot {\mathbf{v}} \hfill \\
\end{gathered} \]

Answer 5

So, if A were A(x,y,z,t) would the answer change to \[\partial A/(\partial t) + grad A \times v\] ?

Answer 6

errr, A(x(t), y(t), z(t), t)

Answer 7

yes! not vector product "x", but scalar product :
\[\begin{gathered}
\frac{d}{{dt}}A\left( {x,y,z,t} \right) = \frac{{\partial A}}{{\partial t}} + \frac{{\partial A}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial A}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial A}}{{\partial z}}\frac{{dz}}{{dt}} = \hfill \\
= \frac{{\partial A}}{{\partial t}} + \nabla A \cdot {\mathbf{v}} \hfill \\
\end{gathered} \]

Answer 8

Thats what I meant. Forgot I was talking in vectors, bad mistake. In my head it was "times v". Thank you very much!

Answer 9

Thank you!