The material derivative Let $u:S(t) \to \mathbb{R}$ be a scalar field on a surface $S(t)$ parametrised by time. The material derivative is
$$Du = u_t + v \cdot \nabla u$$ where $v$ is the velocity. I fail to understand the significance of this.. isn't this just $Du = \frac{du(x,t)}{dt}$ and we apply the chain rule thinking of $x$ to depend on time $t$?
 A: The material derivative $\frac{Du}{Dt}$ represents the rate of change of $u$ experienced by a particular material element, while the partial derivative $\frac{\partial u}{\partial t}$ represents the rate of change of $u$ experienced by a particular position in space. The notation is a little misleading, since $x$ is frequently used to represent both a spatial coordinate while $x(t)$ represents a path. To clarify, let $v(x,t)$ be a velocity field and $\sigma(t)$ be the path of a material element, so that $\frac{d}{dt} \sigma(t) = v(\sigma(t), t)$. If $\sigma(t_0) = x_0$, then 
$$ \frac{Du}{Dt}(x_0,t_0) = \frac{d}{dt}u(\sigma(t),t)|_{t_0} = \frac{\partial u(x_0,t_0)}{\partial t} + \frac{\partial u(x_0,t_0)}{\partial x} \cdot \frac{d\sigma(t)}{dt} = \frac{\partial u(x_0,t_0)}{\partial t} + v(x_0,t_0) \cdot \nabla u(x_0,t_0).$$
The derivative is take along a path passing through $(x_0, t_0)$ with velocity $v(x_0, t_0)$.
A: No, the chain rule thinking of $x = x(t)$ would still give you a derivative that is essentially just a derivative with respect to time.
The material derivative gives you something that depends on both time and space -- for instance, the pressure of a fluid might vary with time, but also on the depth of the fluid element in the tank. Even in a fluid that is invariant w/r.t. time, there will be a pressure gradient that exists based on spatial coordinates. The material derivative gives you the total of both of those factors.
