I'm studying fluid mechanics, and I got the impression that the material derivative is nothing more than "differentiating along a path" and so I got confused on why do we need it. Basically, let $D\subset \mathbb{R}^3$ be the region containing the fluid and let $f : D\times \mathbb{R}\to \mathbb{R}$ be a time dependent function on $D$.
Suppose then $\gamma : I\subset \mathbb{R}\to D$ is a trajectory on the fluid. The material derivative of $f$ along $\gamma$ is defined by
$$\dfrac{D}{Dt}f(\gamma(t),t) = \dfrac{\partial f}{\partial t}(\gamma(t),t) + (\mathbf{u}\cdot \nabla)f(\gamma(t),t)$$
Where $\mathbf{u}$ is the spatial velocity field of the fluid. But that expression is nothing more nothing less than simply differentiating $f(\gamma(t),t)$ with respect to $t$, or better, differentiating the function $f\circ (\gamma, I)$ where $I$ is the interval in $\mathbb{R}$.
Indeed we have
$$\dfrac{d}{dt}f(\gamma(t),t) = \nabla f(\gamma(t),t)\cdot \gamma'(t) + \dfrac{\partial f}{\partial t}(\gamma(t),t) = \dfrac{D}{Dt} f(\gamma(t),t)$$
So since $\dfrac{d}{dt}= \dfrac{D}{Dt}$ why do we need the material derivative? Why do we define it, since in truth it is just the well know derivative of a composition just? What are the advantages of defining it?
EDIT: I think I got it now. When computing $\dfrac{d}{dt}$ as I said a composition is needed, in other words, we need a path. But since all information is contained in $\mathbf{u}$ we can dispose the path by defining:
$$\dfrac{D}{Dt} f(a,t) = \dfrac{\partial f}{\partial t}(a,t) + (\mathbf{u}\cdot \nabla) f(a,t)$$
And that coincides with $\dfrac{d}{dt}$ if $a = \gamma(t_0)$ for some $\gamma$ and some $t_0$. Is that really the point we are making when defining the material derivative? Is the same we would get if we had a path going through there with velocity $\mathbf{u}$ but we dispose the path since $\mathbf{u}$ already contains all needed info.
Best Answer
You've pretty much answered your own question; you've motivated it and derived the expression, just with a different notation.
To rephrase what you've said: the material derivative is the derivative along the path defined by the integral curves of $\vec{u}$. This is useful because this is the path taken by a small element of the fluid.
As a motivating example: imagine a fluid which is incompressible, so any given element has constant density, but is a mixture of things with different densities. This condition is summed up by $\dfrac{D\rho}{Dt}=0.$