Intuition behind the difference between $\frac{\partial f}{\partial t}$ and $\frac{df}{dt}$ Let $f=f(x,y,t)$. Using the definition of the differential $df$, we can easily conclude that
$\frac{df}{dt} = \frac{\partial f}{\partial t} + (\vec{v} \cdot \vec{\nabla})f$
$\hspace{2mm} (1)$
So my question is: We refer to both $\frac{df}{dt}$ and
$\frac{\partial f}{\partial t}$ as a change of the quantity $f$ with respect to $t$. But what does each time derivative symbolize in eq$(1)$? How could I measure them?
(I know the definition of partial derivative and the role of
$(\vec{v} \cdot \vec{\nabla})f$ in eq$(1)$. All I want is to understand the connection the $2$ derivatives in eq$(1)$.)
 A: $\partial f/\partial t$ means "I sit at a fixed point $(x,y)$ and change $t$ a bit", i.e. 
$$f(x,y,t+h) = f(x,y,t) + h \frac{\partial f}{\partial t}(x,y,t) + o(h). $$
$df/dt$ means "I'm on a path $(x(t),y(t))$, and look at how $f$ changes as $t$ changes a bit (and I move a bit as well)", i.e.
$$ f(x(t+h),y(t+h),t+h) = f(x(t)+hx'(t)+o(h),y(t)+hy'(t)+o(h),t+h) \\
= f(x(t),y(t),t)+h\left(x'(t) \frac{\partial f}{\partial x} + y'(t) \frac{\partial f}{\partial y} + \frac{\partial f}{\partial t} \right)(x(t),y(t),t) + o(h). $$
A: *

*With a function $f(x,y,t)$, the notation $\frac{\partial f}{\partial t}$ denotes the change of $f$ with respect to $t$ while leaving $x$ and $y$ constant.

*But if $x$ and $y$ are not held constant, but are themselves dependenat on $t$, we may write $f(x(t),y(t),t)$ and then use differentiation rules to get $\frac{df}{dt}=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$
The mathematical formal way to get out of this ambiguity is to always write the parameters as in
$\frac{\partial f(x,y,t)}{\partial t}$
and
$\frac{\partial f(x(t),y(t),t)}{\partial t}$. But thats a lot to write and we are lazy, so we the short-hand notation $d$ and $\partial$
A: $$\frac{\partial f}{\partial t}$$ is the variation of $f$ due to time alone, while keeping $x,y$ constant, i.e. staying at a fixed location.
$$\frac{df}{dt}$$ is a total variation due to both the effect of time and of displacement. In particular, it $v$ is the velocity of a particle in motion in a fluid, the total derivative describes the variations seen by that particle as it moves. In this case, it is called the material derivative.
A: I think this can be cleared up in a very formal way, but this may be complicated to understand.
So $f$ is a function of three variable, i.e. $f$ has domain $\Bbb R^3$. The symbol $\frac{\partial f}{\partial t}$ just means partial derivative with respect to the third variable. Nothing special here.
To interpret the symbol $\frac{df}{dt}$, we introduce a function $\gamma:\Bbb R \to \Bbb R^3$, written as $\gamma(t)=(x,y,t)$, which specifies how the variables $x,y$ depends on $t$. Then $\frac{df}{dt}$ should be properly written as $\frac{d(f\circ \gamma)}{dt}$. There should be no ambiguity in how to interpret the last symbol as $f\circ \gamma$ has domain and codomain being $\Bbb R$.
The equality $$\frac{d(f\circ \gamma)}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt} +\frac{\partial f}{\partial t}$$ is then a result of chain rule.
A: Regards..Mixalis. 
I may say that the usage of partial derivative $ \frac{\partial f(x,y,t)}{\partial t}$ is treating the function $f$ as multivariable function. $x$ , $y$, and $z$ are independent variables. 
But, in some cases, for example in modelling fluids using partial diff. equation, we would like to express the solution $f$ as
$$f(x(t),y(t),t)$$
The functions $x(t)$ and $y(t)$ are called the characteristic function. So $\frac{df}{dt}$ acts as a derivation of one variable function. You can learn about the simplest PDE that has characteristic method for its solution, which is the advection/translation equation. In this solution of advection eqn, each particles move along the curve $(x(t),y(t))$. So $(x(t),y(t))$ tracks the movement of particles.
If $f(x,y,t)= x + y + t$, then $\frac{\partial f}{\partial t} = 1$. 
But, if we would like to see how $f$ changes in a particular path $(x(t),y(t))$, the derivative $\frac{d f}{dt} = x'(t) + y'(t) + 1$ is used. This measures the rate of change of $f$ along a particular path/trail $(x(t),y(t))$.
Thanks.
