Can anyone help me understand howcome:
$$df(x(t),y(t),z(t),t)/dt = \partial(f)/dt + \nabla(f) \cdot \partial(x,y,z)/dt$$ where $\partial(x,y,z)/dt$ represents of course the velocity what i cant understand really well is the $\partial(f)/dt$. (sorry for the lack of notation, i dont know how to use these tools yet)
what i thought was $(x(t),y(t),z(t),t)=m(t)$ therefore: $$df(m(t))/dt = \nabla(f) \cdot \partial(x(t),y(t),z(t),t)= \partial(f)/dx \cdot \partial(x)/dt +..+ \partial(f)/dz \cdot \partial(z)/dt+ \partial(f)/dt \cdot \partial(t)/dt$$
Is that correct?
Thank you very much
That's It is just a special case for the chain rule with
$$f=f(t,x(t),y(t),...)$$
indeed in this special case, since $x,y,...$ are function of one variable $t$, we have
$$\frac{d f}{d t} =\frac{\partial f}{\partial t} = \frac{\partial f}{\partial t} \cdot \frac{\partial t}{\partial t} + \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial t}+...=\frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} \cdot \frac{dx}{dt}+ \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}+...$$
This special case as total derivative of $f$ with respect to $t$.