Please explain this equation from my textbook to me
- The fundamental equation of hydrodynamics of an ideal fluid (Eulerian equation):$$\rho\frac{\mathrm d \mathbf v}{\mathrm dt}=\mathbf f +\nabla p$$
where $\mathbf f$ represents volume density of mass forces and $\nabla$p is the pressure gradient.
I couldn't find anything related to this formula at Hyper physics.
As mentioned in the comments, this is the Euler equation; more specifically, the incompressible Euler equations. There is a boat-load of information available at the afore-linked Wikipedia entry, as well as on PSE and other sites and books.
The other answer correctly states that this is, "...just the Newton's Second Law of Motion" but I would put forth another, possibly more fundamental, meaning/interpretation to this equation: it is the conservation of momentum for the fluid. I go into the details in this other post, from which I'll copy some of the relevant details:
In hydrodynamics, conservation means that what flows into the control volume is equivalent to the flow out of the control volume. With respect to momentum, we mean precisely that
any change in momentum of the fluid within a control volume is due to the net flow of fluid into the volume and the action of external forces on the fluid within the volume (source)
Which can be mathematically represented as the integrals I give at the link or as the PDE you have (with a minor transformation).
I think part of the problem may have been that you seemed to have interpreted the velocity as the volume.1 I think I can understand that thought, given that $M\approx\rho V$. But you should recognize that $\rho$ and $V$ (volume) are scalar quantities while the force, $\mathbf f$, and pressure gradient, $\nabla p$, are both vector quantities, so the "equation" $scalar\cdot scalar=vector+vector$ cannot hold; so it must be that $\mathbf v$ is also a vector quantity with velocity the more logical choice.
1. The comment in which OP had stated this interpretation of $v$ has since been deleted.
Mike Dunlavey is right in his comment; it's just the Newton's Second Law of Motion.
Take a small cube of water. Since, there are no shear forces in a static fluid, pressure at any place is same in all directions.
The net force in the cube exerted in the faces perpendicular to $x$ axis parallel to the base of the cube (neglecting higher order derivatives of $p$) is given by $(p - \left(p + \partial_x p~\Delta x\right))\Delta y\Delta z$ where '$\partial_x$' denotes the partial derivative with respect to $x\,.$
Similarly, taking into consideration all the other faces, the resultant force is $$-~(\partial_x p + \partial_y p + \partial_z p)~\Delta x\Delta y\Delta z = -~\nabla p ~\Delta x\Delta y\Delta z\,.$$
Then for the cube to be in equilibrium, $$-\nabla p + \mathrm F_\textrm{others} = 0\,.~~~~~~~~~~~~~~~~~\textrm{Newton's Second Law of Motion}$$
If the force can be expressed as the gradient of a scalar potential $\varphi,$ per unit mass, then
$$\underbrace{-\nabla p - \rho\nabla \varphi = 0}_\textrm{equation of hydrostatics}\,.\tag{I}$$
(If $\rho$ is constant, then the second term is a pure gradient term. Hence,
$$\begin{align}\nabla(p +\rho\cdot \varphi) &= 0\\ \implies~~~~~~~~ p + \rho \cdot \varphi &= \textrm{constant} \end{align}$$)
In the general case, the shear stress is non-zero for there exists internal force in the flowing fluid. So, there exists viscous force $\mathbf F_\textrm{viscous force}$. Hence, the equation of motion for an element of fluid per unit volume is given by $$\rho \times (\textrm{acceleration}) = -\nabla p - \rho \nabla \varphi + \mathrm F_\textrm{viscous force} ~~~~\textrm{Newton' s Second Law of Motion}\tag{II}$$
where
$\begin{align}\textrm{acceleration} \equiv \dot{\mathbf v}(x,y,z,t) & = v_x ~\partial_x\mathbf v + v_y ~\partial_y \mathbf v + v_z~\partial_ z \mathbf v + \partial_t \mathbf v \\ & = (\mathbf v\cdot\nabla) ~\mathbf v + \partial_t \mathbf v\tag{III}\,.\end{align}$
Putting $\rm (III)$ in $\rm (II),$ we get
$$\underbrace{\frac{\partial \mathbf v}{\partial t} +(\mathbf v\cdot\nabla )~\mathbf v = -\frac{\nabla p}{\rho} -\nabla \varphi + \frac{\mathbf F_\textrm{viscous force}}{\rho}}_\textrm{General equation of motion of a fluid element per unit volume}\,.\tag{IV}$$
For, ideal "dry liquid", $F_\textrm{viscous force} = 0.$ This concludes OP's result as $\rho ~\dot{\mathbf v} = -~\nabla p + \mathbf F ,$ the latter term happens to be conservative in most cases.
$\bullet$ Lectures on Physics Vol. 2 by Feynman, Leighton, Sands.
