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Lets assume block motion for a fluid. From Navier-Stokes equation we got $\vec \nabla p = \rho (\vec g - \vec a)$

Lets say s is the direction where pressure has the steepest increase.

How do you calculate the change of pressure on direction s ?

According to my book, the answer is:

$$ dp=(\vec \nabla p \cdot\ d\vec s) $$

My question is: Instead of this, shouldn't we simply project the gradient of pressure on an unit vector of path s ?

To calculate work, you do $dW=\vec F \dot\ d\vec r $ , so you’re getting F multiplied by the magnitude of dr on the path, but that’s the definition of work. You apply a force and when you multiply it by the distance you get WORK. In my case, shouldn’t we just do the dot product between the gradient of pressure with the unit vector of path?

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your suggestion is basically what your book says,

$$ {\rm d}p = \nabla p\cdot {\rm d}{\bf s} \tag{1} $$

where ${\rm d}{\bf s} = {\rm d}s \hat{\bf s}$, and $\hat{\bf s}$ is a unit vector along the path ${\bf s}$. So Eq. (1) is the projection of the gradient along the path

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  • $\begingroup$ It seems like work, where you do a line integral, but Work is by definition the Force times the distance. Here I think that we just want to see the magnitude of a vector on other direction, so we would just want to make the dot product between a vector V and a unit vector û, to know the magnitude of V on the direction of U. Help me understanding why I’m wrong. $\endgroup$ – VitorAguiar68 Oct 12 '17 at 7:02
  • $\begingroup$ Gradient of P by itself gives the change of pressure, we would simply point it to the direction S. $\endgroup$ – VitorAguiar68 Oct 12 '17 at 7:29

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