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In my textbook They considered a parallelopiped $ABCDEFGH$ with sides $dx,dy,dz$ parallel to $x,y,z$ axis respectively

  • $\vec V$ represents the vector velocity of the fluid at the centre $P$ of i added this pic because i don't know how to draw that cordinate systema and parallelopiped right there,i'm learning that thing..it will be very helpfull if you tell me more about this,how it works.f the volume (parallelopiped) with components $V_x,V_y,V_z$ along three axis
  • the fluid flows in or out through the all six faces.
  • let us consider the fluid flow through the two opposite faces 1 and 2 of the volume element each being normal to x-axis and has area=dydz
  • The value of the x component of $\vec V$ at the centre of the face 1 and 2 will be different from $V_x$ at the at the centre $P$.

  • $\frac {\partial V_x}{\partial x}$ is the rate of change of $V_x$ along x-axis, then the change in value of $V_x$ in going from $P$ to the centre of vertical faces 1 or 2 = $\frac{\partial V_xdx}{\partial x2}$

  • the value of x-component of the velocity $\vec V$ at the centre of face 1 = $V_x-\frac{1}{2}\frac {\partial V_x}{x}dx$

Now my problem is

  • "The value of the $x$ component of $\vec V$ at the centre of the face 1 and 2 will be different from $V_x$ at the at the centre $P$" Why?

  • $\frac {\partial V_x}{\partial x}$ is the rate of change of $V_x$ along $x$-axis, then the change in value of $V_x$ in going from $P$ to the centre of vertical faces 1 or 2 = $\frac{\partial V_xdx}{\partial x2}$

  • $\frac {\partial V_x}{\partial x}$ is the rate of change of $V$ along x-axis because here water is flowing in all directions and we are concerned with the $x$ axis only, okay. But, change in the value of $V_x$ in going from $p$ to the centre of vertical faces of 1 or 2 = $$\frac{\partial V_xdx}{\partial x2}$$ but how and why. Please explain

  • the value of $x$-component of the velocity $\vec V$ at the centre of face 1 = $V_x-\frac{1}{2}\frac {\partial V_x}{x}dx$

How and why please explain.

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  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/264509/2451 and links therein. $\endgroup$
    – Qmechanic
    Oct 28 '18 at 6:40
  • $\begingroup$ 1. Because $v_x$ changes with $x$. $\endgroup$
    – ProfRob
    Oct 28 '18 at 8:01
  • $\begingroup$ @RobJeffries what?? $\endgroup$
    – Akash
    Oct 28 '18 at 8:02
  • $\begingroup$ @Qmechanici checked out that...i didn't get my answer ..that is not a duplicate $\endgroup$
    – Akash
    Oct 28 '18 at 8:10
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“The value of the 𝑥 component of $\vec{V}$ at the centre of the face 1 and 2 will be different from $𝑉_x$ at the centre 𝑃.”

“The $x$ component of $\vec{V}$” And “$V_x$” are the same thing. This just says $V_x$ is a function, so it’s value changes because $x$ changed going from center to a face. (See note about their bad terminology at the bottom of this.)


Unless otherwise specified, $\vec{V}$, and $V_x$ (the $x$-component of $V$), are specified at the point being considered, point P. Let’s call that point (a,b,c).

The center of face I is at ($a - \frac{dx}{2}$, b, c), and center of II is at ($a + \frac{dx}{2}$, b, c). If y =b and z=c don’t change, then on that line from center of I... to P... to center of II, everything is only a function of x.

Because dx is small, we can assume the change in $V_x$ is just the change in $x$ times $\frac{\partial V_x}{\partial x}$. But notice that the change in $x$ is $\frac{d x}{2}$.

(That’s what $\frac{\partial V_x}{\partial x}$ means: how much $V_x$ changes per change in $x$)


The value of x-component of the velocity 𝑉⃗ at the centre of face 1 $= V_x- \left( \frac{\partial V_x}{\partial x}\right)~\left( \frac{dx}{2}\right)$ How? Why?

Change in $V_x$ going from P to face I

= Change in $V_x$ from $x=a$ to $x=a - \frac{dx}{2}$

= Change in $V_x$ from decreasing x by $ \frac{dx}{2}$

$=~ \left( \frac{\partial V_x}{\partial x}\right)~\left(\Delta x\right)=~ \left( \frac{\partial V_x}{\partial x}\right)~\left(- \frac{dx}{2}\right) $ $$\implies V_{x~\text{(face-}I)}= V_{x~(P)}- \left( \frac{\partial V_x}{\partial x}\right)~\left( \frac{dx}{2}\right)$$

Note, they seem at times to be implying “When I say $V_x$, Im talking about the x-component of $V$, at point P, but when I say ‘the $x$ component of V’ that could be referring to anywhere”. Which is dumb.

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    $\begingroup$ Use \left(content\right) instead of (content) in order to adjust their height to that of the content. Similarly for \left(content\right). $\endgroup$
    – Frobenius
    Sep 3 at 9:34
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    $\begingroup$ @Frobenius Was wondering exactly that. But Im missing something about what you wrote, if we could do an example “( \frac{\partial V_x}{\partial x} ) ” .. you are talking about parenthesis right? $\endgroup$
    – Al Brown
    Sep 3 at 9:38
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    $\begingroup$ Ok I see thanks $\endgroup$
    – Al Brown
    Sep 3 at 9:40
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    $\begingroup$ I edit some parentheses in your question. There are also bigger symbols for example \bigl(content\bigr),\biggl(content\biggr),\Bigl(content\Bigr). $\endgroup$
    – Frobenius
    Sep 3 at 9:42
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    $\begingroup$ @Frobenius Thanks much 🙏🏻👍🏻 $\endgroup$
    – Al Brown
    Sep 3 at 9:43
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The velocity $V$ is actually a vector field i.e it has different values of velocity at different points in space. That is why you get different components at different points.

Rate of change of $x-component$ of velocity in the x- direction is $\frac{\partial V}{\partial x}$. Now since the centre of parallelopiped is $\frac{dx}{2}$ units away from the centres of faces of parallelopiped, we multiply by $\frac{dx}{2} $ to obtain change in velocity.

Using the same logic, since the centre of face 1 is $\frac{dx}{2}$ away from the centre of parallelopiped, we subtract the change in velocity from the velocity at the centre.

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  • $\begingroup$ So,according to your 3rd paragraph, the velocity $V_x$ is changing at a constant rate as ths velocity at face 1 is find out by $V_x-\frac{1}{2}\frac {\partial V_x}{x}dx$. $\endgroup$
    – Akash
    Oct 28 '18 at 23:52

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