Moving Infinite Plate at a Constant Velocity

In summary, the chain rule is used to rearrange terms in order to solve for del u/del t and del^2 u/del y^2. However, there is a inconsistency in how these equations are derived. The solution uses the transformation variable eta to rewrite the terms, but this does not yield the equation in (del u/del t), although this is true for (del^2 u/del y^2). Additionally, the reason for the second derivative of eta being the square of the first derivative is not clear.
  • #1
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Hi all!

I am required to find the velocity distribution of the flow around an infinite plate that suddenly starts moving with a constant speed U_o. The solution has already been worked out, but I still do not understand all of it. The part that is perplexing to me is where they use the transformation variable to rewrite the terms (del u/del t) and (del^2 u/del y^2) with respect to eta.

Simple rearrangement of the terms in eta certainly does not yield the equation in (del u/del t), although this is true for (del^2 u/del y^2). I would think that there would be more of a consistency?

Please help! Thanks!
 

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  • #2
jhuleea said:
The part that is perplexing to me is where they use the transformation variable to rewrite the terms (del u/del t) and (del^2 u/del y^2) with respect to eta.

It's just the chain rule for differentiation

del u/del t = (del u/del eta)(del eta/del t)

and a similar equation for del^2 u/del y^2
 
  • #3
AlephZero said:
It's just the chain rule for differentiation

del u/del t = (del u/del eta)(del eta/del t)

and a similar equation for del^2 u/del y^2

Why of course! Chain Rule, so elementary yet I didn't even consider it. Thanks!

This brings up one more question:

For (del^2 eta/del y^2), why is it that the answer is the square of the first derivative of eta instead of the second derivative of eta? That is,

(del eta/del y) = 1/(2*sqrt(nu*t))

=> (del^2 eta/del y^2) = 0


But according to the solution,

(del^2 eta/del y^2) = [(del eta/del y)]^2 = 1/(4*nu*t)
 
  • #4
[tex]\frac{\partial}{\partial y} = \frac{\partial \eta}{\partial y}\frac{\partial}{\partial \eta}[/tex]

[tex]\frac{\partial^2}{\partial y^2} =
\frac{\partial \eta}{\partial y} \frac{\partial}{\partial \eta}
\left( \frac{\partial \eta}{\partial y} \frac {\partial}{\partial \eta} \right)[/tex]
 
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  • #5
AlephZero said:
[tex]\frac{\partial}{\partial y} = \frac{\partial \eta}{\partial y}\frac{\partial}{\partial \eta}[/tex]

[tex]\frac{\partial^2}{\partial y^2} =
\frac{\partial \eta}{\partial y} \frac{\partial}{\partial \eta}
\left( \frac{\partial \eta}{\partial y} \frac {\partial}{\partial \eta} \right)[/tex]

I'm sorry, but I'm not familiar with your notation. For the expression for
[tex]\frac{\partial}{\partial y}[/tex], you used [tex]\frac{\partial}{\partial \eta}[/tex]. What function are you taking the partial (with respect to eta)??

Also, for your expression of [tex]\frac{\partial^2}{\partial y^2}[/tex], are you just restating what I had posted in my second question? I'm looking for why the partial of second order (that is, [tex]\frac{\partial^2 \eta}{\partial y^2}[/tex]) is not zero, but rather just the square of the the partial of the first order.

Thank you for your patience!
 
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  • #6
I was using [tex]\frac{\partial}{\partial y}[/tex] as an operator, that is something that can operate on any function.

If you don't like that standard calculus notation, then

Differentiate u once:

[tex]\frac{\partial u}{\partial y} = \frac{\partial \eta}{\partial y}\frac{\partial u}{\partial \eta}[/tex]

Then differentiate again:

[tex]\frac{\partial^2 u}{\partial y^2} = \frac{\partial \eta}{\partial y} \partial \left( \frac{\partial \eta}{\partial y} \frac {\partial u}{\partial \eta} \right) / \partial \eta} [/tex]

So, because [tex]\frac{\partial \eta}{\partial y}[/tex] is only a function of [tex]\nu[/tex] and t so it's just a constant when differentiating with respect to [tex]\eta[/tex]

[tex]\frac{\partial^2 u}{\partial y^2} =
\frac{\partial \eta}{\partial y}
\frac{\partial \eta}{\partial y}
\frac {\partial^2 u}{\partial \eta^2}[/tex]
 
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  • #7
A simple example to cut out all the partial differentials, if they are confusing you:

Let y = ax, where a is a constant. Then dy/dx = a

For any function u(x), [tex]\frac{du}{dx} = \frac{dy}{dx} \frac{du}{dy} = a \frac{du}{dy}[/tex]

[tex]\frac{d^2u}{dx^2} = \frac{d}{dx}\left(\frac{du}{dx}\right) = \frac{dy}{dx} \frac{d}{dy}\left(a \frac{du}{dy}\right)[/tex]
[tex] = a^2 \frac{d^2u}{dy^2}[/tex]
 
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  • #8
AlephZero, THANK YOU!

Your explanantion with the example was amazing! I truly appreciate your patience with me, as I realize that my math is a little rusty (since I just started school again recently).

Once again, many thanks!
 

1. What is a moving infinite plate at a constant velocity?

A moving infinite plate at a constant velocity is a theoretical model used in geophysics to study the movement of tectonic plates on the Earth's surface. It assumes that the plate is infinitely large and is moving at a constant speed without any acceleration or deceleration.

2. How does a moving infinite plate at a constant velocity affect the Earth?

The movement of tectonic plates, including a moving infinite plate at a constant velocity, is responsible for various geological processes such as earthquakes, volcanic eruptions, and the formation of mountains. It also plays a role in the distribution of landmasses and ocean basins on the Earth's surface.

3. What factors can affect the velocity of a moving infinite plate?

The velocity of a moving infinite plate can be affected by various factors such as the density and composition of the plate, the strength of the underlying mantle, and the forces generated by the interaction with other plates. These factors can cause variations in the plate's velocity over time.

4. How is the velocity of a moving infinite plate measured?

The velocity of a moving infinite plate can be measured using a variety of techniques such as GPS (Global Positioning System), satellite imagery, and seismology. These methods allow scientists to track the movement of tectonic plates and calculate their velocities.

5. Can the velocity of a moving infinite plate change over time?

Yes, the velocity of a moving infinite plate can change over time. This can be due to changes in the factors that affect its velocity, such as variations in the strength of the underlying mantle or the forces acting on the plate. It can also change due to the interactions and collisions with other tectonic plates.

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