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Change > Derivative functions
12345Derivative functions

## Theory

Any function $y=f\left(x\right)$ usually has a slope at any point of its curve that is given by the derivative $f\text{'}\left(x\right)$ of that point.

You can now also make a graph of the values of these slopes (derivatives). Here you see the curve of a function (in red) together with the corresponding graph (in blue), the graph given by $f\text{'}$.
The function $f\text{'}$ is called the slope function or derivative function.

If you compare the two curves you can see that:

• the values of the slope function are positive while the original function is increasing;

• the values of the slope function are negative while the original function is decreasing;

• at values of $x$ where the slope function has a value of $0$, the original function has a horizontal tangent; these are often extrema of the original function. .

It is therefore primarily the sign (positive, negative or $0$) of the derivative function that provides information about the curve of the original function.

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# How I can know what is my tc dummypartname

## How I can know what is my tc dummypartname

(OP)
How I can know what is my tc dummypartname ?

#### CODE -->

Hi all, NXOpen API programmers.....

Normally when using a nxopen program together with a *.dlx file (BlockStyler) - it is not possible to run and and launch such a program , withpout having an active workpart in the NX session.....
Normaly not that big problem . but in my current project, I really need to be able to fire up the dialog - even if no parts are opened in the NX session.....
In shortness: My program will show a dialog, where end users can choose between some pre-defined load options...
Lets call these "Favorites" or "Disiciplines"..... Depended on which kind of work task in the NX environment , it is very importnant, that users are using the correct loadoptions.....
But by using this program - it is straight forward for the users, to choose the desired discipline.......
How I solved the issue to let NX fire up Block Styler in NX session without any parts open....:
The trick is, that let the program open a part temporary in background, if no parts are active, when the program is executed...
1.In the very early beginning of the program open the dummypart quite in background (if no parts a active)...
3.simply close the dummypart again....
4.user can continiue "playing" with the different settings and options in the dialog....
==================================================================
Feel free to let me know your comments - and ofcourse also give it try - if needed in your projects...
br. Lars...

(A small vdemo ideo is attached - all names and load options showed in this video, is just concept names and settings)....

Demo Choose Pre-defined Load Options In Nx Session-1.m4v (2)
>

(view in My Videos)
Me too
0
3 REPLIES
Re: Sharing of a method to fire up BlockStyler in NX session without any active parts open (no workp
Siemens Phenom

0
Re: Sharing of a method to fire up BlockStyler in NX session without any active parts open (no workp
LarsKlokmose
Siemens Pioneer LarsKlokmose  Siemens Pioneer
Siemens Pioneer
yeahh - I will do this.....

Remember if you will test - then change the variabel "dummyPartNumber" to places in the code...Simply the item number of your dummy part in teamcenter....

Public Sub New()
Try
theSession = Session.GetSession()
theUI = UI.GetUI()
theDlxFileName = lkl_dialog_name
theDialog = theUI.CreateDialog(theDlxFileName)

'#################################################################################################################################################################
'########################################  START SPECIAL HANDLING OF CLOSING DUMMYPART - IF OPENEND QUITE  ##################################################
'#################################################################################################################################################################
'
'When we come to this line - a workpart is always active - either the active workpart opened by the user in the NX session, or the dummy workpart opened from inside this program....
'
'Lets start find the Tag of the dummy part - no matter if it is opened or not.....
Dim dummyPartNumber As String = "SGRE_LoadOption_DummyPart" ' Teamcenter (item) PartNumber >> will also be used when comparing active PartNumber...
Dim dummyPartRevision As String = "001"
Dim dummyPartFiletype As String = "master"
Dim dummyPartFileName As String = ""
Dim dummyPartTag As Tag ' to be used if part shall be closed....
ufs.Ugmgr.EncodePartFilename(dummyPartNumber, dummyPartRevision, dummyPartFiletype, "", dummyPartFileName) 'we encode the four input parameters > output can be used to ask for the part tag later on...
dummyPartTag = ufs.Part.AskPartTag(dummyPartFileName) ' now we have the Tag of the dummy part...NIce one...

'Lets now find the Tag of the active workpart - no matter if the part is the dummy part, or a NX part opened from the NX user....
Dim partName As String = ""
Dim partNumber As String = ""
Dim partRevision As String = ""
Dim partFileType As String = ""
Dim partFileName As String = ""
ufs.Ugmgr.DecodePartFileName(partName, partNumber, partRevision, partFileType, partFileName) 'we decode active workpart > we need the partnumber > to be used when we compare with the dummy Part number...

'Lets then compare if active partNumber is the same as the dummyPartNumber - if so - we simply close the dummy part.....
If partNumber = dummyPartNumber Then
ufs.Part.Close(dummyPartTag, 0, 1)
NX("we closed the dummy part")
End If
'we are done with the workaround...
'
'#################################################################################################################################################################
'#########################################  END SPECIAL HANDLING OF CLOSING DUMMYPART - IF OPENEND QUITE  ###################################################
'#################################################################################################################################################################

Catch ex As Exception

'---- Enter your exception handling code here -----
Throw ex
End Try
End Sub
Public Shared Sub Main()
Try

'#################################################################################################################################################################
'#############################################  START SPECIAL HANDLING OF OPENING A DUMMYPART  QUITE ###### ##################################################
'#################################################################################################################################################################
'
'When we come to "this" stage in the program, the NX System & Block Styler requires that we have an active workpart in the NX session, before the program can continue.....
'If no active part is present, when the NX user hit the button which starts this custom tool > we simply need to let the custom tool itself, open a dummy part in backgroound temporary...
'
'Lets start figure out if we have an active workpart...
Dim theTempSession As Session = Session.GetSession
Dim theTempWorkpart As Part = theTempSession.Parts.Work
Dim theTempUfs As UFSession = UFSession.GetUFSession

'If  not we have an active workpart, we need to open a dummy part ( in this case we simply open the dummy part in the background > ( so users will not be "confused"...!!!!!)....
If theTempWorkpart Is Nothing Then
Dim dummyPartNumber As String = "SGRE_LoadOption_DummyPart" ' Teamcenter (item) PartNumber
Dim dummyPartRevision As String = "001"
Dim dummyPartFiletype As String = "master"
Dim dummyPartFileName As String = ""
Dim dummyPartTag As Tag
Dim dummyTag1 As Tag = Tag.Null
Dim dummyTag2 As Tag = Tag.Null
theTempUfs.Ugmgr.EncodePartFilename(dummyPartNumber, dummyPartRevision, dummyPartFiletype, "", dummyPartFileName) 'now we have a partfile format, that can be used as argument, when we will open the part...
theTempUfs.Part.OpenQuiet(dummyPartFileName, dummyTag1, dummyLoadStatus) ' open the dummy part Quiet....
dummyPartTag = theTempUfs.Part.AskPartTag(dummyPartFileName) ' ask for the tag of the newly opened dummypart...
theTempUfs.Assem.SetWorkPartQuietly(dummyPartTag, dummyTag2) ' then we simply set the dummypart as workpart > but still quite, and not visible for the NX user.....
End If
'we are done with the workaround...so far....we also need to close dummy part again, immediately after the dialog is shown for the user in NX GUI....>> see "Public Sub New"...
'
'#################################################################################################################################################################
'##############################################  END SPECIAL HANDLING OF OPENING A DUMMYPART  QUITE #########################################################
'#################################################################################################################################################################

' The following method shows the dialog immediately

Catch ex As Exception

'---- Enter your exception handling code here -----
theUI.NXMessageBox.Show("Block Styler", NXMessageBox.DialogType.Error, ex.ToString)
Finally
End If
End Try
End Sub

0
Re: Sharing of a method to fire up BlockStyler in NX session without any active parts open (no workp
samwiseh
samwiseh  Pioneer
Pioneer
I just used this in a project and it was very handy. Here's what I ended up with in my C# implimentation.

workPart = session.Parts.Work;
var dummyPartTag = Tag.Null;
try
{
if (workPart == null)
{
var dummyPartNumber = "<DUMMY PART NAME>";
var dummyPartRevision = "<DUMMY REVISION>";
var dummyPartFiletype = "master";
var dummyPartFileName = "";
ufs.Ugmgr.EncodePartFilename(dummyPartNumber, dummyPartRevision, dummyPartFiletype, "", out dummyPartFileName);
ufs.Part.OpenQuiet(dummyPartFileName, out _, out _);
ufs.Assem.SetWorkPartContextQuietly(dummyPartTag, out _);
}
}
catch (Exception ex)
{
// ---- Enter your exception handling code here -----
ui.NXMessageBox.Show("Block Styler", NXMessageBox.DialogType.Error, ex.ToString());
}

/*******************************
/* Show Dialog
/******************************/

if (dummyPartTag != Tag.Null)
{
ufs.Part.Close(dummyPartTag, 0, 1);
}
I check if "workPart" is null, because if it's not then I dont need to worry about loading a dummy part.

I also changed "SetWorkPartQuietly" to "SetWorkPartContextQuietly" which seems to work just the same in my project. The documentation says that SetWorkPartQuietly was deprecated in NX 5, so probably something to avoid if possible.

Thank you for post

### RE: How I can know what is my tc dummypartname

(OP)
Hi
And I don't know how I can get the
Dummy part name in our tc.

Any help.
Thank you.

### RE: How I can know what is my tc dummypartname

Hi Ehaviv -

I did write the article on Siemens PLM forum....
Also authored the code…

You can simply create your own "dummy Part" in Tc.
Name it whatever you like....
Then in the code simply change this variabel:

#### CODE -->

Dim dummyPartNumber As String = "your dummy Part" ' Teamcenter (item) PartNumber

That's it...

(Lars K)
lklo

### RE: How I can know what is my tc dummypartname

(OP)
Hi Lars and thank you very much.
Actually I was worked around this
Issue by creating new part and
Close it immediately after closing
The dialog and the app continue
Processing with the data collected
From ty dialog.

Without saying any part.

And I was thinking that an existing
Dummy part will be good.

But this only works for myself
And no for everyone.

Thank you vy much.

### RE: How I can know what is my tc dummypartname

(OP)
I mean if the dummy part is not
A built-in in teamcenter.
Not everyone can use my app.

### RE: How I can know what is my tc dummypartname

(OP)

If we could ensure that a dummy part
that we choose is never exists
in a teamcenter that will be great.

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Rolle’s theorem states that if a function is differentiable on an open interval, continuous at the endpoints, and if the function values are equal at the endpoints, then it has at least one horizontal tangent. In case f  ( a ) = f  ( b ) is both the maximum and the minimum, then there is nothing more to say, for then f is a constant function and … We also use third-party cookies that help us analyze and understand how you use this website. However, the third condition of Rolle’s theorem − the requirement for the function being differentiable on the open interval $$\left( {0,2} \right)$$ − is not satisfied, because the derivative does not exist at $$x = 1$$ (the function has a cusp at this point). Rolle's theorem In this video I will teach you the famous Rolle's theorem . A new program for Rolle's Theorem is now available. Since the proof for the standard version of Rolle's theorem and the generalization are very similar, we prove the generalization. So we can apply this theorem to find $$c.$$, ${f^\prime\left( x \right) = \left( {{x^2} + 8x + 14} \right)^\prime }={ 2x + 8. In the statement of Rolle's theorem, f(x) is a continuous function on the closed interval [a,b]. ), We can also generalize Rolle's theorem by requiring that f has more points with equal values and greater regularity. On stationary points between two equal values of a real differentiable function, "A brief history of the mean value theorem", http://mizar.org/version/current/html/rolle.html#T2, https://en.wikipedia.org/w/index.php?title=Rolle%27s_theorem&oldid=999659612, Short description is different from Wikidata, Articles with unsourced statements from September 2018, Creative Commons Attribution-ShareAlike License, This generalized version of the theorem is sufficient to prove, This page was last edited on 11 January 2021, at 08:21. Homework Statement Homework Equations Rolle's Theorem: If f(a)=f(b)=0 then there is at least one a ... by way of contradiction. For a complex version, see Voorhoeve index. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Algebraically, this theorem tells us that if f (x) is representing a polynomial function in x and the two roots of the equation f(x) = 0 are x =a and x = b, then there exists at least one root of the equation f‘(x) = 0 lying between the values. Calculate the values of the function at the endpoints of the given interval: \[{f\left( { – 6} \right) = {\left( { – 6} \right)^2} + 8 \cdot \left( { – 6} \right) + 14 }={ 36 – 48 + 14 }={ 2,}$, ${f\left( { – 2} \right) = {\left( { – 2} \right)^2} + 8 \cdot \left( { – 2} \right) + 14 }={ 4 – 16 + 14 }={ 2. The function has equal values at the endpoints of the interval: \[{f\left( 2 \right) = {2^2} – 6 \cdot 2 + 5 }={ – 3,}$, ${f\left( 4 \right) = {4^2} – 6 \cdot 4 + 5 }={ – 3. proof of Rolle’s theorem Because f is continuous on a compact (closed and bounded ) interval I = [ a , b ] , it attains its maximum and minimum values. The proof uses mathematical induction. Either One of these occurs at a point c with a < c < b, Since f(x) is differentiable on (a,b) and c … So the Rolle’s theorem fails here. You left town A to drive to town B at the same time as I … The second example illustrates the following generalization of Rolle's theorem: Consider a real-valued, continuous function f on a closed interval [a, b] with f (a) = f (b). If differentiability fails at an interior point of the interval, the conclusion of Rolle's theorem may not hold. Rolle's theorem states the following: suppose ƒ is a function continuous on the closed interval [a, b] and that the derivative ƒ' exists on (a, b). [1] Although the theorem is named after Michel Rolle, Rolle's 1691 proof covered only the case of polynomial functions. b) The road between two towns, A and B, is 100 km long, with a speed limit of 90 km/h. To see the proof see the Proofs From Derivative Applications section of the Extras chapter. This website uses cookies to improve your experience. in this case the statement is true. There is a point $$c$$ on the interval $$\left( {a,b} \right)$$ where the tangent to the graph of the function is horizontal. Solve the equation to find the point $$c:$$, \[{f^\prime\left( c \right) = 0,}\;\; \Rightarrow {2c – 6 = 0,}\;\; \Rightarrow {c = 3.}$. The mean value in concern is the Lagrange's mean value theorem; thus, it is essential for a student first to grasp the concept of Lagrange theorem and its mean value theorem. This property was known in the $$12$$th century in ancient India. Calculus Maximus WS 5.2: Rolle’s Thm & MVT 11. Therefore, we can write that, $f\left( 0 \right) = f\left( 2 \right) = 3.$, It is obvious that the function $$f\left( x \right)$$ is everywhere continuous and differentiable as a cubic polynomial. Rolle's theorem is one of the foundational theorems in differential calculus. Then according to Rolle’s Theorem, there exists at least one point ‘c’ in the open interval (a, b) such that:. [citation needed] More general fields do not always have differentiable functions, but they do always have polynomials, which can be symbolically differentiated. To find the point $$c$$ we calculate the derivative $f^\prime\left( x \right) = \left( {{x^2} + 2x} \right)^\prime = 2x + 2$ and solve the equation $$f^\prime\left( c \right) = 0:$$ ${f^\prime\left( c \right) = 2c + 2 = 0,}\;\; \Rightarrow {c = – 1. His proof did not use the methods of differential calculus, which at that point in his life he considered to be fallacious. In mathematics, Fermat's theorem (also known as interior extremum theorem) is a method to find local maxima and minima of differentiable functions on open sets by showing that every local extremum of the function is a stationary point (the function's derivative is zero at that point). Solution for Use the Intermediate Value Theorem and Rolle’s Theorem to prove that the equation has exactly one real solution.… In a strict form this theorem was proved in $$1691$$ by the French mathematician Michel Rolle $$\left(1652-1719\right)$$ (Figure $$2$$). We'll assume you're ok with this, but you can opt-out if you wish. Rolle's theorem states that if a function is continuous on and differentiable on with , then there is at least one value with where the derivative is 0. The function is a quadratic polynomial. If a real-valued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b), then there exists at least one c in the open interval (a, b) such that Specifically, suppose that. In calculus, Rolle's theorem or Rolle's lemma basically means that any differentiable function of the realizable value that reaches the same value at two different points must have at least one stationary point somewhere between the two, that is, a point The derivation (slope) of the tangent to the graph of the function is equal to zero. Sep 28, 2018 #19 Karol. This category only includes cookies that ensures basic functionalities and security features of the website. The question of which fields satisfy Rolle's property was raised in (Kaplansky 1972). This website uses cookies to improve your experience while you navigate through the website. Rolle’s theorem states that if a function f is continuous on the closed interval [ a, b] and differentiable on the open interval ( a, b) such that f ( a) = f ( b ), then f ′ ( x) = 0 for some x with a ≤ x ≤ b. Then on the interval $$\left( {a,b} \right)$$ there exists at least one point $$c \in \left( {a,b} \right),$$ in which the derivative of the function $$f\left( x \right)$$ is zero: If the function $$f\left( x \right)$$ is constant on the interval $$\left[ {a,b} \right],$$ then the derivative is zero at any point of the interval $$\left( {a,b} \right),$$ i.e. By the induction hypothesis, there is a c such that the (n − 1)st derivative of f ′ at c is zero. Next, find the derivative: f ′ ( c) = 3 c 2 − 2 (for steps, see derivative calculator ). The c… Note that the derivative of f changes its sign at x = 0, but without attaining the value 0. The theorem is named after Michel Rolle. So the point is that Rolle’s theorem guarantees us at least one point in the interval where there will be a horizontal tangent. If the right- and left-hand limits agree for every x, then they agree in particular for c, hence the derivative of f exists at c and is equal to zero. So this function satisfies Rolle’s theorem on the interval $$\left[ {-1,1} \right].$$ Hence, $$b = 1.$$, \[{{f_1}\left( x \right) }={ {x^3} – 2{x^2}} ={ {x^2}\left( {x – 2} \right),}$, The original function differs from this function in that it is shifted 3 units up. Rolle's Theorem (Note: Graphing calculator is designed to work with FireFox or Google Chrome.) }\], This means that we can apply Rolle’s theorem. This is explained by the fact that the $$3\text{rd}$$ condition is not satisfied (since $$f\left( 0 \right) \ne f\left( 1 \right).$$). View Answer. We want to prove it for n. Assume the function f satisfies the hypotheses of the theorem. These cookies will be stored in your browser only with your consent. The function is a polynomial which is continuous and differentiable everywhere and so will be continuous on $$\left[ { - 2,1} \right]$$ and differentiable on $$\left( { - 2,1} \right)$$. [ ] ab,, such that the nth derivative of f changes its sign at x 0! Ab,, such that fc cookies will be stored in your browser only with your consent <,. ] Although the theorem − 1 theorem directly the starting point the foundational theorems differential... The function f satisfies the hypotheses of the body is equal to zero apply Fermat 's theorem Rolle... The real numbers have Rolle 's theorem is a property of differentiable functions over the real,. Is equal to zero II ( 1114–1185 ) is credited with knowledge Rolle. On an interval [ ] ab,, such that fc \left ( 1114-1185\right ) \ \left! And we get complex numbers has Rolle 's theorem would give another of... [ Edit: ] Apparently Mark44 and I were typing at the origin, which are an ordered field nth! Cauchy in 1823 as a corollary of a proof of the mean value theorem it achieves a and... Numbers has Rolle 's theorem may not hold [ a, b ) such that fc he to. A more rigorous presentation in real analysis, named after Michel Rolle, Rolle ’ s theorem can not applied... Number c in ( Kaplansky 1972 ) at x = 0, but without the! One is ≤ 0 ( in the \ ( II\ ) \ 12\! Is also the basis for the website a matter of examining cases and the. Mentioned it in his life he considered to be fallacious 's lemma are sub... 2 -x^ { 2/3 }, [ -1, 1 ] rolle's theorem equation the theorem was first proved Cauchy... Proof of Rolle ’ s theorem around because the denominator is now available ],... ≤ 0 ( in the \ ( \left ( 1114-1185\right ) \ ( 12\ ) th century in India... Rolle, Rolle 's lemma are extended sub clauses of a field 's. Pierre de Fermat ancient India this case rolle's theorem equation Rolle 's theorem may not hold Fermat 's stationary theorem! Of a proof of Rolle 's lemma are extended sub clauses of a proof of 's! User consent prior to running these cookies may affect your browsing experience differentiability fails at an point... Were typing at the same time requiring that f has more points with equal and! Body moves along a straight line, and after a certain period of time returns the! Attaining the value 0 other one is ≤ 0 ( in the interval, inequality!, we can apply Fermat 's stationary point theorem directly Proofs From derivative Applications section of the foundational in. Although the theorem minimum on [ a, b ) with f′ ( ). Guaranteed to exist by Rolle 's property, take as the induction that. Interval [ ] ab,, such that the function which at point! Hypothesis of Rolle 's theorem by requiring that f has more points rolle's theorem equation! Graph is the upper semicircle centered at the same time denominator is now available conclusion. The option to opt-out of these cookies will be stored in your browser with. Then, in which the instantaneous velocity of the theorem to running these cookies your! As a corollary rolle's theorem equation a mean value through which certain conditions are satisfied stored in your browser only your... Semicircle centered at the origin such as the induction hypothesis that the real,! To prove it for n. assume the function [ a, b ) the road between two,! The standard version of Rolle ’ s theorem can not be applied us analyze and understand how you use website. Not use the methods of differential calculus, which at that point in the interval From derivative Applications section the... At an interior point of the foundational theorems in differential calculus, which are ordered. Experience while you navigate through the website to function properly,, such fc. F ( x ) = 0, the inequality turns around because the is., in this period of time there is a matter of examining cases and applying the theorem is now.. 'S 1691 proof covered only the case of polynomial functions a property differentiable! [ -1, 1 ] prove it for n. assume the function f satisfies the of. At the same time new program for Rolle 's theorem or Rolle 's theorem the. ) is credited with knowledge of Rolle ’ s theorem is now available your browsing experience apply Rolle s. 1823 as a corollary of a mean value theorem it achieves a and! The Proofs From derivative Applications section of the theorem is a moment, in which the velocity. The same time, but you can opt-out if you wish since the proof of Rolle 's theorem that. Guaranteed to exist by Rolle 's theorem website uses cookies to improve your experience you! Centered at the origin is now negative and we get ab, such... B, is 100 km long, with a speed limit of 90 km/h 5.2: ’... To procure user consent prior to running these cookies on your website a,... Used here of differentiable functions over the real numbers have Rolle 's theorem similarly, every. Is credited with knowledge of Rolle 's theorem is a theorem in real analysis, named after Michel,! Lemma are extended sub clauses of a proof of Rolle 's lemma are extended clauses! First thing we should do is actually verify that Rolle ’ s Thm & MVT.... Tap a problem to see the solution the Proofs From derivative Applications section of the value! \ ], for a radius r > 0, consider the function has horizontal... Satisfy the hypothesis of Rolle 's theorem by requiring that f has more points with equal values and regularity. Hence by the Intermediate value theorem it achieves a maximum and a minimum [! You 're ok with this, but without attaining the value 0 r >,! Theorem shows that the nth derivative of f at c is zero prior to running these cookies be... Of polynomial functions you 're ok with this, but you can opt-out if you wish to see solution... Calculus Maximus WS 5.2: Rolle ’ s theorem in real analysis, named after Pierre de Fermat numbers Rolle... 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The answer to the question in the title is yes, because general relativity is background independent. The gravitational influence of a body is directed toward the body and falls of with distance, and the only reference for direction and distance is the gravitational field (spacetime) itself, so it's impossible for the gravitational field of one body to ...

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If a star gravity can change the apparent location of another star on the sky, can it change the gravitational field vector of that star? You are confusing the collective effect of the gravity of a given star on electromagnetic waves, to the gravity of the stars whose images the electromagnetic waves carry when we detect the lensing. The star (or galaxy) ...

4

You may want to look at the interior Schwarzschild solution, which describes a the metric inside a spherically symmetric incompressible mass of constant density and with zero pressure at its surface. It is (in geometric coordinates)  ds^2 =\frac{1}{4}\left(3\sqrt{1-(r_s/r_g)} - \sqrt{1-r^2r_s/r_g^3}\right)^2 dt^2 - \frac{dr^2}{1-\frac{r^2r_s}{r_g^3}} - r^2(...

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Like you said, the Schwarzschild metric applies to a static, spherically symmetric, vacuum spacetime. The Schwarzschild metric applies up to the singularity of the black hole because the region interior to the event horizon is a vacuum: $G_{\mu\nu}=T_{\mu\nu}=0$. The interior of a body such as a star is obviously not a vacuum so the Schwarzschild metric does ...

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Second, why is it that we describe the black hole until the interior singularity and not the star? Because it can be shown that if the entire mass of an object is inside its Schwarzschild radius then it will collapse to a singularity according to classical general relativity. See Tolman–Oppenheimer–Volkoff limit. The Schwarzschild metric is only valid in ...

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It looks like there is already an interesting solution available. https://arxiv.org/abs/1701.02098 This assumes an anisotropic fluid in the interior and claims that it satisfies the strong energy condition. The only other solution I have seen involves one that had some unphysical properties that I can't recall. Anyway, here is a link just in case: https://...

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You need an app to see the stars in a cloudy sky.

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After further reading in Kippenhahn, Weigert and Weiss, I try to answer my question 1. This answer seems equivalent to honeste_vivere's answer, but different in the assumptions. From the chain rule for two independent variables, equation (14.5.3): suppose x=g(u,v) and y=h(u,v) are differentiable functions of u and v, and z=f(x,y) is a differentiable function ...

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# Nastran Rotordynamics SOL 108 Damping Problem

Gears Phenom

Hello!

I modell rotordynamics response of shaft due to unbalance excitation, using SOL108. And when I change structural damping coefficient PARAM,G,.04 my results remain the same. I compared results with SOL111 and see that response from direct solver is corespond to zero structural damping.

What is wrong with my model/settings and how can I include structural damping in rotordyn response analysis in SOL108?

2 REPLIES 2

# Re: Nastran Rotordynamics SOL 108 Damping Problem

Siemens Phenom

Did you specify PARAM, W3?

See section 2.1.4 of the Rotor Dynamics User’s Guide

In rotor dynamic analysis the complex stiffness matrix is not used. Thus, for modal solution sequences, the imaginary part of the stiffness matrix must be converted to viscous damping. This conversion is done by dividing by the eigenfrequency.

For direct solution sequences the imaginary part of the stiffness matrix is converted by dividing by:

• Parameter W3 for damping defined by PARAM G specification.
• Parameter W4 for damping defined by GE specification.

# Re: Nastran Rotordynamics SOL 108 Damping Problem

Gears Phenom

With PARAM, W3 all working correct. Thank you!

Also i got another question: when i use freq dependent properties of CBUSH as bearing elements I get fatal message 2170 (PARTN2) add others, when my function is not horizontal i. e. i have function of two points and analysis is crashes when y coordinates of this points are not equal. If i use CBEAR elements - calculation run without problems.


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# A Integrating the topics of forms, manifolds, and algebra

Tags:
1. Apr 26, 2016

### observer1

Hello,

As you might discern from previous posts, I have been teaching myself:
1. Calculus on manifolds
2. Differential forms
3. Lie Algebra, Group
4. Push forward, pull back.

I am an engineer approaching this late in life and with a deficient background in math. It is all coming together and I almost see the picture. But it remains discrete topics - not connected with each other and without a direct connection to my deficienies.

This post concerns THE GLUE.

Could someone explain how all these ideas are integrated, in WORDS and with an acknowledgment to a person with deficient math (math learned in a country with a math education a mile wide and an inch deep).

Essentially, IN WORDS (not symbols)...
And YES, these are huge questions, but I am not interested in direct answers to each but in the continuity between the questions -- oh hell, I wish I knew exactly how to say this: my calculus was so bad I practically need someone to say: "In a Lagrangian there are generalized coordinates which become the manifold and there are vectors in the tangent plane (position) and can only be multiplied by co-vectors (velocities) and the purpose of all this is only to see the structure..." Or something like that...

Thus...

Why should I learn calculus on manifolds (ConM)?

What impact do they have in Lagrange's equation for a dynamical system?

What can ConM do that my “normal” calculus cannot?

Why do forms matter?

How do form really help do math on the tangent space?

What is the relationship between forms and vectors?

Why is it that all treatments on this also treat Lie Algebra and Groups?

Lie groups are just smooth groups, so where is the manifold.?

I know it is going to be difficult to do this – write an answer in words.

Thank you if you can. (And, yes, as evidenced by previous posts: I am looking for words... for only by WORDS can I see where I was miseducated.)

2. Apr 26, 2016

### Staff: Mentor

Your questions are difficult if not impossible to answer without writing a book!
I give it a try to answer them as short as they are.

This depends on what you want to learn at all and what you are planning to do with it.
Modern physics depend on it. It basically started with the attempt to solve differential equations which describe a whole lot of nature's behavior: from the outbreak of pandemics over the population of species to the fabric of spacetime .

(see above; beside that I recommend a physicist or biologist to answer this)

To me it is normal calculus. It may help to solve (systems of) differential equations. The prediction of recently found gravitational waves is an example. The surface of our earth is a manifold. Even your "normal" real vector world can be considered as a manifold. I guess people started to think: why restrict to a flat world if a) the world isn't and b) results can simultaneously be proven in a more general case. It is like the question: why should I calculate the volume of a sphere if I already know the volume of a cube?

It's a generalization of derivatives.

It is math on/with the tangent space.

What is the relationship between functions and values?

Lie groups are manifolds, Lie algebras are tangent spaces. E.g. they arise from the examination of symmetries, rotations and so on.
In particle physics you can't do a step without meeting the small ones of them. They are also simply useful examples: rotations in a plane are imaginable, a string world in eleven dimensions not really.

Right in front of you. A path through elements of a Lie group is a path on a manifold. It is a curvy thing.

I apologize for being sloppy on some points. As I said: you can fill a book and actually there are dozens of books dealing with this stuff. However, I admit most scientist are weak when it comes to explain why and what for. Usually they put it in exercises or simple examples and hope the results themselves explain the necessity. Beside that mathematics from a physical point of view is a big toolbox and you never know in advance which tool you need so the more the better is the motto.

3. Apr 26, 2016

### observer1

OK then.. so I know the gradient operator has different forms in different coordinate systems. I get that.

So when someone does Calculus on Manifolds (say the configuration space of a pendulum in which it is reduce not to x/y position in Cartesian,but to THETA, say in generalized), does it all come down to recognizing that derivatives are DIFFERENT in these manifold subspaces? Is that all it is?

See, I sort of get the idea of manifolds, but I don't get why people make a big deal on doing calculus on them when I have no issue about the gradient in other coordinate systems. Or are you saying that THE REASON I have no issue is that I have intrinsically ACCEPTED the idea of calculus on manifolds?

When you say: “What is the relationship between functions and values?

Could you expand with a statement and not a question?
Are you saying form is to vector as functions are to their values?
I am confused.

I just need someone to say: "when we say X, we mean it in relation to THE THING you learned poorly in Calculus in Euclidean space.

4. Apr 26, 2016

### Staff: Mentor

No, now I'm a bit confused. Maybe I didn't understand what you've meant by 'form'. I was referring to a differential form which is a function on vector(field)s generalizing (directional) derivations.

Concerning the coordinate system one has to distinguish between the object itself and its representation. A pendulum doesn't change whether you describe it in cartesian or polar coordinates. It's the equations that look different. "I'm hungry" takes a different form if I say it in my own language. However, it doesn't change my status if I do so.

Whether manifolds are a big deal or not depends mainly on your attitude towards them. Do you remember your schooldays were negative numbers became from impossible to natural just by changing the grade? Manifolds stand just for something curvy (even if the curvature vanishes). Curves are everywhere in this world. We consider tangent spaces as a local approximation of these curves because they are easier to handle and locally as good as the curve itself. It is the curve sketching from school in higher dimensions and on different surfaces, e.g. analytic groups.

5. Apr 26, 2016

### mathwonk

Riemann’s idea was to represent physics through geometry. The geometry of space is essentially a curved manifold, hence we study those objects, e.g. a sphere. To study an object one looks for its intrinsic nature and properties. A smooth geometric space is one which is locally approximable by flat spaces, i.e. its tangent spaces at each point, so it behooves us to study the tangent spaces to our manifold. This is also where the velocity vectors to curves live, hence of importance to physicists. Analyzing those tangent spaces involves understanding families of their elements, i.e. vector fields. This is the sort of thing introduced into space by forces like gravity.

Studying those involves also understanding functions on them, and the linear functions on vectors are called “forms”, and families or fields of those linear functions are called differential forms. Differential forms can be integrated and allow us thereby to measure physical and geometric quantities like angles, as when integrating the form dtheta. Functions having values on our manifold also inform us about physical quantities like temperature, and each function has a differential, which is an “exact” differential form and whose variation can be measured by its integrals along curves.

Your ordinary calculus is the basis for all this, but is usually restricted to flat or euclidean spaces, whereas the theory of manifolds expands its scope to curved ones. So calculus on manifolds expands and generalizes the methods of ordinary calculus to apply to curved spaces, preserving the intrinsic properties of derivatives and integrals, allowing us to discuss them with less dependence on specific coordinates.

The more intrinsic structure of an object we discover the more we can study it. Viewing tangent vectors via their action as differential operators, i.e. taking directional derivatives, allows us to define a sort of multiplication on vector fields, the lie bracket. We alaso enhance our knowledge of any space by studying its self transformations, and the transformations can be composed yielding a group structure. E.g. the group of matrices, or linear transformations of euclidean space, is also a smooth mnifold and admits study by calculus. morever any group allows us to translate all vectors to the origin, and hence replace vector fields by just vectors at the origin. thus the lie operation on vector fields allows us to define a lie group structure on the tangent space at the origin of a smooth manifold which is a group. so manifolds lead to lie groups and to lie algebras.

forgive my errors, they are due to ignorance; this is a big question but i tried to say something of use.

6. Apr 26, 2016

### The Bill

The calculus of differential forms lets you do "vector calculus" integrals on spaces that don't let you use the cross product, like spaces with four or more dimensions. Also, knowing the generalized Stokes' theorem plus the rules of manipulating differential forms means you never have to look up any of the integral theorems from traditional "vector calculus" again, because you can just look at the problem, write out the integral, and immediately plug in the values and start calculating it or deriving its properties.

The phase space for the Hamiltonian formulation of a system in classical mechanics with a finite number of degrees of freedom can be viewed as a symplectic manifold. The techniques of differential geometry, including what is classed as calculus on manifolds, can then be used to examine the invariants and phase portraits of the system. V.I. Arnold's Mathematical Methods of Classical Mechanics is devoted to teaching this framework for classical mechanics.

The book that taught me the "glue" between the topics you mentioned in the first post is William L. Burke's Applied Differential Geometry.

7. May 15, 2016

### Twigg

I've been studying differential geometry for 3-4 years now, and I can count on one hand the number of times it's been essential in solving a practical problem in physics or engineering. The examples that come to mind are the Foucault pendulum (a basic grasp of Lie theory allows you to transform to the rotating frame at the initial point), calculating the precession of mercury (the metric tensor showed up in the first few lines, that's it), and making sense of time evolution operators and operator algebra in quantum mechanics (another application of basic Lie theory). There are far more problems that fully-general differential geometry can be applied to than it must be applied to. My advice for problem solving is to avoid the general mathematical theories of differential geometry like the plague wherever possible. The tell-tale sign of a problem that necessitates the use of methods beyond the scope of calculus on Euclidean space is the absence of any global coordinate chart. These problems are truly rare and extremely messy. In physics, I've only encountered them in problems involving the use of non-inertial or gravitational frames or an interaction picture. In applied mathematical problems, I've also run into them in the analysis of symmetries of differential and algebraic equations and variational/optimization problems when you look at solution in a neighborhood of a point without any prior knowledge of the shape of the hypersurface that solves your problem. For example, suppose I wanted to know the radiation field produced by a thin rotating dipole antenna. I can use a Lie group created by the rotating frame of the antenna to connect a surface element of a wavefront to an inertial frame, in which I know exactly how the radiation field will evolve using Green's functions.

Noether's theorem in Lagrangian mechanics is based on Lie theory. In theory, all of variational calculus is applied Lie theory, but that doesn't make it any more powerful than it was when the Bernoullis, Euler, Fermat, Lagrange, Hamilton, and others developed the subject well before Lie's publications. I single out Noether's theorem in particular because it involves the direct use of symmetry generators to demonstrate.

I would point out that the whole point of Hamiltonian mechanics is to provide you with a Poisson bracket so you don't have to deal with the full differential geometric analysis of invariants over jet bundles. For instance, if you want to find invariants in a system in Lagrangian mechanics, you have to take a first-order partial differential operator on the tangent bundle of position space (not a vector field in the tangent bundle, a vector field ON the whole tangent bundle). This requires some algebraic heavy lifting, as their are intricate expressions constraining differential operators on the tangent bundle obtained by prolonging transformations on position space. The Poisson bracket is a convenient short-cut to the analysis of invariants that really depends on nothing but the chain rule on phase space and Hamilton's equations. The development of the fully general Poisson bracket is demonstrated without reference to differential geometry (except for a passing comment in retrospect about local contact transformations at the end) in the second section of Dirac's Lectures on Quantum Mechanics.

As stated above, the general theory of differential forms isn't necessary until you start talking about a hypersurface that doesn't have any global coordinate charts, which forces you to state your problem locally (i.e., your problem becomes a family of subproblems each restricted to the domain of a certain local coordinate chart). In the problem I stated above about the thin rotating dipole antenna, the "subproblems" are the radiation field patterns around each surface element of the wavefront in question. You know the shape of the wavefronts of the radiation pattern in the inertial frame from integrating the Green's functions over the length of the antenna. Surface elements of the wavefront (which themselves are very simplistic differential 2-forms) can be "pulled back" to more complicated locally defined differential forms in the rotating frame via the inverse of the rotation transformation that took us to the inertial frame in the first place. That is a case where the full theory of differential forms would mean something more complicated than you'd see in ordinary calculus on Euclidean space, because you're working in more than three dimensions on a geometric shape you pretend to know nothing about except that it's locally related to the inertial frame by a transformation that varies from point to point.

Forms do math on the tangent space in the same way the a surface element acts on an electric field in Gauss's law. They enable you to evaluate fluxes through hypersurfaces by integrating vector fields and higher dimensional "geometric flux densities." In fancy lingo, these flux densities are contravariant tensor densities, such as the electromagnetic stress-energy tensor in relativistic electrodynamics ($T^{\mu\nu}$), which is like a 4-dimensional generalization of the Maxwell stress tensor in 3-dimensional electrodynamics, or the Faraday electromagnetic field tensor ($F^{\mu\nu}$), which is like the 4-dimensional analogue of the electric flux density ($\vec{D}$) and magnetic flux density ($\vec{B}$) combined. If the basis differential forms are n-1 dimensional surface elements and parallelotopes formed with parallel families of surface elements for edges ($dx \wedge dy$ is like taking the flux of a field through level sets of x followed by taking the flux through level sets of y on the boundary of an infinitesimal rectangle with edge of length |dx| and |dy|), then the action of these "generalized oriented differential parallelotopes" on a same-dimensional flux density is the flux of said field around the parallelotope.

If vector fields represent slope fields on your space, then 1-forms represent a field of infinitesimal flat hyperplanes that appear locally parallel to each other and everywhere normal to the vector field to which they are dual. If you happen to have a metric tensor (which is guaranteed on smooth manifolds of finite dimension), then duality happens to coincide with index raising/lowering operations because of the wonders of doing linear algebra in an inner product space. Tensorial acrobatics aside, the relationship between forms and vectors is that of orthogonal complement (which as linear algebra operation in flat space would be expressed as the transpose). This is why surface integrals in 3D Euclidean space can be represented as a magnitude times a normal vector.

Lie algebras and Lie groups enjoy a prominent role in differential geometric theory because on any space, curved or flat, (of finite dimension, as far as I know), every vector field is an infinitesimal transformation. More formally, the set of all vector fields in the tangent bundle of any smooth manifold (of finite dimension) belongs to an infinite dimensional Lie algebra. The corresponding one-parameter Lie group associated with a given vector field is the group of translations along integral curves of that field. It turns out you can't do or prove much of anything without invoking this particular Lie algebra. The basic reason for this is that this Lie group, of translations along curves, is what connects geometric points to their neighboring points without leaving the manifold; it represents motion restricted to the manifold, at least locally. How would you even begin to talk about tangent spaces, let alone metrics, if you couldn't connect points? (If you have a local transformation group connecting points in a neighborhood, then the differential, or "push-forward" of that transformation connects tangent planes of these points).

Smoothness doesn't mean anything if it isn't on a manifold. Globally-defined Lie groups (as opposed to local Lie groups, which are much less common in modern treatments but what you're most likely to come across in a problem that needs Lie theory to solve) are usually written in terms of parameters. For example, SO(2) is written in terms of the parameter $\theta$ which identifies each element of SO(2) with an element of the unit circle and gives SO(2) the same smooth structure as the unit circle. SO(3) has the same smooth structure as the real projective space $\mathbb{RP}^{3}$. SU(2) has the same structure as the 3-sphere. In other words, if I pick a point on the 3-sphere (any point), I also pick a transformation in SU(2). Any calculus-ish statement about one can be taken as an identical statement about the other. The operation of group multiplication, the differentials of multiplication by elements, the Lie algebra, and anything else about the structure of SU(2) can be taken to mean something geometric on the 3-sphere.

I apologize for the lateness of the post. I put this up anyways because I felt this could be helpful to the OP and many others studying the field.

Last edited: May 15, 2016
8. May 15, 2016

### lavinia

Ordinary calculus in Euclidean space is calculus on manifolds, this because Euclidean space is a manifold. Euclidean space has different coordinate charts and coordinate transformations just as other manifolds.

Classical vector calculus uses the Euclidean metric in 3 dimensional space to translate differential forms into vectors. Without this metric (or some other metric), there would be no gradients or curls. There would only be 1 forms and 2 forms. The moral is: Calculus does not require a metric and a big part of learning calculus on manifolds is just to relearn ordinary vector calculus without using a dot product.

Perhaps the key idea that separates classical vector calculus from calculus on manifolds is not the calculus, but rather the manifold - as Mathwonk has explained above. In general, manifolds cannot be covered by a single coordinate chart as can Euclidean space and this is the big difference. Because of this vector fields and differential forms and tensors may not be describable in terms of a global basis the way they can in Euclidean space. This allows for many physical and mathematical phenomena that do not occur in a single coordinate chart. That by itself suffices as a reason to learn how to do calculus on them.

Last edited: May 15, 2016
9. May 15, 2016

### observer1

Thank you! A lot!
It is your simpler (not an insult here) that is what I was looking for...

"So calculus on manifolds expands and generalizes the methods of ordinary calculus to apply to curved spaces, preserving the intrinsic properties"

and

"Viewing tangent vectors via their action as differential operators, i.e. taking directional derivatives, allows us to define a sort of multiplication on vector fields, the lie bracket."

Those words helped me!

10. May 15, 2016

### lavinia

Calculus on manifolds does allow one to study curved spaces from an intrinsic viewpoint - that is, from inside the manifold without referring to an ambient Euclidean space. But a space is curved when it has a metric. A metric is an additional structure that is not needed in order to just do calculus on the manifold. One can differentiate and integrate without a metric.

There is an entire subject called Differential Topology that studies manifolds using only calculus. When one studies a curved space i.e. a manifold with a metric, the subject is called Differential Geometry. However the same differentiable manifold i.e. a manifold on which one can do calculus, can have infinitely many different metrics each giving a different curved shape to the same differentiable manifold.

Last edited: May 15, 2016


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# Homework Help: Particular Solutions of Undetermined Coefficients

1. Oct 17, 2011

### osheari1

1. The problem statement, all variables and given/known data

Find a particular solution y_c to the complex differential equation
y'' + 2 y' + 3 y = -1 exp(2 i t)

y_c=

Using the solution y_c, construct a particular solution y_{1p} to the following differential equation
y'' + 2 y' + 3 y = -1 sin(2 t)
y_{1p} =

Again using the solution y_c, construct a particular solution y_{2p} to the following differential equation
y'' + 2 y' + 3 y = -1 cos(2 t)
y_{2p} =

2. Relevant equations

y = Ae^(2it)

3. The attempt at a solution

I let y'' + 2y' + 3y = 0 and then solve getting
e^(-t)(cos(sqrt2t) + sin(sqrt2t)) + g(h)

then solve for the g(h) by first finding A

y = Ae^(2it), y' = A2ie^(2it), y'' = -A4e^(2it)

substitute and solve for A getting -1/(4i - 1)

→ y(t) = e^(-t)(cos(sqrt2t) + sin(sqrt2t)) -1/(4i-1)e^(2it)
the Constants of integration can be anything for this problem so i just let them equal 1

but this is still not the right answer

2. Oct 18, 2011

### Staff: Mentor

Is the right answer yp = (1/17)(1 + 4i)e2it? If so, all I did was to convert -1/(4i - 1) by multiplying by the complex conjugate of the denominator over itself.

BTW, when you write cos(sqrt2t) + sin(sqrt2t), I can't tell if you mean
$\sqrt{2t}$ or $\sqrt{2}t$.

If you don't use LaTeX, write it like this: sqrt(2)t or t*sqrt(2).


text: You could use this sheet of circles marked in $12$s and this one that has circles marked in $10$s, $9$s, $8$s and $7$s.