# Fundamental confusions of calculus

Discussion in 'Physics & Math' started by arfa brane, Feb 11, 2012.

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1. ### arfa branecall me arfValued Senior Member

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But when you said $\frac {\partial} {\partial x}$ multiplied by a scalar is not a vector, you said it was "nonsense", you weren't misinterpreting what Penrose was saying? Because that's what he says in his book.

Actually what he says is the scalar should be a function of x (or whatever variable is being varied). But as others have noted that's a mere mathematical detail because a vector multiplied by any scalar is still a vector. That's something I learned in a linear algebra course.

Last edited: Feb 13, 2012

3. ### przyksquishyValued Senior Member

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No, I mean your attitude (talking down to others even when it was your fault you hadn't defined your example properly), and generally misrepresenting and revising history.

Also, asserting errors without being able to point them out (in an unrelated post at that), like you just did here.

I've already explained to you that in both those examples, the total derivative is taken through all the function's parameters. The analogue in your case would be to do something like
$\frac{\mathrm{d}f}{\mathrm{d}\theta} \,=\, \frac{\partial f}{\partial \theta} \,+\, \frac{\partial f}{\partial u} \, \frac{\mathrm{d}u}{\mathrm{d}\theta} \,+\, \frac{\partial f}{\partial v} \, \frac{\mathrm{d}v}{\mathrm{d}\theta} \,.$​
But since you've already said you're not letting $v$ be a function of $\theta$, you're not doing the same thing as in those examples, and they therefore don't support your case.

I said nothing in my original post in this thread that I don't still stand by. If you think I'm getting something "wrong" now that I got "right" before, then either you didn't understand my original post, or you don't understand what I'm saying now, or both.

I wasn't ignoring anything. I'd answered your question. Then I asked you two similar questions. The first was just the partial derivative of an isolated expression. There was simply no $f$, $u$, or $v$, and it was irrelevant how you might have defined $f$ elsewhere. In the second - a problem I gave you remember - there was an $f$, but no $u$ or $v$. Just $\theta$ and $x$.

5. ### TrippyALEA IACTA ESTStaff Member

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Tach, a question, if I may.

How would you apply the chain rule to the following problem:

f(x) = a.sin[sup]2[/sup](bx+c)

I'm not concerned with seeing you work through the problem as a whole at this stage, I would simply like to know how you would define y, and define u.

7. ### AlphaNumericFully ionizedRegistered Senior Member

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So now you're claiming the definition of the partial derivative is wrong? Wow.

The impression I get is my opinion and I'm drawing an analogy between your behaviour and that of other members. Neither of them constitute personal attacks on you.

I question that because if you had working familiarity with Hamiltonian/Lagrangian mechanics you'd know that it contradicts your claims. I gave the pendulum example and it explicitly involves a sin term. You have yet to counter it. Much of the Hamiltonian mechanics can be formalised in terms of differential geometry and the whole tangent/cotangent spaces, which pertain to the $\partial_{x}$ vector space basis elements originally asked about. You didn't grasp that either. All of these things fit together into a nice coherent structure in line with what everyone is telling you.

But why don't you enlighten me as to your experience with Hamiltonian mechanics. Can you tell me the canonical momentum for $T = \frac{1}{2}m\dot{q}^{2}$? Do you need me to give you the definition or can you manage it yourself?

I hardly think saying "You're pulling a Reiku" counts as much of an attack. Besides, you are not exactly on the moral high ground when it comes to attitude and behaviour. I find it somewhat funny that you have no problem being extremely abrasive to others but when someone gets anywhere close to not complementing you you cry foul.

Considering how abrasive you have been in the past if you think my comments shouldn't be allowed then you are being a hypocrite. And no, this isn't an insult, it is a statement of fact in that you try to hold others to standards you do not hold yourself to. Not being chummy is not the same as being rude. A certain amount of opinion and commentary is allowed in discussions. People, including moderators, can say "Come on, you're being a bit thick there" or the like, provided it's infrequent and they go on to justify why they are saying it. If you cannot handle a bit of discussion which isn't all hugs and kisses perhaps the internet isn't for you.

I don't feel you're even reading half of what people say. I discussed this, the difference between explicit and implicit parametrisations and the way they are related to the different derivatives. Rather than discuss that you try to be patronising (which is all the more hypocritical given you complained I was supposedly talking down to you) by repeating (as if people haven't seen it already) the same expression again and again.

I see this thread is going the way of the mirrored wheel one. Regardless of whether you consider it 'unworthy' of me to say or not I'll point out that perhaps it's a bad sign when your significant participation in a thread leads to similar 'discussions' as when Motor Daddy starts talking about relativity or Farsight about electromagnetism.

You've been asked a few questions by myself and Trippy and others. If by morning you're still dancing around crying "The book, the book!" then this thread can be relocked and those of us who actually do this stuff for a living can get back to it.

8. ### TachBannedBanned

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It isn't , but what does this older error of yours have to do with the partial vs. total derivative we are discussing?

9. ### TachBannedBanned

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No, I am simply pointing out that your understanding of the issue is flawed. Twice I showed you the mathematical steps that outline your error.

10. ### TachBannedBanned

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So, why aren't you responding to the errors that I pointed out in this post? That post wasn't even calculus, it was basic algebra.

Which is precisely what I did for Pete's benefit. u is chosen as a function of $\theta$, v is chosen as afunction of $x$ resulting into exactly what I posted in post 18:

$\frac{\mathrm{d}f}{\mathrm{d}\theta} \,=\, \frac{\partial f}{\partial \theta} \,+\, \frac{\partial f}{\partial u} \, \frac{\mathrm{d}u}{\mathrm{d}\theta} \,$​

Before you and the other jumped in with your ideas about partial vs. total derivatives, I was trying to explain to Pete the reason for the absence of the term: $\, \frac{\partial f}{\partial v} \, \frac{\mathrm{d}v}{\mathrm{d}\theta} \,.$

Last edited: Feb 14, 2012
11. ### arfa branecall me arfValued Senior Member

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If it isn't a vector you should be able to revisit Quarkhead's post and explain his mistake to him (and everyone else reading it).

What does a partial derivative like $\frac {\partial} {\partial x}$ of no function have to do with the discussion? I'll let you work it out.

12. ### TachBannedBanned

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5,265
Based on how you phrased the question, it seems that you need an explanation about what is being debated.
Say that you have two definitions of the function f:

1. $f_1(x)=3x+a sin^2(bx+c)$

and :

2. $f_2(x,u)=3x+u$
where $u=a sin^2(bx+c)$

then, the partial derivatives wrt x of the two functions are DIFFERENT (while the total derivatives are the SAME). Why is this?

$\frac{ \partial f_1}{\partial x}=3+ab sin(2(bx+c))$

$\frac{ \partial f_1}{\partial x}=3$

Finally:

$\frac{d f_1}{dx}=\frac{d f_2}{dx}$

There is video that Pete linked in that explains all this very well. There is a page from a very good calculus book that I linked in that explains this very well. Take your choice.

13. ### TrippyALEA IACTA ESTStaff Member

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None of that answered my question, Tach. I don't need an explanation of what is being debated. If I have questions, I will ask them.

I asked you a straightforward question.

Here it is again:

14. ### TachBannedBanned

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You mean that you did not recognise $\frac{ df}{d x}=ab sin(2(bx+c))$ ?

15. ### TrippyALEA IACTA ESTStaff Member

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Yes, I recognized it, but I didn't ask you to complete the derivation, did I?

I simply asked you to define the first step for me - nothing more.

Using the generalized equation I provided you, how would you define u and y to arrive at dy/dx using the chain rule?

That's all I'm interested in at this stage.

16. ### TachBannedBanned

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You see, you still fail to understand that $\lambda \frac {\partial} {\partial x}$ is not a vector because, contrary to your beliefs it doesn't have either sense, nor direction. Now, if , instead of your misgiuded attempt at multiplying $\frac {\partial} {\partial x}$ by a scalar, you tried (as most introductory books show) by a vector, then , you would get a vector, as in:

$\vec{e_x} \frac {\partial} {\partial x}$

The above (and not $\frac {\partial} {\partial x}$ as in your misconceptions) indeed points along the x-axis.

You can even have:

$\vec{e_z} \frac {\partial} {\partial x}$

as a vector. Comes up in the definition of the curl, for example.

Now, that I explained this to you, please let us return to the discussion of partial vs. total derivatives. Thank you.

17. ### TachBannedBanned

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Why are you so interested in this? It is obvious that there is a virtually infinite number of ways of doing that, here are two:

$y=au^2$
$u=sin(bx+c)$

$y=au$
$u=sin^2(bx+c)$

You can even go:

$y=au$
$u=sin^2(v)$
$v=bx+c$

Now that I have answered your questions, what are you attempting to accomplish with your post? Where are you going with it?

18. ### PeteIt's not rocket surgeryRegistered Senior Member

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przyk's numbers don't seem to gel. Try this instead:

\begin{align} f(t) &= \sin^2t \\ g(t) &= \cos^2t \\ a &= b = c = 1 \\ t &= 0 \\ x &= f(t) \\ &= \sin^2t \\ &= 0 \\ y &= g(t) \\ &= \cos^2t \\ &= 1 \end{align}​

What's the temperature of the bug at t=0?

Last edited: Feb 14, 2012
19. ### TachBannedBanned

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It isn't only the numbers that "don't gel", it is the whole approach that is pure BS. But he will not own to it.

Huh? Et tu, Brutus?

20. ### PeteIt's not rocket surgeryRegistered Senior Member

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Whoops!
That should be x = sin^2t.
It's possible that several posters in this thread are thinking exactly that about you.

21. ### arfa branecall me arfValued Senior Member

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--http://en.wikipedia.org/wiki/Partial_derivatives

22. ### TachBannedBanned

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5,265
What does this have to do with my suggestion for you to calculate the total derivative wrt t of the function $T=e^{-z}(....)$?

23. ### przyksquishyValued Senior Member

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*sigh*

You're only delaying the inevitable, you know. $t = 5$, $a = b = c = 1$ and $f(t) = g(t) = \sin(\pi t)^{2}$. What's the bug's temperature?

If I happen to have made any other oversights, due to not thinking this deserves more than about five seconds of my time, I trust you're intelligent enough to figure out the point I was making.

And that may well be the appropriate thing to do. But if you leave $v$ and/or $x$ independent of $\theta$ you're wrong to call that a total derivative.