# Fundamental confusions of calculus

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Tach said:
It is not that I disagree with Penrose, it is that you misinterpret what he's writing.
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.

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You mean like this error-filled post?
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.

Actually, this video makes it clear that what you keep calling partial derivative is in reality the total derivative. So does eq (2) from this textbook. This is really simple stuff, I do not know why you are fighting so hard in admitting that you goofed.
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.

What is most puzzling is that you got this one right in the beginning.
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.

If you insist in ignoring the definition of f as a function of u and v and you keep insisting in making up your own definition, then, yes, your partial derivative is correct.
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 $$x$$.

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.

I didn't say "the book", I pointed out your mistakes very precisely in post 80. It is very simple, really, you clearly do not know the difference between partial and total derivative.

You are calculating the total derivative but you claim it to be the partial derivative. Same mistake as przyk but he's cleverer about it, he's tried (repeatedly) to change the definition of the function f.
So now you're claiming the definition of the partial derivative is wrong? Wow.

Resorting to personal attacks does not further your point, nor is it appropriate given your role as a moderator. You have made a simple math mistake, own it and be done with it.
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.

Yes, quite a bit. I can post some of the work that I've done but what does it have to do with your basic confusion between the definition of partial vs. total derivatives?
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?

Personal attacks are unworthy for you given that you are a moderator. It is very simple, really:
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.

Your error is not really huge, you are mixing the definitions , the video will help you figuring out the difference. I track your error to the fact that you aren't considering the fact that f is a function of $$\theta$$ and $$u(\theta)$$ but rather you lump all the expressions in $$\theta$$ together. przyk is doing the same exact error.
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.

But when you said $$\frac {\partial} {\partial x}$$ multiplied by a scalar is not a vector,

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

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

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.

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.

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.

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,

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: $$\, \frac{\partial f}{\partial v} \, \frac{\mathrm{d}v}{\mathrm{d}\theta} \,.$$

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

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.

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.

[Irrelevant content snipped]

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:
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.

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:

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

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

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

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.

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.

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.

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.

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.

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?

No, I am not. Pete, it is very simple, assume that the temperature is a function of the altitude of the observer above the plane containing the bug, z is NOT the coordinate of the bug, actually it has NOTHING to do with the bug and its motion. For example, the temperature drops exponentially with the increase in altitude.

Try this (I know I am going to regret giving you an example):

$$T(t,x,y,z)=e^{-z}(at+bx+cy)$$
a,b,c are some constants
$$x=f(t)$$
$$y=g(t)$$

f,g are functions differentiable (at least) in the first order

The (x,y) coordinate and its dependence of t is given. The temperature of the bug does not depend on z, the temperature T of the medium is what depends on z.

Really? $$x = 3$$, $$y = 2$$, $$t = 5$$, $$a = b = c = 1$$ and $$f(t) = g(t) = \sin(t/\pi)^{2}$$. What's the bug's temperature?
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?

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przyk's numbers don't seem to gel

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.

$$t = 0 \\ x = f(t)= \cos^2t= 0 \\$$

Huh? Et tu, Brutus?

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

Tach said:
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.

An important example of a function of several variables is the case of a scalar-valued function f(x1,...xn) on a domain in Euclidean space R[sup]n[/sup] (e.g., on R[sup]2[/sup] or R[sup]3[/sup]). In this case f has a partial derivative ∂f/∂x[sub]j[/sub] with respect to each variable x[sub]j[/sub]. At the point a, these partial derivatives define the vector.
--http://en.wikipedia.org/wiki/Partial_derivatives

Whoops!
That should be x = sin^2t.

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

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.
*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.

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: $$\, \frac{\partial f}{\partial v} \, \frac{\mathrm{d}v}{\mathrm{d}\theta} \,.$$
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.

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