# Fundamental confusions of calculus

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There's one other major sticking point between Tach and I that I'd really like confirmed:
If x and $$\theta$$ are independent variables...
$$f(\theta, x) =3 \theta+ sin^2(\theta)+ln(x) \\ \frac{d}{d \theta}f(\theta, x) \mbox{ is meaningless.}$$​

Your claim that $$\frac{d}{d \theta}f(\theta, x)$$ "is meaningless" is not only wrong but borders on pettiness.

I've been led to believe by the aforementioned video, and also by [post=2901729]temur[/post], that the total derivative dx/dt is not meaningful for a function like x = f(y, t) without some expression (even a hypothetical one) relating y and t.

Have I been misled?

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This is why you need to read post 18 before jumping in and claiming that I made an error.
You did make a mistake. The partial derivative expression will have the sin term in there too. If x is independent of theta then in this case the partial and total derivatives will coincide.

Tach;2902967I can assure you that I am not confused at all. The reference I gave lines up with my posts and disproves yours (and the others) claims. [/quote said:
Except it doesn't.

The partial derivative is, qualitatively, "Treat everything else as fixed and just do a normal derivative on this variable". A total derivative is, qualitatively, "Consider all unknowns as functions of this variable and do the chain rule on each one of them". If you explicitly know the function, ie $$u(x) = \sin^{2}x$$ then you can write down it's x derivative as a function of x manifestly. Your reference is about reparametrisations and how to deal with them. I don't deny the equation, I deny your view of it.

Please stop talking down to me. Especially in the context of my having gone out of the way and having explained the above to Pete from base principle.
You haven't explained it from base principle. In fact I've asked you more than once to consider the definition of the partial derivative, which you seem unwilling to do. I'll do it for you since you seem so unwilling to do it from 'base principle'.

$$f(x,y) = 3x+ \sin^{2}x + \ln y$$
$$\partial_{x}f(X,Y) \equiv \lim_{h \to 0} \frac{f(X+h,Y)-f(X,Y)}{h}$$
$$f(X+h,Y) = 3(X+h) + \sin^{2}(X+h) + \ln Y = 3X + 3h + \ln Y + (\sin X \cos h + \cos X \sin h)^{2} = 3X + 3h + \ln Y + (\sin X \cos h + \cos X \sin h)^{2}$$
Therefore
$$f(X+h,Y)-f(X,Y) = 3h + (\sin X \cos h + \cos X \sin h)^{2} - \sin^{2}X$$
Using $$\sin h \sim h$$ and $$\cos h \sim 1$$ we have
$$f(X+h,Y)-f(X,Y) = 3h + (\sin X + h\cos X)^{2} - \sin^{2}X = 3h + 2h\sin X \cos X h^{2}\cos^{2} X$$
Therefore
$$\partial_{x}f(X,Y) = 3 + 2\sin X \cos X = 3+\sin 2X$$

As stated many times. The equation you keep quoting is what happens if you were to turn around and say "But I know that $$y = y(x) = e^{x}$$!" (for instance), "What if I'd put that in to start with?". Well you can either compute $$f(x,e^{x})$$ and then do $$\partial_{x}f$$ or you can just compute $$d_{x}f(x,y)$$ and then put in the relevant expression for $$\partial_{x}y$$ which appears in the formula. Let's see this in action,

$$g(x,y) = 3x + y$$
$$d_{x}g(X,Y) = 3 + \frac{\partial y}{\partial x}|_{x=X}$$
If we know afterwards that $$y(x) = \sin^{2}x$$ then $$\partial_{x}y = \sin 2x$$ and so $$d_{x}g(X,Y) = 3 + \sin 2X$$. But what if we put it in before hand, $$g(x,\sin^{2}x) \equiv h(x)$$ and so $$\partial_{x}g = \partial_{x}h = \partial_{x}(3x+\sin^{2}x)|_{x=X} = 3+\sin 2X$$

That's how the total derivative comes into it. It's a sort of apriori/posteri substitution. The partial derivative when you're working in terms of functions of x (or functions of theta as was in the original example) is the same as if you changed parametrisations, didn't know the parametrisation, did a total derivative and then converted back. By altering the arguments in the way you are you're confusing yourself. However, as I've just illustrated, when put in the proper context the equation you keep pointing at is entirely consistent with the definition, which you misstated, and in complete agreement with what we have been telling you.

Any disagreement should explain why the definition is wrong, because now I've explicitly done it for you, you have no excuse to ignore that. You should also address how your method contradicts all of Lagrangian/Hamiltonian mechanics, another point you keep skipping.

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What in $$f=f(\theta,u,v)$$ is that you fail to see?
Its presence. Because it wasn't there in your example. You never mentioned $$u$$ or $$v$$ or how $$f$$ was supposed to depend on them.

Or is not clear enough to you that $$sin(\theta)$$ is not a variable but a function?
It's neither a variable nor a function. $$\theta$$ is a variable. $$\sin \,:\, \mathbb{R} \to [-1,\, 1]$$ is a function. $$\sin(\theta)$$ is an expression given in terms of a function and a variable, that was part of the larger expression $$3 \theta \,+\, \sin(\theta)^{2} \,+\, \log(x)$$ that you used to define $$f$$.

It was not clear enough that $$u=u(\theta)=sin(\theta)$$ and $$v=v(x)=ln(x)$$?
No, and there's no standard definition or convention that makes it clear.

Either way , your claim that $$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$$ is false since the partial derivative is always taken wrt the direct variable (in this case, $$\theta$$) and not through a function (in this case $$\sin(\theta)$$) .
No, you just made that condition up, and it isn't even well defined. As Guest pointed out, any part of that expression could be considered a function. The mapping $$\theta \mapsto 3\theta$$ is a function for instance. So is $$\theta \mapsto 3\theta \,+\, \sin(\theta)^{2}$$, for that matter.

For the record, here's how Mathematical Analysis (second edition) by Tom Apostol defines partial derivation, on page 115:
Let $$S$$ be an open set in Euclidean space $$\mathbf{R}^{n}$$, and let $$f \,:\, S \to \mathbf{R}$$ be a real-valued function defined on $$S$$. If $$\mathrm{\mathbf{x}} \,=\, (x_{1},\, \ldots ,\, x_{n})$$ and $$\mathbf{c} \,=\, (c_{1},\, \ldots,\, c_{n})$$ are two points of $$S$$ having corresponding coordinates equal except for the kth, that is, if $$x_{i} \,=\, c_{i}$$ for $$i \,\neq\, k$$ and if $$x_{k} \,\neq\, c_{k}$$, then we can consider the limit
$$\lim_{x_{k} \to c_{k}} \, \frac{f(\mathbf{x}) \,-\, f(\mathbf{c})}{x_{k} \,-\, c_{k}} \,.$$​
When this limit exists, it is called the partial derivative of f with respect to the kth coordinate and is denoted by
$$D_{k}f(\mathbf{c}) ,\qquad f_{k}(\mathbf{c}) ,\qquad \frac{\partial f}{\partial x_{k}}(\mathbf{c}) \,,$$​
or by a similar expression.
This makes the partial derivatives completely independent of how you chose to write the definition of the function f. It doesn't matter if you gave the definition of f in terms of other functions.

The moment you start using chain differentiation you cease calculating partial differentials and you are starting to calculate total differentials.
This is also false. For example, for the functions
$$f \,:\, \mathbb{R}^{n} \to \mathbb{R} \,:\, x \mapsto f(x) \,, \\ g \,:\, \mathbb{R}^{m} \to \mathbb{R}^{n} \,:\, y \mapsto g(y) \,,$$​
the chain rule for the partial derivatives of $$h = f \circ g$$ is
$$\frac{\partial h}{\partial y^{k}} \,=\, \sum_{j=1}^{n} \frac{\partial f}{\partial x^{j}} \, \frac{\partial g^{j}}{\partial y^{k}} \,.$$​
The chain rule does not automatically mean you're calculating a total derivative. For the example above that's only the special case where $$m = 1$$.

You did make a mistake. The partial derivative expression will have the sin term in there too.

The textbook says clearly that it will not, because f is a function of $$\theta, u,v$$.

The term in sin comes in only in the total derivative through $$u$$.

I really need to check my understanding, so please be patient with the long post.
Er, nope. As written, $$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$$.
Should that be:
$$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin^2\theta \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, 2\sin\theta\cos\theta$$​
Or am I missing something?
D'oh!

$$\sin 2\theta = 2\sin\theta\cos\theta$$

Problem solved.

Anyone want to take a break from Tach and look at the rest of post 37 for me?

Its presence. Because it wasn't there in your example. You never mentioned $$u$$ or $$v$$ or how $$f$$ was supposed to depend on them.

They are both present, in the post that shows both the theory and the example. See post 18.

Watching this vid helped a lot.

This is an excellent video, it tells you exactly what I have been telling you. Using the exact terminology in the video:

$$\frac{dT}{dt}$$ is the total derivative

$$\frac{\partial T}{\partial t}$$ is the partial derivative

The relationship is:

$$\frac{dT}{dt}=\frac{\partial T}{\partial t} + (\frac{\partial T}{\partial x} \frac{dx}{dt}+\frac{\partial T}{\partial y} \frac{dy}{dt})$$

Now, you can replace $$t$$ with $$\theta$$, $$x$$ with $$u$$ and you will be getting what I have been telling you all along. Pay special attention to the definition of the partial derivative.

Now, post 18 tells you the same exact thing, with a twist. If y is not a function of t then:

$$\frac{dT}{dt}=\frac{\partial T}{\partial t} + \frac{\partial T}{\partial x} \frac{dx}{dt}$$

This is the part that you seem to have trouble with because you are stuck on the idea that the above is somewhat connected to your claim that "$$\frac{dy}{dt}$$ does not make sense".

This is an excellent video, it tells you exactly what I have been telling you.
No, it's a different situation.
The function he's describing is T = f(x, y, t), where x = g(t) and y=h(t).
Note that all the function parameters other than t are dependent on t.
There is no variable independent of t.

Note what he says at 4:25
If you just told me, "big T [temperature] is a function of x, y, and t [time]", and then you said "calculate ordinary d of temperature with respect to time", I'd say "that doesn't make any sense. This isn't a one-variable function. You must mean a partial derivative.
In other words, dT/dt is not meaningful unless all the function parameters are either t or depend on t.
It only becomes meaningful when:
But then you say "Oh wait, I forgot to tell you x and y are functions of t as well."

No, it's a different situation.
The function he's describing is T = f(x, y, t), where x = g(t) and y=h(t).
Note that all the function parameters other than t are dependent on t.
There is no variable independent of t.

I told you that THIS is exactly your problem. Once you manage to get past this hangup, I think that you will understand the rest. There is NOTHING stopping you to calculate the total derivative wrt t if y is not a function of t, you are misinterpreting his words. What he's telling you is that T is a :

-direct function of t

-indirect function of t (through x and y) . It is not mandatory that BOTH x and y depend on t.

Case and point:

$$T=T(t,x,y,z)$$

but the bug moves only in x and y, so

$$\frac{dT}{dt}=\frac{\partial T}{\partial t} + (\frac{\partial T}{\partial x} \frac{dx}{dt}+\frac{\partial T}{\partial y} \frac{dy}{dt})$$

There is no term in z, THOUGH T depends on z.

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Is anyone other than Tach able to help me out with the questions in post 41 and post 37 (only after the second quote)?

I feel they've been lost in the noise.

They are both present, in the post that shows both the theory and the example. See post 18.
Post 18 is here. This is how you define your example:
Example:

$$f=3 \theta+ sin^2(\theta)+ln(x)$$

$$\frac{df}{d \theta}=3+2 sin(\theta) cos (\theta)$$

$$\frac{\partial f}{\partial \theta}=3$$
You do not say here that $$f$$ has parameters $$u$$ and $$v$$. The reader is left to infer that from earlier in the same post. You also do not state how $$f$$ depends on $$u$$ and $$v$$. The reader is left to guess that too.

Incidentally during the five years I spent studying physics at university, including calculus intensive courses like thermodynamics and analytical mechanics, I never saw anyone use the notation
$$f=f(\theta, u(\theta),v,w,...)$$
to indicate a function's parameter dependence.

Post 18 is here. This is how you define your example:

You do not say here that $$f$$ has parameters $$u$$ and $$v$$.

You need to understand that Pete and I have we been going over this for about 50 posts PRIOR to post 18.

The reader is left to infer that from earlier in the same post. You also do not state how $$f$$ depends on $$u$$ and $$v$$. The reader is left to guess that too.

Isn't this clear enough?

Tach said:
In math terms: if $$f=f(\theta, u(\theta),v,w,...)$$

Incidentally during the five years I spent studying physics at university, including calculus intensive courses like thermodynamics and analytical mechanics, I never saw anyone use the notation

to indicate a function's parameter dependence.

This is your problem. So, it all boils down that you did not read the post. Paradoxically, you seemed to have understood it first time.

Incidentally during the five years I spent studying physics at university, including calculus intensive courses like thermodynamics and analytical mechanics, I never saw anyone use the notation

to indicate a function's parameter dependence.
I can see why. I've used that notation myself (don't know if I made it up, or copied it from Tach), and it introduces confusion. I did it in post 37 in this thread. Perhaps that's why no one's answered it. Should I go back and fix it? (edit - done)

What do you do if you have used f(...), g(...), h(...), and you want to describe another function? Just keep climbing through the alphabet, risking ambiguities?

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Isn't this clear enough?
No.
Like przyk said, you haven't described u as a function of theta, or f as a function of u.

Perhaps you meant this, which I believe would be correct:
\begin{align} u &= sin^2\theta \\ f &= 3 \theta + u + ln(x) \\ \frac{\partial f}{\partial \theta} &= 3 \end{align}​
or this (also correct):
\begin{align} u &= sin\theta \\ f &= 3 \theta + u^2 + ln(x) \\ \frac{\partial f}{\partial \theta} &= 3 \end{align}​

...but note that this is also correct:
\begin{align} u &= 3\theta \\ f &= u + sin^2\theta + ln(x) \\ \frac{\partial f}{\partial \theta} &= \sin2\theta \end{align}​

And, as you've been told several times, this is also correct:
\begin{align} u &= 0 \\ f &= u + 3\theta + sin^2\theta + ln(x) \\ \frac{\partial f}{\partial \theta} &= 3 + \sin2\theta \end{align}​

The textbook says clearly that it will not, because f is a function of $$\theta, u,v$$.

The term in sin comes in only in the total derivative through $$u$$.
Tach, I explicitly worked through the definition of the partial derivative for the example you'd given. It came out to include the sin term. I explicitly asked you to address this. I also explicitly asked you to address how your logic completely disagrees with all of Lagrangian and Hamiltonian mechanics. I also gave an explicit example for that.

You are not actually engaging in a discussion here, you're just pointing at the book and saying "Look, look!". Yes, we've looked at it. And several of us have explained to you its context and application, ie the implicit/explicit parametrisation. Some of us use Hamiltonian mechanics a lot. And the distinction between $$\frac{dL}{dt}$$ and $$\frac{\partial L}{\partial t}$$ where $$L = L(q_{i},\dot{q}_{j},t)$$ and $$q_{i} = q_{i}(t)$$ is a very important one. Another example, which przyk brought up, is statistical mechanics, where there's partial and total derivatives everywhere. This isn't some result no one here has any experience with, it's something used a lot.

When you just point at a book and say "Look, look, there's an equation!" you're giving the impression your experience with this stuff isn't on a working level. Of course that isn't a crime but it does call into question whether or not you understand the application of the result. No one is denying the equation, the problem is your understanding of it.

Please address the explicit examples I gave, namely the definition of the partial derivative and the role of the partial derivative in Hamiltonian mechanics. If I have to ask a third time I'm giving you a warning for trolling because avoiding addressing the definition of a mathematical construct is trolling.

Isn't this clear enough?
Absolutely not.

No. Your use of non standard definitions and conventions, and the problems that causes, are your own fault and nobody else's.

I told you that THIS is exactly your problem. Once you manage to get past this hangup, I think that you will understand the rest. There is NOTHING stopping you to calculate the total derivative wrt t if y is not a function of t
Try Mathematica:
"In Mathematica, D(f,x) gives a partial derivative, with all other variables assumed independent of x. Dt(f,x) gives a total derivative, in which all variables are assumed to depend on x."

Tach said:
$$T=T(t,x,y,z)$$

but the bug moves only in x and y, so
...so you're defining the bug's position in z over time, specifically z=0, so dz/dt = 0

If you don't know how the bug moves (or not) in z, then how can you determine dT/dt?

...but note that this is also correct:
\begin{align} u &= 3\theta \\ f &= u + sin^2\theta + ln(x) \\ \frac{\partial f}{\partial \theta} &= \sin2\theta \end{align}​

This is not what the example says, besides it is irrelevant to solving your misunderstandings.

And, as you've been told several times, this is also correct:
\begin{align} u &= 0 \\ f &= u + 3\theta + sin^2\theta + ln(x) \\ \frac{\partial f}{\partial \theta} &= 3 + \sin2\theta \end{align}​

This is not what the example says, besides it is irrelevant to solving your misunderstandings.

You know, if you spent less time on your pettiness relative to notation and more time in trying to solve your lack of understanding as to why the total derivative does not depend on $$v$$, you could have used your time and a much more productive manner.

...so you're defining the bug's position in z over time, specifically z=0, so dz/dt = 0

Nope, I am simply defining T as a function of z, which is very natural, if you stop and think about it.

If you don't know how the bug moves (or not) in z, then how can you determine dT/dt?

The bug moves only in x,y as explained. Do you really want to understand this or do you want to be right no matter what?

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