# Fundamental confusions of calculus

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

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

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Hi rpenner, and thanks.
I don't understand the subscript notation.
Does it just mean that a,b,c,z are constants?

Last edited: Feb 14, 2012

3. ### rpennerFully WiredValued Senior Member

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Yes, that was my intent

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

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Thanks,
Notation isn't my strong point... which is one of the reasons confusion abounds

7. ### TachBannedBanned

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I simply wanted to generate for you a case that :

-helps you understand the difference between total and partial derivative
-helps you understand what happens when not all variables in the function description are functions of the variable (t in this case) you are taking the derivatives

This is the partial derivative wrt t.

This is the total derivative wrt t.

Last edited: Feb 14, 2012
8. ### PeteIt's not rocket surgeryRegistered Senior Member

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I appreciate your generous effort, but forgive me if I am not confident in your teaching ability in this area.
I'm generally inclined to trust the combined credentials of temur, rpenner, przyk, and AlphaNumeric, supported by textbooks and online material.
No, that's another partial derivative.
In that derivative z is an independent variable, treated as constant.

Taking a total derivative would mean treating z as a function of t, not as a constant.
See rpenner's post (also said by przyk, temur, David Metzler in the video, and possibly others):

Last edited: Feb 14, 2012
9. ### AlphaNumericFully ionizedRegistered Senior Member

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You failed to show how my use of the definition was false, despite having been asked more than once. You failed to respond to my question about what the canonical momentum for $T = \frac{1}{2}m\dot{q}^{2}$ is, because it would mean admitting that $p \equiv \frac{\partial}{\partial \dot{q}}\frac{1}{2}m\dot{q}^{2} = m\dot{q}$, which contradicts your claims about partial derivatives and also your claim you understand Hamiltonian mechanics while holding this view.

You are deliberately avoided addressing simple, relevant, direct questions which are based on definitions and found throughout the literature, as well as basic lecture notes.

I would lock this thread but it seems Rpenner and Pete are in the middle of something. You have one last chance. If you respond without addressing my direct questions then this thread will definitely be locked and you given a warning for trolling. You've already had more last chances than you deserve.

10. ### TachBannedBanned

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You got it totally backwards, let's try it one more time: partial derivative means that if $f=f(t,u)$ where $u=u(t)$ then $\frac{\partial f}{\partial t}$ is the partial derivative while $\frac{\partial f}{\partial t}+\frac{\partial f}{\partial u}\frac{du}{dt}$ is the total derivative.Partial derivative means differentiation only wrt the explicit variable (t in the example), total derivative means partial derivative PLUS differentiation (via chain rule) through all variables that are a function of the variable considered. Therefore, contrary to your claims, I know exactly what I am doing and the method described (it is textbook, really) produces the correct answer:

So, $p=\frac{\partial T}{\partial \dot{q}}=m\dot{q}$

The video that Pete linked in is very good, in this respect. I can track your error to the fact that you failed repeatedly to note that in my example, $f=3 \theta +u+v$ and you kept considering that $f=3 \theta +sin^2(\theta)+v$. So, you are getting the wrong partial derivative. Not a big error, but please don't try to pin it on me, own it.

Last edited: Feb 14, 2012
11. ### przyksquishyValued Senior Member

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And? $x$ and $y$ are also not the bug's coordinates. They're general coordinates on the plane.

For the purpose of calculating the bug's temperature from $T(t,\,x,\,y,\,z)$, you have to set $x(t) = f(t)$, $y(t) = g(t)$, and $z(t) = 0$. Then, and only then, are you taking a total derivative.

There is a difference between saying
$\frac{\mathrm{d}T}{\mathrm{d}t} \,=\, \frac{\partial T}{\partial t} \,+\, \frac{\partial T}{\partial x} \, \frac{\mathrm{d}x}{\mathrm{d}t} \,+\, \frac{\partial T}{\partial y} \, \frac{\mathrm{d}y}{\mathrm{d}t}$​
because $\frac{\mathrm{d}z}{\mathrm{d}t} = 0$, which is a total derivative, and
$\Bigl( \frac{\partial T}{\partial t} \Bigr)_{z} \,=\, \frac{\partial T}{\partial t} \,+\, \frac{\partial T}{\partial x} \, \frac{\mathrm{d}x}{\mathrm{d}t} \,+\, \frac{\partial T}{\partial y} \, \frac{\mathrm{d}y}{\mathrm{d}t}$​
(for lack of better notation), because you're leaving $z$ independent, in which case you're still taking a partial derivative, specifically of the function
$(t,\,z) \,\mapsto\, T(t,\, x(t),\, y(t),\, z) \,.$​
To calculate the bug's temperature from $T$, you have to set $z(t) = 0$, and you're therefore in the former case. And you're calculating a total derivative only because you set $z(t) = 0$.

Do you understand this? Because it's the point we've been trying to get through to you.

12. ### TachBannedBanned

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Once again, I am not taking the bug temperature. I am taking the temperature of an arbitrary point in space.

13. ### przyksquishyValued Senior Member

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If you're just doing this then you're not calculating any sort of derivative at all. You're just evaluating $T(t,\,x,\,y,\,z)$ at some arbitrary location and time.

So what the hell are you even talking about?

14. ### Guest254Valued Senior Member

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This really should have ended by now. Tach doesn't know what he's talking about and these two posts cement that fact. No ifs or buts -- he doesn't understand elementary calculus and these two posts aptly demonstrate this unfortunate truth.

However, I don't think I want the thread to end. It's strangely amusing to watch the train wreck unfold!

15. ### TachBannedBanned

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I simply constructed a function for Pete to calculate the total derivative.

16. ### TachBannedBanned

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Of course, in your trolling haste, you failed to notice that $f(\theta,u,v)=3 \theta+u+v$ where $u=sin^2(\theta)$ and $v=ln(x)$.
So, what is $\frac{\partial f}{\partial \theta}$, Guest? Did your degree skirt over partial derivatives?

Last edited: Feb 14, 2012
17. ### rpennerFully WiredValued Senior Member

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Your reference is non-normative. Normative references do not support your assertions and slurs. It's not clear that this reference about a "bug crawling on a plate" supports your assertions, let alone your slurs.

http://www.wolframalpha.com/input/?i=D[ 3 theta + (sin theta)^2 + ln(x), theta ]
http://www.wolframalpha.com/input/?...+ (sin theta)^2 + ln(x) with respect to theta (The show steps button is nice)

Moving the goal posts from $f(\theta,x) = 3 \theta + \sin^2 \theta + \ln x$ to $f(\theta, u, v) = 3 \theta + u + v$ serves no pedagogical purpose.

18. ### TachBannedBanned

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I haven't moved any goal-posts, $f=f(\theta,u(\theta),v(x))$ since the beginning, you arrived to this thread late.
Here is a normative reference I gave earlier.
As for Guest, he has a long history of trolling my posts. He would have had an excuse , except that I have shown the $f=f(\theta,u(\theta),v(x))$ repeatedly. So, he has no excuse. Come to think of it, neither have you since I take it that you understand perfectly the significance of $f=f(\theta,u(\theta),v(x))$.

19. ### TrippyALEA IACTA ESTStaff Member

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See, rightly or wrongly, that's not how I would perform the partial derivative, I, and from what I gather, everybody else who is disagreeing with you, would not re-define the equation that way.

If I was given:
$f(\theta,x)=3 \theta+sin^2(\theta)+ln(x)$

And asked to find the partial derivatives, I would do it using:
$f(\theta)=3 \theta+sin^2(\theta)+k$
{where k=ln(x)}
And:
$f(x)=k+ln(x)$
{where k=$3 \theta+sin^2(\theta)$}

And if I was asked to find the partial derivative $\frac{\partial f}{\partial \theta}$ using the chain rule, I would start with:
$f(\theta)=3 \theta+sin^2(\theta)+k$
{where k=ln(x)}
Then I would set:
$u=sin(\theta)$
Then I would find:
$\frac{df}{du}$ for the equation $f(\theta)=3 \theta+u^2+k$
And find:
$\frac{du}{d \theta}$ for $u=sin(\theta)$
And then proceed to evaluate: $\frac{\partial f}{\partial \theta} = \frac{df}{du}\frac{du}{d \theta}$

But that, to me, seems unneccessary, when the initial problem is relatively uncomplicated.

It seemd to me that what you've done is redefined the problem as being:
$f(\theta,x)=g(\theta)+h(\theta)+i(x)$
And then come to the conclusion that $f'(\theta)=g'(\theta)$ while neglecting to evaluate $h'(\theta)$

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

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Sure, IF you were given the above. But what you are being given is

$f(\theta, u(\theta),v)=3\theta+u+v$ where $u(\theta)=sin^2(\theta)$, $v(x)=ln(x)$

The total derivative will be the same but the partial derivative differs , depending on what you consider that has been given as $f$.

21. ### TrippyALEA IACTA ESTStaff Member

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No, that's exactly what we were given:
Post #18
If you meant something else, perhaps you should be more careful in the future.

22. ### RJBeeryNatural PhilosopherValued Senior Member

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Refusing to acknowledge the error only makes the clanger that much louder, Tach; especially in light or your hyper-critical, condescending nature. Live by the sword...

23. ### TrippyALEA IACTA ESTStaff Member

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There's a greater irony in there.