Definition of derivative - infinitesimal approach, help :)

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The discussion centers on understanding the mathematical definitions of dy, f'(x), and dx in the context of infinitesimal calculus. It clarifies that Δy represents the change in y along a secant line between two points, while dy signifies the change along the tangent line, making dy an infinitesimal change. The expressions Δy = f(x+Δx) - f(x) and dy = f'(x)Δx + εΔx are shown to be equivalent, with the latter including an infinitesimal error term εΔx. The conversation also addresses the treatment of dy/dx as a quotient, suggesting that while it's technically a limit, it can be treated as a quotient in practical applications, though this is not mathematically precise. Ultimately, the discussion emphasizes the complexities and potential confusions in using infinitesimals compared to traditional calculus methods.
  • #31
Hi pondhockey, thank you for the reply. Could you give an example of when treating dx and dy as algebra seems like unsupported or unclear "magic" that substitutes a vague intuition for what is really intended. And would you be able to give an example of what the real way of doing it is? I think this could help me for when I'm going to read physics and chemistry.
 
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  • #32
Christian, here is an answer to one of my questions in a thermodynamics thread; I'll provide the "differential algebra magic" example later.

https://www.physicsforums.com/threa...ntials-in-thermodynamics.791811/#post-5135710

I'm "LATEX challenged" so I'll have to talk around it, but the link to Couchyam's answer is pretty explicit. Briefly, he expands an integral of PdV to what it really means. Note that it is NOT the integral of P with respect to V.
 
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  • #33
OK, here is an example of what I regard to be differential magic. I'm not going to attempt to provide the mathematically correct restatement because I'm still a thermodynamics grasshopper and I'll probably get it wrong. I think it conveys a "vague intuitive" understanding, but not a clear, precise understanding.

From H.Y. McSween, et. al. (2003) Geochemistry, pathways and processes. Columbia Press

Let P be pressure
let dV be a volume change.
let dq be an amount of heat gained.
T is (absolute) temperature
define dS = dq/T
let dE be a change in internal energy
write
dE + PdV<=TdS (p44; take this as given, so we can follow the "logic")

consider only processes for which dP = 0, and use the “fact” that

d(PV) = PdV + VdP (oh, really?! is d now the symbol for a derivative?)

to get

dE +d(PV) <= TdS

d(E+PV) <= TdS (so is this a new differential or a derivative?!)

etc.
 
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  • #34
Thank you for sharing, when I', done with calculus and start reading Thermodynamics, I'll return to this post and it will probably make more sense to me :)
 
  • #35
pondhockey said:
OK, here is an example of what I regard to be differential magic. I'm not going to attempt to provide the mathematically correct restatement because I'm still a thermodynamics grasshopper and I'll probably get it wrong. I think it conveys a "vague intuitive" understanding, but not a clear, precise understanding.

From H.Y. McSween, et. al. (2003) Geochemistry, pathways and processes. Columbia Press

Let P be pressure
let dV be a volume change.
let dq be an amount of heat gained.
T is (absolute) temperature
define dS = dq/T
let dE be a change in internal energy
write
dE + PdV<=TdS (p44; take this as given, so we can follow the "logic")

consider only processes for which dP = 0, and use the “fact” that

d(PV) = PdV + VdP (oh, really?! is d now the symbol for a derivative?)
I don't see anything wrong here. "d" means "differential of", not "derivative of".
pondhockey said:
to get

dE +d(PV) <= TdS

d(E+PV) <= TdS (so is this a new differential or a derivative?!)
Again, d here means differential. The operator is linear, so the differential of a sum of functions is the sum of the differentials of the functions.
 
  • #36
Mark, thank you for the comment.

Can you tell me, in what context d (i.e. the differential) is considered an operator? It is not defined that way in either calculus/advance calculus text that I just now referred back to. In those texts, dx is a real variable, referred to as an independent variable, and dy is defined as f' dx. A comparable "total" differential is defined for functions of two variables. No "operator" or corresponding differential algebra is introduced - only standard algebra of real numbers is used.
 
  • #37
pondhockey said:
Mark, thank you for the comment.

Can you tell me, in what context d (i.e. the differential) is considered an operator? It is not defined that way in either calculus/advance calculus text that I just now referred back to. In those texts, dx is a real variable, referred to as an independent variable, and dy is defined as f' dx. A comparable "total" differential is defined for functions of two variables. No "operator" or corresponding differential algebra is introduced - only standard algebra of real numbers is used.

See https://en.wikipedia.org/wiki/Differential_of_a_function#Properties
 
  • #38
christian0710 said:
Yess I understand it! so if we choose any point fx (1,f(1)), and set dx or dy to any number fx if dx=5, then dy=10 (dy depends on dx) in our example is interpreted as: When the slope of the tangent f'(1)=2 at point (1,f(1)) is multiplied by the horizontal run of 5 units, then we "run along the tangent" and end up at the point (1+dx, f(1) + dy) = (1+5, 2+10) = (6,12) ON the tangent.

So would this be the correct Conclusion: So dy and dx can be calculated for bigger numbers/values of dy and dx, and then we end up on a more distant point on the tangent. However, in practice this is not useful, because we are interested in using dy to approximate the slope of the function f at a point near f'(x). But why bother using this method if we have the derivative of a function? Is it not more practical to get the 100% accurate slope af a point as dy/dx = f'(x)?

I don't know what you mean by 100% accurate, because this is a limit, involving ## \delta ##s and ## \epsilon##s ; unless your graph is a straight line, this is not what I would call 100% accurate.

What you are trying to do is to approximate the _Real_ change in values of the function by the change along the tangent line. This is what dy=f'(x)dx gives you.
 
  • #39
pondhockey said:
Mark, thank you for the comment.

Can you tell me, in what context d (i.e. the differential) is considered an operator? It is not defined that way in either calculus/advance calculus text that I just now referred back to. In those texts, dx is a real variable, referred to as an independent variable, and dy is defined as f' dx. A comparable "total" differential is defined for functions of two variables. No "operator" or corresponding differential algebra is introduced - only standard algebra of real numbers is used.

But the idea is always the same: you have a tangent linear object (line, plane, etc.) that approximates the change of a (differentiable, at least at the point) function , at least near the point in question.
 
  • #40
WWGD said:
But the idea is always the same: you have a tangent linear object (line, plane, etc.) that approximates the change of a (differentiable, at least at the point) function , at least near the point in question.

I agree with this. And I think it begs the question of why mess with differentials in expositions of physics (and in particular thermodynamics.) To me it seems to cloud more issues than it simplifies. It certainly inspires a lot of threads just like this one!
 
  • #41
Mark writes:

..
Mark44 said:

..Thank you for the link. It makes me think that this info should be in an introduction to every physics and thermodynamics text. I've never seen it in an undergraduate math text. There's a culture, in physics, that I find regrettable: the prof used integrals and line (path) integrals in my physics class way before any of us were formally introduced to them. I think that handwaving and intuitive arguments become embedded in the culture and practice of physics.
 

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