How can engineers get away with splitting differentials in dynamics?

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Discussion Overview

The discussion revolves around the mathematical practice of splitting differentials in dynamics, particularly in the context of introductory physics and engineering problems. Participants explore the conditions under which this practice is deemed acceptable and the potential pitfalls associated with it.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • One participant questions the validity of splitting differentials, seeking clarity on why it works under certain conditions and how engineers justify this practice.
  • Another participant notes that physicists often handle differentials casually, highlighting the risk of confusing vector and scalar equations, which could lead to overlooking important mathematical concepts like the Jacobian.
  • A later reply emphasizes the complexity of formally proving the legitimacy of manipulating differentials, suggesting that such proofs can be challenging at the introductory level.
  • Concerns are raised about the potential for errors when casually juggling differentials, indicating that while it may work, it is not without risks.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of splitting differentials, with some acknowledging its utility in specific contexts while others caution against its casual application. The discussion remains unresolved regarding the broader implications of this practice.

Contextual Notes

Participants mention the importance of understanding the mathematical foundations of differentials and the potential for confusion when transitioning between vector and scalar forms. There is an acknowledgment of the limitations of introductory treatments of these concepts.

Trying2Learn
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TL;DR
splitting the differential
In an introductory dynamics textbook, we often see this progression

v = ds/dt ---> dt = ds/v

a = dv/dt ---> dt = dv/a

Equating the dt, we get: vdv=ads

Now my question

On the one hand, this works for certain problems.
On the other hand, this is splitting the differential.

Could someone please explain

Why it works under certain conditions? How engineers get away with this?

If it is poor math to do this: why? Is it because one should never split the differential?

How can engineers get away with this?

I see that it does work, but only in ONE dimension.

This whole issue has always bothered me but I cannot state, with clarity, conviction, precision:
Why it is poor math to do this
Why we can get away with it.
 
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Physicists are usually pretty casual in dealing with differentials :smile: .
In the 'progression' you quote, the expressions at the left are vector equations; the ones on the right are scalar expressions.
Beginning physicists should be careful not to accidentally hop back and forth, or they risk overlooking a Jacobian and other useful mathematical goodies.

##\ ##
 
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Trying2Learn said:
TL;DR Summary: splitting the differential

In an introductory dynamics textbook, we often see this progression

v = ds/dt ---> dt = ds/v

a = dv/dt ---> dt = dv/a

Equating the dt, we get: vdv=ads

Now my question

On the one hand, this works for certain problems.
On the other hand, this is splitting the differential.

Could someone please explain

Why it works under certain conditions? How engineers get away with this?

If it is poor math to do this: why? Is it because one should never split the differential?

How can engineers get away with this?

I see that it does work, but only in ONE dimension.

This whole issue has always bothered me but I cannot state, with clarity, conviction, precision:
Why it is poor math to do this
Why we can get away with it.
A formal proof of why this jugglery with "d"'s actually works is a nightmare because they are not even defined as solids at this level.
$$
\dfrac{ds}{dt}=\lim_{h \to 0}\dfrac{s(t+h)-s(t)}{h}
$$
Now, how would you isolate ##dt## here? I like to avoid such steps by using Weierstraß's formula: $$ s(t+h)= s(t)+ s'(t) \cdot h + o(h)$$ with a remainder ##o(h)## that is quadratic in ##h## so it vanishes fast as ##h## goes to zero. With that formula, Weierstraß has out all the limit stuff in the ##o(h)## term and we can work with them as there was no limit stuff.

You should be careful with
BvU said:
Physicists are usually pretty casual in dealing with differentials :smile: .
because: they have practiced juggling! It can go wrong!
 
fresh_42 said:
It can go wrong!
Tell me something I don't know :smile:

Been there, done that.

##\ ##
 
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