How can engineers get away with splitting differentials in dynamics?

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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!