Can We Take the Derivative of a Unit?

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    Differentiating Units
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Discussion Overview

The discussion revolves around the concept of taking derivatives of physical quantities, specifically focusing on the units of velocity and acceleration. Participants explore whether it is valid to derive units directly and how this relates to the mathematical process of differentiation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions if taking the derivative of the unit of velocity (m/s) can yield a unit of acceleration (m/s²) by simply manipulating the units.
  • Another participant asserts that the units of X/Y are equivalent to the units of dX/dY, suggesting that the process of differentiation does not alter the units involved.
  • There is a correction regarding the sign in the unit manipulation, indicating that the minus sign may not be necessary.
  • Participants discuss the relationship between differentiation and unit extraction, with one stating that performing differentiation is unnecessary if only units are of concern, yet it yields the same result.
  • Clarification is provided about the definition of dX/dY as the limit of ΔX/ΔY, emphasizing that this limit does not affect the units.

Areas of Agreement / Disagreement

Participants generally agree on the relationship between units and differentiation, but there is some uncertainty regarding the necessity of the minus sign and the interpretation of the derivative in terms of units.

Contextual Notes

Some participants express that the concept of derivatives and their units was not thoroughly explained in their previous education, indicating a potential gap in understanding the foundational principles.

cragar
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okay we know that velocity is in meters per second and that acceleration is in
m/(s^2) , so if I take the derivative of velocity with respect to time i get acceleration .
but just looking at the units , if i start with velocity m/s (ms^-1) this might sound crazy but can i just take the derivative of s and get -ms^(-2) , any help would be appreciated.
 
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cragar said:
okay we know that velocity is in meters per second and that acceleration is in
m/(s^2) , so if I take the derivative of velocity with respect to time i get acceleration .
but just looking at the units , if i start with velocity m/s (ms^-1) this might sound crazy but can i just take the derivative of s and get -ms^(-2) , any help would be appreciated.

The units of X/Y are the same as the units of -X/Y. The factor has -1 has no dimension.

Cheers -- sylas
 
so what i did was correct , except for the minus sign .
 
cragar said:
so what i did was correct , except for the minus sign .

Yes. When you obtain the units for a differentiation, you just need to do division. The units of X/Y are the same as the units of dX/dY. If you do a full differentiation and then extract units from the result you will get the same result as simply extracting the units by getting units for X and units for Y, and dividing. Actually carrying out a differentiation is overkill if all you want is units, but it gives the same result.

Cheers -- sylas
 
so we are just dividing everything by dY , in our case s , in all of my physics classes it was never really explained so thanks for taking the time to explain it.
 
cragar said:
so we are just dividing everything by dY , in our case s , in all of my physics classes it was never really explained so thanks for taking the time to explain it.

You're welcome. Remember that dX/dY is defined as the limit of ΔX/ΔY as the small delta changes go towards 0. Taking a limit makes no difference to the units, so the units of the derivative is indeed the units of a division.
 

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