Can We Take the Derivative of a Unit?

[cragar]

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.

 

[sylas]

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

 

[cragar]

so what i did was correct , except for the minus sign .

 

[sylas]

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

 

[cragar]

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.

 

[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.

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.