Derivatives and Integrals of units

[Dougggggg]

I couldn't decide whether to place this in the Physics or the Math section of the forums, deep down it is really a Math question for Physics problems. So mods please move if you feel it would be more appropriate in the Physics section.

So when doing calculations, I always like to make sure my units are behaving correctly. That I am not adding kg to m/s or something weird like that. One thing that I haven't really thought about how it applies is doing calculus type manipulation of units. Like if I take the derivative of a unit of displacement, with respect to a unit of time, I get a unit of displacement over time.

For Integrals if I take a integral of force, with respect to displacement, I get a unit of force times displacement.

Basically, what is the basic method to these kind of things?

 

[CompuChip]

It is exactly as you say.
When you differentiate with respect to x, the units get a factor of 1/[x] (where [x] are x's units).
When you integrate with respect to x, the units get a factor of [x].
For differentiation, this follows simply from the definition, because f'(x) is a limit of the quotient
\frac{\Delta f(x)}{\Delta x}
which has units
\frac{[\Delta f(x)]}{[\Delta x]} = \frac{[f(x)]}{[x]}.

For integration, simply reverse the argument, i.e. something along the lines of
\left[ \frac{d}{dx} \left( \int f(x) \, dx \right) \right] = [ f(x) ]
but it is also
\left[\int f(x) \, dx \right] / [x]
therefore,
\left[ \int f(x) \, dx \right] = [f(x)] [x].

 

[HallsofIvy]

Think of differentiation as being an extension of division and integration as an extension of multiplication.

The units of dy/dx are the units of y divided by the units of x. For example, if y is measured in meters and x is measured in seconds then dy/dx (the rate of change of y with respect to x= rate of change of distance with respect to time) has units of "meters per second".

The units of \int f(x) dx are the units of f(x) times the units of dx (which are the same as the units of x). For example, if f(x) is a "linear mass density" in "kilograms per meter" and x has units of meters, then \int f(x)dx has units of (kilogram/meter)(meter)= kilogram.