# Angular rate of change

1. Oct 30, 2007

### jeffyhow

1. The problem statement, all variables and given/known data

If I have an equation that describes L in terms of some constants and an angle.

e.g. L=C1*$$\sqrt{(C2-Sin[angle])}$$+C1*$$\sqrt{(C2-Cos[angle])}$$

Then, if I take the partial derivative of the above wrt angle, then I would get the unit change in L for a unit change in angle.

However, if I solve the partial derivative at a specific angle, will it give me units of length/degrees or units of length/radians?

The above equation is just an example of one case for the my question.

Last edited: Oct 30, 2007
2. Oct 30, 2007

### Dick

Hard to say with without a context, but probably radians. If you say the derivative of sin(x) is cos(x), then x is in radians. If your angle x is represented in degrees then the derivative of sin(x) is cos(x)*pi/180.

3. Oct 31, 2007

### HallsofIvy

Staff Emeritus
Trig functions (as opposed to values you would use to solve right triangle problems) don't necessairily have anything to do with angles. They are defined in such a way that if you are treating the variables as angles, they would have to be in radians.

I'm not sure why you are talking about "partial derivatives"- there are no partial derivatives. The only independent variable is "angle" so you only need an ordinary derivative.

4. Oct 31, 2007

### Smartass

Depends on how you integrate. If you are taking derivative of six=cosx, then x will be in radians, ie, you answer will be in lenght/rad. Oh and partial derivative, as HallsofIvy said, is not necessary, as only a single variable is involved.