What units will the partial derivative of an angular equation give?

jeffyhow
Messages
1
Reaction score
0

Homework Statement



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:
Physics news on Phys.org
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.
 
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.
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top