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Angular rate of change

  1. Oct 30, 2007 #1
    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*[tex]\sqrt{(C2-Sin[angle])}[/tex]+C1*[tex]\sqrt{(C2-Cos[angle])}[/tex]

    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. jcsd
  3. Oct 30, 2007 #2

    Dick

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    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.
     
  4. Oct 31, 2007 #3

    HallsofIvy

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    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.
     
  5. Oct 31, 2007 #4
    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.
     
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