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Derivative of an Angular Velocity Equation

  1. Dec 22, 2013 #1
    1. The problem statement, all variables and given/known data

    In reviewing my last physics exam I found the following statement in the posted solution:

    ...the derivative of [itex]\sqrt{\frac{3g}{l}(1-cos\theta)}[/itex] is [itex]\frac{3g}{2l}sin\theta[/itex]

    2. Relevant equations

    N/A

    3. The attempt at a solution

    This is from a physics course and I'm not taking the calc co-requisite yet so I'm trying to get by on what I can teach myself, which is inadequate for this task apparently. I don't understand the process of getting [itex]sin\theta[/itex] out of [itex](1-cos\theta)[/itex]
     
  2. jcsd
  3. Dec 22, 2013 #2

    jedishrfu

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    so you're taking the derivative with respect to theta

    Basic diferentiation rules:
    - d/dx (sin x ) = cos x
    - and d/dx (cos x) = - sin x

    And so the d/dtheta (1 - cos theta) = d/dtheta (1) - d/dtheta (cos theta) = 0 - (-sin theta) = sin theta
     
  4. Dec 22, 2013 #3

    SteamKing

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    I don't think you can ignore the square root of the original expression when taking the derivative.
     
  5. Dec 22, 2013 #4

    jedishrfu

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    True, I was only addressing the (1-cos theta) part that the OP said he had trouble with.
     
  6. Dec 28, 2013 #5
    Great, thanks! I understand it now.

    Gotta remember...

    Basic diferentiation rules:
    - d/dx (sin x ) = cos x
    - and d/dx (cos x) = - sin x
     
  7. Dec 28, 2013 #6

    haruspex

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    Well, I'm glad you do because I don't. This is simply not true:
    The derivative of [itex]\frac{3g}{l}(1-cos\theta)[/itex] is [itex]\frac{3g}{l}sin\theta[/itex]

    The derivative of [itex]\sqrt{\frac{3g}{l}(1-cos\theta)}[/itex] is [itex]\frac{3g}{2l}sin(\theta)\left(\frac{3g}{l}(1-cos(\theta))\right)^{-\frac 12} = \frac 12\sqrt{\frac{3g}l\left(1+\cos(\theta)\right)}= \sqrt{\frac{3g}{2l}}\cos(\frac \theta 2)[/itex]
    But I would prefer to simplify first: [itex]\sqrt{\frac{3g}{l}(1-\cos\theta)} = \sqrt{\frac{6g}{l}}\sin(\frac \theta 2)[/itex]
     
  8. Dec 28, 2013 #7
    So here is the source of my question. This is the solution to the physics problem posted by my physics professor at MIT. I probably just posed the question incorrectly, but since what I was really interested in was just "what is the derivative of (1-cos(theta))" I understand now what the answer is.
     

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  9. Dec 28, 2013 #8

    haruspex

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    No, you posed it correctly. Equation (3) in your attachment is wrong.
    Probably something got left out by accident, but without seeing the subsequent equations it's hard to know. E.g this would be correct: ##\dot \theta \ddot \theta = \frac{3g}{2l}\sin(\theta)##
     
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