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Help required to differentiate a function.

  1. Oct 12, 2012 #1
    I need help to take the time derivative of a physics equation.

    The equation:

    ME=1/2 mr^2 (dθ/dt)^2 +mgr θ^2/2

    Where ME is the mechanical energy , m is the mass of the object, r is the radius of the path of the object and θ is the angle around the path.

    Walter Lewin at MIT differentiates it and gets the result:

    mr^2 (dθ/dt) (d^2 θ/dt^2)+mgr θ dθ/dt

    Can someone please explain why this is so. I can't understand how that would be the derivative of the function. Please answer.
     
  2. jcsd
  3. Oct 12, 2012 #2

    haruspex

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    It's standard chain rule. Which part don't you get? If y = y(t), the derivative of y2 is 2y dy/dt
     
  4. Oct 12, 2012 #3
    The derivative of 1/2mr^2
    Would be 2 1/2mr right? Why do we need to use the chain rule?
     
    Last edited: Oct 12, 2012
  5. Oct 12, 2012 #4
    Can you please just explain the entire derivative of the function.
     
  6. Oct 12, 2012 #5
    I get that the derivative of 1/2mr^2=2 1/2 mr r = mr^2, what i don't get is the dtheta/dt part.
     
  7. Oct 12, 2012 #6
    it appears as though r does not depend on t if Walter Lewin at MIT differentiates it correctly.

    The chain rule is required because [itex]\frac{d((f(t))^{2})}{dt} = 2 f \frac{d(f(t))}{dt}[/itex]

    Thus, your derivative of 1/2mr^2 is wrong because you forgot the factor of dr/dt but that isn't really relavent

    What you want to do is take the derivative of the θ stuff. What i mean is

    [itex]\frac{d}{dt} ME = \frac{1}{2}mr^{2} \frac{d}{dt} (\frac{dθ}{dt})^{2} + \frac{1}{2}mgr\frac{d}{dt}θ^{2}[/itex]
     
    Last edited: Oct 12, 2012
  8. Oct 12, 2012 #7
    R is a constant radius, it doesn't depend on t. But how would I do d/dt of (dtheta/dt)^2
     
  9. Oct 12, 2012 #8
    Ok thanks a lot. I understand. Correct me if I'm wrong:
    d/dt of 1/2 mr^2(dθ/dt)^2 is (1/2 mr^2) 2 dθ/dt d^2 θ/dt^2)=mr^2 dθ/dt d^2 θ/dt^2 and the 2 you bring down by the power rule is from dθ/dt squared and not from r square and then derivative of the inside is the second derivative of θ and then the second part d/dt of mgr θ^2/2 is mgr 2 θ/2 dθ/dt = mgr θ dθ/dt.

    Is this correct?
     
  10. Oct 12, 2012 #9

    haruspex

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    Yes.
     
  11. Oct 12, 2012 #10
    Walter Lewin writes all the stuff he write on the board from his book so sometimes what he writes isn't what he says. Over here he said the 2 from the r square would cancel out the 1/2. That confused me because then the derivative of dtheta/dt square becomes 2dtheta/dt dtheta/dt square.
     
  12. Oct 12, 2012 #11
    Thanks for the help.
     
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