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I Derivative of an integral help

  1. May 24, 2016 #1
    OK, I lied a bit. It's not JUST the derivative of an integral. It's the derivative of a cosine of an integral. Solving the problem of the motion of a simple pendulum under a gravitational field using the lagrangian, I came into this mess (which I don't know if it's right):
    l^2*((dθ/dt)/dt)-mg*d(cosθ)/(dθ/dt)=dL/dθ

    The part I want to focus on is d(cosθ)/(dθ/dt). I know d(cosθ) is not a constant, because it depends on dθ/dt, since it's its integral with respect to dt, but how do I find what it is?
     
  2. jcsd
  3. May 24, 2016 #2

    BvU

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    Why don't you show us the full problem statement and the way you found this 'mess'
     
  4. May 24, 2016 #3
    OK, I will, I just have to mention that I meant d(cosθ)/d(dθ/dt), not d(cosθ)/(dθ/dt), that wouldn't make sense anyway.
     
  5. May 24, 2016 #4

    BvU

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    Yes. And then it still doesn't make sense: you also have to mention the same thing for the lefthand side: not (dθ/dt)/dt but d(dθ/dt)/dt :smile:.

    Furthermore -- and this is rather important -- you don't really mean d(cosθ)/(dθ/dt) ##\qquad## (or rather $$ {d\cos\theta \over {d{d\theta\over dt} }} $$ but instead you mean $$
    {\partial \cos\theta \over {\partial {d\theta\over dt} } }$$ The ##\partial## derivative is different from the ##d## derivative (I think it's called the total derivative). Look it up - it appears a lot more in physics. The Euler Lagrange equation is a partial differential equation.

    Meaning you consider ##\mathcal{L}## as a function of independent variables ##q## and ##\dot q## and take partial derivatives (differentiate to the one while keeping the other constant):$$
    { \partial \mathcal{L} \over \partial q} - {d\over dt}\left (\partial L\over\partial \dot q \right ) = 0
    $$And what do you get when you do a partial differentiation wrt ##\dot \theta## of your $$\mathcal L = {m\over 2} \left (\ell\dot\theta\right)^2 - mg\ell\left (1-\cos\theta\right ) \ \ \rm ? $$
     
  6. May 24, 2016 #5
    That is actually exactly what I meant. That symbol is called the partial derivative, but I didn't know how to insert it, and I thought it would be the same, thanks :) I'm gonna work it out.
     
  7. May 24, 2016 #6

    BvU

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    Hint: shouldn't take long !
     
  8. May 24, 2016 #7
    I'm gonna do it when I'm back home, I am on my phone now!
     
  9. May 24, 2016 #8
    I got (m*l^2*(dθ/dt))/2, is that right?

    Edit: for ∂L/∂θ I got -mgl*sinθ, should I worry about the minus sign? Don't worry, it's not for homework, I'm teaching myself classical mechanics.
     
    Last edited: May 24, 2016
  10. May 24, 2016 #9

    BvU

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    no, you still want to differentiate it !
    The hint was a nudge to make you see (which you did) that ##\cos\theta## is not a function of ##\dot\theta## so the partial derivative of ##V## wrt ##\dot\theta## is zero.
    Yes you should. It's the difference beteween a stable equilibrium at ##\theta = 0## and an unstable one.
    Kudos! Not an easy subject all on your own. Good thing there is PF :smile:!
     
  11. May 24, 2016 #10
    Ηmmm... I'm a bit confused now... Are you saying I have to differentiate (m*l^2*(dθ/dt))/2 again?

    So I should have a plus sign instead of a minus one?
     
  12. May 24, 2016 #11
    Ok, I think I found the mistake, I forgot to cancel out the *1/2 with the *2, so in the case of ∂L/∂(dθ/dt) it's just ((m*l^2)/2)*(∂(dθ/dt)^2/∂(dθ/dt)), right? Well, (∂(dθ/dt)^2/∂(dθ/dt))=2*(dθ/dt), so ((m*l^2)/2)*((dθ/dt)^2/(dθ/dt))=m*l^2*(dθ/dt). That's correct, isn't it?
     
  13. May 24, 2016 #12

    BvU

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    Not 'again'. m*l^2*(dθ/dt))/2 is the kinetic energy T in ##\mathcal L = T - V\ ##. You haven't differentiated it yet.
    [edit] My mistake, I misread the /2 for ^2. Anyway, you found it and fixed it.
    Certainly not. You got a minus sign for ##{d\mathcal L\over d\theta}## and you want to keep it. The Euler Lagrange equation is $${ \partial \mathcal{L} \over \partial q} - {d\over dt}\left (\partial L\over\partial \dot q \right ) = 0$$ and if you do it right you automatically end up with ##\ddot \theta + \omega^2 \sin\theta = 0##.

    (or, of course ##- \omega^2 \sin\theta - \ddot \theta = 0## :smile: ) .
     
  14. May 24, 2016 #13
    Ok, I think I've got it: ##\ddot \theta##=-mgl*sinθ. Is that what you mean?
     
  15. May 24, 2016 #14

    BvU

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    Yes! Well done.

    (you lost a factor ml2 on the left, however...)
     
  16. May 24, 2016 #15
    Hmmm... ∂L/∂θ=∂V/∂θ=mgl*(∂(1-cosθ)/∂θ)=-mgl*sinθ, where does l^2 come from?
     
    Last edited: May 24, 2016
  17. May 24, 2016 #16
    Ooooh, I just realized what you meant! Well, since it was ##ml^2 \ddot θ =-mgl*sinθ##, l^2 cancels out with l, and we're left with just l.
     
  18. May 24, 2016 #17
    Oh my god, what a tangle I have gotten myself into! Well, it's the first problem I solve using the lagrangian, so I guess it is to be expected...
    If my calculations are correct, then ##ml^2 \ddot θ =-mgl*sinθ##, which means that ##\ddot θ = -(g*sinθ)/l##, which doesn't sound right at all...
     
  19. May 24, 2016 #18
    Actually, I looked it up and it IS correct! Yay! Thanks a lot for your patience, you were very helpful! However, there's one thing I don't understand: how does g/l become ω^2?
     
  20. May 24, 2016 #19

    BvU

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    For a harmonic oscillator (pendulum with small amplitude, ##\ \theta\approx \sin\theta\ ##) the equation is ##\ddot\theta + \omega^2\theta=0\ ## with solution ##\theta = a \sin\omega t + b\cos\omega t##.

    The period is ##T = {2\pi\over \omega} = 2\pi\sqrt{\ell/g}##.
     
  21. May 24, 2016 #20
    Ah, ok, so it's a special case. However, the equation of motion I found here is for any amplitude, right?

    Also, say I now want to find the x and y components of acceleration and velocity. How do I do that? I had a couple of ideas but I'm not sure if they are going to be fruitful...
     
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