# I Derivative of an integral help

1. May 24, 2016

### Andreas C

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. May 24, 2016

### BvU

Why don't you show us the full problem statement and the way you found this 'mess'

3. May 24, 2016

### Andreas C

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.

4. May 24, 2016

### BvU

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 .

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 $$\mathcal L = {m\over 2} \left (\ell\dot\theta\right)^2 - mg\ell\left (1-\cos\theta\right ) \ \ \rm ?$$

5. May 24, 2016

### Andreas C

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.

6. May 24, 2016

### BvU

Hint: shouldn't take long !

7. May 24, 2016

### Andreas C

I'm gonna do it when I'm back home, I am on my phone now!

8. May 24, 2016

### Andreas C

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
9. May 24, 2016

### BvU

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 !

10. May 24, 2016

### Andreas C

Ηmmm... I'm a bit confused now... Are you saying I have to differentiate (m*l^2*(dθ/dt))/2 again?

11. May 24, 2016

### Andreas C

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?

12. May 24, 2016

### BvU

Not 'again'. m*l^2*(dθ/dt))/2 is the kinetic energy T in $\mathcal L = T - V\$. You haven't differentiated it yet.
 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 $- \omega^2 \sin\theta - \ddot \theta = 0$ ) .

13. May 24, 2016

### Andreas C

Ok, I think I've got it: $\ddot \theta$=-mgl*sinθ. Is that what you mean?

14. May 24, 2016

### BvU

Yes! Well done.

(you lost a factor ml2 on the left, however...)

15. May 24, 2016

### Andreas C

Hmmm... ∂L/∂θ=∂V/∂θ=mgl*(∂(1-cosθ)/∂θ)=-mgl*sinθ, where does l^2 come from?

Last edited: May 24, 2016
16. May 24, 2016

### Andreas C

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.

17. May 24, 2016

### Andreas C

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...

18. May 24, 2016

### Andreas C

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?

19. May 24, 2016

### BvU

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 $T = {2\pi\over \omega} = 2\pi\sqrt{\ell/g}$.

20. May 24, 2016

### Andreas C

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...