# Derivative of an integral help

• I
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?

BvU
Homework Helper
(which I don't know if it's right):
Why don't you show us the full problem statement and the way you found this 'mess'

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

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.

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

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

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.

BvU
Homework Helper
Hint: shouldn't take long !

Hint: shouldn't take long !
I'm gonna do it when I'm back home, I am on my phone now!

Hint: shouldn't take long !

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.

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BvU
BvU
Homework Helper
I got (m*l^2*(dθ/dt))/2, is that right?
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.
for ∂L/∂θ I got -mgl*sinθ, should I worry about the minus sign?
Yes you should. It's the difference beteween a stable equilibrium at ##\theta = 0## and an unstable one.
Don't worry, it's not for homework, I'm teaching myself classical mechanics.
Kudos! Not an easy subject all on your own. Good thing there is PF !

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 !

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

Yes you should.

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?

BvU
Homework Helper
Are you saying I have to differentiate (m*l^2*(dθ/dt))/2 again?
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## ) .

Andreas C
Not 'again'. m*l^2*(dθ/dt))/2 is the kinetic energy T in ##\mathcal L = T - V\ ##. You haven't differentiated it yet.
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## ) .

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

BvU
Homework Helper
Yes! Well done.

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

Andreas C
Yes! Well done.

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

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

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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.

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...

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?

BvU
Homework Helper
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}##.

Andreas C
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}##.

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...

BvU
Homework Helper
However, the equation of motion I found here is for any amplitude, right
the equation of motion is a second order differential equation. A solution needs two boundary conditions, e.g. an amplitude and a phase (or the a and b in the expression above).
Also, say I now want to find the x and y components of acceleration and velocity
For that you need to express x and y in ##\theta## ; rather straightforward !

the equation of motion is a second order differential equation. A solution needs two boundary conditions, e.g. an amplitude and a phase (or the a and b in the expression above).
For that you need to express x and y in ##\theta## ; rather straightforward !

So I have to know the initial θ in the case of a pendulum starting from a stationary position, because that would be the amplitude. What about the phase?

Also, what do you mean by "For that you need to express x and y in ##\theta## ; rather straightforward !"?
Does that mean that ##\ddot x = l*sin\ddotθ##?

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BvU
Homework Helper
So I have to know the initial θ in the case of a pendulum starting from a stationary position, because that would be the amplitude. What about the phase?
Never done a mass and spring oscillator before ?
##\ \ddot\theta = -\omega^2\theta\ ## with ##\ \theta(0) = \theta_0\ ## and ##\ \dot \theta(0) = 0\ ## gives ##\ \theta(t) = \theta_0\cos(\omega t) = \theta_0\sin(\omega t + {\pi\over 2})##
Does that mean that ##\ddot x = l*\sin\ddot\theta \rm ?##
As long as ##\sin\theta\approx\theta## you have ##x=\ell\theta## so ##\dot x = \ell\dot\theta## and ##\ddot x = \ell\ddot\theta## etc. With ##x=\ell\sin\theta## things get more complicated (24A here).

Never done a mass and spring oscillator before ?
##\ \ddot\theta = -\omega^2\theta\ ## with ##\ \theta(0) = \theta_0\ ## and ##\ \dot \theta(0) = 0\ ## gives ##\ \theta(t) = \theta_0\cos(\omega t) = \theta_0\sin(\omega t + {\pi\over 2})##
As long as ##\sin\theta\approx\theta## you have ##x=\ell\theta## so ##\dot x = \ell\dot\theta## and ##\ddot x = \ell\ddot\theta## etc. With ##x=\ell\sin\theta## things get more complicated (24A here).

Yeah, but that would be the phase for sinθ approaching θ, and I'm looking to find the equations for any amplitude, not just very small ones... It also bugs me how for some reason, the angular acceleration always turns out to be negative, shouldn't it be positive when the angular speed is increasing, like after it reaches the maximum height?

BvU