How do I find the linear components of acceleration in a pendulum? I have managed to derive the equation of motion of a simple pendulum under the influence of gravity using the Lagrangian, but since that only tells me what the angular acceleration is, I now want to derive the $x$ and $y$ components of acceleration. The formula I have derived is the following:
$$\ddot θ = -(g/l)\sinθ$$ 
And the $x$ and $y$ are these: $$x = l\sinθ$$ $$y = l(1-\cosθ)$$
So now I'm not sure what to do next... Do I use the chain rule and differentiate $x$ and $y$ with respect to time twice? I don't really know what to do here...
If you can, please give me a hint instead of a full solution, and, if possible, stick to these formulas and Euler-Lagrange equations, instead of a solution that would incorporate something not directly related. 
 A: As you noticed, if we use Euler-Lagrange equation on $L= \frac 1 2 (\dot x^2 + \dot y^2) -mgy$ we get
$$\ddot x=0$$
$$\ddot y = -g$$
Something is clearly missing: gravity is not the only force acting on our mass: we have to take into account the tension of the rod/string. But why doesn't it come out from the equations?
The point is that system only has one degree of freedom: $\theta$. If we use $x$ and $y$ we are using the Lagrangian formalism as if it had two. But it doesn't: one degree of freedom is taken away from the constraint $x^2+y^2=l^2$. So the usual formulation of the EL equation is not going to work properly.
This was the hint. If you want the solution, you'll find it here: http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node90.html
A: Everything is a function of the angle $\theta$ and its derivatives $\dot{\theta}$ and $\ddot{\theta}$. From there use the chain rule of differentiation.
$$\begin{align}
  x & = \ell \sin \theta &  y & = \ell (1-\cos \theta) \\
  \dot{x} & = \ell \dot{\theta} \cos \theta &  y & = \ell \dot{\theta} \sin \theta \\
  \ddot{x} & = \ell \ddot{\theta} \cos\theta -\ell \dot{\theta}^2 \sin\theta & \ddot{y} & = \ell \ddot{\theta} \sin \theta + \ell \dot{\theta}^2 \cos\theta
\end{align} $$
