Finding the Lagrangian from the derivative of position

Question

I have to find the Lagrangian for a system. In the point of interest I have come up with the following position coordinates:

$$x = Rcos(\omega t)+\ell sin(\phi)$$ and $$y = Rsin(\omega t)-\ell cos(\phi)$$ Now, since I want to find the Lagrangian I need to take the derivative with respect to $t$ of both of them (To get the $\dot{x}$ and $\dot{y}$), and plug it into the formula for kinetic energy. Although it seems easy, the result I should get I'm a little confused about.

In my opinion only the first term in each equation is the only ones that is dependent of $t$, but I know that the answer should be:

$$\dot{x} = -R \omega sin(\omega t)+\ell cos(\phi)\dot{\phi}$$ and $$\dot{y} = R \omega cos(\omega t)+\ell sin(\phi)\dot{\phi}$$

So my question is, mathematically, how do I justify, that I just take the derivative of the second term in each equation, as it was dependent of $t$ (Which is obviously is), and the just put a $\dot{\phi}$ outside?

Answer 1

If $\phi$ is a function of $t$, then $x$, for example, is written as $$ x(t)=R\cos(\omega t)+\ell\sin(\phi(t)). $$ Applying the chain rule gives $$ \begin{align} \frac{d}{dt}x(t)&=\frac{d}{dt}R\cos(\omega t)+\frac{d}{d\phi}\ell\sin(\phi)\frac{d}{dt}\phi(t)\\ &=-R\omega\sin(\omega t)+\ell\cos(\phi)\dot{\phi}. \end{align} $$