Very simple Lagrangian mechanics problem

[Je m'appelle]

Homework Statement

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Consider a mass m moving in a frictionless plane that slopes at an angle \alpha with the horizontal.
Write down the Lagrangian \mathcal{L} in terms of coordinates x measured horizontally across the slope, and y,
measured down the slope. (Treat the system as two-dimensional, but include the gravitational potential
energy.) Find the two Lagrange equations and show that they are what you should have expected.

Homework Equations

1. The Lagrangian, in terms of the kinetic energy T and potential energy U

\mathcal{L} = T - U

2. The Lagrange equation, for a generalized coordinate q_i

\frac{\partial \mathcal{L}}{\partial q_i} = \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right)

The Attempt at a Solution

There's one thing I don't understand here, which is the problem statement assuming there are two generalized coordinates — thus two resulting Lagrange equations —, whereas I only see one generalized coordinate y, since x (y) = y \cos \alpha and \alpha is constant.

So the Lagrangian would be

\mathcal{L} = \frac{1}{2}m \left(\dot{y}^2 \cos^2 \alpha + \dot{y}^2 \right) - mg y \sin \alpha

And the corresponding single Lagrange equation would be

\frac{\partial \mathcal{L}}{\partial y} = \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{y}} \right)

My guess would be I'm not treating it as two dimensional? But then how do I proceed to do so? I can't visualize what the problem wants me to do.

EDIT:

I think I have it, maybe I should call y as a 'parameter' and work this problem through \vec{r} = \vec{r}(a(y), b(y)), where a(y) = y \cos \alpha and b(y) = y?

EDIT 2:
Nevermind, I'd still be working solely in terms of y.

 

[DrClaude]

My interpretation of the problem is that x is varies along the width of the plane. In other words, the plane in which the motion happens in the x-y plane, not that x is horizontal and y vertical.