a question related to calculus of variations

Question

Consider a particle with coordinates $(x(t),y(t))$ on a smooth curve $\phi(x,y)=0$. If the particle moves from $(x(0),y(0))$ to $(x(\tau),y(\tau))$ for $\tau >0$ such that its kinetic energy is minimized, then

$(a)$ $\frac{\ddot{x}}{\phi_x}=\frac{\ddot{y}}{\phi_y}$.

$(b)$ $\dot{x}^2(0)+\dot{y}^2(0)=\dot{x}^2(\tau)+\dot{y}^2(\tau)$.

$(c)$ $\dot{x}\phi_x+\dot{y}\phi_y=0$.

$(d)$ $\dot{x}^2(0)=\dot{x}^2(\tau)$.

Now, if we consider this problem as minimizing a functional $J[x(t),y(t)]=\int_0^\tau F(x,\dot{x},y,\dot{y},t)dt$ in two dependent variables representing coordinates and one independent variable representing time, then Euler-Lagrange equation will give a family of extremals from which we can conclude the answer. But I am unable to find a way to relate the K.E. as a functional as written above, and include the curve $\phi(x,y)=0$ in the same. So is there another method for this problem or am I on the right track? Any help will be appreciated.

Answer 1

Hint.

Defining the Lagrangian with $X = (x(t),y(t))$

$$ L(X,\dot X,\lambda) = \frac m2\left(\dot x(t)^2+\dot y(t)^2\right)+\lambda \phi(x(t),y(t)) $$

we have the Euler-Lagrange movement equations

$$ L_X-\frac{d}{dt}\left(L_{X'}\right) = \left\{\begin{array}{rcl}m \ddot x(t)-\lambda\phi_x(x(t),y(t)) & = & 0\\ m \ddot y(t)-\lambda\phi_y(x(t),y(t)) & = & 0\end{array}\right. $$

Also

$$ \frac{d\phi}{dt} = 0 = \frac{\partial\phi}{\partial x}\frac{dx}{dt}+ \frac{\partial\phi}{\partial y}\frac{dy}{dt} = \phi_x\dot x+\phi_y\dot y $$