# a question related to calculus of variations

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.

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

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

• $L=T-V$ where $T$ is K.E. and $V$ is P.E. I'm unabale to understand how P.E.=$-\lambda \phi(x,y)$ ? I know P.E.=$mgy$ where $g$ is gravity and $m$ is mass of the particle. – Empty Nov 23 '19 at 13:28
• $V = 0$ because the particle moves in the plane $x\times y$ with $z = C^{te}$ Regarding $\lambda$ it is a Lagrange multiplier to consider the movement restriction to $\phi(x,y)=0$. The option (a) is true. – Cesareo Nov 23 '19 at 13:42
• So you are saying that total $L$ is the K.E. $T$ ? – Empty Nov 23 '19 at 13:49
• Another question: What about the other options ? Can you please give some hints.? – Empty Nov 23 '19 at 13:50
• Yes. Only kinetic energy. – Cesareo Nov 23 '19 at 13:50