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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.

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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

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

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  • $\begingroup$ $L=T-V$ 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. $\endgroup$ – Empty Nov 23 '19 at 13:28
  • $\begingroup$ $V = 0$ 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. $\endgroup$ – Cesareo Nov 23 '19 at 13:42
  • $\begingroup$ So you are saying that total $L$ is the K.E. $T$ ? $\endgroup$ – Empty Nov 23 '19 at 13:49
  • $\begingroup$ Another question: What about the other options ? Can you please give some hints.? $\endgroup$ – Empty Nov 23 '19 at 13:50
  • $\begingroup$ Yes. Only kinetic energy. $\endgroup$ – Cesareo Nov 23 '19 at 13:50

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