Euler-Lagrange Equation

**MrJack1990** · November 14th 2010, 12:50 AM

Let $f = \frac{\dot{x}^2}{t^2}$

The Euler-Lagrange Equation is

$\frac{\partial f}{\partial x} - \frac{d}{dt} \Big(\frac{\partial f}{\partial \dot{x}} \Big) = 0$

I get

$0 - \frac{d}{dt} \Big(\frac{2 \dot{x}}{t^2} \Big) = 0$ ....(Eqn A)

$\frac{2 \ddot{x}}{t^2} = c$

$\ddot{x} = \frac{1}{2}ct^2$ , $\dot{x} = \frac{1}{6} ct^3$ , $x = \frac{1}{24}ct^4 + k$

but this is wrong. According to solutions

$x = \frac{1}{3}ct^3 + k$ where k is a constant.

I can't see where I have gone wrong from eqn A.

**mr fantastic** · November 14th 2010, 01:06 AM

Originally Posted by MrJack1990

Let $f = \frac{\dot{x}^2}{t^2}$

The Euler-Lagrange Equation is

$\frac{\partial f}{\partial x} - \frac{d}{dt} \Big(\frac{\partial f}{\partial \dot{x}} \Big) = 0$

I get

$0 - \frac{d}{dt} \Big(\frac{2 \dot{x}}{t^2} \Big) = 0$ ....(Eqn A)

$\frac{2 \ddot{x}}{t^2} = c$

$\ddot{x} = \frac{1}{2}ct^2$ , $\dot{x} = \frac{1}{6} ct^3$ , $x = \frac{1}{24}ct^4 + k$

but this is wrong. According to solutions

$x = \frac{1}{3}ct^3 + k$ where k is a constant.

I can't see where I have gone wrong from eqn A.

Integrate both sides wrt t: $\displaystyle \frac{2 \dot{x}}{t^2} = A \Rightarrow \dot{x} = Bt^2$ etc.

**MrJack1990** · November 14th 2010, 01:08 AM

$\frac{2 \ddot{x}}{t^2} = c$

Should this be

$\frac{2 \dot{x}}{t^2} = c$

**mr fantastic** · November 14th 2010, 01:13 AM

Originally Posted by MrJack1990

Should this be

$\frac{2 \dot{x}}{t^2} = c$

Well yes (why would it not?).

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