Finding Extremal - Stuck in the middle

**BrooketheChook** · November 16th 2010, 05:11 PM

I have a Euler-Lagrange equation

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

which for my f is

$\dot{x} + 1 - \frac{d}{dt} (\dot{x} + x + 1) = 0$

The solutions say I should end up with

$\ddot{x} - 1 = 0$

I end up with

$\dot{x} + 1 - \ddot{x} - \dot{x} - x = 0$

$\ddot{x} + x = -1$

I must be doing

$\frac{d}{dt} (\dot{x} + x + 1)$

incorrectly? What am I doing wrong?

**TwistedOne151** · November 16th 2010, 08:41 PM

Yes, you did appear to differentiate incorrectly. Noting that,
$\frac{d}{dt}(\dot{x}+x+1)=\frac{d}{dt}\dot{x}+\fra c{d}{dt}x+\frac{d}{dt}1$ ,
I'd ask you what is $\frac{d}{dt}1$ ?

--Kevin C.

**BrooketheChook** · November 17th 2010, 05:41 AM

Yes, I ended up figuring out where I went wrong. I am stuck on this one however....

$\frac{d}{dt} (1 + 2t^2 \dot{x}) = 0$

I get

$2t^2 \dot{x} = C$

$\dot{x} = \frac{1}{2} C t^{-2}$ ......wondering if $\frac{1}{2}$ should be absorbed by the constant

$x = -C t^{-3} + B$

but the solution is

$x = -At^{-1} + B$

**Ackbeet** · November 17th 2010, 06:06 AM

You're integrating incorrectly. The exponent gets more positive by one when you integrate. It does not become more negative.

**BrooketheChook** · November 17th 2010, 06:14 AM

Cannot believe I missed that.

**Ackbeet** · November 17th 2010, 06:30 AM

I've seen loads of people do that sort of thing. They're studying something more advanced, and they are so focused on learning the more advanced thing that they forget some elementary principle. Ah, well. Have a good one!

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