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mathlete
Nov30-04, 04:48 PM
I have a system of equations here:
\frac{d\theta}{dt} = u-(\frac{1}{u})cos(\theta)
\frac{du}{dt} = -sin(\theta)

It asks to show that C(\theta,u) = u^3-3ucos(\theta). That's fine, if it worked. From looking at it and taking the partial derivatives, it doesn't seem to be a conserved quantity. Any ideas, or am I missing something?

mathlete
Nov30-04, 05:23 PM
Sorry to add it here as well, but I don't want to start a new thread. For this system above, I am supposed to sketch certain solutions in the xy plane (this system is for the motion of a glider, theta is the angle it starts at, u is its initial velocity). What's the code to do this, I can't find it anywhere?

HallsofIvy
Dec1-04, 07:14 AM
We can't answer your first question because you haven't told us what "C(θ,u)
means! Without knowing that, we can't even say if it should be a conserved quantity.

mathlete
Dec1-04, 08:55 AM
We can't answer your first question because you haven't told us what "C(θ,u)
means! Without knowing that, we can't even say if it should be a conserved quantity.
I wasn't told what it means either :cry:

I assumed it was just a Hamiltonian of the system, so I tried taking the partials and it comes out close if you fudge a few numbers or variables here or there, but otherwise I get zilch.

robphy
Dec1-04, 01:01 PM
Given
\frac{d\theta}{dt} = u-(\frac{1}{u})cos(\theta)
and
\frac{du}{dt} = -sin(\theta)
the quantity defined by
C(\theta,u) = u^3-3ucos(\theta)
has the property that
\frac{dC}{dt}=0, i.e., it is unchanged as "t" varies.


\begin{align*}
0
&\stackrel{?}{=}
\frac{d}{dt}\left(u^3-3u \cos\theta \right)\\
&\stackrel{?}{=}
3u^2\dot u-3(\dot u \cos\theta - u\sin\theta\dot\theta)\\
&\stackrel{?}{=}
3u^2[-\sin\theta]-3([-\sin\theta]\cos\theta - u\sin\theta[u-\frac{1}{u}\cos\theta ])\\
&\stackrel{\surd}{=}0
\end{align*}


I was rushing when I did this... Please check.

mathlete
Dec1-04, 02:36 PM
Given
\frac{d\theta}{dt} = u-(\frac{1}{u})cos(\theta)
and
\frac{du}{dt} = -sin(\theta)
the quantity defined by
C(\theta,u) = u^3-3ucos(\theta)
has the property that
\frac{dC}{dt}=0, i.e., it is unchanged as "t" varies.


\begin{align*}
0
&\stackrel{?}{=}
\frac{d}{dt}\left(u^3-3u \cos\theta \right)\\
&\stackrel{?}{=}
3u^2\dot u-3(\dot u \cos\theta - u\sin\theta\dot\theta)\\
&\stackrel{?}{=}
3u^2[-\sin\theta]-3([-\sin\theta]\cos\theta - u\sin\theta[u-\frac{1}{u}\cos\theta ])\\
&\stackrel{\surd}{=}0
\end{align*}


I was rushing when I did this... Please check.
Ah, thanks very much! Much appreciated - I didn't think of it that way :smile:

Have an ideas on the maple one? I know that really isn't as much math oriented it's just that I don't really know how to use Maple :grumpy: