Conserving Quantity in System of Equations: Idea or Miss?

[mathlete]

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]

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]

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]

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

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]

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]

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

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: