Archive for December, 2007

Monday Math 1: The Isoperimetric Problem

December 31, 2007

Isoperimetric Problem

The isoperimetric problem is the problem of finding the closed plane curve with a given perimeter that encloses the greatest area. Below, I will use the calculus of variations in a proof of the long-known solution, the circle.

We let our enclosed region be D, and the enclosing curve ∂D, with orientation on ∂D chosen so that D is to it's left (counterclockwise). We thus want to minimize subject to the constraint . Note that these are quite different kinds of integrals, and cannot be combined directly via the multiplier method of including constraints.

However, using Green's theorem, which says that over a plane region D with (counterclockwise oriented) boundary ∂D we have:

By choosing , and using the fact that the area of D is , we obtain

So now we can combine the minimized quantity and the constraint using a multiplier. As both are integral constraints, the multiplier is a constant λ, so parametrizing our curve ∂D as (x(t),y(t)), so
and we want to find the extrema of

The integrand is , so we have the Euler-Lagrange equation for x:

Now, , so , and

Similarly for y:

So we have and
Squaring and summing these, we obtain:

This is the equation of a circle with radius λ and and center (C2,C1). From our original constraint , we obtain λ=p/2π.

The Pain Beam

December 29, 2007

No, it's not the twisted creation of a mad scientist in a grade B movie, it's here. While the concept and it's execution are pretty cool, scientifically speaking; and while I'm in favor of developing non-lethal weaponry; the abuse potential of something like this is rather worrying.

Orexin A: Substitute for Sleep…

December 29, 2007

This is a very interesting result, but it will be years, if ever, before we see medical use of these results; not to mention that we still don't know what the long-term side-effects may be. Still, anything that might help people with sleep disorders is a good thing, in my opinion.

Pocket Veto

December 29, 2007

Here is an analysis of the constitutionality of President Bush's current “pocket veto.” Look also at the first comment and at this Overlawyered post for info on the President's key objection to the bill.

testing again

December 29, 2007


Physics Friday 1

December 28, 2007

Bead on a hoop:

Let us consider a thin hoop of radius R. On this hoop is strung a small bead of mass m which can slide along the hoop. The bead-hoop interaction is frictionless. If we place the hoop on end, and rotate it about the vertical with a constant angular velocity , what is the equation of motion for the movement of the bead on the hoop? What are the equilibrium positions, and are they stable or unstable equilibria? We decribe the position of the bead on the hoop by the angle from the bottom of the hoop.

Rotating Reference Frame Approach:

We work in the reference frame of the hoop. As this is a rotating reference frame, it is non-inertial and we have fictitious forces to consider, namely the centrifugal and Coriolis forces.

As our bead’s motion is constrained to the plane of the hoop, and the axis of the frame’s rotation is also in that plane, the Coriolis force is perpendicular to the plane, and can be ignored (as it will be entirely canceled by a normal force of the hoop on the bead, and will not affect the motion at all). Thus we have as the only relevant fictitious force the centrifugal force straight away from the axis.

We need only consider the component of net force tangential to the hoop, as only motion along the hoop is possible. The component of gravity tangential to the hoop is, and the component of the centrifugal force is , so

The equation of motion:

If we define , then the above can be written as

Equilibrium angles are those where
which occurs when , so , so long as
Note when , we have only the point , as the two coincide.

Stability: Look at the sign of near the zeroes

1. For , we have only the one equilibrium at . We can see that for <img src=”<theta, 0″> and <img src=”\omega^2\cos\theta-\omega_0^2<\omega^2-\omega_0^2, so <img src=”\ddot{\theta}, and the equilibrium is stable.

2. For \omega_0″>, we have two equilibria, at and at . We can see that for <img src=”;\theta, 0″>. For <img src=”;\theta, 0″>, so 0″>; for <img src=”\theta_0<\theta, <img src=”\omega^2\cos\theta-\omega_0^2, so <img src=”\ddot{\theta}.
and thus the equilibrium at is unstable, and that at is stable.

For a 50 cm diameter hoop, we have

or approximately one revolution a second.

Nintendo should hire this guy…

December 28, 2007

This is very cool:

It demonstrates how powerful the motion-tracking of the Wiimote really is.

This is an interesting video

December 28, 2007

The American Civil War in four minutes.

ScienceDebate 2008 Update

December 28, 2007

There's been a lot of progress on ScienceDebate 2008: check out the links below:,0,2945719.story

Testing 2…

December 10, 2007