Originally posted by CalJust
Four bugs sit on the corners of a square of side one meter.
I posted this originally as a continuation of the "fly and the cysclists". Maybe someone else besides gastel can have a go at it - the former is slightly preoccupied with his last few marriages. Whilst I hae a lot of sympathy for his position, seems to me his mind is not on the job - in a mkanner ...[text shortened]... ot. The question is, however, what is the length of the spiral that the bugs travelled?
Here's my solution:
At the beginning of the problem, the bugs sit at the four corners of a square. Because the bugs are all doing the same thing, the path of each bug is the same just rotated 90 degrees from the last one. Due to this symmetry, the bugs stay in a square configuration. The square itself just rotates and shrinks.
To simplify matters, we'll deal with the first bug only. It's velocity V will always be along the side length connected to the bug in front of it. We can break V into two components: one directed towards the centre of the square (Vr); and the other perpendicular to this (Vtheta). (By breaking the velocity into two parts, it's easy to see that Vr shrinks the square, while Vtheta rotates it.) Because of the square geometry, the values of Vr and Vtheta are as follows:
Vr = Vtheta = V/SQRT(2)
Now we can set up some equations:
(1) dr/dt = -Vr (this is negative because the square is shrinking)
(2) dtheta/dt = Vtheta/r (this is not simply Vtheta, because the rotational effect increases with decreasing r)
Solving (1), we get r = r0 - Vr*t. Subbing this result into (2), we get:
(3) dtheta/dt = Vtheta/(r0 - Vr*t)
Solving (3), we get:
(theta-theta0)/Vtheta = -(1/Vr)*ln[(r0 - Vr*t)/r0]
Since r0 - Vr*t = r, we can simplify and solve for r:
r = r0*e^(-Vr/Vtheta)*(theta-theta0)
Recall that Vr = Vtheta = V/SQRT(2). Subbing this into the above equation we get:
r = r0*e^(theta0-theta)
This gives us the path of the first bug in polar coordinates. The length of this path is the integral of r*dtheta from theta0 to theta, as follows:
L = int[r0*e^(theta0-theta)]dtheta
= -r0*e^(theta0-theta)] from: theta0...theta
= -r0*[e^(theta0-theta) - e^(theta0-theta0)]
= r0*[1 - e^(theta0-theta)]
= r0*[1 - 1/e^(theta-theta0)]
The path length is the limit of this expression as theta tends towards infinity, which becomes:
lim (theta--}inf) r = r0*[1 - 1/inf] = r0*[1 - 0] = r0
Therefore, the path length is equal to r0. Interestingly, the bugs will spiral an infinite number of times, but will do so in a finite amount of time and with a bounded path length.