The tortoise is regarded as a particle, so the basic idea is as follows:
1. The motion of all particles is equivalent, that is, the relative geometric relationship between particles remains unchanged during the motion.
2. The instantaneous motion direction of any particle is the tangent direction of its trajectory curve.
According to the meaning of the question, it is more convenient to describe it in polar coordinates, and to deduce it in rectangular coordinates. Through n particles
* * * The center of the circle is the coordinate origin O, and any particle is selected on the OX axis, so there are:
The rectangular coordinates of the particle Pk labeled K are Pk(xk, yk) and the polar coordinates are Pk(r, θ+k*α), where θ is the particle.
The phase angle of P0, k is the number of particles, and α = 2π/n. Because at any time, the distance from each particle to the origin of coordinates is equal, so
Just like R.
Assume that the initial coordinate of P0 point is P0(R, 0) .................., that is, the initial circle radius is R.
Now study the motion of P0 particle, assuming P0(x, y) at any moment, that is, there is
x = r*cosθ.......dx = cosθ*dr - r*sinθ*dθ
y = r*sinθ.......dy = sinθ*dr + r*cosθ*dθ
The point tracked by P0 is P 1, and its coordinates are
x 1 = r*cos(θ+α)
y 1 = r*sin(θ+α)
The tangent slope of P0 trajectory is:
(y 1 - y)/(x 1 - x) = dy/dx...........................( 1)
(y 1 - y)*dx
= r *[sin(θ+α)-sinθ]*(cosθ* dr-r * sinθ* dθ)
= 2r * sin(α/2)* cos(θ+α/2)*(cosθ* dr-r * sinθ* dθ)
= 2r * sin(π/n)*[cos(θ+π/n)* cosθ* dr-r * cos(θ+π/n)* sinθ* dθ]
(x 1 - x)*dy
= r *[cos(θ+α)-cosθ]*(sinθ* dr+r * cosθ* dθ)
=-2r * sin(π/n)* sin(θ+π/n)*(sinθ* dr+r * cosθ* dθ)
=-2r * sin(π/n)*[sin(θ+π/n)* sinθ* dr+r * sin(θ+π/n)* cosθ* dθ]
Expand (1) to get
(y 1-y)* dx-(x 1-x)* dy = 0
Considering that R is infinitesimal only in the last stage of catching up, r*sin(π/n) can be rounded, and we get:
cos(π/n)*dr + r*sin(π/n)*dθ = 0
dr/r = -tg(π/n)*dθ.................................(2)
(2) Integral of
ln(r) = -tg(π/n)θ + C
Pay attention to the initial conditions, P0 coordinate is (r, 0), we get
R*e^[-tg(π/n)θ].................................(3)
Let K = tg(π/n), which is simplified as
r = R*e(-K*θ)
In order to solve the curve length, the standard formula: (DS) 2 = (DX) 2+(DY) 2 is used for derivation.
ds =[(dr)^2+(r*dθ)^2]^( 1/2)
=[(dr/dθ)^2+r^2]^( 1/2)*dθ
=[k^2+ 1]^( 1/2)*r*dθ
For ds integration, note that the integration interval is (0, ∞).
Define the integral symbol: S[f(x)dx](a, b) represents the definite integral of f(x), and the integral interval is (a, b).
Curve length L = S[ds](0, ∞)
l = r *[ 1+( 1/k)^2]^( 1/2)
Substitute K = tg(π/n) to get.
L = R/sin(π/n).....................................(4)
For the famous four-turtle problem in martin gardner, it is a special case of n = 4, and its total path length is L = √ 2 * r.
That is, the side length of a quadrilateral.