THE BRACHISTOCHRONE PROBLEM; CYCLOIDS

4. THE BRACHISTOCHRONE PROBLEM; CYCLOIDS

We have already mentioned this problem in Section 1. We are given the points (x 1 ,y 1 ) and (x 2 ,y 2 ); we choose axes through the point 1 with the y axis positive downward as shown in Figure

4.1. Our problem is to find the curve joining the two points, down which a bead will slide (from rest) in the least time; that is, we want to minimize

be our reference level for potential energy. Then at the point (x, y) we have

Figure 4.1

kinetic energy = mv 2 = 1 m

2 2 dt potential energy = −mgy.

The sum of the two energies is zero initially and therefore zero at any time since the total energy is constant when there is no friction. Hence we have

mv 2 − mgy = 0 or v =

Then the integral which we want to minimize is

1 x 2 1+y ′2 dt =

ds

ds

dx.

2gy

2g x 1 y

Section 4 The Brachistochrone Problem; Cycloids 483

This is the integral (3.5) in Example 3, Section 3. Then the first integral of the Euler equation is given by (3.6):

√ = c.

x ′2 +1

Solving for x ′ , we get (4.1)

1 − cy This simplifies if we let cy = sin 2θ 2 = 1 2 (1 − cos θ). We find (Problem 1)

The equations for x and y as functions of θ are parametric equations of the curve along which the particle slides in minimum time. Since we have chosen axes to make the curve pass through the origin, x = y = 0 must satisfy the equations of the curve, so c ′ = 0, and we have

We shall now show that these are the parametric equations of a cycloid. Imagine

a circle of radius a (say a wheel) in the (x, y) plane rolling along the x axis. Let it start tangent to the x axis at the origin O in Figure 4.2. Place a mark on the circle at O. As the circle rolls, the mark traces out a cycloid as shown in Figure 4.3. Let point P in Figure 4.2 be the position of the mark when the circle is tangent to the x axis at A; let (x, y) be the coordinates of P . Since the circle rolled, OA = P A = aθ with θ in radians. Then from Figure 4.2 we have

x = OA − P B = aθ − a sin θ = a(θ − sin θ), (4.4)

y = AB = AC − BC = a − a cos θ = a(1 − cos θ).

Figure 4.3

Figure 4.2

484 Calculus of Variations Chapter 9

Equations (4.4) are the parametric equations of a cycloid. Comparing (4.3), we see that the brachistochrone is a cycloid as we claimed. Note that, since we have taken the y axis positive down (Figures 4.1 and 4.4), the circle which generates the brachistochrone rolls along the under side of the x axis.

From either (4.3) or (4.4), we see that all cycloids are similar; that is they differ from each other only in size (determined by a or c) and not in shape. Figure 4.4 is

a sketch of a cycloid for arbitrary a. If the given endpoints for the wire along which

the bead slides are O and P 3 , we see that

the particle slides down to P 2 and back up

Figure 4.4

to P 3 in minimum time! At point P 2 the

circle has rolled halfway around so OA = 1 2 · 2πa = πa. For any point P 1 on arc OP 2 ,P 1 is below the line OP 2 , and the coordinates (x 1 ,y 1 ) of P 1 have

or x 1 /y 1 < π/2. For points like P 3 on P 2 3 /y B, x 3 > π/2, whereas at P 2 , we have x 2 /y 2 = π/2 (Problem 2). Then if the right-hand endpoint is (x, y) and the origin is the left-hand endpoint, we can say that the bead just slides down, or slides down and back up, depending on whether x/y is less than or greater than π/2 (Problem 2).

PROBLEMS, SECTION 4

1. Verify equations (4.2). 2. Show, in Figure 4.4, that for a point like P 3 ,x 3 /y 3 > π/2 and for P 2 ,x 2 /y 2 = π/2. 3. In the brachistochrone problem, show that if the particle is given an initial velocity

v 0 4. Consider a rapid transit system consisting of frictionless tun-

nels bored through the earth between points A and B on the earth’s surface (see figure). The unpowered passenger trains would move under gravity. Using polar coordinates, set up R dt to be minimized to find the path through the earth requiring the least time. See Chapter 6, Problem 8.21, for the potential inside the earth. Find a first integral of the Euler equation. Evaluate the constant

of integration using dr/dθ = 0 when r = r 0 (where r 0 is the deepest point of the tunnel—see figure). Now solve for θ ′ = dθ/dr as a function of r. Substitute this into the integral for t and evaluate the integral to show that the transit time is

s R 2 −r 0 2 Z T=π R . Hint: Find 2 dt.

gR

r=r 0

Evaluate T for r 0 = 0 (path through the center of the earth—see Chapter 8, Problem 5.35); for r 0 = 0.99R. [For more detail, see Am. J. Phys. 34 701–704 (1966).]

In Problems 5 to 7, use Fermat’s principle to find the path followed by a light ray if the index of refraction is proportional to the given function.

5. − x 1/2

6. − (y − 1) 1/2 7. (2x + 5) 1/2

Section 5 Several Dependent Variables; Lagrange’s Equations 485