Brachistochrone Problem

Find the shape of the curve down which a bead sliding from rest and accelerated by gravity will slip (without friction) from one point to another in the least time. The term derives from the Greek betarhoalphachiiotasigmatauomicronsigma (brachistos) "the shortest" and chirhoomicronnuomicronsigma (chronos) "time, delay."

The brachistochrone problem was one of the earliest problems posed in the calculus of variations. Newton was challenged to solve the problem in 1696, and did so the very next day (Boyer and Merzbach 1991, p. 405). In fact, the solution, which is a segment of a cycloid, was found by Leibniz, L'Hospital, Newton, and the two Bernoullis. Johann Bernoulli solved the problem using the analogous one of considering the path of light refracted by transparent layers of varying density (Mach 1893, Gardner 1984, Courant and Robbins 1996). Actually, Johann Bernoulli had originally found an incorrect proof that the curve is a cycloid, and challenged his brother Jakob to find the required curve. When Jakob correctly did so, Johann tried to substitute the proof for his own (Boyer and Merzbach 1991, p. 417).

In the solution, the bead may actually travel uphill along the cycloid for a distance, but the path is nonetheless faster than a straight line (or any other line).

The time to travel from a point P_1 to another point P_2 is given by the integral

 t_(12)=int_(P_1)^(P_2)(ds)/v,
(1)

where s is the arc length and v is the speed. The speed at any point is given by a simple application of conservation of energy equating kinetic energy to gravitational potential energy,

 1/2mv^2=mgy,
(2)

giving

 v=sqrt(2gy).
(3)

Plugging this into (◇) together with the identity

 ds=sqrt(dx^2+dy^2)=sqrt(1+y^('2))dx
(4)

then gives

t_(12)=int_(P_1)^(P_2)(sqrt(1+y^('2)))/(sqrt(2gy))dx
(5)
=int_(P_1)^(P_2)sqrt((1+y^('2))/(2gy))dx.
(6)

The function to be varied is thus

 f=(1+y^('2))^(1/2)(2gy)^(-1/2).
(7)

To proceed, one would normally have to apply the full-blown Euler-Lagrange differential equation

 (partialf)/(partialy)-d/(dx)((partialf)/(partialy^'))=0.
(8)

However, the function f(y,y^') is particularly nice since x does not appear explicitly. Therefore, partialf/partialx=0, and we can immediately use the Beltrami identity

 f-y^'(partialf)/(partialy^')=C.
(9)

Computing

 (partialf)/(partialy^')=y^'(1+y^('2))^(-1/2)(2gy)^(-1/2),
(10)

subtracting y^'(partialf/partialy^') from f, and simplifying then gives

 1/(sqrt(2gy)sqrt(1+y^('2)))=C.
(11)

Squaring both sides and rearranging slightly results in

[1+((dy)/(dx))^2]y=1/(2gC^2)
(12)
=k^2,
(13)

where the square of the old constant C has been expressed in terms of a new (positive) constant k^2. This equation is solved by the parametric equations

x=1/2k^2(theta-sintheta)
(14)
y=1/2k^2(1-costheta),
(15)

which are--lo and behold--the equations of a cycloid.

If kinetic friction is included, the problem can also be solved analytically, although the solution is significantly messier. In that case, terms corresponding to the normal component of weight and the normal component of the acceleration (present because of path curvature) must be included. Including both terms requires a constrained variational technique (Ashby et al. 1975), but including the normal component of weight only gives an approximate solution. The tangent and normal vectors are

T=(dx)/(ds)x^^+(dy)/(ds)y^^
(16)
N=-(dy)/(ds)x^^+(dx)/(ds)y^^,
(17)

gravity and friction are then

F_(gravity)=mgy^^
(18)
F_(friction)=-mu(F_(gravity)N^.)T
(19)
=-mumg(dx)/(ds)T,
(20)

and the components along the curve are

F_(gravity)T^.=mg(dy)/(ds)
(21)
F_(friction)T^.=-mumg(dx)/(ds),
(22)

so Newton's Second Law gives

 m(dv)/(dt)=mg(dy)/(ds)-mumg(dx)/(ds).
(23)

But

(dv)/(dt)=v(dv)/(ds)
(24)
=1/2d/(ds)(v^2)
(25)
 1/2v^2=g(y-mux)
(26)
 v=sqrt(2g(y-mux)),
(27)

so

 t=intsqrt((1+(y^')^2)/(2g(y-mux)))dx.
(28)

Using the Euler-Lagrange differential equation gives

 [1+y^('2)](1+muy^')+2(y-mux)y^('')=0.
(29)

This can be reduced to

 (1+(y^')^2)/((1+muy^')^2)=C/(y-mux).
(30)

Now letting

 y^'=cot(1/2theta),
(31)

the solution is

x=1/2k^2[(theta-sintheta)+mu(1-costheta)]
(32)
y=1/2k^2[(1-costheta)+mu(theta+sintheta)].
(33)

Wolfram Web Resources

Mathematica »

The #1 tool for creating Demonstrations and anything technical.

Wolfram|Alpha »

Explore anything with the first computational knowledge engine.

Wolfram Demonstrations Project »

Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more.

Computerbasedmath.org »

Join the initiative for modernizing math education.

Online Integral Calculator »

Solve integrals with Wolfram|Alpha.

Step-by-step Solutions »

Walk through homework problems step-by-step from beginning to end. Hints help you try the next step on your own.

Wolfram Problem Generator »

Unlimited random practice problems and answers with built-in Step-by-step solutions. Practice online or make a printable study sheet.

Wolfram Education Portal »

Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more.

Wolfram Language »

Knowledge-based programming for everyone.