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A tautochrone or isochrone curve is the curve for which the time taken by a frictionless particle sliding down it under uniform gravity to its lowest point is independent of its starting point. The curve is a cycloid, and the time is equal to π times the square root of the radius over the acceleration of gravity. The tautochrone problem, the attempt to identify this curve, was solved by Christiaan Huygens in 1659. He proved geometrically in his Horologium oscillatorium (The Pendulum Clock, 1673) that the curve was a cycloid. This solution was later used to attack the problem of the brachistochrone curve. Jakob Bernoulli solved the problem using calculus in a paper (Acta Eruditorum, 1690) that saw the first published use of the term integral. Later mathematicians such as Joseph Louis Lagrange and Leonhard Euler looked for an analytical solution to the problem.
"Virtual gravity" solution Perhaps the simplest solution to the tautochrone problem is to note a direct relation between the angle of an incline and the gravity felt by a particle on the incline. A particle on a 90° vertical incline feels the full effect of gravity, while a particle on a horizontal plane feels effectively no gravity. At intermediate angles, the "virtual gravity" felt by the particle is . The first step is to find a "virtual gravity" that produces the desired behavior. The "virtual gravity" required for the tautochrone is simply proportional to the distance remaining to be traveled, which admits a simple solution: rac = - k^2s s = A cos kt , It can be easily verified both that this solution solves the differential equation and that a particle will reach at time from any starting height . The problem is now to construct a curve that will produce a "virtual gravity" proportional to the distance remaining to travel, i.e, a curve that satisfies: g sin heta = - k^2 s , The explicit appearance of the distance remaining is troublesome, but we can differentiate to obtain a more manageable form: g cos heta d heta = - k^2 ds , or ds = - rac cos heta d heta, This equation relates the change in the curve's angle to the change in the distance along the curve. We now use the Pythagorean theorem, the fact that the slope of the curve is equal to the tangent of its angle, and some trigonometric identities to obtain is terms of : egin ds^2 & = & dx^2 + dy^2 \ & = & left ( 1 + left ( rac ight ) ^2 ight ) dx^2 \ & = & ( 1 + an^2 heta ) dx^2 \ & = & sec^2 heta dx^2 \ ds & = & sec heta dx end Substituting this into the first differential equation lets us solve for in terms of : egin ds & = & - rac cos heta d heta \ sec heta dx & = & - rac cos heta d heta \ dx & = & - rac cos^2 heta d heta \ & = & - rac left ( cos 2 heta + 1 ight ) d heta \ x & = & - rac left ( sin 2 heta + 2 heta ight ) + C_x end Likewise, we can also express is terms of and solve for in terms of : egin rac & = & an heta \ dx & = & cot heta dy \ cot heta dy & = & - rac cos^2 heta d heta \ dy & = & - rac sin heta cos heta d heta \ & = & - rac sin 2 heta d heta \ y & = & rac cos 2 heta + C_y end Substituting and , we see that these equations for and are those of a circle rolling along a horizontal line — a cycloid: egin x & = & r ( sin phi + phi ) + C_x \ y & = & r ( cos phi ) + C_y end Solving for and remembering that is the time required for descent, we find the descent time in terms of the radius : egin r & = & rac \ k & = & rac sqrt \ T & = & pi sqrt end (Based loosely on Proctor, pp. 135-139) Abels solution Abel attacked a generalized version of the tautochrone problem (Abel's mechanical problem), namely, given a function that specifies the total time of descent for a given starting height, find an equation of the curve that yields this result. The tautochrone problem is a special case of Abel's mechanical problem when is a constant. Abel's solution begins with the principle of conservation of energy — since the particle is frictionless, and thus loses no energy to heat, its kinetic energy at any point is exactly equal to the difference in potential energy from its starting point. The kinetic energy is , and since the particle is constrained to move along a curve, its velocity is simply , where is the distance measured along the curve. Likewise, the gravitational potential energy gained in falling from an initial height to a height is , thus: egin rac m left ( rac ight ) ^2 & = & mg(y_0-y) \ rac & = & pm sqrt \ dt & = & pm rac \ dt & = & - rac rac dy\ end In the last equation, we've anticipated writing the distance remaining along the curve as a function of height (), recognized that the distance remaining must decrease as time increases (thus the minus sign), and used the chain rule in the form . Now we integrate from to to get the total time required for the particle to fall: T(y_0) = int_^ , dt = rac int_^ rac rac dy, This is called Abel's integral equation and allows us to compute the total time required for a particle to fall along a given curve (for which would be easy to calculate). But Abel's mechanical problem requires the converse — given , we wish to find , from which an equation for the curve would follow in a straightforward manner. To proceed, we note that the integral on the right is the convolution of with and thus take the Laplace transform of both sides: mathcalT(y_0) = rac mathcal left rac{1}{sqrt{y}} ight mathcal left rac{ds}{dy} ight Since , we now have an expression for the Laplace transform of in terms of 's Laplace transform: mathcalleft rac{ds}{dy} ight = sqrt z^ mathcalT(y_0) This is as far as we can go without specifying . Once is known, we can compute its Laplace transform, calculate the Laplace transform of and then take the inverse transform (or try to) to find . For the tautochrone problem, is constant. Since the Laplace transform of 1 is , we continue: egin mathcalleft rac{ds}{dy} ight & = & sqrt z^ mathcalT_0 \ & = & sqrt T_0 z^ \ end Making use again of the Laplace transform above, we invert the transform and conclude: rac = T_0 rac rac It can be shown that the cycloid obeys this equation. (Simmons, Section 54). Bibliography | ||||||||
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