Tautochrone curve

Tautochrone curve


300px|right|thumb|Four points run over a cycloid from different positions, but theyarrive at the bottom at the same time. The blue arrows show the points' acceleration. On the top is the time-position diagram.

A tautochrone or isochrone curve is the curve for which the time taken by an object sliding without friction in 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 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 g sin heta,. 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:

:frac{d^2s}dt}^2} = - 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 s=0, at time frac{pi}{2k} from any starting height A,. 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 = - frac{g}{k^2} 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 ds is terms of dx:

:egin{matrix}ds^2 & = & dx^2 + dy^2 \\ & = & left ( 1 + left ( frac{dy}{dx} ight ) ^2 ight ) dx^2 \\ & = & ( 1 + an^2 heta ) dx^2 \\ & = & sec^2 heta dx^2 \\ ds & = & sec heta dxend{matrix}

Substituting this into the first differential equation lets us solve for x in terms of heta:

:egin{matrix}ds & = & - frac{g}{k^2} cos heta d heta \\sec heta dx & = & - frac{g}{k^2} cos heta d heta \\dx & = & - frac{g}{k^2} cos^2 heta d heta \\ & = & - frac{g}{2 k^2} left ( cos 2 heta + 1 ight ) d heta \\x & = & - frac{g}{4 k^2} left ( sin 2 heta + 2 heta ight ) + C_xend{matrix}

Likewise, we can also express dx in terms of dy and solve for y in terms of heta:

:egin{matrix}frac{dy}{dx} & = & an heta \\ dx & = & cot heta dy \\cot heta dy & = & - frac{g}{k^2} cos^2 heta d heta \\dy & = & - frac{g}{k^2} sin heta cos heta d heta \\ & = & - frac{g}{2k^2} sin 2 heta d heta \\ y & = & frac{g}{4k^2} cos 2 heta + C_yend{matrix}

Substituting phi = - 2 heta, and r = frac{g}{4k^2},, we see that these equations for x and y are those of a circle rolling along a horizontal line — a cycloid:

:egin{matrix}x & = & r ( sin phi + phi ) + C_x \\y & = & r ( cos phi ) + C_y end{matrix}

Solving for k and remembering that T = frac{pi}{2k} is the time required for descent, we find the descent time in terms of the radius r:

:egin{matrix}r & = & frac{g}{4k^2} \\k & = & frac{1}{2} sqrt{frac{g}{r \\T & = & pi sqrt{frac{r}{gend{matrix}

(Based loosely on "Proctor", pp. 135-139)

Abel's solution

Abel attacked a generalized version of the tautochrone problem ("Abel's mechanical problem"), namely, given a function T(y) 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 T(y) 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 frac{1}{2} mv^2, and since the particle is constrained to move along a curve, its velocity is simply frac{ds}{dt}, where s is the distance measured along the curve. Likewise, the gravitational potential energy gained in falling from an initial height y_0, to a height y, is mg(y_0-y),, thus:

:egin{matrix}frac{1}{2} m left ( frac{ds}{dt} ight ) ^2 & = & mg(y_0-y) \\frac{ds}{dt} & = & pm sqrt{2g(y_0-y)} \\dt & = & pm frac{ds}{sqrt{2g(y_0-y) \\dt & = & - frac{1}{sqrt{2g(y_0-y) frac{ds}{dy} dy\\end{matrix}

In the last equation, we've anticipated writing the distance remaining along the curve as a function of height (s(y),), recognized that the distance remaining must decrease as time increases (thus the minus sign), and used the chain rule in the form ds = frac{ds}{dv} dv.

Now we integrate from y=y_0 to y=0 to get the total time required for the particle to fall:

:T(y_0) = int_{y=y_0}^{y=0} , dt = frac{1}{sqrt{2g int_{0}^{y_0} frac{1}{sqrt{y_0-y frac{ds}{dy} 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 frac{ds}{dy} would be easy to calculate). But Abel's mechanical problem requires the converse — given T(y_0),, we wish to find frac{ds}{dy}, 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 frac{ds}{dy} with frac{1}{sqrt{y and thus take the Laplace transform of both sides:

:mathcal{L} [T(y_0)] = frac{1}{sqrt{2g mathcal{L} left [ frac{1}{sqrt{y ight ] mathcal{L} left [ frac{ds}{dy} ight ]

Since mathcal{L} left [ frac{1}{sqrt{y ight ] = sqrt{pi}z^{-frac{1}{2, we now have an expression for the Laplace transform of frac{ds}{dy} in terms of T(y_0),'s Laplace transform:

:mathcal{L}left [ frac{ds}{dy} ight ] = sqrt{frac{2g}{pi z^{frac{1}{2 mathcal{L} [T(y_0)]

This is as far as we can go without specifying T(y_0),. Once T(y_0), is known, we can compute its Laplace transform, calculate the Laplace transform of frac{ds}{dy} and then take the inverse transform (or try to) to find frac{ds}{dy}.

For the tautochrone problem, T(y_0) = T_0, is constant. Since the Laplace transform of 1 is frac{1}{z}, we continue:

:egin{matrix}mathcal{L}left [ frac{ds}{dy} ight ] & = & sqrt{frac{2g}{pi z^{frac{1}{2 mathcal{L} [T_0] \\ & = & sqrt{frac{2g}{pi T_0 z^{-frac{1}{2 \\end{matrix}

Making use again of the Laplace transform above, we invert the transform and conclude:

:frac{ds}{dy} = T_0 frac{sqrt{2g{pi}frac{1}{sqrt{y

It can be shown that the cycloid obeys this equation.

("Simmons", Section 54).

External links

* [http://mathworld.wolfram.com/TautochroneProblem.html Mathworld]

Bibliography

* Simmons, George, "Differential Equations with Applications and Historical Notes," McGraw-Hill, 1972. ISBN 0-07-057540-1.
* Proctor, Richard, [http://historical.library.cornell.edu/cgi-bin/cul.math/docviewer?did=02260001&seq=9 A Treatise on The Cycloid and all forms of Cycloidal Curves] (1878), posted by [http://historical.library.cornell.edu/math/index.html Cornell University Library] .


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  • Tautochrone — Tau to*chrone, n. [Gr. ?, for ? ? the same + ? time: cf. F. tautochrone.] (Math.) A curved line, such that a heavy body, descending along it by the action of gravity, will always arrive at the lowest point in the same time, wherever in the curve… …   The Collaborative International Dictionary of English

  • Tautochrŏne — Tautochrŏne, diejenige Curve, in welcher ein Pendel schwingen muß, damit die Schwingungsdauer immer dieselbe sei, wie groß od. wie klein auch der Schwingungsbogen sei; diese Curve ist die Cycloide (s.d.) …   Pierer's Universal-Lexikon

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  • Curve — For other uses, see Curve (disambiguation). A parabola, a simple example of a curve In mathematics, a curve (also called a curved line in older texts) is, generally speaking, an object similar to a line but which is not required to be straight.… …   Wikipedia

  • tautochrone — ˈtȯd.əˌkrōn noun ( s) Etymology: French, from taut + Greek chronos time : a curve which is a cycloid under a horizontal base and down which the time of descent under gravity from every point to the lowest point is the same …   Useful english dictionary

  • Brachistochrone curve — A Brachistochrone curve (Gr. βραχίστος, brachistos the shortest, χρόνος, chronos time), or curve of fastest descent, is the curve between two points that is covered in the least time by a point like body that starts at the first point with zero… …   Wikipedia

  • tautochronous curve — tautochrona statusas T sritis fizika atitikmenys: angl. tautochrone; tautochronous curve vok. Tautochrone, f rus. таутохрона, f; таутохронная кривая, f pranc. courbe tautochrone, f; ligne tautochrone, f; tautochrone, f …   Fizikos terminų žodynas

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