|
|
| (One intermediate revision by one other user not shown) |
| Line 1: |
Line 1: |
| [[Image:Tautochrone curve.gif|300px|right|thumb|Four points run over a cycloid from different positions, but they
| | re healthy your body will be strong and you have the best chances of adding to your height. The utmost desire of short heighted people is that they may acquire enough height that they would look just as their fellow normal people. Chlorine is also an important mineral and its function is to keep your muscles loose and limber as well as helps in cleaning the body system in order to fight off infections or bacteria that can get you ill. Healthy foods like salads, fruits and milk must be incorporated in your diet. This unit will show students the life cycle of a flower and the parts of a flower. <br><br>History holds testimony to the tallest man in the world: Robert Pershing Wadlow. Your nutrition is one of the most important things to think about when making these type of changes to your frame. Another thing to remember is that every person is not going to increase in height as they expect. This is a very enjoyable exercise and is really good to decompress your spine. This gives a woman the opportunity to dress warmly without sacrificing style and femininity. <br><br>If that bit just described you, then you are in luck. If this is not you, chances are you still could use some work. It should take approximately five to ten minutes to conduct both. In fact, these methods do not have any side effects on you. Just like there are exercises to lose weight, there also exist exercises that will help you grow taller naturally. <br><br>Pull your legs together and lift them over your head until they touch the floor behind your head. The ground being harder makes you back and neck firm which promotes health. than many other inferior products online that contain nothing but L-Arginine. Simply have faith, in many months the mage will surely come back to you. It during the night, the spin relaxes and decompresses, allowing for added height. <br><br>Also, sleep allows the body to grow and develop itself naturally. In an interview with the British Broadcasting Corporation (BBC) at the time of the discovery, researcher Dr Tim Frayling of the Peninsula Medical School in Exeter expressed the widely held belief that more genes will be revealed. Some girls claim platform shoes are rather comfortable, and they most likely are because the entire length of the foot is on a flat plane instead of the heel being higher. The exercises need to be strenuous yet controlled so as to result in greater muscle exhaustion. The pictures of female bodies are seen being displayed all over the place.<br><br>If you are you looking for more about what foods make you grow taller, [http://www.estampas.info/sitemap/ click through the up coming internet page], check out our web-page. |
| arrive at the bottom at the same time. The blue arrows show the points' acceleration along the curve. On the top is the time-position diagram.]]
| |
| | |
| A '''tautochrone''' or '''isochrone curve''' (from Greek prefixes [[wiktionary:/wiki/tauto-|tauto-]] meaning ''same'' or [[wiktionary:/wiki/iso-|iso-]] ''equal'', and [[wiktionary:/wiki/chrono|chrono]] ''time'') 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 [[Pi|π]] times the [[square root]] of the radius over the acceleration of gravity.
| |
| | |
| == The tautochrone problem ==
| |
| | |
| {{quotebox|width=30%|
| |
| quote=It was in the left hand try-pot of the Pequod, with the soapstone diligently circling round me, that I was first indirectly struck by the remarkable fact, that in geometry all bodies gliding along the cycloid, my soapstone for example, will descend from any point in precisely the same time.
| |
| |source=''[[Moby Dick]]'' by [[Herman Melville]], 1851}}
| |
| The tautochrone problem, the attempt to identify this curve, was solved by [[Christiaan Huygens]] in 1659. He proved geometrically in his ''[[Horologium Oscillatorium]]'', originally published in 1673, that the curve was a [[cycloid]].
| |
| | |
| :On a cycloid whose axis is erected on the perpendicular and whose vertex is located at the bottom, the times of descent, in which a body arrives at the lowest point at the vertex after having departed from any point on the cycloid, are equal to each other...<ref>
| |
| {{cite book | last = Blackwell | first = Richard J. | title = Christiaan Huygens' The Pendulum Clock | publisher = Iowa State University Press | date = 1986 | location = Ames, Iowa | isbn = 0-8138-0933-9}} Part II, Proposition XXV, p. 69</ref>
| |
| | |
| Huygens also proved that the time of descent is equal to the time a body takes to fall vertically the same distance as the diameter of the circle which generates the cycloid, multiplied by <sup>π</sup>⁄<sub>2</sub>. In modern terms, this means that the time of descent is <math>\pi \sqrt{r/g}</math>, where ''r'' is the radius of the circle which generates the cycloid and ''g'' is the [[gravity of Earth]].
| |
| | |
| 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]].<ref>{{cite web |url=http://jeff560.tripod.com/i.html|title=Earliest Known Uses of Some of the Words of Mathematics (I)|author=Jeff Miller|date=20 July 2010||accessdate=28 June 2012}}</ref>
| |
| | |
| [[File:CyloidPendulum.png|right|thumb|Schematic of a cycloidal pendulum.]]
| |
| The tautochrone problem was studied more closely when it was realized that a pendulum, which follows a circular path, was not [[isochronous]] and thus his [[pendulum clock]] would keep different time depending on how far the pendulum swung. After determining the correct path, Christiaan Huygens attempted to create pendulum clocks that used a string to suspend the bob and curb cheeks near the top of the string to change the path to the tautochrone curve. These attempts proved to not be useful for a number of reasons. First, the bending of the string causes friction, changing the timing. Second, there were much more significant sources of timing errors that overwhelmed any theoretical improvements that traveling on the tautochrone curve helps. Finally, the "circular error" of a pendulum decreases as length of the swing decreases, so better clock [[escapement]]s could greatly reduce this source of inaccuracy.
| |
| | |
| Later, the mathematicians [[Joseph Louis Lagrange]] and [[Leonhard Euler]] provided an analytical solution to the problem.
| |
| | |
| == Lagrangian solution ==
| |
| | |
| If the particle's position is parametrized by the [[arclength]] s(t) from the lowest point, the kinetic energy is proportional to <math>\scriptstyle \dot{s}^2</math>. The potential energy is proportional to the height y(s). In order to be an isochrone, the Lagrangian must be that of a [[simple harmonic oscillator]]: the height of the curve must be proportional to the arclength squared.
| |
| | |
| ::<math> y(s) = s^2 \,</math>
| |
| | |
| where the constant of proportionality has been set to 1 by changing units of length.
| |
| The differential form of this relation is
| |
| | |
| ::<math> dy = 2s \,ds </math>
| |
| ::<math> dy^2 = 4s^2 \,ds^2 = 4y \,(dx^2 + dy^2) \,</math>
| |
| | |
| Which eliminates ''s'', and leaves a differential equation for ''dx'' and ''dy''. To find the solution, integrate for ''x'' in terms of ''y'':
| |
| | |
| ::<math> {dx \over dy} = {\sqrt{1-4y}\over 2\sqrt{y}} \,</math>
| |
| | |
| ::<math> x = \int \sqrt{1-4u^2} \, du </math>
| |
| | |
| Where <math>\scriptstyle u=\sqrt{y}</math>. This integral is the area under a circle, which can be naturally cut into a triangle and a circular wedge:
| |
| | |
| ::<math> x= {1\over 2} u \sqrt{1-4u^2} + {1\over 4} \sin^{-1}(2u) \,</math>
| |
| ::<math> y= u^2 \,</math>
| |
| | |
| To see that this is a strangely parametrized [[cycloid]], change variables to disentangle the transcendental and algebraic parts: define the angle <math>\theta = \sin^{-1}(2u)</math>.
| |
| | |
| ::<math> 8x = 2\sin(\theta) \cos(\theta) + 2\theta = \sin(2\theta) + 2\theta \,</math>
| |
| ::<math> 8y = 2\sin(\theta)^2 = 1 - \cos(2\theta)\,</math>
| |
| | |
| Which is the standard parametrization, except for the scale of ''x'', ''y'' and ''θ''.
| |
| | |
| =="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 ''θ''. 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:
| |
| | |
| :<math>
| |
| \frac{d^2s}{{dt}^2} = - k^2s
| |
| </math> | |
| | |
| :<math>
| |
| s = A \cos kt \,
| |
| </math>
| |
| | |
| It can be easily verified both that this solution solves the differential equation and that a particle will reach ''s'' = 0 at time ''π''/(2''k'') 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:
| |
| | |
| :<math>
| |
| g \sin \theta = - k^2 s \,
| |
| </math>
| |
| | |
| The explicit appearance of the distance remaining is troublesome, but we can [[derivative|differentiate]] to obtain a more manageable form:
| |
| | |
| :<math>
| |
| g \cos \theta \,d\theta = - k^2 \,ds \,
| |
| </math>
| |
| | |
| or
| |
| | |
| :<math>
| |
| ds = - \frac{g}{k^2} \cos \theta \,d\theta\,
| |
| </math>
| |
| | |
| 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 (trigonometric function)|tangent]] of its angle, and some [[trigonometric]] identities to obtain ''ds'' in terms of ''dx'':
| |
| | |
| :<math>
| |
| \begin{align}
| |
| ds^2 & = dx^2 + dy^2 \\
| |
| \ & = \left ( 1 + \left ( \frac{dy}{dx} \right ) ^2 \right ) \,dx^2 \\
| |
| \ & = ( 1 + \tan^2 \theta )\, dx^2 \\
| |
| \ & = \sec^2 \theta \,dx^2 \\
| |
| ds & = \sec \theta \,dx
| |
| \end{align}
| |
| </math>
| |
| | |
| Substituting this into the first differential equation lets us solve for ''x'' in terms of ''θ'':
| |
| | |
| :<math>
| |
| \begin{align}
| |
| ds & = - \frac{g}{k^2} \cos \theta \,d\theta \\
| |
| \sec\theta \, dx & = - \frac{g}{k^2} \cos \theta\, d\theta \\
| |
| dx & = - \frac{g}{k^2} \cos^2 \theta \,d\theta \\
| |
| & = - \frac{g}{2 k^2} \left ( \cos 2 \theta + 1 \right ) \,d\theta \\
| |
| x & = - \frac{g}{4 k^2} \left ( \sin 2 \theta + 2 \theta \right ) + C_x
| |
| \end{align}
| |
| </math>
| |
| | |
| Likewise, we can also express ''dx'' in terms of ''dy'' and solve for ''y'' in terms of ''θ'':
| |
| | |
| :<math>
| |
| \begin{align}
| |
| \frac{dy}{dx} & = \tan \theta \\
| |
| dx & = \cot \theta \, dy \\
| |
| \cot \theta dy & = - \frac{g}{k^2} \cos^2 \theta \,d\theta \\
| |
| dy & = - \frac{g}{k^2} \sin \theta \cos \theta \,d\theta \\
| |
| & = - \frac{g}{2k^2} \sin 2 \theta \,d\theta \\
| |
| y & = \frac{g}{4k^2} \cos 2 \theta + C_y
| |
| \end{align}
| |
| </math>
| |
| | |
| Substituting <math>\phi = - 2\theta\,</math> and <math>r = \frac{g}{4k^2}\,</math>, we see that these equations for <math>x</math> and <math>y</math> are those of a circle rolling along a horizontal line — a [[cycloid]]:
| |
| | |
| :<math>
| |
| \begin{align}
| |
| x & = r ( \sin \phi + \phi ) + C_x \\
| |
| y & = r ( \cos \phi ) + C_y
| |
| \end{align}
| |
| </math>
| |
| | |
| Solving for ''k'' and remembering that <math>T = \frac{\pi}{2k}</math> is the time required for descent, we find the descent time in terms of the radius ''r'':
| |
| | |
| :<math>
| |
| \begin{align}
| |
| r & = \frac{g}{4k^2} \\
| |
| k & = \frac{1}{2} \sqrt{\frac{g}{r}} \\
| |
| T & = \pi \sqrt{\frac{r}{g}}
| |
| \end{align}
| |
| </math>
| |
| | |
| (Based loosely on ''Proctor'', pp. 135–139)
| |
| | |
| =={{Anchor|Abel problem}}Abel's solution==
| |
| | |
| [[Niels Henrik 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 <math>\frac{1}{2} mv^2</math>, and since the particle is constrained to move along a curve, its velocity is simply <math>\frac{ds}{dt}</math>, where <math>s</math> is the distance measured along the curve. Likewise, the gravitational potential energy gained in falling from an initial height <math>y_0\,</math> to a height <math>y\,</math> is <math>mg(y_0-y)\,</math>, thus:
| |
| | |
| :<math>
| |
| \begin{align}
| |
| \frac{1}{2} m \left ( \frac{ds}{dt} \right ) ^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{align}
| |
| </math>
| |
| | |
| 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 <math>ds = \frac{ds}{dy} dy</math>.
| |
| | |
| Now we integrate from <math>y=y_0</math> to <math>y=0</math> to get the total time required for the particle to fall:
| |
| | |
| :<math>
| |
| 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
| |
| </math>
| |
| | |
| 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 <math>\frac{ds}{dy}</math> would be easy to calculate). But Abel's mechanical problem requires the converse — given <math>T(y_0)\,</math>, we wish to find <math>\frac{ds}{dy}</math>, 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 <math>\frac{ds}{dy}</math> with <math>\frac{1}{\sqrt{y}}</math> and thus take the [[Laplace transform]] of both sides:
| |
| | |
| :<math>
| |
| \mathcal{L}[T(y_0)] = \frac{1}{\sqrt{2g}} \mathcal{L} \left [ \frac{1}{\sqrt{y}} \right ] \mathcal{L} \left [ \frac{ds}{dy} \right ]
| |
| </math>
| |
| | |
| Since <math>\mathcal{L} \left [ \frac{1}{\sqrt{y}} \right ] = \sqrt{\pi}z^{-\frac{1}{2}}</math>, we now have an expression for the Laplace transform of <math>\frac{ds}{dy}</math> in terms of <math>T(y_0)\,</math>'s Laplace transform:
| |
| | |
| :<math>
| |
| \mathcal{L}\left [ \frac{ds}{dy} \right ] = \sqrt{\frac{2g}{\pi}} z^{\frac{1}{2}} \mathcal{L}[T(y_0)]
| |
| </math>
| |
| | |
| This is as far as we can go without specifying <math>T(y_0)\,</math>. Once <math>T(y_0)\,</math> is known, we can compute its Laplace transform, calculate the Laplace transform of <math>\frac{ds}{dy}</math> and then take the inverse transform (or try to) to find <math>\frac{ds}{dy}</math>.
| |
| | |
| For the tautochrone problem, <math>T(y_0) = T_0\,</math> is constant. Since the Laplace transform of 1 is <math>\frac{1}{z}</math>, we continue:
| |
| | |
| :<math>
| |
| \begin{align}
| |
| \mathcal{L}\left [ \frac{ds}{dy} \right ] & = \sqrt{\frac{2g}{\pi}} z^{\frac{1}{2}} \mathcal{L}[T_0] \\
| |
| & = \sqrt{\frac{2g}{\pi}} T_0 z^{-\frac{1}{2}}
| |
| \end{align}
| |
| </math>
| |
| | |
| Making use again of the Laplace transform above, we invert the transform and conclude:
| |
| | |
| :<math>
| |
| \frac{ds}{dy} = T_0 \frac{\sqrt{2g}}{\pi}\frac{1}{\sqrt{y}}
| |
| </math>
| |
| | |
| It can be shown that the cycloid obeys this equation.
| |
| | |
| (''Simmons'', Section 54).
| |
| | |
| ==See also==
| |
| * [[Calculus of variations]]
| |
| * [[Beltrami identity]]
| |
| * [[Cycloid]]
| |
| * [[Catenary]]
| |
| * [[Equations of motion#Equations of uniformly accelerated linear motion|Uniformly accelerated motion]]
| |
| *[[Brachistochrone curve]]
| |
| | |
| ==References==
| |
| <references/>
| |
| | |
| ==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].
| |
| | |
| ==External links==
| |
| *[http://mathworld.wolfram.com/TautochroneProblem.html Mathworld]
| |
| | |
| [[Category:Curves]]
| |
| | |
| [[de:Zykloide#Die Tautochronie der Zykloide]]
| |