Find the arc length of the first arc of the cycloid. Remarkable curves and their properties. Geometric definition of a cycloid

The arc length of a cycloid was first calculated by the English architect and mathematician Wren in 1658. Wren proceeded from mechanical considerations reminiscent of the early work of Torricelli and Roberval. He considered the rotation of a rolling circle through a very small angle near the "lower" point of the generating circle. In order to give Wren's suggestive considerations probative force, one would have to consider a number of auxiliary theorems, and accordingly one would have to expend too much work.

It is much more convenient to use a longer, but gentler path. To do this, you need to consider a special curve that every flat curve has - its sweep.

Consider a convex arc AB of a curved line (Fig. 4.1). Let us imagine that a flexible inextensible thread of the same length as the arc AB itself is attached to the arc AB at point A, and this thread is “wound” on the curve and adjoins it tightly, so that its end coincides with point B. We will “unroll” - straighten the thread, keeping it taut, so that the free part of the CM thread will always be directed tangentially to the arc AB. Under these conditions, the end of the thread will describe some curve. It is this curve that is called a sweep, or, in Latin, involute original curve.

If the arc of the curve is not everywhere convex in one direction, if it, like the curve AB in Fig. 4.2 has a point C at which the tangent to the curve passes from one side to the other (such a point is called an inflection point), then in this case we can also talk about the development of the curve, but the reasoning will have to be a little more complicated.

Imagine that the thread is fixed just at the inflection point C (Fig. 4.2). The thread, winding from the arc BC, will describe the BMP curve - a scan.

Now imagine a thread wound around the arc AC of the original curve, but this thread is already elongated: at point C, a piece of thread CP is attached to it. By winding the elongated ACP thread with the SA curve, we get an RNA arc that, together with the BMP arc, forms a single continuous curve - continuous, but not everywhere smooth: the deflection point C of the original curve will correspond to the tip (return point) of the VMRNA curve: the VMRNA curve will be the evolvent (sweep) of the ICA curve.

These examples helped us get used to the new concepts of evolute and evolvent. Now let's study the unfolding of cycloidal curves.

Studying this or that curve, we often built an auxiliary curve - a "companion" of this curve. So, we cost a sinusoid - a companion of a cycloid. Now, starting from the given cycloid, we construct an auxiliary cycloid inextricably linked with it. It turns out that the joint study of such a pair of cycloids is in some respects easier than the study of one single cycloid. We will call such an auxiliary cycloid an accompanying cycloid.

Consider half of the arch of the cycloid AMB (Fig. 4.3). We should not be embarrassed that this cycloid is located in an unusual way (“upside down”). Let's draw 4 lines parallel to the directing line AK at distances a, 2a, 3a and 4 a. Let's build a generating circle in the position corresponding to the point M (in Fig. 4.3, the center of this circle is indicated by the letter O). The angle of rotation of MON will be denoted by c. Then the segment AN will be equal to bc (the angle u is expressed in radians).

We continue the diameter HT of the generating circle beyond the point T until it intersects (at the point E) with the straight line PP. On TE as a diameter, we construct a circle (with center O 1). Let us construct a tangent at the point M to the cycloid AMB. To do this, the point M must, as we know, be connected to the point T. We continue the tangent MT beyond the point T to the intersection with the auxiliary circle, and we will call the intersection point M 1. It is this point M 1 that we now want to deal with.

We denoted the angle MON by c. Therefore, the angle MTH will be equal to (inscribed angle based on the same arc). Triangle TO 1 M 1, obviously, isosceles. Therefore, not only the angle O 1 TM 1, but also the angle TM 1 O 1 will each be equal. Thus, the share of the angle TO 1 M 1 in the triangle TO 1 M 1 remains exactly p - q radians (recall that the angle 180? is equal to p radians). We also note that the segment NK is obviously equal to b (p - c).

Consider now a circle with a center O 2, shown in Figure 4.3 by a dashed line. From the drawing it is clear what kind of circle it is. If you roll it without sliding along the straight line CB, then its point B will describe the cycloid BB. When the dashed circle rotates through the angle p - c, the center O 2 will come to the point O 1, and the radius O 2 B will take the position O 1 M 1. Thus, the point M 1 constructed by us turns out to be a point of the cycloid BB.

The described construction assigns to each point M of the cycloid AMB a point M 1 of the cycloid VM 1 B. In fig. 4.4 this correspondence is shown more clearly. The cycloid obtained in this way is called the accompanying cycloid. On fig. 4.3 and 4.4, the cycloids depicted by bold dashed lines are accompanying with respect to the cycloids depicted by bold solid lines.

From fig. 4.3 it can be seen that the line MM 1 is the normal at the point M 1 to the accompanying cycloid. Indeed, this line passes through the point M 1 of the cycloid and through the point T of tangency between the generating circle and the directing line (the "lowest" point of the generating circle, as we used to say; now it turned out to be "highest", because the drawing is rotated). But this same line, by construction, is tangent to the "base" of the cycloid AMB. Thus, the original cycloid touches every normal of the accompanying cycloid. It is the envelope for the normals of the accompanying cycloid, i.e. her evolution. And the "accompanying" cycloid turns out to be simply an involute of the original cycloid!

Working on this cumbersome, but essentially simple construction, we proved a remarkable theorem discovered by the Dutch scientist Huygens. Here is the theorem: the evolute of a cycloid is exactly the same cycloid, only shifted.

Having constructed an evolute not to one arch, but to the entire cycloid (which, of course, can only be done mentally), then an evolute to this evolute, etc., we obtain Fig. 4.5, resembling tiles.

Let us pay attention to the fact that when proving Huygens' theorem, we did not use either infinitesimal, or indivisible, or approximate estimates. We did not even use mechanics, although we sometimes used expressions borrowed from mechanics. This proof is entirely in the spirit of the reasoning used by seventeenth-century scientists when they wanted to rigorously substantiate the results obtained with the help of various suggestive considerations.

An important corollary immediately follows from Huygens' theorem. Consider the segment AB in Fig. 4.4. The length of this segment is obviously equal to 4 a. Imagine now that a thread is wound on the arc of the AMB cycloid, fixed at point A and equipped with a pencil at point B. If we “wind” the thread, then the pencil will move along the development of the AMB cycloid, i.e. according to the cycloid BM 1 V. The length of the thread, equal to the length semi-arch of the cycloid, will obviously be equal to the segment AB, i.e., as we have seen, 4 a. Therefore, the length L of the entire arc of the cycloid will be equal to 8 a, and formula L=8 a can now be considered sufficiently rigorously proven.

We calculate the arc length using differential geometry. The solution obtained in this way will be much shorter and easier:

where t?

| r(t)|===2sin

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educational institution

"Belarusian State Pedagogical

Maxim Tank University

Faculty of Physics and Mathematics

Department of Mathematics and Methods of Teaching Mathematics

COURSE WORK ON THE TOPIC

"CYCLOID"

Minsk, 2016

cycloid arch tautochronous pendulum

Introduction
1. Basic properties of the cycloid
2. Geometric definition of a cycloid
3. Area of the arch of the cycloid
4. Arc length of the cycloid arch
5. The volume of the body obtained by rotating the arch of the cycloid
6. The best pendulum
Conclusion
Bibliography

INTRODUCTION

Theme of my term paper- cycloid. This curve is remarkable in many ways. It is also the trace of the point of the rim of the rolling wheel, it is also the curve of oscillations of a constant period, it is also the curve of the fastest descent. Nowadays, cycloidal curves are used in many technical calculations, and knowledge of these curves facilitates the study of machine parts. Without going into details, we mention that the properties of cycloidal curves are used in the construction of gear tooth profiles and in many other technical issues. Even from a purely applied point of view, these curves deserve the most serious attention. Therefore, I considered this topic relevant and interesting to study.

What problems arise in the study of the cycloid? First of all, it is necessary to give it a purely geometric definition, independent of mechanics. Next, you need to study its properties, consider the tangent, calculate the area bounded by the arch of the cycloid and its base, the length of the arc, the volume of the body formed by the rotation of the arch of the cycloid around the guide line.

The course work will consider in detail the tautochronous property of the cycloid and its application to create the best pendulum. The importance of pendulum clocks cannot be underestimated, since until recently they served as the most accurate clocks that ensured the service of time in astronomical observatories.

Another merit of the cycloid, which should be noted, is that it was used by scientists in the development of methods for studying curved lines, which led to the invention of differential and integral calculus. In my work, I propose to compare the calculation of the length of the arc of the cycloid arch, the surface area under the arch and the volumes of bodies formed by the rotation of the cycloid arc before the advent of integral calculus, long and not always absolutely strict, and using integration.

Objective: study of material on the topic "Cycloid"; study of the features of the best pendulum; comparison of the study of curved lines before and after the appearance of integral calculus, calculation of the length of the arc of the cycloid arch, the surface area under the arch and the volumes of bodies formed by the rotation of the cycloid arch.

1. MAIN PROPERTIES OF THE CYCLODY

First you need to find out what kind of curve is called a cycloid.

Consider a circle of radius a centered at point A. Let the circle in question roll without slipping along the OX axis. A curve described by any point on the circle is called cycloid.

This definition of the cycloid has never satisfied scientists: after all, it is based on mechanical concepts - speeds, additions of motions, etc. Therefore, geometers have always sought to give the cycloid a "purely geometric definition." But in order to give such a definition, you must first of all study the basic properties of the cycloid, using its mechanical definition. By choosing the simplest and most characteristic of these properties, we can use it as the basis of a geometric definition.

Let's start by studying the tangent and the normal to the cycloid. What's happened tangent to a crooked line, everyone imagines clearly enough; therefore we will not present it here. normal called the perpendicular to the tangent, restored at the point of contact. On fig. 1.1 shows the tangent and normal to the curve AB at its point M.

Consider a cycloid (Fig. 1.2). The circle rolls in a straight line AB. Let us assume that the vertical radius of the circle, which at the initial moment passed through the lowest point of the cycloid, managed to turn through an angle χ and took up the position OM. In other words, we believe that the segment M o T is such a fraction of the segment M o M 1 that the angle q is from a full turn. In this case, the point M 0 came to the point M.

The point M is the point of the cycloid that interests us.

Arrow Oh represents the speed of the center of the rolling circle. All points of the circle, including point M, have the same horizontal speed. But, in addition, point M takes part in the rotation of the circle. The speed MC, which the point M on the circle receives during this rotation, is directed tangentially MS 1 to the circle, i.e. perpendicular to the radius OM. And since in this case, the MS speed is equal in magnitude to the MP speed (i.e., the speed HE). Therefore, the parallelogram of velocities in the case of our movement will be a rhombus (the rhombus of the MSKR in Fig. 1.2). The MK diagonal of this rhombus will just give us a tangent to the cycloid.

All of the above makes it possible to solve the following construction problem: given the directing line AB of the cycloid, the radius r of the generating circle, and the point M belonging to the cycloid (Fig. 1.2). It is required to construct a tangent of the MK to the cycloid.

Having a point M, we can easily build a generating circle, in its position when the point on the circle falls into M. To do this, we first find the center O using a radius MO= r (point O must lie on a straight line parallel to AB, at a distance r from it). Then we construct a segment MP of arbitrary length, parallel to the guide line. Next we build a line MS 1 , perpendicular to OM. On this line, we lay off from the point M segment MC equal to MP. On the MS and MP, as on the sides, we build a rhombus. The diagonal of this rhombus will be the tangent to the cycloid at the point M.

This construction is purely geometric, although we obtained it using the concepts of mechanics. Now we can say goodbye to mechanics and get further consequences without her help. Let's start with a simple theorem.

Theorem 1. Angle between tangent to cycloid (at an arbitrary point) and directing line is equal to the complement of 90° half the angle of rotation of the radius of the generating circle.

In other words, in fig. 1.2

? KLT equal to or

We will now prove this equality. To shorten speech, we will agree to call the angle φ of rotation of the radius of the generating circle the "basic angle". This means that the MOT angle in Fig. 1.2 - the main angle. We will assume that the main angle is acute. For the case when the rolling circle makes more than a quarter of a full turn, the proof will be similar.

Consider the angle CMP. CM side is perpendicular OM(the tangent to the circle is perpendicular to the radius). Side MP (horizontal) perpendicular to FROM(towards the vertical). But the angle MOP, by the condition, is acute, and the angle CMP is obtuse. So the corners ILO and SMR add up to 180° (angles with mutually perpendicular sides, one of which is acute and the other obtuse).

So, the angle CMP is equal to 180° -ts. But, as you know, the diagonal of a rhombus bisects the angle at the vertex. Therefore, Hugo

KMR = 90° -,

Q.E.D.

Let us now turn our attention to the normal to the cycloid. Let's depict the left side of Fig. 1.2 is larger, and let's draw a normal ME (ME ? MK; rice. 1.3).

From fig. 1.3 it follows that the angle EMP is equal to the difference of the angles KME and KMR, i.e. equals 90° - ? KMP.

But we have just proved that the corner itself KMR equals 90° -

Thus, we get:

? RME= 90° - ? KMR= 90° - (90° -) =

We have proved a simple but useful theorem. Let's formulate it:

Theorem 2. Angle between the normal to the cycloid (at any point in it) and the guide line is equal to half of the "main angle".

Let us connect the "point (T) of the generating circle now with the point M" (the "current" point of the cycloid) with the "lower" one (with the point of contact of the generating circle and the directing line - Fig. 1.3). Triangle MOT is obviously isosceles (OM and FROM are the radii of the generating circle). The sum of the angles at the base of this triangle is equal to 180 ° - u, and each of the angles at the base is a rug of this sum. So, ? OMT= 90° - .

Pay attention to the corner RMT. It is equal to the angle difference HTA and OMR. We have now seen that ? OMT equal to 90° - ; as for the angle OMP, it is not difficult to find out what it is equal to. After all, the corner OMRequal to the angle DOM(internal cross-lying angles when parallel).

It is immediately obvious that ? DOM equal to 90 ° - c. Means, ? OMP= = 90° - c. Thus, we get:

RMT = ? HTA - ? OMR \u003d 90 ° - - (90 ° - c) \u003d.

It turns out a wonderful result: the angle RMT turns out to be equal to the angle PME (by Theorem 2). Therefore, direct ME and MT will merge! Our rice. 1.3 is not done quite right! The correct arrangement of lines is given in fig. 1.4.

Let us formulate the result obtained in the form of Theorem 3.

Theorem 3 (the first main property of the cycloid). The normal to the cycloid passes through the "lower" point of the generating circle.

This theorem has a simple corollary. The angle between a tangent and a normal is, by definition, a right angle. This is the angle inscribed in the circumference of the generating circle. Therefore, it must be based on the diameter of the circle. So, TT 1 -- diameter, and T 1 -- "upper" point of the generating circle. Let us formulate the result obtained.

Consequence (the second main property of the cycloid). The tangent to the cycloid passes through the "upper" point of the generating circle.

To explain this property, we need to construct a cycloid.

The construction of the cycloid is carried out in the following sequence:

1. On the guide horizontal line, a segment AA 12 is laid, equal to the length of the generating circle of radius r, (2pr);

2. Construct a generating circle of radius r, so that the directing line is tangent to it at point A;

3. The circle and segment AA 12 are divided into several equal parts, for example 12;

4. From the division points 1 1 , 2 1 , ...12 1 restore perpendiculars to the intersection with the continuation of the horizontal axis of the circle at points 0 1 , 0 2 , ...0 12 ;

5. From the division points of the circle 1, 2, ... 12, horizontal straight lines are drawn, on which serifs are made with arcs of a circle of radius r;

6. The obtained points A 1 , A 2 , ... A 12 belong to the cycloid.

On fig. 1.6 the base of the cycloid is divided into 6 equal parts;

The greater the number of divisions, the more accurate the drawing will be. At each point of the cycloid constructed by us, we draw a tangent, connecting the point of the curve with the "upper" point of the generating circle. In our drawing, we got seven tangents (two of them are vertical). Now drawing the cycloid by hand, we will take care that it really touches each of these tangents: this will significantly increase the accuracy of the drawing. In this case, the cycloid itself will go around all these tangents).

Let's carry out on the same fig. 1.6 normals at all found points of the cycloid. In total there will be five normals, not counting the guide. It is possible to construct by hand the envelope of these normals. If instead of six we took 12 or 16 division points, then there would be more normals in the drawing, and the envelope would be outlined more clearly. Such an envelope of all normals plays important role when studying the properties of any curved line. In the case of a cycloid, a curious fact is revealed: the envelope of the normals of the cycloid is exactly the same cycloid, only shifted by 2 a down and on ra to the right. This fact is characteristic of the cycloid.

2. GEOMETRIC DEFINITION OF A CYCLODY

We now give a definition of the cycloid as the locus of points without using mechanics. It's easiest to do so. Consider an arbitrary line AB(we will conditionally consider its direction to be horizontal) and a point on it M 0 . Next, consider all possible circles of a certain radius that are tangent to this line and located on one side of it. On every circle from the point T touching it with a straight line AB set aside (in the direction of the point M 0 ) arc tm, equal in length to the segment M 0 T. Locus of points M(taken on all the circles we have mentioned) and will be a cycloid.

Let us establish one more important property of the cycloid and try to use it as the basis for the study of this curve.

Consider a triangle MTT 1 (Fig. 2.1), formed by the vertical diameter of the generating circle, the tangent to the cycloid and the normal to it.

Injection MT 1 T, as inscribed in a circle, is equal to half the central angle based on the same arc, that is, equal to. Let's spend MK||AB and ME?AB. The segment ME will play a significant role in the future, so we will give it a name and designation: we will call it the “height” of the point M of the cycloid and denote it by the letter h. So the height of the point M the cycloids are its distance from the directing line.

Let's pay attention to the angle KMT. It is equal to the angle MT 1 T. From a triangle TMT 1 we get:

MT = 2 asin and from the TKM triangle:

CT = MT sin.

Comparing these results and noticing that CT = h, we finally get:

h = 2 a sin 2 .

We expressed the height of the point M in terms of the angle between the tangent at the point M and the vertical (we still consider the direction of the straight line AB to be the horizontal). Now let's express the sine of this angle in terms of "height". We get, obviously:

where through k the value constant for the given cycloid is denoted . We state the result obtained in the theorem.

Theorem 4. The sine of the angle between the tangent to the cycloid at point M and the vertical is proportional to the square root of the “height” of point M.

Obviously, any cycloid possesses this property. The question arises: to what extent does this property characterize exactly the cycloid: will any curve possessing this property necessarily be a cycloid? It can be proved that this will be the case, -- that the following (inverse) theorem is also true:

Theorem 5. Given a line AB and a point M, then the only curve that satisfies the conditions of Theorem 4 and passes through the point M is a cycloid.

Moreover, the radius of the generating circle of this cycloid is related to the coefficient k, which is mentioned in Theorem 4, by the following relationship:.

It is also worth paying attention to another remarkable curve, which is called companion of the cycloid.

Consider a cycloid (Figure 2.2). From its point M we lower the perpendicular to the vertical diameter of the generating circle. Let's get the point P. Let's do this construction for all points of the cycloid without exception.

Then the point P will describe some curve. This curve is called the companion of the cycloid.

Consider a cycloid, a point M on it and the corresponding point P on the satellite (Fig. 2.3). The center of the generating circle will be denoted by the letter Q. Then we will have:

QP=QM cos?MQP= a cos(180 0 -ts)=- a cosц=- a sin(90 0 -ts)= a sin(c -90 0).

Let us draw the locus of centers of the generating circle (straight line XX 1 ). from point M 0 put off by AB section M 0 K, equal. Let's spend KY ? XX 1 . The point of intersection of these lines will be denoted by the letter O. Section M 0 R on the directing straight line from the tip of the cycloid to the point of contact of the generating circle is equal to a c, where c is the main angle MQR, expressed in radians. Section OQ on the horizontal axis XX 1 equals M 0 R - M 0 K=a(c -), and the segment QP equals a sin? PMQ, i.e. equal to the sine of the angle (c -), multiplied by the radius a.

So from the point O horizontally, segments equal in length to the arcs of a circle are plotted, and along the vertical lines of sines of the angles corresponding to these arcs. This is the construction of an ordinary sinusoid known to us.

Means, The companion of a cycloid is called a sinusoid.

We will not delve into the study of the properties of this truly remarkable curve, we will only note as a fact that the area bounded by the companion of one arch of the cycloid and its base is equal to twice the area of the generating circle.

3. SQUARE OF THE ARCH OF THE CYCLODY

The first mention of the calculation of the area enclosed between the arch of the cycloid and its base is in the works of Viviani and Torricelli. At the same time, they used a special technique, which was called the "method of indivisibles." This method consists in dividing a curvilinear figure into infinitely thin strips, the area of which is relatively easy to find, and then these areas are added up. This technique led to the appearance of integral calculus half a century later.

Consider a figure bounded by a cycloid arch and a sinusoid. In figure 3.1, this figure, consisting of two petals, is circled with a thick line. Let's calculate its area.

First of all, let's build a mirror image of the right petal of the figure relative to the directing straight line AB (this reflection is given in Figure 4.1 by a dashed line). Let us then move this dashed curve upwards to the left and apply it to the left petal so that the arcs of the sinusoids included in the contour of each of the petals coincide. We get a convex figure, shaded in figure 3.1 and depicted separately in fig. 3.2. This figure is called the figure of Roberval. Let's establish the most important properties of this figure.

1. The convex figure M 0 RLM is equal in size to the two-petal figure depicted by the thick line in Fig. 3.1. This can be seen from the fact that it is "composed" of the same petals.

2. Any horizontal chord of a convex figure is equal to twice the chord of a petal at the same distance from AB. Indeed, the chords CE and PH (Fig. 3.1) of the right petal, equidistant from the generating circle, are equally distant from the center. So CT \u003d CE \u003d PH \u003d P 1 H 1 \u003d TL.

This gives an important result: the chord MR of a convex figure (Fig. 3.2) is equal to the chord of the generating circle SK, located at the same distance from the directing line.

Let us now consider the convex figure of Roberval and the circle tangent to the same lines AB and A 1 B 1, and connect the points of their intersection with the circle and with the contour of the convex figure in succession by straight line segments, as shown in the figure. The inscribed polygons thus obtained (HLMNPQRSTK and H 1 L 1 M 1 N 1 P 1 Q 1 R 1 S 1 T 1 K 1) will be called "corresponding" polygons on a series of trapezoids (and triangles). The areas of the “corresponding” trapezoids in the circle and in the Robervel figure, for example, NPRS and N 1 P 1 R 1 S 1, are equal, because these trapezoids, respectively, have the same lower bases, upper bases (corresponding chords) and heights. On fig. 3.2 equal-sized corresponding trapezoids are covered with the same hatching.

We will now indefinitely increase the number of "intermediate" straight lines parallel to AB, so that the distance between any neighboring pair tends to zero. Then in a circle we get a series of inscribed polygons, the number of sides of which increases indefinitely, and each of the sides tends to zero. We know that the areas S n of these polygons are limited by the area of a circle:

lim S n=p a 2 .

How will the sequence of polygons inscribed in a convex Roberval figure behave in this case? Square? n consecutive inscribed polygons will tend to area? figures of Roberval. It is known that if two variables retain correspondingly equal values for all their changes and one of them tends to a certain limit, then the other tends to the same limit. But each polygon inscribed in the Roberval figure is equal in size to the corresponding polygon inscribed in the circle. We therefore conclude that the limit of the areas of polygons inscribed in a Roberval figure is equal to the limit of the areas of the corresponding polygons inscribed in a circle; which means that the area of the convex Roberval figure is equal to the area of the generating circle:

From this we obtain an immediate consequence: the area of the two-petal figure is equal to the area of the generating circle.

Let's now look at Figure 3.1. The area of the figure AOTPBKA, as we have seen, is equal to twice the area of the generating circle. We have just determined the area of the two-petal figure: it is equal to the area of the generating circle. Hence, the area bounded by the arc of the cycloid and its base is equal to three times the area of the generating circle.

Now let's find the area enclosed between the arc of the cycloid and its base using differential geometry.

Where t? .

Let's find the derivative

4. LENGTH OF THE ARCH OF THE CYCLOYID

It is much more convenient to use a longer, but gentler path. To do this, you need to consider a special curve that every flat curve has - its sweep.

Imagine that the thread is fixed just at the inflection point C (Fig. 4.2). The thread, winding from the arc BC, will describe the BMP curve - a scan.

These examples helped us get used to the new concepts of evolute and evolvent. Now let's study the unfolding of cycloidal curves.

Having constructed an evolute not to one arch, but to the entire cycloid (which, of course, can only be done mentally), then an evolute to this evolute, etc., we obtain Fig. 4.5, resembling tiles.

An important corollary immediately follows from Huygens' theorem. Consider the segment AB in Fig. 4.4. The length of this segment is obviously equal to 4 a. Imagine now that a thread is wound on the arc of the AMB cycloid, fixed at point A and equipped with a pencil at point B. If we “wind” the thread, then the pencil will move along the development of the AMB cycloid, i.e. along the cycloid BM 1 B. The length of the thread, equal to the length of the semi-arch of the cycloid, will obviously be equal to the segment AB, i.e., as we have seen, 4 a. Therefore, the length L of the entire arc of the cycloid will be equal to 8 a, and formula L=8 a can now be considered sufficiently rigorously proven.

We calculate the arc length using differential geometry. The solution obtained in this way will be much shorter and easier:

where t?

r(t)=

| r(t)|===2sin

5. VOLUME OF THE BODY OBTAINED BY THE ROTATION OF THE ARCH OF THE CYCLOID

Let us find the volume of the body generated by the rotation of the cycloid arch around its base. Roberval found it by breaking the resulting egg-shaped body (Fig. 5.1) into infinitely thin layers, inscribing cylinders into these layers and adding up their volumes. The proof is long, tedious, and not entirely rigorous. Therefore, to calculate it, we turn to higher mathematics. Let us set the cycloid equation parametrically.

In integral calculus, when studying volumes, he uses the following remark:

If the curve bounding the curvilinear trapezoid is given by parametric equations and the functions in these equations satisfy the conditions of the theorem on the change of variable in a certain integral, then the volume of the body of rotation of the trapezoid around the Ox axis will be calculated by the formula:

Let's use this formula to find the volume we need.

In the same way, we calculate the surface of this body.

L=((x,y): x=a(t - sin t), y=a(1 - cost), 0 ? t ? 2р)

In integral calculus, there is the following formula for finding the surface area of a body of revolution around the x-axis of a curve specified on a segment parametrically (t 0 ?t ?t 1):

Applying this formula to our cycloid equation, we get:

Consider also another surface generated by the rotation of the cycloid arc. To do this, we will build a mirror reflection of the cycloid arch relative to its base, and we will rotate the oval figure formed by the cycloid and its reflection around the KT axis (Fig. 5.2)

First, let's find the volume of the body formed by the rotation of the cycloid arch around the KT axis. Its volume will be calculated by the formula (*):

Thus, we calculated the volume of half of this turnip body. Then the total volume will be

To find the surface area of this body of revolution using the integral, it is also necessary to split it in half horizontally and consider its upper part.

So the surface area of the resulting body is

6. THE BEST PENDULUM

Watching the swinging chandelier in the temple, Galileo discovered that the time of the full swing of the chandelier, i.e. the time after which it will return to its original position (the so-called oscillation period), was the same for both large and small spans. This observation led Galileo to think that a swinging body (a pendulum) could be used to control the running of a clock.

Galileo himself failed to make a clock with a pendulum, and it soon became clear that his observations were inaccurate. More accurate observations have shown that the period of oscillation of the pendulum is the greater, the larger the swing; but due to the inevitable friction of the axis and air resistance, the swing of an ordinary pendulum is constantly decreasing, which means that its period of oscillation will also decrease. Clock with an ordinary pendulum - otherwise called circular pendulum(because each point of it describes an arc of a circle) cannot go correctly.

Huygens figured out how to make a circular pendulum so that it has a constant swing. But he also solved another interesting problem - he answered the question of which curve the point should move so that the period of its oscillations does not depend on the amplitude. He came up with a design that moved the center of gravity of the pendulum along this curve.

Let's start with a device that ensures the correct movement of a clock with a circular pendulum. Gear A(Fig. 6.1) be driven by a chain with a weight V at the end. A gear is mounted on the axle of this wheel, tightly connected with it. This gear drives the hands of the clock, and therefore it is necessary that the wheel A moved evenly.

But the weight V, like any body, under the influence of gravity will move accelerated, imparting accelerated rotation to the wheel A. The pendulum should eliminate the difficulty MM.

Anchor WITH lying in the plane of the wheel A, tightly connected to the pendulum MM, MM pendulum itself lies behind the plane of the drawing and therefore it is drawn with a dotted line. Anchor equipped with teeth H and TO.

At the moment shown in Fig. 6.1 wheel A held by the left prong H anchors WITH. When the pendulum swings to the left, the prong H the anchor will release the captured cog of the wheel, and the wheel will turn, but only by the crawler, because the cog TO anchors will fall into the gap between the teeth of the wheel and delay it. When the pendulum then swings to the right again, the prong on that side will be held back by the anchor. So, with each full swing of the pendulum (back and forth), the wheel will turn exactly one tooth, i.e. to a certain fraction of the circle. The movement of the wheel will be strictly uniform. Anchor teeth, as seen from Fig. 6.1 are cut obliquely, so that the tooth of the wheel, which was held back by the anchor and released again, must slide on the oblique surface of the armature tooth. As a result, the anchor will give the pendulum a small push. These rhythmic pushes will make up for the loss of energy that the pendulum expends in overcoming friction and air resistance. Therefore, the swing of the pendulum will not decrease. Thus, the weight imparts energy to both the wheels of the clock and to pendulum, pendulum also regulates the clock.

What if the clock stops? It is not difficult to use them: it is enough to lift the weight and swing the pendulum. But at the same time, the swing swing may turn out to be different, and the clock will go evenly, but incorrectly (will go ahead or start to lag behind). Huygens came up with a device that allows you to easily adjust the course of the clock. But Huygens, as a true scientist, was interested in the question: what should be a "perfect" pendulum, a pendulum whose swing time does not depend on the magnitude of the swing? Let us consider in detail how Huygens solved this problem.

The word "tautochrone" means "uniform". This is how Huygens called the curve, which he began to look for, i.e. such a curve along which the center of gravity of the pendulum must move so that the period of its swing does not depend on the magnitude of the swing. The search was crowned with success: the mysterious tautochrone turned out to be a shortly studied cycloid. In doing so, Huygens showed exceptional wit. Suffice it to say that the doctrine of evolutes was created in the process of solving precisely this problem.

Huygens reasoned as follows. Imagine a groove in the form of a cycloid, as shown in Fig. 6.2.

A heavy ball rolls along this groove M. We will consider the ideal case - the case when there is no friction and air resistance.

Denote the cusp points of the cycloid by M 0 and M? 0 , and the radius of the generating circle through a. Draw a circle of radius a tangent to the cycloid at the vertex (circle centered O) and generating a circle in the position corresponding to the point M cycloids (given by the dashed line). Let's say that we put the ball at the point M 1 groove and let it go without a push. Under the influence of gravity, it will roll down. Let's study his movement.

What will be the speed of the ball when it reaches the point M cycloids? It's easy to calculate. Descending from the point M 1 point M, the ball will use up some potential energy. This energy loss is equal to the product of the weight of the ball mg(m -- ball weight, g-- acceleration of gravity) to "loss of height", i.e. on the difference in the height of the ball in positions M 1 and M, and heights are counted from some certain level e.g. from ground level. From whatever level the heights are counted, their difference in our case will be equal to the segment NM. So, the potential energy loss of the ball will be equal to mg· HM.

But by virtue of the law of conservation of energy, the lost potential energy of the ball will turn into the kinetic energy of its movement, which, as you know, is equal to, if we denote the still unknown speed of the ball through. Equating this kinetic energy with the lost potential energy, we obtain the equation

from which we immediately find the value of the desired speed

The direction of this velocity is also easy to determine. It will be directed tangent to the cycloid, i.e. along the chord ML(Fig. 6.2), where L- the "lowest" point of the generating circle.

We will be interested not so much in the speed itself as in its vertical projection, i.e. "the speed of the ball's descent", the rate of change of its height. This vertical projection is easy to calculate: it is equal to, where is the angle between the chord ML and vertical. Chord AT circle with center O, is obviously equal and parallel to the chord ML, and therefore the angle LMP equal to the angle CAT, which is noted in Fig. 6.2. So:

We will compare uneven motion along a cycloid with uniform motion along a circle. To this end, we construct an auxiliary circle as follows: through the vertex A cycloid is drawn perpendicular AD(diameter of circle with center O), and through the starting point M 1 of the movement of the ball, a parallel M 1 B is drawn to its base. Let the point of intersection of these parallels and perpendiculars be denoted by the letter V. The circle built on AB, as on the diameter, and will be the required auxiliary circle. It is not yet clear how exactly it is better than other circles.

Let's start with the fact that we will connect the vertical component of the speed of the ball with the elements of the auxiliary circle. We have:

because NM = VC. From a triangle ACT we get:

But AT=2a cos , and therefore

We substitute the found cosine value into the expression for MR marked with an asterisk (*). We get:

The last root is equal to the average proportional between the segments VC and AK, i.e. between the segments of the hypotenuse AB triangle ABC, into which the latter is divided by the height SC. But this average is proportional, according to the well-known theorem on proportional lines in right triangle, is equal to the height SC:

VK·AK=SK 2 .

Therefore, for the vertical component MR the velocity of the ball along the cycloid we finally obtain:

MP=· KS.

Quantities a and g given to us from the very beginning and are not connected with a point M, nor with its initial position M 1 . Thus, the movement of the ball along the cycloid is completely determined by the chord KS auxiliary circle, i.e. end point position WITH on this circle.

Consider the uniform motion of a point WITH along an auxiliary circle with an angular velocity of radians per second, i.e. degrees per second. In this case, the speed of the point WITH along a circle will be equal to the product of the radius of the circle and the angular velocity, expressed in radians (per second), i.e. is equal to

How fast is the point going down? WITH, How fast does its distance from the straight line change? M 0 M? 0 with uniform motion of the point WITH around the circumference? This is easy to calculate.

Speed the movement of a circle point is directed tangentially to the circle, i.e. perpendicular to the radius. Its projection onto the vertical is equal to the speed itself multiplied by the cosine of the angle Fig. 6.3. But the angle is obviously equal to the angle CSR 1: both are obtained by subtracting the angle O 1 CE from a right angle. Cosine of an angle CSR 1 equals . For vertical projection of speed uniform motion around the circle we find:

A remarkable result is obtained: when a point moves uniformly along a circle, its projection onto the vertical moves in exactly the same way as the projection onto the vertical of a ball rolling along a cycloid. The projections of both velocities at any moment of time are equal to each other. But it follows from this that the point of the circle from V v A and a ball on a cycloid of M 1 in A will come at the same time. This time is easy to determine. We have already said that a point on the auxiliary circle makes radians a second, in other words, it will turn one radian in seconds, and radians in. Exactly the same time is needed for our ball to roll down the cycloid from the point M 1 point A. It will take him the same time to rise by inertia to the point M? 1 , the same - to go down again, and the same - to rise and return to the starting position (to the point M one). This means that the time of full oscillation of the ball (oscillation period) will be equal to:

This is a very wonderful formula. We see that the period of motion of the ball along the cycloidal groove is completely determined by the dimensions of the groove (the radius of the generating circle of the cycloid) and the acceleration of gravity. Point position M 1 on the cycloid, its distance from the straight line M 0 M? 0 it does not matter. From whatever point of the cycloid the ball starts moving, the period of its oscillation will be the same.

Huygens thought about how to use the tautochronous property of the cycloid to construct a "perfect" pendulum. How to make the pendulum ball move tautochronously without resorting to grooves and similar devices with a lot of friction? Thinking about this, Huygens came to the concepts of evolute and involute.

Let's make a template consisting of two identical semi-arches of the cycloid having a common cusp O(Fig. 6.4). We denote the radius of the generating circle, as always, by a. We will strengthen the template vertically, and at the cusp O tie a thread with a length equal to 4 a-- i.e. twice the diameter of the generating circle of the cycloid. free end thread T give it a heavy ball.

The ball will describe during its movement the unfolding of the cycloid ASOEV, because the thread will wrap around the template. But the unfolding of the cycloid is exactly the same cycloid. So the curve VMTRA, along which the ball moves, will be a cycloid generated by a circle of radius a.

If we place the ball at an arbitrary point M and leave it to itself, it will begin to oscillate, and the period of these oscillations will not depend on the choice of the point M. Even if, under the influence of friction and air resistance, the swing of oscillations decreases, the time of oscillation of the pendulum remains unchanged. Truly this pendulum will be tautochronous!

Let us now consider small oscillations of the pendulum along the arc AB cycloids (Fig. 6.5). If these oscillations are very small, then the influence of the guide template will not be practically felt and the pendulum will move almost like an ordinary pendulum with a length l=4a, suspended at a point O. Path AB cycloidal pendulum will practically not differ from the path CE circular pendulum length 4 a. This means that the period of small oscillations of an ordinary circular pendulum with a length l=4a will not practically differ from the period of the cycloidal pendulum. Entering into the formula

with which we met above, instead of a equal to it, we obtain the expression for the period of small oscillations of a circular pendulum in terms of its length:

CONCLUSION

In the process of completing my course work, I studied materials on the topic of the cycloid, studied the features of the best pendulum, compared a rather elegant, but not very simple study of the cycloid before the advent of integral calculus, with the simplest and most familiar, studied in differential geometry and mathematical analysis; once again convinced of the need to study these disciplines. As it turned out, the cycloid has a huge practical application not only in mathematics, but also in technological calculations, in physics.

The work on the study of this topic turned out to be quite exciting and interesting.

BIBLIOGRAPHY

1. Berman G.N. Cycloid. -M., 2007. -113s.

2. Savelov A.A. Flat curves. - M., 1960. - 293 p.

3. Fikhtengolts G.M. Basics mathematical analysis. -M., 2005, v.2. -464 p.

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LEmniskates
Equation in polar coordinates:
r 2 = a 2 cos2θ

(x 2 + y 2) 2 = a 2 (x 2 - y 2)

Angle between AB" or A"B and the x-axis = 45 o

The area of one loop \u003d a 2 / 2

CYCLOID

Area of one arc = 3πa 2

Arc length of one arch = 8a

This is the curve described by the point P on a circle of radius a, which rolls along the x axis.

HYPOCYCLOIDS WITH FOUR POINTS
Equation in rectangular coordinates:
x 2/3 + y 2/3 = a 2/3

Equations in parametric form:

Area enclosed by curve = 3πa 2 /8

Whole curve arc length = 6a

This is a curve described by a point P on a circle of radius a/4, which rolls inside a circle of radius a.

CARDIOID
Equation: r = a(1 + cosθ)

Area bounded by curve = 3πa 2 /2

Curve arc length = 8a

This is the curve described by the point P on a circle of radius a, which rolls outside the circle of radius a. This curve is also a special case of Pascal's snail.

CHAIN LINE
The equation:
y = a(e x/a + e -x/a)/2 = acosh(x/a)

This is the curve that would be followed by a chain suspended vertically from point A to point B.

THREE-PETAL ROSE
Equation: r = acos3θ

The equation r = acos3θ is like a curve obtained by rotating counterclockwise along a curve of 30 o or π/6 radians.

In general, r = acosnθ or r = asinnθ has n lobes if n is odd.

FOUR-PETAL ROSE
Equation: r = acos2θ

The equation r = asin2θ is like a curve obtained by rotating counterclockwise along a 45o or π/4 radian curve.

In general, r = acosnθ or r = asinnθ has 2n petals if n is even.

EPICYCLOID
Parametric equations:

This is the curve described by a point P on a circle of radius b as it rolls along the outside of a circle of radius a. The cardioid is a special case of the epicycloid.

GENERAL HYPOCYCLOID
Parametric equations:

This is the curve described by a point P on a circle of radius b as it rolls along the outside of a circle of radius a.

If b = a/4, the curve is a hypocycloid with four cusps.

Trochoid
Parametric equations:

This is the curve described by a point P at a distance b from the center of a circle with radius a as it rolls along the x-axis.
If b is a shortened cycloid.
If b > a, the curve has the shape shown in fig. 11-11 and called trochod.
If b = a, the curve is a cycloid.

TRAKTRICE
Parametric equations:

This is the curve described by the end point P of a stretched string of length PQ when the other end Q moves along the x-axis.

VERZIERA (VERZIERA) AGNESI (SOMETIMES AGNESIA CURLS)
Rectangular equation: y = 8a 3 /(x 2 + 4a 2)

Parametric equations:

B. In the figure, a variable line OA intersecting y = 2a and a circle with radius a centered on (0,a) at A and B, respectively. Any point P on the "curl" is determined by constructing lines parallel to the x and y axes, and through B and A, respectively, and defining the intersection point P.

Cartesian sheet
Equation in rectangular coordinates:
x 3 + y 3 = 3axy

Parametric equations:

Loop area 3a 2 /2

Asymptote equation: x + y + a = 0.

CIRCLE EVOLVENT
Parametric equations:

This is the curve described by the end point P of the string as it unwinds from a circle of radius a.

EVOLVENT ELLIPSE
Equation in rectangular coordinates:
(ax) 2/3 + (by) 2/3 = (a 2 - b 2) 2/3

Parametric equations:

This curve is the envelope normal to the ellipse x 2 /a 2 + y 2 /b 2 = 1.

OVALS OF CASINI
Polar equation: r 4 + a 4 - 2a 2 r 2 cos2θ = b 4 .

This is a curve described by a point P such that the product of its distance from two fixed points [distance 2a to the side] is a constant b 2 .

Curve as in the figures below when b a respectively.

If b = a, the curve is lemniscate

Pascal's snail
Polar equation: r = b + acosθ

Let OQ be a line joining the center O to any point Q on a circle of diameter a passing through O. Then the curve is the focus of all points P such that PQ = b.

The curve shown in the figures below when b > a or b

CISSOID OF DIOCLE
Equation in rectangular coordinates: y 2 = x 3 /(2a - x)

Parametric equations:

It is a curve described by a point P such that the distance OP = the distance RS. Used in task doubling cube, i.e. finding the side of a cube that has twice the volume of a given cube

ARCHIMEDES SPIRAL
Polar equation: r = aθ

Remember-those oran-same-layer-mass-co-ka-ta-fo-you - light-from-ra-zha-te-whether, attached-la-u-schi-e-sya to the knitting needles ve-lo-si-ped-no-go-ko-le-sa? At-cre-pim ka-ta-fot to sa-mo-mu rim-du ko-le-sa and follow his tra-ek-to-ri-she. According to the lu-chen-nye curves, they are attached to the family of the cyclo-id.

At the same time, Ko-le-so is called-zy-va-et-sya about-from-in-dia-circle (or circle-stu) cyclo-lo-and-dy.

But come on, let's go back to our century and re-re-sya-dem on more modern tech-no-ku. On the way, bye-ka, ka-mu-shek fell, someone got stuck in a pro-tek-to-re ko-le-sa. Pro-ver-now-shis a few laps with a co-le-som, where-yes-le-tit ka-men, when are you-cheat from the pro-tek-to-ra? Against the right-le-tion of the movement of the mo-the-cycle-la or on the right-le-tion?

As a matter of fact, free movement of the body na-chi-na-et-sya according to ka-sa-tel-noy to that tra-ek-that-rii, according to which then swarm it moved. Ka-sa-tel-naya to cyclo-lo-and-de always on-right-le-on on-right-le-tion of movement and passes through the upper point ku pro-from-in-dia-circle-no-sti. In the right direction, the movement of the same-le-tit and our ka-mu-shek.

Do you remember how you rode in the puddles in your childhood on a bicycle-lo-si-pe-de without a rear-wing? Mok-para-lo-ka on your back -ta-ta.

The 17th century is the century of cyclo-and-dy. The best scientists have studied its amazing properties.

Some kind of tra-ek-to-riya brings the body, moves under the action of the force of gravity, from one point in another for a short tea-neck time? This would be one of the first tasks of that na-at-ki, someone-paradise this hour no-sits on-the-name of va-ri-a-tsi-on-noe number.

Mi-ni-mi-zi-ro-vat (or mak-si-mi-zi-ro-vat) can be different things - the length of the way, speed, time. In the za-da-che about bra-hi-hundred-chrone mi-ni-mi-zi-ru-et-x name-but time -call-em: Greek βράχιστος - the smallest, χρόνος - time).

The first thing that comes to mind is a direct tra-ek-to-riya. Come on, let’s also look at the re-re-ver-well-thuyu cyclo-lo-and-du with the point of return-vra-ta in the upper from for-given-th- receipt. And, following Ga-li-leo Ga-li-le-em, - four-vert-tin-ku circle-no-sti, connecting-nya-yu-shchy on-shi dots.

For some reason, Ga-li-leo Ga-li-ley ras-smat-ri-val four-vert-tin-ku circles and considered that this was the best in the sense le time-me-no tra-ek-that-riya descent-ka? He entered lo-ma-nye into it and noted that with an increase in the number of links, the time will decrease. From-here-yes, Ga-li-lei naturally went to the circle, but made the wrong conclusion that this tra-something -ria is the best among all possible. As we can see, the best-shey tra-ek-to-ri-she is-la-et-sya cyclo-lo-and-yes.

Through two given points, it is possible to pass a single cycle-lo-and-du with the condition that at the top point there is toch-ka vra-ta tsik-lo-and-dy. And yes, when the cycle-lo-and-de comes-ho-dit-sya under-no-mother-sya to pass through the second point, she will still be Cree -howl nai-sko-rei-she-go descent!

Another one-to-the-beauti-si-vaya for-da-cha, connected with the cyclo-lo-and-doy, is for-da-cha about ta-at-that-chron. Translated from Greek, ταύτίς means “the same”, χρόνος, as we already know - “time”.

Let's make three one-on-one mountains with pro-fi-lem in the form of cyclo-lo-and-dy, so that the ends of the mountains are owl-pa-da-li and ras-po-la-ga-lis at the top of the cycle-lo-and-dy. Let's put three bo-ba on different you-so-you and let's-dim from-mash-ku. Udi-vi-tel-ney-shiy fact - everyone would come down one-now-men-but!

In the winter, you can build a mountain of ice in the yard and check this property live.

For-da-cha about ta-at-that-chrono-st-it in na-ho-de-nii such a crooked howl that, na-chi-naya from any-bo-go-initial- but in the same way, the time to descend to a given point will be the same.

Hri-sti-en Guy-gens do-ka-zal, that the only ta-at-that-chron-noy is-la-et-sya cyclo-lo-and-yes.

Ko-nech-but, Guy-gen-sa is not in-te-re-so-val descent along the icy mountains. At that time, scientists didn’t have such a growth for-no-mother-for-y-for-mi out of love for art. For-yes-chi, some-rye studied, was-ho-di-li from life and for-pro-owls of those-no-ki of that time-me-no. In the 17th century, co-ver-sha-yut-sya is already distant sea plans. Shi-ro-tu mo-rya-ki can you determine-de-lyat already to-a-hundred-accurate-but-accurate-but, but surprise-vi-tel-but that you don’t know how to determine -de-lyat with-everything. And one of the pre-la-gav-shih-s-so-bos from-me-re-niya shi-ro-you was os-no-van on na-li-chii accurate chrono-no-met- ditch.

The first, who for-du-small de-lat ma-yat-no-no-know-s, someone would be accurate, was Ga-li-leo Ga-li-ley . However, at that moment when he na-chi-na-et them re-a-li-zo-vy-vat, he is already old, he is blind, and for the remaining year the scientist does not have time to make a watch in his life. He za-ve-shcha-et this son-well, however, that honey-lit and na-chi-na-et for-no-mother-sya ma-yat-no-one, the same only pe- red death and not having time to re-a-li-zo-vat for-we-sat down. The next sign of the swarm was Christi-sti-an Guy-gens.

He remarked that re-ri-od-ko-le-ba-niya usually-go ma-yat-ni-ka, ras-smat-ri-vav-she-go-sya Ga-li- le-em, for-wee-sit from from-to-the-beginning-but-of-the-lo-same, i.e. from am-pli-tu-dy. For-thinking about how-to-va should be a tra-ek-that-riya of the movement of a load, so that the time of ka-che-tion on it does not -se-lo from am-pli-tu-dy, he re-sha-et for-da-chu about ta-at-that-chron. But how to move the load along the cycle-lo-and-de? Pe-re-vo-dya theo-re-ti-che-research-before-va-tion into the practical-ti-che-plane, Guy-gens de-la-et “cheeks” , on someone-rye na-ma-you-va-et-sya ve-roar-ka ma-yat-no-ka, and re-sha-et a few more ma-te-ma-ti-che tasks. It does-ka-zy-va-et that the “cheeks” should have the profile of the same cycle-lo-and-dy, the same way that evo-lu-that cycle-lo-and-dy yav-la-et-sya cycle-lo-and-yes with the same pa-ra-met-ra-mi.

In addition to that, the pre-lo-female Guy-gen-som construction of the cyclo-lo-and-far-no-go ma-yat-no-ka poses-in-la-et by -count the length of the cycle-lo-and-dy. If there is a blue-no-dot-ku, the length-for-some-swarm is equal to four-you-rem ra-di-u-sam about-from-in-dya-sche-go circle, mak-si-mal-but from-clo-thread, then its end will be at the point of re-re-se-che-niya “cheeks” and cyclo-lo-and-dy-tra- ek-to-rii, i.e. at the top of the cyclo-lo-and-dy-“cheek-ki”. Since this is in-lo-vi-on the length of ar-ki cyclo-and-dy, then the full length is equal to eight ra-di-u-sam pro-of-in- uncle-th circle.

Christ-sti-an Guy-gens made a cyclic-lo-and-distant ma-yat-nik, and hours with him pro-ho-di-whether is-py-ta-niya in the sea pu-te-she-stvi-yah, but did not come. However, just like watches with the usual ma-yat-no-one for these purposes.

From-what, one-to-one, still su-sche-stu-yut cha-so-me-ha-bottom-we with the usual-but-vein-ny ma-yat-no-one ? If you look at it, then with small from-clo-not-no-yah, like the red-no-go ma-yat-no-ka, “cheeks” cyclo-lo- and-far-but-go ma-yat-no-ka almost no eye-zy-va-yut influence. Correspondingly, the movement along the cyclic-lo-and-de and around the circumference with small from-clo-no-no-yah is almost owl-pa- yes-yut.

Cyclomid (from the Greek KhklpaydYut - round) - a flat transcendental curve. A cycloid is defined kinematically as the trajectory of a fixed point of a generating circle of radius r rolling without slipping in a straight line.

Equations

Let us take the horizontal coordinate axis as a straight line along which the generating circle of radius r rolls.

The cycloid is described by parametric equations

Equation in Cartesian coordinates:

The cycloid can be obtained as a solution to the differential equation:

Properties

· Cycloid -- periodic function along the abscissa, with a period of 2pr. It is convenient to take singular points (customs points) of the form t = 2рk, where k is an arbitrary integer, as the boundaries of the period.
· To draw a tangent to the cycloid at its arbitrary point A, it is enough to connect this point with the top point of the generating circle. Connecting A to the lowest point of the generating circle, we get the normal.
· The length of the cycloid arch is 8r. This property was discovered by Christopher Wren (1658).
· The area under each arch of the cycloid is three times larger than the area of the generating circle. Torricelli claims that this fact was discovered by Galileo.
· The radius of curvature of the first arch of the cycloid is equal to.

· An "inverted" cycloid is a curve of steepest descent (brachistochrone). Moreover, it also has the property of tautochronism: a heavy body placed at any point of the cycloid arc reaches the horizontal in the same time.
· The oscillation period of a material point sliding along an inverted cycloid does not depend on the amplitude, this fact was used by Huygens to create an accurate mechanical watch.
· The evolution of a cycloid is a cycloid that is congruent to the original one, namely, it is parallel shifted so that the vertices turn into “points”.
Details of machines that perform simultaneously uniform rotational and forward movement, describe cycloid curves (cycloid, epicycloid, hypocycloid, trochoid, astroid) (cf. the construction of Bernoulli's lemniscate).