Please weave the fourier transform or anything from https://en.wikipedia.org/wiki/Fourier_transform into method 4 of the attached document. Write the answer under method 4 of the given document in approximately 2-3 pages. The objective of the document and everything is already stated.

Abstract
The purpose of this investigation is to create a road function for a rolling equilateral triangle. Based on three major assumptions which are an inverted catenary road, a centre of the triangle that remains constant during the rolling process, and rolling in a clockwise direction parallel to the x-axis, it has been made possible to compute the road function. Different methods can be used to create the function. However, this paper has only utilized five of the methods which are geometric calculus, polar functions, equating slopes, tentative, and linear transformations. All of these methods are different, but they have been unified by the three assumptions and the coordinates systems. Overall, the conclusion of the road function and the equilateral triangle suggests that the combination would be best for an industrial setting. The investigation has concluded that a road which is not smooth can only be used in such as setting due to factors such as friction, the weight of the load, and others.
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Road Function for a Rolling Equilateral Triangle
Since time immemorial, people have been drawn to the fact that a circular object, for instance, a car’s wheel, as the only thing that can roll on the solid ground. It is evident that the circular properties of the wheel allow it to roll on the ground smoothly. However, there are instances where the road is not always smooth, and it becomes difficult for the circular wheel to keep up with the conditions of the road. Additionally, there are circumstances when the wheel ends up bursting. However, does that dictate the fact that the car will completely remain stationary until the wheel is repaired? Nonetheless, this paper will try to introduce a new shape to the scene; an equilateral triangle. From a mathematical point of view, the paper will incorporate different principles to obtain the road function for a rolling equilateral triangle.
Statement of Task
The purpose of this paper is to investigate methods to obtain the road function for a rolling equilateral triangle.
Plan of Investigation
The investigation will be carried out using four different methods. They are geometric calculus, polar functions, equating slopes, and linear transformations.
 The road is an inverted catenary function. This function’s shape to similar to an inverted U-like shape which also resembles that of a parabolic arch. In this case, the road is made up of many joint pieces of inverted catenaries.
 The centre of the triangle must remain at a constant height during the rolling. The implication of this assumption is to ease the calculation process; if the centre is regularly shifting, then a uniform road function cannot be established. Ideally, the road function will utilise the triangle’s centre for the most of the different methods and models that will be introduced later in the paper.
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 Lastly, the triangle rolls clockwise and parallel to the x-axis. Despite the fact that
Figure 1.1: Inverted catenary road and a rotating equilateral triangle.
Method 1: Geometric Calculus
Let f(x) be the road function for the equilateral triangle. The equilateral triangle ΔABC has lengths of side s
Let g(x)=OP be the distance from the centre of the triangle to the point of contact with the road function (x,f(x))
Let l(x)=CP be the length that the triangle has rolled between the intervals x=0 and the point of contact’s x-coordinates assuming that the triangle starts rolling pointy side down with f(0)=0
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The figure below will illustrate these conditions
The above dimensions were obtained as illustrated in the diagram below in order to make calculations for the distances from the orthocentre of the ΔABC in the figure above which will be required to proceed with the paper.
In ΔOEC, cos 30=√32=???? ⇒ OC = 2??√3=?√3×√3√3=?√33
In ΔOBE, sin 30=12=????⇒OE=OB2=s√36
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The entire idea of rolling revolves around the key idea that the centre of the triangle must remain at a constant height throughout the rolling Therefore: f(x) +g(x) =constant, x∈R Reverting to the assumption where the triangle is pointy side down before rolling, this means that f(0)=0 and g(0)=?√33. This implies that for an equilateral triangle with side length s f(x)+g(x)=?√33………………………….(i) In order to obtain the road function of the equilateral triangle, g(x) and l(x) will have to be eliminated. We can do this firstly by using the Pythagoras Theorem to express g(x) in terms of l(x) ?(?)=√(?√36)2+(?2−?(?))2………….(ii) On substituting the value of g(x) of equation (ii) into equation (i), we get ?(?)+√(?√36)2+(?2−?(?))2=?√33 By making l(x) the subject of the formula ?(?)=16(3?−√3√3?2−8?√3?(?)+12?(?)2) But we know that l(x) is also an arc length because it is equal to the length that the triangle has rolled over the road function. Hence we can use the handy arc length formula below to proceed. ?(?)=∫√1+?′(?)2?0 By having eliminated l(x) and g(x), we have obtained the differential equation below ∫√1+?′(?)2?0=16(3?−√3√3?2−8?√3?(?)+12?(?)2)
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The above equation can be simplified by differentiating both sides with respect to x ???(∫√1+?′(?)2)?0=???(16(3?−√3√3?2−8?√3?(?)+12?(?)2)) √1+?′(?)2=???(√3√12?(?)2+3?2−8?√3?(?)6+?2) =√36???(√12?(?)2+3?2−8?√3?(?)) =√36×12(12?(?)2+3?2−8?√3?(?)−12×(24?(?)?′(?)−8?√3?′(?) =√3(24?(?)?′(?)−8?√3?′(?))12√12?(?)2+3?2−8?√3?(?) ∴√1+?′(?)2=√3(24?(?)?′(?)−8?√3?′(?))12√12?(?)2+3?2−8?√3?(?) Now we make f’(x) the subject of the formula and obtain: ?′(?)=−√−−12?(?)2+8?√3?(?)−3?2?2 It is impossible to proceed without assigning a value to s, say s=1, then, ?′(?)=−√12?(?)2−8√3?(?)+3 We then obtain the first order non-linear differential ordinary equation below ?′(?)√12?(?)2−8√3?(?)+3=−1 On solving using a differential equation solving software, due to its non-linearity, we obtain the hyperbolic function below: ?(?)=−?−2√3(?−?)+3?2√3(?−?)+4√3+2412
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Where c and d are the constants of integration. However, we have the initial condition where f(0)=0, which we will apply to the above equation to get ?(?)=−?−2√3(?−1)+3?2√3(?−1)+4√312 It can be observed that the portion of the function in between the two roots resembles our road function. So we can restrict the graph to the interval {0, 1.625} and replicate this n number of times on the x-axis, thus obtaining the domain {0, 1.625n}. The road obtained will look like the one below.
Method 2: Polar Functions
Let the road function be f (x) and let the wheel be described in polar form ?(?)=? . The distance of the center to the point of contact and the depth of the road at any point must match. Let the segment joining the center of the wheel to the point on the wheel touching the road at x make the angle ? with horizontal axis. Also ?(0)=?2
This further indicates that
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?[?(?)]=−?(?)−(?)
The road length is expressed as ∫√1+[?′(?)]2 ???0
For the wheel, it is; ∫√?(?)2+?′(?)22 ???(?)0
On equating both sides and ????.?.?.? 1+[?′(?)]2=(????)2[?(?)2+(????)2]
From equation (i), we can differentiate ?.??.? I to yield ????.????=−?′(?)
Hence; 1+[?′(?)]2=(????)2.?(?)2+[?′(?)]2
Which simplifies to ????=1?(?)
Hence, we obtain a differential equation which relates to the rolling function ?(?) to the shape of the wheel ?(?).
By separation of variables into ?(?).??=??, and using the initial conditions; ?=∫?(?)?−?2 ??
If we can solve this for ?(?), we obtain the shape of the road, since ?(?)=−?[?(?)]
Method 3: Equating Slopes
The long integrals can be avoided by simply matching the tangent slopes. Since the equation of the tangent line will be equal to the equation of the edge of the triangle, they can be equated to investigate the relationship −???=[?(?)]=?[?(?)]
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Method 4: Tentative
The equation of the edge in contact with the road on a Cartesian plane is at a 120° and -120° difference with the other two edges not in contact. The initial condition equations being: ?1=√3?
?2=ℎ (height of triangle) ?3=−√3? [?1?2?3]→[ ?1?1?2?2?2?2] =
This can be used to verify the shape of the road
Method 5: Linear Transformations
According to Rowland (2018), a linear transformation can be described as a linear map where there is a movement from point A to B. For instance, the linear transformation between two vectors q and w can be referred to as a map. This transformation can be described as follows: ?:?→?
The first transformation would be:
?(?1+?2). This equation describes any transformation for the vectors that are associated with q
Additionally, when it comes to a scalar; ?(∝1)=∝?(?)
The relationship between the linear transformation and the road function is that whenever the two vectors are on the same dimension, there is a high chance that the
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transformation will be invertible. In this case, the road can be considered to be vector q and the triangular wheel to be vector w.
The product of the transformation and its inverse would result in 1 (?×?−1=1)
This could be further elaborated on the road function such that a single transformation (or movement) of the wheel will cancel out with the inverse transformation of the road resulting to 1. This would, therefore, allow the triangular wheel to move on the inverted catenary road continuously.
Discussion
Mathematics has been involved in the entire process of finding the road function for a rolling equilateral triangle. Before mathematical investigation came the plan of investigation where three crucial assumptions were made which were used in drawing the analysis provided in the mathematical investigation section. First, the road is an inverted catenary function. This means that the road used to carry out the analysis is not straight, rather it has the shape of an inverted U. Additionally, the triangle to be used is an equilateral triangle. The centre of the triangular wheel will have an equidistant regardless of the vertex. The last assumption made was that the triangle rolls in a clockwise direction and parallel to the x-axis. This last assumption was to ensure that there is consistency in the investigation. If the triangle is rotating in both directions, then it would have been more complicated to come up with the road functions.
Geometric calculus is the first method which has been used to formulate the road function for a rolling equilateral triangle. In this method, the principles of integration and differentiation have been utilized (Macdonald, 2012). The geometric calculus is recommended for such an analysis since it incorporates different theories to the study. For instance, both differentiation and integration have been used. As suggested earlier, the centre of the triangle must remain at a constant height during the rolling process. Thus, the road
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function is bound to the function f(x) +g(x) =constant, x∈R. The geometric calculus has also been used to draw the shape of the road using a domain of {0, 1.625n}.
The polar function, as explained by Shroyer (2011), is a coordinate system where every point on the given plane is determined from a specific point. Additionally, an angle is included from the direction of the specific point. The polar function is quite similar to the geometric calculus method. The only difference is that the polar function uses the coordinates system which has been ignored by the earlier method.
The equating slopes is also a method that primarily focuses on the stationary centre of the equilateral triangles so as to make the process simple. In this case, it is assumed that the points presented by both the equilateral triangle and the road form a Cartesian plane. The coordinate system is then solved by equating the points of the x-axis of the road to the y-axis of the equilateral triangle. Additionally, the equation of the tangent line of the will be equal to the equation of the edge of the triangle. Thus, the equation [?(?)]=?[?(?)]?
The fourth method, tentative, has utilized the principles of a Cartesian plane and the angle exhibited by the equilateral triangle after upon coming into contact with the road. It was observed that the angles formed are 120° and -120°.
Lastly, the linear transformation was used to create the road function. The transformation makes use of the principles of a vector. If the road is one of the vectors and the triangular wheel is the other, the movement is made possible since the vector movement cancel each other out. Thus, the wheel can freely move on the inverted catenary.
Limitations of the Study
First, the function formulated would be difficult to use on an actual road. There are a number of reasons as to why this is impossible. The assumption was that the wheel could only move in a clockwise direction. This means that if that is the direction in which the wheel moves forwards, then it will be unable to move backwards as well. Additionally, there is a
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high chance that an equilateral triangle wheel of the same size to a circular wheel would be much slower on the road compared to a circular one. This is due to the fact that the distance from the centre of the triangle to one of its vertex is longer compared to the same distance but to one of the sides from the centre. A circular wheel has an equal radius thus it will move much faster.
Additionally, there is a high chance that the friction produced by the equilateral triangle and the road is expected to high (360 Create Ltd, 2018). This is due to the fact that none of the surfaces is smooth. More so, the equilateral triangle which has sharp corners. Even if the investigation was made to be practical, there is still a high chance that it would not be long-lasting since the more the friction, the faster the wheels will wear out.
Due to the sharpness of the vertices of the equilateral triangle, there is only a maximum weight that the wheel can manage to carry. A heavy load would eventually wear and tear the vertices of the triangle making movement difficult and almost impossible.
Conclusion and Recommendation
The investigation carried out shows that it is possible for an equilateral triangle wheel to move freely on the road provided that the road is an inverted catenary. The world is used to the typical circular wheel which, of course, is used in 99% of the settings. Additionally, the rolling of the triangle is possible if the three assumptions are observed. The first being that the road should be an inverted catenary function. The other two assumptions are related to the triangle. They are the centre of the triangle must remain at a constant height during the rolling process, and the triangle rolls clockwise and parallel to the x-axis.
The construction of the road function has been made possible utilizing the concepts of mathematics which are geometric calculus, the polar functions, equating slopes, tentative, and linear transformation. A deeper analysis shows that all of the five different methods based on
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the assumptions made earlier. However, the geometric calculus was the most suitable for the investigation since it was more detailed compared to the rest.
Today, the wheel has adopted the circular shape and all the vehicles, mostly personal and commercial, use these kinds of wheels. Thus, it would be impossible to introduce the triangular wheel while not having considered the fact that the road would not be smooth. Thus, the road analyzed in this paper would be suitable in an industrial setting. For instance, a conveyer belt could use triangular wheels on an inverted catenary to move products from one end to the other.
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References 360 Create Ltd, h. (2018). Friction and road surfaces – Road surfaces- Forces – Science – Pupils – 3M UK Streetwise – Bright thinking on our roads. Retrieved from http://www.3m.co.uk/intl/uk/3Mstreetwise/pupils-friction-roads.htm
Macdonald, A. (2012). Vector and Geometric Calculus. CreateSpace Independent Publishing Platform. Rowland, T. (2018). Linear Transformation — from Wolfram MathWorld. Retrieved from http://mathworld.wolfram.com/LinearTransformation.html Shroyer, S. (2011). The Polar Fuction: A Single Model Analysis. Stephen John Shroyer.