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My inspiration for this math exploration arose after spending a couple of weeks on the topic calculus. I was always wondering the math higher level syllabus, which it does not seem applicable in our daily life.
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“ Amazing as always, gave her a week to finish a big assignment and came through way ahead of time. ”
Therefore, I often argue with my teacher, why do we have to learn mathematics if it is not applicable in the real world. However, when I do more research, I have discovered mathematics exist with physics do lie around us. As a result, I want to apply what I have learnt in class that is relevant to real-life situation, in an attempt to justify what I was learning was real.
When I travelled to Japan, I saw numerous overhead cables (shown in fig.1).
With these in mind, I have studied these curves insightfully in respect to their characteristics, and discovered these curves are named as catenary— a hyperbolic cosine function. Not to mention, when I was queuing up at school tuck shop, I have found another example of catenary— metal stanchion chains.
Nowadays, the transmission of electrical power has become an essential component in our daily life. Abundant overhead cables are always suspended outreach of people, however there are potential danger which cables may hand closely to the ground, poles may collapse or fall over. Hence, putting people in danger. In addition, these transmission cables must be maintained and replaced by specialists. In an effort to minimize the cost of replacement whilst having a safe as well as long-lasting cables, they are required to understand how these cables behave when handing between the transmission poles.
Therefore, this exploration focuses on determining the minimum length of a hanging cable required to hang between two vertical poles.
With the intention of calculating the length of the hanging chain, hyperbolic function and Newton’s method are required. However, the IB syllabus do not encounter these topics, And that’s why I found challenging and have chosen for my math exploration.
My aim in this exploration is to apply mathematics outside of the lessons and to delve into calculus insightfully— the hyperbolic function and the Newton’s method. This math exploration attempts to determine the minimum length of a handing chain required to hang effectively between poles. Throughout this study, I will write equations describing it and solve it exactly, hence we can have an insight of how mathematics theories, physics and engineering works in our life.
Before proceeding to examine the catenary, it is necessary to introduce the hyperbolic function and Newton’s method.
Hyperbolic functions are applicable to mathematics, physics and engineering; they are used to describe the formation of satellite rings around the plantlets, the construction of suspension bridge, for example. What’s more, the most famous application is the use of hyperbolic cosine to describe the shape of a hanging chain.
Hyperbolic functions have similar names to the trigonometric functions, but with the letter “h” added to each name. In addition, it defines in terms of exponential functions, with sinhx= (e^x-e^(-x))/2 and coshx= (e^x+e^(-x))/2. The signs sinh and cosh are called hyperbolic sine and hyperbolic cosine respectively. The hyperbolic function is related to hyperbola, thus we can use the individual coordinates x and y of the points on the unit hyperbola to define sinh and cosh.
The equation of hyperbola is given as x^2-y^2 = 1. Refer to fig. 3, the graph of hyperbola has two oblique asymptotes— y=x and y=-x; in addition they are perpendicular. Furthermore, the hyperbolic function takes area as their argument, namely, hyperbolic angle (t). Yet, it has nothing to do with angles. In fact, the size of the hyperbolic angle is equivalent to the twice of the area of the sector (shaded part.)
Hyperbolic angle = 2×t/2=t t∈[-∞,∞]
Therefore, it provides the x and y coordinates as
x=cosh(t) and y=sinh (t ) t∈[-∞,∞]
Defining f(x)=coshx
The function is defined by the formula:
cosh(x= (e^x+e^(-x))/2) for every real number x
The domain of the function is (-∞, ∞), while the range is [1, ∞)
To sketch the graph of coshx. First, calculate the value of cosh0. When x=0,e^x=1.
Next, we can sketch the two exponential functions by rewriting coshx as
coshx= e^x/2+ e^(-x)/2
Two exponential functions— y=e^(-x)/2 and y=e^x/2
As x becomes larger (x > 0), e^x will increase rapidly; while e^(-x) decrease rapidly. That is to say, e^x/2 gets very large; while e^(-x)/2 gets very small as x increases. Therefore, as x gets larger, coshx gets nearer to e^x/2. We can interpret as
coshx≈e^x/2 for large x
Note that the graph of coshx will always stay above the graph of e^x/2. Even though e^(-x)/2 gets smaller, it is always greater than zero.
On the other hand, as x becomes more negative (x < 0), e^(-x) will increase rapidly; while e^x decrease rapidly. That is to say, e^(-x)/2 gets very large; while e^x/2 gets very small as x increases. Therefore, as x gets smaller, coshx gets nearer to e^(-x)/2. We can also interpret as
coshx≈e^(-x)/2 for negative x
Again, note that the graph of coshx will always stay above the graph of e^(-x)/2. Despite e^x/2 gets smaller, it is always greater than zero.
With this information, the graph of coshx (purple line) can be sketched; the shape of coshx is named as catenary.
coshx =cosh((-x))
An alternative method— algebraic expression is used to confirm “coshx is an even function
cosh((-x)=) (e^(-x)+e^x)/2
=(e^x+e^(-x))/2
=cosh x
To simplify complicated hyperbolic expressions, identities are often used. Hence, in this section, I am going to introduce and prove one of the identities— cosh^2 x-sinh^2 x = 1.
Proof.
cosh^2 x-sinh^2 x= ((e^x+e^(-x))/2 )^2-((e^x+e^(-x))/2 )^2
= (((e)^x+e^(-x) )^2)/4-(((e)^x-e^(-x) )^2)/4
= (e^2x+2+e^(-2x)-(e^2x-2 + e^(-2x)))/4
= 4/4
= 1
To have a better knowledge on catenary, we must able to distinguish between a catenary and a parabola, since they have a similar “U” shape. Even the famous Italian scientist Galileo Galilei has mixed up with these two definitions, such that he believed the shape of a hanging cable would form a parabola, but instead it is actually a catenary . Hence, in this section I will explain their own mathematics descriptions and the shape of it.
The word “catenary” is derived from the Latin word “catena” which means “chain.” A catenary described a uniform chain hanging from two supports under a uniform gravitational field. In fact, a catenary describes the chain hanging freely under its own weight with no load adding on it , while a parabola describes the chain carries a uniform load along its length . For example, a suspension bridge holds its own weight forms a catenary (shown in fig. 6.) Whilst a suspended deck bridge forms a parabola due to the weight of the bridges it carries.
In mathematical description, parabola gives an a quadratic equation of y=ax^2, while the catenary gives an equation of y=acosh(x/a), where a is a constant— a scaling factor of the curve. For instance, a low scaling factor will have a deeper curve.
Newton’s method is used to find the approximation of a root of polynomial equation f(x)=0. The theory of this method is based on the geometry of a curve, using the tangent lines to the curve. Consider if x_0 is not a root of a polynomial equation, but it is pretty close to a root; then by sliding down the tangent line at x_0 to generate an approximation of the actual root.
We seek root for f(x)=0; where the red point is the actual root. First, we start with an initial guess x_0, we then calculate f(x_0 ). Apparently, the initial guess is probably not equal to zero; hence, we will use the tangent line (blue line) to f(x_0 ) at x_0. This tangent line will intersect the x-axis and give a coordinate of (x_1,0). Thus, this point will be much closer to the actual root than x_0 does. Similarly, we then proceed the same method by substituting x_(1 ),(x)_2,x_(3 ),… to the equation f(x), ultimately it will approach to the actual root (red point). Repeat this process until two successive approximations agree to a certain number of decimal places. Hence, the Newton’s method can be used to provide an approximation of the actual root.
To sum up, the key calculation of Newton’s method is to find the tangent line and x-intercept.
Hence, the equation is given as
(y- y_0)=f^' (x_0 )(x-x_0 )
y=f^' (x_0 )(x-x_0 )+y_0
Since we are looking for x¬-¬intercept, let y = 0,
0=f^' (x_0 )(x-x_0 )+y_0
x =x_0-y_0/(f'(x_0))
x =x_0-(f(x_0))/(f'(x_0))
Formally, the general Newton’s method is as follows
As mentioned in section (c), a catenary has an equation defined by hyperbolic cosine and a scaling factor— y=acosh(x/a). The scaling factor suggests the ratio between the horizontal tension on the cable and the weight of the cable per length . However, both of these values are unknown, thus Newton’s method was used to approximate the scaling factor in order to determine the length of the cable.
Assume the cable hangs between the two poles with an equal height of 60m, similar. The centre of the cable is 40m above the ground and the distance between the poles are 110m. Find the length of the hanging cable in terms of T/w , where T/w is the catenary constant (a=T/w.)
T is the horizontal tension on the cable, while w is the weight of the cable. Use Newton’s method to approximate T/w to the nearest 0.001; and find the length of the cable to the nearest 0.1m.
The following are the equations needed to solve the problem.
Catenary equation:
Arc length formula: l=∫_a^b(√(1+([f^' (x)])^2 ) dx ) (ii)
Newton’s method formula: x_(n+1)= x_n- (g(x))/(g^' (x)) (iii)
With the aim to simplify the equation, a single variable a=T/w is substituted into the catenary equation (i).
Assume each pole is located at a distance of 55m away from the midpoint, in which it is assumed to be the y-axis. While, x-axis touches the lowest point of the hanging cable.
First, find the derivative of the catenary equation,
Second, determine the length of the cable by using arc length formula,
l=∫_a^b(√(1+([f^' (x)])^2 ) dx )
Next, substitute f^' (x) into the equation,
Then, substitute the identity cosh^2 x-sinh^2 x = 1 into the equation (vi)
As mentioned in section (b), coshx is an even function, hence the equation(vii) is equivalent to
Given integration formula
Substitute u=x/a → du/dx=1/a → dx=a∙du
l=' 2 ∙ a' ∫_0^55coshu du
l= 2 '∙ [a ∙ sinh ' u]_0^55
l= 2 '∙ [a ∙ sinh(' □( x/a) )]_0^55
l= 2 ∙a∙'sinh(' □( 55/a) ) (viii)
Lastly, substitute a=T/w in the equation (viii)
l= 2 T/w 'sinh(' □(55 ∙ w/T) ) (ix)
Since T/w was not given in the problem statement, Newton’s method is used to approximate a value for T/w. Hence, function p(θ) is defined, such that the root of p(θ) can provide the value for T/w.
Arrange the formula of Newton’s method in terms of θ (θ=T/w ),
θ_(n+1)=θ_n-(p(θ_n))/(p'(θ_n)) (x)
Since the height of the poles are 60m tall and are located at f(-55) and f(55). The centre of the cable is 40m above the ground. Thus, it gives an equation,
f(55)-f(0)=60-40
f(55)-f(0)=20
a cosh((□(55/a))-a cosh((□(0/a))=20 ))
a cosh((□(55/a))-a(=20 ))
0 =a+20-a cosh(□(55/a)) (xi)
Hence, let
p(θ)= θ+20-θ cosh(□(55/θ)) (xii)
Next, find the derivative of function (p(θ)) (xi),
p'(θ)=d/dθ [θ+20-θ cosh(□(55/θ)) ]
=d/dθ(θ)+d/dθ(20)-d/dθ [θ cosh(□(55/θ)) ]
=1+0-{d/dθ (θ)∙cosh((□(55/θ))+d/dθ [cosh(□(55/θ)) ](∙θ) ) }
Given integration formula
=1-cosh((□(55/θ))+[-sinh((55/θ) ∙ d/dθ (□(55/θ)) ∙ θ)) ])
=1-cosh((□(55/θ))-sinh((55/θ) ∙ 1/θ^2 ) )∙-55 ∙ θ
= 1-cosh((□(55/θ))+) (55 sinh(□(55/θ)))/θ (xiii)
Hence, substitute the equations (xi) and (xii) into equation (x),
θ_(n+1)=θ_n-( θ_n+20-θ_n cosh(□(55/θ_n )))/(1-cosh((□(55/θ_n ))+) (55 sinh(□(55/θ_n )))/θ_n )
I have started the calculations with an initial value of θ_n=15 for the Newton’s method. The results are shown below
| n | θ_(n+1) |
|---|---|
| 1 | 19.86989 |
| 2 | 27.82294 |
| 3 | 40.44806 |
| 4 | 57.27616 |
| 5 | 72.11235 |
| 6 | 78.12323 |
| 7 | 78.74394 |
| 8 | 78.74949 |
| 9 | 78.74949 |
| 10 | 78.74949 |
As the problem statement required the value correct to the nearest 0.001m, thus, the approximation value is obtained as 78.749. Hence, the value of T/w is 78.749. In order to obtain an accurate value, I have plot the function p(θ) on graph for double checking. Turn out, both methods have arrived the same value, which indicated the value is reliable.
Lastly, substitute this value into equation (ix) to determine the length of the cable.
l= 2 T/w 'sinh(' □(55 ∙ w/T) )
l= 2∙(78.749)∙sinh(55∙□(1/78.749))
l= 119.2m
Therefore, the length of the hanging cable is 119.2m.
To summaries, we have successfully determine the length of a hanging cable which satisfies with the aim. Though exploring the hyperbolic function especially for hyperbolic cosine (cosh), hyperbolic identity and also Newton’s method. More importantly, we are now able to distinguish between a parabola and a catenary, in terms of the graphical and algebraic features. Lastly, this exploration has allowed me to explore mathematics on my own and able to link to real life application.
Future studies could extend this exploration by considering factors like unequal pole heights, wind loads, and the actual physical properties of cables. Such investigations would offer a more comprehensive understanding of the complexities involved in engineering applications of the catenary curve.
The primary limitation of this study lies in its simplification of real-world conditions, such as uniform cable weight and constant tension. To enhance accuracy, future research could incorporate varying environmental and material factors, employing more advanced mathematical models and computational methods.
Exploration of the Catenary Curve in Hyperbolic Function. (2024, Feb 22). Retrieved from https://studymoose.com/document/exploration-of-the-catenary-curve-in-hyperbolic-function
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