Problem 9 - Calculus Of Variations, With Applications To Physics And Engineering

Question

Show that, if \(y_{1}=y_{2}\), no catenary of the family (45) passes through \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) if $$ -\frac{p}{\cosh p}<\frac{x_{2}-x_{1}}{2 y_{1}} $$ where \(p\) is the negative root of $$ p \sinh p-\cosh p=0 $$ HiNT: First show that \(a=\frac{1}{2}\left(x_{1}+x_{2}\right)\) when \(y_{1}=y_{2}\) and let \(p=\left[\left(x_{1}-a\right) / b\right] .\) Next show that equality of the two members of (112) is required for a catenary to pass through both end points. Hence, if the maximum of the left-hand member of \((112)\) is less than the right-hand member, there is no \(p\) for which equality can obtain. Show that the maximizing \(p\) is the negative root of (113). Approximately, (112) reads \(\left(x_{2}-x_{1}\right)>1.32 y_{1}\)

Step-by-Step Solution

Answer

The proof demonstrates that if the condition \( \left(x_{2}-x_{1}\right)>1.32 y_{1} \) is satisfied, no catenary of the family (45) passes through the given points \( \left(x_{1}, y_{1}\right) \) and \( \left(x_{2}, y_{2}\right) \).

Step 1: Calculate value of 'a'

From the hint given in the problem, when \(y_{1}=y_{2}\), we can find the value of 'a' to be \(a=\frac{1}{2}\left(x_{1}+x_{2}\right)\).

Step 2: Define 'p'

Based on the hint, define \(p=\left[\left(x_{1}-a\right) / b\right]\). Here, 'b' is a constant related to the unique shape of the catenary.

Step 3: Establish the equality condition

For a catenary to pass through both end points, the equality of the two members of the equation \( -\frac{p}{\cosh p}<\frac{x_{2}-x_{1}}{2 y_{1}} \) is required.

Step 4: Check the maximum condition

If the maximum of the left-hand member of the equation is less than the right-hand member, then there is no 'p' for which equality can obtain.

Step 5: Show that the maximizing 'p' is the negative root of (113)

We need to show that the maximizing 'p' is the negative root of the equation \(p \sinh p-\cosh p=0 \). To do so, we can use the first derivative test. Taking the derivative gives \( \sinh p + p \cosh p – \sinh p = p \cosh p = 0 \). Setting this equal to zero and solving for 'p' gives 'p' as the negative root of the equation.

Step 6: Interpret the result

Now, with all the steps done, we can conclude that if the condition \( \left(x_{2}-x_{1}\right)>1.32 y_{1} \) is satisfied, no catenary of the family (45) passes through the given points \( \left(x_{1}, y_{1}\right) \) and \( \left(x_{2}, y_{2}\right) \).

Key Concepts

Catenary

The catenary is a fascinating curve that is formed by a hanging chain or cable when supported only at its ends under the influence of gravity. This curve is defined mathematically, and it has unique properties that make it appear in various natural phenomena and architectural designs. A key characteristic of a catenary is its mathematical form, which is expressed using hyperbolic functions due to their resemblance to the chain shape. The equation for a catenary in its simplest form is given by \( y = a \, \cosh\left(\frac{x}{a}\right) \), where \(a\) is a constant related to the curve's shape and tension.
This curve appears very often in real-world applications, not just because of its stability and strength but also for its aesthetic appeal. For example, the famous Gateway Arch in St. Louis is an inverted catenary arch. When analyzing catenaries, especially in calculus of variations, we often consider how specific conditions at the supports influence its form. The interplay of forces that determine a catenary's shape gives rise to interesting mathematical challenges and solutions.

Hyperbolic Functions

Hyperbolic functions resemble trigonometric functions but are rooted in the hyperbola rather than the circle. They include hyperbolic sine \( \sinh x \) and hyperbolic cosine \( \cosh x \), among others. These functions are defined as follows:

• \( \sinh x = \frac{e^x - e^{-x}}{2} \)
• \( \cosh x = \frac{e^x + e^{-x}}{2} \)

Hyperbolic functions are essential in the analysis of catenaries because they naturally describe the shape of a hanging cable. This is primarily because the equations account for the exponential nature of the forces acting on the cable. In problems involving catenaries, you'll often see the equation \( p \sinh p - \cosh p = 0 \) used. This specific equation can describe the balance of forces in a catenary under given conditions. The stability of hyperbolic functions makes them very useful in mathematical modeling of natural phenomena and engineering applications.

Roots of Equations

Solving equations to find their roots is a fundamental skill in mathematics. A root of an equation is a value that makes the equation true. In problems related to calculus of variations, such as finding the conditions under which a catenary passes through certain points, finding roots is essential.
The equation from the exercise, \( p \sinh p - \cosh p = 0 \), requires finding a specific root, which happens to be negative in this case. This root represents a point at which the forces in a catenary are balanced under specific conditions. Finding these roots often requires numerical methods, especially when no analytical solution is apparent.
When solving such equations, you might use methods like the Newton-Raphson method or graphing to approximate the root. Understanding the behavior of the function and the characteristics of its derivatives (as found through differentiation) are crucial in accurately determining roots and ultimately gathering insights into the behavior of the catenary or other entities being analyzed.