Problem 55 - Calculus

Question

A cable hanging under its own weight has slope \(S=d y / d x\) that satisfes \(d S / d x=c \sqrt{1+S^{2}}\). The constant \(c\) is the ratio of cable density to tension. (a) Show that \(S=\sinh c x\) satisfes the equation. (b) Integrate \(d y / d x=\sinh c x\) to find the cable height \(y(x)\) if \(y(0)=1 / c\) (c) Sketch the cable hanging between \(x=-L\) and \(x=L\) and find how far it sags down at \(x=0\).

Step-by-Step Solution

Answer

(a) Verified. (b) Height is \(y(x) = \frac{1}{c} \cosh(cx)\). (c) Sag at \(x=0\) is \(\frac{1}{c}\).

Step 1: Verify that S=\sinh(cx) satisfies the equation

Start by substituting \(S = \sinh(cx)\) into the given equation. Calculate \(d S/dx = c \cosh(cx)\) and \(\sqrt{1 + S^2} = \cosh(cx)\) because \(1 + \sinh^2(cx) = \cosh^2(cx)\). Substituting these into the equation \(dS/dx = c \sqrt{1+S^2}\) gives \(c \cosh(cx) = c \cosh(cx)\), which holds true. Therefore, \(S = \sinh(cx)\) satisfies the equation.

Step 2: Integrate to find y(x)

Given \(d y/dx = \sinh(cx)\), integrate both sides to find \(y(x)\). The integral of \(\sinh(cx)\) is \(1/c \cdot \cosh(cx) + C\). Using the initial condition \(y(0) = 1/c\), substitute \(x = 0\) to find \(C\): \(1/c = 1/c \cdot \cosh(0) + C\) leads to \(C = 0\). Thus, \(y(x) = \frac{1}{c} \cosh(cx)\).

Step 3: Sketch and find the sag at x=0

The function \(y(x) = \frac{1}{c} \cosh(cx)\) represents the hyperbolic cosine curve, which is symmetric about the y-axis. Sketching this curve for \(x\) ranging from \(-L\) to \(L\) shows that it attains its minimum at \(x = 0\). The sag or dip at \(x = 0\) is the value of \(y(0)\), which is \(\frac{1}{c} \cosh(0) = \frac{1}{c}\).

Key Concepts

Catenary Curve

When a flexible chain or cable hangs freely under its own weight between two fixed points, the curve it forms is called a **catenary curve**. A catenary curve represents the shape of a hanging chain or cable when it is supported only at its ends and is acted upon by a uniform gravitational force.
The mathematical form of the catenary curve is expressed through the hyperbolic cosine function, which is similar in shape to a parabola but derived from hyperbolic functions. Unlike a simple parabola, a catenary takes into account the uniform distribution of weight along the curve.
**Key Properties of Catenary Curves:**

• They are described by the equation: \[ y(x) = rac{1}{c} \cosh(cx) \]
• The highest point is at the center, and it is symmetrical about the vertical axis.
• These curves are fundamental in understanding structures like suspension bridges and arches.

Integration

Integration is a fundamental concept in calculus, used to find such things as areas under curves and solutions to differential equations. In this exercise, the process of integration is used to find the equation that describes the height of the cable, \( y(x) \) from its slope \( S = \sinh(cx) \).
Integration involves finding the antiderivative of a given function. In our case, we are asked to integrate \( \sinh(cx) \) which requires the use of standard integration techniques.
**Steps to Integrate**:

• Recognize the function to be integrated: \( \int \sinh(cx) \, dx \).
• The antiderivative of \( \sinh(cx) \) is \( \frac{1}{c} \cosh(cx) \).
• Use the initial condition to calculate any constants: The function is given as \( y(0) = \frac{1}{c} \), solving for any additional constants.

Differential Equations

Differential equations are equations that relate a function with its derivatives. They are crucial in describing various natural phenomena involving dynamically changing systems.
In this specific problem, we start with the differential equation \( \frac{dS}{dx} = c \sqrt{1+S^2} \), which describes the slope of the cable.
**Understanding Differential Equations**:

• A **differential equation** defines a relationship between a function and its derivative, providing a description of a physical or mathematical system in terms of rates of change.
• Solving the equation involves finding a function \( S \) that satisfies it, which in this problem leads to the solution \( S = \sinh(cx) \).
• Such equations are pivotal for expressing complex problems in engineering and physics.

Initial Conditions

Initial conditions in a mathematical problem specify the value of a function at a particular point. They are crucial in actions like solving differential equations, as they allow determining the specific solution from among the possible ones.
In our example, the initial condition is given as \( y(0) = \frac{1}{c} \). This condition helps us determine the constant \( C \) when integrating.
**Why Initial Conditions Matter**:

• They allow the determination of specific solutions by providing a starting point.
• Initially known values are crucial in applications such as physics for determining accurate predictions of system behaviors over time.
• With initial conditions applied, we can solve the integration resulting in a precise function to describe the physical scenario.

Through these steps, let's solve problems depicted by differential equations precisely, using the initial conditions to ground our solution specifically to the scenario.