Question

For the following exercises, use this scenario: A cable hanging under its own weight has a slope \(S=d y / d x\) that satisfies \(d S / d x=c \sqrt{1+S^{2}} .\) The constant \(c\) is the ratio of cable density to tension.Sketch the cable and determine how far down it sags at \(x=0\).

Short answer

The cable reaches its lowest point modeled by \( y(0) = \frac{1}{c}\cosh(C) + D \) at \( x=0 \).

Step-by-step solution

Understanding the Equation

The original equation given is \( \frac{dS}{dx} = c \sqrt{1 + S^2} \). This represents a differential equation describing the slope \( S \) of the cable with respect to \( x \). Our task is to solve this equation to find an expression for \( S(x) \) and eventually the shape of the cable \( y(x) \).

Solve the Differential Equation

First, notice that the equation is separable. Rewriting, we have:\[\frac{dS}{\sqrt{1+S^2}} = c\, dx\]By integrating both sides, we can find \( S(x) \). The integral of the left side is \( \sinh^{-1}(S) \) and the integral of the right side is \( cx \), giving us:\[\sinh^{-1}(S) = cx + C\]where \( C \) is the constant of integration. Solving for \( S \), we get \( S = \sinh(cx + C) \).

Integrate to Find Cable Shape

Since \( S(x) = \frac{dy}{dx} = \sinh(cx + C) \), integrate \( S \) with respect to \( x \) to find \( y(x) \):\[y = \int \sinh(cx + C) \, dx\]The result of this integration is:\[y(x) = \frac{1}{c}\cosh(cx + C) + D\]where \( D \) is a new constant of integration representing vertical translation.

Determine Cable Sag at x = 0

To find the sag at \(x = 0\), focus on the expression \(y(x) = \frac{1}{c}\cosh(cx + C) + D\). Input \(x = 0\):\[y(0) = \frac{1}{c}\cosh(C) + D\]At \(x = 0\), the cable is at its lowest point, so \(y(0)\) gives the sag from the anchor point. The actual sag depends on the specific values of \(C\), \(D\), and \(c\), which relate to initial conditions.

Key concepts

Differential Equation

A differential equation is a type of mathematical equation that involves unknown functions and their derivatives. In this exercise, the differential equation describes the behavior of the cable as it hangs under its own weight. The equation given is \( \frac{dS}{dx} = c \sqrt{1 + S^2} \), where \( S \) is the slope of the cable, \( x \) is the horizontal distance, and \( c \) is a constant related to the cable's properties. Differential equations like these are crucial because they can model various physical phenomena, such as motion, growth, and in this case, the shape of a hanging cable.
By solving this differential equation, we aim to discover how the slope of the cable changes with respect to \( x \) and ultimately find the shape of the cable \( y(x) \). This involves a sequence of integration and handling of constants, which are determined by specific initial conditions.

Cable Mechanics

Cable mechanics refers to the study of how cables behave under forces such as gravity and tension. In this problem, the cable hangs naturally under its own weight, and we're interested in its curve and sag. The cable's shape is governed by mechanics principles and mathematical equations.
The slope \( S \) represents the steepness of the cable at any point. As the problem states, the slope satisfies the differential equation due to the natural forces acting on it. The constant \( c \) embodies the relationship between the cable's density and the tension within. It essentially tells us how much one unit of horizontal force influences the cable's slope. Mechanics of cables can apply to engineering tasks, such as designing suspension bridges that need to support significant weight while maintaining structural integrity.

Hyperbolic Functions

Hyperbolic functions are analogies of trigonometric functions but are based on hyperbolas rather than circles. Some common hyperbolic functions include hyperbolic sine \( \sinh(x) \) and hyperbolic cosine \( \cosh(x) \). These functions describe the behavior of the cable in the problem.
The function \( \sinh(x) \) is similar to the sine function and \( \cosh(x) \) to the cosine function, but with differences that make them particularly useful for modeling catenary curves, like the hanging cable. When we solve the original equation, we obtain \( S(x) = \sinh(cx + C) \) and \( y(x) = \frac{1}{c} \cosh(cx + C) + D \), indicating how hyperbolic functions naturally describe the cable's shape.

Separable Equations

Separable equations are a type of differential equation that can be rewritten, allowing the separation of variables. This makes them easier to solve. The key here is to manipulate the equation so that one variable and its differential are on one side, and the other variable's differential is on the opposite side.
In our cable exercise, the equation \( \frac{dS}{dx} = c \sqrt{1 + S^2} \) becomes separable when we adjust it to \( \frac{dS}{\sqrt{1+S^2}} = c \, dx \). This form allows us to integrate both sides independently, find an expression for \( S \), and then again find the overall shape of the cable by further integration. Separable equations are a powerful tool in calculus for solving many types of differential equations efficiently.