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.\text { Show that } S=\sinh (c x) \text { satisfies this equation. }

Short answer

The function \( S = \sinh(cx) \) satisfies the differential equation.

Step-by-step solution

Understand the Problem

We are given the equation \( \frac{dS}{dx} = c \sqrt{1 + S^2} \). We need to show that the function \( S = \sinh(cx) \) is a solution to this equation.

Define Hyperbolic Sine Function

The hyperbolic sine function \( \sinh(x) \) is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \). Its derivative is \( \cosh(x) \), where \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).

Differentiate \( S = \sinh(cx) \)

Since \( S = \sinh(cx) \), differentiate it with respect to \( x \) to get \( \frac{dS}{dx} = c \cosh(cx) \).

Express \( \sqrt{1 + S^2} \)

We know \( \sinh^2(x) = \cosh^2(x) - 1 \), so we have \( 1 + \sinh^2(cx) = \cosh^2(cx) \). Thus, \( \sqrt{1 + S^2} = \cosh(cx) \).

Substitute and Verify

Substitute \( S = \sinh(cx) \) into the original differential equation \( \frac{dS}{dx} = c \sqrt{1 + S^2} \). We have \( c \cosh(cx) = c \cosh(cx) \), which holds true. This shows \( S = \sinh(cx) \) satisfies the equation.

Key concepts

Differential Equations

Differential equations are fundamental in describing change over time or space and are prominent in various fields such as physics and engineering. They involve derivatives, which provide information about rates of change. In this scenario, we are dealing with a first-order differential equation: \( \frac{dS}{dx} = c \sqrt{1 + S^2} \). This equation expresses how the slope \( S \) of a cable under tension and its own weight changes as you move along the length of the cable.

To solve this equation, one typically looks for a function \( S(x) \) that satisfies the relationship between \( dS/dx \) and \( S \). This equation indicates that the change in slope depends not just on the position along the cable, but is also influenced by the term \( \sqrt{1 + S^2} \), encapsulating the cable's physical properties through the constant \( c \). Solving such equations often relies on clever substitutions or known functions that simplify the differential equation.

Hyperbolic Functions

Hyperbolic functions, much like trigonometric functions, are based on exponential functions. They are defined mathematically using exponential equations and serve as useful solutions for differential equations involving shapes like catenaries or curves typical in hyperbolic geometry. Among the commonly used hyperbolic functions are hyperbolic sine and cosine.

The hyperbolic sine function \( \sinh(x) \) is expressed as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \). It has the special property of defining the shape of a hanging cable or suspended chain, known as a catenary. Its derivative, the hyperbolic cosine \( \cosh(x) = \frac{e^x + e^{-x}}{2} \), maintains the connection between hyperbolic functions by acting similarly to cosine in the trigonometric context.

• \( \sinh^2(x) + 1 = \cosh^2(x) \) is a crucial identity often used in these computations.
• Recognizing and manipulating hyperbolic functions helps verify specific solutions to differential equations as seen in this exercise.

Solution Verification

Solution verification is an essential part of solving differential equations. It involves confirming that a potential solution actually satisfies the original equation. For the problem at hand, we need to confirm that \( S = \sinh(cx) \) is indeed the solution to the differential equation \( \frac{dS}{dx} = c \sqrt{1 + S^2} \).

First, differentiate \( S = \sinh(cx) \) to obtain \( \frac{dS}{dx} = c \cosh(cx) \). Next, express \( \sqrt{1 + S^2} \) using the identity \( \sinh^2(cx) = \cosh^2(cx) - 1 \), which simplifies to \( \sqrt{1 + S^2} = \cosh(cx) \). By substituting these into the differential equation, you see both sides equal \( c \cosh(cx) \), confirming \( S = \sinh(cx) \) as a true solution.

• Verification is crucial as it ensures the solution is correct and applicable.
• This process also solidifies understanding of the relation between the function, its derivatives, and the structure of the differential equation.