   Chapter 3.11, Problem 52E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Using principles from physics it can be shown that when a cable is hung between two poles, it takes the shape of a curve y = f(x) that satisfies the differential equation d 2 y d x 2 = ρ g T 1 + ( d y d x ) 2 where ρ is the linear density of the cable, g is the acceleration due to gravity, T is the tension in the cable at its lowest point, and the coordinate system is chosen appropriately. Verify that the function y = f ( x ) = T ρ g cosh ( ρ g x t ) is a solution of this differential equation.

To determine

To verify: The given function is a solution of the given differential equation.

Explanation

Given:

The differential equation is, d2ydx2=ρgT1+(dydx)2.

The function y=Tρgcosh(ρgxT).

Formula used:

The derivative hyperbolic functions ddx(coshx)=sinhx,ddx(sinhx)=coshx

Calculation:

Consider the function y=Tρgcosh(ρgxT).

Differentiate with respect to x,

ddx(y)=ddx(Tρgcosh(ρgxT))=Tρgddx(cosh(ρgxT))

Let u=ρgxT.

dydx=Tρgddx(coshu)

Use chain rule and simplify the terms,

dydx=Tρg[ddu(coshu)dudx]=Tρg[sinhududx]

Substitute u=ρgxT,

dydx=Tρg[sinh(ρgxT)ddx(ρgxT)]=Tρg[sinh(

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