Question

a) The figure shows a telephone wire hanging between two poles at \({\rm{x = - b}}\)and\({\rm{x = b}}\). It takes the shape of a catenary with equation \({\rm{y = c + acosh(x/a)}}\). Find the length of the wire.

b) Suppose two telephone poles are \({\rm{50ft}}\)apart and the length of the wire between the poles is\({\rm{51ft}}\). If the lowest point of the wire must be\({\rm{20ft}}\) above the ground, how high up on each pole should the wire be attached?

Short answer

a) The length of the telephone wire is\(2a\sinh (b/a)\).

b) Telephone wire should be attached to \(24.36ft\)above the ground.

Step-by-step solution

Arc Length Function

Arc length function is useful to measure the arc length of a curve from a particular starting to any other point on the curve.

Arc length calculated as \(L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx\)

Given parameters

(a) Telephone wire hanging between two poles at \(x = - b\) and \(x = b\).

The parabolic trajectory of the wire is \(y = c + a\cosh (x/a)\).

(b) Telephone poles are \(50ft\)apart and the length of the wire between the poles is\(51ft\)Lowest point of the wire is \(20ft\)above the ground.

Calculating arc length

(a)

As \(y = c + a\cosh (x/a)\)

\(\begin{aligned}{}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}(c + a\cosh (x/a))\\ &= 0 + a \cdot \frac{1}{a}\sinh (x/a)\\ &= \sinh (x/a)\end{aligned}\)

Compute arc length by formula \(L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx\)

\(\begin{aligned}L &= \int_{ - b}^b {\sqrt {1 + {{(\sinh (x/a))}^2}} } dx\\ &= \int_{ - b}^b {\sqrt {1 + {{\sinh }^2}(x/a)} } dx\\ &= \int_{ - b}^b {\sqrt {{{\cosh }^2}(x/a)} } dx\\ &= \int_{ - b}^b {\cosh } (x/a)dx\\ &= (a\sinh (x/a))_{ - b}^b\\ &= 2a\sinh (b/a)\end{aligned}\)

Therefore, the length of the telephone wire is\(2a\sinh (b/a)\).

Calculating values of a, b and c for maximum height of the wire

(b)

Calculate minimum height of the wire as wire is lowest at \(x = 0\).

\(\begin{aligned}y(0) &= c + a\cosh (0/a)\\ &= c + a\end{aligned}\)

Wire is \(20ft\)above the ground; i.e., \(y(0) = 20\)

\(\therefore c + a = 20{\rm{ }}\_\_\_\_\_\_\_\_\_\_\_Equation{\rm{ }}1\)

From part (a), length of the wire is \(2a\sinh (b/a)\)and it is given as \(51ft\).

\(\therefore 2a\sinh (b/a) = 51{\rm{ }}\_\_\_\_\_Equation{\rm{ }}2\)

As two poles are \(50ft\)apart,

Therefore,

\(2b = 50 \Rightarrow b = 25\)

Plot graph for \(y = 2x\sinh (25/x)\)and \(y = 51\) and find x-coordinates of the points of intersection.

\(\therefore \)a ≈ ~72.4

From Equation 1,

\(\begin{aligned}c &= 20 - a\\ &= 20 - 72.4\\ &= - 52.4\end{aligned}\)

Substitute values of a, b and c in given parabolic trajectory of the wire-

\(\begin{aligned}y &= c + a\cosh (x/a)\\ &= - 52.4 + 72.4\cosh (x/72.4)\end{aligned}\)

To find how high the pole should be attached, find the value of \(y(b)\):

\(\begin{aligned}y(b) &= - 52.4 + 72.4\cosh (25/72.4)\\ &\approx 24.36ft\end{aligned}\)

Therefore, telephone wire should be attached to \(24.36ft\)above the ground.