A telephone line hanging between two poles.

A telephone line hangs between two poles 14 meters apart in the shape of a catenary $y = 20cosh(\frac{x}{20})-15 $ , where x and y are measured in meters. Find the length of telephone wire needed between the two poles. Note that the formula for arc length is $L = \int_{a}^{b} \sqrt{1+f' (x))^2}dx$ , that coshx and sinhx are each others derivative, and that $cosh^{2} x-sinh^{2} x=1$ .

Answer

Answers can be viewed only if
  1. The questioner was satisfied and accepted the answer, or
  2. The answer was disputed, but the judge evaluated it as 100% correct.
View the answer

1 Attachment

The answer is accepted.