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 $cosh^{2} x-sinh^{2} x=1$ .
Wcramsay
73
Answer
Answers can only be viewed under the following conditions:
- The questioner was satisfied with and accepted the answer, or
- The answer was evaluated as being 100% correct by the judge.
1 Attachment
Aman R
649
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 1546 views
- $7.00
Related Questions
- Calculus Questions - Domains; Limits; Derivatives; Integrals
- Taylor Polynom/Lagrange form om the remainder term.
- Integration headache, please help.
- Beginner Differential Equations - Growth Rate Question
- Easy money (basic calc)
- Evaluate $\int \sqrt{\tan x} dx$
- Optimisation Problem
- Parametric, Polar, and Vector-Valued Equations for Kav10
