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
- 743 views
- $7.00
Related Questions
- Get the volume and surface area of the paraboloid $z=4-x^2-y^2$ cut by the plane $z=4-2x$
- Notation question. Where does the x in the denominator come from?
- Calculus 1
- Help
- Obtaining the absolute velocity of a moving train based on angle of raindrops with respect to vertical axis
- Is $\int_0^{\infty}\frac{x+3}{x^2+\cos x}$ convergent?
- Integrate $\int_0^1\int_{\sqrt{x}}^{1}e^{y^3}dydx$
- Determine where the following function is discontinuous
