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 $cosh^{2} xsinh^{2} x=1$ .
Wcramsay
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Aman R
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