User:Editeur24/hyperbolic

hyperbolic functions

Examples Hyperbolic functions are widely used in engineering. The equation for a catenary, the shape of a rope suspended from two ends, is f(x) = a cosh(x/a) - a if the ends are x = - b and x = + b at equal heights and the lowest point is f(x) = 0. Suppose the constant is a =  10. The value x =  0 is where the rope hangs lowest, since it reaches its minimum where f'(x) = 0 and if f'(x) = 10 sinh(x/10) [1/10] = sinh(x/10) = 0 we need sinh(x) = 0. We can calculate how much lower the middle is than the ends using f(b) = 10 cosh(b/10) - 10, so if b = 5, the height of each end is f(5) = 1.28.

The equation for the velocity in meters per second of an idealized surface wave travelling across a lake is

$v={\sqrt {\left({\frac {g\lambda }{2\pi }}\right)\tanh \left({\frac {2\pi d}{\lambda }}\right)}},$

where g = 9.8 m/s$^{2}$ is gravity's acceleration, $\lambda$ is the wavelength from crest to crest in meters, and d is the depth of undisturbed water in meters. Thus, if the wavelength is 12 meters and the velocity is 4 meters per second, we can find the depth of the lake from

$4={\sqrt {\left({\frac {9.8\cdot 12}{2\pi }}\right)(\tanh \left({\frac {2\pi d}{12}}\right)}},$

which implies that

$d=\left({\frac {6}{\pi }}\right)(\tanh ^{-1}\left({\frac {8\pi }{29.4}}\right)\approx 2.4\;meters.$