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