Use the definition of sinh x and cosh x to show that cosh^2 x - sinh^2 x = 1. Ve

Question

Use the definition of sinh x and cosh x to show that cosh^2 x - sinh^2 x = 1. Verify that the derivatives of the other four hyperbolic functions are analogous to the derivatives of the ordinary trigonometric functions: Hyperbolic trigonometric functions are useful in solving differential equations. Show that the functions y = A sinh kx and y = B cosh kx, where A, B, and k are constants, satisfy the equation y(x) - ky(x) = 0. 9. A well-known application of hyperbolic trigonometric functions arises in the field of statics. A flexible hanging chain or cable, supported at its ends, and acted on only by the gravitation force (its own weight) takes the shape of a catenary, described by the function y = a cosh (x/a). Graph the catenary for a = 0.5. 1, and 2. (The term catenary comes from the Latin for chain; however, Thomas Jefferson is credited with introducing the English word.) 10. The Gateway Arch in St. Louis is shaped like an inverted catenary, 630 feet wide at the base and 630 feet high at its center. The function

Explanation / Answer

8)

y = Asinh(kx)

y' = A*cosh(kx)*k = Ak*cosh(kx)

y'' = Ak*sinh(kx)*k

y'' = Ak^2*sinh(kx)

Diff equation is ----> y'' - k^2*y = 0 ---> I beloeve that this is what it has to be, it must not be k*y(x)

Plug in results :

Ak^2sinh(kx) - k^2*sinh(kx) = 0

Ak^2sinh(kx) - Ak^2sinh(kx) = 0

0 = 0

Hence proved!

y = Bcosh(kx)

y' = Bk*sinh(kx)

y'' = Bk^2*cosh(kx)

y'' - k^2*y = 0

Bk^2*cosh(kx) - k^2*Bcosh(kx) = 0

Bk^2*cosh(kx) - Bk^2*cosh(kx) = 0

0 = 0

Hence proved that y = Asinh(kx) and Bcosh(kx) are solutions

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