   Chapter 3.11, Problem 49E

Chapter
Section
Textbook Problem

If a water wave with length L. moves with velocity v in a body of water with depth d, then v = d L 2 π tan ( 2 π d L ) where g is the acceleration due to gravity. (See Figure 5.) Explain why the approximation v ≈ g L 2 π is appropriate in deep water.

To determine

To explain: The approximation vgL2π

Explanation

Given:

The length of the water wave is L and d is the depth of the water.

The velocity v=gL2πtanh(2πdL) .

Formula used:

The hyperbolic functions tanhx=sinhxcoshx,sinhx=exex2and coshx=ex+ex2 .

Calculation:

The velocity of the water wave v=gL2πtanh(2πdL).

Here, the hyperbolic function occur tanh(2πdL) and the value of 2πdL is very large if water is deep.

Obtain the approximation value of the velocity when 2πdL is large value.

The value of limxtanhx is computed as follows,

tanhx=sinhxcoshx                           [Qtanhx=sinhxcoshx]=exex2ex+ex2                        [sinhx=exex2coshx=ex+e

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