How Do You Calculate the Rate of Change of Wave Frequency with Water Depth?

In summary: I am very rusty on my derivatives. But I will say that it doesn't make any sense to integrate something while looking for the derivative of a function. It appears to be a nasty little hyperbolic function...You may want to repost this in the homework or math sections for a bit more help.cunhasb - you seem to be on the right track. Make sure constants are correct, and remembertanh x = sinh x/ cosh x = sech x/csch x, csch x = 1/sinh x, sech x = 1/cosh x
  • #1
cunhasb
12
0
Could anyone give a hand in solving this problem?

In shallow ocean water, the frequency of the wave action (number of wave crests passing a given point per second) is given by f= [(g/(2PiL))tanh(2Pih/L)]^1/2, where g is the acceleration due to gravity, L is the distance between wave crests, and h is the depth of the water. Get an equation for the rate of change of f with respect to h assuming g and L are constants.

Well this is what I've gotten so far...I'm not sure if it is right or not...

df/dh=1/2((g/2PiL) tanh(2Pih/L))^-1/2 * ((g/L^2)sech^2(2Pih/L))
=((g/L^2)sech^2(2Pih/L))/2√((g/2PiL)tanh(2Pih/L))

well seems that the final answer is

(gPi/2L^3)^1/2 csch^1/2(2Pih/L)sech^3/2(2Pih/L)

So, I have no idea how this answer was derived, looking at it seems that they integrated it (sech^3/2)... is that so?

It would be of great help you anyone could give a hand on this one...


Thank you so much... guys...
 
Engineering news on Phys.org
  • #2
I am very rusty on my derivatives. But I will say that it doesn't make any sense to integrate something while looking for the derivative of a function. It appears to be a nasty little hyperbolic function...You may want to repost this in the homework or math sections for a bit more help.
 
  • #3
cunhasb - you seem to be on the right track. Make sure constants are correct, and remember

tanh x = sinh x/ cosh x = sech x/csch x, csch x = 1/sinh x, sech x = 1/cosh x

--------------------------------------

let f = [itex](\frac{g}{2\pi L} tanh (\frac{2\pi h}{L})^{1/2}[/itex]

or f = [itex] (a g)^{1/2}[/itex], so

df/dh = f ' = 1/2 a1/2 g-1/2 g '

now if g = tanh (bh) , then g ' = b sech2 (bh) where b = [itex]\frac{2\pi}{L}[/itex]

so

f ' = 1/2 [itex] (\frac{g}{2\pi L})^{1/2} (tanh (\frac{2\pi}{L}))^{-1/2}\,sech^2 (\frac{2\pi h}{L})\,\,\frac{2\pi}{L} [/itex]
 
  • #4
thank you guys... Sorry not to thank you before... but I was away for a few days...

Thank you so much...
 

FAQ: How Do You Calculate the Rate of Change of Wave Frequency with Water Depth?

What are derivatives hyper functions?

Derivatives hyper functions are a type of mathematical function used in the field of functional analysis. They are defined as a collection of functions that are obtained by taking derivatives of an underlying function with respect to a variable.

How are derivatives hyper functions different from regular derivatives?

While regular derivatives are defined for smooth functions, derivatives hyper functions can be defined for functions that are not necessarily smooth. They are also more general in nature and can be used to solve problems where regular derivatives fail.

What is the importance of derivatives hyper functions in mathematics?

Derivatives hyper functions are important in mathematics because they allow us to extend the concept of derivatives to a larger class of functions. This allows us to solve more complex problems and obtain more precise results in various fields of mathematics.

How are derivatives hyper functions used in physics?

In physics, derivatives hyper functions are used to solve problems where the traditional concept of derivatives fails. They are particularly useful in solving problems related to fundamental principles such as the conservation of energy and momentum.

Are there any real-world applications of derivatives hyper functions?

Yes, derivatives hyper functions have many real-world applications, including in finance, signal processing, and image processing. They are also used in engineering and physics to model complex systems and solve problems that cannot be solved with traditional derivatives.

Back
Top