Question

If a water wave with lengthLmoves with velocityvacross

a body of water with depthd, as in the figure on page 496 , then

\[{v^2} = \frac{{gL}}{{2\pi }}\tanh \frac{{2\pi d}}{L}\]

(a) If the water is deep, show that\[v \approx \sqrt {gL/(2\pi )} \].

(b) If the water is shallow, use the Maclaurin series for tanh often used. A better approximation is obtained by using to show that\[v \approx \sqrt {gd} \]. (Thus in shallow water the velocity of wave tends to be independent of the length of the wave.)

(c) Use the Alternating Series Estimation Theorem to show that if\[L > 10d\], then the estimate\[{v^2} \approx gd\]is accurate to within\[0.014{\rm{gL}}\].

Short answer

Part a)

When "the water is deep," we can approximate v²with this limit.

We have:

\[{v^2} \approx gL/(2\pi ) \to v \approx \sqrt {gL/(2\pi )} \]

Part b)

In shallow water d is small and by using the Maclaurin series for tanh it simplifies to \[ \approx \sqrt {gd} \].

Part c)Bye the Alternating Series Estimation Theorem to showL > 10d a decrease in L would increase the number in front of the gL, while a increase in Lwould decrease it, so the estimate is within 0.014gL when L > 10d

Step-by-step solution

Concept of Maclaurin series

If 0 is that the point where the derivatives are considered, a Taylor series is additionally called a Maclaurin series, after Colin Maclaurin, who made extensive use of this special case of Taylor series within the 18th century.
Taylor polynomials are approximations of a function, which become generally better as n increases.

Given parameters

An equation \[{v^2} = \frac{{gL}}{{2\pi }}\tanh \frac{{2\pi d}}{L}\]

Prove \[v \approx \sqrt {gL/(2\pi )} \]

Part a)

In order to simulate that "the water is deep, we are looking for what happens with\[{v^2}\]when\[d\]is a very large number. That is, we are looking at:

\[\begin{array}{l}\mathop {\lim }\limits_{d \to \infty } {v^2} = \mathop {\lim }\limits_{d \to \infty } \frac{{gL}}{{2\pi }}\tanh \frac{{2\pi d}}{L}\\ = \frac{{gL}}{{2\pi }}\mathop {\lim }\limits_{d \to \infty } \tanh \frac{{2\pi d}}{L}\\ = \frac{{gL}}{{2\pi }} \cdot 1\\ = \frac{{gL}}{{2\pi }}\end{array}\]

Therefore, approximated\[{v^2}\]with this limit.

We have:

\[{v^2} \approx gL/(2\pi ) \to v \approx \sqrt {gL/(2\pi )} \]

Using the Maclaurin series for tanh

Part b)

Find the Maclaurin series for\[\tanh x\]

\[\begin{array}{*{20}{c}}{f(x)}&{ = \tanh x}&{f(0)}&{ = 0}\\{{f^\prime }(x)}&{ = {{{\mathop{\rm sech}\nolimits} }^2}x}&{{f^\prime }(0)}&{ = 1}\\{{f^{\prime \prime }}(x)}&{ = - 2\tanh x{{{\mathop{\rm sech}\nolimits} }^2}x}&{{f^{\prime \prime }}(0)}&{ = 0}\\{{f^{\prime \prime \prime }}(x)}&{ = 2(\cosh 2x - 2){{{\mathop{\rm sech}\nolimits} }^4}x}&{{f^{\prime \prime \prime }}(0)}&{ = - 2}\\{{f^{(4)}}(x)}&{ = - 2(\sinh (3x) - 11\sinh x){{{\mathop{\rm sech}\nolimits} }^5}x}&{{f^{(4)}}(0)}&{ = 0}\\{\tanh x}&{ \approx 0 + x + \frac{0}{{2!}}{x^2} - \frac{2}{{3!}}{x^3} + 0}&{}&{}\\{}&{ \approx x - \frac{1}{3}{x^3}}&{}&{}\end{array}\]

Step 2

Letting\[x = \frac{{2\pi d}}{L}\]

\[\tanh \frac{{2\pi d}}{L} \approx \frac{{2\pi d}}{L} - \frac{1}{3}{\left( {\frac{{2\pi d}}{L}} \right)^3}\]

Now put it into the equation

\[\begin{array}{l}{v^2} = \frac{{gL}}{{2\pi }}\tanh \frac{{2\pi d}}{L}\\ \approx \frac{{gL}}{{2\pi }}\left( {\frac{{2\pi d}}{L} - \frac{1}{3}{{\left( {\frac{{2\pi d}}{L}} \right)}^3}} \right)\\ \approx \frac{{gL}}{{2\pi }}\left( {\frac{{2\pi }}{L}\left[ {d - \frac{1}{3}{{\left( {\frac{{2\pi d}}{L}} \right)}^2}} \right]} \right)\\ \approx g\left( {d - \frac{1}{3}{{\left( {\frac{{2\pi d}}{L}} \right)}^2}} \right)\end{array}\]

In shallow water\[d\]is small and it simplifies to

\[\begin{array}{l}{v^2} \approx g(d - 0)\\ \approx gd\\v \approx \sqrt {gd} \end{array}\]

Using alternating series estimation to show \[T \approx 2\pi \sqrt {\frac{L}{g}} \left( {1 + \frac{1}{4}{k^2}} \right)\]

Part c)

From part (b), the Maclaurin series for tanh x

\[\begin{array}{l}\tanh x \approx x - \frac{1}{3}{x^3}{v^2}\\ = \frac{{gL}}{{2\pi }}\tanh \frac{{2\pi d}}{L}\\ = \frac{{gL}}{{2\pi }}\left( {\frac{{2\pi d}}{L} - \frac{1}{3}{{\left( {\frac{{2\pi d}}{L}} \right)}^3}} \right)\\ = g\left( {d - \frac{1}{3}{{\left( {\frac{{2\pi d}}{L}} \right)}^2}} \right)\\ = gd - \frac{g}{3}{\left( {\frac{{2\pi d}}{L}} \right)^2}\end{array}\]

For shallow water:

\[{v^2} \approx gd\]

We dropped the\[\frac{1}{3}{\left( {\frac{{2\pi d}}{L}} \right)^3}\]part for the shallow water estimation. We can use\[\frac{{gL}}{{2\pi }} \cdot \frac{1}{3}{\left( {\frac{{2\pi d}}{L}} \right)^3}\]as the next term for the Alternating Series Estimation Theorem.

\[\begin{array}{l}\frac{{gL}}{{2\pi }} \cdot \frac{1}{3}{\left( {\frac{{2\pi d}}{L}} \right)^3}\\ = \frac{{4{\pi ^2}gL}}{3}{\left( {\frac{d}{L}} \right)^3}\\ = \frac{{4{\pi ^2}}}{3}{\left( {\frac{d}{L}} \right)^3}gL\end{array}\]

If L > 10d then \[\frac{d}{L} < \frac{1}{{10}}\].

Let's plug in\[\frac{d}{L} = 0.1\]

\[\frac{{4{\pi ^2}}}{3}{(0.1)^3}gL \approx 0.0132gL\]

So, the estimate is within 0.014gL when L > 10d.