Chapter 6.4, Problem 43E

Surfing the Perfect Wave For a wave to be surfable, it can’t break all at once. Robert Gura and Tony Bowen have shown that a wave has a surfable shoulder if it hits the shoreline at an angle θ given by

$\theta ={\sin }^{-1}\left(\frac{1}{2n+1\tan \beta }\right)$

where β is the angle at which the beach slopes down and where n = 0, 1, 2, ….

1. (a) For β = 10°, find θ when n = 3.
2. (b) For β = 15°, find θ when n = 2, 3, and 4. Explain why the formula does not give a value for θ when n = 0 or 1.

To determine

To Find: The angle $\theta$ at which the wave is surfable.

Answer to Problem 43E

The angle at which a wave has a surfable shoulder is $\theta =54.1°$.

Explanation of Solution

Given:

The angle at which a wave has a surfable shoulder is $\theta ={\sin }^{-1}\left(\frac{1}{\left(2n+1\right)\tan \beta }\right)$.

The value of $\beta =10°$ and $n=3$.

Calculation:

The formula is,

$\theta ={\sin }^{-1}\left(\frac{1}{\left(2n+1\right)\tan \beta }\right)$

Substitute $10°$ for $\beta$ and $3$ for $n$,

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{1}{\left(2\left(3\right)+1\right)\tan 10°}\right)\\ ={\sin }^{-1}\left(\frac{1}{7\times 0.18}\right)\left(\tan 10°=0.18\right)\\ ={\sin }^{-1}\left(\frac{1}{1.26}\right)\\ \approx 54.1°\end{array}$

Thus, the angle at which a wave has a surfable shoulder is $\theta =54.1°$.

To determine

To Find: The angle $\theta$ at which the wave is surfable.

Answer to Problem 43E

The angle at which a wave has a surfable shoulder is $\theta =48.26°$.

Explanation of Solution

Given:

The angle at which a wave has a surfable shoulder is $\theta ={\sin }^{-1}\left(\frac{1}{\left(2n+1\right)\tan \beta }\right)$.

The value of $\beta =15°$ and $n=2$.

Calculation:

The formula is,

$\theta ={\sin }^{-1}\left(\frac{1}{\left(2n+1\right)\tan \beta }\right)$ (I)

Substitute $15°$ for $\beta$ and $2$ for $n$,

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{1}{\left(2\left(2\right)+1\right)\tan 15°}\right)\\ ={\sin }^{-1}\left(\frac{1}{5\times 0.18}\right)\left(\tan 15°=0.26\right)\\ ={\sin }^{-1}\left(\frac{1}{1.34}\right)\\ =48.26\end{array}$

The approximate value is,

$\theta =48.3$.

Thus, the angle at which a wave has a surfable shoulder is $48.3°$.

Substitute $15°$ for $\beta$ and $3$ for $n$ in equation (I) to find value of $\theta$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{1}{\left(2\left(3\right)+1\right)\tan 15°}\right)\\ ={\sin }^{-1}\left(\frac{1}{7\times 0.26}\right)\left(\tan 15°=0.26\right)\\ ={\sin }^{-1}\left(\frac{1}{1.87}\right)\\ =32.23°\end{array}$

The approximate value is,

$\theta =32.2°$.

Thus, the angle at which a wave has a surfable shoulder is $32.2°$.

Substitute $15°$ for $\beta$ and $4$ for $n$ in equation (I) to find the next value of $\theta$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{1}{\left(2\left(4\right)+1\right)\tan 15°}\right)\\ ={\sin }^{-1}\left(\frac{1}{9\times 0.26}\right)\left(\tan 15°=0.26\right)\\ ={\sin }^{-1}\left(\frac{1}{2.41}\right)\\ =24.51°\end{array}$

The approximate value is,

$\theta =24.5°$.

Thus, the angle at which a wave has a surfable shoulder is $24.5°$.

To determine

To show: The formula $\theta ={\sin }^{-1}\left(\frac{1}{\left(2n+1\right)\tan \beta }\right)$ does not give a value at $n=0,1$.

Explanation of Solution

The angle at which the wave hits the shoreline is,

$\theta ={\sin }^{-1}\left(\frac{1}{\left(2n+1\right)\tan \beta }\right)$

Substitute $0$ for $n$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{1}{\left(2\left(0\right)+1\right)\tan 15°}\right)\\ ={\sin }^{-1}\left(\frac{1}{\tan 15°}\right)\left(\because \tan 15°=0.26\right)\\ ={\sin }^{-1}\left(\frac{1}{0.26}\right)\end{array}$

This function of ${\sin }^{-1}$ is not defined as it is outside the interval $\left[-1,1\right]$. ${\sin }^{-1}$ is not defined outside of this interval.

Again substitute $1$ for $n$ in the above formula,

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{1}{\left(2\left(1\right)+1\right)\tan 15°}\right)\\ ={\sin }^{-1}\left(\frac{1}{3\times \tan 15°}\right)\left(\because \tan 15°=0.26\right)\\ ={\sin }^{-1}\left(\frac{1}{3\times 0.26}\right)\\ ={\sin }^{-1}\left(\frac{1}{0.78}\right)\end{array}$

This function of ${\sin }^{-1}$ is not defined as it is outside the interval $\left[-1,1\right]$. ${\sin }^{-1}$ is not defined outside of this interval.

Thus, The formula $\theta ={\sin }^{-1}\left(\frac{1}{\left(2n+1\right)\tan \beta }\right)$ does not give a value at $n=0,1$.