Surf's Up

A surfer is riding along a crest of a sinusoidal wave given by the function $y = 2·sin(x−t),$ where $y$ is the height above sea level in meters, $x$ is horizontal position in meters, and $t$ is time in seconds. Based on this, the surfer knows that her speed—and the speed of the wave—is $1$ meter per second.

But hold on—it appears that another wave is approaching from the opposite direction. This wave has half the amplitude but twice the speed: $y = sin(x+2t).$

The two waves, the forward and the backward, ”interfere,” meaning they combine. In other words, the height of the water at position $x$ and time $t$ is given by $y = 2·sin(x−t) + sin(x+2t).$ Nevertheless, the surfer stays atop a crest of the combined wave. As the wave moves along, what are the surfer’s slowest and fastest speeds, in meters per second?

Let's assume that at time $t = 0$, that the surfer is perch above the crest of the wave at $x(0)=\frac{\pi}{2},$ with $y(0)=3.$ Since at this point we have $\frac{\partial y}{\partial x}=0,$ for any $t \gt 0,$ we can take the derivative of $y$ with respect to $x$ and solve for $x$ in terms of time $t.$ That is, for any t, we have \begin{align*}\frac{\partial y}{\partial x} &= 2 \cos ( x - t ) + \cos (x + 2t) \\ &= 2 \cos x \cos t + 2 \sin x \sin t + \cos x \cos (2t) - \sin x \sin (2t) \\ &= \left( \cos (2t) + 2 \cos t \right) \cos x - \left( \sin (2t) - 2 \sin t \right) \sin x.\end{align*} So by setting $\frac{\partial y}{\partial x} = 0,$ we get $$\tan x = \frac{ \cos (2t) + 2\cos t }{ \sin (2t) - 2\sin t }.$$ One way to solve this equation is to set $$x(t) = \tan^{-1} \left( \frac{ \cos (2t) + 2\cos t }{ \sin (2t) - 2\sin t } \right);$$ however, this will have a bunch of nasty jump discontinuities that likely would be fairly gnarly for the surfer to deal with. Instead, let's resolve this by ensuring that $x(t)$ is a nondecreasing, continuous function of $t,$ which would lead us to $$x(t) = \pi \left\lceil \frac{t}{\pi} \right\rceil + \tan^{-1} \left( \frac{\cos (2t) + 2\cos t }{ \sin (2t) - 2\sin t } \right).$$

Since we want to compute the surfer's horizontal speed at various times, we can ignore the almost everywhere constant initial term, $\pi \lceil t/\pi \rceil,$ and calculate \begin{align*}\dot{x}(t) = \frac{d}{dt} x(t) &= \frac{1}{1 + \left( \frac{\cos (2t) + 2 \cos t }{\sin (2t) - 2 \sin t} \right)^2} \frac{d}{dt} \left( \frac{ \cos (2t) + 2\cos t}{\sin (2t) - 2\sin t} \right).\end{align*} We first see that \begin{align*}\frac{d}{dt} \left( \frac{\cos (2t) + 2\cos t}{\sin (2t) - 2\sin t} \right) &= \frac{ (\sin (2t) - 2\sin t) (-2\sin (2t) - 2\sin t) - (\cos (2t) + 2\cos t) (2\cos (2t) - 2 \cos t)}{ (\sin(2t) - 2 \sin t)^2} \\ &= \frac{ -2 \sin^2 (2t) + 2 \sin t \sin (2t) + 4 \sin^2 t - 2 \cos^2 (2t) - 2 \cos t \cos (2t) + 4 \cos^2 (2t)}{ (\sin (2t) - 2 \sin t)^2} \\ &= \frac{ -2 - 2 \left( \cos t \cos (2t) - \sin t \sin (2t) \right) + 4 }{ (\sin (2t) - 2 \sin t)^2} \\ &= \frac{ 2( 1 - \cos (3t))}{ (\sin (2t) - 2 \sin t)^2}. \end{align*} Secondly, we see that \begin{align*}\frac{1}{1 + \left( \frac{ \cos (2t) + 2 \cos t }{ \sin (2t) - 2 \sin t } \right)^2} &= \frac{ (\sin (2t) - 2 \sin t)^2}{ (\sin (2t) - 2 \sin t)^2 + (\cos (2t) - 2 \cos t)^2 } \\ &= \frac{ (\sin (2t) - 2 \sin t)^2}{ \sin^2 (2t) - 4 \sin t \sin (2t) + 4 \sin^2 t + \cos^2 (2t) + 4 \cos t \cos (2t) + 4 \cos^2 t} \\ &= \frac{( \sin (2t) - 2 \sin t)^2}{ 1 + 4 \left( \cos t \cos (2t) - \sin t \sin (2t) \right) + 4 } \\ &= \frac{ (\sin (2t) - 2 \sin t)^2}{ 5 + 4 \cos (3t)}.\end{align*} Therefore, putting everything together we get $$\dot{x}(t) = \frac{ (\sin (2t) - 2 \sin t)^2 }{ 5 + 4 \cos (3t)} \frac{ 2 ( 1 - \cos (3t) )}{ (\sin (2t) - 2 \sin t)^2} = \frac{ 2(1 - \cos (3t))}{ 5 + 4 \cos (3t) }.$$

Ok, so far so good for our surfer. We see that since $5 + 4 \cos (3t) \geq 1$ for all $t \leq 0,$ that the denominator of her speed is always well defined. Furthermore, we see that if we define $$f(s) = \frac{ 2(1-s) }{5 + 4s}$$ then we have $$f'(s) = \frac{-18}{(5+4s)^2} \leq 0$$ for all $s \in \mathbb{R},$ so $f$ is a monotone decreasing function of $s.$ Therefore, in particular, since $-1 \leq \cos (3t) \leq 1$ for all $t \geq 0,$ we see that if $-1 \leq s = \cos (3t) \leq 1,$ then we have $$0 =\frac{ 2 ( 1 - 1 )}{ 5 + 4 \cdot 1} = \dot{x} \left( \frac{1}{3} \cos^{-1} 1 \right) \leq \dot{x} (t) \leq \dot{x} \left( \frac{1}{3} \cos^{-1} (-1) \right) = \frac{ 2(1 - (-1))}{ 5 + 4 \cdot -1} = 4.$$ So the surfer's slowest and fastest speeds in meters per second are 0 and 4, respectively.