The surfer rides the crest of the combined wave, $y = 2\sin(x−t) + \sin(x+2t),$ for a very long time. What is her average speed, in meters per second?
As we saw in the Classic answer, the surfer's horizontal speed with as a function of time is given by $$\dot{x}(t) = \frac{ 2(1-\cos (3t))}{5 + 4 \cos(3t)},$$ while her horizontal location $$x(t) = \pi \left\lceil \frac{x}{\pi} \right\rceil + \tan^{-1} \left( \frac{\cos (2t) + 2\cos t}{\sin (2t) - 2 \sin t} \right).$$
There are obviously two ways of measuring the average speed: (a) integrate the periodic function $\dot{x}(t)$ over some large time period from $0$ to $T \gg 1;$ or (b) rely on the relationship between distance and average speed to take $\bar{x} = (x(T) - x(0)) / T$ for some large $T \gg 1.$ For kicks, let's do both!
Since $\dot{x}$ is periodic with period $2\pi/3,$ we get that \begin{align*}\bar{x} &= \frac{1}{T} \int_0^T \dot{x}(t)\, dt\\ &=\frac{1}{T} \left( \int_0^{\frac{2\pi}{3} \lfloor 3T/2\pi \rfloor} \dot{x}(t) \,dt + \int_{\frac{2\pi}{3} \lfloor 3T/2\pi \rfloor}^T \dot{x}(t) \,dt \right).\end{align*} Firstly, we see that \begin{align*}I = \int_0^{2\pi/3} \dot{x}(t) \,dt &= 2 \int_0^{\pi/3} \frac{2(1-\cos (3t))}{5 + 4 \cos (3t)} \,dt \\ &= 2\int_0^\infty \frac{ 2 \left( 1 - \frac{1 - s^2}{1+ s^2} \right)}{ 5 + 4 \frac{1-s^2}{1+s^2} } \, \frac{ 2 ds }{ 3 ( 1 + s^2 ) } \\ &= \frac{16}{3} \int_0^\infty \frac{s^2}{(9+s^2)(1+s^2)} \,ds,\end{align*} where we have taken advantage of the powerful and mysterious tangent half-angle substitution $s = \tan \frac{3t}{2},$ where $\cos 3t = \frac{1 - s^2}{1+ s^2}$ and $dt = \frac{2ds}{3(1+s^2)}.$ Using the method of partial fractions we see that $$\frac{16}{3} \frac{s^2}{(9+s^2)(1+s^2)} = \frac{6}{9+s^2} - \frac{2}{3(1+s^2)},$$ and since $\int \frac{du}{a^2+u^2} = \frac{1}{a} \tan^{-1} (u/a) + C,$ so we have \begin{align*}I &= \frac{16}{3} \int_0^\infty \frac{ s^2 }{ (9+s^2) (1+s^2) } \, ds = \int_0^\infty \frac{6}{9+s^2} \,ds - \frac{2}{3} \int_0^\infty \frac{ds}{1+s^2} \\ &= 6 \left[ \frac{1}{3} \tan^{-1} \frac{s}{3} \right]_0^\infty - \frac{2}{3} \left[ \tan^{-1} s \right]_0^\infty \\ &= 2 \left( \frac{\pi}{2} - 0 \right) - \frac{2}{3} \left( \frac{\pi}{2} - 0 \right) \\ &= \frac{2\pi}{3}.\end{align*} We have $$\int_0^{\frac{2\pi}{3} \lfloor 3T/2\pi \rfloor} \dot{x}(t) \,dt = \left\lfloor \frac{3T}{2\pi} \right\rfloor \int_0^{2\pi/3} \dot{x}(t)\,dt = \left\lfloor \frac{3T}{2\pi} \right\rfloor I = \frac{2\pi}{3} \left\lfloor \frac{3T}{2\pi} \right\rfloor.$$ Since we already saw that $0 \leq \dot{x}(t) \leq 4,$ we have \begin{align*}\bar{x} &=\frac{1}{T} \left( \int_0^{\frac{2\pi}{3} \lfloor 3T/2\pi \rfloor} \dot{x}(t) \,dt + \int_{\frac{2\pi}{3} \lfloor 3T/2\pi \rfloor}^T \dot{x}(t) \,dt \right) \\ & \geq \frac{1}{T} \left( \frac{2\pi}{3} \left\lfloor \frac{3T}{2\pi} \right\rfloor + 0 \right) = \frac{2\pi}{3T} \left\lfloor \frac{3T}{2\pi} \right\rfloor =: m(T).\end{align*} Additionally, since $0 \leq T - \left\lfloor \frac{3T}{2\pi} \right\rfloor \frac{2\pi}{3} \leq \frac{2\pi}{3},$ we have $$\bar{x} \leq \frac{1}{T} \left( \frac{2\pi}{3} \left\lfloor \frac{3T}{2\pi} \right\rfloor + 4 \left( T - \left\lfloor \frac{3T}{2\pi} \right\rfloor \frac{2\pi}{3} \right) \right) \leq \frac{2\pi}{3T} \left\lfloor \frac{3T}{2\pi} \right\rfloor + \frac{8\pi}{3T} =: M(T).$$ Since we further have that for any $z \gt 0$ that $1 - \frac{1}{z} \leq \frac{\lfloor z \rfloor}{z} \leq 1,$ we have that $\lim_{T \to \infty} m(T) = \lim_{T \to \infty} M(T) = 1,$ so by the Squeeze Theorem, we have that the surfer's average speed is $\hat{x} = 1$ m/s, as $T \to \infty.$
Alternatively, let's try to directly bound $(x(T) - x(0))/ T$ for large $T \gg 1.$ First, let's look at the behavior of $\dot{x}$ on $[0, 2\pi/3].$ We see that there are two solutions to $\dot{x} = 1,$ namely $t_1 = \frac{2\pi}{9}$ and $t_2 = \frac{4\pi}{9}.$ The tangent lines at $t_1$ and $t_2$ are given by $\dot{x}(t_1) (t-t_1) + x(t_1) = t + \pi/3$ and $\dot{x}(t_2) (t-t_2) + x(t_2) = t + 2\pi/3,$ respectively. Since we can differentiate again to obtain $$\ddot{x}(t) = \frac{54 \sin (3t)}{(5+4\cos(3t))^2},$$ we see that $x(t)$ is concave up from $[0, \pi/3]$ and concave down from $[\pi/3, 2\pi/3].$ Since the concavity of a function determines whether tangent lines lie above or below the curves, we see that we have $t+\pi/3 \leq x(t) \leq t + 2\pi/3,$ at least for $[0, 2\pi/3];$ however, due to the periodicity of $x$, we can extend this result to show that for all $t \geq 0$ we have $$ t + \frac{\pi}{3} \leq x(t) \leq t + \frac{2\pi}{3}.$$ Therefore, we have $$1 - \frac{\pi}{6T} = \frac{ T + \frac{\pi}{3} - \frac{\pi}{2} }{T} \leq \bar{x} = \frac{x(T) - x(0)}{T} \leq \frac{T + \frac{2\pi}{3} - \frac{\pi}{2}}{T} = 1 + \frac{\pi}{6T}.$$ Again, since both the upper and lower bounds converge to $1$, we see that as $T \to \infty,$ the surfers average speed will converge to $1$ m/s.
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