### July 7th Riddler Express: The Beach Puzzle¶

The puzzle states that two swimmers leave from a beach at identical speeds, with Swimmer $A$ moving orthogonally from the beach front and Swimmer $B$, starting 100 yards away, swimming always in the direction of Swimmer $B$. The question then is, when Swimmer $B$ is eventually directly behind Swimmer $A$ how far apart will they be?

This is a bit of a trick question, since always moving directly towards Swimmer $A$ means Swimmer $B$ will never be exactly behind him, but we'll think about it in the limit. The answer is that Swimmer $B$ will be 50 yards behind Swimmer $A$ when he is directly in his wake.

I solved the problem by numerical simulation. It is easy to set up a system of difference equations that describe the relationship between Swimmer $B$'s position at time $t + \Delta t$, given his and Swimmer $A$'s positions at time $t$. Here I've assumed that the beach front is the $y$-axis, Swimmer $B$ begins at the origin, Swimmer $A$ begins at the position $(0,100)$, and they both travel at a speed of $\nu$. We then have

\begin{eqnarray} \nonumber x_B(t + \Delta t) &=& x_B(t) + \nu ~\Delta t ~u_x(t) \\ \nonumber y_B(t + \Delta t) &=& y_B(t) + \nu ~\Delta t ~u_y(t) \\ \end{eqnarray}

where $u_x(t)$ and $u_y(t)$ are the $x$- and $y$-components, respectively, of the unit vector describing the direction of travel of Swimmer $B$ at time $t$. These quantities can be expressed based on the positions of the two swimmers at time $t$. Note that the position of Swimmer $A$ is described by

\begin{eqnarray} \nonumber x_A(t) &=& \nu ~ t \\ \nonumber y_A(t) &=& 100 \end{eqnarray}

We then have

\begin{eqnarray} \nonumber u_x(t) &=& \frac{\nu~t - x_B(t)}{\sqrt{[\nu~t-x_B(t)]^2 + [100-y_B(t)]^2}} \\ \nonumber u_y(t) &=& \frac{100 - y_B(t)}{\sqrt{[\nu~t-x_B(t)]^2 + [100-y_B(t)]^2}} \\ \end{eqnarray}

Our final set of difference equations is then

\begin{eqnarray} \nonumber x_B(t + \Delta t) &=& x_B(t) + \nu ~ \Delta t ~ \frac{\nu~t - x_B(t)}{\sqrt{[\nu~t-x_B(t)]^2 + [100-y_B(t)]^2}} \\ \nonumber y_B(t + \Delta t) &=& y_B(t) + \nu ~ \Delta t ~ \frac{100 - y_B(t)}{\sqrt{[\nu~t-x_B(t)]^2 + [100-y_B(t)]^2}} \\ \end{eqnarray}

Starting from their initial positions we can then step through time until some suitable stopping criterion is reached (I chose to stop when the $x$-component of the direction vector was within $10^{-14}$ of 1, indicating that Swimmer $B$ was nearly swimming horizontally) and check the distance between $x_B(t)$ and $x_A(t)$.

The following Julia function runs the simulation for given speed and time-step size and returns the resulting distance.

In [1]:
using PyPlot

function swimmer_sim(speed, dt, show_plot=true)
# ---------------------------------------------
# Function to simulate the paths of Swimmers
# ---------------------------------------------

# Initial time, position, and direction parameters
xB, yB, t, ux, uy = 0.0, 0.0, 0.0, 0.0, 1.0

# Arrays to store history of swimmer B for plotting
xpos, ypos = [], []

# Step through time until B's direction is nearly horiztonal
while abs(1-ux) > 1e-14
vx, vy = speed*t-xB, 100-yB
ux, uy = vx/norm([vx,vy],2), vy/norm([vx,vy],2)
xB, yB = xB + speed*dt*ux, yB + speed*dt*uy
t += dt
push!(xpos,xB)
push!(ypos,yB)
end

# Compute and print final distance
final_dist = speed*t - xB
print("distance between swimmers = ", final_dist,"\n")

# Make plot of position histories
if show_plot
plot([0,speed*t],[100,100],"b--", label="Swimmer A")
plot(xpos, ypos, "k", label="Swimmer B")
plot(xB, yB, "ko")
plot(speed*t, 100, "bo");
legend(loc="lower right")
end

return final_dist

end;

Here is the result and picture when the swimmers are traveling at $\nu = 5$ yards per $\texttt{<pick your favorite time unit>}$

In [2]:
swimmer_sim(5.0, 0.001);
distance between swimmers = 50.001250028202435

We should probably also check that we obtain the same result regardless of the speeds of the swimmers.

In [3]:
# Test for speed = 1, 2, 3, 4, and 5
for speed in [ii for ii=1:5]
swimmer_sim(speed, 0.001, false);
end
distance between swimmers = 50.000249991526516
distance between swimmers = 50.00050000180863
distance between swimmers = 50.00075000816571
distance between swimmers = 50.001000018359946
distance between swimmers = 50.001250028202435

Finally, we should check that the resulting distance from the simulation is in fact tending to $50$ as $\Delta t \rightarrow 0$

In [4]:
# Test for dt = .1, .01, .001, .0001, and .00001
for dt in [0.1^ii for ii=1:5]
swimmer_sim(5.0, dt, false);
end
distance between swimmers = 50.12526106941539
distance between swimmers = 50.01250260484869
distance between swimmers = 50.00125002820232
distance between swimmers = 50.00012502579966
distance between swimmers = 50.00001276351054

and it is!