Math Puzzles

Solution to The February/March, 2007 Problem of the Month

A sea otter is 10 meters from a very long and straight river, with a large river bank. We call this point (0,10). A wolf suddenly appears at the point (-7, 13). The otter heads directly to a point on the river bank to jump in to safety. The wolf heads to the exact same point that the otter is heading toward, hoping to get there first. If the sea otter runs at 8 m/s and the wolf runs at 12 m/s, find the point (to three decimal places) that the sea otter should run to, so that it reaches that point with the most time to spare. That is, find the point on the river bank so that if both animals head toward that point, the difference between the the wolf's time and the otter's time is a maximum. (See diagram below.)

Also, for a given otter speed, v_o, a given initial starting point for the otter, (0,y_o), and a given wolf speed, v_w, find the envelope of safety for the otter. That is, find the equation of the curve so that if the wolf is on the outside of the curve, the otter can reach a point on the river bank before the wolf. (It will be a tie if the wolf starts on the curve.)

For the specific problem, we want to find a so that

Dt = ((a + 7)² + 13²)^½/12 - (a² + 10²)^½/8
is a maximum. Taking the derivative of the above function (wrt a), and setting it equal to zero, we have
(a + 7)/[12((a + 7)² + 13²)^½] - a/[8(a² + 10²)^½] = 0.
Solving numerically, we get a ~ 5.0977 meters, i.e., 5.0977 meters down the shoreline, and the otter beats the wolf by about .0768 seconds.

In general, we want to find all (x,y) so that Dt is a maximum, and also Dt = 0. The answer is an (upper half) ellipse.
For, the two equations we want to solve are

(a + x)/[v_w((a + x)² + y²)^½] - a/[v_o(a² + y_o²)^½] = 0
and
Dt = ((a + x)² + y²)^½/v_w - (a² + y_o²)^½/v_o = 0.
We want to eliminate a, and form an equation with x and y with only the constants y_o, v_o, and v_w. From the first equation,
(a + x)²/[v_w²((a + x)² + y²)] = a²/[v_o²(a² + y_o²)] so (a + x)²v_o²(a² + y_o²)/[a²v_w²] = ((a + x)² + y²).
And from the second equation ((a + x)² + y²) = v_w²(a² + y_o²)/v_o² . Therefore (a + x)²v_o²(a² + y_o²)/[a²v_w²] = v_w²(a² + y_o²)/v_o² or (a + x)²v_o²/[a²v_w²] = v_w²/v_o² so x = av_w²/v_o² - a = a(v_w² - v_o²)/v_o².
Conversely, a = xv_o²/(v_w² - v_o²), and (a + x) = xv_w²/(v_w² - v_o²).
Plugging this into what we got from the second equation, we have (x²v_w⁴/(v_w² - v_o²)² + y²) = (x²v_o⁴/(v_w² - v_o²)² + y_o²)v_w²/v_o², or [v_w⁴v_o²/(v_w² - v_o²)²]x² + v_o²y² = [v_o⁴v_w²/(v_w² - v_o²)²]x² + v_w²y_o² so [v_w²v_o²/(v_w² - v_o²)]x² + v_o²y² = v_w²y_o². Set r = v_w/v_o, the ratio of their respective speeds, so that we can write the final equation as x²/[y_o²(r² - 1)] + y²/[y_o²r²] = 1. An ellipse!