TREASURE BENEATH THE SEA Teacher’s Guide – Getting Started

Purpose In this lesson, students help the crew of a shipwreck recovery team minimize the amount of work done to remove treasure chests lost at sea. The divers must move the chests to a rope that is between their locations and comes from descending from the recovery team’s boat above. The captain of the boat’s crew insists on placing the rope in one spot; he doesn’t want to waste time and money moving it each time a chest is collected. Prerequisites Basic algebra skills and some geometry. Materials Needed Required: Large, flat pieces of cardboard, string, small weights, and scissors (to pierce cardboard). Suggested: Rulers or straightedges, compasses, geometry software. Optional: None. Worksheet 1 Guide The next three pages constitute the first day’s work. A physical model of the situation can be constructed in the classroom out of cardboard. To do this, cut two holes in the cardboard 40cm apart from each other and thread two strings through the holes. Tie them together above the cardboard so that the lengths of the strings below the knot are equal. Set the cardboard between two posts or two tables so that it is level and the weights can hang freely. Students can experiment with using equal weights at the ends of each string and with different weights at the end of each string. The position of the knot should help them figure out where to position the rope from the boat. Worksheet 2 Guide The fourth page constitutes the second day’s work. The cardboard model above should be modified to fit the “three chests” problem (shown to the right). Students will need to experiment using different combinations of weights: all three the same, two the same and one different, and all three different. Students learn the precise definition of work and will modify their ideas about how work should be defined from the first day to use this mathematical definition.

TREASURE BENEATH THE SEA A shipwreck containing treasure chests filled with gold and silver was recently discovered in the Bermuda Triangle. Underwater photos have revealed two treasure chests, spaced 40 meters apart, and it is up to you to figure out how best to retrieve them. Your boat has a rope that can be lowered and tied around the treasure chests, but your captain insists he doesn’t want to sail back and forth all day. He says that you have to choose one place to lower the rope, and then you can swim down with it and start collecting the treasure.

What is the best way to collect the treasure without upsetting the captain?

1. What information is necessary to have before a mathematical model is constructed? What variables do you have to consider? What variables should be not be taken into account?

2. Call the treasure chest on the left Chest 1 and the treasure chest on the right Chest 2. If you know the distance from Chest 1 to the rope, how can you determine the distance from Chest 2 to the rope?

3. When you dive down, you will have to move the chests over to the rope. What’s a good way to measure the amount of work done to move the chests? What variables should be taken into account in this measurement? Is there a way to consider all of these variables together?

4. You estimate that each of the chests has the same weight. You want the total work you do to move both chests to be as little as possible. Where should you lower the rope? How would your answer change if the chests weighed different amounts? Use a cardboard model to experiment and test your ideas.

5. When you dive to the bottom, you find that there is an additional third treasure chest! Fortunately, they are all lined up in a row. If you still want to minimize the work in moving all the chests, where do you place the rope now assuming that they each weigh the same? Provide a mathematical explanation for your reasoning.

In mathematics and physics work has a very precise meaning. Work is the amount of energy transferred by a force acting over a certain distance. Here, “energy transferred by a force” means the same thing as “weight”. The equation is: Work = Force× Distance. 6. You did so well on your first dive that your captain is bringing you to another site. This site has 3 treasure chests, all equal weights, but they don’t lie even almost in a line. How do you minimize the amount of work done to move the chests to the rope?

7. If the chests in question 6 did not all have the same weight, how would the model change? Modify the cardboard model for this physical situation. What happens? Can you give a mathematical explanation for what is going on?

8. What are the differences between a physical model and a mathematical model? What are some advantages and disadvantages of each?

TREASURE BENEATH THE SEA Teacher’s Guide – Possible Solutions

The solutions shown represent only one possible solution method. Please evaluate students’ solution methods on the basis of mathematical validity. 1. The weight of the chests and the distance between the chests and the rope are the two variables that need to be considered. On the other hand, the length of the rope and the depth are not variables that need to be considered in this model. 2. If the distance between Chest 1 and the rope is x then the distance between the rope and Chest 2 is (40 – x) meters. 3. Work Force Distance. In this example, weight is an appropriate substitution for force. However, you could also measure the amount of work done in time, energy expended by the divers, or even cost of the entire operation. 4. When the chests weigh the same, it doesn’t matter where the rope is positioned, as long as it is between the two chests. Yet, when the weights are varied, it is most efficient to place the rope directly above the heavier of the two chests. 5. Placing the rope over the middle chest will provide for the smallest amount of work. Explanations will vary but they are all valid if they conclude the correct placement. 6. Three points that are not collinear will create a triangle. The Fermat point is the point at which the lines drawn from the vertices create the shortest path. The angle between any two of these three lines is 120 degrees. 7. The weights on the model would need to be adjusted accordingly. The knot would be pulled towards the heaviest weight and shifted near the second heaviest weight. As a possible extension, you may want to try using one set of three fixed weights and see what occurs, although note that each combination of weights will have a unique point. 8. Answers will vary. CCSS Addressed G-MG.3, A-CED.1

TREASURE BENEATH THE SEA Teacher’s Guide – Extending the Model

You may wish to consider the extension to two dimensions with three locations. You would expect that if the three locations are close to being in a straight line that the solution will look similar to the one-dimensional case. First look at the problem if the three weights w1, w2 and w3 are equal, and are located at points P1, P2, and P3, respectively. If the triangle formed by the Pi’s is sufficiently obtuse, then the optimal location for the rope is at the vertex of the obtuse angle. This is true, as long as the obtuse angle is at least 120 degrees. If all angles are less than 120 degrees, the optimum location for the boom is the point P inside the triangle at which PP1, PP2 and PP3 meet at 120 degrees. There are a number of nice geometric proofs of this, but the easiest one is by physics. In order to see this, we might as well assume that the weights at the three locations are general rather than equal. Imagine a piece of Plexiglass, or a sheet of wood, and drill holes at the three locations of P1, P2, and P3. Tie three pieces of string together at a point P, and run the three strings – one through each hole – and attach the weights, wi, to the strings going through their respective Pi. Let the configuration go. It will settle into a configuration of minimum potential energy, and it follows with a little energy argument that this will minimize the sum of the three products w1, w2 and w3 multiplied by the distance PPi. If one of the three wi’s is much more than the sum of the others, then P will be pulled to Pi, and you get the same end point problem as before. There is a geometric construction for the general 3-point case, which uses Ptolemy’s Theorem. If you have more than three points, locating a single P, which minimizes the sum of the distances is a classic problem on which there is a literature, but nothing especially pretty. If you have four points and you want the shortest network connecting them, that’s a different problem. There is a nice literature going back to Gauss, who put the solution into a letter to Schumacher but didn’t publish. He got interested in it because his son was working for the Duchy of Hanover who were planning their first railroad. But there was earlier interest in the problem for planning canals in England. But as I say, it is not the same problem unless n=3.