The NRICH asks an interesting math question at this Link. Let's say you're a spider, and you see a fly across a room. What would be the shortest, and therefore fastest, route you would have to crawl to catch the fly?

The spider is in a room, so it's 3-D. The usual 2-D graph of x and y, rise over run, doesn't setup the problem accurately. There is width, height, and length: an x, y, and z coordinates respectively. But even though the spider and the fly lives in a 3-D world, we can break down the problem into a 2-D one, since the spider can only crawl on the wall in a 2-D way - side-to-side (x), up-and-down (y).

We are given that the room is 4 meters wide, 2.5 meters high, and 5 meters long. The spider is in the middle of the smallest wall on one side of the room. The fly is on the opposite, smallest wall and it's position is 1.5 meters high and 0.5 meters away from the adjacent wall next to the window. Below is how I visualized the room, and how I labeled the coordinates. The room is not drawn to scale.

Then, I thought how to move the spider from wall to wall to get to the fly in the shortest distance. If the spider goes towards the back wall, then the spider would have to cross the window to get to the fly, which is 3.5 m away from the fly (4-0.5 = 3.5). So, the better way for the spider to travel is to move towards the front wall. Next, I imagined the fly on the spider's wall in it's (0.5, 1.5) position. Eventually, the spider has to get to the height of 1.5 meters, so I figured it should do so while traveling to the edge of the front wall. That way, the spider will only need to crawl in a straight line to get to the fly. Below is the route indicated by the arrows.

Since the spider is crawling a straight line on the front wall and the opposite wall, then the spider will be crawling a distance of 0.5 + 5 = 5.5 meters for those two walls. I needed to use the distance formula to calculate the diagonal distance the spider crawls to get to the edge of the front wall at the height of 1.5 meters.

Distance Formula |

The coordinate of the spot I want the spider to crawl to is (0, 1.5, 5), since x=0 is the edge of the front wall, 1.5 is the height of the fly, and 5 is the z coordinate for end wall. I don't need to use the z-coordinate, 5, because the spider is moving in 2-D, from side-to-side and up-and-down. My calculation using the distance formula is below.

This distance is approximately 2.02 meters. Therefore, the total distance traveled for the spider to reach the fly is 0.5+5+2.02= 7.52 meters. Then the problem asks another question: If the fly starts moving towards the floor (I'm assuming the fly moving straight down), then at what point should the spider change it's route to catch the fly?

Not knowing when the fly starts moving downwards, and where the spider is on it's route, I decided to redo the problem with the fly already at the bottom of the wall. Then I thought I would do the problem in the same way as above. But then I had an Eureka! moment. I was already breaking down spider's movements or the walls of the room into 2-D. The two end walls had (x, y) coordinates and the front and back walls had (z, y) coordinates. So, I decided to remove the back wall, and flatten the 3 remaining walls into one large wall. Then, the spider just needs to travel diagonally to get to the fly, instead of changing direction to go straight once the spider reaches an edge. Above is how I visualized the new problem (the fly is moving straight downward).

Now, we have just have (x,y) coordinates to deal with. The fly's new coordinate is (9.5, 0) and the spider's new coordinate is (2, 1.25). I used the distance formula again to calculate the distance the spider will travel on the diagonal line. Below is the calculation.

The answer is approximately 10.68 meters. Since, I thought of a new way to look at the problem I decided to re-calculate the distance of the first problem using this method. So, the spider will, again, move diagonally only, and not change direction to go straight. Below is the new visual for the first problem.

So, the spider has the coordinates of (2, 1.25) and the fly has the coordinates of (9.5, 1.5). The calculation for distance using the distance formula is below.
The answer is approximately 10.61 meters. So, it would seem that the best route was my first one. Going diagonally all the way across adds distance instead of shortening it. So, going diagonally over a shorter distance and then going in a straight line to the final destination point is the best route. But the fastest route, instead of going around the room, is to go through the room. So, what if, instead of a spider that has to crawl over a 2-D space (side-to-side/up-and-down), there is a dragonfly that can fly through a 3-D one (side-to-side/up-and-down/forward-and-backward)?

3-D Extension

I was very interested in exploring 3-D geometry, and the first thing I wanted to learn was how to obtain a line in 3-D space. So, I watched David Butler's "Example: Finding the equation of the line in 3D through two points" YouTube Video from Maths Learning Centre UofA channel, and applied it to this problem. My work is below.

Then, I wanted to find the distance the dragonfly would have to travel to reach the fly. My professor explained I could use the distance formula with the z-coordinates added. My work is below:

The square root is about 5.226. So, the dragonfly would have to fly about 5.226 meters to get to the fly.

Take-Away

A great problem, like this one, inspires the imagination. The original problem, even though the setup is in 3-D, asks a 2-D question. The concept that needs to be understood is that a spider can only move in two ways on the wall or plane. The problem can be thought of and solved in different ways, as shown above. So, the original problem has lots to explore, but it also offers more. As soon as I read the problem, my mind went straight to "But what if the spider could fly?", turning the 2-D question into a 3-D one, and then I sought out ways to answer my question. Curiosity is an indication of interest and readiness to learn something new. Offering opportunities for students to imagine "what if" is the best way to engage and encourage their learning. It will motivate them to learn on their own, fostering a sense of responsibility for their education. And as a teacher, that is the most important skill I can encourage in my students.

Figures created using GeoGebra and Microsoft Word. I do not have any affiliation with NRICH, GeoGebra, or David Butler, and none have endorsed my work. I am a college student.

Love it! Great discussion and think aloud and strong consolidation. Got me thinking about doing this in GGB 3D, too. 5Cs +

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