Friday, April 22, 2016

Connecting the distance formula to the equation of a circle

Today was focused on establishing a connection between calculating distances and the equation of a circle. I am hoping to build the foundation that enables students to derive the equation of a circle.

There are two different online resources that I could use to work on this. I elected to track along with material found on the Great Minds site. This site does require registration, but the teacher and student resources are free in pdf format. I am using chapter 17 of the circle unit as a basis for my lessons. The other resource would be the circle material from the Mathematics Vision Project. A colleague has elected to follow this tract and found teacher resource material for these lessons.

To start things off, I presented to different distance calculation problems. These tied in directly with the practice we had last class. Students seemed comfortable calculating out the required distances and had no issues or questions.

Next, I gave students a whiteboard grid, a compass, and tissue paper. I told students to draw a point near the center of the tissue paper; this was their given fixed point. I then had students measure off five units on the whiteboard grid using the compass. Next, they drew a circle of radius five units centered on their point.

Surprisingly, this took a lot longer than I expected. Students had trouble getting the measurement correct, or they tore their tissue paper as they attempted to draw their circle, or the tissue paper slid as they were drawing. Finally, we had our circles.

I asked students to place their circle such that the center was at the origin. I was projecting up an example to make clear what should be done. Next, I labeled the points (0, 5) and (5, 0) on the graph. Although these points didn't actually form triangles, we could think of one side length as a length of zero and one with a length of five.

I wrote out the equations (0 - 0)2 + (5 - 0)2 = 52 and (5 - 0)2 + (0 - 0)2 = 5to represent the process we actually use to calculate the distance between these points. I then asked students to use this as a model and to identify other points on the circle that aligned with grid intersections on the graph. I used a radius of 5 specifically because the points with a combination of 3 and 4 (either positive and negative or in reverse order) would meet this criteria.

Students were able to find the points but struggled on the process idea. Many were confused because they basically were asking, "Wouldn't the result just be a distance of 5?" I explained to them we wanted to focus on the process they used versus the result that came out.

This idea bothered students. Working with the class this year, I have seen that students tend not to think about what they are doing and just do something. Something as simplistic as finding the distance between two points is difficult for students to break down as to what they are actually doing.

Many students came to the conclusion that they were counting grid lines to find the lengths. I finally had to return the opening problem of finding the distance between the points (7, 15) and (3, 9). When asked how they found the side lengths of their right triangles, they said they counted grids. I changed the points to (70, 150) and (30, 90) and asked if they would still count grid lines? They all responded that they wouldn't. I pushed on what would they do?

Finally, students started to realize they would take the difference between the x-coordinate values and then the difference between the y-coordinate values. I wrote out  (70 - 30)2 + (150 - 90)2 to represent what they had just said.

At this point the light bulbs seemed to turn on. The class was able to go back and write out equations like (-3 - 0)2 + (4 - 0)2 = 52 and (3 - 0)2 + (4 - 0)2 = 52.  Then questions started coming. When would you do something like (-3 - 4)2 + (4 - 3)2  or (7 - 3)2 + (15 - 9)2? Other students responded that those would represent the distance between the points (-3, 4) and (4, 3) or the distance between (7, 15) and (3, 9). Another student asked why all the results were equaling 5? Another student responded that the circle had a radius of 5 and all points on the circle are the same distance from the center.

It took time to get to this point but students were getting the point. I drew a point on the circle and labeled the point (x, y). For homework, I asked students to use what we had just worked on to write out how they would calculate the distance from this point to the center of the circle located at (0,0).

Next class we'll focus on solidifying the idea that the equation of a circle centered at (0, 0) is given by x2 + y2 = r2. From there, we'll move to looking at what happens when a circle is not centered at the origin.

No comments:

Post a Comment