Counting Out πr 2. Teacher Lab Discussion. Overview. Picture, Data Table, and Graph. Part I Middle Counting Length/Area Out πrinvestigation

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1 5 6 7 Middle Counting Length/rea Out πrinvestigation, page 1 of 7 Counting Out πr Teacher Lab Discussion Figure 1 Overview In this experiment we study the relationship between the radius of a circle and its area. In the Circumference vs. Diameter experiment we found that the plot of circumference vs. diameter was a straight line with a slope of π. For Counting Out πr you can use the same circles, or you can use new circles. In either case, find the area () and the radius (R) of each circle. Since the radius is somewhat easier to measure than we treat R as the manipulated variable (Question 1) and as the responding variable. The principal controlled variable is the shape of the figures: they are all circles (Question ). Picture, Data Table, and Graph Part I picture of the experiment is shown in Figure 1. Both and R are measured independently as the children in the picture are doing. They are using tops of bottles and cans to trace the circles directly onto 1-sq-cm graph paper. They use a ruler to find R but they have to count sq cm to find, the area. We remind you how to do this in Figure. First the whole square centimeters are numbered. In this case there is only one whole square centimeter, so we number it 1. Next, we must piece together parts of square centimeters to make Figure ¹ ₂ 3 4&5

2 Counting Out πr, page of 7 Figure 3 Figure 5 R in cm in sq cm 1 cm R =.5 cm < 1 sq cm R =.5 cm <.5 sq cm R =.1 cm <.4 sq cm R =.5 cm <.5 sq cm whole ones. The pieces on the right labeled together make the second square centimeter. The top left piece and the bottom middle piece combine to make the third square centimeter. Continuing in this way we find that we have 5 whole square centimeters and a half left over for a total of 5.5 square centimeters. Do not let your students use a formula no matter what they may have learned in math about πr. We have found that even gifted students who know the πr formula backwards and forwards are surprised and amazed when they discover that the ratio of area to radius squared is the same for every circle! Typical data for the circles is shown in Figure 3, and a plot of vs R is shown in Figure 4. Note that the curve shown in Figure 4 goes through (,) (Question 3). We can see that this is correct by thinking about circles with small but non-zero radii. Figure 5 shows circles of increasingly smaller radius. From this figure we see that as the radius gets smaller and smaller so does the area. When the radius becomes cm, whatever that means, then the area is square centimeters. This is why the curve should go through (,). The curve is not a straight line(question 4); this reflects the fact that the ratio /R is not constant. Indeed the curve gets steeper as R increases. One goal of this experiment is to produce a related graph that is a straight line through the origin. Figure 4 Figure in sq cm 4 3 in sq cm R in cm R in cm

3 Counting Out πr, page 3 of 7 Figure 7 in (sq cm) Extension Extension 1 extrapolate: just get a ruler. curve is much more difficult to extend. Figure 7 shows two possible ways to extend the vs. R curve. Which is correct? It is very hard to tell just by looking. These different extensions will lead to very different answers to extrapolation questions. For example, Question 7 asks for the area of a circle with a radius of 5 cm. From Extension 1 we get that the area is 7 sq cm. From Extension we find = 98 sq cm. This disparity will get worse the further we try to extrapolate. Question 8 asks for the area of a circle with radius 6 cm. gain the different extensions of the curve lead to different answers. From Extension 1 we get that the area is about 95 sq cm. From extension we find = 14 sq cm. Let the children put their results for Questions 7 and 8 on the board and have them discuss why the answers vary so much. This is an outstanding opportunity to have the children check their predictions by actually drawing circles with radii of 5 cm and 6 cm and then counting sq cm. Comprehension Questions Part I R (cm) Even though the curve is not a straight line we can still use it to answer certain questions. For example, in Question 5 we ask for the area of a circle with a radius of 3 cm. Figure 6 shows how to interpolate to find that is about 8 sq cm. Question 6 asks for the radius of a circle with an area of 1.5 sq cm. Figure 6 shows that R = cm. So we see that interpolation is not more difficult for a curve than it was for straight lines. Trouble occurs when we try to extrapolate. The problem with extrapolation is, how do we extend the curve? straight line is easy to So we see that since the curve is not a straight line, extrapolation will be tricky and most likely will not be very accurate. s noted above, your students should check some of their predictions. Question 9 and Question 1 require this. In Question 9, the circles have radii 1 cm and 5 cm, and areas of about 3.5 sq cm and 79 sq cm. So, when they find their area from their curves, then in Question 1a they should get rea of circle 1 = 3 to 3 1 sq cm and rea of circle = 7 sq cm to 98 sq cm Now counting out sq cm they should get for Question 1b: rea of circle 1 ~ 3 1 sq cm and rea of circle ~ 8 sq cm

4 Counting Out πr, page 4 of 7 Now they compare the two by calculating % differences. Most should find that the % differences for circle 1 is around 1% or so, but in the class, the % difference for circle might range from 1% to as much as 7%! Straightening Out the Curve Part II We now go a step further and ask if there might be a way to do more accurate extrapolation. Here we apply a fundamental problem solving technique: change the problem into one that we already know how to solve! Perhaps if we plot versus R we can get a straight line through (,). This is the point of Table and Graph in the student worksheets. There are, of course, good reasons to suspect that is proportional to R. For some children it might be best just to appeal to authority: you (the teacher) think it s a good idea. The truly inquisitive can be led to the idea of squaring the radius by looking at the data. Note that when the radius is roughly doubled (from 1.3 cm to.5 cm), the area increases by a factor of about four (from 5.5 sq cm to sq cm). When the radius is roughly tripled (from 1.3 cm to 4 cm), the area increases by a factor of about nine (from 5.5 sq cm to sq cm). Clearly the area is not proportional to the radius. But something funny is happening with squares! fter all, four is the square of two and nine is the square of three. This leads us to guess the area might be proportional to the square of the radius. To find out, we add a column for R to the data table and the children plot vs R. s you can see from Figure 8, the resulting curve is not the same as the curve for graph I (Question 11) and indeed is a straight line (Question 1). Hooray! It is much easier to extrapolate a straight line than a curve. There is a problem, however. Our straight line is vs R, but we are interested in R. This complicates things a bit. For instance, Question 13 in the Figure 8 in sq cm R in cm R in sq cm in sq cm R in sq cm Comprehension Questions requires the children to use Graph II to find when R = 3 cm. We must first square R = 3 cm to obtain R = 9 sq cm. Then we can interpolate, as shown in Figure 8, to obtain = 8 sq cm, which agrees with what we found earlier by interpolating with Graph I. Question 14 in the Comprehension Questions asks the children to use Graph II again, this time to find the area of a circle with radius 5 cm. Exactly the same procedure as in Question 13 will give = 79 sq cm. Question 14 is designed to drive home the point that extrapolation is more accurate with Graph II than with Graph I. In Question 15 the children compare their answers to Question 7 (Graph I) and Question 14 (Graph II) with an exact answer provided by the teacher, 78.5 sq cm. To find the percent difference between their answers and the correct answer the children should subtract the smaller of these numbers from the larger, divide that difference by the correct answer, and multiply by 1. For example, suppose a child 3

5 Counting Out πr, page 5 of 7 finds from Graph I that the area of a circle with radius 5 cm is 98 sq cm. The percent difference, then, is 98 sq cm 78.5 sq cm 78.5 sq cm 1 5% This is a pretty large error. The same child, however, might have found from Graph II that the area of a circle with radius 5 cm is 79 sq cm. In this case the percent error is 79 sq cm 78.5 sq cm 78.5 sq cm 1 =.6% This is much better, so we conclude in Question 16 that Graph II is much better for extrapolating than Graph I but that both are about the same for interpolation. Similarly we can solve for R if we know, but we must extract a square root. This used to be painful in the days before pocket calculators, but now it is easy. For example, Question 17 asks for the radii of circles with areas of 7 sq cm and 1 sq cm. From Figure 8 we see that R is sq cm when = 7 sq cm. Taking the square root of sq cm (with a calculator) we find R = 4.7 cm. Similarly, when = 1 sq cm, then R is 3 sq cm and so R = 5.7 cm. From the fact that the curve for vs. R is a straight line through (,) we know that /R is a constant ratio. The students are asked to find this constant ratio in Question 18. But what is the value of the constant? If you let the children compare the ratios they each get from their best fit lines you will see that the values are close to each other. Using our data we have /R = 3.1 sq cm/1 sq cm. If we had super accurate measuring instruments we would discover that every circle, no matter what, lies on the same line and that /R is a very famous number, pi. Pi is the name of a letter in the Greek alphabet that looks like this: π. So we have the ratio for all circles π = R In Question 19 the children are asked to find the percentage difference between their value of π and a precise value. The fact that the ratio /R and the ratio C/D (Circumference/Diameter) are both equal to the exact same number, π = , is an amazing fact that was proved by the Greek mathematician rchimedes over years ago. When you think about it, why should these ratios, /R and C/D, be related? Proving they are the same was only one of rchimedes many accomplishments. Solving circle problems is not trivial. For example in Question we ask them to find the radius of a rocket nozzle with an area 3 sq cm. We know the ratio /R is a constant, namely π. Hence we must solve 3 sq cm R R = R = sq cm 3.14 = sq cm But we are interested in R not R, so we must extract a square root. We find R = 3.9 cm. We can also solve for area if we know the radius. Thus in Question 1 we ask the children to find the area of a circle with radius 5 cm. In Question 1a we ask the children to use 3 to estimate the area and not to use their calculators. So they have to use / R = 3. But our ratio requires R, and we only know R. Squaring is easy, however, and we quickly find R = (5cm) = 65 sq cm. So we have

6 Counting Out πr, page 6 of 7 R = 3 65 sq cm = 3 = 3 65 sq cm = 1875 sq cm In Question 1b we use the constant ratio /R = to solve for the unknown. Repeating the above steps we have: R = sq cm = Thus = (65 sq cm) = sq cm, where we have written to three decimal places. What if we make our measurements using different units? Will we still get the same ratio? The short answer is yes. We found the ratio /R to have units (sq cm)/(sq cm), but these are the same and so they cancel. So really /R is a pure number. The units will always cancel as long as R and are measured in compatible units, for example, in inches and square inches, or miles and square miles, or meters and square meters. Question in the Comprehension Questions uses this fact. When computing the area of a circle of radius 5 m we can use the same /R ratio as before. So, noting that (5 m) = 65 sq m, we solve: R = sq m = 3.14 = 3.14 (65 sq m) = sq m t this point, we may introduce the well-known formula, = πr. This is the answer to Question 3. So we see that what seemed so simple, the formula = πr, is really not so simple at all, but is actually at the end of a fairly long road of mathematical analysis. We feel that it is far better to develop the formula for the area of a circle by having your students measure area and radius directly, squaring the radius, and plotting vs. R than to have the children memorize it. They will be far more likely to understand and retain the formula if they do. Thinking, according to Dewey, is the intentional endeavor to discover specific connections between something which we do and the consequences which result, so that the two become continuous (in Democracy and Education by John Dewey, 1916). Consequently, the students must be active and productive instead of passive and reproductive learners. Using the equation, the children are asked to find the area of a circle of D = 1 cm. Finding R = 5 cm, = πr = 3.14 (5) sq cm = 78.5 sq cm In Question 4 we conclude with a multistep logic problem. What will it cost to seed a circular area of R = 4 m if a pound of grass seed cost $.5 and covers an area of 1 sq cm. First, we find the area to be covered: = πr = 3.14 (4 m) = 4 sq m Second, we see how many pounds of grass seed we need: N pounds = 1 pound 1 sq m = N 4 sq m N =.4 pounds

7 Counting Out πr, page 7 of 7 Third, we determine the cost Summary $ N = $. 1 pound = $.4 $ = $156. Counting πr is a very important experiment which starts by determining to R independently, plots the relationship, and studies how to interpolate and extrapolate a nonlinear curve. We then learn how to straighten that curve and make more accurate predictions. The children have to learn to take square roots with their calculators and do simple proportions. The experiment is very formal operational but comes down to earth as the children derive π from their data. That s an awful lot of stuff for a week, but with their strong foundation in TIMS the children should be able to handle it. Materials per Team 3 cans or tops small, medium, and large 1 ruler 1 centimeter graph paper -centimeter graph paper