5 6 7 Middle olume Length/olume vs. Diameter, Investigation page 1 of olume vs. Diameter Teacher Lab Discussion Overview Figure 1 In this experiment we investigate the relationship between the diameter of a sphere and its volume. The children have to use all their length and volume measuring skills as they work with spheres of various sizes. To do the experiment, you need spheres of three different diameters. We like steel ones that come in.5 inch, 1 inch, and 1.5 inch sizes, though of course we measure them in centimeters. We also use ", ", and 1" Lucite spheres in a second part of the experiment. We want spheres of two different materials to determine whether the material has any effect on the diameter vs. volume curve. Picture, Data Table, and Graph As usual, identifying variables is a key step. In this experiment it is somewhat arbitrary which variable is designated the manipulated and which the responding. We have chosen to call the diameter, D, the manipulated variable (Question 1), and the volume,, the responding variable (Question ). There are several reasons for this: it may be easier to measure the diameter than the volume; length is a more fundamental variable than volume; and we want a graph with diameter on the horizontal axis. The most important controlled variable is the shape of the objects: they are all spheres. (In the Comprehension Questions the students are asked to predict the volume of spheres of different materials but the same diameter; in this situation the diameter is the controlled variable, the material is the manipulated variable, and the volume is again the responding variable.) As usual the children have to draw a labeled picture of the experiment. They should identify the two variables and also how they measured them. As we discuss below, the measuring process is not trivial. Our picture is shown in Figure 1. When doing this experiment the students are told to be as accurate as possible. Their first challenge is to measure the diameter of a sphere. The diameter measurement is a little tricky since you can t get the sphere to lie flat on the meterstick. The children might place the sphere on the meterstick and try to eyeball the diameter; see Figure for a top view. This does not give a very accurate measurement. Whenever we measure a small object there is a difficulty getting an accurate measurement. In the TIMS Tutor. The Concept Figure 5 1 15 5 cm
olume vs. Diameter, page of Figure Figure 4 5 1 15 5 cm 5 1 15 5 cm oflength we explain a general method for measuring the length of small things. To summarize, we measure the length of a small object by taking a sufficiently large number of the objects, lining them up, measuring the length of the group, and finally dividing by the number of objects. For spheres this is a particular problem since they do not stay in a nice line. To solve this problem we line up two metersticks parallel to each other with the desired number of spheres between. The smaller the sphere, the more will be needed to achieve acceptable accuracy. The metersticks make a little track so we can make a small train of spheres. Now we put one block at either end of the train and measure the distance between the blocks to get the length of the train; see Figure. Dividing the length of the train by the number of spheres in the train gives us the diameter of one sphere. Because this experiment stresses accuracy, we suggest using this technique to find D for all sphere sizes. The children s answer to Question should reflect the above discussion. the bottom of the cylinder. Now use the 1 cc graduated cylinder to measure out 1 cc of water and add it to the 5 cc graduated cylinder, as shown in Figure 5. The students can now read off the combined volume of water and spheres in the cylinder and subtract 1 cc to find the volume of the spheres. We have a similar problem in measuring for small spheres as we do for measuring D. In order to be accurate one must use several spheres at once and then divide the total volume of the, say, 5 spheres, by 5 to find the volume of one. Here again, the smaller the sphere, the more will be necessary for accurate results. It is essential that the water level cross at least one graduation on the cylinder. One or two spheres are enough for the Figure 5 The next challenge is to find the volumes of the spheres. Carelessly dropping a steel ball into glass graduated cylinders has knocked out the bottom of many a cylinder. You also have to be careful not to splash water out of the graduated cylinder when you add the spheres. This is almost impossible for the large sphere, so we have to be a bit clever. First we carefully roll the spheres into the 5 cc graduated cylinder, as shown in Figure 4. Tilting the cylinder will prevent the spheres from damaging
olume vs. Diameter, page of large sphere, but ten or more of the smallest spheres are recommended. This requires that the children share materials. Thus children should bring these ideas into their answer for Question 4. The data table is given in Figure 6. A quick look at the data reveals the obvious but important fact that as D increases so does. We say there is a positive correlation between the variables. In fact, as D increases, increases faster and faster. Exploring the functional relationship between and D is the heart of this experiment. Indeed, one thing the children can immediately see is that D and are not proportional. As the children should explain in Question 5, If I increase D from 1. cm to.5 cm, which is almost double, the volume increases by! So D and are not proportional. Another way to see the functional relationship is to plot the data. Our graph is shown in Figure 7. We ask in Question 6 if the curve should go through (,). As always this special data point is worth some thought. To answer the question we must ask, what happens as D goes to? The sphere obviously shrinks in all directions and so must go to, too. So, the curve should go through (,). Plotting this free data point should help us answer Question 7. That plus the answer to Question 5 should be enough for the student to tell you that the curve is not a straight line. When the data points do not lie on straight line we have to decide what to do about drawing a curve. Figure 6 Table I: Steel Spheres Size of sphere D Diameter in cm olume in cc Small 1. cm 1.1 cc Medium.5 cm. cc Large. cm cc Many children (and adults) will just connect the dots with straight line segments. This method will not produce accurate interpolations or extrapolations. Mother Nature does not arrange her variables so they are related by straight line segments. Rather, she relates variables so that the data lies on nice smooth curves. Another wrong approach to plotting D vs. is to fit a straight line to the data, even though the fit is very bad. In one sense this is better than connecting the dots because it is at least a smooth curve, but the line is almost useless for interpolation and even worse for extrapolation. We have discussed this problem in the experiment Area vs. Perimeter. So the children should draw the best smooth curve that passes through or near the data points. See Figure 7 for our D versus curve. Now interpolation will be easy, extrapolation difficult, and proportional reasoning impossible. The children should learn this. Figure 7 in cubic cm 44 4 6 4 16 1 4 Steel Lucite 1 4 5 6 7 D in cm
olume vs. Diameter, page 4 of Comprehension Questions We begin with a very fundamental question. We ask in Question what would happen if lucite spheres were used instead of steel. We hope the children will answer, Nothing would be different. The data points would be on the same curve. Using the TIMS Lucite spheres, ", ", and 1", we ask the children to check their prediction. It s basically the same experiment they have just completed only now they should do it on their own. No help! The data table is shown in Figure and the results are plotted as open circles in Figure 7. Just as we predicted they lie precisely on the curve for steel. This point is driven home in the next three questions. In Question 9 the children find the volume of a cm steel sphere. By using interpolation the volume is about 4.5 cubic cm. What about a cm lead sphere (Question 1) and a cm wooden sphere (Question 11)? In both cases the answer is the same as Question 9, equals about 4.5 cc. Question 1 focuses on a common misconception, namely that changing the shape of an object changes its volume. The volume of a piece of clay is independent of its shape. Reshaping a clay sphere of diameter.5 cm into a worm or other shape will not change the volume, which will remain about cc. The only way to change the volume is by removing or adding clay. This idea is the focus of the TIMS experiment olume vs. Shape. In Question 1 we use the graph to find D for a sphere of = 15 cc. We can read this directly off the curve to get an answer of approximately.1 cm. Note that there is a range of acceptable answers; any answer within a few tenths of centimeters should be considered correct, depending on the curve as drawn. Such interpolation is not difficult even though the best fit curve is not a straight line. In Question 14 we ask the children to use their graph to find when D = 6 cm. Now we have a problem. While it is very easy to extrapolate a straight line curve, it is rather difficult to extrapolate a curvy curve. We just don t have any guidance as to how curvy to make the curve when we extrapolate. Your students should discover this by experience. Figure 9 shows two plausible extensions of the curve. Extension 1 yields = cc when D = 6 cm; extension yields Figure 9 1 1 11 14 96 Steel Spheres Extension Figure Table III: Lucite Spheres Size D in cm in cubic cm in cc 7 64 56 4 4 Extension 1 inch.95 cm.5 cc 4 4 inch 1.9 cm. cc 16 1 inch.5 cm. cc 1 4 5 6 7 D in cm
olume vs. Diameter, page 5 of = 1 cc. It is difficult to tell which of these answers is accurate, if either. The further we try to extrapolate the less sure we are confident in our answer. The exact value of is 11 cc. The children are asked to use that data to see how far off the prediction is compared to 11 cc. Using our upper guess of 15 cc, then Figure 1 D in cm 1. Table III D in cc. in cc 1 % diff. = 1 cc 11 cc 11 cc 16. 1 16.5. 15.6 54. Steel Spheres.7 while our lower guess was off by 14 % diff. = = 4 cc 11 cc 11 cc - 6 1 You might ask your students to compare their answers to Questions 1 and 14 by writing them on the board: this could lead to some very productive discussions of the relative difficulty of interpolation and extrapolation for curves and straight lines. Since the D vs. curve is not a straight line through (,) we know the ratio /D is not constant. The students are asked to verify this in Question 15. Since the ratio /D is not constant we cannot use it for solving for unknown volumes or diameters by proportional reasoning. If is not proportional to D, what is it proportional to? Because three dimensions are involved, we would not be remiss in suspecting that is proportional to D, that is, we suspect /D is constant. This means if we plot versus D, we should get a straight line. The children tackle this idea in Question 16 where they determine D, record D and, and plot vs D. The data table and graph are shown in Figure 1. The analysis parallels that in Counting Out πr. See the TIMS Tutor 9. Straightening Out the Curve for a general discussion of this process. The in cc 1 1 6 4 4 6 1 1 14 16 1 D in cc data should fall on a straight line, thus verifying our supposition (Question 16b). This straight line can be used for interpolation and extrapolation, though more work is involved than before. Question 17 requires that the volume of a sphere with diameter cm be found from the D vs. straight line. The first problem is that we are given D but our graph has D on the horizontal axis. Hence we must cube cm to find that D is 7 cc. Then, as shown in Figure 11, interpolation reveals is about 14 cc. Question 1 asks for an extrapolation to D = 6 cm, that is, D = 16 cc. Figure 11 shows is about 11 cc. Compare this to the answers above from Figure 9. It turns out one of our extensions was low and one was high. We did not bend up enough! This is typical of extending curved lines.
olume vs. Diameter, page 6 of Figure 11 16 Steel Spheres 14 1 in cc 1 6 4 4 6 1 1 14 16 1 D in cc If we know we can also find D using Figure 11, but again more work is necessary. Question 19 asks for the diameter of a sphere with volume 5 cc. Figure 11 shows how to find D which is about 1 cc. The problem is, we want D, not D. One way to find D given D is illustrated in Figure 1. The idea is to find the answer by educated guessing, using a calculator. The first guess is D = 1 cm. This is not a very good guess since we have D = 1 cc whereas we want D = 1 cc, but at least it is a start. Since 1 cc is much less than 1 cc, we double D to cm. This gives D = cc, which is better but still way too small. Eventually we overshoot 1 cc, so we backtrack. Continuing in this way we can find as many digits of accuracy as we like. Given our general level of precision, we are satisfied with digits of accuracy in D so that we stop when we find D = 4.6 cm. Another quicker though more mysterious way to find D is to extract the cube root of 1 cc by using the Y X key on a scientific calculator. Enter 1 Y X (.) = and you get 4.6 cm for D. Thus interpolation and extrapolation are possible with the graph of D vs., though compared to experiments like The Bouncing Ball and Circumference vs. Diameter we must work harder because of the two-step logic required. A further problem with extrapolation is that D quickly runs off the graph. For instance, since our last data Figure 1 D in cm 1 4 6 5 4.5 4.6 4.7 D in cc 1 64 51 16 15 91 97 14
olume vs. Diameter, page 7 of point is for D =. cm, we might reasonably like to extrapolate to spheres about twice as big, that is, to D = cm. When we cube cm, however, we have D = 51 cc, which is way off our graph. This motivates us to try proportional reasoning. Since the graph of D vs. is a straight line through the origin we know that the ratio /D is a constant. Question asks for this constant. The best fit line in Figure 1 goes through the origin and the point ( cc,1 cc), so the slope is 1 cc/ cc or. (You may notice that this number is very close to π/6.) Once we have the ratio we can solve problems with bigger numbers. Questions 1 and require this. Question 1 asks for the volume of a sphere with diameter 1 cm. Cubing 1 cm we have D = 1 cc. Using /D = we solve D = 1 1 cc = 1 = 5 cc To find the volume of the museum s globe (Question ), the student must realize that the shift in units to m from cm is easy. Noting that (4m) = 64, cu m, we have D = 1 64, cu m = 1 =, cu m Question 4 involves rephrasing the statement / D = : D = 1 = D This is only an approximation. The exact formula for in terms of D is Question asks for the diameter of a sphere with volume cc. We have = πd 6 Note that this is equivalent to the more familiar D = 1 cc D = 1 D = 4 cc Again we have the problem with having to find D knowing D. Either the method of Figure 1 or the Y X key on the calculator gives D = 15.9 cm. = 4πR Questions 5 and 6 are special challenges. The chocolate bars in Question 5 have a volume of (5 cm) ( cm) (1 cm) = 15, cc A ball 1 cm in diameter has a volume of roughly.5 cc. So, to find the number of chocolate balls we set up the ratio
olume vs. Diameter, page of Figure 1 Summary olume vs. Diameter is a wonderful experiment. The children get a chance to practice basic length and volume measuring skills while they learn to handle a nonlinear function. The mathematical techniques and insights they acquire will be invaluable in their further studies. Materials per Team 1 m N = 1 ball.5 cc = N 15, cc 1 ball N = 15, cc.5 cc =, balls several " steel spheres 1" steel spheres 1 1.5" steel spheres several " Lucite spheres several " Lucite spheres 1" Lucite spheres 1 cc graduated cylinder 5 cc graduated cylinder metersticks Hence about, balls can be made from each bar. (,647 is the answer if you use π/6 instead of in the formula for.) Question 6 involves reshaping a cube 1 meter on an edge into a sphere. That the sphere will have a diameter more than 1 m can be seen by thinking about the volume of a sphere of diameter 1 m: such a sphere will fit inside a cube 1 m on an edge with room to spare, as shown in Figure 1. We solve D = 1 1 cu m = 1 D D = cu m Once again, taking a cube root we find D = 1.6 m