Volume and Area Ratios with Geogebra Libuse Samkova
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10 North American GeoGebra Journal (ISSN: 2162-3856) Vol. 2, No. 1, 2013 Volume and area ratios with GeoGebra Libuse Samkova Abstract: This article was motivated by student work in a biological laboratory where a question 1 was asked regarding the liquid level corresponding to 5 of the volume of a laboratory flask. In this article, we use GeoGebra to explore such questions, analyzing volume and area ratios for containers of various shapes. We present various dynamic GeoGebra models including one of the experiment which initiated the article. Keywords: GeoGebra, modeling, algebraic proof, physics 1. INTRODUCTION moment, the volume of the liquid in the cylindrical por- 1 tion of the flask). Note that 5 of the ball volume does not 1 This article was motivated by student work in a biological correspond to 5 of the ball height. laboratory during the demonstration of a candle experi- The volume of a ball with radius r is given by a formula ment similar to that suggested illustrated in Fig. 1. 4 3 3 pr . The liquid in the ball reaches an unknown height h, and its volume equals the volume of the shaded re- gion shown in Fig. 2 (which we refer to as a spherical cap). Fig 1: An illustration of a candle experiment (Gallástegui 2011). The student experimenter ended the presentation by stat- Fig 2: Front views of the liquid in the ball (left), and of ing that “the liquid in the flask occupies 20% of the flask the spherical cap (right). volume,” and the audience nodded in agreement. Is it re- 1 A basic volume formula for a spherical cap with height h ally so easy to determine 5 of the flask volume? Con- sider the following steps required to complete the calcula- and base radius a is given by tion. 1 2 2 Vcap = ph(3a + h ) (Weisstein, 2012a) (1) 1 6 • We have to determine 5 of the volume of the flask whose shape consists of a ball and a cylinder linked From Pythagorean theorem we have together. a2 = r2 − (r − h)2 = r2 − r2 + 2rh − h2 = 2rh − h2 (2) • We have to take into account the volume of the sub- which implies that merged part of the candle. 1 V = ph(6rh − 3h2 + h2) cap 6 2. BALL CASE 1 = ph(6rh − 2h2) 6 Let’s start with a simplified version of the problem by cal- 1 2 1 = ph (3r − h) (Weisstein, 2012a) (3) culating 5 of the volume of a ball (and ignoring, for the 3 Libuse Samkova is an associate professor in the Department For clarity in subsequent calculations, we express the height of Mathematics at the University of South Bohemia, Czech Re- of the liquid as a k-multiple of the height of the ball, i.e., public (e-mail: [email protected]). h = k · 2r. Then, This study was funded by a European FP7 project: The FI- BONACCI Project - Large scale dissemination of inquiry based 1 4 V = p(k · 2r)2(3r − k · 2r) = pk2r3(3 − 2k) (4) science and mathematics education (No. 244684). cap 3 3 North American GeoGebra Journal (ISSN: 2162-3856) Vol. 2, No. 1, 2013 11 Next, we solve the following equation to find a liquid level 1 corresponding to 5 of the ball volume. 1 V = ·V (5) cap 5 ball 4 1 4 pk2r3(3 − 2k) = · pr3 (6) 3 5 3 1 k2(3 − 2k) = (7) 5 1 2k3 − 3k2 + = 0 (8) 5 1 Note that the ratio 5 appears as a constant term in the re- sulting cubic equation, as shown in (8). This equation does not have an easy manual solution, but Fig 4: Graphical solution using sliders. we can solve it graphically with help of GeoGebra. We 3 2 1 generate the graph of a function f (x) = 2x − 3x + 5 and construct the x-coordinate of the intersection of the graph and the x-axis. This coordinate is the required solution to the cubic equation (8). 1 Fig 5: Dynamic model illustrating a solution for 5 of ball volume. The model is based on the graphical solution from Fig. 4. First we create a circle with radius r and sliders for m and n. For given m and n, we define height ratio k as 3 2 m the solution to the equation 2x − 3x + n = 0. Then we create a circumcular arc with a height k · 2r, and adjust its opacity to 80 percent to simulate liquid. Fig 3: Graphical solution to 2x3 − 3x2 + 1 = 0. With this dynamic model we can investigate various vol- 5 1 ume ratios. Note that the liquid level for 10 of the volume 1 As shown in Fig. 3, the solution is approximately 0.29. reaches approximately 5 of the ball height. This corre- What does it mean? If we have a ball with diameter 10 sponds to the numerical solution in Fig. 4. cm, then the height of the ball is also 10 cm, and the liquid 1 occupying 5 of the ball volume reaches a height of 0:29 · 10 cm ≈ 2:9 cm. 3. BACK TO THE EXPERIMENT 2.1. Generalizing the ball case 3.1. Building the flask with liquid Next, we generalize the problem for arbitrary ratio m of n With some work, a dynamic sketch modeling the liquid ball volume. The corresponding k now satisfies the equa- in the flask from the original candle experiment can be tion 2k3 −3k2 + m = 0. As shown in Fig. 4, we use sliders n constructed. We do so first without the candle, answer- m and n to generate function f (x) = 2x3 − 3x2 + m , then n ing the question “What does 1 of the flask volume look - as before - we approximate the solution using the inter- 5 like?” Again, the model is constructed dynamically, with section tool. sliders used to modify flask proportions (r controls the ra- Alternatively, we can create a dynamic sketch modeling dius of the cylinder; r, the radius of the ball; h, the height the liquid in the ball, as shown in Fig. 5. (Editor’s note: A of the cylinder). These features are illustrated in Fig. 6. copy of the sketch, Fig5.ggb, is provided alongside the pdf (Editor’s note: A copy of the sketch, Fig6.ggb, is provided version of this paper.) alongside the pdf version of this paper.) 12 North American GeoGebra Journal (ISSN: 2162-3856) Vol. 2, No. 1, 2013 Fig 6: A dynamic model of the flask volume Fig 7: A dynamic model to the candle experiment. 3.2. Adding the candle Building on the previous sketch, we create a model of the experiment with a candle inside the flask. The con- The volume of a cone with base radius r and height h is 1 2 struction requires knowledge of the Archimedes’ principle given by a formula 3 pr h. The liquid in the cone reaches of buoyancy which states that the buoyant force exerted an unknown height hc, and its volume equals the volume on a solid floating in the liquid is equal to the weight of of a conical frustum with height hc, and base radius r. The the volume of the liquid which is displaced by the solid, top radius of the conical frustum is not obvious, but it can i.e., be derived from hc, r, and h. Let’s express hc as a k- multiple of the height of the cone, i.e., hc = k · h. Then Vsolid · g · rsolid = Vsubmerged part · g · rliquid (9) the similarity of triangles shown in Fig. 8 gives the top rsolid radius of the conical frustum equal r(1 − k). Vsubmerged part = Vsolid · (10) rliquid see (Weisstein 2012c and Burley, et al. 1997). The liquid in our experiment is the water, with density 1000 kg/m3. The density of the candle is determined by an auxiliary experiment as 950 kg/m3. This means that 950 V = V · (11) submerged part candle 1000 Vsubmerged part = 0:95 ·Vcandle (12) Note that (12) implies the following: Fig 8: A front view of the liquid in the cone (left), and the • If 95% of the candle height is less than the depth of two similar triangles (middle and right). the water, then the candle floats in the water, and the height of the submerged part of the candle equals the 95% of the candle height. Weisstein (2012b) notes that the volume of a conical frus- tum can be expressed as • Otherwise the candle stands on the bottom. We use sliders to model changes in candle proportions (c h 1 for the height of the candle, c for base radius). See the V = phr2 + r r + r2 (13) r f rust 3 base base top top final illustration in Fig. 7. (Editor’s note: A copy of the sketch, Fig7.ggb, is provided alongside the pdf version of this paper.) Thus, in our particular case: 1 2 2 2 2 4. FURTHER MODELS OF VOLUME AND AREA Vf rust = pkh r + r (1 − k) + r (1 − k) RATIOS 3 1 = pr2kh(3 − 3k + k2) (14) 3 A cone provides an interesting, and more challenging, con- text for exploring volume ratios. Unlike the ball, a cone does not satisfy the rule that half of the volume corre- m sponds to half of the height. We seek a liquid level corresponding to n of the cone vol- North American GeoGebra Journal (ISSN: 2162-3856) Vol. 2, No. 1, 2013 13 ume. We accomplish this in the following manner. REFERENCES m V = ·V (15) f rust n cone Gallástegui, J.