Making Implicit Multivariable Calculus ...

Making Implicit Multivariable Calculus Representations Explicit: A Clinical Study Daniel McGee Executive Director Kentucky Center for Mathematics MEP-475, Northern Kentucky University Highland Heights, KY 41099 [email protected] Deborah Moore-Russo SUNY-Buffalo [email protected] Rafael Martinez-Planell UPR-Mayaguez [email protected]

Abstract Reviewing numerous textbooks, it was found that in both the differential and integral calculus, textbook authors commonly assume that: a) students can generalize associations between representations in 2D to associations between representations of the same mathematical concept in 3D on their own and b) explicit discussions of these representations are not necessary. For example, in the presentation of 3D derivatives, it is assumed that students will understand a 3D slope without it ever being explicitly presented. As preliminary interviews indicated that such assumptions may be erroneous, materials that explicitly use representations commonly omitted in textbook presentations of the differential and integral calculus of functions of two and three variables were created and then used in class. This paper discusses the classroom activities that were created to include these missing representations as well as the results when they are implemented in classroom instruction.

Introduction By the time that students study two-dimensional (2D) derivatives for the first time, students are expected to have seen and studied 2D slopes as a fundamental aspect of 2D linear relations and of 2D linear functions. An understanding of the geometry and algebra of 2D slopes was implicit to the study of derivatives in the six commonly used calculus textbooks we reviewed [1, 3-7]. In these same texts, the nature of the discussion of 3D derivatives indicates that an understanding of slopes in 3D is also implicit to the study of derivatives in 3D. However, it is quite common for textbooks [1, 3-7] to introduce 3D derivatives without ever presenting a discussion or even a single representation, such as those shown in Table 1, for directional slopes on a plane or the slope from one point to another in a 3D coordinate system. All of these reviewed texts discussed planes before presenting derivatives. However, they were discussed exclusively in the context of vectors, dot products and cross products, which may represent a lost opportunity to extend the concepts of slopes and linearity from 2D to 3D. To illustrate this, the diagram of a plane in Table 1 and the activity presented in Appendix B introduce the terms ∆𝑥, ∆𝑦, 𝑚𝑥 , 𝑚𝑦 , ∆𝑧𝑥 , ∆𝑧𝑦 and ∆𝑧 in the context of movement and slopes on a plane. The following are some ways that using these representations has significantly impacted topics of differential calculus. 1. Understanding 𝑚𝑥 as the slope in the x direction of a plane provides context for 𝜕𝑓 as both the slope of a tangent line to a surface in the x direction and as 𝑚𝑥 on 𝑑𝑥 𝜕𝑓

a tangent plane. The same is true with 𝑚𝑦 and 𝑑𝑦

2. Understanding (𝑧 − 𝑧0 ) = 𝑚𝑥 (𝑥 − 𝑥0 ) + 𝑚𝑦 (𝑦 − 𝑦0 ) as a plane with slopes 𝑚𝑥 and 𝑚𝑦 that passes through the point (𝑥0 , 𝑦0 , 𝑧0 ) lends clarity to (𝑧 − 𝑧0 ) = 𝑓𝑥 (𝑥0 , 𝑦0 )(𝑥 − 𝑥0 ) + 𝑓𝑦 (𝑥0 , 𝑦0 )(𝑦 − 𝑦0 ) as the formula for the tangent plane to a surface at the point (𝑥0 , 𝑦0 , 𝑧0 ) 3. Generalizing from Appendix B, the vertical rise associated with movement on a plane over over the vector < ∆𝑥, ∆𝑦 > in the xy plane can be expressed by ∆𝑧 = 𝑚𝑥 ∆𝑥 + 𝑚𝑦 ∆𝑦. This lends clarity to the differential as the vertical rise on the tangent plane given by 𝑑𝑓(𝑥0 , 𝑦0 ) = 𝑓𝑥 (𝑥0 , 𝑦0 )𝑑𝑥 + 𝑓𝑦 (𝑥0 , 𝑦0 )𝑑𝑦. 4. Generalizing from Appendix B, the slope on a plane can be expressed as 𝑟𝑖𝑠𝑒 𝑚𝑥 𝑢1 + 𝑚𝑦 𝑢2 𝑚 = = 𝑟𝑢𝑛 �𝑢1 2 + 𝑢2 2 If < 𝑢1 , 𝑢2 > is a unit vector and the plane is a tangent plane then 𝑟𝑖𝑠𝑒 𝑓𝑥 (𝑥0 , 𝑦0 )𝑢1 + 𝑓𝑦 (𝑥0 , 𝑦0 )𝑢2 𝑢1 𝐷�𝑢1 � 𝑓(𝑥0 , 𝑦0 ) = = = ∇𝑓(𝑥0 , 𝑦0 ) ∙ �𝑢 � 2 𝑟𝑢𝑛 𝑢2 �𝑢1 2 + 𝑢2 2

Slope from one point to another

Directional slopes on a plane

Table 1: Representations associated with slopes in 3D

With integral topics of Multivariable Calculus, there seems to be a tendency to assume that representations and discussions that were necessary in 2D are no longer necessary in 3D. To trace the path from a numerical Riemann sum approximating the area under a curve to a definite integral representing the precise area, every textbook reviewed [1, 37] presented the 2D geometric, numerical and algebraic representations that are shown in Table 2. Before presenting double integrals, texts sometimes explored numerical approximations for a volume under a surface. However, not a single text continued to present representations 1, 2 and 3 in Table 2 throughout rectangular, polar, cubic, cylindrical and spherical coordinates. From either rectangular or polar coordinates onward, all of these texts begin their discussions with the 3D equivalent of representation 4 in Table 2: An expression such as ∑𝑛𝑖=1 ∑𝑚 𝑗=1 𝑓(𝑟, 𝜃)(𝑟)𝑑𝑟𝑑𝜃 or a similar expression for rectangular, cubic, cylindrical or spherical. However, our experience has shown that continuing to present representation 1, 2 and 3 in Table 2 may still be necessary throughout all integral topics of Multivariable Calculus and that presentations that begin with representation 4 in Table 2 may be insufficient for students to comprehend the associated integral topic. The omission of these representations associated with derivatives and integrals is quite ubiquitous and correspondingly affects a vast number of Multivariable Calculus students. This article first presents some activities that have been created to include these representations in Multivariable Calculus discussions. It then explores the possibility that the following curricular modifications may be helpful to Multivariable Calculus students: • Explicit discussion of 3D slopes accompanied by their appropriate geometric representations and • Inclusion of a numerical representation and an expanded sum representation before proceeding through integral topics associated with rectangular, polar, cubic, cylindrical and polar coordinates. 1. Geometric representation for an approximation of the area under y = x2 between x = 1 and x = 5 using four rectangles and the midpoint rule for the height of each rectangle. 2. Numerical representation for the approximation of the area. 3. Algebraic expanded sum representation for the approximation of the area. 4. Algebraic sum with sigma representation for the approximation of the area. 5. Algebraic sum with sigma representation for the precise value of the area. 6. Algebraic integral representation for the precise value of the area.

(1.5)2*1 + (2.5)2*1+(3.5)2*1+(4.5)2*1 𝑥1 2 ∆𝑥 + 𝑥2 2 ∆𝑥 + 𝑥3 2 ∆𝑥 + 𝑥4 2 ∆𝑥 4

� 𝑥𝑖 2 ∆𝑥 𝑖=1

𝑛

lim �� 𝑥𝑖 2 ∆𝑥�

𝑛→∞

𝑖=1 5

� 𝑥 2 𝑑𝑥 1

Table 2: 2D representation associated with a simple area

Classroom Activities The classroom activities presented in this study were designed to be used in small groups with a physical manipulative (referred to as a “kit” in class and in the appendices) that previously had been found to be useful for geometric representations in a 3D coordinate system [2]. However, these activities can easily be adapted for use with any physical or computer system for geometric representations in 3D. Some adaptions in Spanish can be found at http://pegasus.uprm.edu/rmplanell. One slope activity and one integral activity are presented in this section. Appendix A provides an additional slope activity while Appendix B provides an additional integral activity. A complete set of the slope and integral activities from this study can be found at http://kymath.org/calc3. Sample Slope Activity: Find the slope of the segment from (1, 1, 0) to (4, 5, 6). Find the run: Solution: The run is the distance from (1, 1, 0) and (4, 5, 0) in the xy plane (see diagram to the right). Using the Pythagorean Theorem, it can be determined that the run is 5. Find the rise: Solution: The rise is the distance from (4, 5, 0) to (4, 5, 6) (see diagram to the right) which is a vertical pole of height 6.

Find the slope: Solution: The slope can now be visualized by placing the right triangle with vertices (1, 1, 0), (4, 5, 0), and (4, 5, 6) along with its associated rise and run in three dimensional space (see diagram to the right). 𝑟𝑖𝑠𝑒 6 𝑚= = 𝑟𝑢𝑛 5

Sample Integral Activity: Use the following steps to find the volume below the surface 𝑧=

𝑥2 2

+ 𝑦 2 + 4 and above the region 0 ≤ 𝑥 ≤ 4, 0 ≤ 𝑦 ≤ 4 on the xy plane.

a. Draw the region on the xy plane and use a computer, calculator or physical manipulative to visualize the surface over the region. Solution shown in the figure to the right. b. If there are two divisions in x, two divisions in y and the volume is to be approximated using the middle value for x and y in each division, find x1 , x2 , y1 , y2 , ∆x , and ∆y : Solution: x1 = 1, x2 = 3, y1 = 1, y2 = 3, ∆x = 2 and ∆y = 2 c. For each of the four divisions, indicate the point and associated height that will be used to approximate the volume for that

division and identify the four cubes with which we will approximate the volume: Solution shown in figure to the right. d. Use the values obtained in part c to fill in the following table with numerical values. Note: In sections d and e, the table is initially empty and students fill it in. Division Length Width Height Volume 1 2 2 5.5 2 x 2 x 5.5 2 2 2 9.5 2 x 2 x 9.5 3 2 2 13.5 2 x 2 x 13.5 4 2 2 17.5 2 x 2 x 17.5 e. Fill in the same table below using x1 , x2 , y1 , y2 , ∆x , and ∆y, instead of numerical values. The divisions should not change between this and the previous table. Division Length Width Height Volume 2 2 2 1 ∆x 0.5x1 + y1 + 4 (0.5x1 + y12 + 4)∆x∆y ∆y 2 ∆y ∆x 0.5x22 + y12 + 4 (0.5x22 + y12 + 4)∆x∆y 3 ∆y ∆x 0.5x12 + y22 + 4 (0.5x12 + y22 + 4)∆x∆y 2 2 4 ∆y ∆x 0.5x2 + y2 + 4 (0.5x22 + y22 + 4)∆x∆y f. Use the two preceding tables to represent the same approximation for the volume (i) numerically, (ii) symbolically in expanded form and (iii) symbolically in sigma form. (2)(2)(9.5) (2)(2)(13.5) volume ≈ (2)(2)(5.5) + + 2 𝑥 𝑥1 2 2 + 𝑦1 2 + 4� ∆𝑥∆𝑦 + � + 𝑦1 2 + 4� ∆𝑥∆𝑦� volume ≈ �� 2 2 𝑥1 2 𝑥2 2 + �� + 𝑦2 2 + 4� ∆𝑥∆𝑦 + � + 𝑦2 2 + 4� ∆𝑥∆𝑦� 2 2

+ (2)(2)(17.5)

2 𝑥𝑖 2 𝑥𝑖 2 + 𝑦1 2 + 4� ∆𝑥∆𝑦 + � � + 𝑦2 2 + 4� ∆𝑥∆𝑦 2 2 𝑖=1 𝑖=1 2 2 𝑥𝑖 2 volume ≈ � � � + 𝑦𝑗 2 + 4� ∆𝑥∆𝑦 2 𝑗=1 𝑖=1

volume ≈ �

2

�

g. Generalize the approximation in part f to find expressions for the precise value of the volume. 4 4 𝟐 𝑛 𝑚 𝑥𝑖 2 𝒙 2 𝑣𝑜𝑙𝑢𝑚𝑒 = lim lim � � � + 𝑦𝑗 + 4� ∆𝑥∆𝑦 = � � + 𝒚𝟐 + 𝟒𝑑𝑥𝑑𝑦 ∆𝑥→0 ∆𝑦→0 𝟐 𝑖=1 𝑗=1 2 0 0

Methodology In the spring of 2012, a random section of 36 students enrolled in a Multivariable Calculus course was selected to use experimental materials associated with topics of integration. In the fall of 2012 another section of 36 students was selected to use experimental materials associated with 3D slopes. For comparison, the control sections selected were considered in light of all available measures, and no significant difference between the two groups was found. Also, the professors teaching the control and experimental sections each had a minimum of 12 years of experience teaching the course and are consistently rated in the top quarter of student evaluations of their instruction in this course. The results of the study were based on interviews with and common exam questions administered to students in both the experimental and control groups. It was found that a more thorough understanding of slope and integral representations allowed for more rapid discussions of related topics. Hence, the extra time dedicated to

group activities in the experimental sections was made up in this way and the traditional classroom syllabus was covered by both the experimental and control groups. The assignments were the same for the two groups, although there were supplemental problems for the experimental group associated with the experimental materials. Results were gathered based on: • Classroom observations made by a separate observer (not the professor) placed in the experimental class to observe the small group experimental activities. • Interviews conducted with students from both the experimental and control groups. 3 A students and 3 C students were interviewed in both the control and experimental groups to assess their mastery of 3D differential topics. 4 A students and 2 C students from the control group and 2 A students 1 B student and 2 C students were interviewed from the experimental group to assess their mastery of 3D integral topics. An “A student” refers to a student with an average over 89% on classroom evaluations. A “B Student” refers to a student with an average between 80% and 89% and a “C student” refers to a student with an average between 70% and 79%. • Common exam questions on traditional Multivariable Calculus topics were given to both the experimental and the control group students. Results Classroom Observations Perhaps the most notable observation regarding slopes in 3D was that when asked to find the slope of the segment from (1,2,1) to (3,2,5), every small group initially relapsed to ∆𝑦 𝑚 = ∆𝑥 = 0 even though the in-class presentation had emphasized that this formula did not work in three dimensions. As the slope was clearly not equal to zero, some groups were able to self-correct. Others self-corrected after some gentle scaffolding by the instructor through exploratory questions. However, two groups needed to be directly informed that the slope was not zero before they were able to get beyond this formula. The activity that the students seemed to enjoy the most (and gain the most from) involved placing a plane in 3D with intercept (0,0,2), mx =1 and my =2. The concepts associated with slopes seemed to “fall in place” for the students during the tactile act of tilting a plane to have different slopes in different directions. As with the sample integral activity and the activity shown in Appendix B, all of the integral activities involved: a. Finding a numerical approximation, b. Expressing the numerical approximation symbolically without sigma notation and c. Expressing the numerical approximation symbolically with sigma notation. It was noteworthy that most students seemed to initially disassociate numerical Riemann sums from the same Riemann sum expressed in symbolic form. The repetition of steps a, b and c throughout rectangular, polar, cubic, cylindrical and spherical coordinates appeared to be very productive in reinforcing that steps a, b and c can generate different representations for the same approximation. The act of physically identifying each division and representing its information both numerically and symbolically appeared very helpful in reinforcing the relationship between a symbolic Riemann Sum in sigma form and its associated meaning in a real world situation. It was also found during integral activities that students frequently struggled with the geometry associated with volumes in different coordinate systems. Identifying each division and then generating numerical approximations for each division’s volume or mass in polar (𝑓�𝑟𝑖 , 𝜃𝑗 �(𝑟𝑖 )∆𝑟∆𝜃), cylindrical (𝑓�𝑟𝑖 , 𝜃𝑗 , 𝑧𝑘 �(𝑟𝑖 )∆𝑧∆𝑟∆𝜃) and spherical (𝑓�𝜌𝑖 , 𝜃𝑗 , ∅𝑘 �sin(∅𝑘 )(𝜌𝑖 2 )∆𝜌∆∅∆𝜃) coordinates

was also very helpful in associating meaning with integral expressions. In the forthcoming sections, it will be seen that for the most part the experimental group significantly outperformed the control group in both interviews and common exam questions. The classroom observations provide considerable insight as to why this occurred as: • The experimental group appeared to gain the ability to spontaneously generate representations associated with 3D slopes in order to make sense of problems associated with 3D derivatives. Without this ability, it was found that the control group would often struggle. • The experimental students seemed to gain the ability to fluidly move between integrals, sigma approximations and an associated set of divisions where an analysis of each division could provide insight into the meaning of the integral. The control group seemed far less adept both at correctly associating an integral with its associated divisions and at analyzing the information contained in each division. Interviews with the Experimental and Control Groups The question-by-question results and the common tasks presented to select students from both the experimental and control groups can be found in Appendices C and D. These files also provide question-by-question results for the interviews. Table 3 presents a summary of the results of these interviews. Nature of Interview Tasks 3D Slopes (Seen only by the experimental group) Standard 3D Derivatives (Seen by both groups) 3D Integral Representations (Seen only by the experimental group) Standard 3D Integral Representations (Seen by both groups) Table 3:

Control Group 4% (n = 6) 23% (n = 6) 15% (n = 6) 29% (n = 6)

Experimental Group 59% (n = 6) 58% (n = 6) 78% (n = 5) 80% (n = 5)

Percentage of tasks performed correctly in interviews

While students in the experimental group were quite successful finding the slope between two points in 3D, three students in the control group (2 C students and 1 A student) appeared to have no concept of a slope in 3D. When prodded, they all knew that a partial derivative in 3D represented a slope. However, both the algebra and geometry of a 3D slope seemed to “draw a blank” for them. Their facial expressions and their comments “I’m used to slopes in 2D,” “I have never seen this before,” and “Formulas are a lot easier” reflected confusion not only with the procedures of how to solve the problem but with respect to the very essence of the task. The remaining three students in the control group (2 A students and 1 C student) all struggled; however, when prodded, all three drew a line in the air that appeared consistent with the slope between (1,1,2) and (3,1,6). So, even if they could not find a value, they seemed to have some concept of the slope that was placed before them. Only one A student in the control group was able to obtain the slope between the points (1,1,2) and (3,1,6). This student was unable to find the slope during the interview but after working with the rest of the questions suddenly seemed to have figured it out as he was leaving the interview. When asked why he only was able to obtain the slope at the close of the interview he responded, “I kind of knew the answer before but didn’t want to say it.” For students with a similar mistrust of the unfamiliar, explicit

support for underlying ideas may be helpful. As the results in Table 3 indicate, while the experimental group was very successful with integral topics, there was considerable confusion in the control group. One C student from the control group frequently tried to make sense of 3D integral concepts by reverting to representations that she was familiar with in 2D. When asked for an approximation for a volume in sigma notation, she first wrote down ∑∞ 𝑖=1 𝑓(𝑥𝑖 )∆𝑥 to represent an area in 2D and then tried to extend it to volumes in 3D by writing down ∑∞ 𝑖=1 𝑓(𝑥, 𝑦)∆𝑥∆𝑦. When 𝑘𝑔 presented with the units x = meters, y = meters, 𝑓(𝑥, 𝑦) = 𝑚2 and asked to find the units 9

6

associated with∫1 ∫2 𝑑𝑦 𝑑𝑥, she wrote down 𝑦 = 𝑚𝑒𝑡𝑒𝑟𝑠 →

𝑑𝑦 𝑚𝑒𝑡𝑒𝑟𝑠 𝑚𝑒𝑡𝑒𝑟𝑠 = →� 𝑑𝑡 = 𝑚𝑒𝑡𝑒𝑟𝑠. 𝑑𝑡 𝑠𝑒𝑐𝑜𝑛𝑑 𝑠𝑒𝑐𝑜𝑛𝑑

While she was unsuccessful with both of these problems, she seemed quite aware that what she was doing was incorrect and was very frustrated by her inability to use well understood 2D expressions to solve 3D problems. One of the most successful A students in the control group was able to move fluidly between geometric representations and their associated definite integral. However, he could work with neither numeric nor symbolic Riemann Sums. When asked if Riemann Sums were associated with definite integrals, he responded that “supposedly they are as they’re the basis for the whole theory of integrals but it’s really complicated.” There is little doubt that this student had pieced together a limited perspective that allowed him to solve many problems but he didn’t have a solid conceptual foundation for this. However, the expectation that he would implicitly come to understand concepts associated with Riemann Sums even though their associated representations were not presented explicitly was clearly not happening. Common Exam Questions given to both the Experimental and Control Groups To obtain another measure of whether the experimental materials are helpful with traditional multivariable differentiation and integration topics, traditional exam questions that could be given on any Multivariable Calculus exam were administered to all the students in both the experimental group and the control group. These exams were part of the students’ grades; so it is probably that the students performed to the best of their abilities on the questions. The average scores on these common exam questions are presented in Table 4. Common Questions Administered During the Fall 2012 Semester 1. If f is represented by the surface to the right, A. Draw the cross sections x = 0 and y = 0. B. Identify the signs of the following derivatives where 𝑢 �⃗ is in the direction −𝚤⃗ − 𝚥⃗. a. 𝑓𝑥 (2,2) b. 𝑓𝑦 (2, −1) c. Du�⃗ f(2,2) 2. If f(x,y) = sin(x2y), find formulas for the following: a. 𝑓𝑥 (𝑥, 𝑦) b. 𝑓𝑦 (𝑥, 𝑦)

Control Group (n=32)

Experimental Group (n=36)

53% (n=32)

76%* (n=36)

88%** (n=32)

83% (n=36)

3. If the function f is represented by the table to the right a. Find the best approximations for 𝑓𝑥 (1,1)and 𝑓𝑦 (1,1) b. Find the formula for the tangent plane to f at the point (1,1,3) and use it to approximate f(1.1, 1.2) 4. Find the volume over the xy-plane and between the surfaces y =0 and z = 10 – x2- y. 5. Find the volume over the plane z = 2, below the surface z = 9 – x2 and bounded by the planes y =-1 and y =4.

y =1 y =3

x =1 3 7

x =3 5 4

37% (n=32)

61%* (n=36)

26% (n=68)

53%* (n=36)

47% (n=68)

75%* (n=36)

*p 0.5 Table 4: Common exam questions

Discussion Both the interviews and common exam questions would indicate that it may be beneficial in topics of Multivariable Calculus to explicitly present representations that are implicitly assumed to be understood by students. The classroom observations indicated that activities that explicitly presented 3D slopes seem to clear up misunderstandings and greatly improve performance on differential topics of Multivariable Calculus. Appropriate activities that continue to present the numerical representation and the expanded sum algebraic representation of Riemann Sums throughout rectangular, polar, cubic, cylindrical and spherical coordinates also seemed to clear up many misunderstandings associated with integral topics of Multivariable Calculus. Moreover, interviews with the control group showed that, without such activities, many of these misconceptions were not cleared up and the conceptual foundation for the concepts was not solidly formed. Hence, while there is certainly room for a variety of presentations and approaches to Multivariable Calculus, it may be worth paying attention to the need to make explicit what is assumed to be implicitly understood by students so that they are able to make connections between various representations (and coordinate systems). In doing so, the students are more likely to develop the fundamental tool set that they need to truly understand the concepts and succeed in subsequent mathematics. Acknowledgement The material in this article is based upon work supported by the National Science Foundation (NSF) under Grant NSF-DUE 0941877. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author and do not necessarily reflect the views of the NSF. References 1. Edwards, H., & Penney, D. (2002). Calculus: Early Transcendentals 6th Edition, Prentice Hall. 2. McGee, D. Moore-Russo D, Ebersole D, Lomen D, Marin M. Using Physical Manipulatives in the Multivariable Calculus Classroom, Primus, 22(4), 2012, 265-283. 3. Rodriguez, P. (2008). Calculus for the Biological Sciences. United States: Wiley. 4. Stewart, J. (2006), Calculus: Early Transcendentals 6th Edition, United States: Thomson-Brooks/Cole. 5. Strauss, M., Bradley, G., & Smith, K. (2002).Calculus 3rd Edition. United States: Prentice Hall

6. Swokowski, E., Olinick, M., & Pence, D. (1992). Calculus 6th Edition, United States: PWS. 7. Warner, S., & Costenoble S. (2007). Finite Mathematics and Applied Calculus 4th Edition, United States: Thomson-Brooks/Cole.

Appendix A: Two Sample Group Activities from Fall 2012 using 3D slopes Activity 1. Given a plane with 𝑚𝑥 = 1 and 𝑚𝑦 = 2, find the slope in the direction of the vector . We will refer to this as 𝑚

Solution: 1. Place a plane consistent with these 3D slopes and identify the right triangle whose rise and run are associated with this slope.

2. Obtain the rise for ∆𝑥 = 2, ∆𝑦 = 3.

• • •

∆𝑥 = 2, 𝑚𝑥 = 1 → ∆𝑧𝑥 = 2 ∆𝑦 = 3, 𝑚𝑦 = 2 → ∆𝑧𝑦 = 6 𝑟𝑖𝑠𝑒 = ∆𝑧 = ∆𝑧𝑥 + ∆𝑧𝑦 = 2 + 6 = 8

3. Obtain the run for ∆x = 2, ∆y = 3 .

Using the Pythagorean Theorem and the above right triangle, 𝑟𝑢𝑛 = √22 + 32 = √13.

4. Calculate the slope. 𝑚=

8 𝑟𝑖𝑠𝑒 = 𝑟𝑢𝑛 √13

Appendix B: A Sample Experimental Group Activity from Spring 2012 Using the Missing Integral Representations Sample Activity with Solution: Use Riemann Sums with two divisions in 𝑟 and 𝜃 to 𝜋 3𝜋 approximate the volume over the region 2 ≤ 𝑟 ≤ 6, 4 ≤ 𝜃 ≤ 4 in the xy plane and below the plane 𝑧 = 𝑥 + 8 = 𝑟𝑐𝑜𝑠(𝜃) + 8 . Use the smallest value of each variable to represent a given division. Solution: Step 1: Identify the domain associated with the volume to be obtained.

Domain seen in 3D

Domain seen in the xy plane

Step 2: Divide 𝑟 and 𝜃 into two parts identifying 𝑟1 , 𝑟2 , 𝜃1 , 𝜃2 , ∆𝑟 and ∆𝜃 .

Divisions seen in 3D Divisions seen in the xy plane In divisions 1 and 2, r goes from 2 to 4 and in divisions 3 and 4 r goes from 4 to 6. In divisions 1 𝜋 𝜋 𝜋 3𝜋 and 3, 𝜃 goes from 4 to 2 and in divisions 2 and 4, 𝜃 goes from 2 to 4 . Hence, 𝑟1 = 2, 𝑟2 = 𝜋

4, 𝜃1 = 4 , 𝜃2 =

𝜋 , ∆𝑟 2

= 2 and ∆𝜃 =

𝜋 4

Step 3: Determine the numeric volume of each division that will result when using the minimum value of each variable to obtain the length width and height: For all four divisions, ∆𝜃 =

𝜋 4

and ∆𝑟 = 2. In divisions 1 and 2, r goes from 2 to 4 and in

divisions 3 and 4 r goes from 4 to 6. In divisions 1 and 3, 𝜃 goes from

𝜋 4

𝜋

to 2 and in divisions 2 and

𝜋

3𝜋

4, 𝜃 goes from 2 to 4 . In class we determined that if ∆𝑟 and ∆𝜃 are small, we can approximate the area of a polar rectangle by treating it as a rectangle with width ∆𝑟 and length 𝑟∆𝜃. Hence for each division we can approximate the length and height using the smallest value for 𝑟 and 𝜃 found in each region as shown in the following diagram.

The length, width, height and volume of each division are summarized in the following table: Division 1 2 3 4

Length 𝜋 𝜋 (2) � � = 4 2 𝜋 𝜋 (2) � � = 4 2 𝜋 (4) � � = 𝜋 4 𝜋 (4) � � = 𝜋 4

Width 2 2 2 2

Height 𝜋 2 cos � � + 8 4 𝜋 2 cos � � + 8 2 𝜋 4 cos � � + 8 4 𝜋 4 cos � � + 8 2

Volume 𝜋(√2 + 8) 8𝜋

2𝜋(√2 + 8) 16𝜋

Step 4: Add the volumes of the four divisions to approximate the total volume of the solid: 𝜋�√2 + 8� + 8𝜋 + 2𝜋�√2 + 8� + 16 𝜋

Step 5: Find the length, width, height and volume of each division using 𝑟1 , 𝑟2 , 𝜃1 , 𝜃2 , ∆𝑟 and ∆𝜃. Using the same Table we used in Step 3 with symbolic values produces: Division 1 2 3 4

Length (𝑟1 )(∆𝜃) (𝑟1 )(∆𝜃) (𝑟2 )(∆𝜃) (𝑟2 )(∆𝜃)

Width ∆𝑟 ∆𝑟 ∆𝑟 ∆𝑟

Height 𝑟1 cos(𝜃1 ) + 8 𝑟1 cos(𝜃2 ) + 8 𝑟2 cos(𝜃1 ) + 8 𝑟2 cos(𝜃2 ) + 8

Volume (𝑟1 cos(𝜃1 ) + 8)(𝑟1 )(∆𝑟∆𝜃) (𝑟1 cos(𝜃2 ) + 8)(𝑟1 )(∆𝑟∆𝜃) (𝑟2 cos(𝜃1 ) + 8)(𝑟2 )(∆𝑟∆𝜃) (𝑟2 cos(𝜃2 ) + 8)(𝑟2 )(∆𝑟∆𝜃)

Step 6: Add the volumes of the four divisions in the table to approximate the total volume of the

solid using 𝑟1 , 𝑟2 , 𝜃1 , 𝜃2 , ∆𝑟 and ∆𝜃. 𝑣𝑜𝑙𝑢𝑚𝑒 ≈ (𝑟1 cos(𝜃1 ) + 8)(𝑟1 )(∆𝑟∆𝜃) + (𝑟1 cos(𝜃2 ) + 8)(𝑟1 )(∆𝑟∆𝜃) +(𝑟2 cos(𝜃1 ) + 8)(𝑟1 )(∆𝑟∆𝜃) + (𝑟2 cos(𝜃2 ) + 8)(𝑟1 )(∆𝑟∆𝜃) Step 7: Express the approximate volume in Step 6 as a double summation:

𝑣𝑜𝑙𝑢𝑚𝑒 ≈ [(𝑟1 cos(𝜃1 ) + 8)(𝑟1 )(∆𝑟∆𝜃) + (𝑟1 cos(𝜃2 ) + 8)(𝑟1 )(∆𝑟∆𝜃)] +[(𝑟2 cos(𝜃1 ) + 8)(𝑟2 )(∆𝑟∆𝜃) + (𝑟2 cos(𝜃2 ) + 8)(𝑟2 )(∆𝑟∆𝜃)]

𝑣𝑜𝑙𝑢𝑚𝑒 ≈ �

2

(𝑟1 cos�𝜃𝑗 � + 8)(𝑟1 )(∆𝑟∆𝜃) + �

𝑗=1

𝑣𝑜𝑙𝑢𝑚𝑒 ≈ �

2

𝑖=1

�

2

2

(𝑟2 cos�𝜃𝑗 � + 8)(𝑟2 )(∆𝑟∆𝜃)

𝑗=1

(𝑟𝑖 cos�𝜃𝑗 � + 8)(𝑟𝑖 )(∆𝑟∆𝜃)

𝑗=1

Step 8: From the class discussion, we know that as ∆𝑟 → 0 and ∆𝜃 → 0, this approximation becomes precise so generalize the double sum in step 7 in order to take the limit to find the precise volume of the solid: 𝑣𝑜𝑙𝑢𝑚𝑒 = lim lim � ∆𝑟 →0 ∆𝜃→0 6

3𝜋 4

𝑛

𝑖=1

�

𝑚

(𝑟𝑖 cos�𝜃𝑗 � + 8)(𝑟𝑖 )(∆𝜃∆𝑟)

𝑗=1

� � (𝑟 cos(𝜃) + 8)(𝑟)𝑑𝑟𝑑𝜃 2

𝜋 4

Appendix C: Interview Tasks and Results involving Slopes - Fall Semester 2012 Task 1 [Parts 1 and 2 – converting from a geometric to a numerical representation, Part 3 – converting from a geometric to an algebraic representation] Part 1. The points (1,1,2) and (3,1,6) will be shown using the kit. Can you find the slope of the line that goes from (1,1,2) to (3,1,6)? Part 2. If the student correctly answered Part 1, the points (1,1,1) and (4,5,6) were placed on the kit and this question was asked: Can you find the slope of the line that goes from (1,1,1) to (4,5,6)? Part 3. 2 points are placed on the kit and it is indicated that we don’t know the scale. The first point is (x0,y0,z0) and the second point is (x1,y1,z1). What is the slope between the two points? Task 1 Part 1

Control Group Results 5 Unsuccessful (AACCC) 1 Successful (A)

Part 2

6 Unsuccessful (AAACCC)

Part 3

6 Unsuccessful (AAACCC)

Experimental Group Results 1 Unsuccessful (C) 5 Successful (AAACC) 2 Unsuccessful (CC) 4 Successful (AAAC) 2 Unsuccessful (CC) 4 Successful (AAAC)

Table C1. Results for the Experimental and Control Groups for Task 1

Task 2 [Parts 1 and 2 – converting from a geometric to a numeric representation] A plane is placed in front of the students (using the 3D manipulative) that has slope 1 in the x direction and 2 in the y direction: Part 1. What is the slope if we walk in the direction of the vector on this plane? Part 2. What is the slope if we walk in the direction of the vector on this plane?

(Note: If the students did not understand “x” and “y” directions, the interviewer then tried again with alternate language such as “East”, “the direction of the vector ”, etc. finally drawing the direction on the xy plane.)

Task 2

Control Group Results

Experimental Group Results 2 Unsuccessful (CC)

Part 1

6 Unsuccessful (AAACCC)

4 Successful (AAAC) 4 Unsuccessful (ACCC)

Part 2

6 Unsuccessful (AAACCC)

2 Successful (AA)

Table C2. Results for the Experimental and Control Groups for Task 2

Task 3 [Parts 1 and 2 – converting from an algebraic to a numeric representation] A plane has the formula z = 2x + 3y + 1 Part 1. What is the slope if we walk in the y direction on this plane? Part 2. What is the slope if we walk in the direction of the vector on this plane? Note: If the students did not understand vector directions, the interviewer then tried again with alternate language such as “East”, “the x direction”, etc. finally drawing the direction on the xy plane. Task 3 Part 1 Part 2

Control Group Results 4 Unsuccessful (ACCC) 2 Successful (AA) 6 Unsuccessful (AAACCC)

Experimental Group Results 1 Unsuccessful (C) 5 Successful (AAACC) 4 Unsuccessful (ACCC) 2 Successful (AA)

Table C3. Results for the Experimental and Control Groups for Task 3

Task 4 [converting from a verbal to an algebraic representation] If we walk 2 meters eastward on a plane we rise 10 meters and if we walk 3 meters northward we rise 12 meters. If the plane passes through the origin (0,0,0), what is the formula for the plane? Task 4 Obtained Slopes

Control Group Results 6 Unsuccessful (AAACCC)

Experimental Group Results 1 Unsuccessful (C)

Obtained Formula

6 Unsuccessful (AAACCC)

5 Successful (AAAC) 3 Unsuccessful (CCC) 3 Successful (AAA)

Table C4. Results for the Experimental and Control Groups for Task 4

Task 5 [converting from a geometric to an algebraic representation] A plane with mx = 2, my = 1 and intercept (0,0,1) is placed on the kit. Find the formula for this function. Control Group Results

Task 5 Obtained surface with mx

6 Unsuccessful (AAACCC)

Obtained surface with my

6 Unsuccessful (AAACCC)

Obtained formula

6 Unsuccessful (AAACCC)

Experimental Group Results 4 Unsuccessful (ACCC) 2 Successful (AA) 4 Unsuccessful (ACCC) 2 Successful (AA) 4 Unsuccessful (ACCC) 2 Successful (AA)

Table C5. Results for the Experimental and Control Groups for Task 5

Task 6 [Part 1 – moving from one geometric representation to another; Part 2 – converting from a geometric to a numeric representation] If f is represented by the surface (shown using the kit) to the left,

Part 1. A. Find a point where fx is (i) equal to zero, (ii) greater than zero and (iii) less than zero. B. Find a point with Df (i) equal to zero, (ii) greater than zero and (iii) less than zero. Part 2. Label the point (1,1,f(1,1)) on the kit. A. Approximate fx(1,1). B. Approximate Df(1,1) where goes in the direction . Note: If the student placed a tangent line in the right direction and tried to get rise/run, this was accepted as correct, as was practically anything that was remotely possible along these lines..

Task 6 Part 1A Part 1B

Control Group Results 3 Unsuccessful (ACC) 3 Successful (AAC) 4 Unsuccessful (ACCC) 2 Successful (AA)

Part 2A

6 Unsuccessful (AAACCC)

Part 2B

6 Unsuccessful (AAACCC)

Experimental Group Results 1 Unsuccessful (C) 5 Successful (AAACC) 2 Unsuccessful (CC) 4 Successful (AAAC) 3 Unsuccessful (ACC) 3 Successful (AAC) 3 Unsuccessful (ACC) 3 Successful (AAC)

Table C6. Results for the Experimental and Control Groups for Task 6

Task 7 [Part 1 – moving from one numeric representation to another; Part 2 – converting from a numerical to an algebraic representation] Y 1

3

5

2

1

5

1

x 5

4

3

2

8

5

3

5

If the function f is represented by the above table, Part 1. Find the best approximation of fx(5,1) Part 2. If g is the tangent plane of f at the point where (x,y) = (2,1), find a formula for g.

Part 1

Control Group Results 5 Unsuccessful (AACCC) 1 Successful (A)

Part 2

6 Unsuccessful (AAACCC)

Task 7

Experimental Group Results 2 Unsuccessful (CC) 4 Successful (AAAC) 3 Unsuccessful (CCC) 3 Successful (AAA)

Table C7. Results for the Experimental and Control Groups for Task 7

Task 8 [Part 1 – converting from an algebraic to a numeric representation; Part 2 – converting from an algebraic to a geometric representation] The function f is represented by the formula f(x,y) = x2y3 and we are interested in fx(2,1) Part 1. Find the value of fx(2,1).

Part 2. With the axes above, draw the cross section needed to visualize fx(2,1) in 2-D. Label the axes and the formula of this cross section. Draw the line the slope of which is associated with fx(2,1) Control Group Results 1 Unsuccessful (C) 5 Successful (AAACC) 5 Unsuccessful (AAACC) 1 Successful (C)

Task 8 Part 1 Part 2

Experimental Group Results 1 Unsuccessful (C) 5 Successful (AAACC) 5 Unsuccessful (AAACC) 1 Successful (A)

Table C8. Results for the Experimental and Control Groups for Task 8

Task 9 [converting from an algebraic to a numeric representation] The function f is represented by the formula f(x,y)=x2 + y2 and we are interested in approximating fx(1,0). Fill in the following table as we use successively closer approximations for the indicated derivative (a limit). After you fill this table out, try to guess fx(1,0)

Two Points used

(1,0) and ( ,

Run

1

)

(1,0) and ( , 0.1

)

(1,0) and ( , 0.01

Rise Approximation of fy(1,0) Table C9. Table to be filled out by students for Task 9

Task 9 Identified Points

Control Group Results 6 Unsuccessful (AAACCC)

Experimental Group Results 2 Unsuccessful (CC) 4 Successful (AAAC)

)

Identified Rise

6 Unsuccessful (AAACCC)

Approximated Derivative

6 Unsuccessful (AAACCC)

3 Unsuccessful (CCC) 3 Successful (AAA) 3 Unsuccessful (CCC) 3 Successful (AAA)

Table C10. Results for the Experimental and Control Groups for Task 9

Task 10 [converting from a numeric to a geometric representation] Given the linear function f is a plane, show using the 3D kit, a plane consistent with fx(1,1) = 2 and fy(1,1) = -2. Task 10

Control Group Results

Obtained surface with fx(1,1) = 2

6 Unsuccessful (AAACCC)

Obtained surface with fy(1,1) = -2

6 Unsuccessful (AAACCC)

Obtained surface consistent with both

6 Unsuccessful (AAACCC)

Experimental Group Results 2 Unsuccessful (CC) 4 Successful (AAAC) 2 Unsuccessful (CC) 4 Successful (AAAC) 2 Unsuccessful (CC) 4 Successful (AAAC)

Table C11. Results for the Experimental and Control Groups for Task 10

Appendix D: Interview Tasks and Results for Integral Representations – Spring Semester 2012 The following tasks were presented to six students in the control group and five students in the experimental group. In the following tables that record the results, the final course grade for each student associated with a result is presented in parentheses. Task 1 [converting from a geometric to a symbolic representation]

x = meters y = meters z = f(x,y) = height over a point (x,y) in meters. What is an expression for the volume under the surface f and above the shaded region? (Students were physically shown a surface over the region.) All efforts were made in Task 1 to assure that students visualized the solid associated with the geometric representation. The results are presented in Table D1. Task 1

Control Group (4 A, 2 C students) Unsuccessful (AA) Successful (AACC)

Experimental Group (2 A, 1B, 2 C students) Unsuccessful (C) Successful (AABC)

Table D1: The results of Interview Task 1.

Task 2 [converting from a symbolic representation to geometric and verbal representations] x = meters y = meters Given

Part 1. Interviewer asks: What are the units of this integral? Part 2. Interviewer asks: Is it possible to find a result for the integral without evaluating it? It should be noted that students could interpret the integral as an area, or could recognize 9 6 9 6 that∫1 ∫2 𝑑𝑦𝑑𝑥 = ∫1 ∫2 (1)𝑑𝑦𝑑𝑥 could interpret the integral as the volume under the surface z = 1 over the indicated region. The goal was to see if students would recognize that f was not a part of this problem. The results are presented in Table D2. Task 2

Control Group (4 A, 2 C students)

Experimental Group (2 A, 1B, 2 C students)

Part 1 Part 2

Unsuccessful (AAACC) Successful (A) Unsuccessful (AAACC) Successful (A)

Successful (AABCC) Unsuccessful (C) Successful (AABC)

Table D2: The results of interview Task 2.

Task 3 [converting from a geometric to a numeric representation] The height of a surface at a given point (x,y)is given by the formula f(x,y) = x + y + 1 and we wish to find the volume below this surface and over the region shown below.

It should be noted that the students had a large sheet of paper with a very large representation of the region, and they were physically shown the plane z = x + y + 1 over this region to assure that the geometric representation was involved. Part 1. Interviewer: We wish to approximate the volume using 2 divisions in x and 2 divisions in y, can you draw the associated divisions and indicate the length, width and height of each division if we use the smallest height in each region?

Part 2. Interviewer: Now we wish to approximate the same volume using 3 divisions in x and 3 divisions in y, can you draw the associated divisions and indicate the length, width and height of each division if we use the smallest height in each region?

Part 3. Interviewer: Between Part 1 and Part 2, which approximation is better? Part 4. Interviewer: Is there something we can we do to make these approximations even more precise? Task 3, Parts 3 and 4 were to determine whether students understand the underlying limits that are used to transition from the sum with sigma representation to the definite integral representation. Part 5. Interviewer: We’re going to go back to your approximation in Part 1. Can this approximation be expressed using sigma notation? If so, how? If not, why not? Task 3, Part 5 was to determine whether students could convert from a numerical representation to the sum with sigma representation. Any double sigma expression that could represent the numerical approximation was considered a success. Task 3, Part 6 was based on the answer to the student had given to Part 5. a. If the student had used sigma, then the interviewer asked the following questions: i. How does this sum in sigma form change if we use the approximation in B instead of the approximation in A? ii. How can we make the approximation more and more precise in sigma form? iii. How do we obtain the precise value for the volume from this sigma form? b. If the student did not use the sigma, then the interviewer asked the following questions: i. If we had been able to put this in sigma notation, would there be any advantage?

ii. iii. iv.

If response is “no” then leave the topic. If response is “yes” then the interviewer asked: What the advantage would be? How would we make the expression more precise in sigma form? How do we obtain the precise value of the volume from the sigma form?

The purpose of Task 3, Part 6 was to determine whether the concept, which was established in Parts 3 and 4 of the task (that more divisions make the approximation more precise), could be used to relating the sum with sigma representation to the definite integral representation. Any scheme taking the limit of a double sigma to arrive at a double integral was considered a success. The results of Task 3 are presented in Table D3. Task 3A 3B 3C 3D 3E 3F

Control Group (4 A, 2 C students) Unsuccessful (ACC) Successful (AAA) Unsuccessful (AACC) Successful (AA) Unsuccessful (ACC) Successful (AAA) Unsuccessful (ACC) Successful (AAA) Unsuccessful (AAAAC) Successful (C) Unsuccessful (AAAACC)

Experimental Group (2 A, 1B, 2 C students) Successful (AABCC) Successful (AABCC) Successful (AABCC) Successful (AABCC) Unsuccessful (A) Successful (ABCC) Unsuccessful (A) Successful (ABCC)

Table D3: The results of interview Task 3.

Task 4: [converting a verbal to a numeric representation and a verbal representation to a symbolic integral representation] In the region below x= km, y = km and the density of people is defined by f(y)= 9-y2persons/km2. We are interested in the number of people that live in the region. In the interview students were provided an explanation that perhaps there is a river that runs where y < 0 and as people prefer to live near the river, the density of people in this region is greater the closer we are to the river.

Part 1. Interviewer: How would you approximate the number of people that live in this region? Any Riemann sum was considered a success for this part of the task. Part 2. If the student performs divisions in both x and y, the interviewer asked: Do we have to divide in both x and y? Some prompting was provided for Task 4, Part 2 but only a Riemann sum with divisions in y and not in x was considered a success. If the students provided such a Riemann sum in Part 1 then Part 2 was automatically considered a success. Part 3. Interviewer: Can you provide an expression for the precise number of people? Any correct integral expression was considered a success for Part 3 of the Task. Part 4. If the student provides a double integral in Part 3, the interviewer asked if the precise number of people could be expressed as a single integral. As with Part 2, some prompting was provided for Part 4 but only a single integral expression in terms of y was considered a success. If the student provided the correct single integral in Part 3, Part 4 was automatically considered a success. The results are presented in Table D4. Task 4

Control Group (4 A, 2 C students)

Part 1

Unsuccessful (AAAACC)

Part 2

Unsuccessful (AAAACC)

Part 3 Part 4

Unsuccessful (AAAC) Successful (AC) Unsuccessful (AAACC) Successful (A)

Experimental Group (2 A, 1B, 2 C students) Unsuccessful (BC) Successful (AAC) Unsuccessful (ABC) Successful (AC) Unsuccessful (C) Successful (AABC) Unsuccessful (ABC) Successful (AC)

Table D4: The results of interview Task 4.